Integrand size = 22, antiderivative size = 55 \[ \int (c+d x)^2 \sec ^2(a+b x) \tan (a+b x) \, dx=-\frac {d^2 \log (\cos (a+b x))}{b^3}+\frac {(c+d x)^2 \sec ^2(a+b x)}{2 b}-\frac {d (c+d x) \tan (a+b x)}{b^2} \]
Time = 0.65 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.20 \[ \int (c+d x)^2 \sec ^2(a+b x) \tan (a+b x) \, dx=\frac {b^2 (c+d x)^2 \sec ^2(a+b x)-2 b d (c+d x) \sec (a) \sec (a+b x) \sin (b x)-2 d^2 (\log (\cos (a+b x))+b x \tan (a))}{2 b^3} \]
(b^2*(c + d*x)^2*Sec[a + b*x]^2 - 2*b*d*(c + d*x)*Sec[a]*Sec[a + b*x]*Sin[ b*x] - 2*d^2*(Log[Cos[a + b*x]] + b*x*Tan[a]))/(2*b^3)
Time = 0.34 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {4909, 3042, 4672, 25, 3042, 3956}
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.
\(\displaystyle \int (c+d x)^2 \tan (a+b x) \sec ^2(a+b x) \, dx\) |
\(\Big \downarrow \) 4909 |
\(\displaystyle \frac {(c+d x)^2 \sec ^2(a+b x)}{2 b}-\frac {d \int (c+d x) \sec ^2(a+b x)dx}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(c+d x)^2 \sec ^2(a+b x)}{2 b}-\frac {d \int (c+d x) \csc \left (a+b x+\frac {\pi }{2}\right )^2dx}{b}\) |
\(\Big \downarrow \) 4672 |
\(\displaystyle \frac {(c+d x)^2 \sec ^2(a+b x)}{2 b}-\frac {d \left (\frac {d \int -\tan (a+b x)dx}{b}+\frac {(c+d x) \tan (a+b x)}{b}\right )}{b}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {(c+d x)^2 \sec ^2(a+b x)}{2 b}-\frac {d \left (\frac {(c+d x) \tan (a+b x)}{b}-\frac {d \int \tan (a+b x)dx}{b}\right )}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(c+d x)^2 \sec ^2(a+b x)}{2 b}-\frac {d \left (\frac {(c+d x) \tan (a+b x)}{b}-\frac {d \int \tan (a+b x)dx}{b}\right )}{b}\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle \frac {(c+d x)^2 \sec ^2(a+b x)}{2 b}-\frac {d \left (\frac {d \log (\cos (a+b x))}{b^2}+\frac {(c+d x) \tan (a+b x)}{b}\right )}{b}\) |
((c + d*x)^2*Sec[a + b*x]^2)/(2*b) - (d*((d*Log[Cos[a + b*x]])/b^2 + ((c + d*x)*Tan[a + b*x])/b))/b
3.3.93.3.1 Defintions of rubi rules used
Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d *x], x]]/d, x] /; FreeQ[{c, d}, x]
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[((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]
Result contains complex when optimal does not.
Time = 1.53 (sec) , antiderivative size = 149, normalized size of antiderivative = 2.71
method | result | size |
risch | \(\frac {2 i d^{2} x}{b^{2}}+\frac {2 i d^{2} a}{b^{3}}+\frac {2 b \,d^{2} x^{2} {\mathrm e}^{2 i \left (x b +a \right )}+4 b c d x \,{\mathrm e}^{2 i \left (x b +a \right )}+2 b \,c^{2} {\mathrm e}^{2 i \left (x b +a \right )}-2 i d^{2} x \,{\mathrm e}^{2 i \left (x b +a \right )}-2 i c d \,{\mathrm e}^{2 i \left (x b +a \right )}-2 i d^{2} x -2 i d c}{b^{2} \left ({\mathrm e}^{2 i \left (x b +a \right )}+1\right )^{2}}-\frac {d^{2} \ln \left ({\mathrm e}^{2 i \left (x b +a \right )}+1\right )}{b^{3}}\) | \(149\) |
derivativedivides | \(\frac {\frac {a^{2} d^{2}}{2 b^{2} \cos \left (x b +a \right )^{2}}-\frac {a c d}{b \cos \left (x b +a \right )^{2}}-\frac {2 a \,d^{2} \left (\frac {x b +a}{2 \cos \left (x b +a \right )^{2}}-\frac {\tan \left (x b +a \right )}{2}\right )}{b^{2}}+\frac {c^{2}}{2 \cos \left (x b +a \right )^{2}}+\frac {2 c d \left (\frac {x b +a}{2 \cos \left (x b +a \right )^{2}}-\frac {\tan \left (x b +a \right )}{2}\right )}{b}+\frac {d^{2} \left (\frac {\left (x b +a \right )^{2}}{2 \cos \left (x b +a \right )^{2}}-\left (x b +a \right ) \tan \left (x b +a \right )-\ln \left (\cos \left (x b +a \right )\right )\right )}{b^{2}}}{b}\) | \(165\) |
default | \(\frac {\frac {a^{2} d^{2}}{2 b^{2} \cos \left (x b +a \right )^{2}}-\frac {a c d}{b \cos \left (x b +a \right )^{2}}-\frac {2 a \,d^{2} \left (\frac {x b +a}{2 \cos \left (x b +a \right )^{2}}-\frac {\tan \left (x b +a \right )}{2}\right )}{b^{2}}+\frac {c^{2}}{2 \cos \left (x b +a \right )^{2}}+\frac {2 c d \left (\frac {x b +a}{2 \cos \left (x b +a \right )^{2}}-\frac {\tan \left (x b +a \right )}{2}\right )}{b}+\frac {d^{2} \left (\frac {\left (x b +a \right )^{2}}{2 \cos \left (x b +a \right )^{2}}-\left (x b +a \right ) \tan \left (x b +a \right )-\ln \left (\cos \left (x b +a \right )\right )\right )}{b^{2}}}{b}\) | \(165\) |
2*I*d^2/b^2*x+2*I*d^2/b^3*a+2*(b*d^2*x^2*exp(2*I*(b*x+a))+2*b*c*d*x*exp(2* I*(b*x+a))+b*c^2*exp(2*I*(b*x+a))-I*d^2*x*exp(2*I*(b*x+a))-I*c*d*exp(2*I*( b*x+a))-I*d^2*x-I*d*c)/b^2/(exp(2*I*(b*x+a))+1)^2-d^2/b^3*ln(exp(2*I*(b*x+ a))+1)
Time = 0.26 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.56 \[ \int (c+d x)^2 \sec ^2(a+b x) \tan (a+b x) \, dx=\frac {b^{2} d^{2} x^{2} + 2 \, b^{2} c d x - 2 \, d^{2} \cos \left (b x + a\right )^{2} \log \left (-\cos \left (b x + a\right )\right ) + b^{2} c^{2} - 2 \, {\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right )}{2 \, b^{3} \cos \left (b x + a\right )^{2}} \]
1/2*(b^2*d^2*x^2 + 2*b^2*c*d*x - 2*d^2*cos(b*x + a)^2*log(-cos(b*x + a)) + b^2*c^2 - 2*(b*d^2*x + b*c*d)*cos(b*x + a)*sin(b*x + a))/(b^3*cos(b*x + a )^2)
\[ \int (c+d x)^2 \sec ^2(a+b x) \tan (a+b x) \, dx=\int \left (c + d x\right )^{2} \tan {\left (a + b x \right )} \sec ^{2}{\left (a + b x \right )}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 988 vs. \(2 (53) = 106\).
Time = 0.31 (sec) , antiderivative size = 988, normalized size of antiderivative = 17.96 \[ \int (c+d x)^2 \sec ^2(a+b x) \tan (a+b x) \, dx=\text {Too large to display} \]
1/2*(c^2*tan(b*x + a)^2 - 2*a*c*d*tan(b*x + a)^2/b + a^2*d^2*tan(b*x + a)^ 2/b^2 + 4*(4*(b*x + a)*cos(2*b*x + 2*a)^2 + 4*(b*x + a)*sin(2*b*x + 2*a)^2 + (2*(b*x + a)*cos(2*b*x + 2*a) + sin(2*b*x + 2*a))*cos(4*b*x + 4*a) + 2* (b*x + a)*cos(2*b*x + 2*a) + (2*(b*x + a)*sin(2*b*x + 2*a) - cos(2*b*x + 2 *a) - 1)*sin(4*b*x + 4*a) - sin(2*b*x + 2*a))*c*d/((2*(2*cos(2*b*x + 2*a) + 1)*cos(4*b*x + 4*a) + cos(4*b*x + 4*a)^2 + 4*cos(2*b*x + 2*a)^2 + sin(4* b*x + 4*a)^2 + 4*sin(4*b*x + 4*a)*sin(2*b*x + 2*a) + 4*sin(2*b*x + 2*a)^2 + 4*cos(2*b*x + 2*a) + 1)*b) - 4*(4*(b*x + a)*cos(2*b*x + 2*a)^2 + 4*(b*x + a)*sin(2*b*x + 2*a)^2 + (2*(b*x + a)*cos(2*b*x + 2*a) + sin(2*b*x + 2*a) )*cos(4*b*x + 4*a) + 2*(b*x + a)*cos(2*b*x + 2*a) + (2*(b*x + a)*sin(2*b*x + 2*a) - cos(2*b*x + 2*a) - 1)*sin(4*b*x + 4*a) - sin(2*b*x + 2*a))*a*d^2 /((2*(2*cos(2*b*x + 2*a) + 1)*cos(4*b*x + 4*a) + cos(4*b*x + 4*a)^2 + 4*co s(2*b*x + 2*a)^2 + sin(4*b*x + 4*a)^2 + 4*sin(4*b*x + 4*a)*sin(2*b*x + 2*a ) + 4*sin(2*b*x + 2*a)^2 + 4*cos(2*b*x + 2*a) + 1)*b^2) + (8*(b*x + a)^2*c os(2*b*x + 2*a)^2 + 8*(b*x + a)^2*sin(2*b*x + 2*a)^2 + 4*(b*x + a)^2*cos(2 *b*x + 2*a) + 4*((b*x + a)^2*cos(2*b*x + 2*a) + (b*x + a)*sin(2*b*x + 2*a) )*cos(4*b*x + 4*a) - (2*(2*cos(2*b*x + 2*a) + 1)*cos(4*b*x + 4*a) + cos(4* b*x + 4*a)^2 + 4*cos(2*b*x + 2*a)^2 + sin(4*b*x + 4*a)^2 + 4*sin(4*b*x + 4 *a)*sin(2*b*x + 2*a) + 4*sin(2*b*x + 2*a)^2 + 4*cos(2*b*x + 2*a) + 1)*log( cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 + 2*cos(2*b*x + 2*a) + 1) + 4*(...
Leaf count of result is larger than twice the leaf count of optimal. 4331 vs. \(2 (53) = 106\).
Time = 1.10 (sec) , antiderivative size = 4331, normalized size of antiderivative = 78.75 \[ \int (c+d x)^2 \sec ^2(a+b x) \tan (a+b x) \, dx=\text {Too large to display} \]
1/2*(b^2*d^2*x^2*tan(1/2*b*x)^4*tan(1/2*a)^4 + 2*b^2*c*d*x*tan(1/2*b*x)^4* tan(1/2*a)^4 + 2*b^2*d^2*x^2*tan(1/2*b*x)^4*tan(1/2*a)^2 + 2*b^2*d^2*x^2*t an(1/2*b*x)^2*tan(1/2*a)^4 + b^2*c^2*tan(1/2*b*x)^4*tan(1/2*a)^4 + 4*b^2*c *d*x*tan(1/2*b*x)^4*tan(1/2*a)^2 + 4*b*d^2*x*tan(1/2*b*x)^4*tan(1/2*a)^3 + 4*b^2*c*d*x*tan(1/2*b*x)^2*tan(1/2*a)^4 + 4*b*d^2*x*tan(1/2*b*x)^3*tan(1/ 2*a)^4 - d^2*log(4*(tan(1/2*b*x)^4*tan(1/2*a)^4 - 2*tan(1/2*b*x)^4*tan(1/2 *a)^2 - 8*tan(1/2*b*x)^3*tan(1/2*a)^3 - 2*tan(1/2*b*x)^2*tan(1/2*a)^4 + ta n(1/2*b*x)^4 + 8*tan(1/2*b*x)^3*tan(1/2*a) + 20*tan(1/2*b*x)^2*tan(1/2*a)^ 2 + 8*tan(1/2*b*x)*tan(1/2*a)^3 + tan(1/2*a)^4 - 2*tan(1/2*b*x)^2 - 8*tan( 1/2*b*x)*tan(1/2*a) - 2*tan(1/2*a)^2 + 1)/(tan(1/2*b*x)^4*tan(1/2*a)^4 + 2 *tan(1/2*b*x)^4*tan(1/2*a)^2 + 2*tan(1/2*b*x)^2*tan(1/2*a)^4 + tan(1/2*b*x )^4 + 4*tan(1/2*b*x)^2*tan(1/2*a)^2 + tan(1/2*a)^4 + 2*tan(1/2*b*x)^2 + 2* tan(1/2*a)^2 + 1))*tan(1/2*b*x)^4*tan(1/2*a)^4 + b^2*d^2*x^2*tan(1/2*b*x)^ 4 + 4*b^2*d^2*x^2*tan(1/2*b*x)^2*tan(1/2*a)^2 + 2*b^2*c^2*tan(1/2*b*x)^4*t an(1/2*a)^2 + 4*b*c*d*tan(1/2*b*x)^4*tan(1/2*a)^3 + b^2*d^2*x^2*tan(1/2*a) ^4 + 2*b^2*c^2*tan(1/2*b*x)^2*tan(1/2*a)^4 + 4*b*c*d*tan(1/2*b*x)^3*tan(1/ 2*a)^4 + 2*b^2*c*d*x*tan(1/2*b*x)^4 - 4*b*d^2*x*tan(1/2*b*x)^4*tan(1/2*a) + 8*b^2*c*d*x*tan(1/2*b*x)^2*tan(1/2*a)^2 - 24*b*d^2*x*tan(1/2*b*x)^3*tan( 1/2*a)^2 + 2*d^2*log(4*(tan(1/2*b*x)^4*tan(1/2*a)^4 - 2*tan(1/2*b*x)^4*tan (1/2*a)^2 - 8*tan(1/2*b*x)^3*tan(1/2*a)^3 - 2*tan(1/2*b*x)^2*tan(1/2*a)...
Time = 27.52 (sec) , antiderivative size = 150, normalized size of antiderivative = 2.73 \[ \int (c+d x)^2 \sec ^2(a+b x) \tan (a+b x) \, dx=-\frac {\frac {{\left (c+d\,x\right )}^2}{b}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,{\left (c+d\,x\right )}^2}{b}}{2\,{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+{\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}+1}+\frac {d^2\,x\,2{}\mathrm {i}}{b^2}+\frac {b\,c^2+2\,b\,c\,d\,x-c\,d\,2{}\mathrm {i}+b\,d^2\,x^2-d^2\,x\,2{}\mathrm {i}}{b^2\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )}-\frac {d^2\,\ln \left ({\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\mathrm {e}}^{b\,x\,2{}\mathrm {i}}+1\right )}{b^3} \]