Integrand size = 20, antiderivative size = 211 \[ \int (c+d x)^2 (a+b \tanh (e+f x))^2 \, dx=-\frac {b^2 (c+d x)^2}{f}+\frac {a^2 (c+d x)^3}{3 d}-\frac {2 a b (c+d x)^3}{3 d}+\frac {b^2 (c+d x)^3}{3 d}+\frac {2 b^2 d (c+d x) \log \left (1+e^{2 (e+f x)}\right )}{f^2}+\frac {2 a b (c+d x)^2 \log \left (1+e^{2 (e+f x)}\right )}{f}+\frac {b^2 d^2 \operatorname {PolyLog}\left (2,-e^{2 (e+f x)}\right )}{f^3}+\frac {2 a b d (c+d x) \operatorname {PolyLog}\left (2,-e^{2 (e+f x)}\right )}{f^2}-\frac {a b d^2 \operatorname {PolyLog}\left (3,-e^{2 (e+f x)}\right )}{f^3}-\frac {b^2 (c+d x)^2 \tanh (e+f x)}{f} \]
-b^2*(d*x+c)^2/f+1/3*a^2*(d*x+c)^3/d-2/3*a*b*(d*x+c)^3/d+1/3*b^2*(d*x+c)^3 /d+2*b^2*d*(d*x+c)*ln(1+exp(2*f*x+2*e))/f^2+2*a*b*(d*x+c)^2*ln(1+exp(2*f*x +2*e))/f+b^2*d^2*polylog(2,-exp(2*f*x+2*e))/f^3+2*a*b*d*(d*x+c)*polylog(2, -exp(2*f*x+2*e))/f^2-a*b*d^2*polylog(3,-exp(2*f*x+2*e))/f^3-b^2*(d*x+c)^2* tanh(f*x+e)/f
Time = 1.81 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.12 \[ \int (c+d x)^2 (a+b \tanh (e+f x))^2 \, dx=\frac {1}{3} \left (\frac {2 b \left (\frac {f x \left (3 b d \left (-2 c e^{2 e}+d x\right )+2 a f \left (-3 c^2 e^{2 e}+3 c d x+d^2 x^2\right )\right )}{1+e^{2 e}}+3 d x (b d+a f (2 c+d x)) \log \left (1+e^{-2 (e+f x)}\right )+3 c (b d+a c f) \log \left (1+e^{2 (e+f x)}\right )\right )}{f^2}-\frac {3 b d (b d+2 a f (c+d x)) \operatorname {PolyLog}\left (2,-e^{-2 (e+f x)}\right )}{f^3}-\frac {3 a b d^2 \operatorname {PolyLog}\left (3,-e^{-2 (e+f x)}\right )}{f^3}-\frac {3 b^2 (c+d x)^2 \text {sech}(e) \text {sech}(e+f x) \sinh (f x)}{f}+x \left (3 c^2+3 c d x+d^2 x^2\right ) \left (a^2+b^2+2 a b \tanh (e)\right )\right ) \]
((2*b*((f*x*(3*b*d*(-2*c*E^(2*e) + d*x) + 2*a*f*(-3*c^2*E^(2*e) + 3*c*d*x + d^2*x^2)))/(1 + E^(2*e)) + 3*d*x*(b*d + a*f*(2*c + d*x))*Log[1 + E^(-2*( e + f*x))] + 3*c*(b*d + a*c*f)*Log[1 + E^(2*(e + f*x))]))/f^2 - (3*b*d*(b* d + 2*a*f*(c + d*x))*PolyLog[2, -E^(-2*(e + f*x))])/f^3 - (3*a*b*d^2*PolyL og[3, -E^(-2*(e + f*x))])/f^3 - (3*b^2*(c + d*x)^2*Sech[e]*Sech[e + f*x]*S inh[f*x])/f + x*(3*c^2 + 3*c*d*x + d^2*x^2)*(a^2 + b^2 + 2*a*b*Tanh[e]))/3
Time = 0.66 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {3042, 4205, 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.
| \(\displaystyle \int (c+d x)^2 (a+b \tanh (e+f x))^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (c+d x)^2 (a-i b \tan (i e+i f x))^2dx\) |
| \(\Big \downarrow \) 4205 |
| \(\displaystyle \int \left (a^2 (c+d x)^2+2 a b (c+d x)^2 \tanh (e+f x)+b^2 (c+d x)^2 \tanh ^2(e+f x)\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a^2 (c+d x)^3}{3 d}+\frac {2 a b d (c+d x) \operatorname {PolyLog}\left (2,-e^{2 (e+f x)}\right )}{f^2}+\frac {2 a b (c+d x)^2 \log \left (e^{2 (e+f x)}+1\right )}{f}-\frac {2 a b (c+d x)^3}{3 d}-\frac {a b d^2 \operatorname {PolyLog}\left (3,-e^{2 (e+f x)}\right )}{f^3}+\frac {2 b^2 d (c+d x) \log \left (e^{2 (e+f x)}+1\right )}{f^2}-\frac {b^2 (c+d x)^2 \tanh (e+f x)}{f}-\frac {b^2 (c+d x)^2}{f}+\frac {b^2 (c+d x)^3}{3 d}+\frac {b^2 d^2 \operatorname {PolyLog}\left (2,-e^{2 (e+f x)}\right )}{f^3}\) |
-((b^2*(c + d*x)^2)/f) + (a^2*(c + d*x)^3)/(3*d) - (2*a*b*(c + d*x)^3)/(3* d) + (b^2*(c + d*x)^3)/(3*d) + (2*b^2*d*(c + d*x)*Log[1 + E^(2*(e + f*x))] )/f^2 + (2*a*b*(c + d*x)^2*Log[1 + E^(2*(e + f*x))])/f + (b^2*d^2*PolyLog[ 2, -E^(2*(e + f*x))])/f^3 + (2*a*b*d*(c + d*x)*PolyLog[2, -E^(2*(e + f*x)) ])/f^2 - (a*b*d^2*PolyLog[3, -E^(2*(e + f*x))])/f^3 - (b^2*(c + d*x)^2*Tan h[e + f*x])/f
3.1.59.3.1 Defintions of rubi rules used
Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.) , x_Symbol] :> Int[ExpandIntegrand[(c + d*x)^m, (a + b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[m, 0] && IGtQ[n, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(541\) vs. \(2(205)=410\).
Time = 0.36 (sec) , antiderivative size = 542, normalized size of antiderivative = 2.57
| method | result | size |
| risch | \(\frac {4 b a \,d^{2} e^{2} x}{f^{2}}-\frac {4 b c d a \,e^{2}}{f^{2}}-\frac {4 b \,e^{2} a \,d^{2} \ln \left ({\mathrm e}^{f x +e}\right )}{f^{3}}+\frac {2 b a \,d^{2} \ln \left (1+{\mathrm e}^{2 f x +2 e}\right ) x^{2}}{f}+\frac {2 b^{2} \left (x^{2} d^{2}+2 c d x +c^{2}\right )}{f \left (1+{\mathrm e}^{2 f x +2 e}\right )}-\frac {4 b a \,c^{2} \ln \left ({\mathrm e}^{f x +e}\right )}{f}+\frac {2 b^{2} c d \ln \left (1+{\mathrm e}^{2 f x +2 e}\right )}{f^{2}}-\frac {4 b^{2} c d \ln \left ({\mathrm e}^{f x +e}\right )}{f^{2}}+d \,b^{2} c \,x^{2}+b^{2} c^{2} x +a^{2} d c \,x^{2}+a^{2} c^{2} x -\frac {a b \,d^{2} \operatorname {polylog}\left (3, -{\mathrm e}^{2 f x +2 e}\right )}{f^{3}}+\frac {d^{2} b^{2} x^{3}}{3}+\frac {b^{2} c^{3}}{3 d}+\frac {a^{2} d^{2} x^{3}}{3}+\frac {a^{2} c^{3}}{3 d}-\frac {2 b^{2} d^{2} x^{2}}{f}-\frac {2 b^{2} d^{2} e^{2}}{f^{3}}+\frac {2 b a \,d^{2} \operatorname {polylog}\left (2, -{\mathrm e}^{2 f x +2 e}\right ) x}{f^{2}}+\frac {2 b c d a \operatorname {polylog}\left (2, -{\mathrm e}^{2 f x +2 e}\right )}{f^{2}}-\frac {4 b^{2} d^{2} e x}{f^{2}}+\frac {b^{2} d^{2} \operatorname {polylog}\left (2, -{\mathrm e}^{2 f x +2 e}\right )}{f^{3}}+\frac {8 b a \,d^{2} e^{3}}{3 f^{3}}+\frac {4 b^{2} e \,d^{2} \ln \left ({\mathrm e}^{f x +e}\right )}{f^{3}}+\frac {2 b^{2} d^{2} \ln \left (1+{\mathrm e}^{2 f x +2 e}\right ) x}{f^{2}}+\frac {2 b a \,c^{2} \ln \left (1+{\mathrm e}^{2 f x +2 e}\right )}{f}-\frac {2 d^{2} a b \,x^{3}}{3}+\frac {2 c^{3} a b}{3 d}-2 d a b c \,x^{2}+2 a b \,c^{2} x -\frac {8 b c d a e x}{f}+\frac {4 b c d a \ln \left (1+{\mathrm e}^{2 f x +2 e}\right ) x}{f}+\frac {8 b e c d a \ln \left ({\mathrm e}^{f x +e}\right )}{f^{2}}\) | \(542\) |
4/f^2*b*a*d^2*e^2*x-4/f^2*b*c*d*a*e^2-4/f^3*b*e^2*a*d^2*ln(exp(f*x+e))+2/f *b*a*d^2*ln(1+exp(2*f*x+2*e))*x^2+2/f*b^2*(d^2*x^2+2*c*d*x+c^2)/(1+exp(2*f *x+2*e))-4/f*b*a*c^2*ln(exp(f*x+e))+2/f^2*b^2*c*d*ln(1+exp(2*f*x+2*e))-4/f ^2*b^2*c*d*ln(exp(f*x+e))+d*b^2*c*x^2+b^2*c^2*x+a^2*d*c*x^2+a^2*c^2*x-a*b* d^2*polylog(3,-exp(2*f*x+2*e))/f^3+1/3*d^2*b^2*x^3+1/3/d*b^2*c^3+1/3*a^2*d ^2*x^3+1/3*a^2/d*c^3-2/f*b^2*d^2*x^2-2/f^3*b^2*d^2*e^2+2/f^2*b*a*d^2*polyl og(2,-exp(2*f*x+2*e))*x+2/f^2*b*c*d*a*polylog(2,-exp(2*f*x+2*e))-4/f^2*b^2 *d^2*e*x+b^2*d^2*polylog(2,-exp(2*f*x+2*e))/f^3+8/3/f^3*b*a*d^2*e^3+4/f^3* b^2*e*d^2*ln(exp(f*x+e))+2/f^2*b^2*d^2*ln(1+exp(2*f*x+2*e))*x+2/f*b*a*c^2* ln(1+exp(2*f*x+2*e))-2/3*d^2*a*b*x^3+2/3/d*c^3*a*b-2*d*a*b*c*x^2+2*a*b*c^2 *x-8/f*b*c*d*a*e*x+4/f*b*c*d*a*ln(1+exp(2*f*x+2*e))*x+8/f^2*b*e*c*d*a*ln(e xp(f*x+e))
Result contains complex when optimal does not.
Time = 0.28 (sec) , antiderivative size = 2123, normalized size of antiderivative = 10.06 \[ \int (c+d x)^2 (a+b \tanh (e+f x))^2 \, dx=\text {Too large to display} \]
1/3*((a^2 - 2*a*b + b^2)*d^2*f^3*x^3 + 3*(a^2 - 2*a*b + b^2)*c*d*f^3*x^2 - 4*a*b*d^2*e^3 + 3*(a^2 - 2*a*b + b^2)*c^2*f^3*x + 6*b^2*d^2*e^2 - 6*(2*a* b*c^2*e - b^2*c^2)*f^2 + ((a^2 - 2*a*b + b^2)*d^2*f^3*x^3 - 4*a*b*d^2*e^3 - 12*a*b*c^2*e*f^2 + 6*b^2*d^2*e^2 - 3*(2*b^2*d^2*f^2 - (a^2 - 2*a*b + b^2 )*c*d*f^3)*x^2 + 12*(a*b*c*d*e^2 - b^2*c*d*e)*f - 3*(4*b^2*c*d*f^2 - (a^2 - 2*a*b + b^2)*c^2*f^3)*x)*cosh(f*x + e)^2 + 2*((a^2 - 2*a*b + b^2)*d^2*f^ 3*x^3 - 4*a*b*d^2*e^3 - 12*a*b*c^2*e*f^2 + 6*b^2*d^2*e^2 - 3*(2*b^2*d^2*f^ 2 - (a^2 - 2*a*b + b^2)*c*d*f^3)*x^2 + 12*(a*b*c*d*e^2 - b^2*c*d*e)*f - 3* (4*b^2*c*d*f^2 - (a^2 - 2*a*b + b^2)*c^2*f^3)*x)*cosh(f*x + e)*sinh(f*x + e) + ((a^2 - 2*a*b + b^2)*d^2*f^3*x^3 - 4*a*b*d^2*e^3 - 12*a*b*c^2*e*f^2 + 6*b^2*d^2*e^2 - 3*(2*b^2*d^2*f^2 - (a^2 - 2*a*b + b^2)*c*d*f^3)*x^2 + 12* (a*b*c*d*e^2 - b^2*c*d*e)*f - 3*(4*b^2*c*d*f^2 - (a^2 - 2*a*b + b^2)*c^2*f ^3)*x)*sinh(f*x + e)^2 + 12*(a*b*c*d*e^2 - b^2*c*d*e)*f + 6*(2*a*b*d^2*f*x + 2*a*b*c*d*f + b^2*d^2 + (2*a*b*d^2*f*x + 2*a*b*c*d*f + b^2*d^2)*cosh(f* x + e)^2 + 2*(2*a*b*d^2*f*x + 2*a*b*c*d*f + b^2*d^2)*cosh(f*x + e)*sinh(f* x + e) + (2*a*b*d^2*f*x + 2*a*b*c*d*f + b^2*d^2)*sinh(f*x + e)^2)*dilog(I* cosh(f*x + e) + I*sinh(f*x + e)) + 6*(2*a*b*d^2*f*x + 2*a*b*c*d*f + b^2*d^ 2 + (2*a*b*d^2*f*x + 2*a*b*c*d*f + b^2*d^2)*cosh(f*x + e)^2 + 2*(2*a*b*d^2 *f*x + 2*a*b*c*d*f + b^2*d^2)*cosh(f*x + e)*sinh(f*x + e) + (2*a*b*d^2*f*x + 2*a*b*c*d*f + b^2*d^2)*sinh(f*x + e)^2)*dilog(-I*cosh(f*x + e) - I*s...
\[ \int (c+d x)^2 (a+b \tanh (e+f x))^2 \, dx=\int \left (a + b \tanh {\left (e + f x \right )}\right )^{2} \left (c + d x\right )^{2}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 411 vs. \(2 (203) = 406\).
Time = 0.28 (sec) , antiderivative size = 411, normalized size of antiderivative = 1.95 \[ \int (c+d x)^2 (a+b \tanh (e+f x))^2 \, dx=\frac {1}{3} \, a^{2} d^{2} x^{3} + a^{2} c d x^{2} + b^{2} c^{2} {\left (x + \frac {e}{f} - \frac {2}{f {\left (e^{\left (-2 \, f x - 2 \, e\right )} + 1\right )}}\right )} + a^{2} c^{2} x + b^{2} c d {\left (\frac {f x^{2} + {\left (f x^{2} e^{\left (2 \, e\right )} - 4 \, x e^{\left (2 \, e\right )}\right )} e^{\left (2 \, f x\right )}}{f e^{\left (2 \, f x + 2 \, e\right )} + f} + \frac {2 \, \log \left ({\left (e^{\left (2 \, f x + 2 \, e\right )} + 1\right )} e^{\left (-2 \, e\right )}\right )}{f^{2}}\right )} + \frac {2 \, a b c^{2} \log \left (\cosh \left (f x + e\right )\right )}{f} + \frac {{\left (2 \, f^{2} x^{2} \log \left (e^{\left (2 \, f x + 2 \, e\right )} + 1\right ) + 2 \, f x {\rm Li}_2\left (-e^{\left (2 \, f x + 2 \, e\right )}\right ) - {\rm Li}_{3}(-e^{\left (2 \, f x + 2 \, e\right )})\right )} a b d^{2}}{f^{3}} + \frac {{\left (2 \, a b d^{2} f + b^{2} d^{2} f\right )} x^{3} + 6 \, {\left (a b c d f + b^{2} d^{2}\right )} x^{2} + {\left (6 \, a b c d f x^{2} e^{\left (2 \, e\right )} + {\left (2 \, a b d^{2} f e^{\left (2 \, e\right )} + b^{2} d^{2} f e^{\left (2 \, e\right )}\right )} x^{3}\right )} e^{\left (2 \, f x\right )}}{3 \, {\left (f e^{\left (2 \, f x + 2 \, e\right )} + f\right )}} + \frac {{\left (2 \, a b c d f + b^{2} d^{2}\right )} {\left (2 \, f x \log \left (e^{\left (2 \, f x + 2 \, e\right )} + 1\right ) + {\rm Li}_2\left (-e^{\left (2 \, f x + 2 \, e\right )}\right )\right )}}{f^{3}} - \frac {2 \, {\left (2 \, a b d^{2} f^{3} x^{3} + 3 \, {\left (2 \, a b c d f + b^{2} d^{2}\right )} f^{2} x^{2}\right )}}{3 \, f^{3}} \]
1/3*a^2*d^2*x^3 + a^2*c*d*x^2 + b^2*c^2*(x + e/f - 2/(f*(e^(-2*f*x - 2*e) + 1))) + a^2*c^2*x + b^2*c*d*((f*x^2 + (f*x^2*e^(2*e) - 4*x*e^(2*e))*e^(2* f*x))/(f*e^(2*f*x + 2*e) + f) + 2*log((e^(2*f*x + 2*e) + 1)*e^(-2*e))/f^2) + 2*a*b*c^2*log(cosh(f*x + e))/f + (2*f^2*x^2*log(e^(2*f*x + 2*e) + 1) + 2*f*x*dilog(-e^(2*f*x + 2*e)) - polylog(3, -e^(2*f*x + 2*e)))*a*b*d^2/f^3 + 1/3*((2*a*b*d^2*f + b^2*d^2*f)*x^3 + 6*(a*b*c*d*f + b^2*d^2)*x^2 + (6*a* b*c*d*f*x^2*e^(2*e) + (2*a*b*d^2*f*e^(2*e) + b^2*d^2*f*e^(2*e))*x^3)*e^(2* f*x))/(f*e^(2*f*x + 2*e) + f) + (2*a*b*c*d*f + b^2*d^2)*(2*f*x*log(e^(2*f* x + 2*e) + 1) + dilog(-e^(2*f*x + 2*e)))/f^3 - 2/3*(2*a*b*d^2*f^3*x^3 + 3* (2*a*b*c*d*f + b^2*d^2)*f^2*x^2)/f^3
\[ \int (c+d x)^2 (a+b \tanh (e+f x))^2 \, dx=\int { {\left (d x + c\right )}^{2} {\left (b \tanh \left (f x + e\right ) + a\right )}^{2} \,d x } \]
Timed out. \[ \int (c+d x)^2 (a+b \tanh (e+f x))^2 \, dx=\int {\left (a+b\,\mathrm {tanh}\left (e+f\,x\right )\right )}^2\,{\left (c+d\,x\right )}^2 \,d x \]