\(\int x^3 \tanh ^2(a+b x) \, dx\) [83]

Optimal result

Integrand size = 12, antiderivative size = 89 \[ \int x^3 \tanh ^2(a+b x) \, dx=-\frac {x^3}{b}+\frac {x^4}{4}+\frac {3 x^2 \log \left (1+e^{2 (a+b x)}\right )}{b^2}+\frac {3 x \operatorname {PolyLog}\left (2,-e^{2 (a+b x)}\right )}{b^3}-\frac {3 \operatorname {PolyLog}\left (3,-e^{2 (a+b x)}\right )}{2 b^4}-\frac {x^3 \tanh (a+b x)}{b} \] Output:

-x^3/b+1/4*x^4+3*x^2*ln(1+exp(2*b*x+2*a))/b^2+3*x*polylog(2,-exp(2*b*x+2*a 
))/b^3-3/2*polylog(3,-exp(2*b*x+2*a))/b^4-x^3*tanh(b*x+a)/b
 

Mathematica [A] (verified)

Time = 1.02 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.11 \[ \int x^3 \tanh ^2(a+b x) \, dx=\frac {1}{4} \left (-\frac {12 x \operatorname {PolyLog}\left (2,-e^{-2 (a+b x)}\right )}{b^3}-\frac {6 \operatorname {PolyLog}\left (3,-e^{-2 (a+b x)}\right )}{b^4}+x^2 \left (\frac {8 x}{b+b e^{2 a}}+x^2+\frac {12 \log \left (1+e^{-2 (a+b x)}\right )}{b^2}-\frac {4 x \text {sech}(a) \text {sech}(a+b x) \sinh (b x)}{b}\right )\right ) \] Input:

Integrate[x^3*Tanh[a + b*x]^2,x]
 

Output:

((-12*x*PolyLog[2, -E^(-2*(a + b*x))])/b^3 - (6*PolyLog[3, -E^(-2*(a + b*x 
))])/b^4 + x^2*((8*x)/(b + b*E^(2*a)) + x^2 + (12*Log[1 + E^(-2*(a + b*x)) 
])/b^2 - (4*x*Sech[a]*Sech[a + b*x]*Sinh[b*x])/b))/4
 

Rubi [C] (verified)

Result contains complex when optimal does not.

Time = 0.64 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.27, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {3042, 25, 4203, 15, 26, 3042, 26, 4201, 2620, 3011, 2720, 7143}

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.

\(\displaystyle \int x^3 \tanh ^2(a+b x) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int -x^3 \tan (i a+i b x)^2dx\)

\(\Big \downarrow \) 25

\(\displaystyle -\int x^3 \tan (i a+i b x)^2dx\)

\(\Big \downarrow \) 4203

\(\displaystyle -\frac {3 i \int i x^2 \tanh (a+b x)dx}{b}+\int x^3dx-\frac {x^3 \tanh (a+b x)}{b}\)

\(\Big \downarrow \) 15

\(\displaystyle -\frac {3 i \int i x^2 \tanh (a+b x)dx}{b}-\frac {x^3 \tanh (a+b x)}{b}+\frac {x^4}{4}\)

\(\Big \downarrow \) 26

\(\displaystyle \frac {3 \int x^2 \tanh (a+b x)dx}{b}-\frac {x^3 \tanh (a+b x)}{b}+\frac {x^4}{4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 \int -i x^2 \tan (i a+i b x)dx}{b}-\frac {x^3 \tanh (a+b x)}{b}+\frac {x^4}{4}\)

\(\Big \downarrow \) 26

\(\displaystyle -\frac {3 i \int x^2 \tan (i a+i b x)dx}{b}-\frac {x^3 \tanh (a+b x)}{b}+\frac {x^4}{4}\)

\(\Big \downarrow \) 4201

\(\displaystyle -\frac {3 i \left (2 i \int \frac {e^{2 (a+b x)} x^2}{1+e^{2 (a+b x)}}dx-\frac {i x^3}{3}\right )}{b}-\frac {x^3 \tanh (a+b x)}{b}+\frac {x^4}{4}\)

\(\Big \downarrow \) 2620

\(\displaystyle -\frac {3 i \left (2 i \left (\frac {x^2 \log \left (e^{2 (a+b x)}+1\right )}{2 b}-\frac {\int x \log \left (1+e^{2 (a+b x)}\right )dx}{b}\right )-\frac {i x^3}{3}\right )}{b}-\frac {x^3 \tanh (a+b x)}{b}+\frac {x^4}{4}\)

\(\Big \downarrow \) 3011

\(\displaystyle -\frac {3 i \left (2 i \left (\frac {x^2 \log \left (e^{2 (a+b x)}+1\right )}{2 b}-\frac {\frac {\int \operatorname {PolyLog}\left (2,-e^{2 (a+b x)}\right )dx}{2 b}-\frac {x \operatorname {PolyLog}\left (2,-e^{2 (a+b x)}\right )}{2 b}}{b}\right )-\frac {i x^3}{3}\right )}{b}-\frac {x^3 \tanh (a+b x)}{b}+\frac {x^4}{4}\)

\(\Big \downarrow \) 2720

\(\displaystyle -\frac {3 i \left (2 i \left (\frac {x^2 \log \left (e^{2 (a+b x)}+1\right )}{2 b}-\frac {\frac {\int e^{-2 (a+b x)} \operatorname {PolyLog}\left (2,-e^{2 (a+b x)}\right )de^{2 (a+b x)}}{4 b^2}-\frac {x \operatorname {PolyLog}\left (2,-e^{2 (a+b x)}\right )}{2 b}}{b}\right )-\frac {i x^3}{3}\right )}{b}-\frac {x^3 \tanh (a+b x)}{b}+\frac {x^4}{4}\)

\(\Big \downarrow \) 7143

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

Input:

Int[x^3*Tanh[a + b*x]^2,x]
 

Output:

x^4/4 - ((3*I)*((-1/3*I)*x^3 + (2*I)*((x^2*Log[1 + E^(2*(a + b*x))])/(2*b) 
 - (-1/2*(x*PolyLog[2, -E^(2*(a + b*x))])/b + PolyLog[3, -E^(2*(a + b*x))] 
/(4*b^2))/b)))/b - (x^3*Tanh[a + b*x])/b
 

Defintions of rubi rules used

rule 15

Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ 
{a, m}, x] && NeQ[m, -1]
 

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 26

Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a])   I 
nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
 

rule 2620

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ 
((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp 
[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si 
mp[d*(m/(b*f*g*n*Log[F]))   Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x 
)))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
 

rule 2720

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] 
   Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct 
ionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ 
[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) 
*(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
 

rule 3011

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.) 
*(x_))^(m_.), x_Symbol] :> Simp[(-(f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + 
b*x)))^n]/(b*c*n*Log[F])), x] + Simp[g*(m/(b*c*n*Log[F]))   Int[(f + g*x)^( 
m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e 
, f, g, n}, x] && GtQ[m, 0]
 

rule 3042

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

rule 4201

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (Complex[0, fz_])*(f_.)*(x_)], x 
_Symbol] :> Simp[(-I)*((c + d*x)^(m + 1)/(d*(m + 1))), x] + Simp[2*I   Int[ 
(c + d*x)^m*(E^(2*((-I)*e + f*fz*x))/(1 + E^(2*((-I)*e + f*fz*x)))), x], x] 
 /; FreeQ[{c, d, e, f, fz}, x] && IGtQ[m, 0]
 

rule 4203

Int[((c_.) + (d_.)*(x_))^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symb 
ol] :> Simp[b*(c + d*x)^m*((b*Tan[e + f*x])^(n - 1)/(f*(n - 1))), x] + (-Si 
mp[b*d*(m/(f*(n - 1)))   Int[(c + d*x)^(m - 1)*(b*Tan[e + f*x])^(n - 1), x] 
, x] - Simp[b^2   Int[(c + d*x)^m*(b*Tan[e + f*x])^(n - 2), x], x]) /; Free 
Q[{b, c, d, e, f}, x] && GtQ[n, 1] && GtQ[m, 0]
 

rule 7143

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S 
ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d 
, e, n, p}, x] && EqQ[b*d, a*e]
 

Maple [A] (verified)

Time = 0.48 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.40

method	result	size
risch	\(\frac {x^{4}}{4}+\frac {2 x^{3}}{\left ({\mathrm e}^{2 b x +2 a}+1\right ) b}-\frac {6 a^{2} \ln \left ({\mathrm e}^{b x +a}\right )}{b^{4}}-\frac {2 x^{3}}{b}+\frac {6 a^{2} x}{b^{3}}+\frac {4 a^{3}}{b^{4}}+\frac {3 x^{2} \ln \left ({\mathrm e}^{2 b x +2 a}+1\right )}{b^{2}}+\frac {3 x \operatorname {polylog}\left (2, -{\mathrm e}^{2 b x +2 a}\right )}{b^{3}}-\frac {3 \operatorname {polylog}\left (3, -{\mathrm e}^{2 b x +2 a}\right )}{2 b^{4}}\)	\(125\)

Input:

int(x^3*tanh(b*x+a)^2,x,method=_RETURNVERBOSE)
 

Output:

1/4*x^4+2*x^3/(exp(2*b*x+2*a)+1)/b-6/b^4*a^2*ln(exp(b*x+a))-2/b*x^3+6/b^3* 
a^2*x+4/b^4*a^3+3*x^2*ln(exp(2*b*x+2*a)+1)/b^2+3*x*polylog(2,-exp(2*b*x+2* 
a))/b^3-3/2*polylog(3,-exp(2*b*x+2*a))/b^4
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 721, normalized size of antiderivative = 8.10 \[ \int x^3 \tanh ^2(a+b x) \, dx =\text {Too large to display} \] Input:

integrate(x^3*tanh(b*x+a)^2,x, algorithm="fricas")
 

Output:

1/4*(b^4*x^4 - 8*a^3 + (b^4*x^4 - 8*b^3*x^3 - 8*a^3)*cosh(b*x + a)^2 + 2*( 
b^4*x^4 - 8*b^3*x^3 - 8*a^3)*cosh(b*x + a)*sinh(b*x + a) + (b^4*x^4 - 8*b^ 
3*x^3 - 8*a^3)*sinh(b*x + a)^2 + 24*(b*x*cosh(b*x + a)^2 + 2*b*x*cosh(b*x 
+ a)*sinh(b*x + a) + b*x*sinh(b*x + a)^2 + b*x)*dilog(I*cosh(b*x + a) + I* 
sinh(b*x + a)) + 24*(b*x*cosh(b*x + a)^2 + 2*b*x*cosh(b*x + a)*sinh(b*x + 
a) + b*x*sinh(b*x + a)^2 + b*x)*dilog(-I*cosh(b*x + a) - I*sinh(b*x + a)) 
+ 12*(a^2*cosh(b*x + a)^2 + 2*a^2*cosh(b*x + a)*sinh(b*x + a) + a^2*sinh(b 
*x + a)^2 + a^2)*log(cosh(b*x + a) + sinh(b*x + a) + I) + 12*(a^2*cosh(b*x 
 + a)^2 + 2*a^2*cosh(b*x + a)*sinh(b*x + a) + a^2*sinh(b*x + a)^2 + a^2)*l 
og(cosh(b*x + a) + sinh(b*x + a) - I) + 12*(b^2*x^2 + (b^2*x^2 - a^2)*cosh 
(b*x + a)^2 + 2*(b^2*x^2 - a^2)*cosh(b*x + a)*sinh(b*x + a) + (b^2*x^2 - a 
^2)*sinh(b*x + a)^2 - a^2)*log(I*cosh(b*x + a) + I*sinh(b*x + a) + 1) + 12 
*(b^2*x^2 + (b^2*x^2 - a^2)*cosh(b*x + a)^2 + 2*(b^2*x^2 - a^2)*cosh(b*x + 
 a)*sinh(b*x + a) + (b^2*x^2 - a^2)*sinh(b*x + a)^2 - a^2)*log(-I*cosh(b*x 
 + a) - I*sinh(b*x + a) + 1) - 24*(cosh(b*x + a)^2 + 2*cosh(b*x + a)*sinh( 
b*x + a) + sinh(b*x + a)^2 + 1)*polylog(3, I*cosh(b*x + a) + I*sinh(b*x + 
a)) - 24*(cosh(b*x + a)^2 + 2*cosh(b*x + a)*sinh(b*x + a) + sinh(b*x + a)^ 
2 + 1)*polylog(3, -I*cosh(b*x + a) - I*sinh(b*x + a)))/(b^4*cosh(b*x + a)^ 
2 + 2*b^4*cosh(b*x + a)*sinh(b*x + a) + b^4*sinh(b*x + a)^2 + b^4)
 

Sympy [F]

\[ \int x^3 \tanh ^2(a+b x) \, dx=\int x^{3} \tanh ^{2}{\left (a + b x \right )}\, dx \] Input:

integrate(x**3*tanh(b*x+a)**2,x)
 

Output:

Integral(x**3*tanh(a + b*x)**2, x)
 

Maxima [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.21 \[ \int x^3 \tanh ^2(a+b x) \, dx=-\frac {2 \, x^{3}}{b} + \frac {b x^{4} e^{\left (2 \, b x + 2 \, a\right )} + b x^{4} + 8 \, x^{3}}{4 \, {\left (b e^{\left (2 \, b x + 2 \, a\right )} + b\right )}} + \frac {3 \, {\left (2 \, b^{2} x^{2} \log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + 2 \, b x {\rm Li}_2\left (-e^{\left (2 \, b x + 2 \, a\right )}\right ) - {\rm Li}_{3}(-e^{\left (2 \, b x + 2 \, a\right )})\right )}}{2 \, b^{4}} \] Input:

integrate(x^3*tanh(b*x+a)^2,x, algorithm="maxima")
 

Output:

-2*x^3/b + 1/4*(b*x^4*e^(2*b*x + 2*a) + b*x^4 + 8*x^3)/(b*e^(2*b*x + 2*a) 
+ b) + 3/2*(2*b^2*x^2*log(e^(2*b*x + 2*a) + 1) + 2*b*x*dilog(-e^(2*b*x + 2 
*a)) - polylog(3, -e^(2*b*x + 2*a)))/b^4
 

Giac [F]

\[ \int x^3 \tanh ^2(a+b x) \, dx=\int { x^{3} \tanh \left (b x + a\right )^{2} \,d x } \] Input:

integrate(x^3*tanh(b*x+a)^2,x, algorithm="giac")
 

Output:

integrate(x^3*tanh(b*x + a)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int x^3 \tanh ^2(a+b x) \, dx=\int x^3\,{\mathrm {tanh}\left (a+b\,x\right )}^2 \,d x \] Input:

int(x^3*tanh(a + b*x)^2,x)
 

Output:

int(x^3*tanh(a + b*x)^2, x)
 

Reduce [F]

\[ \int x^3 \tanh ^2(a+b x) \, dx=\frac {-24 e^{2 b x +2 a} \left (\int \frac {x^{2}}{e^{4 b x +4 a}+2 e^{2 b x +2 a}+1}d x \right ) b^{3}-24 e^{2 b x +2 a} \left (\int \frac {x}{e^{4 b x +4 a}+2 e^{2 b x +2 a}+1}d x \right ) b^{2}+6 e^{2 b x +2 a} \mathrm {log}\left (e^{2 b x +2 a}+1\right )+e^{2 b x +2 a} b^{4} x^{4}-12 e^{2 b x +2 a} b x -24 \left (\int \frac {x^{2}}{e^{4 b x +4 a}+2 e^{2 b x +2 a}+1}d x \right ) b^{3}-24 \left (\int \frac {x}{e^{4 b x +4 a}+2 e^{2 b x +2 a}+1}d x \right ) b^{2}+6 \,\mathrm {log}\left (e^{2 b x +2 a}+1\right )+b^{4} x^{4}+8 b^{3} x^{3}+12 b^{2} x^{2}}{4 b^{4} \left (e^{2 b x +2 a}+1\right )} \] Input:

int(x^3*tanh(b*x+a)^2,x)
 

Output:

( - 24*e**(2*a + 2*b*x)*int(x**2/(e**(4*a + 4*b*x) + 2*e**(2*a + 2*b*x) + 
1),x)*b**3 - 24*e**(2*a + 2*b*x)*int(x/(e**(4*a + 4*b*x) + 2*e**(2*a + 2*b 
*x) + 1),x)*b**2 + 6*e**(2*a + 2*b*x)*log(e**(2*a + 2*b*x) + 1) + e**(2*a 
+ 2*b*x)*b**4*x**4 - 12*e**(2*a + 2*b*x)*b*x - 24*int(x**2/(e**(4*a + 4*b* 
x) + 2*e**(2*a + 2*b*x) + 1),x)*b**3 - 24*int(x/(e**(4*a + 4*b*x) + 2*e**( 
2*a + 2*b*x) + 1),x)*b**2 + 6*log(e**(2*a + 2*b*x) + 1) + b**4*x**4 + 8*b* 
*3*x**3 + 12*b**2*x**2)/(4*b**4*(e**(2*a + 2*b*x) + 1))