3.200 \(\int \frac {\cosh ^4(x)}{(a+b \sinh (x))^2} \, dx\)

Optimal. Leaf size=94 \[ \frac {3 x \left (2 a^2+b^2\right )}{2 b^4}+\frac {6 a \sqrt {a^2+b^2} \tanh ^{-1}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{b^4}-\frac {3 \cosh (x) (2 a-b \sinh (x))}{2 b^3}-\frac {\cosh ^3(x)}{b (a+b \sinh (x))} \]

[Out]

3/2*(2*a^2+b^2)*x/b^4-3/2*cosh(x)*(2*a-b*sinh(x))/b^3-cosh(x)^3/b/(a+b*sinh(x))+6*a*arctanh((b-a*tanh(1/2*x))/
(a^2+b^2)^(1/2))*(a^2+b^2)^(1/2)/b^4

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Rubi [A]  time = 0.22, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {2693, 2865, 2735, 2660, 618, 206} \[ \frac {3 x \left (2 a^2+b^2\right )}{2 b^4}+\frac {6 a \sqrt {a^2+b^2} \tanh ^{-1}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{b^4}-\frac {3 \cosh (x) (2 a-b \sinh (x))}{2 b^3}-\frac {\cosh ^3(x)}{b (a+b \sinh (x))} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[x]^4/(a + b*Sinh[x])^2,x]

[Out]

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

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])

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 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 2693

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

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]

Rubi steps

\begin {align*} \int \frac {\cosh ^4(x)}{(a+b \sinh (x))^2} \, dx &=-\frac {\cosh ^3(x)}{b (a+b \sinh (x))}+\frac {3 \int \frac {\cosh ^2(x) \sinh (x)}{a+b \sinh (x)} \, dx}{b}\\ &=-\frac {3 \cosh (x) (2 a-b \sinh (x))}{2 b^3}-\frac {\cosh ^3(x)}{b (a+b \sinh (x))}+\frac {(3 i) \int \frac {i a b-i \left (2 a^2+b^2\right ) \sinh (x)}{a+b \sinh (x)} \, dx}{2 b^3}\\ &=\frac {3 \left (2 a^2+b^2\right ) x}{2 b^4}-\frac {3 \cosh (x) (2 a-b \sinh (x))}{2 b^3}-\frac {\cosh ^3(x)}{b (a+b \sinh (x))}-\frac {\left (3 a \left (a^2+b^2\right )\right ) \int \frac {1}{a+b \sinh (x)} \, dx}{b^4}\\ &=\frac {3 \left (2 a^2+b^2\right ) x}{2 b^4}-\frac {3 \cosh (x) (2 a-b \sinh (x))}{2 b^3}-\frac {\cosh ^3(x)}{b (a+b \sinh (x))}-\frac {\left (6 a \left (a^2+b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a+2 b x-a x^2} \, dx,x,\tanh \left (\frac {x}{2}\right )\right )}{b^4}\\ &=\frac {3 \left (2 a^2+b^2\right ) x}{2 b^4}-\frac {3 \cosh (x) (2 a-b \sinh (x))}{2 b^3}-\frac {\cosh ^3(x)}{b (a+b \sinh (x))}+\frac {\left (12 a \left (a^2+b^2\right )\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 )}{b^4}\\ &=\frac {3 \left (2 a^2+b^2\right ) x}{2 b^4}+\frac {6 a \sqrt {a^2+b^2} \tanh ^{-1}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{b^4}-\frac {3 \cosh (x) (2 a-b \sinh (x))}{2 b^3}-\frac {\cosh ^3(x)}{b (a+b \sinh (x))}\\ \end {align*}

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Mathematica [C]  time = 4.46, size = 660, normalized size = 7.02 \[ \frac {\cosh ^3(x) \left (-12 a \left (a^2+b^2\right ) \sqrt {1+i \sinh (x)} (a+b \sinh (x)) \tanh ^{-1}\left (\frac {\sqrt {a-i b} \sqrt {-\frac {b (\sinh (x)+i)}{a-i b}}}{\sqrt {a+i b} \sqrt {-\frac {b (\sinh (x)-i)}{a+i b}}}\right )+\sqrt {a+i b} \sqrt {-\frac {b (\sinh (x)-i)}{a+i b}} \left (6 (-1)^{3/4} a \sqrt {b} \left (2 a^2+i a b+b^2\right ) \sin ^{-1}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {a-i b} \sqrt {-\frac {b (\sinh (x)+i)}{a-i b}}}{\sqrt {b}}\right )+6 (-1)^{3/4} b^{3/2} \left (2 a^2+i a b+b^2\right ) \sinh (x) \sin ^{-1}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {a-i b} \sqrt {-\frac {b (\sinh (x)+i)}{a-i b}}}{\sqrt {b}}\right )-2 \sqrt {a-i b} \left (3 a^3+3 i a^2 b+a b^2+i b^3\right ) \sqrt {1+i \sinh (x)} \sqrt {-\frac {b (\sinh (x)+i)}{a-i b}}+b^2 \sqrt {a-i b} (a+i b) \sqrt {1+i \sinh (x)} \sinh ^2(x) \sqrt {-\frac {b (\sinh (x)+i)}{a-i b}}-3 a b \sqrt {a-i b} (a+i b) \sqrt {1+i \sinh (x)} \sinh (x) \sqrt {-\frac {b (\sinh (x)+i)}{a-i b}}\right )+12 a \sqrt {a-i b} (a+i b)^{3/2} \sqrt {1+i \sinh (x)} (a+b \sinh (x)) \tanh ^{-1}\left (\frac {\sqrt {-\frac {b (\sinh (x)+i)}{a-i b}}}{\sqrt {-\frac {b (\sinh (x)-i)}{a+i b}}}\right )\right )}{2 b (a-i b)^{3/2} (a+i b)^{5/2} \sqrt {1+i \sinh (x)} \left (-\frac {b (\sinh (x)-i)}{a+i b}\right )^{3/2} \left (-\frac {b (\sinh (x)+i)}{a-i b}\right )^{3/2} (a+b \sinh (x))} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[x]^4/(a + b*Sinh[x])^2,x]

[Out]

(Cosh[x]^3*(12*a*Sqrt[a - I*b]*(a + I*b)^(3/2)*ArcTanh[Sqrt[-((b*(I + Sinh[x]))/(a - I*b))]/Sqrt[-((b*(-I + Si
nh[x]))/(a + I*b))]]*Sqrt[1 + I*Sinh[x]]*(a + b*Sinh[x]) - 12*a*(a^2 + b^2)*ArcTanh[(Sqrt[a - I*b]*Sqrt[-((b*(
I + Sinh[x]))/(a - I*b))])/(Sqrt[a + I*b]*Sqrt[-((b*(-I + Sinh[x]))/(a + I*b))])]*Sqrt[1 + I*Sinh[x]]*(a + b*S
inh[x]) + Sqrt[a + I*b]*Sqrt[-((b*(-I + Sinh[x]))/(a + I*b))]*(6*(-1)^(3/4)*a*Sqrt[b]*(2*a^2 + I*a*b + b^2)*Ar
cSin[((1/2 + I/2)*Sqrt[a - I*b]*Sqrt[-((b*(I + Sinh[x]))/(a - I*b))])/Sqrt[b]] + 6*(-1)^(3/4)*b^(3/2)*(2*a^2 +
I*a*b + b^2)*ArcSin[((1/2 + I/2)*Sqrt[a - I*b]*Sqrt[-((b*(I + Sinh[x]))/(a - I*b))])/Sqrt[b]]*Sinh[x] - 2*Sqr
t[a - I*b]*(3*a^3 + (3*I)*a^2*b + a*b^2 + I*b^3)*Sqrt[1 + I*Sinh[x]]*Sqrt[-((b*(I + Sinh[x]))/(a - I*b))] - 3*
a*Sqrt[a - I*b]*(a + I*b)*b*Sqrt[1 + I*Sinh[x]]*Sinh[x]*Sqrt[-((b*(I + Sinh[x]))/(a - I*b))] + Sqrt[a - I*b]*(
a + I*b)*b^2*Sqrt[1 + I*Sinh[x]]*Sinh[x]^2*Sqrt[-((b*(I + Sinh[x]))/(a - I*b))])))/(2*(a - I*b)^(3/2)*(a + I*b
)^(5/2)*b*Sqrt[1 + I*Sinh[x]]*(-((b*(-I + Sinh[x]))/(a + I*b)))^(3/2)*(-((b*(I + Sinh[x]))/(a - I*b)))^(3/2)*(
a + b*Sinh[x]))

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fricas [B]  time = 0.54, size = 833, normalized size = 8.86 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)^4/(a+b*sinh(x))^2,x, algorithm="fricas")

[Out]

1/8*(b^3*cosh(x)^6 + b^3*sinh(x)^6 - 6*a*b^2*cosh(x)^5 + 6*(b^3*cosh(x) - a*b^2)*sinh(x)^5 - (16*a^2*b + b^3 -
12*(2*a^2*b + b^3)*x)*cosh(x)^4 + (15*b^3*cosh(x)^2 - 30*a*b^2*cosh(x) - 16*a^2*b - b^3 + 12*(2*a^2*b + b^3)*
x)*sinh(x)^4 + 6*a*b^2*cosh(x) + 8*(2*a^3 + 2*a*b^2 + 3*(2*a^3 + a*b^2)*x)*cosh(x)^3 + 4*(5*b^3*cosh(x)^3 - 15
*a*b^2*cosh(x)^2 + 4*a^3 + 4*a*b^2 + 6*(2*a^3 + a*b^2)*x - (16*a^2*b + b^3 - 12*(2*a^2*b + b^3)*x)*cosh(x))*si
nh(x)^3 + b^3 - (32*a^2*b + 17*b^3 + 12*(2*a^2*b + b^3)*x)*cosh(x)^2 + (15*b^3*cosh(x)^4 - 60*a*b^2*cosh(x)^3
- 32*a^2*b - 17*b^3 - 6*(16*a^2*b + b^3 - 12*(2*a^2*b + b^3)*x)*cosh(x)^2 - 12*(2*a^2*b + b^3)*x + 24*(2*a^3 +
2*a*b^2 + 3*(2*a^3 + a*b^2)*x)*cosh(x))*sinh(x)^2 + 24*(a*b*cosh(x)^4 + a*b*sinh(x)^4 + 2*a^2*cosh(x)^3 - a*b
*cosh(x)^2 + 2*(2*a*b*cosh(x) + a^2)*sinh(x)^3 + (6*a*b*cosh(x)^2 + 6*a^2*cosh(x) - a*b)*sinh(x)^2 + 2*(2*a*b*
cosh(x)^3 + 3*a^2*cosh(x)^2 - a*b*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*cos
h(x)^2 + b*sinh(x)^2 + 2*a*cosh(x) + 2*(b*cosh(x) + a)*sinh(x) - b)) + 2*(3*b^3*cosh(x)^5 - 15*a*b^2*cosh(x)^4
- 2*(16*a^2*b + b^3 - 12*(2*a^2*b + b^3)*x)*cosh(x)^3 + 3*a*b^2 + 12*(2*a^3 + 2*a*b^2 + 3*(2*a^3 + a*b^2)*x)*
cosh(x)^2 - (32*a^2*b + 17*b^3 + 12*(2*a^2*b + b^3)*x)*cosh(x))*sinh(x))/(b^5*cosh(x)^4 + b^5*sinh(x)^4 + 2*a*
b^4*cosh(x)^3 - b^5*cosh(x)^2 + 2*(2*b^5*cosh(x) + a*b^4)*sinh(x)^3 + (6*b^5*cosh(x)^2 + 6*a*b^4*cosh(x) - b^5
)*sinh(x)^2 + 2*(2*b^5*cosh(x)^3 + 3*a*b^4*cosh(x)^2 - b^5*cosh(x))*sinh(x))

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giac [B]  time = 0.18, size = 178, normalized size = 1.89 \[ \frac {3 \, {\left (2 \, a^{2} + b^{2}\right )} x}{2 \, b^{4}} - \frac {3 \, {\left (a^{3} + a b^{2}\right )} \log \left (\frac {{\left | 2 \, b e^{x} + 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{x} + 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}} b^{4}} + \frac {b^{2} e^{\left (2 \, x\right )} - 8 \, a b e^{x}}{8 \, b^{4}} + \frac {{\left (6 \, a b^{2} e^{x} + b^{3} + 8 \, {\left (2 \, a^{3} + a b^{2}\right )} e^{\left (3 \, x\right )} - {\left (32 \, a^{2} b + 17 \, b^{3}\right )} e^{\left (2 \, x\right )}\right )} e^{\left (-2 \, x\right )}}{8 \, {\left (b e^{\left (2 \, x\right )} + 2 \, a e^{x} - b\right )} b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)^4/(a+b*sinh(x))^2,x, algorithm="giac")

[Out]

3/2*(2*a^2 + b^2)*x/b^4 - 3*(a^3 + a*b^2)*log(abs(2*b*e^x + 2*a - 2*sqrt(a^2 + b^2))/abs(2*b*e^x + 2*a + 2*sqr
t(a^2 + b^2)))/(sqrt(a^2 + b^2)*b^4) + 1/8*(b^2*e^(2*x) - 8*a*b*e^x)/b^4 + 1/8*(6*a*b^2*e^x + b^3 + 8*(2*a^3 +
a*b^2)*e^(3*x) - (32*a^2*b + 17*b^3)*e^(2*x))*e^(-2*x)/((b*e^(2*x) + 2*a*e^x - b)*b^4)

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maple [B]  time = 0.08, size = 290, normalized size = 3.09 \[ \frac {2 a \tanh \left (\frac {x}{2}\right )}{b^{2} \left (a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-2 \tanh \left (\frac {x}{2}\right ) b -a \right )}+\frac {2 \tanh \left (\frac {x}{2}\right )}{\left (a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-2 \tanh \left (\frac {x}{2}\right ) b -a \right ) a}+\frac {2 a^{2}}{b^{3} \left (a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-2 \tanh \left (\frac {x}{2}\right ) b -a \right )}+\frac {2}{b \left (a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-2 \tanh \left (\frac {x}{2}\right ) b -a \right )}-\frac {6 a \sqrt {a^{2}+b^{2}}\, \arctanh \left (\frac {2 a \tanh \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{b^{4}}+\frac {1}{2 b^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}+\frac {1}{2 b^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {2 a}{b^{3} \left (\tanh \left (\frac {x}{2}\right )-1\right )}-\frac {3 \ln \left (\tanh \left (\frac {x}{2}\right )-1\right ) a^{2}}{b^{4}}-\frac {3 \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2 b^{2}}-\frac {1}{2 b^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}+\frac {1}{2 b^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )}-\frac {2 a}{b^{3} \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {3 \ln \left (\tanh \left (\frac {x}{2}\right )+1\right ) a^{2}}{b^{4}}+\frac {3 \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{2 b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)^4/(a+b*sinh(x))^2,x)

[Out]

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

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maxima [B]  time = 0.42, size = 176, normalized size = 1.87 \[ -\frac {6 \, a b^{2} e^{\left (-x\right )} - b^{3} + {\left (32 \, a^{2} b + 17 \, b^{3}\right )} e^{\left (-2 \, x\right )} + 8 \, {\left (2 \, a^{3} + a b^{2}\right )} e^{\left (-3 \, x\right )}}{8 \, {\left (b^{5} e^{\left (-2 \, x\right )} + 2 \, a b^{4} e^{\left (-3 \, x\right )} - b^{5} e^{\left (-4 \, x\right )}\right )}} - \frac {3 \, \sqrt {a^{2} + b^{2}} a \log \left (\frac {b e^{\left (-x\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-x\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{b^{4}} - \frac {8 \, a e^{\left (-x\right )} + b e^{\left (-2 \, x\right )}}{8 \, b^{3}} + \frac {3 \, {\left (2 \, a^{2} + b^{2}\right )} x}{2 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)^4/(a+b*sinh(x))^2,x, algorithm="maxima")

[Out]

-1/8*(6*a*b^2*e^(-x) - b^3 + (32*a^2*b + 17*b^3)*e^(-2*x) + 8*(2*a^3 + a*b^2)*e^(-3*x))/(b^5*e^(-2*x) + 2*a*b^
4*e^(-3*x) - b^5*e^(-4*x)) - 3*sqrt(a^2 + b^2)*a*log((b*e^(-x) - a - sqrt(a^2 + b^2))/(b*e^(-x) - a + sqrt(a^2
+ b^2)))/b^4 - 1/8*(8*a*e^(-x) + b*e^(-2*x))/b^3 + 3/2*(2*a^2 + b^2)*x/b^4

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mupad [B]  time = 0.88, size = 256, normalized size = 2.72 \[ \frac {{\mathrm {e}}^{2\,x}}{8\,b^2}-\frac {{\mathrm {e}}^{-2\,x}}{8\,b^2}-\frac {\frac {2\,\left (a^4\,b^2+2\,a^2\,b^4+b^6\right )}{b^4\,\left (a^2\,b+b^3\right )}-\frac {2\,{\mathrm {e}}^x\,\left (a^5\,b^2+2\,a^3\,b^4+a\,b^6\right )}{b^5\,\left (a^2\,b+b^3\right )}}{2\,a\,{\mathrm {e}}^x-b+b\,{\mathrm {e}}^{2\,x}}+\frac {x\,\left (6\,a^2+3\,b^2\right )}{2\,b^4}-\frac {a\,{\mathrm {e}}^x}{b^3}-\frac {a\,{\mathrm {e}}^{-x}}{b^3}-\frac {3\,a\,\ln \left (\frac {6\,a\,{\mathrm {e}}^x\,\left (a^2+b^2\right )}{b^5}-\frac {6\,a\,\left (b-a\,{\mathrm {e}}^x\right )\,\sqrt {a^2+b^2}}{b^5}\right )\,\sqrt {a^2+b^2}}{b^4}+\frac {3\,a\,\ln \left (\frac {6\,a\,\left (b-a\,{\mathrm {e}}^x\right )\,\sqrt {a^2+b^2}}{b^5}+\frac {6\,a\,{\mathrm {e}}^x\,\left (a^2+b^2\right )}{b^5}\right )\,\sqrt {a^2+b^2}}{b^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)^4/(a + b*sinh(x))^2,x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)**4/(a+b*sinh(x))**2,x)

[Out]

Timed out

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