3.3.29 \(\int \frac {\tanh ^3(x)}{a+b \sinh (x)} \, dx\) [229]

Optimal. Leaf size=88 \[ \frac {b \left (3 a^2+b^2\right ) \text {ArcTan}(\sinh (x))}{2 \left (a^2+b^2\right )^2}+\frac {a^3 \log (\cosh (x))}{\left (a^2+b^2\right )^2}-\frac {a^3 \log (a+b \sinh (x))}{\left (a^2+b^2\right )^2}+\frac {\text {sech}^2(x) (a-b \sinh (x))}{2 \left (a^2+b^2\right )} \]

[Out]

1/2*b*(3*a^2+b^2)*arctan(sinh(x))/(a^2+b^2)^2+a^3*ln(cosh(x))/(a^2+b^2)^2-a^3*ln(a+b*sinh(x))/(a^2+b^2)^2+1/2*
sech(x)^2*(a-b*sinh(x))/(a^2+b^2)

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Rubi [A]
time = 0.14, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {2800, 1661, 815, 649, 209, 266} \begin {gather*} \frac {b \left (3 a^2+b^2\right ) \text {ArcTan}(\sinh (x))}{2 \left (a^2+b^2\right )^2}+\frac {\text {sech}^2(x) (a-b \sinh (x))}{2 \left (a^2+b^2\right )}-\frac {a^3 \log (a+b \sinh (x))}{\left (a^2+b^2\right )^2}+\frac {a^3 \log (\cosh (x))}{\left (a^2+b^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Tanh[x]^3/(a + b*Sinh[x]),x]

[Out]

(b*(3*a^2 + b^2)*ArcTan[Sinh[x]])/(2*(a^2 + b^2)^2) + (a^3*Log[Cosh[x]])/(a^2 + b^2)^2 - (a^3*Log[a + b*Sinh[x
]])/(a^2 + b^2)^2 + (Sech[x]^2*(a - b*Sinh[x]))/(2*(a^2 + b^2))

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 649

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[(-a)*c]

Rule 815

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x)^m*((f + g*x)/(a + c*x^2)), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 1661

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(d +
e*x)^m*Pq, a + c*x^2, x], f = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 0], g = Coeff[Polyn
omialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 1]}, Simp[(a*g - c*f*x)*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1)))
, x] + Dist[1/(2*a*c*(p + 1)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*ExpandToSum[(2*a*c*(p + 1)*Q)/(d + e*x)^m +
(c*f*(2*p + 3))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] &
& LtQ[p, -1] && ILtQ[m, 0]

Rule 2800

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[(x^p*(a + x)^m)/(b^2 - x^2)^((p + 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2
- b^2, 0] && IntegerQ[(p + 1)/2]

Rubi steps

\begin {align*} \int \frac {\tanh ^3(x)}{a+b \sinh (x)} \, dx &=\text {Subst}\left (\int \frac {x^3}{(a+x) \left (-b^2-x^2\right )^2} \, dx,x,b \sinh (x)\right )\\ &=\frac {\text {sech}^2(x) (a-b \sinh (x))}{2 \left (a^2+b^2\right )}-\frac {\text {Subst}\left (\int \frac {\frac {a b^4}{a^2+b^2}+\frac {b^2 \left (2 a^2+b^2\right ) x}{a^2+b^2}}{(a+x) \left (-b^2-x^2\right )} \, dx,x,b \sinh (x)\right )}{2 b^2}\\ &=\frac {\text {sech}^2(x) (a-b \sinh (x))}{2 \left (a^2+b^2\right )}-\frac {\text {Subst}\left (\int \left (\frac {2 a^3 b^2}{\left (a^2+b^2\right )^2 (a+x)}-\frac {b^2 \left (3 a^2 b^2+b^4+2 a^3 x\right )}{\left (a^2+b^2\right )^2 \left (b^2+x^2\right )}\right ) \, dx,x,b \sinh (x)\right )}{2 b^2}\\ &=-\frac {a^3 \log (a+b \sinh (x))}{\left (a^2+b^2\right )^2}+\frac {\text {sech}^2(x) (a-b \sinh (x))}{2 \left (a^2+b^2\right )}+\frac {\text {Subst}\left (\int \frac {3 a^2 b^2+b^4+2 a^3 x}{b^2+x^2} \, dx,x,b \sinh (x)\right )}{2 \left (a^2+b^2\right )^2}\\ &=-\frac {a^3 \log (a+b \sinh (x))}{\left (a^2+b^2\right )^2}+\frac {\text {sech}^2(x) (a-b \sinh (x))}{2 \left (a^2+b^2\right )}+\frac {a^3 \text {Subst}\left (\int \frac {x}{b^2+x^2} \, dx,x,b \sinh (x)\right )}{\left (a^2+b^2\right )^2}+\frac {\left (b^2 \left (3 a^2+b^2\right )\right ) \text {Subst}\left (\int \frac {1}{b^2+x^2} \, dx,x,b \sinh (x)\right )}{2 \left (a^2+b^2\right )^2}\\ &=\frac {b \left (3 a^2+b^2\right ) \tan ^{-1}(\sinh (x))}{2 \left (a^2+b^2\right )^2}+\frac {a^3 \log (\cosh (x))}{\left (a^2+b^2\right )^2}-\frac {a^3 \log (a+b \sinh (x))}{\left (a^2+b^2\right )^2}+\frac {\text {sech}^2(x) (a-b \sinh (x))}{2 \left (a^2+b^2\right )}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.14, size = 153, normalized size = 1.74 \begin {gather*} -\frac {b \text {ArcTan}(\sinh (x))}{2 \left (a^2+b^2\right )}+\frac {\left (a^3-i \left (2 a^2 b+b^3\right )\right ) \log (i-\sinh (x))}{2 \left (a^2+b^2\right )^2}+\frac {\left (a^3+i \left (2 a^2 b+b^3\right )\right ) \log (i+\sinh (x))}{2 \left (a^2+b^2\right )^2}-\frac {a^3 \log (a+b \sinh (x))}{\left (a^2+b^2\right )^2}+\frac {a \text {sech}^2(x)}{2 \left (a^2+b^2\right )}-\frac {b \text {sech}(x) \tanh (x)}{2 \left (a^2+b^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Tanh[x]^3/(a + b*Sinh[x]),x]

[Out]

-1/2*(b*ArcTan[Sinh[x]])/(a^2 + b^2) + ((a^3 - I*(2*a^2*b + b^3))*Log[I - Sinh[x]])/(2*(a^2 + b^2)^2) + ((a^3
+ I*(2*a^2*b + b^3))*Log[I + Sinh[x]])/(2*(a^2 + b^2)^2) - (a^3*Log[a + b*Sinh[x]])/(a^2 + b^2)^2 + (a*Sech[x]
^2)/(2*(a^2 + b^2)) - (b*Sech[x]*Tanh[x])/(2*(a^2 + b^2))

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Maple [A]
time = 0.58, size = 166, normalized size = 1.89

method	result	size
default	\(\frac {\frac {2 \left (\left (\frac {1}{2} a^{2} b +\frac {1}{2} b^{3}\right ) \left (\tanh ^{3}\left (\frac {x}{2}\right )\right )+\left (-a^{3}-a \,b^{2}\right ) \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )+\left (-\frac {1}{2} a^{2} b -\frac {1}{2} b^{3}\right ) \tanh \left (\frac {x}{2}\right )\right )}{\left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{2}}+a^{3} \ln \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )+\left (3 a^{2} b +b^{3}\right ) \arctan \left (\tanh \left (\frac {x}{2}\right )\right )}{a^{4}+2 a^{2} b^{2}+b^{4}}-\frac {8 a^{3} \ln \left (a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-2 b \tanh \left (\frac {x}{2}\right )-a \right )}{8 a^{4}+16 a^{2} b^{2}+8 b^{4}}\)	\(166\)
risch	\(\frac {{\mathrm e}^{x} \left (-b \,{\mathrm e}^{2 x}+2 a \,{\mathrm e}^{x}+b \right )}{\left (1+{\mathrm e}^{2 x}\right )^{2} \left (a^{2}+b^{2}\right )}+\frac {3 i \ln \left ({\mathrm e}^{x}+i\right ) a^{2} b}{2 \left (a^{4}+2 a^{2} b^{2}+b^{4}\right )}+\frac {i \ln \left ({\mathrm e}^{x}+i\right ) b^{3}}{2 a^{4}+4 a^{2} b^{2}+2 b^{4}}+\frac {\ln \left ({\mathrm e}^{x}+i\right ) a^{3}}{a^{4}+2 a^{2} b^{2}+b^{4}}-\frac {3 i \ln \left ({\mathrm e}^{x}-i\right ) a^{2} b}{2 \left (a^{4}+2 a^{2} b^{2}+b^{4}\right )}-\frac {i \ln \left ({\mathrm e}^{x}-i\right ) b^{3}}{2 \left (a^{4}+2 a^{2} b^{2}+b^{4}\right )}+\frac {\ln \left ({\mathrm e}^{x}-i\right ) a^{3}}{a^{4}+2 a^{2} b^{2}+b^{4}}-\frac {a^{3} \ln \left ({\mathrm e}^{2 x}+\frac {2 a \,{\mathrm e}^{x}}{b}-1\right )}{a^{4}+2 a^{2} b^{2}+b^{4}}\)	\(245\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/(a^4+2*a^2*b^2+b^4)*(((1/2*a^2*b+1/2*b^3)*tanh(1/2*x)^3+(-a^3-a*b^2)*tanh(1/2*x)^2+(-1/2*a^2*b-1/2*b^3)*tanh
(1/2*x))/(tanh(1/2*x)^2+1)^2+1/2*a^3*ln(tanh(1/2*x)^2+1)+1/2*(3*a^2*b+b^3)*arctan(tanh(1/2*x)))-8*a^3/(8*a^4+1
6*a^2*b^2+8*b^4)*ln(a*tanh(1/2*x)^2-2*b*tanh(1/2*x)-a)

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Maxima [A]
time = 0.50, size = 160, normalized size = 1.82 \begin {gather*} -\frac {a^{3} \log \left (-2 \, a e^{\left (-x\right )} + b e^{\left (-2 \, x\right )} - b\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {a^{3} \log \left (e^{\left (-2 \, x\right )} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {{\left (3 \, a^{2} b + b^{3}\right )} \arctan \left (e^{\left (-x\right )}\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {b e^{\left (-x\right )} - 2 \, a e^{\left (-2 \, x\right )} - b e^{\left (-3 \, x\right )}}{a^{2} + b^{2} + 2 \, {\left (a^{2} + b^{2}\right )} e^{\left (-2 \, x\right )} + {\left (a^{2} + b^{2}\right )} e^{\left (-4 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a^3*log(-2*a*e^(-x) + b*e^(-2*x) - b)/(a^4 + 2*a^2*b^2 + b^4) + a^3*log(e^(-2*x) + 1)/(a^4 + 2*a^2*b^2 + b^4)
- (3*a^2*b + b^3)*arctan(e^(-x))/(a^4 + 2*a^2*b^2 + b^4) - (b*e^(-x) - 2*a*e^(-2*x) - b*e^(-3*x))/(a^2 + b^2
+ 2*(a^2 + b^2)*e^(-2*x) + (a^2 + b^2)*e^(-4*x))

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 655 vs. \(2 (85) = 170\).
time = 0.45, size = 655, normalized size = 7.44 \begin {gather*} -\frac {{\left (a^{2} b + b^{3}\right )} \cosh \left (x\right )^{3} + {\left (a^{2} b + b^{3}\right )} \sinh \left (x\right )^{3} - 2 \, {\left (a^{3} + a b^{2}\right )} \cosh \left (x\right )^{2} - {\left (2 \, a^{3} + 2 \, a b^{2} - 3 \, {\left (a^{2} b + b^{3}\right )} \cosh \left (x\right )\right )} \sinh \left (x\right )^{2} - {\left ({\left (3 \, a^{2} b + b^{3}\right )} \cosh \left (x\right )^{4} + 4 \, {\left (3 \, a^{2} b + b^{3}\right )} \cosh \left (x\right ) \sinh \left (x\right )^{3} + {\left (3 \, a^{2} b + b^{3}\right )} \sinh \left (x\right )^{4} + 3 \, a^{2} b + b^{3} + 2 \, {\left (3 \, a^{2} b + b^{3}\right )} \cosh \left (x\right )^{2} + 2 \, {\left (3 \, a^{2} b + b^{3} + 3 \, {\left (3 \, a^{2} b + b^{3}\right )} \cosh \left (x\right )^{2}\right )} \sinh \left (x\right )^{2} + 4 \, {\left ({\left (3 \, a^{2} b + b^{3}\right )} \cosh \left (x\right )^{3} + {\left (3 \, a^{2} b + b^{3}\right )} \cosh \left (x\right )\right )} \sinh \left (x\right )\right )} \arctan \left (\cosh \left (x\right ) + \sinh \left (x\right )\right ) - {\left (a^{2} b + b^{3}\right )} \cosh \left (x\right ) + {\left (a^{3} \cosh \left (x\right )^{4} + 4 \, a^{3} \cosh \left (x\right ) \sinh \left (x\right )^{3} + a^{3} \sinh \left (x\right )^{4} + 2 \, a^{3} \cosh \left (x\right )^{2} + a^{3} + 2 \, {\left (3 \, a^{3} \cosh \left (x\right )^{2} + a^{3}\right )} \sinh \left (x\right )^{2} + 4 \, {\left (a^{3} \cosh \left (x\right )^{3} + a^{3} \cosh \left (x\right )\right )} \sinh \left (x\right )\right )} \log \left (\frac {2 \, {\left (b \sinh \left (x\right ) + a\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) - {\left (a^{3} \cosh \left (x\right )^{4} + 4 \, a^{3} \cosh \left (x\right ) \sinh \left (x\right )^{3} + a^{3} \sinh \left (x\right )^{4} + 2 \, a^{3} \cosh \left (x\right )^{2} + a^{3} + 2 \, {\left (3 \, a^{3} \cosh \left (x\right )^{2} + a^{3}\right )} \sinh \left (x\right )^{2} + 4 \, {\left (a^{3} \cosh \left (x\right )^{3} + a^{3} \cosh \left (x\right )\right )} \sinh \left (x\right )\right )} \log \left (\frac {2 \, \cosh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) - {\left (a^{2} b + b^{3} - 3 \, {\left (a^{2} b + b^{3}\right )} \cosh \left (x\right )^{2} + 4 \, {\left (a^{3} + a b^{2}\right )} \cosh \left (x\right )\right )} \sinh \left (x\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \left (x\right )^{4} + 4 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \left (x\right ) \sinh \left (x\right )^{3} + {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sinh \left (x\right )^{4} + a^{4} + 2 \, a^{2} b^{2} + b^{4} + 2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \left (x\right )^{2} + 2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4} + 3 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \left (x\right )^{2}\right )} \sinh \left (x\right )^{2} + 4 \, {\left ({\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \left (x\right )^{3} + {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \left (x\right )\right )} \sinh \left (x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((a^2*b + b^3)*cosh(x)^3 + (a^2*b + b^3)*sinh(x)^3 - 2*(a^3 + a*b^2)*cosh(x)^2 - (2*a^3 + 2*a*b^2 - 3*(a^2*b
+ b^3)*cosh(x))*sinh(x)^2 - ((3*a^2*b + b^3)*cosh(x)^4 + 4*(3*a^2*b + b^3)*cosh(x)*sinh(x)^3 + (3*a^2*b + b^3)
*sinh(x)^4 + 3*a^2*b + b^3 + 2*(3*a^2*b + b^3)*cosh(x)^2 + 2*(3*a^2*b + b^3 + 3*(3*a^2*b + b^3)*cosh(x)^2)*sin
h(x)^2 + 4*((3*a^2*b + b^3)*cosh(x)^3 + (3*a^2*b + b^3)*cosh(x))*sinh(x))*arctan(cosh(x) + sinh(x)) - (a^2*b +
b^3)*cosh(x) + (a^3*cosh(x)^4 + 4*a^3*cosh(x)*sinh(x)^3 + a^3*sinh(x)^4 + 2*a^3*cosh(x)^2 + a^3 + 2*(3*a^3*co
sh(x)^2 + a^3)*sinh(x)^2 + 4*(a^3*cosh(x)^3 + a^3*cosh(x))*sinh(x))*log(2*(b*sinh(x) + a)/(cosh(x) - sinh(x)))
- (a^3*cosh(x)^4 + 4*a^3*cosh(x)*sinh(x)^3 + a^3*sinh(x)^4 + 2*a^3*cosh(x)^2 + a^3 + 2*(3*a^3*cosh(x)^2 + a^3
)*sinh(x)^2 + 4*(a^3*cosh(x)^3 + a^3*cosh(x))*sinh(x))*log(2*cosh(x)/(cosh(x) - sinh(x))) - (a^2*b + b^3 - 3*(
a^2*b + b^3)*cosh(x)^2 + 4*(a^3 + a*b^2)*cosh(x))*sinh(x))/((a^4 + 2*a^2*b^2 + b^4)*cosh(x)^4 + 4*(a^4 + 2*a^2
*b^2 + b^4)*cosh(x)*sinh(x)^3 + (a^4 + 2*a^2*b^2 + b^4)*sinh(x)^4 + a^4 + 2*a^2*b^2 + b^4 + 2*(a^4 + 2*a^2*b^2
+ b^4)*cosh(x)^2 + 2*(a^4 + 2*a^2*b^2 + b^4 + 3*(a^4 + 2*a^2*b^2 + b^4)*cosh(x)^2)*sinh(x)^2 + 4*((a^4 + 2*a^
2*b^2 + b^4)*cosh(x)^3 + (a^4 + 2*a^2*b^2 + b^4)*cosh(x))*sinh(x))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\tanh ^{3}{\left (x \right )}}{a + b \sinh {\left (x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)**3/(a+b*sinh(x)),x)

[Out]

Integral(tanh(x)**3/(a + b*sinh(x)), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 211 vs. \(2 (85) = 170\).
time = 0.42, size = 211, normalized size = 2.40 \begin {gather*} -\frac {a^{3} b \log \left ({\left | -b {\left (e^{\left (-x\right )} - e^{x}\right )} + 2 \, a \right |}\right )}{a^{4} b + 2 \, a^{2} b^{3} + b^{5}} + \frac {a^{3} \log \left ({\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 4\right )}{2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}} + \frac {{\left (\pi + 2 \, \arctan \left (\frac {1}{2} \, {\left (e^{\left (2 \, x\right )} - 1\right )} e^{\left (-x\right )}\right )\right )} {\left (3 \, a^{2} b + b^{3}\right )}}{4 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}} - \frac {a^{3} {\left (e^{\left (-x\right )} - e^{x}\right )}^{2} - 2 \, a^{2} b {\left (e^{\left (-x\right )} - e^{x}\right )} - 2 \, b^{3} {\left (e^{\left (-x\right )} - e^{x}\right )} - 4 \, a b^{2}}{2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} {\left ({\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a^3*b*log(abs(-b*(e^(-x) - e^x) + 2*a))/(a^4*b + 2*a^2*b^3 + b^5) + 1/2*a^3*log((e^(-x) - e^x)^2 + 4)/(a^4 +
2*a^2*b^2 + b^4) + 1/4*(pi + 2*arctan(1/2*(e^(2*x) - 1)*e^(-x)))*(3*a^2*b + b^3)/(a^4 + 2*a^2*b^2 + b^4) - 1/2
*(a^3*(e^(-x) - e^x)^2 - 2*a^2*b*(e^(-x) - e^x) - 2*b^3*(e^(-x) - e^x) - 4*a*b^2)/((a^4 + 2*a^2*b^2 + b^4)*((e
^(-x) - e^x)^2 + 4))

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Mupad [B]
time = 2.28, size = 291, normalized size = 3.31 \begin {gather*} \frac {\frac {2\,\left (a^3+a\,b^2\right )}{{\left (a^2+b^2\right )}^2}-\frac {{\mathrm {e}}^x\,\left (a^2\,b+b^3\right )}{{\left (a^2+b^2\right )}^2}}{{\mathrm {e}}^{2\,x}+1}-\frac {\frac {2\,a}{a^2+b^2}-\frac {2\,b\,{\mathrm {e}}^x}{a^2+b^2}}{2\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^{4\,x}+1}+\frac {\ln \left (1+{\mathrm {e}}^x\,1{}\mathrm {i}\right )\,\left (2\,a+b\,1{}\mathrm {i}\right )}{2\,\left (a^2+a\,b\,2{}\mathrm {i}-b^2\right )}-\frac {a^3\,\ln \left (b^7\,{\mathrm {e}}^{2\,x}-16\,a^6\,b-b^7-6\,a^2\,b^5-9\,a^4\,b^3+32\,a^7\,{\mathrm {e}}^x+6\,a^2\,b^5\,{\mathrm {e}}^{2\,x}+9\,a^4\,b^3\,{\mathrm {e}}^{2\,x}+2\,a\,b^6\,{\mathrm {e}}^x+16\,a^6\,b\,{\mathrm {e}}^{2\,x}+12\,a^3\,b^4\,{\mathrm {e}}^x+18\,a^5\,b^2\,{\mathrm {e}}^x\right )}{a^4+2\,a^2\,b^2+b^4}+\frac {\ln \left ({\mathrm {e}}^x+1{}\mathrm {i}\right )\,\left (b+a\,2{}\mathrm {i}\right )}{2\,\left (a^2\,1{}\mathrm {i}+2\,a\,b-b^2\,1{}\mathrm {i}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)^3/(a + b*sinh(x)),x)

[Out]

((2*(a*b^2 + a^3))/(a^2 + b^2)^2 - (exp(x)*(a^2*b + b^3))/(a^2 + b^2)^2)/(exp(2*x) + 1) - ((2*a)/(a^2 + b^2) -
(2*b*exp(x))/(a^2 + b^2))/(2*exp(2*x) + exp(4*x) + 1) + (log(exp(x)*1i + 1)*(2*a + b*1i))/(2*(a*b*2i + a^2 -
b^2)) - (a^3*log(b^7*exp(2*x) - 16*a^6*b - b^7 - 6*a^2*b^5 - 9*a^4*b^3 + 32*a^7*exp(x) + 6*a^2*b^5*exp(2*x) +
9*a^4*b^3*exp(2*x) + 2*a*b^6*exp(x) + 16*a^6*b*exp(2*x) + 12*a^3*b^4*exp(x) + 18*a^5*b^2*exp(x)))/(a^4 + b^4 +
2*a^2*b^2) + (log(exp(x) + 1i)*(a*2i + b))/(2*(2*a*b + a^2*1i - b^2*1i))

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