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*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]))

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Rubi [A]  time = 0.220092, antiderivative size = 94, normalized size of antiderivative = 1., 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 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 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 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 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{\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*}

Mathematica [C]  time = 6.278, size = 2901, normalized size = 30.86 \[ \text{Result too large to show} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*Cosh[x]^3*((I*b*(((-I)*b)/(a - I*b) - (b*Sinh[x])/(a - I*b))^(5/2)*((I*b)/(a + I*b) - (b*Sinh[x])/(a + I
*b))^(5/2))/((((-I)*a*b)/(a - I*b) - b^2/(a - I*b))*(((-I)*a*b)/(a + I*b) + b^2/(a + I*b))*(a + b*Sinh[x])) -
((16*Sqrt[2]*(a - I*b)*b^2*(((-I)*b)/(a - I*b) - (b*Sinh[x])/(a - I*b))^(5/2)*Sqrt[(I*b)/(a + I*b) - (b*Sinh[x
])/(a + I*b)]*(1 - ((I/2)*(a - I*b)*(((-I)*b)/(a - I*b) - (b*Sinh[x])/(a - I*b)))/b)^(5/2)*((5*(1/(2*(1 - ((I/
2)*(a - I*b)*(((-I)*b)/(a - I*b) - (b*Sinh[x])/(a - I*b)))/b)^2) + (1 - ((I/2)*(a - I*b)*(((-I)*b)/(a - I*b) -
(b*Sinh[x])/(a - I*b)))/b)^(-1)))/8 + (((15*I)/32)*b^3*(((-I)*(a - I*b)*(((-I)*b)/(a - I*b) - (b*Sinh[x])/(a
- I*b)))/b + ((a - I*b)^2*(((-I)*b)/(a - I*b) - (b*Sinh[x])/(a - I*b))^2)/(3*b^2) + ((-1)^(1/4)*Sqrt[2]*Sqrt[a
- I*b]*ArcSin[((-1)^(1/4)*Sqrt[a - I*b]*Sqrt[((-I)*b)/(a - I*b) - (b*Sinh[x])/(a - I*b)])/(Sqrt[2]*Sqrt[b])]*
Sqrt[((-I)*b)/(a - I*b) - (b*Sinh[x])/(a - I*b)])/(Sqrt[b]*Sqrt[1 - ((I/2)*(a - I*b)*(((-I)*b)/(a - I*b) - (b*
Sinh[x])/(a - I*b)))/b])))/((a - I*b)^3*(((-I)*b)/(a - I*b) - (b*Sinh[x])/(a - I*b))^3*(1 - ((I/2)*(a - I*b)*(
((-I)*b)/(a - I*b) - (b*Sinh[x])/(a - I*b)))/b)^2)))/(5*(a + I*b)*(a^2 + b^2)*Sqrt[((-I)*(a + I*b)*((I*b)/(a +
I*b) - (b*Sinh[x])/(a + I*b)))/b]) - (3*a*b^2*((-4*Sqrt[2]*(((-I)*b)/(a - I*b) - (b*Sinh[x])/(a - I*b))^(3/2)
*Sqrt[(I*b)/(a + I*b) - (b*Sinh[x])/(a + I*b)]*(1 - ((I/2)*(a - I*b)*(((-I)*b)/(a - I*b) - (b*Sinh[x])/(a - I*
b)))/b)^(5/2)*((3/(4*(1 - ((I/2)*(a - I*b)*(((-I)*b)/(a - I*b) - (b*Sinh[x])/(a - I*b)))/b)^2) + (1 - ((I/2)*(
a - I*b)*(((-I)*b)/(a - I*b) - (b*Sinh[x])/(a - I*b)))/b)^(-1))/2 - (3*b^2*(((-I)*(a - I*b)*(((-I)*b)/(a - I*b
) - (b*Sinh[x])/(a - I*b)))/b + ((-1)^(1/4)*Sqrt[2]*Sqrt[a - I*b]*ArcSin[((-1)^(1/4)*Sqrt[a - I*b]*Sqrt[((-I)*
b)/(a - I*b) - (b*Sinh[x])/(a - I*b)])/(Sqrt[2]*Sqrt[b])]*Sqrt[((-I)*b)/(a - I*b) - (b*Sinh[x])/(a - I*b)])/(S
qrt[b]*Sqrt[1 - ((I/2)*(a - I*b)*(((-I)*b)/(a - I*b) - (b*Sinh[x])/(a - I*b)))/b])))/(8*(a - I*b)^2*(((-I)*b)/
(a - I*b) - (b*Sinh[x])/(a - I*b))^2*(1 - ((I/2)*(a - I*b)*(((-I)*b)/(a - I*b) - (b*Sinh[x])/(a - I*b)))/b)^2)
))/(3*(a + I*b)*Sqrt[((-I)*(a + I*b)*((I*b)/(a + I*b) - (b*Sinh[x])/(a + I*b)))/b]) - (I*((I*a*b)/(a - I*b) +
b^2/(a - I*b))*(((-I)*((I*a*b)/(a - I*b) + b^2/(a - I*b))*(((-I)*((I*a*b)/(a + I*b) - b^2/(a + I*b))*(((-2*I)*
((I*a*b)/(a + I*b) - b^2/(a + I*b))*ArcTan[(Sqrt[((-I)*a*b)/(a + I*b) + b^2/(a + I*b)]*Sqrt[((-I)*b)/(a - I*b)
- (b*Sinh[x])/(a - I*b)])/(Sqrt[(I*a*b)/(a - I*b) + b^2/(a - I*b)]*Sqrt[(I*b)/(a + I*b) - (b*Sinh[x])/(a + I*
b)])])/(b*Sqrt[(I*a*b)/(a - I*b) + b^2/(a - I*b)]*Sqrt[((-I)*a*b)/(a + I*b) + b^2/(a + I*b)]) + ((2*I)*Sqrt[a
- I*b]*ArcTanh[(Sqrt[a - I*b]*Sqrt[((-I)*b)/(a - I*b) - (b*Sinh[x])/(a - I*b)])/(Sqrt[a + I*b]*Sqrt[(I*b)/(a +
I*b) - (b*Sinh[x])/(a + I*b)])])/(Sqrt[a + I*b]*b)))/b + ((2*I)*Sqrt[2]*(a - I*b)*Sqrt[((-I)*b)/(a - I*b) - (
b*Sinh[x])/(a - I*b)]*Sqrt[(I*b)/(a + I*b) - (b*Sinh[x])/(a + I*b)]*(1 - ((I/2)*(a - I*b)*(((-I)*b)/(a - I*b)
- (b*Sinh[x])/(a - I*b)))/b)^(3/2)*(-(((-1)^(3/4)*Sqrt[b]*ArcSin[((-1)^(1/4)*Sqrt[a - I*b]*Sqrt[((-I)*b)/(a -
I*b) - (b*Sinh[x])/(a - I*b)])/(Sqrt[2]*Sqrt[b])])/(Sqrt[2]*Sqrt[a - I*b]*Sqrt[((-I)*b)/(a - I*b) - (b*Sinh[x]
)/(a - I*b)]*(1 - ((I/2)*(a - I*b)*(((-I)*b)/(a - I*b) - (b*Sinh[x])/(a - I*b)))/b)^(3/2))) + 1/(2*(1 - ((I/2)
*(a - I*b)*(((-I)*b)/(a - I*b) - (b*Sinh[x])/(a - I*b)))/b))))/((a + I*b)*b*Sqrt[((-I)*(a + I*b)*((I*b)/(a + I
*b) - (b*Sinh[x])/(a + I*b)))/b])))/b - (4*Sqrt[2]*Sqrt[((-I)*b)/(a - I*b) - (b*Sinh[x])/(a - I*b)]*Sqrt[(I*b)
/(a + I*b) - (b*Sinh[x])/(a + I*b)]*(1 - ((I/2)*(a - I*b)*(((-I)*b)/(a - I*b) - (b*Sinh[x])/(a - I*b)))/b)^(5/
2)*((-3*(-1)^(3/4)*Sqrt[b]*ArcSin[((-1)^(1/4)*Sqrt[a - I*b]*Sqrt[((-I)*b)/(a - I*b) - (b*Sinh[x])/(a - I*b)])/
(Sqrt[2]*Sqrt[b])])/(4*Sqrt[2]*Sqrt[a - I*b]*Sqrt[((-I)*b)/(a - I*b) - (b*Sinh[x])/(a - I*b)]*(1 - ((I/2)*(a -
I*b)*(((-I)*b)/(a - I*b) - (b*Sinh[x])/(a - I*b)))/b)^(5/2)) + (3/(2*(1 - ((I/2)*(a - I*b)*(((-I)*b)/(a - I*b
) - (b*Sinh[x])/(a - I*b)))/b)^2) + (1 - ((I/2)*(a - I*b)*(((-I)*b)/(a - I*b) - (b*Sinh[x])/(a - I*b)))/b)^(-1
))/4))/((a + I*b)*Sqrt[((-I)*(a + I*b)*((I*b)/(a + I*b) - (b*Sinh[x])/(a + I*b)))/b])))/b))/(a^2 + b^2))/((((-
I)*a*b)/(a - I*b) - b^2/(a - I*b))*(((-I)*a*b)/(a + I*b) + b^2/(a + I*b)))))/((1 - (a + b*Sinh[x])/(a - I*b))^
(3/2)*(1 - (a + b*Sinh[x])/(a + I*b))^(3/2))

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Maple [B]  time = 0.058, size = 290, normalized size = 3.1 \begin{align*} -{\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 ) ^{-1}}-2\,{\frac{a}{{b}^{3} \left ( \tanh \left ( x/2 \right ) +1 \right ) }}+3\,{\frac{\ln \left ( \tanh \left ( x/2 \right ) +1 \right ){a}^{2}}{{b}^{4}}}+{\frac{3}{2\,{b}^{2}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }+{\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 ) ^{-1}}+2\,{\frac{a}{{b}^{3} \left ( \tanh \left ( x/2 \right ) -1 \right ) }}-3\,{\frac{\ln \left ( \tanh \left ( x/2 \right ) -1 \right ){a}^{2}}{{b}^{4}}}-{\frac{3}{2\,{b}^{2}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) }+2\,{\frac{a\tanh \left ( x/2 \right ) }{{b}^{2} \left ( a \left ( \tanh \left ( x/2 \right ) \right ) ^{2}-2\,\tanh \left ( x/2 \right ) b-a \right ) }}+2\,{\frac{\tanh \left ( x/2 \right ) }{ \left ( a \left ( \tanh \left ( x/2 \right ) \right ) ^{2}-2\,\tanh \left ( x/2 \right ) b-a \right ) a}}+2\,{\frac{{a}^{2}}{{b}^{3} \left ( a \left ( \tanh \left ( x/2 \right ) \right ) ^{2}-2\,\tanh \left ( x/2 \right ) b-a \right ) }}+2\,{\frac{1}{b \left ( a \left ( \tanh \left ( x/2 \right ) \right ) ^{2}-2\,\tanh \left ( x/2 \right ) b-a \right ) }}-6\,{\frac{a\sqrt{{a}^{2}+{b}^{2}}}{{b}^{4}}{\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(cosh(x)^4/(a+b*sinh(x))^2,x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.1482, size = 2184, normalized size = 23.23 \begin{align*} \text{result too large to display} \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Giac [B]  time = 1.25865, size = 240, normalized size = 2.55 \begin{align*} \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}} \end{align*}

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)