3.827 $$\int \frac{\sinh (x)}{a+b \sinh (x)+c \sinh ^2(x)} \, dx$$

Optimal. Leaf size=280 $\frac{\sqrt{2} \left (\frac{b}{\sqrt{4 a c-b^2}}+i\right ) \tan ^{-1}\left (\frac{\tanh \left (\frac{x}{2}\right ) \sqrt{4 a c-b^2}-i b \tanh \left (\frac{x}{2}\right )+2 i c}{\sqrt{2} \sqrt{i b \sqrt{4 a c-b^2}-2 c (a-c)+b^2}}\right )}{\sqrt{i b \sqrt{4 a c-b^2}-2 c (a-c)+b^2}}+\frac{\sqrt{2} \left (-\frac{b}{\sqrt{4 a c-b^2}}+i\right ) \tan ^{-1}\left (\frac{2 i c-\tanh \left (\frac{x}{2}\right ) \left (\sqrt{4 a c-b^2}+i b\right )}{\sqrt{2} \sqrt{-i b \sqrt{4 a c-b^2}-2 c (a-c)+b^2}}\right )}{\sqrt{-i b \sqrt{4 a c-b^2}-2 c (a-c)+b^2}}$

[Out]

(Sqrt[2]*(I + b/Sqrt[-b^2 + 4*a*c])*ArcTan[((2*I)*c - I*b*Tanh[x/2] + Sqrt[-b^2 + 4*a*c]*Tanh[x/2])/(Sqrt[2]*S
qrt[b^2 - 2*(a - c)*c + I*b*Sqrt[-b^2 + 4*a*c]])])/Sqrt[b^2 - 2*(a - c)*c + I*b*Sqrt[-b^2 + 4*a*c]] + (Sqrt[2]
*(I - b/Sqrt[-b^2 + 4*a*c])*ArcTan[((2*I)*c - (I*b + Sqrt[-b^2 + 4*a*c])*Tanh[x/2])/(Sqrt[2]*Sqrt[b^2 - 2*(a -
c)*c - I*b*Sqrt[-b^2 + 4*a*c]])])/Sqrt[b^2 - 2*(a - c)*c - I*b*Sqrt[-b^2 + 4*a*c]]

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Rubi [A]  time = 0.723667, antiderivative size = 280, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 4, integrand size = 17, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.235, Rules used = {3256, 2660, 618, 204} $\frac{\sqrt{2} \left (\frac{b}{\sqrt{4 a c-b^2}}+i\right ) \tan ^{-1}\left (\frac{\tanh \left (\frac{x}{2}\right ) \sqrt{4 a c-b^2}-i b \tanh \left (\frac{x}{2}\right )+2 i c}{\sqrt{2} \sqrt{i b \sqrt{4 a c-b^2}-2 c (a-c)+b^2}}\right )}{\sqrt{i b \sqrt{4 a c-b^2}-2 c (a-c)+b^2}}+\frac{\sqrt{2} \left (-\frac{b}{\sqrt{4 a c-b^2}}+i\right ) \tan ^{-1}\left (\frac{2 i c-\tanh \left (\frac{x}{2}\right ) \left (\sqrt{4 a c-b^2}+i b\right )}{\sqrt{2} \sqrt{-i b \sqrt{4 a c-b^2}-2 c (a-c)+b^2}}\right )}{\sqrt{-i b \sqrt{4 a c-b^2}-2 c (a-c)+b^2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(Sqrt[2]*(I + b/Sqrt[-b^2 + 4*a*c])*ArcTan[((2*I)*c - I*b*Tanh[x/2] + Sqrt[-b^2 + 4*a*c]*Tanh[x/2])/(Sqrt[2]*S
qrt[b^2 - 2*(a - c)*c + I*b*Sqrt[-b^2 + 4*a*c]])])/Sqrt[b^2 - 2*(a - c)*c + I*b*Sqrt[-b^2 + 4*a*c]] + (Sqrt[2]
*(I - b/Sqrt[-b^2 + 4*a*c])*ArcTan[((2*I)*c - (I*b + Sqrt[-b^2 + 4*a*c])*Tanh[x/2])/(Sqrt[2]*Sqrt[b^2 - 2*(a -
c)*c - I*b*Sqrt[-b^2 + 4*a*c]])])/Sqrt[b^2 - 2*(a - c)*c - I*b*Sqrt[-b^2 + 4*a*c]]

Rule 3256

Int[sin[(d_.) + (e_.)*(x_)]^(m_.)*((a_.) + (b_.)*sin[(d_.) + (e_.)*(x_)]^(n_.) + (c_.)*sin[(d_.) + (e_.)*(x_)]
^(n2_.))^(p_), x_Symbol] :> Int[ExpandTrig[sin[d + e*x]^m*(a + b*sin[d + e*x]^n + c*sin[d + e*x]^(2*n))^p, x],
x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IntegersQ[m, n, p]

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 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])

Rubi steps

\begin{align*} \int \frac{\sinh (x)}{a+b \sinh (x)+c \sinh ^2(x)} \, dx &=-\left (i \int \left (\frac{1+\frac{i b}{\sqrt{-b^2+4 a c}}}{-i b-\sqrt{-b^2+4 a c}-2 i c \sinh (x)}+\frac{1-\frac{i b}{\sqrt{-b^2+4 a c}}}{-i b+\sqrt{-b^2+4 a c}-2 i c \sinh (x)}\right ) \, dx\right )\\ &=-\left (\left (i-\frac{b}{\sqrt{-b^2+4 a c}}\right ) \int \frac{1}{-i b-\sqrt{-b^2+4 a c}-2 i c \sinh (x)} \, dx\right )-\left (i+\frac{b}{\sqrt{-b^2+4 a c}}\right ) \int \frac{1}{-i b+\sqrt{-b^2+4 a c}-2 i c \sinh (x)} \, dx\\ &=-\left (\left (2 \left (i-\frac{b}{\sqrt{-b^2+4 a c}}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-i b-\sqrt{-b^2+4 a c}-4 i c x-\left (-i b-\sqrt{-b^2+4 a c}\right ) x^2} \, dx,x,\tanh \left (\frac{x}{2}\right )\right )\right )-\left (2 \left (i+\frac{b}{\sqrt{-b^2+4 a c}}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-i b+\sqrt{-b^2+4 a c}-4 i c x-\left (-i b+\sqrt{-b^2+4 a c}\right ) x^2} \, dx,x,\tanh \left (\frac{x}{2}\right )\right )\\ &=\left (4 \left (i-\frac{b}{\sqrt{-b^2+4 a c}}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-8 \left (b^2-2 (a-c) c-i b \sqrt{-b^2+4 a c}\right )-x^2} \, dx,x,-4 i c+2 \left (i b+\sqrt{-b^2+4 a c}\right ) \tanh \left (\frac{x}{2}\right )\right )+\left (4 \left (i+\frac{b}{\sqrt{-b^2+4 a c}}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-8 \left (b^2-2 (a-c) c+i b \sqrt{-b^2+4 a c}\right )-x^2} \, dx,x,-4 i c+2 \left (i b-\sqrt{-b^2+4 a c}\right ) \tanh \left (\frac{x}{2}\right )\right )\\ &=\frac{\sqrt{2} \left (i+\frac{b}{\sqrt{-b^2+4 a c}}\right ) \tan ^{-1}\left (\frac{2 i c-\left (i b-\sqrt{-b^2+4 a c}\right ) \tanh \left (\frac{x}{2}\right )}{\sqrt{2} \sqrt{b^2-2 (a-c) c+i b \sqrt{-b^2+4 a c}}}\right )}{\sqrt{b^2-2 (a-c) c+i b \sqrt{-b^2+4 a c}}}+\frac{\sqrt{2} \left (i-\frac{b}{\sqrt{-b^2+4 a c}}\right ) \tan ^{-1}\left (\frac{2 i c-\left (i b+\sqrt{-b^2+4 a c}\right ) \tanh \left (\frac{x}{2}\right )}{\sqrt{2} \sqrt{b^2-2 (a-c) c-i b \sqrt{-b^2+4 a c}}}\right )}{\sqrt{b^2-2 (a-c) c-i b \sqrt{-b^2+4 a c}}}\\ \end{align*}

Mathematica [A]  time = 0.480967, size = 244, normalized size = 0.87 $\frac{\sqrt{2} \left (\frac{\left (\sqrt{b^2-4 a c}-b\right ) \tan ^{-1}\left (\frac{\tanh \left (\frac{x}{2}\right ) \left (\sqrt{b^2-4 a c}-b\right )+2 c}{\sqrt{2 b \sqrt{b^2-4 a c}+4 c (a-c)-2 b^2}}\right )}{\sqrt{b \sqrt{b^2-4 a c}+2 c (a-c)-b^2}}+\frac{\left (\sqrt{b^2-4 a c}+b\right ) \tan ^{-1}\left (\frac{2 c-\tanh \left (\frac{x}{2}\right ) \left (\sqrt{b^2-4 a c}+b\right )}{\sqrt{2} \sqrt{-b \sqrt{b^2-4 a c}+2 c (a-c)-b^2}}\right )}{\sqrt{-b \sqrt{b^2-4 a c}+2 c (a-c)-b^2}}\right )}{\sqrt{b^2-4 a c}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(Sqrt[2]*(((-b + Sqrt[b^2 - 4*a*c])*ArcTan[(2*c + (-b + Sqrt[b^2 - 4*a*c])*Tanh[x/2])/Sqrt[-2*b^2 + 4*(a - c)*
c + 2*b*Sqrt[b^2 - 4*a*c]]])/Sqrt[-b^2 + 2*(a - c)*c + b*Sqrt[b^2 - 4*a*c]] + ((b + Sqrt[b^2 - 4*a*c])*ArcTan[
(2*c - (b + Sqrt[b^2 - 4*a*c])*Tanh[x/2])/(Sqrt[2]*Sqrt[-b^2 + 2*(a - c)*c - b*Sqrt[b^2 - 4*a*c]])])/Sqrt[-b^2
+ 2*(a - c)*c - b*Sqrt[b^2 - 4*a*c]]))/Sqrt[b^2 - 4*a*c]

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Maple [C]  time = 0.035, size = 70, normalized size = 0.3 \begin{align*} 2\,\sum _{{\it \_R}={\it RootOf} \left ( a{{\it \_Z}}^{4}-2\,b{{\it \_Z}}^{3}+ \left ( -2\,a+4\,c \right ){{\it \_Z}}^{2}+2\,b{\it \_Z}+a \right ) }{\frac{{\it \_R}\,\ln \left ( \tanh \left ( x/2 \right ) -{\it \_R} \right ) }{2\,{{\it \_R}}^{3}a-3\,b{{\it \_R}}^{2}-2\,{\it \_R}\,a+4\,c{\it \_R}+b}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sum(_R/(2*_R^3*a-3*_R^2*b-2*_R*a+4*_R*c+b)*ln(tanh(1/2*x)-_R),_R=RootOf(a*_Z^4-2*b*_Z^3+(-2*a+4*c)*_Z^2+2*b*
_Z+a))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sinh \left (x\right )}{c \sinh \left (x\right )^{2} + b \sinh \left (x\right ) + a}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sinh(x)/(c*sinh(x)^2 + b*sinh(x) + a), x)

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Fricas [B]  time = 4.15167, size = 7015, normalized size = 25.05 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(2)*sqrt((2*a^2 + b^2 - 2*a*c + (a^2*b^2 + b^4 - 4*a*c^3 + (8*a^2 + b^2)*c^2 - 2*(2*a^3 + 3*a*b^2)*c)
*sqrt(b^2/(a^4*b^2 + 2*a^2*b^4 + b^6 - 4*a*c^5 + (16*a^2 + b^2)*c^4 - 12*(2*a^3 + a*b^2)*c^3 + 2*(8*a^4 + 11*a
^2*b^2 + b^4)*c^2 - 4*(a^5 + 3*a^3*b^2 + 2*a*b^4)*c)))/(a^2*b^2 + b^4 - 4*a*c^3 + (8*a^2 + b^2)*c^2 - 2*(2*a^3
+ 3*a*b^2)*c))*log(4*a*b*c*cosh(x) + 4*a*b*c*sinh(x) + 2*a*b^2 + sqrt(2)*(a*b^3 + 4*a*b*c^2 - (4*a^2*b + b^3)
*c + (a^3*b^3 + a*b^5 - 4*a*b*c^4 + (4*a^2*b + b^3)*c^3 + (4*a^3*b - 5*a*b^3)*c^2 - (4*a^4*b + 5*a^2*b^3 - b^5
)*c)*sqrt(b^2/(a^4*b^2 + 2*a^2*b^4 + b^6 - 4*a*c^5 + (16*a^2 + b^2)*c^4 - 12*(2*a^3 + a*b^2)*c^3 + 2*(8*a^4 +
11*a^2*b^2 + b^4)*c^2 - 4*(a^5 + 3*a^3*b^2 + 2*a*b^4)*c)))*sqrt((2*a^2 + b^2 - 2*a*c + (a^2*b^2 + b^4 - 4*a*c^
3 + (8*a^2 + b^2)*c^2 - 2*(2*a^3 + 3*a*b^2)*c)*sqrt(b^2/(a^4*b^2 + 2*a^2*b^4 + b^6 - 4*a*c^5 + (16*a^2 + b^2)*
c^4 - 12*(2*a^3 + a*b^2)*c^3 + 2*(8*a^4 + 11*a^2*b^2 + b^4)*c^2 - 4*(a^5 + 3*a^3*b^2 + 2*a*b^4)*c)))/(a^2*b^2
+ b^4 - 4*a*c^3 + (8*a^2 + b^2)*c^2 - 2*(2*a^3 + 3*a*b^2)*c)) + 2*(a^3*b^2 + a*b^4 - 4*a^2*c^3 + (8*a^3 + a*b^
2)*c^2 - 2*(2*a^4 + 3*a^2*b^2)*c)*sqrt(b^2/(a^4*b^2 + 2*a^2*b^4 + b^6 - 4*a*c^5 + (16*a^2 + b^2)*c^4 - 12*(2*a
^3 + a*b^2)*c^3 + 2*(8*a^4 + 11*a^2*b^2 + b^4)*c^2 - 4*(a^5 + 3*a^3*b^2 + 2*a*b^4)*c))) + 1/2*sqrt(2)*sqrt((2*
a^2 + b^2 - 2*a*c + (a^2*b^2 + b^4 - 4*a*c^3 + (8*a^2 + b^2)*c^2 - 2*(2*a^3 + 3*a*b^2)*c)*sqrt(b^2/(a^4*b^2 +
2*a^2*b^4 + b^6 - 4*a*c^5 + (16*a^2 + b^2)*c^4 - 12*(2*a^3 + a*b^2)*c^3 + 2*(8*a^4 + 11*a^2*b^2 + b^4)*c^2 - 4
*(a^5 + 3*a^3*b^2 + 2*a*b^4)*c)))/(a^2*b^2 + b^4 - 4*a*c^3 + (8*a^2 + b^2)*c^2 - 2*(2*a^3 + 3*a*b^2)*c))*log(4
*a*b*c*cosh(x) + 4*a*b*c*sinh(x) + 2*a*b^2 - sqrt(2)*(a*b^3 + 4*a*b*c^2 - (4*a^2*b + b^3)*c + (a^3*b^3 + a*b^5
- 4*a*b*c^4 + (4*a^2*b + b^3)*c^3 + (4*a^3*b - 5*a*b^3)*c^2 - (4*a^4*b + 5*a^2*b^3 - b^5)*c)*sqrt(b^2/(a^4*b^
2 + 2*a^2*b^4 + b^6 - 4*a*c^5 + (16*a^2 + b^2)*c^4 - 12*(2*a^3 + a*b^2)*c^3 + 2*(8*a^4 + 11*a^2*b^2 + b^4)*c^2
- 4*(a^5 + 3*a^3*b^2 + 2*a*b^4)*c)))*sqrt((2*a^2 + b^2 - 2*a*c + (a^2*b^2 + b^4 - 4*a*c^3 + (8*a^2 + b^2)*c^2
- 2*(2*a^3 + 3*a*b^2)*c)*sqrt(b^2/(a^4*b^2 + 2*a^2*b^4 + b^6 - 4*a*c^5 + (16*a^2 + b^2)*c^4 - 12*(2*a^3 + a*b
^2)*c^3 + 2*(8*a^4 + 11*a^2*b^2 + b^4)*c^2 - 4*(a^5 + 3*a^3*b^2 + 2*a*b^4)*c)))/(a^2*b^2 + b^4 - 4*a*c^3 + (8*
a^2 + b^2)*c^2 - 2*(2*a^3 + 3*a*b^2)*c)) + 2*(a^3*b^2 + a*b^4 - 4*a^2*c^3 + (8*a^3 + a*b^2)*c^2 - 2*(2*a^4 + 3
*a^2*b^2)*c)*sqrt(b^2/(a^4*b^2 + 2*a^2*b^4 + b^6 - 4*a*c^5 + (16*a^2 + b^2)*c^4 - 12*(2*a^3 + a*b^2)*c^3 + 2*(
8*a^4 + 11*a^2*b^2 + b^4)*c^2 - 4*(a^5 + 3*a^3*b^2 + 2*a*b^4)*c))) - 1/2*sqrt(2)*sqrt((2*a^2 + b^2 - 2*a*c - (
a^2*b^2 + b^4 - 4*a*c^3 + (8*a^2 + b^2)*c^2 - 2*(2*a^3 + 3*a*b^2)*c)*sqrt(b^2/(a^4*b^2 + 2*a^2*b^4 + b^6 - 4*a
*c^5 + (16*a^2 + b^2)*c^4 - 12*(2*a^3 + a*b^2)*c^3 + 2*(8*a^4 + 11*a^2*b^2 + b^4)*c^2 - 4*(a^5 + 3*a^3*b^2 + 2
*a*b^4)*c)))/(a^2*b^2 + b^4 - 4*a*c^3 + (8*a^2 + b^2)*c^2 - 2*(2*a^3 + 3*a*b^2)*c))*log(4*a*b*c*cosh(x) + 4*a*
b*c*sinh(x) + 2*a*b^2 + sqrt(2)*(a*b^3 + 4*a*b*c^2 - (4*a^2*b + b^3)*c - (a^3*b^3 + a*b^5 - 4*a*b*c^4 + (4*a^2
*b + b^3)*c^3 + (4*a^3*b - 5*a*b^3)*c^2 - (4*a^4*b + 5*a^2*b^3 - b^5)*c)*sqrt(b^2/(a^4*b^2 + 2*a^2*b^4 + b^6 -
4*a*c^5 + (16*a^2 + b^2)*c^4 - 12*(2*a^3 + a*b^2)*c^3 + 2*(8*a^4 + 11*a^2*b^2 + b^4)*c^2 - 4*(a^5 + 3*a^3*b^2
+ 2*a*b^4)*c)))*sqrt((2*a^2 + b^2 - 2*a*c - (a^2*b^2 + b^4 - 4*a*c^3 + (8*a^2 + b^2)*c^2 - 2*(2*a^3 + 3*a*b^2
)*c)*sqrt(b^2/(a^4*b^2 + 2*a^2*b^4 + b^6 - 4*a*c^5 + (16*a^2 + b^2)*c^4 - 12*(2*a^3 + a*b^2)*c^3 + 2*(8*a^4 +
11*a^2*b^2 + b^4)*c^2 - 4*(a^5 + 3*a^3*b^2 + 2*a*b^4)*c)))/(a^2*b^2 + b^4 - 4*a*c^3 + (8*a^2 + b^2)*c^2 - 2*(2
*a^3 + 3*a*b^2)*c)) - 2*(a^3*b^2 + a*b^4 - 4*a^2*c^3 + (8*a^3 + a*b^2)*c^2 - 2*(2*a^4 + 3*a^2*b^2)*c)*sqrt(b^2
/(a^4*b^2 + 2*a^2*b^4 + b^6 - 4*a*c^5 + (16*a^2 + b^2)*c^4 - 12*(2*a^3 + a*b^2)*c^3 + 2*(8*a^4 + 11*a^2*b^2 +
b^4)*c^2 - 4*(a^5 + 3*a^3*b^2 + 2*a*b^4)*c))) + 1/2*sqrt(2)*sqrt((2*a^2 + b^2 - 2*a*c - (a^2*b^2 + b^4 - 4*a*c
^3 + (8*a^2 + b^2)*c^2 - 2*(2*a^3 + 3*a*b^2)*c)*sqrt(b^2/(a^4*b^2 + 2*a^2*b^4 + b^6 - 4*a*c^5 + (16*a^2 + b^2)
*c^4 - 12*(2*a^3 + a*b^2)*c^3 + 2*(8*a^4 + 11*a^2*b^2 + b^4)*c^2 - 4*(a^5 + 3*a^3*b^2 + 2*a*b^4)*c)))/(a^2*b^2
+ b^4 - 4*a*c^3 + (8*a^2 + b^2)*c^2 - 2*(2*a^3 + 3*a*b^2)*c))*log(4*a*b*c*cosh(x) + 4*a*b*c*sinh(x) + 2*a*b^2
- sqrt(2)*(a*b^3 + 4*a*b*c^2 - (4*a^2*b + b^3)*c - (a^3*b^3 + a*b^5 - 4*a*b*c^4 + (4*a^2*b + b^3)*c^3 + (4*a^
3*b - 5*a*b^3)*c^2 - (4*a^4*b + 5*a^2*b^3 - b^5)*c)*sqrt(b^2/(a^4*b^2 + 2*a^2*b^4 + b^6 - 4*a*c^5 + (16*a^2 +
b^2)*c^4 - 12*(2*a^3 + a*b^2)*c^3 + 2*(8*a^4 + 11*a^2*b^2 + b^4)*c^2 - 4*(a^5 + 3*a^3*b^2 + 2*a*b^4)*c)))*sqrt
((2*a^2 + b^2 - 2*a*c - (a^2*b^2 + b^4 - 4*a*c^3 + (8*a^2 + b^2)*c^2 - 2*(2*a^3 + 3*a*b^2)*c)*sqrt(b^2/(a^4*b^
2 + 2*a^2*b^4 + b^6 - 4*a*c^5 + (16*a^2 + b^2)*c^4 - 12*(2*a^3 + a*b^2)*c^3 + 2*(8*a^4 + 11*a^2*b^2 + b^4)*c^2
- 4*(a^5 + 3*a^3*b^2 + 2*a*b^4)*c)))/(a^2*b^2 + b^4 - 4*a*c^3 + (8*a^2 + b^2)*c^2 - 2*(2*a^3 + 3*a*b^2)*c)) -
2*(a^3*b^2 + a*b^4 - 4*a^2*c^3 + (8*a^3 + a*b^2)*c^2 - 2*(2*a^4 + 3*a^2*b^2)*c)*sqrt(b^2/(a^4*b^2 + 2*a^2*b^4
+ b^6 - 4*a*c^5 + (16*a^2 + b^2)*c^4 - 12*(2*a^3 + a*b^2)*c^3 + 2*(8*a^4 + 11*a^2*b^2 + b^4)*c^2 - 4*(a^5 + 3
*a^3*b^2 + 2*a*b^4)*c)))

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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)/(a+b*sinh(x)+c*sinh(x)**2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sinh \left (x\right )}{c \sinh \left (x\right )^{2} + b \sinh \left (x\right ) + a}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sinh(x)/(c*sinh(x)^2 + b*sinh(x) + a), x)