3.698 \(\int \frac {\sinh ^3(x)}{(a \cosh (x)+b \sinh (x))^2} \, dx\)

Optimal. Leaf size=195 \[ -\frac {a^3}{b^3 (a+b)^2 \left (1-\tanh \left (\frac {x}{2}\right )\right )}+\frac {a^3}{b^3 (a-b)^2 \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {2 a^2 \left (a+b \tanh \left (\frac {x}{2}\right )\right )}{\left (a^2-b^2\right )^2 \left (a \tanh ^2\left (\frac {x}{2}\right )+a+2 b \tanh \left (\frac {x}{2}\right )\right )}+\frac {3 a^2 b \tan ^{-1}\left (\frac {a \sinh (x)+b \cosh (x)}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac {\left (2 a^2+b^2\right ) \cosh (x)}{b^4-a^2 b^2}+\frac {a \left (a^2+2 b^2\right ) \sinh (x)}{b^3 \left (a^2-b^2\right )} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 1.22, antiderivative size = 301, normalized size of antiderivative = 1.54, number of steps used = 16, number of rules used = 10, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {4401, 2637, 2638, 6742, 638, 618, 204, 3100, 3074, 206} \[ \frac {3 a^3 \sinh (x)}{b^3 \left (a^2-b^2\right )}-\frac {3 a^2 \cosh (x)}{b^2 \left (a^2-b^2\right )}+\frac {2 a^2 \left (a+b \tanh \left (\frac {x}{2}\right )\right )}{\left (a^2-b^2\right )^2 \left (a \tanh ^2\left (\frac {x}{2}\right )+a+2 b \tanh \left (\frac {x}{2}\right )\right )}-\frac {a^3}{b^3 (a+b)^2 \left (1-\tanh \left (\frac {x}{2}\right )\right )}+\frac {a^3}{b^3 (a-b)^2 \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {2 a^2 \left (3 a^2-b^2\right ) \tan ^{-1}\left (\frac {a \tanh \left (\frac {x}{2}\right )+b}{\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{5/2}}+\frac {2 a^2 b \tan ^{-1}\left (\frac {a \tanh \left (\frac {x}{2}\right )+b}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}-\frac {3 a^2 \tan ^{-1}\left (\frac {a \sinh (x)+b \cosh (x)}{\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{3/2}}-\frac {2 a \sinh (x)}{b^3}+\frac {\cosh (x)}{b^2} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]^3/(a*Cosh[x] + b*Sinh[x])^2,x]

[Out]

(-3*a^2*ArcTan[(b*Cosh[x] + a*Sinh[x])/Sqrt[a^2 - b^2]])/(b*(a^2 - b^2)^(3/2)) + (2*a^2*b*ArcTan[(b + a*Tanh[x
/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(5/2) + (2*a^2*(3*a^2 - b^2)*ArcTan[(b + a*Tanh[x/2])/Sqrt[a^2 - b^2]])/(b*
(a^2 - b^2)^(5/2)) + Cosh[x]/b^2 - (3*a^2*Cosh[x])/(b^2*(a^2 - b^2)) - (2*a*Sinh[x])/b^3 + (3*a^3*Sinh[x])/(b^
3*(a^2 - b^2)) - a^3/(b^3*(a + b)^2*(1 - Tanh[x/2])) + a^3/((a - b)^2*b^3*(1 + Tanh[x/2])) + (2*a^2*(a + b*Tan
h[x/2]))/((a^2 - b^2)^2*(a + 2*b*Tanh[x/2] + a*Tanh[x/2]^2))

Rule 204

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

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 638

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b*d - 2*a*e + (2*c*d -
b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] - Dist[((2*p + 3)*(2*c*d - b*e))/((p + 1)*(b^2
- 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^
2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3074

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

Rule 3100

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

Rule 4401

Int[u_, x_Symbol] :> With[{v = ExpandTrig[u, x]}, Int[v, x] /; SumQ[v]] /;  !InertTrigFreeQ[u]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.46, size = 205, normalized size = 1.05 \[ \frac {-b \cosh (x) \left ((a-b)^{3/2} (a+b)^2 \sinh (x)-6 a^3 \sqrt {a+b} \tan ^{-1}\left (\frac {a \tanh \left (\frac {x}{2}\right )+b}{\sqrt {a-b} \sqrt {a+b}}\right )\right )+a \left (a^2 \sqrt {a-b} (a+b)-2 b^2 \sqrt {a-b} (a+b) \sinh ^2(x)+6 a b^2 \sqrt {a+b} \sinh (x) \tan ^{-1}\left (\frac {a \tanh \left (\frac {x}{2}\right )+b}{\sqrt {a-b} \sqrt {a+b}}\right )\right )+a \sqrt {a-b} \left (a^3+a^2 b+a b^2+b^3\right ) \cosh ^2(x)}{(a-b)^{5/2} (a+b)^3 (a \cosh (x)+b \sinh (x))} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[x]^3/(a*Cosh[x] + b*Sinh[x])^2,x]

[Out]

(a*Sqrt[a - b]*(a^3 + a^2*b + a*b^2 + b^3)*Cosh[x]^2 - b*Cosh[x]*(-6*a^3*Sqrt[a + b]*ArcTan[(b + a*Tanh[x/2])/
(Sqrt[a - b]*Sqrt[a + b])] + (a - b)^(3/2)*(a + b)^2*Sinh[x]) + a*(a^2*Sqrt[a - b]*(a + b) + 6*a*b^2*Sqrt[a +
b]*ArcTan[(b + a*Tanh[x/2])/(Sqrt[a - b]*Sqrt[a + b])]*Sinh[x] - 2*Sqrt[a - b]*b^2*(a + b)*Sinh[x]^2))/((a - b
)^(5/2)*(a + b)^3*(a*Cosh[x] + b*Sinh[x]))

________________________________________________________________________________________

fricas [B]  time = 0.49, size = 1633, normalized size = 8.37 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^3/(a*cosh(x)+b*sinh(x))^2,x, algorithm="fricas")

[Out]

[1/2*(a^5 + a^4*b - 2*a^3*b^2 - 2*a^2*b^3 + a*b^4 + b^5 + (a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*
cosh(x)^4 + 4*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)*sinh(x)^3 + (a^5 - a^4*b - 2*a^3*b^2
+ 2*a^2*b^3 + a*b^4 - b^5)*sinh(x)^4 + 6*(a^5 - a*b^4)*cosh(x)^2 + 6*(a^5 - a*b^4 + (a^5 - a^4*b - 2*a^3*b^2
+ 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^2)*sinh(x)^2 - 6*((a^3*b + a^2*b^2)*cosh(x)^3 + 3*(a^3*b + a^2*b^2)*cosh(x)
*sinh(x)^2 + (a^3*b + a^2*b^2)*sinh(x)^3 + (a^3*b - a^2*b^2)*cosh(x) + (a^3*b - a^2*b^2 + 3*(a^3*b + a^2*b^2)*
cosh(x)^2)*sinh(x))*sqrt(-a^2 + b^2)*log(((a + b)*cosh(x)^2 + 2*(a + b)*cosh(x)*sinh(x) + (a + b)*sinh(x)^2 -
2*sqrt(-a^2 + b^2)*(cosh(x) + sinh(x)) - a + b)/((a + b)*cosh(x)^2 + 2*(a + b)*cosh(x)*sinh(x) + (a + b)*sinh(
x)^2 + a - b)) + 4*((a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^3 + 3*(a^5 - a*b^4)*cosh(x))*s
inh(x))/((a^7 + a^6*b - 3*a^5*b^2 - 3*a^4*b^3 + 3*a^3*b^4 + 3*a^2*b^5 - a*b^6 - b^7)*cosh(x)^3 + 3*(a^7 + a^6*
b - 3*a^5*b^2 - 3*a^4*b^3 + 3*a^3*b^4 + 3*a^2*b^5 - a*b^6 - b^7)*cosh(x)*sinh(x)^2 + (a^7 + a^6*b - 3*a^5*b^2
- 3*a^4*b^3 + 3*a^3*b^4 + 3*a^2*b^5 - a*b^6 - b^7)*sinh(x)^3 + (a^7 - a^6*b - 3*a^5*b^2 + 3*a^4*b^3 + 3*a^3*b^
4 - 3*a^2*b^5 - a*b^6 + b^7)*cosh(x) + (a^7 - a^6*b - 3*a^5*b^2 + 3*a^4*b^3 + 3*a^3*b^4 - 3*a^2*b^5 - a*b^6 +
b^7 + 3*(a^7 + a^6*b - 3*a^5*b^2 - 3*a^4*b^3 + 3*a^3*b^4 + 3*a^2*b^5 - a*b^6 - b^7)*cosh(x)^2)*sinh(x)), 1/2*(
a^5 + a^4*b - 2*a^3*b^2 - 2*a^2*b^3 + a*b^4 + b^5 + (a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x
)^4 + 4*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)*sinh(x)^3 + (a^5 - a^4*b - 2*a^3*b^2 + 2*a
^2*b^3 + a*b^4 - b^5)*sinh(x)^4 + 6*(a^5 - a*b^4)*cosh(x)^2 + 6*(a^5 - a*b^4 + (a^5 - a^4*b - 2*a^3*b^2 + 2*a^
2*b^3 + a*b^4 - b^5)*cosh(x)^2)*sinh(x)^2 - 12*((a^3*b + a^2*b^2)*cosh(x)^3 + 3*(a^3*b + a^2*b^2)*cosh(x)*sinh
(x)^2 + (a^3*b + a^2*b^2)*sinh(x)^3 + (a^3*b - a^2*b^2)*cosh(x) + (a^3*b - a^2*b^2 + 3*(a^3*b + a^2*b^2)*cosh(
x)^2)*sinh(x))*sqrt(a^2 - b^2)*arctan(sqrt(a^2 - b^2)/((a + b)*cosh(x) + (a + b)*sinh(x))) + 4*((a^5 - a^4*b -
2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^3 + 3*(a^5 - a*b^4)*cosh(x))*sinh(x))/((a^7 + a^6*b - 3*a^5*b^2
- 3*a^4*b^3 + 3*a^3*b^4 + 3*a^2*b^5 - a*b^6 - b^7)*cosh(x)^3 + 3*(a^7 + a^6*b - 3*a^5*b^2 - 3*a^4*b^3 + 3*a^3*
b^4 + 3*a^2*b^5 - a*b^6 - b^7)*cosh(x)*sinh(x)^2 + (a^7 + a^6*b - 3*a^5*b^2 - 3*a^4*b^3 + 3*a^3*b^4 + 3*a^2*b^
5 - a*b^6 - b^7)*sinh(x)^3 + (a^7 - a^6*b - 3*a^5*b^2 + 3*a^4*b^3 + 3*a^3*b^4 - 3*a^2*b^5 - a*b^6 + b^7)*cosh(
x) + (a^7 - a^6*b - 3*a^5*b^2 + 3*a^4*b^3 + 3*a^3*b^4 - 3*a^2*b^5 - a*b^6 + b^7 + 3*(a^7 + a^6*b - 3*a^5*b^2 -
3*a^4*b^3 + 3*a^3*b^4 + 3*a^2*b^5 - a*b^6 - b^7)*cosh(x)^2)*sinh(x))]

________________________________________________________________________________________

giac [A]  time = 0.13, size = 174, normalized size = 0.89 \[ \frac {6 \, a^{2} b \arctan \left (\frac {a e^{x} + b e^{x}}{\sqrt {a^{2} - b^{2}}}\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} - b^{2}}} + \frac {e^{x}}{2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )}} + \frac {5 \, a^{3} e^{\left (2 \, x\right )} + 3 \, a^{2} b e^{\left (2 \, x\right )} + 3 \, a b^{2} e^{\left (2 \, x\right )} + b^{3} e^{\left (2 \, x\right )} + a^{3} + a^{2} b - a b^{2} - b^{3}}{2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (a e^{\left (3 \, x\right )} + b e^{\left (3 \, x\right )} + a e^{x} - b e^{x}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^3/(a*cosh(x)+b*sinh(x))^2,x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.28, size = 164, normalized size = 0.84 \[ -\frac {1}{\left (a +b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {2 a^{2} \tanh \left (\frac {x}{2}\right ) b}{\left (a -b \right )^{2} \left (a +b \right )^{2} \left (a +2 \tanh \left (\frac {x}{2}\right ) b +a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )\right )}+\frac {2 a^{3}}{\left (a -b \right )^{2} \left (a +b \right )^{2} \left (a +2 \tanh \left (\frac {x}{2}\right ) b +a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )\right )}+\frac {6 a^{2} b \arctan \left (\frac {2 a \tanh \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2} \sqrt {a^{2}-b^{2}}}+\frac {1}{\left (a -b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)^3/(a*cosh(x)+b*sinh(x))^2,x)

[Out]

-1/(a+b)^2/(tanh(1/2*x)-1)+2*a^2/(a-b)^2/(a+b)^2/(a+2*tanh(1/2*x)*b+a*tanh(1/2*x)^2)*tanh(1/2*x)*b+2*a^3/(a-b)
^2/(a+b)^2/(a+2*tanh(1/2*x)*b+a*tanh(1/2*x)^2)+6*a^2/(a-b)^2/(a+b)^2*b/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tanh(1/
2*x)+2*b)/(a^2-b^2)^(1/2))+1/(a-b)^2/(tanh(1/2*x)+1)

________________________________________________________________________________________

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)^3/(a*cosh(x)+b*sinh(x))^2,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?

________________________________________________________________________________________

mupad [B]  time = 1.79, size = 255, normalized size = 1.31 \[ \frac {{\mathrm {e}}^{-x}}{2\,{\left (a-b\right )}^2}+\frac {{\mathrm {e}}^x}{2\,{\left (a+b\right )}^2}+\frac {6\,\mathrm {atan}\left (\frac {a^2\,b\,{\mathrm {e}}^x\,\sqrt {a^{10}-5\,a^8\,b^2+10\,a^6\,b^4-10\,a^4\,b^6+5\,a^2\,b^8-b^{10}}}{a^5\,\sqrt {a^4\,b^2}-b^5\,\sqrt {a^4\,b^2}+2\,a^2\,b^3\,\sqrt {a^4\,b^2}-2\,a^3\,b^2\,\sqrt {a^4\,b^2}+a\,b^4\,\sqrt {a^4\,b^2}-a^4\,b\,\sqrt {a^4\,b^2}}\right )\,\sqrt {a^4\,b^2}}{\sqrt {a^{10}-5\,a^8\,b^2+10\,a^6\,b^4-10\,a^4\,b^6+5\,a^2\,b^8-b^{10}}}+\frac {2\,a^3\,{\mathrm {e}}^x}{{\left (a+b\right )}^2\,{\left (a-b\right )}^2\,\left (a-b+{\mathrm {e}}^{2\,x}\,\left (a+b\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)^3/(a*cosh(x) + b*sinh(x))^2,x)

[Out]

exp(-x)/(2*(a - b)^2) + exp(x)/(2*(a + b)^2) + (6*atan((a^2*b*exp(x)*(a^10 - b^10 + 5*a^2*b^8 - 10*a^4*b^6 + 1
0*a^6*b^4 - 5*a^8*b^2)^(1/2))/(a^5*(a^4*b^2)^(1/2) - b^5*(a^4*b^2)^(1/2) + 2*a^2*b^3*(a^4*b^2)^(1/2) - 2*a^3*b
^2*(a^4*b^2)^(1/2) + a*b^4*(a^4*b^2)^(1/2) - a^4*b*(a^4*b^2)^(1/2)))*(a^4*b^2)^(1/2))/(a^10 - b^10 + 5*a^2*b^8
- 10*a^4*b^6 + 10*a^6*b^4 - 5*a^8*b^2)^(1/2) + (2*a^3*exp(x))/((a + b)^2*(a - b)^2*(a - b + exp(2*x)*(a + b))
)

________________________________________________________________________________________

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(sinh(x)**3/(a*cosh(x)+b*sinh(x))**2,x)

[Out]

Timed out

________________________________________________________________________________________