3.228 \(\int \frac{\tanh ^4(x)}{a+b \sinh (x)} \, dx\)

Optimal. Leaf size=124 \[ -\frac{2 a^4 \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}-\frac{a \tanh ^3(x)}{3 \left (a^2+b^2\right )}-\frac{a^3 \tanh (x)}{\left (a^2+b^2\right )^2}+\frac{b \text{sech}^3(x)}{3 \left (a^2+b^2\right )}-\frac{a^2 b \text{sech}(x)}{\left (a^2+b^2\right )^2}-\frac{b \text{sech}(x)}{a^2+b^2} \]

[Out]

(-2*a^4*ArcTanh[(b - a*Tanh[x/2])/Sqrt[a^2 + b^2]])/(a^2 + b^2)^(5/2) - (a^2*b*Sech[x])/(a^2 + b^2)^2 - (b*Sec
h[x])/(a^2 + b^2) + (b*Sech[x]^3)/(3*(a^2 + b^2)) - (a^3*Tanh[x])/(a^2 + b^2)^2 - (a*Tanh[x]^3)/(3*(a^2 + b^2)
)

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Rubi [A]  time = 0.184068, antiderivative size = 124, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 9, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.692, Rules used = {2727, 2607, 30, 2606, 3767, 8, 2660, 618, 206} \[ -\frac{2 a^4 \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}-\frac{a \tanh ^3(x)}{3 \left (a^2+b^2\right )}-\frac{a^3 \tanh (x)}{\left (a^2+b^2\right )^2}+\frac{b \text{sech}^3(x)}{3 \left (a^2+b^2\right )}-\frac{a^2 b \text{sech}(x)}{\left (a^2+b^2\right )^2}-\frac{b \text{sech}(x)}{a^2+b^2} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[x]^4/(a + b*Sinh[x]),x]

[Out]

(-2*a^4*ArcTanh[(b - a*Tanh[x/2])/Sqrt[a^2 + b^2]])/(a^2 + b^2)^(5/2) - (a^2*b*Sech[x])/(a^2 + b^2)^2 - (b*Sec
h[x])/(a^2 + b^2) + (b*Sech[x]^3)/(3*(a^2 + b^2)) - (a^3*Tanh[x])/(a^2 + b^2)^2 - (a*Tanh[x]^3)/(3*(a^2 + b^2)
)

Rule 2727

Int[((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[a/(a^2 - b^
2), Int[(g*Tan[e + f*x])^p/Sin[e + f*x]^2, x], x] + (-Dist[(b*g)/(a^2 - b^2), Int[(g*Tan[e + f*x])^(p - 1)/Cos
[e + f*x], x], x] - Dist[(a^2*g^2)/(a^2 - b^2), Int[(g*Tan[e + f*x])^(p - 2)/(a + b*Sin[e + f*x]), x], x]) /;
FreeQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[2*p] && GtQ[p, 1]

Rule 2607

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 30

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

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 206

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

Rubi steps

\begin{align*} \int \frac{\tanh ^4(x)}{a+b \sinh (x)} \, dx &=-\frac{a \int \text{sech}^2(x) \tanh ^2(x) \, dx}{a^2+b^2}+\frac{a^2 \int \frac{\tanh ^2(x)}{a+b \sinh (x)} \, dx}{a^2+b^2}+\frac{b \int \text{sech}(x) \tanh ^3(x) \, dx}{a^2+b^2}\\ &=-\frac{a^3 \int \text{sech}^2(x) \, dx}{\left (a^2+b^2\right )^2}+\frac{a^4 \int \frac{1}{a+b \sinh (x)} \, dx}{\left (a^2+b^2\right )^2}+\frac{\left (a^2 b\right ) \int \text{sech}(x) \tanh (x) \, dx}{\left (a^2+b^2\right )^2}-\frac{(i a) \operatorname{Subst}\left (\int x^2 \, dx,x,i \tanh (x)\right )}{a^2+b^2}+\frac{b \operatorname{Subst}\left (\int \left (-1+x^2\right ) \, dx,x,\text{sech}(x)\right )}{a^2+b^2}\\ &=-\frac{b \text{sech}(x)}{a^2+b^2}+\frac{b \text{sech}^3(x)}{3 \left (a^2+b^2\right )}-\frac{a \tanh ^3(x)}{3 \left (a^2+b^2\right )}-\frac{\left (i a^3\right ) \operatorname{Subst}(\int 1 \, dx,x,-i \tanh (x))}{\left (a^2+b^2\right )^2}+\frac{\left (2 a^4\right ) \operatorname{Subst}\left (\int \frac{1}{a+2 b x-a x^2} \, dx,x,\tanh \left (\frac{x}{2}\right )\right )}{\left (a^2+b^2\right )^2}-\frac{\left (a^2 b\right ) \operatorname{Subst}(\int 1 \, dx,x,\text{sech}(x))}{\left (a^2+b^2\right )^2}\\ &=-\frac{a^2 b \text{sech}(x)}{\left (a^2+b^2\right )^2}-\frac{b \text{sech}(x)}{a^2+b^2}+\frac{b \text{sech}^3(x)}{3 \left (a^2+b^2\right )}-\frac{a^3 \tanh (x)}{\left (a^2+b^2\right )^2}-\frac{a \tanh ^3(x)}{3 \left (a^2+b^2\right )}-\frac{\left (4 a^4\right ) \operatorname{Subst}\left (\int \frac{1}{4 \left (a^2+b^2\right )-x^2} \, dx,x,2 b-2 a \tanh \left (\frac{x}{2}\right )\right )}{\left (a^2+b^2\right )^2}\\ &=-\frac{2 a^4 \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}-\frac{a^2 b \text{sech}(x)}{\left (a^2+b^2\right )^2}-\frac{b \text{sech}(x)}{a^2+b^2}+\frac{b \text{sech}^3(x)}{3 \left (a^2+b^2\right )}-\frac{a^3 \tanh (x)}{\left (a^2+b^2\right )^2}-\frac{a \tanh ^3(x)}{3 \left (a^2+b^2\right )}\\ \end{align*}

Mathematica [A]  time = 0.351928, size = 108, normalized size = 0.87 \[ \frac{-a \left (4 a^2+b^2\right ) \tanh (x)-3 b \left (2 a^2+b^2\right ) \text{sech}(x)+\frac{6 a^4 \tan ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{-a^2-b^2}}\right )}{\sqrt{-a^2-b^2}}+\left (a^2+b^2\right ) \text{sech}^3(x) (a \sinh (x)+b)}{3 \left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[x]^4/(a + b*Sinh[x]),x]

[Out]

((6*a^4*ArcTan[(b - a*Tanh[x/2])/Sqrt[-a^2 - b^2]])/Sqrt[-a^2 - b^2] - 3*b*(2*a^2 + b^2)*Sech[x] + (a^2 + b^2)
*Sech[x]^3*(b + a*Sinh[x]) - a*(4*a^2 + b^2)*Tanh[x])/(3*(a^2 + b^2)^2)

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Maple [A]  time = 0.039, size = 163, normalized size = 1.3 \begin{align*} 2\,{\frac{-{a}^{3} \left ( \tanh \left ( x/2 \right ) \right ) ^{5}-{a}^{2}b \left ( \tanh \left ( x/2 \right ) \right ) ^{4}+ \left ( -10/3\,{a}^{3}-4/3\,a{b}^{2} \right ) \left ( \tanh \left ( x/2 \right ) \right ) ^{3}+ \left ( -4\,{a}^{2}b-2\,{b}^{3} \right ) \left ( \tanh \left ( x/2 \right ) \right ) ^{2}-{a}^{3}\tanh \left ( x/2 \right ) -5/3\,{a}^{2}b-2/3\,{b}^{3}}{ \left ({a}^{2}+{b}^{2} \right ) ^{2} \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{3}}}+32\,{\frac{{a}^{4}}{ \left ( 16\,{a}^{4}+32\,{a}^{2}{b}^{2}+16\,{b}^{4} \right ) \sqrt{{a}^{2}+{b}^{2}}}{\it Artanh} \left ( 1/2\,{\frac{2\,a\tanh \left ( x/2 \right ) -2\,b}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)^4/(a+b*sinh(x)),x)

[Out]

2/(a^2+b^2)^2*(-a^3*tanh(1/2*x)^5-a^2*b*tanh(1/2*x)^4+(-10/3*a^3-4/3*a*b^2)*tanh(1/2*x)^3+(-4*a^2*b-2*b^3)*tan
h(1/2*x)^2-a^3*tanh(1/2*x)-5/3*a^2*b-2/3*b^3)/(tanh(1/2*x)^2+1)^3+32*a^4/(16*a^4+32*a^2*b^2+16*b^4)/(a^2+b^2)^
(1/2)*arctanh(1/2*(2*a*tanh(1/2*x)-2*b)/(a^2+b^2)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^4/(a+b*sinh(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.24552, size = 2944, normalized size = 23.74 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^4/(a+b*sinh(x)),x, algorithm="fricas")

[Out]

-1/3*(6*(2*a^4*b + 3*a^2*b^3 + b^5)*cosh(x)^5 + 6*(2*a^4*b + 3*a^2*b^3 + b^5)*sinh(x)^5 - 8*a^5 - 10*a^3*b^2 -
2*a*b^4 - 6*(2*a^5 + 3*a^3*b^2 + a*b^4)*cosh(x)^4 - 6*(2*a^5 + 3*a^3*b^2 + a*b^4 - 5*(2*a^4*b + 3*a^2*b^3 + b
^5)*cosh(x))*sinh(x)^4 + 4*(4*a^4*b + 5*a^2*b^3 + b^5)*cosh(x)^3 + 4*(4*a^4*b + 5*a^2*b^3 + b^5 + 15*(2*a^4*b
+ 3*a^2*b^3 + b^5)*cosh(x)^2 - 6*(2*a^5 + 3*a^3*b^2 + a*b^4)*cosh(x))*sinh(x)^3 - 12*(a^5 + a^3*b^2)*cosh(x)^2
- 12*(a^5 + a^3*b^2 - 5*(2*a^4*b + 3*a^2*b^3 + b^5)*cosh(x)^3 + 3*(2*a^5 + 3*a^3*b^2 + a*b^4)*cosh(x)^2 - (4*
a^4*b + 5*a^2*b^3 + b^5)*cosh(x))*sinh(x)^2 - 3*(a^4*cosh(x)^6 + 6*a^4*cosh(x)*sinh(x)^5 + a^4*sinh(x)^6 + 3*a
^4*cosh(x)^4 + 3*a^4*cosh(x)^2 + 3*(5*a^4*cosh(x)^2 + a^4)*sinh(x)^4 + a^4 + 4*(5*a^4*cosh(x)^3 + 3*a^4*cosh(x
))*sinh(x)^3 + 3*(5*a^4*cosh(x)^4 + 6*a^4*cosh(x)^2 + a^4)*sinh(x)^2 + 6*(a^4*cosh(x)^5 + 2*a^4*cosh(x)^3 + a^
4*cosh(x))*sinh(x))*sqrt(a^2 + b^2)*log((b^2*cosh(x)^2 + b^2*sinh(x)^2 + 2*a*b*cosh(x) + 2*a^2 + b^2 + 2*(b^2*
cosh(x) + a*b)*sinh(x) - 2*sqrt(a^2 + b^2)*(b*cosh(x) + b*sinh(x) + a))/(b*cosh(x)^2 + b*sinh(x)^2 + 2*a*cosh(
x) + 2*(b*cosh(x) + a)*sinh(x) - b)) + 6*(2*a^4*b + 3*a^2*b^3 + b^5)*cosh(x) + 6*(2*a^4*b + 3*a^2*b^3 + b^5 +
5*(2*a^4*b + 3*a^2*b^3 + b^5)*cosh(x)^4 - 4*(2*a^5 + 3*a^3*b^2 + a*b^4)*cosh(x)^3 + 2*(4*a^4*b + 5*a^2*b^3 + b
^5)*cosh(x)^2 - 4*(a^5 + a^3*b^2)*cosh(x))*sinh(x))/((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*cosh(x)^6 + 6*(a^6 +
3*a^4*b^2 + 3*a^2*b^4 + b^6)*cosh(x)*sinh(x)^5 + (a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*sinh(x)^6 + a^6 + 3*a^4*b
^2 + 3*a^2*b^4 + b^6 + 3*(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*cosh(x)^4 + 3*(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6
+ 5*(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*cosh(x)^2)*sinh(x)^4 + 4*(5*(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*cosh(x
)^3 + 3*(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*cosh(x))*sinh(x)^3 + 3*(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*cosh(x)
^2 + 3*(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6 + 5*(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*cosh(x)^4 + 6*(a^6 + 3*a^4*b
^2 + 3*a^2*b^4 + b^6)*cosh(x)^2)*sinh(x)^2 + 6*((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*cosh(x)^5 + 2*(a^6 + 3*a^4
*b^2 + 3*a^2*b^4 + b^6)*cosh(x)^3 + (a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*cosh(x))*sinh(x))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tanh ^{4}{\left (x \right )}}{a + b \sinh{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)**4/(a+b*sinh(x)),x)

[Out]

Integral(tanh(x)**4/(a + b*sinh(x)), x)

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Giac [A]  time = 1.19481, size = 266, normalized size = 2.15 \begin{align*} \frac{a^{4} \log \left (\frac{{\left | 2 \, b e^{x} + 2 \, a - 2 \, \sqrt{a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{x} + 2 \, a + 2 \, \sqrt{a^{2} + b^{2}} \right |}}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt{a^{2} + b^{2}}} - \frac{2 \,{\left (6 \, a^{2} b e^{\left (5 \, x\right )} + 3 \, b^{3} e^{\left (5 \, x\right )} - 6 \, a^{3} e^{\left (4 \, x\right )} - 3 \, a b^{2} e^{\left (4 \, x\right )} + 8 \, a^{2} b e^{\left (3 \, x\right )} + 2 \, b^{3} e^{\left (3 \, x\right )} - 6 \, a^{3} e^{\left (2 \, x\right )} + 6 \, a^{2} b e^{x} + 3 \, b^{3} e^{x} - 4 \, a^{3} - a b^{2}\right )}}{3 \,{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}{\left (e^{\left (2 \, x\right )} + 1\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^4/(a+b*sinh(x)),x, algorithm="giac")

[Out]

a^4*log(abs(2*b*e^x + 2*a - 2*sqrt(a^2 + b^2))/abs(2*b*e^x + 2*a + 2*sqrt(a^2 + b^2)))/((a^4 + 2*a^2*b^2 + b^4
)*sqrt(a^2 + b^2)) - 2/3*(6*a^2*b*e^(5*x) + 3*b^3*e^(5*x) - 6*a^3*e^(4*x) - 3*a*b^2*e^(4*x) + 8*a^2*b*e^(3*x)
+ 2*b^3*e^(3*x) - 6*a^3*e^(2*x) + 6*a^2*b*e^x + 3*b^3*e^x - 4*a^3 - a*b^2)/((a^4 + 2*a^2*b^2 + b^4)*(e^(2*x) +
1)^3)