3.60 \(\int \frac {\sinh ^4(x)}{a+b \text {sech}(x)} \, dx\)

Optimal. Leaf size=132 \[ -\frac {2 b (a-b)^{3/2} (a+b)^{3/2} \tan ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{a^5}-\frac {\sinh ^3(x) (4 b-3 a \cosh (x))}{12 a^2}+\frac {\sinh (x) \left (8 b \left (a^2-b^2\right )-a \left (3 a^2-4 b^2\right ) \cosh (x)\right )}{8 a^4}+\frac {x \left (3 a^4-12 a^2 b^2+8 b^4\right )}{8 a^5} \]

[Out]

1/8*(3*a^4-12*a^2*b^2+8*b^4)*x/a^5-2*(a-b)^(3/2)*b*(a+b)^(3/2)*arctan((a-b)^(1/2)*tanh(1/2*x)/(a+b)^(1/2))/a^5
+1/8*(8*b*(a^2-b^2)-a*(3*a^2-4*b^2)*cosh(x))*sinh(x)/a^4-1/12*(4*b-3*a*cosh(x))*sinh(x)^3/a^2

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Rubi [A]  time = 0.37, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {3872, 2865, 2735, 2659, 205} \[ \frac {x \left (-12 a^2 b^2+3 a^4+8 b^4\right )}{8 a^5}+\frac {\sinh (x) \left (8 b \left (a^2-b^2\right )-a \left (3 a^2-4 b^2\right ) \cosh (x)\right )}{8 a^4}-\frac {2 b (a-b)^{3/2} (a+b)^{3/2} \tan ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{a^5}-\frac {\sinh ^3(x) (4 b-3 a \cosh (x))}{12 a^2} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]^4/(a + b*Sech[x]),x]

[Out]

((3*a^4 - 12*a^2*b^2 + 8*b^4)*x)/(8*a^5) - (2*(a - b)^(3/2)*b*(a + b)^(3/2)*ArcTan[(Sqrt[a - b]*Tanh[x/2])/Sqr
t[a + b]])/a^5 + ((8*b*(a^2 - b^2) - a*(3*a^2 - 4*b^2)*Cosh[x])*Sinh[x])/(8*a^4) - ((4*b - 3*a*Cosh[x])*Sinh[x
]^3)/(12*a^2)

Rule 205

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

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*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 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 2865

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[(g*(g*Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1)*(b*c*(m + p + 1) -
a*d*p + b*d*(m + p)*Sin[e + f*x]))/(b^2*f*(m + p)*(m + p + 1)), x] + Dist[(g^2*(p - 1))/(b^2*(m + p)*(m + p +
1)), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^m*Simp[b*(a*d*m + b*c*(m + p + 1)) + (a*b*c*(m + p + 1
) - d*(a^2*p - b^2*(m + p)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2,
0] && GtQ[p, 1] && NeQ[m + p, 0] && NeQ[m + p + 1, 0] && IntegerQ[2*m]

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {\sinh ^4(x)}{a+b \text {sech}(x)} \, dx &=-\int \frac {\cosh (x) \sinh ^4(x)}{-b-a \cosh (x)} \, dx\\ &=-\frac {(4 b-3 a \cosh (x)) \sinh ^3(x)}{12 a^2}+\frac {\int \frac {\left (-a b+\left (3 a^2-4 b^2\right ) \cosh (x)\right ) \sinh ^2(x)}{-b-a \cosh (x)} \, dx}{4 a^2}\\ &=\frac {\left (8 b \left (a^2-b^2\right )-a \left (3 a^2-4 b^2\right ) \cosh (x)\right ) \sinh (x)}{8 a^4}-\frac {(4 b-3 a \cosh (x)) \sinh ^3(x)}{12 a^2}-\frac {\int \frac {-a b \left (5 a^2-4 b^2\right )+\left (3 a^4-12 a^2 b^2+8 b^4\right ) \cosh (x)}{-b-a \cosh (x)} \, dx}{8 a^4}\\ &=\frac {\left (3 a^4-12 a^2 b^2+8 b^4\right ) x}{8 a^5}+\frac {\left (8 b \left (a^2-b^2\right )-a \left (3 a^2-4 b^2\right ) \cosh (x)\right ) \sinh (x)}{8 a^4}-\frac {(4 b-3 a \cosh (x)) \sinh ^3(x)}{12 a^2}+\frac {\left (b \left (a^2-b^2\right )^2\right ) \int \frac {1}{-b-a \cosh (x)} \, dx}{a^5}\\ &=\frac {\left (3 a^4-12 a^2 b^2+8 b^4\right ) x}{8 a^5}+\frac {\left (8 b \left (a^2-b^2\right )-a \left (3 a^2-4 b^2\right ) \cosh (x)\right ) \sinh (x)}{8 a^4}-\frac {(4 b-3 a \cosh (x)) \sinh ^3(x)}{12 a^2}+\frac {\left (2 b \left (a^2-b^2\right )^2\right ) \operatorname {Subst}\left (\int \frac {1}{-a-b-(a-b) x^2} \, dx,x,\tanh \left (\frac {x}{2}\right )\right )}{a^5}\\ &=\frac {\left (3 a^4-12 a^2 b^2+8 b^4\right ) x}{8 a^5}-\frac {2 (a-b)^{3/2} b (a+b)^{3/2} \tan ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{a^5}+\frac {\left (8 b \left (a^2-b^2\right )-a \left (3 a^2-4 b^2\right ) \cosh (x)\right ) \sinh (x)}{8 a^4}-\frac {(4 b-3 a \cosh (x)) \sinh ^3(x)}{12 a^2}\\ \end {align*}

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Mathematica [A]  time = 0.74, size = 219, normalized size = 1.66 \[ \frac {36 a^4 x+3 a^4 \sinh (4 x)-8 a^3 b \sinh (3 x)-144 a^2 b^2 x+24 a b \left (5 a^2-4 b^2\right ) \sinh (x)-24 a^2 \left (a^2-b^2\right ) \sinh (2 x)+\frac {192 b^5 \tan ^{-1}\left (\frac {(b-a) \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}-\frac {384 a^2 b^3 \tan ^{-1}\left (\frac {(b-a) \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+\frac {192 a^4 b \tan ^{-1}\left (\frac {(b-a) \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+96 b^4 x}{96 a^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(36*a^4*x - 144*a^2*b^2*x + 96*b^4*x + (192*a^4*b*ArcTan[((-a + b)*Tanh[x/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2
] - (384*a^2*b^3*ArcTan[((-a + b)*Tanh[x/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] + (192*b^5*ArcTan[((-a + b)*Tan
h[x/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] + 24*a*b*(5*a^2 - 4*b^2)*Sinh[x] - 24*a^2*(a^2 - b^2)*Sinh[2*x] - 8*
a^3*b*Sinh[3*x] + 3*a^4*Sinh[4*x])/(96*a^5)

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fricas [B]  time = 0.45, size = 1812, normalized size = 13.73 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/192*(3*a^4*cosh(x)^8 + 3*a^4*sinh(x)^8 - 8*a^3*b*cosh(x)^7 + 8*(3*a^4*cosh(x) - a^3*b)*sinh(x)^7 - 24*(a^4
- a^2*b^2)*cosh(x)^6 + 4*(21*a^4*cosh(x)^2 - 14*a^3*b*cosh(x) - 6*a^4 + 6*a^2*b^2)*sinh(x)^6 + 24*(3*a^4 - 12*
a^2*b^2 + 8*b^4)*x*cosh(x)^4 + 24*(5*a^3*b - 4*a*b^3)*cosh(x)^5 + 24*(7*a^4*cosh(x)^3 - 7*a^3*b*cosh(x)^2 + 5*
a^3*b - 4*a*b^3 - 6*(a^4 - a^2*b^2)*cosh(x))*sinh(x)^5 + 8*a^3*b*cosh(x) + 2*(105*a^4*cosh(x)^4 - 140*a^3*b*co
sh(x)^3 - 180*(a^4 - a^2*b^2)*cosh(x)^2 + 12*(3*a^4 - 12*a^2*b^2 + 8*b^4)*x + 60*(5*a^3*b - 4*a*b^3)*cosh(x))*
sinh(x)^4 - 3*a^4 - 24*(5*a^3*b - 4*a*b^3)*cosh(x)^3 + 8*(21*a^4*cosh(x)^5 - 35*a^3*b*cosh(x)^4 - 15*a^3*b + 1
2*a*b^3 - 60*(a^4 - a^2*b^2)*cosh(x)^3 + 12*(3*a^4 - 12*a^2*b^2 + 8*b^4)*x*cosh(x) + 30*(5*a^3*b - 4*a*b^3)*co
sh(x)^2)*sinh(x)^3 + 24*(a^4 - a^2*b^2)*cosh(x)^2 + 12*(7*a^4*cosh(x)^6 - 14*a^3*b*cosh(x)^5 - 30*(a^4 - a^2*b
^2)*cosh(x)^4 + 2*a^4 - 2*a^2*b^2 + 12*(3*a^4 - 12*a^2*b^2 + 8*b^4)*x*cosh(x)^2 + 20*(5*a^3*b - 4*a*b^3)*cosh(
x)^3 - 6*(5*a^3*b - 4*a*b^3)*cosh(x))*sinh(x)^2 - 192*((a^2*b - b^3)*cosh(x)^4 + 4*(a^2*b - b^3)*cosh(x)^3*sin
h(x) + 6*(a^2*b - b^3)*cosh(x)^2*sinh(x)^2 + 4*(a^2*b - b^3)*cosh(x)*sinh(x)^3 + (a^2*b - b^3)*sinh(x)^4)*sqrt
(-a^2 + b^2)*log((a^2*cosh(x)^2 + a^2*sinh(x)^2 + 2*a*b*cosh(x) - a^2 + 2*b^2 + 2*(a^2*cosh(x) + a*b)*sinh(x)
+ 2*sqrt(-a^2 + b^2)*(a*cosh(x) + a*sinh(x) + b))/(a*cosh(x)^2 + a*sinh(x)^2 + 2*b*cosh(x) + 2*(a*cosh(x) + b)
*sinh(x) + a)) + 8*(3*a^4*cosh(x)^7 - 7*a^3*b*cosh(x)^6 - 18*(a^4 - a^2*b^2)*cosh(x)^5 + 12*(3*a^4 - 12*a^2*b^
2 + 8*b^4)*x*cosh(x)^3 + 15*(5*a^3*b - 4*a*b^3)*cosh(x)^4 + a^3*b - 9*(5*a^3*b - 4*a*b^3)*cosh(x)^2 + 6*(a^4 -
a^2*b^2)*cosh(x))*sinh(x))/(a^5*cosh(x)^4 + 4*a^5*cosh(x)^3*sinh(x) + 6*a^5*cosh(x)^2*sinh(x)^2 + 4*a^5*cosh(
x)*sinh(x)^3 + a^5*sinh(x)^4), 1/192*(3*a^4*cosh(x)^8 + 3*a^4*sinh(x)^8 - 8*a^3*b*cosh(x)^7 + 8*(3*a^4*cosh(x)
- a^3*b)*sinh(x)^7 - 24*(a^4 - a^2*b^2)*cosh(x)^6 + 4*(21*a^4*cosh(x)^2 - 14*a^3*b*cosh(x) - 6*a^4 + 6*a^2*b^
2)*sinh(x)^6 + 24*(3*a^4 - 12*a^2*b^2 + 8*b^4)*x*cosh(x)^4 + 24*(5*a^3*b - 4*a*b^3)*cosh(x)^5 + 24*(7*a^4*cosh
(x)^3 - 7*a^3*b*cosh(x)^2 + 5*a^3*b - 4*a*b^3 - 6*(a^4 - a^2*b^2)*cosh(x))*sinh(x)^5 + 8*a^3*b*cosh(x) + 2*(10
5*a^4*cosh(x)^4 - 140*a^3*b*cosh(x)^3 - 180*(a^4 - a^2*b^2)*cosh(x)^2 + 12*(3*a^4 - 12*a^2*b^2 + 8*b^4)*x + 60
*(5*a^3*b - 4*a*b^3)*cosh(x))*sinh(x)^4 - 3*a^4 - 24*(5*a^3*b - 4*a*b^3)*cosh(x)^3 + 8*(21*a^4*cosh(x)^5 - 35*
a^3*b*cosh(x)^4 - 15*a^3*b + 12*a*b^3 - 60*(a^4 - a^2*b^2)*cosh(x)^3 + 12*(3*a^4 - 12*a^2*b^2 + 8*b^4)*x*cosh(
x) + 30*(5*a^3*b - 4*a*b^3)*cosh(x)^2)*sinh(x)^3 + 24*(a^4 - a^2*b^2)*cosh(x)^2 + 12*(7*a^4*cosh(x)^6 - 14*a^3
*b*cosh(x)^5 - 30*(a^4 - a^2*b^2)*cosh(x)^4 + 2*a^4 - 2*a^2*b^2 + 12*(3*a^4 - 12*a^2*b^2 + 8*b^4)*x*cosh(x)^2
+ 20*(5*a^3*b - 4*a*b^3)*cosh(x)^3 - 6*(5*a^3*b - 4*a*b^3)*cosh(x))*sinh(x)^2 + 384*((a^2*b - b^3)*cosh(x)^4 +
4*(a^2*b - b^3)*cosh(x)^3*sinh(x) + 6*(a^2*b - b^3)*cosh(x)^2*sinh(x)^2 + 4*(a^2*b - b^3)*cosh(x)*sinh(x)^3 +
(a^2*b - b^3)*sinh(x)^4)*sqrt(a^2 - b^2)*arctan(-(a*cosh(x) + a*sinh(x) + b)/sqrt(a^2 - b^2)) + 8*(3*a^4*cosh
(x)^7 - 7*a^3*b*cosh(x)^6 - 18*(a^4 - a^2*b^2)*cosh(x)^5 + 12*(3*a^4 - 12*a^2*b^2 + 8*b^4)*x*cosh(x)^3 + 15*(5
*a^3*b - 4*a*b^3)*cosh(x)^4 + a^3*b - 9*(5*a^3*b - 4*a*b^3)*cosh(x)^2 + 6*(a^4 - a^2*b^2)*cosh(x))*sinh(x))/(a
^5*cosh(x)^4 + 4*a^5*cosh(x)^3*sinh(x) + 6*a^5*cosh(x)^2*sinh(x)^2 + 4*a^5*cosh(x)*sinh(x)^3 + a^5*sinh(x)^4)]

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giac [A]  time = 0.14, size = 197, normalized size = 1.49 \[ \frac {3 \, a^{3} e^{\left (4 \, x\right )} - 8 \, a^{2} b e^{\left (3 \, x\right )} - 24 \, a^{3} e^{\left (2 \, x\right )} + 24 \, a b^{2} e^{\left (2 \, x\right )} + 120 \, a^{2} b e^{x} - 96 \, b^{3} e^{x}}{192 \, a^{4}} + \frac {{\left (3 \, a^{4} - 12 \, a^{2} b^{2} + 8 \, b^{4}\right )} x}{8 \, a^{5}} + \frac {{\left (8 \, a^{3} b e^{x} - 3 \, a^{4} - 24 \, {\left (5 \, a^{3} b - 4 \, a b^{3}\right )} e^{\left (3 \, x\right )} + 24 \, {\left (a^{4} - a^{2} b^{2}\right )} e^{\left (2 \, x\right )}\right )} e^{\left (-4 \, x\right )}}{192 \, a^{5}} - \frac {2 \, {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \arctan \left (\frac {a e^{x} + b}{\sqrt {a^{2} - b^{2}}}\right )}{\sqrt {a^{2} - b^{2}} a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*(3*a^3*e^(4*x) - 8*a^2*b*e^(3*x) - 24*a^3*e^(2*x) + 24*a*b^2*e^(2*x) + 120*a^2*b*e^x - 96*b^3*e^x)/a^4 +
1/8*(3*a^4 - 12*a^2*b^2 + 8*b^4)*x/a^5 + 1/192*(8*a^3*b*e^x - 3*a^4 - 24*(5*a^3*b - 4*a*b^3)*e^(3*x) + 24*(a^
4 - a^2*b^2)*e^(2*x))*e^(-4*x)/a^5 - 2*(a^4*b - 2*a^2*b^3 + b^5)*arctan((a*e^x + b)/sqrt(a^2 - b^2))/(sqrt(a^2
- b^2)*a^5)

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maple [B]  time = 0.12, size = 488, normalized size = 3.70 \[ \frac {3 \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{8 a}-\frac {1}{8 a \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {3}{8 a \left (\tanh \left (\frac {x}{2}\right )-1\right )}-\frac {3 \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{8 a}+\frac {1}{8 a \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {3}{8 a \left (\tanh \left (\frac {x}{2}\right )+1\right )}-\frac {2 b \arctan \left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a \sqrt {\left (a +b \right ) \left (a -b \right )}}+\frac {4 b^{3} \arctan \left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{3} \sqrt {\left (a +b \right ) \left (a -b \right )}}-\frac {2 b^{5} \arctan \left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{5} \sqrt {\left (a +b \right ) \left (a -b \right )}}+\frac {1}{4 a \left (\tanh \left (\frac {x}{2}\right )-1\right )^{4}}+\frac {1}{2 a \left (\tanh \left (\frac {x}{2}\right )-1\right )^{3}}-\frac {1}{4 a \left (\tanh \left (\frac {x}{2}\right )+1\right )^{4}}+\frac {1}{2 a \left (\tanh \left (\frac {x}{2}\right )+1\right )^{3}}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )+1\right ) b^{4}}{a^{5}}-\frac {b}{a^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {b^{2}}{2 a^{3} \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {b^{3}}{a^{4} \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {b}{3 a^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )^{3}}+\frac {b^{2}}{2 a^{3} \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {b^{3}}{a^{4} \left (\tanh \left (\frac {x}{2}\right )-1\right )}-\frac {b}{a^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {b}{2 a^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}+\frac {b^{2}}{2 a^{3} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}+\frac {3 \ln \left (\tanh \left (\frac {x}{2}\right )-1\right ) b^{2}}{2 a^{3}}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right ) b^{4}}{a^{5}}+\frac {b}{3 a^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{3}}-\frac {b}{2 a^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {b^{2}}{2 a^{3} \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {3 \ln \left (\tanh \left (\frac {x}{2}\right )+1\right ) b^{2}}{2 a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/8/a*ln(tanh(1/2*x)+1)-1/8/a/(tanh(1/2*x)-1)^2-3/8/a/(tanh(1/2*x)-1)-3/8/a*ln(tanh(1/2*x)-1)+1/8/a/(tanh(1/2*
x)+1)^2-3/8/a/(tanh(1/2*x)+1)-2*b/a/((a+b)*(a-b))^(1/2)*arctan((a-b)*tanh(1/2*x)/((a+b)*(a-b))^(1/2))+4*b^3/a^
3/((a+b)*(a-b))^(1/2)*arctan((a-b)*tanh(1/2*x)/((a+b)*(a-b))^(1/2))-2*b^5/a^5/((a+b)*(a-b))^(1/2)*arctan((a-b)
*tanh(1/2*x)/((a+b)*(a-b))^(1/2))+1/4/a/(tanh(1/2*x)-1)^4+1/2/a/(tanh(1/2*x)-1)^3-1/4/a/(tanh(1/2*x)+1)^4+1/2/
a/(tanh(1/2*x)+1)^3+1/a^5*ln(tanh(1/2*x)+1)*b^4-1/a^2/(tanh(1/2*x)+1)*b+1/2/a^3/(tanh(1/2*x)+1)*b^2+1/a^4/(tan
h(1/2*x)+1)*b^3+1/3/a^2/(tanh(1/2*x)+1)^3*b+1/2/a^3/(tanh(1/2*x)-1)*b^2+1/a^4/(tanh(1/2*x)-1)*b^3-1/a^2/(tanh(
1/2*x)-1)*b+1/2/a^2/(tanh(1/2*x)-1)^2*b+1/2/a^3/(tanh(1/2*x)-1)^2*b^2+3/2/a^3*ln(tanh(1/2*x)-1)*b^2-1/a^5*ln(t
anh(1/2*x)-1)*b^4+1/3/a^2/(tanh(1/2*x)-1)^3*b-1/2/a^2/(tanh(1/2*x)+1)^2*b-1/2/a^3/(tanh(1/2*x)+1)^2*b^2-3/2/a^
3*ln(tanh(1/2*x)+1)*b^2

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

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mupad [B]  time = 2.01, size = 275, normalized size = 2.08 \[ \frac {{\mathrm {e}}^{4\,x}}{64\,a}-\frac {{\mathrm {e}}^{-4\,x}}{64\,a}+\frac {x\,\left (3\,a^4-12\,a^2\,b^2+8\,b^4\right )}{8\,a^5}-\frac {{\mathrm {e}}^{-x}\,\left (5\,a^2\,b-4\,b^3\right )}{8\,a^4}+\frac {{\mathrm {e}}^{-2\,x}\,\left (a^2-b^2\right )}{8\,a^3}-\frac {{\mathrm {e}}^{2\,x}\,\left (a^2-b^2\right )}{8\,a^3}+\frac {b\,{\mathrm {e}}^{-3\,x}}{24\,a^2}-\frac {b\,{\mathrm {e}}^{3\,x}}{24\,a^2}+\frac {{\mathrm {e}}^x\,\left (5\,a^2\,b-4\,b^3\right )}{8\,a^4}+\frac {b\,\ln \left (\frac {2\,{\mathrm {e}}^x\,\left (a^4\,b-2\,a^2\,b^3+b^5\right )}{a^6}-\frac {2\,b\,{\left (a+b\right )}^{3/2}\,\left (a+b\,{\mathrm {e}}^x\right )\,{\left (b-a\right )}^{3/2}}{a^6}\right )\,{\left (a+b\right )}^{3/2}\,{\left (b-a\right )}^{3/2}}{a^5}-\frac {b\,\ln \left (\frac {2\,{\mathrm {e}}^x\,\left (a^4\,b-2\,a^2\,b^3+b^5\right )}{a^6}+\frac {2\,b\,{\left (a+b\right )}^{3/2}\,\left (a+b\,{\mathrm {e}}^x\right )\,{\left (b-a\right )}^{3/2}}{a^6}\right )\,{\left (a+b\right )}^{3/2}\,{\left (b-a\right )}^{3/2}}{a^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)^4/(a + b/cosh(x)),x)

[Out]

exp(4*x)/(64*a) - exp(-4*x)/(64*a) + (x*(3*a^4 + 8*b^4 - 12*a^2*b^2))/(8*a^5) - (exp(-x)*(5*a^2*b - 4*b^3))/(8
*a^4) + (exp(-2*x)*(a^2 - b^2))/(8*a^3) - (exp(2*x)*(a^2 - b^2))/(8*a^3) + (b*exp(-3*x))/(24*a^2) - (b*exp(3*x
))/(24*a^2) + (exp(x)*(5*a^2*b - 4*b^3))/(8*a^4) + (b*log((2*exp(x)*(a^4*b + b^5 - 2*a^2*b^3))/a^6 - (2*b*(a +
b)^(3/2)*(a + b*exp(x))*(b - a)^(3/2))/a^6)*(a + b)^(3/2)*(b - a)^(3/2))/a^5 - (b*log((2*exp(x)*(a^4*b + b^5
- 2*a^2*b^3))/a^6 + (2*b*(a + b)^(3/2)*(a + b*exp(x))*(b - a)^(3/2))/a^6)*(a + b)^(3/2)*(b - a)^(3/2))/a^5

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sinh ^{4}{\relax (x )}}{a + b \operatorname {sech}{\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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