\(\int (c+d x)^3 (a+b \tanh (e+f x)) \, dx\) [53]

Optimal result

Integrand size = 18, antiderivative size = 137 \[ \int (c+d x)^3 (a+b \tanh (e+f x)) \, dx=\frac {a (c+d x)^4}{4 d}-\frac {b (c+d x)^4}{4 d}+\frac {b (c+d x)^3 \log \left (1+e^{2 (e+f x)}\right )}{f}+\frac {3 b d (c+d x)^2 \operatorname {PolyLog}\left (2,-e^{2 (e+f x)}\right )}{2 f^2}-\frac {3 b d^2 (c+d x) \operatorname {PolyLog}\left (3,-e^{2 (e+f x)}\right )}{2 f^3}+\frac {3 b d^3 \operatorname {PolyLog}\left (4,-e^{2 (e+f x)}\right )}{4 f^4} \] Output:

1/4*a*(d*x+c)^4/d-1/4*b*(d*x+c)^4/d+b*(d*x+c)^3*ln(1+exp(2*f*x+2*e))/f+3/2 
*b*d*(d*x+c)^2*polylog(2,-exp(2*f*x+2*e))/f^2-3/2*b*d^2*(d*x+c)*polylog(3, 
-exp(2*f*x+2*e))/f^3+3/4*b*d^3*polylog(4,-exp(2*f*x+2*e))/f^4
 

Mathematica [A] (verified)

Time = 0.46 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.18 \[ \int (c+d x)^3 (a+b \tanh (e+f x)) \, dx=\frac {1}{4} \left (\frac {2 b (c+d x)^4}{d \left (1+e^{2 e}\right )}+\frac {4 b (c+d x)^3 \log \left (1+e^{-2 (e+f x)}\right )}{f}-\frac {3 b d \left (2 f^2 (c+d x)^2 \operatorname {PolyLog}\left (2,-e^{-2 (e+f x)}\right )+d \left (2 f (c+d x) \operatorname {PolyLog}\left (3,-e^{-2 (e+f x)}\right )+d \operatorname {PolyLog}\left (4,-e^{-2 (e+f x)}\right )\right )\right )}{f^4}+x \left (4 c^3+6 c^2 d x+4 c d^2 x^2+d^3 x^3\right ) (a+b \tanh (e))\right ) \] Input:

Integrate[(c + d*x)^3*(a + b*Tanh[e + f*x]),x]
 

Output:

((2*b*(c + d*x)^4)/(d*(1 + E^(2*e))) + (4*b*(c + d*x)^3*Log[1 + E^(-2*(e + 
 f*x))])/f - (3*b*d*(2*f^2*(c + d*x)^2*PolyLog[2, -E^(-2*(e + f*x))] + d*( 
2*f*(c + d*x)*PolyLog[3, -E^(-2*(e + f*x))] + d*PolyLog[4, -E^(-2*(e + f*x 
))])))/f^4 + x*(4*c^3 + 6*c^2*d*x + 4*c*d^2*x^2 + d^3*x^3)*(a + b*Tanh[e]) 
)/4
 

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 137, 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.

\(\displaystyle \int (c+d x)^3 (a+b \tanh (e+f x)) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (c+d x)^3 (a-i b \tan (i e+i f x))dx\)

\(\Big \downarrow \) 4205

\(\displaystyle \int \left (a (c+d x)^3+b (c+d x)^3 \tanh (e+f x)\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {a (c+d x)^4}{4 d}-\frac {3 b d^2 (c+d x) \operatorname {PolyLog}\left (3,-e^{2 (e+f x)}\right )}{2 f^3}+\frac {3 b d (c+d x)^2 \operatorname {PolyLog}\left (2,-e^{2 (e+f x)}\right )}{2 f^2}+\frac {b (c+d x)^3 \log \left (e^{2 (e+f x)}+1\right )}{f}-\frac {b (c+d x)^4}{4 d}+\frac {3 b d^3 \operatorname {PolyLog}\left (4,-e^{2 (e+f x)}\right )}{4 f^4}\)

Input:

Int[(c + d*x)^3*(a + b*Tanh[e + f*x]),x]
 

Output:

(a*(c + d*x)^4)/(4*d) - (b*(c + d*x)^4)/(4*d) + (b*(c + d*x)^3*Log[1 + E^( 
2*(e + f*x))])/f + (3*b*d*(c + d*x)^2*PolyLog[2, -E^(2*(e + f*x))])/(2*f^2 
) - (3*b*d^2*(c + d*x)*PolyLog[3, -E^(2*(e + f*x))])/(2*f^3) + (3*b*d^3*Po 
lyLog[4, -E^(2*(e + f*x))])/(4*f^4)
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4205

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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(469\) vs. \(2(127)=254\).

Time = 0.74 (sec) , antiderivative size = 470, normalized size of antiderivative = 3.43

method	result	size
risch	\(-\frac {2 b \,d^{3} e^{3} x}{f^{3}}-\frac {3 b c \,d^{2} \operatorname {polylog}\left (3, -{\mathrm e}^{2 f x +2 e}\right )}{2 f^{3}}+\frac {3 b \,c^{2} d \operatorname {polylog}\left (2, -{\mathrm e}^{2 f x +2 e}\right )}{2 f^{2}}+\frac {b \,d^{3} \ln \left (1+{\mathrm e}^{2 f x +2 e}\right ) x^{3}}{f}+\frac {3 b \,d^{3} \operatorname {polylog}\left (2, -{\mathrm e}^{2 f x +2 e}\right ) x^{2}}{2 f^{2}}-\frac {3 b \,d^{3} \operatorname {polylog}\left (3, -{\mathrm e}^{2 f x +2 e}\right ) x}{2 f^{3}}+\frac {2 b \,d^{3} e^{3} \ln \left ({\mathrm e}^{f x +e}\right )}{f^{4}}+\frac {4 b c \,d^{2} e^{3}}{f^{3}}-\frac {3 b \,c^{2} d \,e^{2}}{f^{2}}+d^{2} a c \,x^{3}+\frac {d^{3} a \,x^{4}}{4}-\frac {d^{3} b \,x^{4}}{4}+\frac {a \,c^{4}}{4 d}+\frac {c^{4} b}{4 d}+\frac {3 b \,d^{3} \operatorname {polylog}\left (4, -{\mathrm e}^{2 f x +2 e}\right )}{4 f^{4}}-\frac {6 b \,c^{2} d e x}{f}+\frac {6 b c \,d^{2} e^{2} x}{f^{2}}+\frac {6 b \,c^{2} d e \ln \left ({\mathrm e}^{f x +e}\right )}{f^{2}}+\frac {3 b c \,d^{2} \ln \left (1+{\mathrm e}^{2 f x +2 e}\right ) x^{2}}{f}+\frac {3 b \,c^{2} d \ln \left (1+{\mathrm e}^{2 f x +2 e}\right ) x}{f}-\frac {6 b c \,d^{2} e^{2} \ln \left ({\mathrm e}^{f x +e}\right )}{f^{3}}-\frac {3 b \,d^{3} e^{4}}{2 f^{4}}-\frac {2 b \,c^{3} \ln \left ({\mathrm e}^{f x +e}\right )}{f}+\frac {b \,c^{3} \ln \left (1+{\mathrm e}^{2 f x +2 e}\right )}{f}+\frac {3 b c \,d^{2} \operatorname {polylog}\left (2, -{\mathrm e}^{2 f x +2 e}\right ) x}{f^{2}}-d^{2} b c \,x^{3}+\frac {3 d a \,c^{2} x^{2}}{2}-\frac {3 d b \,c^{2} x^{2}}{2}+a \,c^{3} x +b \,c^{3} x\)	\(470\)

Input:

int((d*x+c)^3*(a+b*tanh(f*x+e)),x,method=_RETURNVERBOSE)
 

Output:

-2/f^3*b*d^3*e^3*x-3/2/f^3*b*c*d^2*polylog(3,-exp(2*f*x+2*e))+3/2/f^2*b*c^ 
2*d*polylog(2,-exp(2*f*x+2*e))+1/f*b*d^3*ln(1+exp(2*f*x+2*e))*x^3+3/2/f^2* 
b*d^3*polylog(2,-exp(2*f*x+2*e))*x^2-3/2/f^3*b*d^3*polylog(3,-exp(2*f*x+2* 
e))*x+2/f^4*b*d^3*e^3*ln(exp(f*x+e))+4/f^3*b*c*d^2*e^3-3/f^2*b*c^2*d*e^2+d 
^2*a*c*x^3+1/4*d^3*a*x^4-1/4*d^3*b*x^4+1/4/d*a*c^4+1/4/d*c^4*b+3/4*b*d^3*p 
olylog(4,-exp(2*f*x+2*e))/f^4-6/f*b*c^2*d*e*x+6/f^2*b*c*d^2*e^2*x+6/f^2*b* 
c^2*d*e*ln(exp(f*x+e))+3/f*b*c*d^2*ln(1+exp(2*f*x+2*e))*x^2+3/f*b*c^2*d*ln 
(1+exp(2*f*x+2*e))*x-6/f^3*b*c*d^2*e^2*ln(exp(f*x+e))-3/2/f^4*b*d^3*e^4-2/ 
f*b*c^3*ln(exp(f*x+e))+1/f*b*c^3*ln(1+exp(2*f*x+2*e))+3/f^2*b*c*d^2*polylo 
g(2,-exp(2*f*x+2*e))*x-d^2*b*c*x^3+3/2*d*a*c^2*x^2-3/2*d*b*c^2*x^2+a*c^3*x 
+b*c^3*x
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.12 (sec) , antiderivative size = 583, normalized size of antiderivative = 4.26 \[ \int (c+d x)^3 (a+b \tanh (e+f x)) \, dx=\frac {{\left (a - b\right )} d^{3} f^{4} x^{4} + 4 \, {\left (a - b\right )} c d^{2} f^{4} x^{3} + 6 \, {\left (a - b\right )} c^{2} d f^{4} x^{2} + 4 \, {\left (a - b\right )} c^{3} f^{4} x + 24 \, b d^{3} {\rm polylog}\left (4, i \, \cosh \left (f x + e\right ) + i \, \sinh \left (f x + e\right )\right ) + 24 \, b d^{3} {\rm polylog}\left (4, -i \, \cosh \left (f x + e\right ) - i \, \sinh \left (f x + e\right )\right ) + 12 \, {\left (b d^{3} f^{2} x^{2} + 2 \, b c d^{2} f^{2} x + b c^{2} d f^{2}\right )} {\rm Li}_2\left (i \, \cosh \left (f x + e\right ) + i \, \sinh \left (f x + e\right )\right ) + 12 \, {\left (b d^{3} f^{2} x^{2} + 2 \, b c d^{2} f^{2} x + b c^{2} d f^{2}\right )} {\rm Li}_2\left (-i \, \cosh \left (f x + e\right ) - i \, \sinh \left (f x + e\right )\right ) - 4 \, {\left (b d^{3} e^{3} - 3 \, b c d^{2} e^{2} f + 3 \, b c^{2} d e f^{2} - b c^{3} f^{3}\right )} \log \left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right ) + i\right ) - 4 \, {\left (b d^{3} e^{3} - 3 \, b c d^{2} e^{2} f + 3 \, b c^{2} d e f^{2} - b c^{3} f^{3}\right )} \log \left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right ) - i\right ) + 4 \, {\left (b d^{3} f^{3} x^{3} + 3 \, b c d^{2} f^{3} x^{2} + 3 \, b c^{2} d f^{3} x + b d^{3} e^{3} - 3 \, b c d^{2} e^{2} f + 3 \, b c^{2} d e f^{2}\right )} \log \left (i \, \cosh \left (f x + e\right ) + i \, \sinh \left (f x + e\right ) + 1\right ) + 4 \, {\left (b d^{3} f^{3} x^{3} + 3 \, b c d^{2} f^{3} x^{2} + 3 \, b c^{2} d f^{3} x + b d^{3} e^{3} - 3 \, b c d^{2} e^{2} f + 3 \, b c^{2} d e f^{2}\right )} \log \left (-i \, \cosh \left (f x + e\right ) - i \, \sinh \left (f x + e\right ) + 1\right ) - 24 \, {\left (b d^{3} f x + b c d^{2} f\right )} {\rm polylog}\left (3, i \, \cosh \left (f x + e\right ) + i \, \sinh \left (f x + e\right )\right ) - 24 \, {\left (b d^{3} f x + b c d^{2} f\right )} {\rm polylog}\left (3, -i \, \cosh \left (f x + e\right ) - i \, \sinh \left (f x + e\right )\right )}{4 \, f^{4}} \] Input:

integrate((d*x+c)^3*(a+b*tanh(f*x+e)),x, algorithm="fricas")
 

Output:

1/4*((a - b)*d^3*f^4*x^4 + 4*(a - b)*c*d^2*f^4*x^3 + 6*(a - b)*c^2*d*f^4*x 
^2 + 4*(a - b)*c^3*f^4*x + 24*b*d^3*polylog(4, I*cosh(f*x + e) + I*sinh(f* 
x + e)) + 24*b*d^3*polylog(4, -I*cosh(f*x + e) - I*sinh(f*x + e)) + 12*(b* 
d^3*f^2*x^2 + 2*b*c*d^2*f^2*x + b*c^2*d*f^2)*dilog(I*cosh(f*x + e) + I*sin 
h(f*x + e)) + 12*(b*d^3*f^2*x^2 + 2*b*c*d^2*f^2*x + b*c^2*d*f^2)*dilog(-I* 
cosh(f*x + e) - I*sinh(f*x + e)) - 4*(b*d^3*e^3 - 3*b*c*d^2*e^2*f + 3*b*c^ 
2*d*e*f^2 - b*c^3*f^3)*log(cosh(f*x + e) + sinh(f*x + e) + I) - 4*(b*d^3*e 
^3 - 3*b*c*d^2*e^2*f + 3*b*c^2*d*e*f^2 - b*c^3*f^3)*log(cosh(f*x + e) + si 
nh(f*x + e) - I) + 4*(b*d^3*f^3*x^3 + 3*b*c*d^2*f^3*x^2 + 3*b*c^2*d*f^3*x 
+ b*d^3*e^3 - 3*b*c*d^2*e^2*f + 3*b*c^2*d*e*f^2)*log(I*cosh(f*x + e) + I*s 
inh(f*x + e) + 1) + 4*(b*d^3*f^3*x^3 + 3*b*c*d^2*f^3*x^2 + 3*b*c^2*d*f^3*x 
 + b*d^3*e^3 - 3*b*c*d^2*e^2*f + 3*b*c^2*d*e*f^2)*log(-I*cosh(f*x + e) - I 
*sinh(f*x + e) + 1) - 24*(b*d^3*f*x + b*c*d^2*f)*polylog(3, I*cosh(f*x + e 
) + I*sinh(f*x + e)) - 24*(b*d^3*f*x + b*c*d^2*f)*polylog(3, -I*cosh(f*x + 
 e) - I*sinh(f*x + e)))/f^4
 

Sympy [F]

\[ \int (c+d x)^3 (a+b \tanh (e+f x)) \, dx=\int \left (a + b \tanh {\left (e + f x \right )}\right ) \left (c + d x\right )^{3}\, dx \] Input:

integrate((d*x+c)**3*(a+b*tanh(f*x+e)),x)
 

Output:

Integral((a + b*tanh(e + f*x))*(c + d*x)**3, x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 304 vs. \(2 (126) = 252\).

Time = 0.13 (sec) , antiderivative size = 304, normalized size of antiderivative = 2.22 \[ \int (c+d x)^3 (a+b \tanh (e+f x)) \, dx=\frac {1}{4} \, a d^{3} x^{4} + \frac {1}{4} \, b d^{3} x^{4} + a c d^{2} x^{3} + b c d^{2} x^{3} + \frac {3}{2} \, a c^{2} d x^{2} + \frac {3}{2} \, b c^{2} d x^{2} + a c^{3} x + \frac {b c^{3} \log \left (\cosh \left (f x + e\right )\right )}{f} + \frac {3 \, {\left (2 \, f x \log \left (e^{\left (2 \, f x + 2 \, e\right )} + 1\right ) + {\rm Li}_2\left (-e^{\left (2 \, f x + 2 \, e\right )}\right )\right )} b c^{2} d}{2 \, f^{2}} + \frac {3 \, {\left (2 \, f^{2} x^{2} \log \left (e^{\left (2 \, f x + 2 \, e\right )} + 1\right ) + 2 \, f x {\rm Li}_2\left (-e^{\left (2 \, f x + 2 \, e\right )}\right ) - {\rm Li}_{3}(-e^{\left (2 \, f x + 2 \, e\right )})\right )} b c d^{2}}{2 \, f^{3}} + \frac {{\left (4 \, f^{3} x^{3} \log \left (e^{\left (2 \, f x + 2 \, e\right )} + 1\right ) + 6 \, f^{2} x^{2} {\rm Li}_2\left (-e^{\left (2 \, f x + 2 \, e\right )}\right ) - 6 \, f x {\rm Li}_{3}(-e^{\left (2 \, f x + 2 \, e\right )}) + 3 \, {\rm Li}_{4}(-e^{\left (2 \, f x + 2 \, e\right )})\right )} b d^{3}}{3 \, f^{4}} - \frac {b d^{3} f^{4} x^{4} + 4 \, b c d^{2} f^{4} x^{3} + 6 \, b c^{2} d f^{4} x^{2}}{2 \, f^{4}} \] Input:

integrate((d*x+c)^3*(a+b*tanh(f*x+e)),x, algorithm="maxima")
 

Output:

1/4*a*d^3*x^4 + 1/4*b*d^3*x^4 + a*c*d^2*x^3 + b*c*d^2*x^3 + 3/2*a*c^2*d*x^ 
2 + 3/2*b*c^2*d*x^2 + a*c^3*x + b*c^3*log(cosh(f*x + e))/f + 3/2*(2*f*x*lo 
g(e^(2*f*x + 2*e) + 1) + dilog(-e^(2*f*x + 2*e)))*b*c^2*d/f^2 + 3/2*(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*c*d^2/f^3 + 1/3*(4*f^3*x^3*log(e^(2*f*x + 2*e) + 1) 
+ 6*f^2*x^2*dilog(-e^(2*f*x + 2*e)) - 6*f*x*polylog(3, -e^(2*f*x + 2*e)) + 
 3*polylog(4, -e^(2*f*x + 2*e)))*b*d^3/f^4 - 1/2*(b*d^3*f^4*x^4 + 4*b*c*d^ 
2*f^4*x^3 + 6*b*c^2*d*f^4*x^2)/f^4
 

Giac [F]

\[ \int (c+d x)^3 (a+b \tanh (e+f x)) \, dx=\int { {\left (d x + c\right )}^{3} {\left (b \tanh \left (f x + e\right ) + a\right )} \,d x } \] Input:

integrate((d*x+c)^3*(a+b*tanh(f*x+e)),x, algorithm="giac")
 

Output:

integrate((d*x + c)^3*(b*tanh(f*x + e) + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int (c+d x)^3 (a+b \tanh (e+f x)) \, dx=\int \left (a+b\,\mathrm {tanh}\left (e+f\,x\right )\right )\,{\left (c+d\,x\right )}^3 \,d x \] Input:

int((a + b*tanh(e + f*x))*(c + d*x)^3,x)
 

Output:

int((a + b*tanh(e + f*x))*(c + d*x)^3, x)
 

Reduce [F]

\[ \int (c+d x)^3 (a+b \tanh (e+f x)) \, dx=\frac {8 e^{2 e} \left (\int \frac {e^{2 f x} x^{3}}{e^{2 f x +2 e}+1}d x \right ) b \,d^{3} f +24 e^{2 e} \left (\int \frac {e^{2 f x} x^{2}}{e^{2 f x +2 e}+1}d x \right ) b c \,d^{2} f +24 e^{2 e} \left (\int \frac {e^{2 f x} x}{e^{2 f x +2 e}+1}d x \right ) b \,c^{2} d f +4 \,\mathrm {log}\left (e^{2 f x +2 e}+1\right ) b \,c^{3}+4 a \,c^{3} f x +6 a \,c^{2} d f \,x^{2}+4 a c \,d^{2} f \,x^{3}+a \,d^{3} f \,x^{4}-4 b \,c^{3} f x -6 b \,c^{2} d f \,x^{2}-4 b c \,d^{2} f \,x^{3}-b \,d^{3} f \,x^{4}}{4 f} \] Input:

int((d*x+c)^3*(a+b*tanh(f*x+e)),x)
 

Output:

(8*e**(2*e)*int((e**(2*f*x)*x**3)/(e**(2*e + 2*f*x) + 1),x)*b*d**3*f + 24* 
e**(2*e)*int((e**(2*f*x)*x**2)/(e**(2*e + 2*f*x) + 1),x)*b*c*d**2*f + 24*e 
**(2*e)*int((e**(2*f*x)*x)/(e**(2*e + 2*f*x) + 1),x)*b*c**2*d*f + 4*log(e* 
*(2*e + 2*f*x) + 1)*b*c**3 + 4*a*c**3*f*x + 6*a*c**2*d*f*x**2 + 4*a*c*d**2 
*f*x**3 + a*d**3*f*x**4 - 4*b*c**3*f*x - 6*b*c**2*d*f*x**2 - 4*b*c*d**2*f* 
x**3 - b*d**3*f*x**4)/(4*f)