Integrand size = 18, antiderivative size = 103 \[ \int (c+d x)^2 (a+b \tanh (e+f x)) \, dx=\frac {a (c+d x)^3}{3 d}-\frac {b (c+d x)^3}{3 d}+\frac {b (c+d x)^2 \log \left (1+e^{2 (e+f x)}\right )}{f}+\frac {b d (c+d x) \operatorname {PolyLog}\left (2,-e^{2 (e+f x)}\right )}{f^2}-\frac {b d^2 \operatorname {PolyLog}\left (3,-e^{2 (e+f x)}\right )}{2 f^3} \] Output:
1/3*a*(d*x+c)^3/d-1/3*b*(d*x+c)^3/d+b*(d*x+c)^2*ln(1+exp(2*f*x+2*e))/f+b*d *(d*x+c)*polylog(2,-exp(2*f*x+2*e))/f^2-1/2*b*d^2*polylog(3,-exp(2*f*x+2*e ))/f^3
Time = 0.34 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.43 \[ \int (c+d x)^2 (a+b \tanh (e+f x)) \, dx=\frac {1}{6} \left (\frac {b e^{2 e} \left (\frac {4 e^{-2 e} (c+d x)^3}{d}+\frac {6 \left (1+e^{-2 e}\right ) (c+d x)^2 \log \left (1+e^{-2 (e+f x)}\right )}{f}-\frac {3 d \left (1+e^{-2 e}\right ) \left (2 f (c+d x) \operatorname {PolyLog}\left (2,-e^{-2 (e+f x)}\right )+d \operatorname {PolyLog}\left (3,-e^{-2 (e+f x)}\right )\right )}{f^3}\right )}{1+e^{2 e}}+2 x \left (3 c^2+3 c d x+d^2 x^2\right ) (a+b \tanh (e))\right ) \] Input:
Integrate[(c + d*x)^2*(a + b*Tanh[e + f*x]),x]
Output:
((b*E^(2*e)*((4*(c + d*x)^3)/(d*E^(2*e)) + (6*(1 + E^(-2*e))*(c + d*x)^2*L og[1 + E^(-2*(e + f*x))])/f - (3*d*(1 + E^(-2*e))*(2*f*(c + d*x)*PolyLog[2 , -E^(-2*(e + f*x))] + d*PolyLog[3, -E^(-2*(e + f*x))]))/f^3))/(1 + E^(2*e )) + 2*x*(3*c^2 + 3*c*d*x + d^2*x^2)*(a + b*Tanh[e]))/6
Time = 0.44 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, 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.
| \(\displaystyle \int (c+d x)^2 (a+b \tanh (e+f x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (c+d x)^2 (a-i b \tan (i e+i f x))dx\) |
| \(\Big \downarrow \) 4205 |
| \(\displaystyle \int \left (a (c+d x)^2+b (c+d x)^2 \tanh (e+f x)\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a (c+d x)^3}{3 d}+\frac {b d (c+d x) \operatorname {PolyLog}\left (2,-e^{2 (e+f x)}\right )}{f^2}+\frac {b (c+d x)^2 \log \left (e^{2 (e+f x)}+1\right )}{f}-\frac {b (c+d x)^3}{3 d}-\frac {b d^2 \operatorname {PolyLog}\left (3,-e^{2 (e+f x)}\right )}{2 f^3}\) |
Input:
Int[(c + d*x)^2*(a + b*Tanh[e + f*x]),x]
Output:
(a*(c + d*x)^3)/(3*d) - (b*(c + d*x)^3)/(3*d) + (b*(c + d*x)^2*Log[1 + E^( 2*(e + f*x))])/f + (b*d*(c + d*x)*PolyLog[2, -E^(2*(e + f*x))])/f^2 - (b*d ^2*PolyLog[3, -E^(2*(e + f*x))])/(2*f^3)
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. \(289\) vs. \(2(97)=194\).
Time = 0.55 (sec) , antiderivative size = 290, normalized size of antiderivative = 2.82
| method | result | size |
| risch | \(\frac {d^{2} a \,x^{3}}{3}-\frac {d^{2} b \,x^{3}}{3}+d a c \,x^{2}-d b c \,x^{2}+a \,c^{2} x +b \,c^{2} x +\frac {a \,c^{3}}{3 d}+\frac {b \,c^{3}}{3 d}-\frac {2 b \,c^{2} \ln \left ({\mathrm e}^{f x +e}\right )}{f}+\frac {b \,c^{2} \ln \left (1+{\mathrm e}^{2 f x +2 e}\right )}{f}+\frac {2 b \,d^{2} e^{2} x}{f^{2}}+\frac {4 b \,d^{2} e^{3}}{3 f^{3}}+\frac {b \,d^{2} \operatorname {polylog}\left (2, -{\mathrm e}^{2 f x +2 e}\right ) x}{f^{2}}-\frac {2 b d c \,e^{2}}{f^{2}}+\frac {2 b d c \ln \left (1+{\mathrm e}^{2 f x +2 e}\right ) x}{f}+\frac {b d c \operatorname {polylog}\left (2, -{\mathrm e}^{2 f x +2 e}\right )}{f^{2}}+\frac {4 b d c e \ln \left ({\mathrm e}^{f x +e}\right )}{f^{2}}-\frac {2 b \,d^{2} e^{2} \ln \left ({\mathrm e}^{f x +e}\right )}{f^{3}}+\frac {b \,d^{2} \ln \left (1+{\mathrm e}^{2 f x +2 e}\right ) x^{2}}{f}-\frac {b \,d^{2} \operatorname {polylog}\left (3, -{\mathrm e}^{2 f x +2 e}\right )}{2 f^{3}}-\frac {4 b d c e x}{f}\) | \(290\) |
Input:
int((d*x+c)^2*(a+b*tanh(f*x+e)),x,method=_RETURNVERBOSE)
Output:
1/3*d^2*a*x^3-1/3*d^2*b*x^3+d*a*c*x^2-d*b*c*x^2+a*c^2*x+b*c^2*x+1/3/d*a*c^ 3+1/3/d*b*c^3-2/f*b*c^2*ln(exp(f*x+e))+1/f*b*c^2*ln(1+exp(2*f*x+2*e))+2/f^ 2*b*d^2*e^2*x+4/3/f^3*b*d^2*e^3+1/f^2*b*d^2*polylog(2,-exp(2*f*x+2*e))*x-2 /f^2*b*d*c*e^2+2/f*b*d*c*ln(1+exp(2*f*x+2*e))*x+1/f^2*b*d*c*polylog(2,-exp (2*f*x+2*e))+4/f^2*b*d*c*e*ln(exp(f*x+e))-2/f^3*b*d^2*e^2*ln(exp(f*x+e))+1 /f*b*d^2*ln(1+exp(2*f*x+2*e))*x^2-1/2*b*d^2*polylog(3,-exp(2*f*x+2*e))/f^3 -4/f*b*d*c*e*x
Result contains complex when optimal does not.
Time = 0.10 (sec) , antiderivative size = 364, normalized size of antiderivative = 3.53 \[ \int (c+d x)^2 (a+b \tanh (e+f x)) \, dx=\frac {{\left (a - b\right )} d^{2} f^{3} x^{3} + 3 \, {\left (a - b\right )} c d f^{3} x^{2} + 3 \, {\left (a - b\right )} c^{2} f^{3} x - 6 \, b d^{2} {\rm polylog}\left (3, i \, \cosh \left (f x + e\right ) + i \, \sinh \left (f x + e\right )\right ) - 6 \, b d^{2} {\rm polylog}\left (3, -i \, \cosh \left (f x + e\right ) - i \, \sinh \left (f x + e\right )\right ) + 6 \, {\left (b d^{2} f x + b c d f\right )} {\rm Li}_2\left (i \, \cosh \left (f x + e\right ) + i \, \sinh \left (f x + e\right )\right ) + 6 \, {\left (b d^{2} f x + b c d f\right )} {\rm Li}_2\left (-i \, \cosh \left (f x + e\right ) - i \, \sinh \left (f x + e\right )\right ) + 3 \, {\left (b d^{2} e^{2} - 2 \, b c d e f + b c^{2} f^{2}\right )} \log \left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right ) + i\right ) + 3 \, {\left (b d^{2} e^{2} - 2 \, b c d e f + b c^{2} f^{2}\right )} \log \left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right ) - i\right ) + 3 \, {\left (b d^{2} f^{2} x^{2} + 2 \, b c d f^{2} x - b d^{2} e^{2} + 2 \, b c d e f\right )} \log \left (i \, \cosh \left (f x + e\right ) + i \, \sinh \left (f x + e\right ) + 1\right ) + 3 \, {\left (b d^{2} f^{2} x^{2} + 2 \, b c d f^{2} x - b d^{2} e^{2} + 2 \, b c d e f\right )} \log \left (-i \, \cosh \left (f x + e\right ) - i \, \sinh \left (f x + e\right ) + 1\right )}{3 \, f^{3}} \] Input:
integrate((d*x+c)^2*(a+b*tanh(f*x+e)),x, algorithm="fricas")
Output:
1/3*((a - b)*d^2*f^3*x^3 + 3*(a - b)*c*d*f^3*x^2 + 3*(a - b)*c^2*f^3*x - 6 *b*d^2*polylog(3, I*cosh(f*x + e) + I*sinh(f*x + e)) - 6*b*d^2*polylog(3, -I*cosh(f*x + e) - I*sinh(f*x + e)) + 6*(b*d^2*f*x + b*c*d*f)*dilog(I*cosh (f*x + e) + I*sinh(f*x + e)) + 6*(b*d^2*f*x + b*c*d*f)*dilog(-I*cosh(f*x + e) - I*sinh(f*x + e)) + 3*(b*d^2*e^2 - 2*b*c*d*e*f + b*c^2*f^2)*log(cosh( f*x + e) + sinh(f*x + e) + I) + 3*(b*d^2*e^2 - 2*b*c*d*e*f + b*c^2*f^2)*lo g(cosh(f*x + e) + sinh(f*x + e) - I) + 3*(b*d^2*f^2*x^2 + 2*b*c*d*f^2*x - b*d^2*e^2 + 2*b*c*d*e*f)*log(I*cosh(f*x + e) + I*sinh(f*x + e) + 1) + 3*(b *d^2*f^2*x^2 + 2*b*c*d*f^2*x - b*d^2*e^2 + 2*b*c*d*e*f)*log(-I*cosh(f*x + e) - I*sinh(f*x + e) + 1))/f^3
\[ \int (c+d x)^2 (a+b \tanh (e+f x)) \, dx=\int \left (a + b \tanh {\left (e + f x \right )}\right ) \left (c + d x\right )^{2}\, dx \] Input:
integrate((d*x+c)**2*(a+b*tanh(f*x+e)),x)
Output:
Integral((a + b*tanh(e + f*x))*(c + d*x)**2, x)
Time = 0.13 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.74 \[ \int (c+d x)^2 (a+b \tanh (e+f x)) \, dx=\frac {1}{3} \, a d^{2} x^{3} + \frac {1}{3} \, b d^{2} x^{3} + a c d x^{2} + b c d x^{2} + a c^{2} x + \frac {b c^{2} \log \left (\cosh \left (f x + e\right )\right )}{f} + \frac {{\left (2 \, f x \log \left (e^{\left (2 \, f x + 2 \, e\right )} + 1\right ) + {\rm Li}_2\left (-e^{\left (2 \, f x + 2 \, e\right )}\right )\right )} b c d}{f^{2}} + \frac {{\left (2 \, f^{2} x^{2} \log \left (e^{\left (2 \, f x + 2 \, e\right )} + 1\right ) + 2 \, f x {\rm Li}_2\left (-e^{\left (2 \, f x + 2 \, e\right )}\right ) - {\rm Li}_{3}(-e^{\left (2 \, f x + 2 \, e\right )})\right )} b d^{2}}{2 \, f^{3}} - \frac {2 \, {\left (b d^{2} f^{3} x^{3} + 3 \, b c d f^{3} x^{2}\right )}}{3 \, f^{3}} \] Input:
integrate((d*x+c)^2*(a+b*tanh(f*x+e)),x, algorithm="maxima")
Output:
1/3*a*d^2*x^3 + 1/3*b*d^2*x^3 + a*c*d*x^2 + b*c*d*x^2 + a*c^2*x + b*c^2*lo g(cosh(f*x + e))/f + (2*f*x*log(e^(2*f*x + 2*e) + 1) + dilog(-e^(2*f*x + 2 *e)))*b*c*d/f^2 + 1/2*(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)))*b*d^2/f^3 - 2/3*(b*d^2*f^3 *x^3 + 3*b*c*d*f^3*x^2)/f^3
\[ \int (c+d x)^2 (a+b \tanh (e+f x)) \, dx=\int { {\left (d x + c\right )}^{2} {\left (b \tanh \left (f x + e\right ) + a\right )} \,d x } \] Input:
integrate((d*x+c)^2*(a+b*tanh(f*x+e)),x, algorithm="giac")
Output:
integrate((d*x + c)^2*(b*tanh(f*x + e) + a), x)
Timed out. \[ \int (c+d x)^2 (a+b \tanh (e+f x)) \, dx=\int \left (a+b\,\mathrm {tanh}\left (e+f\,x\right )\right )\,{\left (c+d\,x\right )}^2 \,d x \] Input:
int((a + b*tanh(e + f*x))*(c + d*x)^2,x)
Output:
int((a + b*tanh(e + f*x))*(c + d*x)^2, x)
\[ \int (c+d x)^2 (a+b \tanh (e+f x)) \, dx=\frac {6 e^{2 e} \left (\int \frac {e^{2 f x} x^{2}}{e^{2 f x +2 e}+1}d x \right ) b \,d^{2} f +12 e^{2 e} \left (\int \frac {e^{2 f x} x}{e^{2 f x +2 e}+1}d x \right ) b c d f +3 \,\mathrm {log}\left (e^{2 f x +2 e}+1\right ) b \,c^{2}+3 a \,c^{2} f x +3 a c d f \,x^{2}+a \,d^{2} f \,x^{3}-3 b \,c^{2} f x -3 b c d f \,x^{2}-b \,d^{2} f \,x^{3}}{3 f} \] Input:
int((d*x+c)^2*(a+b*tanh(f*x+e)),x)
Output:
(6*e**(2*e)*int((e**(2*f*x)*x**2)/(e**(2*e + 2*f*x) + 1),x)*b*d**2*f + 12* e**(2*e)*int((e**(2*f*x)*x)/(e**(2*e + 2*f*x) + 1),x)*b*c*d*f + 3*log(e**( 2*e + 2*f*x) + 1)*b*c**2 + 3*a*c**2*f*x + 3*a*c*d*f*x**2 + a*d**2*f*x**3 - 3*b*c**2*f*x - 3*b*c*d*f*x**2 - b*d**2*f*x**3)/(3*f)