Integrand size = 28, antiderivative size = 475 \[ \int \frac {(e+f x)^3 \sec ^2(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {2 i (e+f x)^3}{3 a d}-\frac {i f (e+f x)^2 \arctan \left (e^{i (c+d x)}\right )}{a d^2}+\frac {f^3 \text {arctanh}(\sin (c+d x))}{a d^4}+\frac {2 f (e+f x)^2 \log \left (1+e^{2 i (c+d x)}\right )}{a d^2}+\frac {f^3 \log (\cos (c+d x))}{a d^4}+\frac {i f^2 (e+f x) \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{a d^3}-\frac {i f^2 (e+f x) \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{a d^3}-\frac {2 i f^2 (e+f x) \operatorname {PolyLog}\left (2,-e^{2 i (c+d x)}\right )}{a d^3}-\frac {f^3 \operatorname {PolyLog}\left (3,-i e^{i (c+d x)}\right )}{a d^4}+\frac {f^3 \operatorname {PolyLog}\left (3,i e^{i (c+d x)}\right )}{a d^4}+\frac {f^3 \operatorname {PolyLog}\left (3,-e^{2 i (c+d x)}\right )}{a d^4}-\frac {f^2 (e+f x) \sec (c+d x)}{a d^3}-\frac {f (e+f x)^2 \sec ^2(c+d x)}{2 a d^2}-\frac {(e+f x)^3 \sec ^3(c+d x)}{3 a d}+\frac {f^2 (e+f x) \tan (c+d x)}{a d^3}+\frac {2 (e+f x)^3 \tan (c+d x)}{3 a d}+\frac {f (e+f x)^2 \sec (c+d x) \tan (c+d x)}{2 a d^2}+\frac {(e+f x)^3 \sec ^2(c+d x) \tan (c+d x)}{3 a d} \]
-I*f^2*(f*x+e)*polylog(2,I*exp(I*(d*x+c)))/a/d^3-I*f*(f*x+e)^2*arctan(exp( I*(d*x+c)))/a/d^2+f^3*arctanh(sin(d*x+c))/a/d^4+2*f*(f*x+e)^2*ln(1+exp(2*I *(d*x+c)))/a/d^2+f^3*ln(cos(d*x+c))/a/d^4-2*I*f^2*(f*x+e)*polylog(2,-exp(2 *I*(d*x+c)))/a/d^3+I*f^2*(f*x+e)*polylog(2,-I*exp(I*(d*x+c)))/a/d^3-2/3*I* (f*x+e)^3/a/d-f^3*polylog(3,-I*exp(I*(d*x+c)))/a/d^4+f^3*polylog(3,I*exp(I *(d*x+c)))/a/d^4+f^3*polylog(3,-exp(2*I*(d*x+c)))/a/d^4-f^2*(f*x+e)*sec(d* x+c)/a/d^3-1/2*f*(f*x+e)^2*sec(d*x+c)^2/a/d^2-1/3*(f*x+e)^3*sec(d*x+c)^3/a /d+f^2*(f*x+e)*tan(d*x+c)/a/d^3+2/3*(f*x+e)^3*tan(d*x+c)/a/d+1/2*f*(f*x+e) ^2*sec(d*x+c)*tan(d*x+c)/a/d^2+1/3*(f*x+e)^3*sec(d*x+c)^2*tan(d*x+c)/a/d
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(1173\) vs. \(2(475)=950\).
Time = 8.38 (sec) , antiderivative size = 1173, normalized size of antiderivative = 2.47 \[ \int \frac {(e+f x)^3 \sec ^2(c+d x)}{a+a \sin (c+d x)} \, dx =\text {Too large to display} \]
((d^3*(e + f*x)^3)/(-I + E^(I*c)) + 3*d^2*f*(e + f*x)^2*Log[1 - I/E^(I*(c + d*x))] + 6*f^2*(I*d*(e + f*x)*PolyLog[2, I/E^(I*(c + d*x))] + f*PolyLog[ 3, I/E^(I*(c + d*x))]))/(2*a*d^4) - (f*(Cos[c] + I*Sin[c])*(5*d^2*e^2*x*Co s[c] + 4*f^2*x*Cos[c] + 5*d^2*e*f*x^2*Cos[c] + (5*d^2*f^2*x^3*(Cos[c] - I* Sin[c]))/3 - (5*I)*d^2*e^2*x*Sin[c] - (4*I)*f^2*x*Sin[c] - (5*I)*d^2*e*f*x ^2*Sin[c] + 10*e*f*PolyLog[2, (-I)*Cos[c + d*x] - Sin[c + d*x]]*(Cos[c] - I*(1 + Sin[c])) + 10*f^2*x*PolyLog[2, (-I)*Cos[c + d*x] - Sin[c + d*x]]*(C os[c] - I*(1 + Sin[c])) - 10*d*e*f*x*Log[1 + I*Cos[c + d*x] + Sin[c + d*x] ]*(Cos[c] - I*Sin[c])*(Cos[c] + I*(1 + Sin[c])) - 5*d*f^2*x^2*Log[1 + I*Co s[c + d*x] + Sin[c + d*x]]*(Cos[c] - I*Sin[c])*(Cos[c] + I*(1 + Sin[c])) - ((5*d^2*e^2 + 4*f^2)*Log[Cos[c + d*x] + I*(1 + Sin[c + d*x])]*(Cos[c] - I *Sin[c])*(Cos[c] + I*(1 + Sin[c])))/d - (10*f^2*PolyLog[3, (-I)*Cos[c + d* x] - Sin[c + d*x]]*(Cos[c] - I*Sin[c])*(Cos[c] + I*(1 + Sin[c])))/d + (5*d ^2*e^2 + 4*f^2)*x*(I*Cos[c] + Sin[c])*(Cos[c] + I*(1 + Sin[c]))))/(2*a*d^3 *(Cos[c] + I*(1 + Sin[c]))) + (e^3*Sin[(d*x)/2] + 3*e^2*f*x*Sin[(d*x)/2] + 3*e*f^2*x^2*Sin[(d*x)/2] + f^3*x^3*Sin[(d*x)/2])/(2*a*d*(Cos[c/2] - Sin[c /2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])) + (e^3*Sin[(d*x)/2] + 3*e^ 2*f*x*Sin[(d*x)/2] + 3*e*f^2*x^2*Sin[(d*x)/2] + f^3*x^3*Sin[(d*x)/2])/(3*a *d*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2])^3) + (- (d*e^3*Cos[c/2]) - 3*e^2*f*Cos[c/2] - 3*d*e^2*f*x*Cos[c/2] - 6*e*f^2*x*...
Time = 2.59 (sec) , antiderivative size = 474, normalized size of antiderivative = 1.00, number of steps used = 22, number of rules used = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {5042, 3042, 4674, 3042, 4672, 25, 3042, 3956, 4202, 2620, 3011, 2720, 4909, 3042, 4674, 3042, 4257, 4669, 3011, 2720, 7143}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {(e+f x)^3 \sec ^2(c+d x)}{a \sin (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 5042 |
\(\displaystyle \frac {\int (e+f x)^3 \sec ^4(c+d x)dx}{a}-\frac {\int (e+f x)^3 \sec ^3(c+d x) \tan (c+d x)dx}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int (e+f x)^3 \csc \left (c+d x+\frac {\pi }{2}\right )^4dx}{a}-\frac {\int (e+f x)^3 \sec ^3(c+d x) \tan (c+d x)dx}{a}\) |
\(\Big \downarrow \) 4674 |
\(\displaystyle \frac {\frac {f^2 \int (e+f x) \sec ^2(c+d x)dx}{d^2}+\frac {2}{3} \int (e+f x)^3 \sec ^2(c+d x)dx-\frac {f (e+f x)^2 \sec ^2(c+d x)}{2 d^2}+\frac {(e+f x)^3 \tan (c+d x) \sec ^2(c+d x)}{3 d}}{a}-\frac {\int (e+f x)^3 \sec ^3(c+d x) \tan (c+d x)dx}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {f^2 \int (e+f x) \csc \left (c+d x+\frac {\pi }{2}\right )^2dx}{d^2}+\frac {2}{3} \int (e+f x)^3 \csc \left (c+d x+\frac {\pi }{2}\right )^2dx-\frac {f (e+f x)^2 \sec ^2(c+d x)}{2 d^2}+\frac {(e+f x)^3 \tan (c+d x) \sec ^2(c+d x)}{3 d}}{a}-\frac {\int (e+f x)^3 \sec ^3(c+d x) \tan (c+d x)dx}{a}\) |
\(\Big \downarrow \) 4672 |
\(\displaystyle \frac {\frac {f^2 \left (\frac {f \int -\tan (c+d x)dx}{d}+\frac {(e+f x) \tan (c+d x)}{d}\right )}{d^2}+\frac {2}{3} \left (\frac {3 f \int -(e+f x)^2 \tan (c+d x)dx}{d}+\frac {(e+f x)^3 \tan (c+d x)}{d}\right )-\frac {f (e+f x)^2 \sec ^2(c+d x)}{2 d^2}+\frac {(e+f x)^3 \tan (c+d x) \sec ^2(c+d x)}{3 d}}{a}-\frac {\int (e+f x)^3 \sec ^3(c+d x) \tan (c+d x)dx}{a}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {f^2 \left (\frac {(e+f x) \tan (c+d x)}{d}-\frac {f \int \tan (c+d x)dx}{d}\right )}{d^2}+\frac {2}{3} \left (\frac {(e+f x)^3 \tan (c+d x)}{d}-\frac {3 f \int (e+f x)^2 \tan (c+d x)dx}{d}\right )-\frac {f (e+f x)^2 \sec ^2(c+d x)}{2 d^2}+\frac {(e+f x)^3 \tan (c+d x) \sec ^2(c+d x)}{3 d}}{a}-\frac {\int (e+f x)^3 \sec ^3(c+d x) \tan (c+d x)dx}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {f^2 \left (\frac {(e+f x) \tan (c+d x)}{d}-\frac {f \int \tan (c+d x)dx}{d}\right )}{d^2}+\frac {2}{3} \left (\frac {(e+f x)^3 \tan (c+d x)}{d}-\frac {3 f \int (e+f x)^2 \tan (c+d x)dx}{d}\right )-\frac {f (e+f x)^2 \sec ^2(c+d x)}{2 d^2}+\frac {(e+f x)^3 \tan (c+d x) \sec ^2(c+d x)}{3 d}}{a}-\frac {\int (e+f x)^3 \sec ^3(c+d x) \tan (c+d x)dx}{a}\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle \frac {\frac {2}{3} \left (\frac {(e+f x)^3 \tan (c+d x)}{d}-\frac {3 f \int (e+f x)^2 \tan (c+d x)dx}{d}\right )+\frac {f^2 \left (\frac {f \log (\cos (c+d x))}{d^2}+\frac {(e+f x) \tan (c+d x)}{d}\right )}{d^2}-\frac {f (e+f x)^2 \sec ^2(c+d x)}{2 d^2}+\frac {(e+f x)^3 \tan (c+d x) \sec ^2(c+d x)}{3 d}}{a}-\frac {\int (e+f x)^3 \sec ^3(c+d x) \tan (c+d x)dx}{a}\) |
\(\Big \downarrow \) 4202 |
\(\displaystyle -\frac {\int (e+f x)^3 \sec ^3(c+d x) \tan (c+d x)dx}{a}+\frac {\frac {2}{3} \left (\frac {(e+f x)^3 \tan (c+d x)}{d}-\frac {3 f \left (\frac {i (e+f x)^3}{3 f}-2 i \int \frac {e^{2 i (c+d x)} (e+f x)^2}{1+e^{2 i (c+d x)}}dx\right )}{d}\right )+\frac {f^2 \left (\frac {f \log (\cos (c+d x))}{d^2}+\frac {(e+f x) \tan (c+d x)}{d}\right )}{d^2}-\frac {f (e+f x)^2 \sec ^2(c+d x)}{2 d^2}+\frac {(e+f x)^3 \tan (c+d x) \sec ^2(c+d x)}{3 d}}{a}\) |
\(\Big \downarrow \) 2620 |
\(\displaystyle -\frac {\int (e+f x)^3 \sec ^3(c+d x) \tan (c+d x)dx}{a}+\frac {\frac {2}{3} \left (\frac {(e+f x)^3 \tan (c+d x)}{d}-\frac {3 f \left (\frac {i (e+f x)^3}{3 f}-2 i \left (\frac {i f \int (e+f x) \log \left (1+e^{2 i (c+d x)}\right )dx}{d}-\frac {i (e+f x)^2 \log \left (1+e^{2 i (c+d x)}\right )}{2 d}\right )\right )}{d}\right )+\frac {f^2 \left (\frac {f \log (\cos (c+d x))}{d^2}+\frac {(e+f x) \tan (c+d x)}{d}\right )}{d^2}-\frac {f (e+f x)^2 \sec ^2(c+d x)}{2 d^2}+\frac {(e+f x)^3 \tan (c+d x) \sec ^2(c+d x)}{3 d}}{a}\) |
\(\Big \downarrow \) 3011 |
\(\displaystyle -\frac {\int (e+f x)^3 \sec ^3(c+d x) \tan (c+d x)dx}{a}+\frac {\frac {2}{3} \left (\frac {(e+f x)^3 \tan (c+d x)}{d}-\frac {3 f \left (\frac {i (e+f x)^3}{3 f}-2 i \left (\frac {i f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,-e^{2 i (c+d x)}\right )}{2 d}-\frac {i f \int \operatorname {PolyLog}\left (2,-e^{2 i (c+d x)}\right )dx}{2 d}\right )}{d}-\frac {i (e+f x)^2 \log \left (1+e^{2 i (c+d x)}\right )}{2 d}\right )\right )}{d}\right )+\frac {f^2 \left (\frac {f \log (\cos (c+d x))}{d^2}+\frac {(e+f x) \tan (c+d x)}{d}\right )}{d^2}-\frac {f (e+f x)^2 \sec ^2(c+d x)}{2 d^2}+\frac {(e+f x)^3 \tan (c+d x) \sec ^2(c+d x)}{3 d}}{a}\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle -\frac {\int (e+f x)^3 \sec ^3(c+d x) \tan (c+d x)dx}{a}+\frac {\frac {2}{3} \left (\frac {(e+f x)^3 \tan (c+d x)}{d}-\frac {3 f \left (\frac {i (e+f x)^3}{3 f}-2 i \left (\frac {i f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,-e^{2 i (c+d x)}\right )}{2 d}-\frac {f \int e^{-2 i (c+d x)} \operatorname {PolyLog}\left (2,-e^{2 i (c+d x)}\right )de^{2 i (c+d x)}}{4 d^2}\right )}{d}-\frac {i (e+f x)^2 \log \left (1+e^{2 i (c+d x)}\right )}{2 d}\right )\right )}{d}\right )+\frac {f^2 \left (\frac {f \log (\cos (c+d x))}{d^2}+\frac {(e+f x) \tan (c+d x)}{d}\right )}{d^2}-\frac {f (e+f x)^2 \sec ^2(c+d x)}{2 d^2}+\frac {(e+f x)^3 \tan (c+d x) \sec ^2(c+d x)}{3 d}}{a}\) |
\(\Big \downarrow \) 4909 |
\(\displaystyle -\frac {\frac {(e+f x)^3 \sec ^3(c+d x)}{3 d}-\frac {f \int (e+f x)^2 \sec ^3(c+d x)dx}{d}}{a}+\frac {\frac {2}{3} \left (\frac {(e+f x)^3 \tan (c+d x)}{d}-\frac {3 f \left (\frac {i (e+f x)^3}{3 f}-2 i \left (\frac {i f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,-e^{2 i (c+d x)}\right )}{2 d}-\frac {f \int e^{-2 i (c+d x)} \operatorname {PolyLog}\left (2,-e^{2 i (c+d x)}\right )de^{2 i (c+d x)}}{4 d^2}\right )}{d}-\frac {i (e+f x)^2 \log \left (1+e^{2 i (c+d x)}\right )}{2 d}\right )\right )}{d}\right )+\frac {f^2 \left (\frac {f \log (\cos (c+d x))}{d^2}+\frac {(e+f x) \tan (c+d x)}{d}\right )}{d^2}-\frac {f (e+f x)^2 \sec ^2(c+d x)}{2 d^2}+\frac {(e+f x)^3 \tan (c+d x) \sec ^2(c+d x)}{3 d}}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\frac {(e+f x)^3 \sec ^3(c+d x)}{3 d}-\frac {f \int (e+f x)^2 \csc \left (c+d x+\frac {\pi }{2}\right )^3dx}{d}}{a}+\frac {\frac {2}{3} \left (\frac {(e+f x)^3 \tan (c+d x)}{d}-\frac {3 f \left (\frac {i (e+f x)^3}{3 f}-2 i \left (\frac {i f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,-e^{2 i (c+d x)}\right )}{2 d}-\frac {f \int e^{-2 i (c+d x)} \operatorname {PolyLog}\left (2,-e^{2 i (c+d x)}\right )de^{2 i (c+d x)}}{4 d^2}\right )}{d}-\frac {i (e+f x)^2 \log \left (1+e^{2 i (c+d x)}\right )}{2 d}\right )\right )}{d}\right )+\frac {f^2 \left (\frac {f \log (\cos (c+d x))}{d^2}+\frac {(e+f x) \tan (c+d x)}{d}\right )}{d^2}-\frac {f (e+f x)^2 \sec ^2(c+d x)}{2 d^2}+\frac {(e+f x)^3 \tan (c+d x) \sec ^2(c+d x)}{3 d}}{a}\) |
\(\Big \downarrow \) 4674 |
\(\displaystyle -\frac {\frac {(e+f x)^3 \sec ^3(c+d x)}{3 d}-\frac {f \left (\frac {f^2 \int \sec (c+d x)dx}{d^2}+\frac {1}{2} \int (e+f x)^2 \sec (c+d x)dx-\frac {f (e+f x) \sec (c+d x)}{d^2}+\frac {(e+f x)^2 \tan (c+d x) \sec (c+d x)}{2 d}\right )}{d}}{a}+\frac {\frac {2}{3} \left (\frac {(e+f x)^3 \tan (c+d x)}{d}-\frac {3 f \left (\frac {i (e+f x)^3}{3 f}-2 i \left (\frac {i f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,-e^{2 i (c+d x)}\right )}{2 d}-\frac {f \int e^{-2 i (c+d x)} \operatorname {PolyLog}\left (2,-e^{2 i (c+d x)}\right )de^{2 i (c+d x)}}{4 d^2}\right )}{d}-\frac {i (e+f x)^2 \log \left (1+e^{2 i (c+d x)}\right )}{2 d}\right )\right )}{d}\right )+\frac {f^2 \left (\frac {f \log (\cos (c+d x))}{d^2}+\frac {(e+f x) \tan (c+d x)}{d}\right )}{d^2}-\frac {f (e+f x)^2 \sec ^2(c+d x)}{2 d^2}+\frac {(e+f x)^3 \tan (c+d x) \sec ^2(c+d x)}{3 d}}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\frac {(e+f x)^3 \sec ^3(c+d x)}{3 d}-\frac {f \left (\frac {f^2 \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{d^2}+\frac {1}{2} \int (e+f x)^2 \csc \left (c+d x+\frac {\pi }{2}\right )dx-\frac {f (e+f x) \sec (c+d x)}{d^2}+\frac {(e+f x)^2 \tan (c+d x) \sec (c+d x)}{2 d}\right )}{d}}{a}+\frac {\frac {2}{3} \left (\frac {(e+f x)^3 \tan (c+d x)}{d}-\frac {3 f \left (\frac {i (e+f x)^3}{3 f}-2 i \left (\frac {i f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,-e^{2 i (c+d x)}\right )}{2 d}-\frac {f \int e^{-2 i (c+d x)} \operatorname {PolyLog}\left (2,-e^{2 i (c+d x)}\right )de^{2 i (c+d x)}}{4 d^2}\right )}{d}-\frac {i (e+f x)^2 \log \left (1+e^{2 i (c+d x)}\right )}{2 d}\right )\right )}{d}\right )+\frac {f^2 \left (\frac {f \log (\cos (c+d x))}{d^2}+\frac {(e+f x) \tan (c+d x)}{d}\right )}{d^2}-\frac {f (e+f x)^2 \sec ^2(c+d x)}{2 d^2}+\frac {(e+f x)^3 \tan (c+d x) \sec ^2(c+d x)}{3 d}}{a}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle -\frac {\frac {(e+f x)^3 \sec ^3(c+d x)}{3 d}-\frac {f \left (\frac {1}{2} \int (e+f x)^2 \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {f^2 \text {arctanh}(\sin (c+d x))}{d^3}-\frac {f (e+f x) \sec (c+d x)}{d^2}+\frac {(e+f x)^2 \tan (c+d x) \sec (c+d x)}{2 d}\right )}{d}}{a}+\frac {\frac {2}{3} \left (\frac {(e+f x)^3 \tan (c+d x)}{d}-\frac {3 f \left (\frac {i (e+f x)^3}{3 f}-2 i \left (\frac {i f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,-e^{2 i (c+d x)}\right )}{2 d}-\frac {f \int e^{-2 i (c+d x)} \operatorname {PolyLog}\left (2,-e^{2 i (c+d x)}\right )de^{2 i (c+d x)}}{4 d^2}\right )}{d}-\frac {i (e+f x)^2 \log \left (1+e^{2 i (c+d x)}\right )}{2 d}\right )\right )}{d}\right )+\frac {f^2 \left (\frac {f \log (\cos (c+d x))}{d^2}+\frac {(e+f x) \tan (c+d x)}{d}\right )}{d^2}-\frac {f (e+f x)^2 \sec ^2(c+d x)}{2 d^2}+\frac {(e+f x)^3 \tan (c+d x) \sec ^2(c+d x)}{3 d}}{a}\) |
\(\Big \downarrow \) 4669 |
\(\displaystyle \frac {\frac {2}{3} \left (\frac {(e+f x)^3 \tan (c+d x)}{d}-\frac {3 f \left (\frac {i (e+f x)^3}{3 f}-2 i \left (\frac {i f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,-e^{2 i (c+d x)}\right )}{2 d}-\frac {f \int e^{-2 i (c+d x)} \operatorname {PolyLog}\left (2,-e^{2 i (c+d x)}\right )de^{2 i (c+d x)}}{4 d^2}\right )}{d}-\frac {i (e+f x)^2 \log \left (1+e^{2 i (c+d x)}\right )}{2 d}\right )\right )}{d}\right )+\frac {f^2 \left (\frac {f \log (\cos (c+d x))}{d^2}+\frac {(e+f x) \tan (c+d x)}{d}\right )}{d^2}-\frac {f (e+f x)^2 \sec ^2(c+d x)}{2 d^2}+\frac {(e+f x)^3 \tan (c+d x) \sec ^2(c+d x)}{3 d}}{a}-\frac {\frac {(e+f x)^3 \sec ^3(c+d x)}{3 d}-\frac {f \left (\frac {1}{2} \left (-\frac {2 f \int (e+f x) \log \left (1-i e^{i (c+d x)}\right )dx}{d}+\frac {2 f \int (e+f x) \log \left (1+i e^{i (c+d x)}\right )dx}{d}-\frac {2 i (e+f x)^2 \arctan \left (e^{i (c+d x)}\right )}{d}\right )+\frac {f^2 \text {arctanh}(\sin (c+d x))}{d^3}-\frac {f (e+f x) \sec (c+d x)}{d^2}+\frac {(e+f x)^2 \tan (c+d x) \sec (c+d x)}{2 d}\right )}{d}}{a}\) |
\(\Big \downarrow \) 3011 |
\(\displaystyle \frac {\frac {2}{3} \left (\frac {(e+f x)^3 \tan (c+d x)}{d}-\frac {3 f \left (\frac {i (e+f x)^3}{3 f}-2 i \left (\frac {i f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,-e^{2 i (c+d x)}\right )}{2 d}-\frac {f \int e^{-2 i (c+d x)} \operatorname {PolyLog}\left (2,-e^{2 i (c+d x)}\right )de^{2 i (c+d x)}}{4 d^2}\right )}{d}-\frac {i (e+f x)^2 \log \left (1+e^{2 i (c+d x)}\right )}{2 d}\right )\right )}{d}\right )+\frac {f^2 \left (\frac {f \log (\cos (c+d x))}{d^2}+\frac {(e+f x) \tan (c+d x)}{d}\right )}{d^2}-\frac {f (e+f x)^2 \sec ^2(c+d x)}{2 d^2}+\frac {(e+f x)^3 \tan (c+d x) \sec ^2(c+d x)}{3 d}}{a}-\frac {\frac {(e+f x)^3 \sec ^3(c+d x)}{3 d}-\frac {f \left (\frac {1}{2} \left (\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d}-\frac {i f \int \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )dx}{d}\right )}{d}-\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{d}-\frac {i f \int \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )dx}{d}\right )}{d}-\frac {2 i (e+f x)^2 \arctan \left (e^{i (c+d x)}\right )}{d}\right )+\frac {f^2 \text {arctanh}(\sin (c+d x))}{d^3}-\frac {f (e+f x) \sec (c+d x)}{d^2}+\frac {(e+f x)^2 \tan (c+d x) \sec (c+d x)}{2 d}\right )}{d}}{a}\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\frac {2}{3} \left (\frac {(e+f x)^3 \tan (c+d x)}{d}-\frac {3 f \left (\frac {i (e+f x)^3}{3 f}-2 i \left (\frac {i f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,-e^{2 i (c+d x)}\right )}{2 d}-\frac {f \int e^{-2 i (c+d x)} \operatorname {PolyLog}\left (2,-e^{2 i (c+d x)}\right )de^{2 i (c+d x)}}{4 d^2}\right )}{d}-\frac {i (e+f x)^2 \log \left (1+e^{2 i (c+d x)}\right )}{2 d}\right )\right )}{d}\right )+\frac {f^2 \left (\frac {f \log (\cos (c+d x))}{d^2}+\frac {(e+f x) \tan (c+d x)}{d}\right )}{d^2}-\frac {f (e+f x)^2 \sec ^2(c+d x)}{2 d^2}+\frac {(e+f x)^3 \tan (c+d x) \sec ^2(c+d x)}{3 d}}{a}-\frac {\frac {(e+f x)^3 \sec ^3(c+d x)}{3 d}-\frac {f \left (\frac {1}{2} \left (\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d}-\frac {f \int e^{-i (c+d x)} \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )de^{i (c+d x)}}{d^2}\right )}{d}-\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{d}-\frac {f \int e^{-i (c+d x)} \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )de^{i (c+d x)}}{d^2}\right )}{d}-\frac {2 i (e+f x)^2 \arctan \left (e^{i (c+d x)}\right )}{d}\right )+\frac {f^2 \text {arctanh}(\sin (c+d x))}{d^3}-\frac {f (e+f x) \sec (c+d x)}{d^2}+\frac {(e+f x)^2 \tan (c+d x) \sec (c+d x)}{2 d}\right )}{d}}{a}\) |
\(\Big \downarrow \) 7143 |
\(\displaystyle \frac {\frac {f^2 \left (\frac {f \log (\cos (c+d x))}{d^2}+\frac {(e+f x) \tan (c+d x)}{d}\right )}{d^2}+\frac {2}{3} \left (\frac {(e+f x)^3 \tan (c+d x)}{d}-\frac {3 f \left (\frac {i (e+f x)^3}{3 f}-2 i \left (\frac {i f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,-e^{2 i (c+d x)}\right )}{2 d}-\frac {f \operatorname {PolyLog}\left (3,-e^{2 i (c+d x)}\right )}{4 d^2}\right )}{d}-\frac {i (e+f x)^2 \log \left (1+e^{2 i (c+d x)}\right )}{2 d}\right )\right )}{d}\right )-\frac {f (e+f x)^2 \sec ^2(c+d x)}{2 d^2}+\frac {(e+f x)^3 \tan (c+d x) \sec ^2(c+d x)}{3 d}}{a}-\frac {\frac {(e+f x)^3 \sec ^3(c+d x)}{3 d}-\frac {f \left (\frac {1}{2} \left (-\frac {2 i (e+f x)^2 \arctan \left (e^{i (c+d x)}\right )}{d}+\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d}-\frac {f \operatorname {PolyLog}\left (3,-i e^{i (c+d x)}\right )}{d^2}\right )}{d}-\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{d}-\frac {f \operatorname {PolyLog}\left (3,i e^{i (c+d x)}\right )}{d^2}\right )}{d}\right )+\frac {f^2 \text {arctanh}(\sin (c+d x))}{d^3}-\frac {f (e+f x) \sec (c+d x)}{d^2}+\frac {(e+f x)^2 \tan (c+d x) \sec (c+d x)}{2 d}\right )}{d}}{a}\) |
(-1/2*(f*(e + f*x)^2*Sec[c + d*x]^2)/d^2 + ((e + f*x)^3*Sec[c + d*x]^2*Tan [c + d*x])/(3*d) + (f^2*((f*Log[Cos[c + d*x]])/d^2 + ((e + f*x)*Tan[c + d* x])/d))/d^2 + (2*((-3*f*(((I/3)*(e + f*x)^3)/f - (2*I)*(((-1/2*I)*(e + f*x )^2*Log[1 + E^((2*I)*(c + d*x))])/d + (I*f*(((I/2)*(e + f*x)*PolyLog[2, -E ^((2*I)*(c + d*x))])/d - (f*PolyLog[3, -E^((2*I)*(c + d*x))])/(4*d^2)))/d) ))/d + ((e + f*x)^3*Tan[c + d*x])/d))/3)/a - (((e + f*x)^3*Sec[c + d*x]^3) /(3*d) - (f*((f^2*ArcTanh[Sin[c + d*x]])/d^3 + (((-2*I)*(e + f*x)^2*ArcTan [E^(I*(c + d*x))])/d + (2*f*((I*(e + f*x)*PolyLog[2, (-I)*E^(I*(c + d*x))] )/d - (f*PolyLog[3, (-I)*E^(I*(c + d*x))])/d^2))/d - (2*f*((I*(e + f*x)*Po lyLog[2, I*E^(I*(c + d*x))])/d - (f*PolyLog[3, I*E^(I*(c + d*x))])/d^2))/d )/2 - (f*(e + f*x)*Sec[c + d*x])/d^2 + ((e + f*x)^2*Sec[c + d*x]*Tan[c + d *x])/(2*d)))/d)/a
3.3.75.3.1 Defintions of rubi rules used
Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ ((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp [((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si mp[d*(m/(b*f*g*n*Log[F])) Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x )))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct ionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ [{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) *(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.) *(x_))^(m_.), x_Symbol] :> Simp[(-(f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + b*x)))^n]/(b*c*n*Log[F])), x] + Simp[g*(m/(b*c*n*Log[F])) Int[(f + g*x)^( m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e , f, g, n}, x] && GtQ[m, 0]
Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d *x], x]]/d, x] /; FreeQ[{c, d}, x]
Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[I *((c + d*x)^(m + 1)/(d*(m + 1))), x] - Simp[2*I Int[(c + d*x)^m*(E^(2*I*( e + f*x))/(1 + E^(2*I*(e + f*x)))), x], x] /; FreeQ[{c, d, e, f}, x] && IGt Q[m, 0]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Int[csc[(e_.) + Pi*(k_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol ] :> Simp[-2*(c + d*x)^m*(ArcTanh[E^(I*k*Pi)*E^(I*(e + f*x))]/f), x] + (-Si mp[d*(m/f) Int[(c + d*x)^(m - 1)*Log[1 - E^(I*k*Pi)*E^(I*(e + f*x))], x], x] + Simp[d*(m/f) Int[(c + d*x)^(m - 1)*Log[1 + E^(I*k*Pi)*E^(I*(e + f*x ))], x], x]) /; FreeQ[{c, d, e, f}, x] && IntegerQ[2*k] && IGtQ[m, 0]
Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp [(-(c + d*x)^m)*(Cot[e + f*x]/f), x] + Simp[d*(m/f) Int[(c + d*x)^(m - 1) *Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((c_.) + (d_.)*(x_))^(m_), x_Symbo l] :> Simp[(-b^2)*(c + d*x)^m*Cot[e + f*x]*((b*Csc[e + f*x])^(n - 2)/(f*(n - 1))), x] + (-Simp[b^2*d*m*(c + d*x)^(m - 1)*((b*Csc[e + f*x])^(n - 2)/(f^ 2*(n - 1)*(n - 2))), x] + Simp[b^2*d^2*m*((m - 1)/(f^2*(n - 1)*(n - 2))) Int[(c + d*x)^(m - 2)*(b*Csc[e + f*x])^(n - 2), x], x] + Simp[b^2*((n - 2)/ (n - 1)) Int[(c + d*x)^m*(b*Csc[e + f*x])^(n - 2), x], x]) /; FreeQ[{b, c , d, e, f}, x] && GtQ[n, 1] && NeQ[n, 2] && GtQ[m, 1]
Int[((c_.) + (d_.)*(x_))^(m_.)*Sec[(a_.) + (b_.)*(x_)]^(n_.)*Tan[(a_.) + (b _.)*(x_)]^(p_.), x_Symbol] :> Simp[(c + d*x)^m*(Sec[a + b*x]^n/(b*n)), x] - Simp[d*(m/(b*n)) Int[(c + d*x)^(m - 1)*Sec[a + b*x]^n, x], x] /; FreeQ[{ a, b, c, d, n}, x] && EqQ[p, 1] && GtQ[m, 0]
Int[(((e_.) + (f_.)*(x_))^(m_.)*Sec[(c_.) + (d_.)*(x_)]^(n_.))/((a_) + (b_. )*Sin[(c_.) + (d_.)*(x_)]), x_Symbol] :> Simp[1/a Int[(e + f*x)^m*Sec[c + d*x]^(n + 2), x], x] - Simp[1/b Int[(e + f*x)^m*Sec[c + d*x]^(n + 1)*Tan [c + d*x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && IGtQ[m, 0] && EqQ[a ^2 - b^2, 0]
Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d , e, n, p}, x] && EqQ[b*d, a*e]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1134 vs. \(2 (438 ) = 876\).
Time = 0.86 (sec) , antiderivative size = 1135, normalized size of antiderivative = 2.39
5/a/d^3*f^2*e*ln(1-I*exp(I*(d*x+c)))*c+5/a/d^2*f^2*e*ln(1-I*exp(I*(d*x+c)) )*x+8/a/d^3*e*f^2*c*ln(exp(I*(d*x+c)))-4/a/d^3*c*f^2*e*ln(1+exp(2*I*(d*x+c )))+3/2/a/d^2*f^3*ln(1+I*exp(I*(d*x+c)))*x^2-4/3*I/a/d*x^3*f^3-2*I/a/d^4*f ^3*arctan(exp(I*(d*x+c)))+8/3*I/a/d^4*c^3*f^3+3/a/d^2*e*f^2*ln(1+I*exp(I*( d*x+c)))*x+3/a/d^3*e*f^2*ln(1+I*exp(I*(d*x+c)))*c-I/a/d^2*e^2*f*arctan(exp (I*(d*x+c)))-5*I/a/d^3*e*f^2*polylog(2,I*exp(I*(d*x+c)))-3*I/a/d^3*e*f^2*p olylog(2,-I*exp(I*(d*x+c)))-4*I/a/d^3*e*f^2*c^2-I/a/d^4*f^3*c^2*arctan(exp (I*(d*x+c)))+4*I/a/d^3*c^2*f^3*x-5*I/a/d^3*f^3*polylog(2,I*exp(I*(d*x+c))) *x-3*I/a/d^3*f^3*polylog(2,-I*exp(I*(d*x+c)))*x-4*I/a/d*e*f^2*x^2-3/2/a/d^ 4*c^2*f^3*ln(1+I*exp(I*(d*x+c)))+2*I/a/d^3*e*f^2*c*arctan(exp(I*(d*x+c)))- 8*I/a/d^2*e*f^2*c*x+1/a/d^4*f^3*ln(1+exp(2*I*(d*x+c)))-2/a/d^4*f^3*ln(exp( I*(d*x+c)))-1/3*(6*I*f^3*x+6*I*e*f^2*exp(2*I*(d*x+c))+12*I*d^2*e*f^2*x^2+6 *I*d*e*f^2*x*exp(I*(d*x+c))+6*I*d*e*f^2*x*exp(3*I*(d*x+c))+6*I*e*f^2+24*d^ 2*e*f^2*x^2*exp(I*(d*x+c))+24*d^2*e^2*f*x*exp(I*(d*x+c))+3*I*d*f^3*x^2*exp (I*(d*x+c))+3*I*d*f^3*x^2*exp(3*I*(d*x+c))+8*d^2*e^3*exp(I*(d*x+c))+4*I*d^ 2*e^3+8*d^2*f^3*x^3*exp(I*(d*x+c))+6*I*f^3*x*exp(2*I*(d*x+c))+3*I*d*e^2*f* exp(3*I*(d*x+c))+4*I*d^2*f^3*x^3+6*f^3*x*exp(3*I*(d*x+c))+6*e*f^2*exp(3*I* (d*x+c))+6*f^3*x*exp(I*(d*x+c))+6*e*f^2*exp(I*(d*x+c))+12*I*d^2*e^2*f*x+3* I*d*e^2*f*exp(I*(d*x+c)))/(-I+exp(I*(d*x+c)))/(exp(I*(d*x+c))+I)^3/d^3/a+3 *f^3*polylog(3,-I*exp(I*(d*x+c)))/a/d^4-4/a/d^4*f^3*c^2*ln(exp(I*(d*x+c...
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1531 vs. \(2 (426) = 852\).
Time = 0.36 (sec) , antiderivative size = 1531, normalized size of antiderivative = 3.22 \[ \int \frac {(e+f x)^3 \sec ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Too large to display} \]
1/12*(4*d^3*f^3*x^3 + 12*d^3*e*f^2*x^2 + 12*d^3*e^2*f*x + 4*d^3*e^3 - 4*(2 *d^3*f^3*x^3 + 6*d^3*e*f^2*x^2 + 2*d^3*e^3 + 3*d*e*f^2 + 3*(2*d^3*e^2*f + d*f^3)*x)*cos(d*x + c)^2 - 6*(d^2*f^3*x^2 + 2*d^2*e*f^2*x + d^2*e^2*f)*cos (d*x + c) - 18*((-I*d*f^3*x - I*d*e*f^2)*cos(d*x + c)*sin(d*x + c) + (-I*d *f^3*x - I*d*e*f^2)*cos(d*x + c))*dilog(I*cos(d*x + c) + sin(d*x + c)) - 3 0*((I*d*f^3*x + I*d*e*f^2)*cos(d*x + c)*sin(d*x + c) + (I*d*f^3*x + I*d*e* f^2)*cos(d*x + c))*dilog(I*cos(d*x + c) - sin(d*x + c)) - 18*((I*d*f^3*x + I*d*e*f^2)*cos(d*x + c)*sin(d*x + c) + (I*d*f^3*x + I*d*e*f^2)*cos(d*x + c))*dilog(-I*cos(d*x + c) + sin(d*x + c)) - 30*((-I*d*f^3*x - I*d*e*f^2)*c os(d*x + c)*sin(d*x + c) + (-I*d*f^3*x - I*d*e*f^2)*cos(d*x + c))*dilog(-I *cos(d*x + c) - sin(d*x + c)) + 3*((5*d^2*e^2*f - 10*c*d*e*f^2 + (5*c^2 + 4)*f^3)*cos(d*x + c)*sin(d*x + c) + (5*d^2*e^2*f - 10*c*d*e*f^2 + (5*c^2 + 4)*f^3)*cos(d*x + c))*log(cos(d*x + c) + I*sin(d*x + c) + I) + 9*((d^2*e^ 2*f - 2*c*d*e*f^2 + c^2*f^3)*cos(d*x + c)*sin(d*x + c) + (d^2*e^2*f - 2*c* d*e*f^2 + c^2*f^3)*cos(d*x + c))*log(cos(d*x + c) - I*sin(d*x + c) + I) + 15*((d^2*f^3*x^2 + 2*d^2*e*f^2*x + 2*c*d*e*f^2 - c^2*f^3)*cos(d*x + c)*sin (d*x + c) + (d^2*f^3*x^2 + 2*d^2*e*f^2*x + 2*c*d*e*f^2 - c^2*f^3)*cos(d*x + c))*log(I*cos(d*x + c) + sin(d*x + c) + 1) + 9*((d^2*f^3*x^2 + 2*d^2*e*f ^2*x + 2*c*d*e*f^2 - c^2*f^3)*cos(d*x + c)*sin(d*x + c) + (d^2*f^3*x^2 + 2 *d^2*e*f^2*x + 2*c*d*e*f^2 - c^2*f^3)*cos(d*x + c))*log(I*cos(d*x + c) ...
\[ \int \frac {(e+f x)^3 \sec ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\int \frac {e^{3} \sec ^{2}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx + \int \frac {f^{3} x^{3} \sec ^{2}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx + \int \frac {3 e f^{2} x^{2} \sec ^{2}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx + \int \frac {3 e^{2} f x \sec ^{2}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \]
(Integral(e**3*sec(c + d*x)**2/(sin(c + d*x) + 1), x) + Integral(f**3*x**3 *sec(c + d*x)**2/(sin(c + d*x) + 1), x) + Integral(3*e*f**2*x**2*sec(c + d *x)**2/(sin(c + d*x) + 1), x) + Integral(3*e**2*f*x*sec(c + d*x)**2/(sin(c + d*x) + 1), x))/a
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 5130 vs. \(2 (426) = 852\).
Time = 1.12 (sec) , antiderivative size = 5130, normalized size of antiderivative = 10.80 \[ \int \frac {(e+f x)^3 \sec ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Too large to display} \]
1/12*(24*c^2*e*f^2*(sin(d*x + c)/(cos(d*x + c) + 1) + 3*sin(d*x + c)^2/(co s(d*x + c) + 1)^2 + 3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 1)/(a*d^2 + 2* a*d^2*sin(d*x + c)/(cos(d*x + c) + 1) - 2*a*d^2*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - a*d^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + 6*(4*(8*(d*x + c) *cos(d*x + c) - sin(3*d*x + 3*c) - sin(d*x + c))*cos(4*d*x + 4*c) + 16*(2* d*x + 4*(d*x + c)*sin(d*x + c) + 2*c + cos(d*x + c))*cos(3*d*x + 3*c) + 8* cos(3*d*x + 3*c)^2 + 8*cos(d*x + c)^2 + 5*(2*(2*sin(3*d*x + 3*c) + 2*sin(d *x + c) + 1)*cos(4*d*x + 4*c) - cos(4*d*x + 4*c)^2 - 4*cos(3*d*x + 3*c)^2 - 8*cos(3*d*x + 3*c)*cos(d*x + c) - 4*cos(d*x + c)^2 - 4*(cos(3*d*x + 3*c) + cos(d*x + c))*sin(4*d*x + 4*c) - sin(4*d*x + 4*c)^2 - 4*(2*sin(d*x + c) + 1)*sin(3*d*x + 3*c) - 4*sin(3*d*x + 3*c)^2 - 4*sin(d*x + c)^2 - 4*sin(d *x + c) - 1)*log(cos(d*x + c)^2 + sin(d*x + c)^2 + 2*sin(d*x + c) + 1) + 3 *(2*(2*sin(3*d*x + 3*c) + 2*sin(d*x + c) + 1)*cos(4*d*x + 4*c) - cos(4*d*x + 4*c)^2 - 4*cos(3*d*x + 3*c)^2 - 8*cos(3*d*x + 3*c)*cos(d*x + c) - 4*cos (d*x + c)^2 - 4*(cos(3*d*x + 3*c) + cos(d*x + c))*sin(4*d*x + 4*c) - sin(4 *d*x + 4*c)^2 - 4*(2*sin(d*x + c) + 1)*sin(3*d*x + 3*c) - 4*sin(3*d*x + 3* c)^2 - 4*sin(d*x + c)^2 - 4*sin(d*x + c) - 1)*log(cos(d*x + c)^2 + sin(d*x + c)^2 - 2*sin(d*x + c) + 1) + 4*(4*d*x + 8*(d*x + c)*sin(d*x + c) + 4*c + cos(3*d*x + 3*c) + cos(d*x + c))*sin(4*d*x + 4*c) - 4*(16*(d*x + c)*cos( d*x + c) - 4*sin(d*x + c) - 1)*sin(3*d*x + 3*c) + 8*sin(3*d*x + 3*c)^2 ...
\[ \int \frac {(e+f x)^3 \sec ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\int { \frac {{\left (f x + e\right )}^{3} \sec \left (d x + c\right )^{2}}{a \sin \left (d x + c\right ) + a} \,d x } \]
Timed out. \[ \int \frac {(e+f x)^3 \sec ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Hanged} \]