3.171 \(\int \frac{\sinh ^6(c+d x)}{a+b \sinh ^3(c+d x)} \, dx\)

Optimal. Leaf size=328 \[ -\frac{2 a^{4/3} \tanh ^{-1}\left (\frac{\sqrt [3]{b}-\sqrt [3]{a} \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^{2/3}+b^{2/3}}}\right )}{3 b^2 d \sqrt{a^{2/3}+b^{2/3}}}-\frac{2 (-1)^{2/3} a^{4/3} \tan ^{-1}\left (\frac{\sqrt [6]{-1} \left (\sqrt [6]{-1} \sqrt [3]{b}+i \sqrt [3]{a} \tanh \left (\frac{1}{2} (c+d x)\right )\right )}{\sqrt{\sqrt [3]{-1} a^{2/3}-(-1)^{2/3} b^{2/3}}}\right )}{3 b^2 d \sqrt{\sqrt [3]{-1} a^{2/3}-(-1)^{2/3} b^{2/3}}}-\frac{2 (-1)^{2/3} a^{4/3} \tan ^{-1}\left (\frac{\sqrt [6]{-1} \left ((-1)^{5/6} \sqrt [3]{b}+i \sqrt [3]{a} \tanh \left (\frac{1}{2} (c+d x)\right )\right )}{\sqrt{\sqrt [3]{-1} a^{2/3}-b^{2/3}}}\right )}{3 b^2 d \sqrt{\sqrt [3]{-1} a^{2/3}-b^{2/3}}}-\frac{a x}{b^2}+\frac{\cosh ^3(c+d x)}{3 b d}-\frac{\cosh (c+d x)}{b d} \]

[Out]

-((a*x)/b^2) - (2*(-1)^(2/3)*a^(4/3)*ArcTan[((-1)^(1/6)*((-1)^(1/6)*b^(1/3) + I*a^(1/3)*Tanh[(c + d*x)/2]))/Sq
rt[(-1)^(1/3)*a^(2/3) - (-1)^(2/3)*b^(2/3)]])/(3*Sqrt[(-1)^(1/3)*a^(2/3) - (-1)^(2/3)*b^(2/3)]*b^2*d) - (2*(-1
)^(2/3)*a^(4/3)*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*d) - (2*a^(4/3)*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*d) - Cosh[c + d*x]/(b*d) + Cosh[c + d*x]^3/(3
*b*d)

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Rubi [A]  time = 0.738527, antiderivative size = 328, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {3220, 2633, 3213, 2660, 618, 204} \[ -\frac{2 a^{4/3} \tanh ^{-1}\left (\frac{\sqrt [3]{b}-\sqrt [3]{a} \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^{2/3}+b^{2/3}}}\right )}{3 b^2 d \sqrt{a^{2/3}+b^{2/3}}}-\frac{2 (-1)^{2/3} a^{4/3} \tan ^{-1}\left (\frac{\sqrt [6]{-1} \left (\sqrt [6]{-1} \sqrt [3]{b}+i \sqrt [3]{a} \tanh \left (\frac{1}{2} (c+d x)\right )\right )}{\sqrt{\sqrt [3]{-1} a^{2/3}-(-1)^{2/3} b^{2/3}}}\right )}{3 b^2 d \sqrt{\sqrt [3]{-1} a^{2/3}-(-1)^{2/3} b^{2/3}}}-\frac{2 (-1)^{2/3} a^{4/3} \tan ^{-1}\left (\frac{\sqrt [6]{-1} \left ((-1)^{5/6} \sqrt [3]{b}+i \sqrt [3]{a} \tanh \left (\frac{1}{2} (c+d x)\right )\right )}{\sqrt{\sqrt [3]{-1} a^{2/3}-b^{2/3}}}\right )}{3 b^2 d \sqrt{\sqrt [3]{-1} a^{2/3}-b^{2/3}}}-\frac{a x}{b^2}+\frac{\cosh ^3(c+d x)}{3 b d}-\frac{\cosh (c+d x)}{b d} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[c + d*x]^6/(a + b*Sinh[c + d*x]^3),x]

[Out]

-((a*x)/b^2) - (2*(-1)^(2/3)*a^(4/3)*ArcTan[((-1)^(1/6)*((-1)^(1/6)*b^(1/3) + I*a^(1/3)*Tanh[(c + d*x)/2]))/Sq
rt[(-1)^(1/3)*a^(2/3) - (-1)^(2/3)*b^(2/3)]])/(3*Sqrt[(-1)^(1/3)*a^(2/3) - (-1)^(2/3)*b^(2/3)]*b^2*d) - (2*(-1
)^(2/3)*a^(4/3)*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*d) - (2*a^(4/3)*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*d) - Cosh[c + d*x]/(b*d) + Cosh[c + d*x]^3/(3
*b*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 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 3213

Int[((a_) + (b_.)*((c_.)*sin[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> Int[ExpandTrig[(a + b*(c*sin[e + f*
x])^n)^p, x], x] /; FreeQ[{a, b, c, e, f, n}, x] && (IGtQ[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 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 ^6(c+d x)}{a+b \sinh ^3(c+d x)} \, dx &=-\int \left (\frac{a}{b^2}-\frac{\sinh ^3(c+d x)}{b}-\frac{a^2}{b^2 \left (a+b \sinh ^3(c+d x)\right )}\right ) \, dx\\ &=-\frac{a x}{b^2}+\frac{a^2 \int \frac{1}{a+b \sinh ^3(c+d x)} \, dx}{b^2}+\frac{\int \sinh ^3(c+d x) \, dx}{b}\\ &=-\frac{a x}{b^2}+\frac{a^2 \int \left (\frac{\sqrt [6]{-1}}{3 a^{2/3} \left (\sqrt [6]{-1} \sqrt [3]{a}-i \sqrt [3]{b} \sinh (c+d x)\right )}+\frac{\sqrt [6]{-1}}{3 a^{2/3} \left (\sqrt [6]{-1} \sqrt [3]{a}+\sqrt [6]{-1} \sqrt [3]{b} \sinh (c+d x)\right )}+\frac{\sqrt [6]{-1}}{3 a^{2/3} \left (\sqrt [6]{-1} \sqrt [3]{a}+(-1)^{5/6} \sqrt [3]{b} \sinh (c+d x)\right )}\right ) \, dx}{b^2}-\frac{\operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cosh (c+d x)\right )}{b d}\\ &=-\frac{a x}{b^2}-\frac{\cosh (c+d x)}{b d}+\frac{\cosh ^3(c+d x)}{3 b d}+\frac{\left (\sqrt [6]{-1} a^{4/3}\right ) \int \frac{1}{\sqrt [6]{-1} \sqrt [3]{a}-i \sqrt [3]{b} \sinh (c+d x)} \, dx}{3 b^2}+\frac{\left (\sqrt [6]{-1} a^{4/3}\right ) \int \frac{1}{\sqrt [6]{-1} \sqrt [3]{a}+\sqrt [6]{-1} \sqrt [3]{b} \sinh (c+d x)} \, dx}{3 b^2}+\frac{\left (\sqrt [6]{-1} a^{4/3}\right ) \int \frac{1}{\sqrt [6]{-1} \sqrt [3]{a}+(-1)^{5/6} \sqrt [3]{b} \sinh (c+d x)} \, dx}{3 b^2}\\ &=-\frac{a x}{b^2}-\frac{\cosh (c+d x)}{b d}+\frac{\cosh ^3(c+d x)}{3 b d}-\frac{\left (2 (-1)^{2/3} a^{4/3}\right ) \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 d}-\frac{\left (2 (-1)^{2/3} a^{4/3}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt [6]{-1} \sqrt [3]{a}+2 \sqrt [3]{-1} \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 d}-\frac{\left (2 (-1)^{2/3} a^{4/3}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt [6]{-1} \sqrt [3]{a}-2 (-1)^{2/3} \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 d}\\ &=-\frac{a x}{b^2}-\frac{\cosh (c+d x)}{b d}+\frac{\cosh ^3(c+d x)}{3 b d}+\frac{\left (4 (-1)^{2/3} a^{4/3}\right ) \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 d}+\frac{\left (4 (-1)^{2/3} a^{4/3}\right ) \operatorname{Subst}\left (\int \frac{1}{-4 \sqrt [3]{-1} \left (a^{2/3}+b^{2/3}\right )-x^2} \, dx,x,-2 (-1)^{2/3} \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 d}+\frac{\left (4 (-1)^{2/3} a^{4/3}\right ) \operatorname{Subst}\left (\int \frac{1}{-4 \left (\sqrt [3]{-1} a^{2/3}-(-1)^{2/3} b^{2/3}\right )-x^2} \, dx,x,2 \sqrt [3]{-1} \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 d}\\ &=-\frac{a x}{b^2}+\frac{2 (-1)^{2/3} a^{4/3} \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 d}-\frac{2 (-1)^{2/3} a^{4/3} \tan ^{-1}\left (\frac{\sqrt [3]{-1} \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}-(-1)^{2/3} b^{2/3}}}\right )}{3 \sqrt{\sqrt [3]{-1} a^{2/3}-(-1)^{2/3} b^{2/3}} b^2 d}-\frac{2 a^{4/3} \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 d}-\frac{\cosh (c+d x)}{b d}+\frac{\cosh ^3(c+d x)}{3 b d}\\ \end{align*}

Mathematica [C]  time = 0.349243, size = 168, normalized size = 0.51 \[ \frac{8 a^2 \text{RootSum}\left [8 \text{$\#$1}^3 a+\text{$\#$1}^6 b-3 \text{$\#$1}^4 b+3 \text{$\#$1}^2 b-b\& ,\frac{2 \text{$\#$1} \log \left (-\text{$\#$1} \sinh \left (\frac{1}{2} (c+d x)\right )+\text{$\#$1} \cosh \left (\frac{1}{2} (c+d x)\right )-\sinh \left (\frac{1}{2} (c+d x)\right )-\cosh \left (\frac{1}{2} (c+d x)\right )\right )+\text{$\#$1} c+\text{$\#$1} d x}{\text{$\#$1}^4 b-2 \text{$\#$1}^2 b+4 \text{$\#$1} a+b}\& \right ]-12 a c-12 a d x-9 b \cosh (c+d x)+b \cosh (3 (c+d x))}{12 b^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[c + d*x]^6/(a + b*Sinh[c + d*x]^3),x]

[Out]

(-12*a*c - 12*a*d*x - 9*b*Cosh[c + d*x] + b*Cosh[3*(c + d*x)] + 8*a^2*RootSum[-b + 3*b*#1^2 + 8*a*#1^3 - 3*b*#
1^4 + b*#1^6 & , (c*#1 + d*x*#1 + 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)/(b + 4*a*#1 - 2*b*#1^2 + b*#1^4) & ])/(12*b^2*d)

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Maple [C]  time = 0.066, size = 259, normalized size = 0.8 \begin{align*}{\frac{1}{3\,bd} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}}-{\frac{1}{2\,bd} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}-{\frac{1}{2\,bd} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{a}{d{b}^{2}}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-{\frac{1}{3\,bd} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-3}}-{\frac{1}{2\,bd} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}+{\frac{1}{2\,bd} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+{\frac{a}{d{b}^{2}}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }-{\frac{{a}^{2}}{3\,d{b}^{2}}\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}}^{4}-2\,{{\it \_R}}^{2}+1}{{{\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)^6/(a+b*sinh(d*x+c)^3),x)

[Out]

1/3/d/b/(tanh(1/2*d*x+1/2*c)+1)^3-1/2/d/b/(tanh(1/2*d*x+1/2*c)+1)^2-1/2/d/b/(tanh(1/2*d*x+1/2*c)+1)-1/d*a/b^2*
ln(tanh(1/2*d*x+1/2*c)+1)-1/3/d/b/(tanh(1/2*d*x+1/2*c)-1)^3-1/2/d/b/(tanh(1/2*d*x+1/2*c)-1)^2+1/2/d/b/(tanh(1/
2*d*x+1/2*c)-1)+1/d*a/b^2*ln(tanh(1/2*d*x+1/2*c)-1)-1/3/d*a^2/b^2*sum((_R^4-2*_R^2+1)/(_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))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} 8 \, a^{2} \int \frac{e^{\left (3 \, d x + 3 \, c\right )}}{b^{3} e^{\left (6 \, d x + 6 \, c\right )} - 3 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 8 \, a b^{2} e^{\left (3 \, d x + 3 \, c\right )} + 3 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} - b^{3}}\,{d x} - \frac{{\left (24 \, a d x e^{\left (3 \, d x + 3 \, c\right )} - b e^{\left (6 \, d x + 6 \, c\right )} + 9 \, b e^{\left (4 \, d x + 4 \, c\right )} + 9 \, b e^{\left (2 \, d x + 2 \, c\right )} - b\right )} e^{\left (-3 \, d x - 3 \, c\right )}}{24 \, b^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)^6/(a+b*sinh(d*x+c)^3),x, algorithm="maxima")

[Out]

8*a^2*integrate(e^(3*d*x + 3*c)/(b^3*e^(6*d*x + 6*c) - 3*b^3*e^(4*d*x + 4*c) + 8*a*b^2*e^(3*d*x + 3*c) + 3*b^3
*e^(2*d*x + 2*c) - b^3), x) - 1/24*(24*a*d*x*e^(3*d*x + 3*c) - b*e^(6*d*x + 6*c) + 9*b*e^(4*d*x + 4*c) + 9*b*e
^(2*d*x + 2*c) - b)*e^(-3*d*x - 3*c)/(b^2*d)

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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)^6/(a+b*sinh(d*x+c)^3),x, algorithm="fricas")

[Out]

Timed out

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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(d*x+c)**6/(a+b*sinh(d*x+c)**3),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sinh \left (d x + c\right )^{6}}{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)^6/(a+b*sinh(d*x+c)^3),x, algorithm="giac")

[Out]

integrate(sinh(d*x + c)^6/(b*sinh(d*x + c)^3 + a), x)