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

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

[Out]

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

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Rubi [A]  time = 1.22499, 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 4401

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

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 6742

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

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 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 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 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 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 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{\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 = 0.4389, size = 205, normalized size = 1.05 $\frac{a \sqrt{a-b} \left (a^2 b+a^3+a b^2+b^3\right ) \cosh ^2(x)+a \left (a^2 \sqrt{a-b} (a+b)-2 b^2 \sqrt{a-b} (a+b) \sinh ^2(x)+6 a b^2 \sqrt{a+b} \sinh (x) \tan ^{-1}\left (\frac{a \tanh \left (\frac{x}{2}\right )+b}{\sqrt{a-b} \sqrt{a+b}}\right )\right )-b \cosh (x) \left ((a-b)^{3/2} (a+b)^2 \sinh (x)-6 a^3 \sqrt{a+b} \tan ^{-1}\left (\frac{a \tanh \left (\frac{x}{2}\right )+b}{\sqrt{a-b} \sqrt{a+b}}\right )\right )}{(a-b)^{5/2} (a+b)^3 (a \cosh (x)+b \sinh (x))}$

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]))

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Maple [A]  time = 0.072, size = 164, normalized size = 0.8 \begin{align*}{\frac{1}{ \left ( a-b \right ) ^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}+2\,{\frac{{a}^{2}\tanh \left ( x/2 \right ) b}{ \left ( a-b \right ) ^{2} \left ( a+b \right ) ^{2} \left ( a+2\,\tanh \left ( x/2 \right ) b+a \left ( \tanh \left ( x/2 \right ) \right ) ^{2} \right ) }}+2\,{\frac{{a}^{3}}{ \left ( a-b \right ) ^{2} \left ( a+b \right ) ^{2} \left ( a+2\,\tanh \left ( x/2 \right ) b+a \left ( \tanh \left ( x/2 \right ) \right ) ^{2} \right ) }}+6\,{\frac{{a}^{2}b}{ \left ( a-b \right ) ^{2} \left ( a+b \right ) ^{2}\sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tanh \left ( x/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }-{\frac{1}{ \left ( a+b \right ) ^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}} \end{align*}

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)

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.0956, size = 3664, normalized size = 18.79 \begin{align*} \text{result too large to display} \end{align*}

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))]

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

[Out]

Timed out

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Giac [A]  time = 1.15241, size = 235, normalized size = 1.21 \begin{align*} \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 )}} \end{align*}

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))