\(\int \frac {\tanh ^3(c+d x)}{a+b \sinh (c+d x)} \, dx\) [418]

   Optimal result
   Rubi [A] (verified)
   Mathematica [C] (verified)
   Maple [A] (verified)
   Fricas [B] (verification not implemented)
   Sympy [F]
   Maxima [A] (verification not implemented)
   Giac [B] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 21, antiderivative size = 120 \[ \int \frac {\tanh ^3(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {b \left (3 a^2+b^2\right ) \arctan (\sinh (c+d x))}{2 \left (a^2+b^2\right )^2 d}+\frac {a^3 \log (\cosh (c+d x))}{\left (a^2+b^2\right )^2 d}-\frac {a^3 \log (a+b \sinh (c+d x))}{\left (a^2+b^2\right )^2 d}+\frac {\text {sech}^2(c+d x) (a-b \sinh (c+d x))}{2 \left (a^2+b^2\right ) d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2800, 1661, 815, 649, 209, 266} \[ \int \frac {\tanh ^3(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {b \left (3 a^2+b^2\right ) \arctan (\sinh (c+d x))}{2 d \left (a^2+b^2\right )^2}+\frac {\text {sech}^2(c+d x) (a-b \sinh (c+d x))}{2 d \left (a^2+b^2\right )}-\frac {a^3 \log (a+b \sinh (c+d x))}{d \left (a^2+b^2\right )^2}+\frac {a^3 \log (\cosh (c+d x))}{d \left (a^2+b^2\right )^2} \]

[In]

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

[Out]

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

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*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x^3}{(a+x) \left (-b^2-x^2\right )^2} \, dx,x,b \sinh (c+d x)\right )}{d} \\ & = \frac {\text {sech}^2(c+d x) (a-b \sinh (c+d x))}{2 \left (a^2+b^2\right ) d}-\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 (c+d x)\right )}{2 b^2 d} \\ & = \frac {\text {sech}^2(c+d x) (a-b \sinh (c+d x))}{2 \left (a^2+b^2\right ) d}-\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 (c+d x)\right )}{2 b^2 d} \\ & = -\frac {a^3 \log (a+b \sinh (c+d x))}{\left (a^2+b^2\right )^2 d}+\frac {\text {sech}^2(c+d x) (a-b \sinh (c+d x))}{2 \left (a^2+b^2\right ) d}+\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 (c+d x)\right )}{2 \left (a^2+b^2\right )^2 d} \\ & = -\frac {a^3 \log (a+b \sinh (c+d x))}{\left (a^2+b^2\right )^2 d}+\frac {\text {sech}^2(c+d x) (a-b \sinh (c+d x))}{2 \left (a^2+b^2\right ) d}+\frac {a^3 \text {Subst}\left (\int \frac {x}{b^2+x^2} \, dx,x,b \sinh (c+d x)\right )}{\left (a^2+b^2\right )^2 d}+\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 (c+d x)\right )}{2 \left (a^2+b^2\right )^2 d} \\ & = \frac {b \left (3 a^2+b^2\right ) \arctan (\sinh (c+d x))}{2 \left (a^2+b^2\right )^2 d}+\frac {a^3 \log (\cosh (c+d x))}{\left (a^2+b^2\right )^2 d}-\frac {a^3 \log (a+b \sinh (c+d x))}{\left (a^2+b^2\right )^2 d}+\frac {\text {sech}^2(c+d x) (a-b \sinh (c+d x))}{2 \left (a^2+b^2\right ) d} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.35 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.27 \[ \int \frac {\tanh ^3(c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {b \left (a^2+b^2\right ) \arctan (\sinh (c+d x))-\left (a^3-i \left (2 a^2 b+b^3\right )\right ) \log (i-\sinh (c+d x))-\left (a^3+i \left (2 a^2 b+b^3\right )\right ) \log (i+\sinh (c+d x))+2 a^3 \log (a+b \sinh (c+d x))-a \left (a^2+b^2\right ) \text {sech}^2(c+d x)+b \left (a^2+b^2\right ) \text {sech}(c+d x) \tanh (c+d x)}{2 \left (a^2+b^2\right )^2 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.43 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.75

method result size
derivativedivides \(\frac {-\frac {8 a^{3} \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right )}{8 a^{4}+16 a^{2} b^{2}+8 b^{4}}+\frac {\frac {2 \left (\left (\frac {1}{2} a^{2} b +\frac {1}{2} b^{3}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (-a^{3}-a \,b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (-\frac {1}{2} a^{2} b -\frac {1}{2} b^{3}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+a^{3} \ln \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )+\left (3 a^{2} b +b^{3}\right ) \arctan \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{4}+2 a^{2} b^{2}+b^{4}}}{d}\) \(210\)
default \(\frac {-\frac {8 a^{3} \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right )}{8 a^{4}+16 a^{2} b^{2}+8 b^{4}}+\frac {\frac {2 \left (\left (\frac {1}{2} a^{2} b +\frac {1}{2} b^{3}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (-a^{3}-a \,b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (-\frac {1}{2} a^{2} b -\frac {1}{2} b^{3}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+a^{3} \ln \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )+\left (3 a^{2} b +b^{3}\right ) \arctan \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{4}+2 a^{2} b^{2}+b^{4}}}{d}\) \(210\)
risch \(-\frac {2 d^{2} a^{3} x}{a^{4} d^{2}+2 a^{2} b^{2} d^{2}+b^{4} d^{2}}-\frac {2 d \,a^{3} c}{a^{4} d^{2}+2 a^{2} b^{2} d^{2}+b^{4} d^{2}}+\frac {2 a^{3} x}{a^{4}+2 a^{2} b^{2}+b^{4}}+\frac {2 a^{3} c}{d \left (a^{4}+2 a^{2} b^{2}+b^{4}\right )}+\frac {{\mathrm e}^{d x +c} \left (-b \,{\mathrm e}^{2 d x +2 c}+2 a \,{\mathrm e}^{d x +c}+b \right )}{d \left (a^{2}+b^{2}\right ) \left (1+{\mathrm e}^{2 d x +2 c}\right )^{2}}+\frac {3 i \ln \left ({\mathrm e}^{d x +c}+i\right ) a^{2} b}{2 \left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) d}+\frac {i \ln \left ({\mathrm e}^{d x +c}+i\right ) b^{3}}{2 \left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) d}+\frac {\ln \left ({\mathrm e}^{d x +c}+i\right ) a^{3}}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) d}-\frac {3 i \ln \left ({\mathrm e}^{d x +c}-i\right ) a^{2} b}{2 \left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) d}-\frac {i \ln \left ({\mathrm e}^{d x +c}-i\right ) b^{3}}{2 \left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) d}+\frac {\ln \left ({\mathrm e}^{d x +c}-i\right ) a^{3}}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) d}-\frac {a^{3} \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 a \,{\mathrm e}^{d x +c}}{b}-1\right )}{d \left (a^{4}+2 a^{2} b^{2}+b^{4}\right )}\) \(441\)

[In]

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

[Out]

1/d*(-8*a^3/(8*a^4+16*a^2*b^2+8*b^4)*ln(tanh(1/2*d*x+1/2*c)^2*a-2*b*tanh(1/2*d*x+1/2*c)-a)+2/(a^4+2*a^2*b^2+b^
4)*(((1/2*a^2*b+1/2*b^3)*tanh(1/2*d*x+1/2*c)^3+(-a^3-a*b^2)*tanh(1/2*d*x+1/2*c)^2+(-1/2*a^2*b-1/2*b^3)*tanh(1/
2*d*x+1/2*c))/(1+tanh(1/2*d*x+1/2*c)^2)^2+1/2*a^3*ln(1+tanh(1/2*d*x+1/2*c)^2)+1/2*(3*a^2*b+b^3)*arctan(tanh(1/
2*d*x+1/2*c))))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 896 vs. \(2 (117) = 234\).

Time = 0.28 (sec) , antiderivative size = 896, normalized size of antiderivative = 7.47 \[ \int \frac {\tanh ^3(c+d x)}{a+b \sinh (c+d x)} \, dx=\text {Too large to display} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {\tanh ^3(c+d x)}{a+b \sinh (c+d x)} \, dx=\int \frac {\tanh ^{3}{\left (c + d x \right )}}{a + b \sinh {\left (c + d x \right )}}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.81 \[ \int \frac {\tanh ^3(c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {a^{3} \log \left (-2 \, a e^{\left (-d x - c\right )} + b e^{\left (-2 \, d x - 2 \, c\right )} - b\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d} + \frac {a^{3} \log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d} - \frac {{\left (3 \, a^{2} b + b^{3}\right )} \arctan \left (e^{\left (-d x - c\right )}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d} - \frac {b e^{\left (-d x - c\right )} - 2 \, a e^{\left (-2 \, d x - 2 \, c\right )} - b e^{\left (-3 \, d x - 3 \, c\right )}}{{\left (a^{2} + b^{2} + 2 \, {\left (a^{2} + b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (a^{2} + b^{2}\right )} e^{\left (-4 \, d x - 4 \, c\right )}\right )} d} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 279 vs. \(2 (117) = 234\).

Time = 0.34 (sec) , antiderivative size = 279, normalized size of antiderivative = 2.32 \[ \int \frac {\tanh ^3(c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {\frac {4 \, a^{3} b \log \left ({\left | b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 2 \, a \right |}\right )}{a^{4} b + 2 \, a^{2} b^{3} + b^{5}} - \frac {2 \, a^{3} \log \left ({\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} + 4\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {{\left (\pi + 2 \, \arctan \left (\frac {1}{2} \, {\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )} e^{\left (-d x - c\right )}\right )\right )} {\left (3 \, a^{2} b + b^{3}\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {2 \, {\left (a^{3} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} + 2 \, a^{2} b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 2 \, b^{3} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} - 4 \, a b^{2}\right )}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} {\left ({\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} + 4\right )}}}{4 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 3.22 (sec) , antiderivative size = 381, normalized size of antiderivative = 3.18 \[ \int \frac {\tanh ^3(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {\frac {2\,\left (a^3+a\,b^2\right )}{d\,{\left (a^2+b^2\right )}^2}-\frac {{\mathrm {e}}^{c+d\,x}\,\left (a^2\,b+b^3\right )}{d\,{\left (a^2+b^2\right )}^2}}{{\mathrm {e}}^{2\,c+2\,d\,x}+1}-\frac {\frac {2\,a}{d\,\left (a^2+b^2\right )}-\frac {2\,b\,{\mathrm {e}}^{c+d\,x}}{d\,\left (a^2+b^2\right )}}{2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1}+\frac {\ln \left (1+{\mathrm {e}}^{c+d\,x}\,1{}\mathrm {i}\right )\,\left (2\,a+b\,1{}\mathrm {i}\right )}{2\,\left (d\,a^2+2{}\mathrm {i}\,d\,a\,b-d\,b^2\right )}+\frac {\ln \left ({\mathrm {e}}^{c+d\,x}+1{}\mathrm {i}\right )\,\left (b+a\,2{}\mathrm {i}\right )}{2\,\left (1{}\mathrm {i}\,d\,a^2+2\,d\,a\,b-1{}\mathrm {i}\,d\,b^2\right )}-\frac {a^3\,\ln \left (32\,a^7\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-b^7-6\,a^2\,b^5-9\,a^4\,b^3-16\,a^6\,b+b^7\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+16\,a^6\,b\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+12\,a^3\,b^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c+18\,a^5\,b^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c+6\,a^2\,b^5\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+9\,a^4\,b^3\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+2\,a\,b^6\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\right )}{d\,a^4+2\,d\,a^2\,b^2+d\,b^4} \]

[In]

int(tanh(c + d*x)^3/(a + b*sinh(c + d*x)),x)

[Out]

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