Integrand size = 22, antiderivative size = 162 \[ \int (c+d x) \csc ^2(a+b x) \sec ^3(a+b x) \, dx=-\frac {3 i d x \arctan \left (e^{i (a+b x)}\right )}{b}-\frac {d \text {arctanh}(\cos (a+b x))}{b^2}+\frac {3 c \text {arctanh}(\sin (a+b x))}{2 b}-\frac {3 (c+d x) \csc (a+b x)}{2 b}+\frac {3 i d \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{2 b^2}-\frac {3 i d \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{2 b^2}-\frac {d \sec (a+b x)}{2 b^2}+\frac {(c+d x) \csc (a+b x) \sec ^2(a+b x)}{2 b} \]
-3*I*d*x*arctan(exp(I*(b*x+a)))/b-d*arctanh(cos(b*x+a))/b^2+3/2*c*arctanh( sin(b*x+a))/b-3/2*(d*x+c)*csc(b*x+a)/b+3/2*I*d*polylog(2,-I*exp(I*(b*x+a)) )/b^2-3/2*I*d*polylog(2,I*exp(I*(b*x+a)))/b^2-1/2*d*sec(b*x+a)/b^2+1/2*(d* x+c)*csc(b*x+a)*sec(b*x+a)^2/b
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 6.96 (sec) , antiderivative size = 669, normalized size of antiderivative = 4.13 \[ \int (c+d x) \csc ^2(a+b x) \sec ^3(a+b x) \, dx=\frac {d \left (a \cos \left (\frac {1}{2} (a+b x)\right )-(a+b x) \cos \left (\frac {1}{2} (a+b x)\right )\right ) \csc \left (\frac {1}{2} (a+b x)\right )}{2 b^2}-\frac {c \csc (a+b x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},2,\frac {1}{2},\sin ^2(a+b x)\right )}{b}-\frac {d \log \left (\cos \left (\frac {1}{2} (a+b x)\right )\right )}{b^2}+\frac {d \log \left (\sin \left (\frac {1}{2} (a+b x)\right )\right )}{b^2}-\frac {3 d x \left (a \log \left (1-\tan \left (\frac {1}{2} (a+b x)\right )\right )-a \log \left (1+\tan \left (\frac {1}{2} (a+b x)\right )\right )-i \left (\log \left (1+i \tan \left (\frac {1}{2} (a+b x)\right )\right ) \log \left (\left (\frac {1}{2}-\frac {i}{2}\right ) \left (1+\tan \left (\frac {1}{2} (a+b x)\right )\right )\right )+\operatorname {PolyLog}\left (2,\frac {1}{2} \left ((1+i)-(1-i) \tan \left (\frac {1}{2} (a+b x)\right )\right )\right )\right )+i \left (\log \left (1-i \tan \left (\frac {1}{2} (a+b x)\right )\right ) \log \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \left (1+\tan \left (\frac {1}{2} (a+b x)\right )\right )\right )+\operatorname {PolyLog}\left (2,\left (-\frac {1}{2}-\frac {i}{2}\right ) \left (i+\tan \left (\frac {1}{2} (a+b x)\right )\right )\right )\right )-i \left (\log \left (1-i \tan \left (\frac {1}{2} (a+b x)\right )\right ) \log \left (\left (-\frac {1}{2}+\frac {i}{2}\right ) \left (-1+\tan \left (\frac {1}{2} (a+b x)\right )\right )\right )+\operatorname {PolyLog}\left (2,\frac {1}{2} \left ((1+i)+(1-i) \tan \left (\frac {1}{2} (a+b x)\right )\right )\right )\right )+i \left (\log \left (1+i \tan \left (\frac {1}{2} (a+b x)\right )\right ) \log \left (\left (-\frac {1}{2}-\frac {i}{2}\right ) \left (-1+\tan \left (\frac {1}{2} (a+b x)\right )\right )\right )+\operatorname {PolyLog}\left (2,\frac {1}{2} \left ((1-i)+(1+i) \tan \left (\frac {1}{2} (a+b x)\right )\right )\right )\right )\right )}{2 b \left (a-i \log \left (1-i \tan \left (\frac {1}{2} (a+b x)\right )\right )+i \log \left (1+i \tan \left (\frac {1}{2} (a+b x)\right )\right )\right )}+\frac {d x}{4 b \left (\cos \left (\frac {1}{2} (a+b x)\right )-\sin \left (\frac {1}{2} (a+b x)\right )\right )^2}-\frac {d \sin \left (\frac {1}{2} (a+b x)\right )}{2 b^2 \left (\cos \left (\frac {1}{2} (a+b x)\right )-\sin \left (\frac {1}{2} (a+b x)\right )\right )}-\frac {d x}{4 b \left (\cos \left (\frac {1}{2} (a+b x)\right )+\sin \left (\frac {1}{2} (a+b x)\right )\right )^2}+\frac {d \sin \left (\frac {1}{2} (a+b x)\right )}{2 b^2 \left (\cos \left (\frac {1}{2} (a+b x)\right )+\sin \left (\frac {1}{2} (a+b x)\right )\right )}+\frac {d \sec \left (\frac {1}{2} (a+b x)\right ) \left (a \sin \left (\frac {1}{2} (a+b x)\right )-(a+b x) \sin \left (\frac {1}{2} (a+b x)\right )\right )}{2 b^2} \]
(d*(a*Cos[(a + b*x)/2] - (a + b*x)*Cos[(a + b*x)/2])*Csc[(a + b*x)/2])/(2* b^2) - (c*Csc[a + b*x]*Hypergeometric2F1[-1/2, 2, 1/2, Sin[a + b*x]^2])/b - (d*Log[Cos[(a + b*x)/2]])/b^2 + (d*Log[Sin[(a + b*x)/2]])/b^2 - (3*d*x*( a*Log[1 - Tan[(a + b*x)/2]] - a*Log[1 + Tan[(a + b*x)/2]] - I*(Log[1 + I*T an[(a + b*x)/2]]*Log[(1/2 - I/2)*(1 + Tan[(a + b*x)/2])] + PolyLog[2, ((1 + I) - (1 - I)*Tan[(a + b*x)/2])/2]) + I*(Log[1 - I*Tan[(a + b*x)/2]]*Log[ (1/2 + I/2)*(1 + Tan[(a + b*x)/2])] + PolyLog[2, (-1/2 - I/2)*(I + Tan[(a + b*x)/2])]) - I*(Log[1 - I*Tan[(a + b*x)/2]]*Log[(-1/2 + I/2)*(-1 + Tan[( a + b*x)/2])] + PolyLog[2, ((1 + I) + (1 - I)*Tan[(a + b*x)/2])/2]) + I*(L og[1 + I*Tan[(a + b*x)/2]]*Log[(-1/2 - I/2)*(-1 + Tan[(a + b*x)/2])] + Pol yLog[2, ((1 - I) + (1 + I)*Tan[(a + b*x)/2])/2])))/(2*b*(a - I*Log[1 - I*T an[(a + b*x)/2]] + I*Log[1 + I*Tan[(a + b*x)/2]])) + (d*x)/(4*b*(Cos[(a + b*x)/2] - Sin[(a + b*x)/2])^2) - (d*Sin[(a + b*x)/2])/(2*b^2*(Cos[(a + b*x )/2] - Sin[(a + b*x)/2])) - (d*x)/(4*b*(Cos[(a + b*x)/2] + Sin[(a + b*x)/2 ])^2) + (d*Sin[(a + b*x)/2])/(2*b^2*(Cos[(a + b*x)/2] + Sin[(a + b*x)/2])) + (d*Sec[(a + b*x)/2]*(a*Sin[(a + b*x)/2] - (a + b*x)*Sin[(a + b*x)/2]))/ (2*b^2)
Time = 0.38 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.10, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {4920, 2009}
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.
| \(\displaystyle \int (c+d x) \csc ^2(a+b x) \sec ^3(a+b x) \, dx\) |
| \(\Big \downarrow \) 4920 |
| \(\displaystyle -d \int \left (\frac {\csc (a+b x) \sec ^2(a+b x)}{2 b}+\frac {3 \text {arctanh}(\sin (a+b x))}{2 b}-\frac {3 \csc (a+b x)}{2 b}\right )dx+\frac {3 (c+d x) \text {arctanh}(\sin (a+b x))}{2 b}-\frac {3 (c+d x) \csc (a+b x)}{2 b}+\frac {(c+d x) \csc (a+b x) \sec ^2(a+b x)}{2 b}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -d \left (\frac {3 i x \arctan \left (e^{i (a+b x)}\right )}{b}+\frac {\text {arctanh}(\cos (a+b x))}{b^2}+\frac {3 x \text {arctanh}(\sin (a+b x))}{2 b}-\frac {3 i \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{2 b^2}+\frac {3 i \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{2 b^2}+\frac {\sec (a+b x)}{2 b^2}\right )+\frac {3 (c+d x) \text {arctanh}(\sin (a+b x))}{2 b}-\frac {3 (c+d x) \csc (a+b x)}{2 b}+\frac {(c+d x) \csc (a+b x) \sec ^2(a+b x)}{2 b}\) |
(3*(c + d*x)*ArcTanh[Sin[a + b*x]])/(2*b) - (3*(c + d*x)*Csc[a + b*x])/(2* b) + ((c + d*x)*Csc[a + b*x]*Sec[a + b*x]^2)/(2*b) - d*(((3*I)*x*ArcTan[E^ (I*(a + b*x))])/b + ArcTanh[Cos[a + b*x]]/b^2 + (3*x*ArcTanh[Sin[a + b*x]] )/(2*b) - (((3*I)/2)*PolyLog[2, (-I)*E^(I*(a + b*x))])/b^2 + (((3*I)/2)*Po lyLog[2, I*E^(I*(a + b*x))])/b^2 + Sec[a + b*x]/(2*b^2))
3.4.19.3.1 Defintions of rubi rules used
Int[Csc[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sec[(a_.) + (b _.)*(x_)]^(p_.), x_Symbol] :> Module[{u = IntHide[Csc[a + b*x]^n*Sec[a + b* x]^p, x]}, Simp[(c + d*x)^m u, x] - Simp[d*m Int[(c + d*x)^(m - 1)*u, x ], x]] /; FreeQ[{a, b, c, d}, x] && IntegersQ[n, p] && GtQ[m, 0] && NeQ[n, p]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 343 vs. \(2 (139 ) = 278\).
Time = 0.90 (sec) , antiderivative size = 344, normalized size of antiderivative = 2.12
| method | result | size |
| risch | \(-\frac {i \left (3 d x b \,{\mathrm e}^{5 i \left (x b +a \right )}+3 c b \,{\mathrm e}^{5 i \left (x b +a \right )}+2 d x b \,{\mathrm e}^{3 i \left (x b +a \right )}+2 c b \,{\mathrm e}^{3 i \left (x b +a \right )}-i d \,{\mathrm e}^{5 i \left (x b +a \right )}+3 d x b \,{\mathrm e}^{i \left (x b +a \right )}+3 c b \,{\mathrm e}^{i \left (x b +a \right )}+i d \,{\mathrm e}^{i \left (x b +a \right )}\right )}{b^{2} \left ({\mathrm e}^{2 i \left (x b +a \right )}+1\right )^{2} \left ({\mathrm e}^{2 i \left (x b +a \right )}-1\right )}-\frac {3 i c \arctan \left ({\mathrm e}^{i \left (x b +a \right )}\right )}{b}+\frac {3 i d a \arctan \left ({\mathrm e}^{i \left (x b +a \right )}\right )}{b^{2}}-\frac {d \ln \left ({\mathrm e}^{i \left (x b +a \right )}+1\right )}{b^{2}}+\frac {d \ln \left ({\mathrm e}^{i \left (x b +a \right )}-1\right )}{b^{2}}-\frac {3 d \ln \left (1+i {\mathrm e}^{i \left (x b +a \right )}\right ) a}{2 b^{2}}-\frac {3 d \ln \left (1+i {\mathrm e}^{i \left (x b +a \right )}\right ) x}{2 b}+\frac {3 d \ln \left (1-i {\mathrm e}^{i \left (x b +a \right )}\right ) x}{2 b}+\frac {3 d \ln \left (1-i {\mathrm e}^{i \left (x b +a \right )}\right ) a}{2 b^{2}}+\frac {3 i d \operatorname {dilog}\left (1+i {\mathrm e}^{i \left (x b +a \right )}\right )}{2 b^{2}}-\frac {3 i d \operatorname {dilog}\left (1-i {\mathrm e}^{i \left (x b +a \right )}\right )}{2 b^{2}}\) | \(344\) |
-I/b^2/(exp(2*I*(b*x+a))+1)^2/(exp(2*I*(b*x+a))-1)*(3*d*x*b*exp(5*I*(b*x+a ))+3*c*b*exp(5*I*(b*x+a))+2*d*x*b*exp(3*I*(b*x+a))+2*c*b*exp(3*I*(b*x+a))- I*d*exp(5*I*(b*x+a))+3*d*x*b*exp(I*(b*x+a))+3*c*b*exp(I*(b*x+a))+I*d*exp(I *(b*x+a)))-3*I/b*c*arctan(exp(I*(b*x+a)))+3*I/b^2*a*d*arctan(exp(I*(b*x+a) ))-d/b^2*ln(exp(I*(b*x+a))+1)+d/b^2*ln(exp(I*(b*x+a))-1)-3/2/b^2*d*ln(1+I* exp(I*(b*x+a)))*a-3/2/b*d*ln(1+I*exp(I*(b*x+a)))*x+3/2/b*d*ln(1-I*exp(I*(b *x+a)))*x+3/2/b^2*d*ln(1-I*exp(I*(b*x+a)))*a+3/2*I/b^2*d*dilog(1+I*exp(I*( b*x+a)))-3/2*I/b^2*d*dilog(1-I*exp(I*(b*x+a)))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 592 vs. \(2 (132) = 264\).
Time = 0.32 (sec) , antiderivative size = 592, normalized size of antiderivative = 3.65 \[ \int (c+d x) \csc ^2(a+b x) \sec ^3(a+b x) \, dx=\frac {-3 i \, d \cos \left (b x + a\right )^{2} {\rm Li}_2\left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) \sin \left (b x + a\right ) - 3 i \, d \cos \left (b x + a\right )^{2} {\rm Li}_2\left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) \sin \left (b x + a\right ) + 3 i \, d \cos \left (b x + a\right )^{2} {\rm Li}_2\left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) \sin \left (b x + a\right ) + 3 i \, d \cos \left (b x + a\right )^{2} {\rm Li}_2\left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) \sin \left (b x + a\right ) + 3 \, {\left (b c - a d\right )} \cos \left (b x + a\right )^{2} \log \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) \sin \left (b x + a\right ) - 3 \, {\left (b c - a d\right )} \cos \left (b x + a\right )^{2} \log \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right ) \sin \left (b x + a\right ) - 2 \, d \cos \left (b x + a\right )^{2} \log \left (\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right ) \sin \left (b x + a\right ) + 3 \, {\left (b d x + a d\right )} \cos \left (b x + a\right )^{2} \log \left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) - 3 \, {\left (b d x + a d\right )} \cos \left (b x + a\right )^{2} \log \left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) + 3 \, {\left (b d x + a d\right )} \cos \left (b x + a\right )^{2} \log \left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) - 3 \, {\left (b d x + a d\right )} \cos \left (b x + a\right )^{2} \log \left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) + 2 \, d \cos \left (b x + a\right )^{2} \log \left (-\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right ) \sin \left (b x + a\right ) + 3 \, {\left (b c - a d\right )} \cos \left (b x + a\right )^{2} \log \left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) \sin \left (b x + a\right ) - 3 \, {\left (b c - a d\right )} \cos \left (b x + a\right )^{2} \log \left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right ) \sin \left (b x + a\right ) + 2 \, b d x - 6 \, {\left (b d x + b c\right )} \cos \left (b x + a\right )^{2} - 2 \, d \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 2 \, b c}{4 \, b^{2} \cos \left (b x + a\right )^{2} \sin \left (b x + a\right )} \]
1/4*(-3*I*d*cos(b*x + a)^2*dilog(I*cos(b*x + a) + sin(b*x + a))*sin(b*x + a) - 3*I*d*cos(b*x + a)^2*dilog(I*cos(b*x + a) - sin(b*x + a))*sin(b*x + a ) + 3*I*d*cos(b*x + a)^2*dilog(-I*cos(b*x + a) + sin(b*x + a))*sin(b*x + a ) + 3*I*d*cos(b*x + a)^2*dilog(-I*cos(b*x + a) - sin(b*x + a))*sin(b*x + a ) + 3*(b*c - a*d)*cos(b*x + a)^2*log(cos(b*x + a) + I*sin(b*x + a) + I)*si n(b*x + a) - 3*(b*c - a*d)*cos(b*x + a)^2*log(cos(b*x + a) - I*sin(b*x + a ) + I)*sin(b*x + a) - 2*d*cos(b*x + a)^2*log(1/2*cos(b*x + a) + 1/2)*sin(b *x + a) + 3*(b*d*x + a*d)*cos(b*x + a)^2*log(I*cos(b*x + a) + sin(b*x + a) + 1)*sin(b*x + a) - 3*(b*d*x + a*d)*cos(b*x + a)^2*log(I*cos(b*x + a) - s in(b*x + a) + 1)*sin(b*x + a) + 3*(b*d*x + a*d)*cos(b*x + a)^2*log(-I*cos( b*x + a) + sin(b*x + a) + 1)*sin(b*x + a) - 3*(b*d*x + a*d)*cos(b*x + a)^2 *log(-I*cos(b*x + a) - sin(b*x + a) + 1)*sin(b*x + a) + 2*d*cos(b*x + a)^2 *log(-1/2*cos(b*x + a) + 1/2)*sin(b*x + a) + 3*(b*c - a*d)*cos(b*x + a)^2* log(-cos(b*x + a) + I*sin(b*x + a) + I)*sin(b*x + a) - 3*(b*c - a*d)*cos(b *x + a)^2*log(-cos(b*x + a) - I*sin(b*x + a) + I)*sin(b*x + a) + 2*b*d*x - 6*(b*d*x + b*c)*cos(b*x + a)^2 - 2*d*cos(b*x + a)*sin(b*x + a) + 2*b*c)/( b^2*cos(b*x + a)^2*sin(b*x + a))
\[ \int (c+d x) \csc ^2(a+b x) \sec ^3(a+b x) \, dx=\int \left (c + d x\right ) \csc ^{2}{\left (a + b x \right )} \sec ^{3}{\left (a + b x \right )}\, dx \]
\[ \int (c+d x) \csc ^2(a+b x) \sec ^3(a+b x) \, dx=\int { {\left (d x + c\right )} \csc \left (b x + a\right )^{2} \sec \left (b x + a\right )^{3} \,d x } \]
1/4*(8*(b*d*x + b*c)*cos(3*b*x + 3*a)*sin(2*b*x + 2*a) - 4*(d*cos(5*b*x + 5*a) - d*cos(b*x + a) - 3*(b*d*x + b*c)*sin(5*b*x + 5*a) - 2*(b*d*x + b*c) *sin(3*b*x + 3*a) - 3*(b*d*x + b*c)*sin(b*x + a))*cos(6*b*x + 6*a) - 4*(d* cos(4*b*x + 4*a) - d*cos(2*b*x + 2*a) + 3*(b*d*x + b*c)*sin(4*b*x + 4*a) - 3*(b*d*x + b*c)*sin(2*b*x + 2*a) - d)*cos(5*b*x + 5*a) + 4*(d*cos(b*x + a ) + 2*(b*d*x + b*c)*sin(3*b*x + 3*a) + 3*(b*d*x + b*c)*sin(b*x + a))*cos(4 *b*x + 4*a) - 4*(d*cos(b*x + a) + 3*(b*d*x + b*c)*sin(b*x + a))*cos(2*b*x + 2*a) - 4*d*cos(b*x + a) + 12*(b^2*d*cos(6*b*x + 6*a)^2 + b^2*d*cos(4*b*x + 4*a)^2 + b^2*d*cos(2*b*x + 2*a)^2 + b^2*d*sin(6*b*x + 6*a)^2 + b^2*d*si n(4*b*x + 4*a)^2 - 2*b^2*d*sin(4*b*x + 4*a)*sin(2*b*x + 2*a) + b^2*d*sin(2 *b*x + 2*a)^2 + 2*b^2*d*cos(2*b*x + 2*a) + b^2*d + 2*(b^2*d*cos(4*b*x + 4* a) - b^2*d*cos(2*b*x + 2*a) - b^2*d)*cos(6*b*x + 6*a) - 2*(b^2*d*cos(2*b*x + 2*a) + b^2*d)*cos(4*b*x + 4*a) + 2*(b^2*d*sin(4*b*x + 4*a) - b^2*d*sin( 2*b*x + 2*a))*sin(6*b*x + 6*a))*integrate((x*cos(2*b*x + 2*a)*cos(b*x + a) + x*sin(2*b*x + 2*a)*sin(b*x + a) + x*cos(b*x + a))/(cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 + 2*cos(2*b*x + 2*a) + 1), x) + 3*(b*c*cos(6*b*x + 6*a )^2 + b*c*cos(4*b*x + 4*a)^2 + b*c*cos(2*b*x + 2*a)^2 + b*c*sin(6*b*x + 6* a)^2 + b*c*sin(4*b*x + 4*a)^2 - 2*b*c*sin(4*b*x + 4*a)*sin(2*b*x + 2*a) + b*c*sin(2*b*x + 2*a)^2 + 2*b*c*cos(2*b*x + 2*a) + b*c + 2*(b*c*cos(4*b*x + 4*a) - b*c*cos(2*b*x + 2*a) - b*c)*cos(6*b*x + 6*a) - 2*(b*c*cos(2*b*x...
\[ \int (c+d x) \csc ^2(a+b x) \sec ^3(a+b x) \, dx=\int { {\left (d x + c\right )} \csc \left (b x + a\right )^{2} \sec \left (b x + a\right )^{3} \,d x } \]
Timed out. \[ \int (c+d x) \csc ^2(a+b x) \sec ^3(a+b x) \, dx=\text {Hanged} \]