3.175 \(\int \frac{\sinh ^2(c+d x)}{a+b \sinh ^3(c+d x)} \, dx\)

Optimal. Leaf size=262 \[ -\frac{2 \tanh ^{-1}\left (\frac{\sqrt [3]{b}-\sqrt [3]{a} \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^{2/3}+b^{2/3}}}\right )}{3 b^{2/3} d \sqrt{a^{2/3}+b^{2/3}}}-\frac{2 \tan ^{-1}\left (\frac{(-1)^{5/6} \left (\sqrt [6]{-1} \sqrt [3]{b}+i \sqrt [3]{a} \tanh \left (\frac{1}{2} (c+d x)\right )\right )}{\sqrt{-(-1)^{2/3} a^{2/3}-b^{2/3}}}\right )}{3 b^{2/3} d \sqrt{-(-1)^{2/3} a^{2/3}-b^{2/3}}}-\frac{2 \tan ^{-1}\left (\frac{\sqrt [6]{-1} \left ((-1)^{5/6} \sqrt [3]{b}+i \sqrt [3]{a} \tanh \left (\frac{1}{2} (c+d x)\right )\right )}{\sqrt{\sqrt [3]{-1} a^{2/3}-b^{2/3}}}\right )}{3 b^{2/3} d \sqrt{\sqrt [3]{-1} a^{2/3}-b^{2/3}}} \]

[Out]

(-2*ArcTan[((-1)^(5/6)*((-1)^(1/6)*b^(1/3) + I*a^(1/3)*Tanh[(c + d*x)/2]))/Sqrt[-((-1)^(2/3)*a^(2/3)) - b^(2/3
)]])/(3*Sqrt[-((-1)^(2/3)*a^(2/3)) - b^(2/3)]*b^(2/3)*d) - (2*ArcTan[((-1)^(1/6)*((-1)^(5/6)*b^(1/3) + I*a^(1/
3)*Tanh[(c + d*x)/2]))/Sqrt[(-1)^(1/3)*a^(2/3) - b^(2/3)]])/(3*Sqrt[(-1)^(1/3)*a^(2/3) - b^(2/3)]*b^(2/3)*d) -
(2*ArcTanh[(b^(1/3) - a^(1/3)*Tanh[(c + d*x)/2])/Sqrt[a^(2/3) + b^(2/3)]])/(3*Sqrt[a^(2/3) + b^(2/3)]*b^(2/3)
*d)

________________________________________________________________________________________

Rubi [A]  time = 0.270745, antiderivative size = 262, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {3220, 2660, 618, 206, 204} \[ -\frac{2 \tanh ^{-1}\left (\frac{\sqrt [3]{b}-\sqrt [3]{a} \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^{2/3}+b^{2/3}}}\right )}{3 b^{2/3} d \sqrt{a^{2/3}+b^{2/3}}}-\frac{2 \tan ^{-1}\left (\frac{(-1)^{5/6} \left (\sqrt [6]{-1} \sqrt [3]{b}+i \sqrt [3]{a} \tanh \left (\frac{1}{2} (c+d x)\right )\right )}{\sqrt{-(-1)^{2/3} a^{2/3}-b^{2/3}}}\right )}{3 b^{2/3} d \sqrt{-(-1)^{2/3} a^{2/3}-b^{2/3}}}-\frac{2 \tan ^{-1}\left (\frac{\sqrt [6]{-1} \left ((-1)^{5/6} \sqrt [3]{b}+i \sqrt [3]{a} \tanh \left (\frac{1}{2} (c+d x)\right )\right )}{\sqrt{\sqrt [3]{-1} a^{2/3}-b^{2/3}}}\right )}{3 b^{2/3} d \sqrt{\sqrt [3]{-1} a^{2/3}-b^{2/3}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*ArcTan[((-1)^(5/6)*((-1)^(1/6)*b^(1/3) + I*a^(1/3)*Tanh[(c + d*x)/2]))/Sqrt[-((-1)^(2/3)*a^(2/3)) - b^(2/3
)]])/(3*Sqrt[-((-1)^(2/3)*a^(2/3)) - b^(2/3)]*b^(2/3)*d) - (2*ArcTan[((-1)^(1/6)*((-1)^(5/6)*b^(1/3) + I*a^(1/
3)*Tanh[(c + d*x)/2]))/Sqrt[(-1)^(1/3)*a^(2/3) - b^(2/3)]])/(3*Sqrt[(-1)^(1/3)*a^(2/3) - b^(2/3)]*b^(2/3)*d) -
(2*ArcTanh[(b^(1/3) - a^(1/3)*Tanh[(c + d*x)/2])/Sqrt[a^(2/3) + b^(2/3)]])/(3*Sqrt[a^(2/3) + b^(2/3)]*b^(2/3)
*d)

Rule 3220

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_.), x_Symbol] :> Int[ExpandTr
ig[sin[e + f*x]^m*(a + b*sin[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, e, f}, x] && IntegersQ[m, p] && (EqQ[n, 4]
|| GtQ[p, 0] || (EqQ[p, -1] && IntegerQ[n]))

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])

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 ^2(c+d x)}{a+b \sinh ^3(c+d x)} \, dx &=-\int \left (\frac{i}{3 b^{2/3} \left (-i \sqrt [3]{a}-i \sqrt [3]{b} \sinh (c+d x)\right )}+\frac{i}{3 b^{2/3} \left (\sqrt [6]{-1} \sqrt [3]{a}-i \sqrt [3]{b} \sinh (c+d x)\right )}+\frac{i}{3 b^{2/3} \left ((-1)^{5/6} \sqrt [3]{a}-i \sqrt [3]{b} \sinh (c+d x)\right )}\right ) \, dx\\ &=-\frac{i \int \frac{1}{-i \sqrt [3]{a}-i \sqrt [3]{b} \sinh (c+d x)} \, dx}{3 b^{2/3}}-\frac{i \int \frac{1}{\sqrt [6]{-1} \sqrt [3]{a}-i \sqrt [3]{b} \sinh (c+d x)} \, dx}{3 b^{2/3}}-\frac{i \int \frac{1}{(-1)^{5/6} \sqrt [3]{a}-i \sqrt [3]{b} \sinh (c+d x)} \, dx}{3 b^{2/3}}\\ &=-\frac{2 \operatorname{Subst}\left (\int \frac{1}{-i \sqrt [3]{a}-2 \sqrt [3]{b} x-i \sqrt [3]{a} x^2} \, dx,x,\tan \left (\frac{1}{2} (i c+i d x)\right )\right )}{3 b^{2/3} d}-\frac{2 \operatorname{Subst}\left (\int \frac{1}{\sqrt [6]{-1} \sqrt [3]{a}-2 \sqrt [3]{b} x+\sqrt [6]{-1} \sqrt [3]{a} x^2} \, dx,x,\tan \left (\frac{1}{2} (i c+i d x)\right )\right )}{3 b^{2/3} d}-\frac{2 \operatorname{Subst}\left (\int \frac{1}{(-1)^{5/6} \sqrt [3]{a}-2 \sqrt [3]{b} x+(-1)^{5/6} \sqrt [3]{a} x^2} \, dx,x,\tan \left (\frac{1}{2} (i c+i d x)\right )\right )}{3 b^{2/3} d}\\ &=\frac{4 \operatorname{Subst}\left (\int \frac{1}{-4 \left (\sqrt [3]{-1} a^{2/3}-b^{2/3}\right )-x^2} \, dx,x,-2 \sqrt [3]{b}+2 \sqrt [6]{-1} \sqrt [3]{a} \tan \left (\frac{1}{2} (i c+i d x)\right )\right )}{3 b^{2/3} d}+\frac{4 \operatorname{Subst}\left (\int \frac{1}{4 \left (a^{2/3}+b^{2/3}\right )-x^2} \, dx,x,-2 \sqrt [3]{b}-2 i \sqrt [3]{a} \tan \left (\frac{1}{2} (i c+i d x)\right )\right )}{3 b^{2/3} d}+\frac{4 \operatorname{Subst}\left (\int \frac{1}{4 \left ((-1)^{2/3} a^{2/3}+b^{2/3}\right )-x^2} \, dx,x,-2 \sqrt [3]{b}+2 (-1)^{5/6} \sqrt [3]{a} \tan \left (\frac{1}{2} (i c+i d x)\right )\right )}{3 b^{2/3} d}\\ &=\frac{2 \tan ^{-1}\left (\frac{\sqrt [3]{b}+\sqrt [3]{-1} \sqrt [3]{a} \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{-(-1)^{2/3} a^{2/3}-b^{2/3}}}\right )}{3 \sqrt{-(-1)^{2/3} a^{2/3}-b^{2/3}} b^{2/3} d}+\frac{2 \tan ^{-1}\left (\frac{\sqrt [3]{b}-(-1)^{2/3} \sqrt [3]{a} \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{\sqrt [3]{-1} a^{2/3}-b^{2/3}}}\right )}{3 \sqrt{\sqrt [3]{-1} a^{2/3}-b^{2/3}} b^{2/3} d}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt [3]{b}-\sqrt [3]{a} \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^{2/3}+b^{2/3}}}\right )}{3 \sqrt{a^{2/3}+b^{2/3}} b^{2/3} d}\\ \end{align*}

Mathematica [C]  time = 0.17752, size = 275, normalized size = 1.05 \[ \frac{\text{RootSum}\left [8 \text{$\#$1}^3 a+\text{$\#$1}^6 b-3 \text{$\#$1}^4 b+3 \text{$\#$1}^2 b-b\& ,\frac{2 \text{$\#$1}^4 \log \left (-\text{$\#$1} \sinh \left (\frac{1}{2} (c+d x)\right )+\text{$\#$1} \cosh \left (\frac{1}{2} (c+d x)\right )-\sinh \left (\frac{1}{2} (c+d x)\right )-\cosh \left (\frac{1}{2} (c+d x)\right )\right )-4 \text{$\#$1}^2 \log \left (-\text{$\#$1} \sinh \left (\frac{1}{2} (c+d x)\right )+\text{$\#$1} \cosh \left (\frac{1}{2} (c+d x)\right )-\sinh \left (\frac{1}{2} (c+d x)\right )-\cosh \left (\frac{1}{2} (c+d x)\right )\right )+\text{$\#$1}^4 c-2 \text{$\#$1}^2 c+\text{$\#$1}^4 d x-2 \text{$\#$1}^2 d x+2 \log \left (-\text{$\#$1} \sinh \left (\frac{1}{2} (c+d x)\right )+\text{$\#$1} \cosh \left (\frac{1}{2} (c+d x)\right )-\sinh \left (\frac{1}{2} (c+d x)\right )-\cosh \left (\frac{1}{2} (c+d x)\right )\right )+c+d x}{4 \text{$\#$1}^2 a+\text{$\#$1}^5 b-2 \text{$\#$1}^3 b+\text{$\#$1} b}\& \right ]}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

RootSum[-b + 3*b*#1^2 + 8*a*#1^3 - 3*b*#1^4 + b*#1^6 & , (c + d*x + 2*Log[-Cosh[(c + d*x)/2] - Sinh[(c + d*x)/
2] + Cosh[(c + d*x)/2]*#1 - Sinh[(c + d*x)/2]*#1] - 2*c*#1^2 - 2*d*x*#1^2 - 4*Log[-Cosh[(c + d*x)/2] - Sinh[(c
+ d*x)/2] + Cosh[(c + d*x)/2]*#1 - Sinh[(c + d*x)/2]*#1]*#1^2 + c*#1^4 + d*x*#1^4 + 2*Log[-Cosh[(c + d*x)/2]
- Sinh[(c + d*x)/2] + Cosh[(c + d*x)/2]*#1 - Sinh[(c + d*x)/2]*#1]*#1^4)/(b*#1 + 4*a*#1^2 - 2*b*#1^3 + b*#1^5)
& ]/(6*d)

________________________________________________________________________________________

Maple [C]  time = 0.042, size = 78, normalized size = 0.3 \begin{align*} -{\frac{4}{3\,d}\sum _{{\it \_R}={\it RootOf} \left ( a{{\it \_Z}}^{6}-3\,a{{\it \_Z}}^{4}-8\,b{{\it \_Z}}^{3}+3\,a{{\it \_Z}}^{2}-a \right ) }{\frac{{{\it \_R}}^{2}}{{{\it \_R}}^{5}a-2\,{{\it \_R}}^{3}a-4\,{{\it \_R}}^{2}b+{\it \_R}\,a}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -{\it \_R} \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/3/d*sum(_R^2/(_R^5*a-2*_R^3*a-4*_R^2*b+_R*a)*ln(tanh(1/2*d*x+1/2*c)-_R),_R=RootOf(_Z^6*a-3*_Z^4*a-8*_Z^3*b+
3*_Z^2*a-a))

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sinh \left (d x + c\right )^{2}}{b \sinh \left (d x + c\right )^{3} + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sinh(d*x + c)^2/(b*sinh(d*x + c)^3 + a), x)

________________________________________________________________________________________

Fricas [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(d*x+c)^2/(a+b*sinh(d*x+c)^3),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sinh \left (d x + c\right )^{2}}{b \sinh \left (d x + c\right )^{3} + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sinh(d*x + c)^2/(b*sinh(d*x + c)^3 + a), x)