3.172 \(\int \frac{\sinh ^5(c+d x)}{a+b \sinh ^3(c+d x)} \, dx\)

Optimal. Leaf size=295 \[ \frac{2 a \tanh ^{-1}\left (\frac{\sqrt [3]{b}-\sqrt [3]{a} \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^{2/3}+b^{2/3}}}\right )}{3 b^{5/3} d \sqrt{a^{2/3}+b^{2/3}}}+\frac{2 a \tan ^{-1}\left (\frac{(-1)^{5/6} \left (\sqrt [6]{-1} \sqrt [3]{b}+i \sqrt [3]{a} \tanh \left (\frac{1}{2} (c+d x)\right )\right )}{\sqrt{-(-1)^{2/3} a^{2/3}-b^{2/3}}}\right )}{3 b^{5/3} d \sqrt{-(-1)^{2/3} a^{2/3}-b^{2/3}}}+\frac{2 a \tan ^{-1}\left (\frac{\sqrt [6]{-1} \left ((-1)^{5/6} \sqrt [3]{b}+i \sqrt [3]{a} \tanh \left (\frac{1}{2} (c+d x)\right )\right )}{\sqrt{\sqrt [3]{-1} a^{2/3}-b^{2/3}}}\right )}{3 b^{5/3} d \sqrt{\sqrt [3]{-1} a^{2/3}-b^{2/3}}}+\frac{\sinh (c+d x) \cosh (c+d x)}{2 b d}-\frac{x}{2 b} \]

[Out]

-x/(2*b) + (2*a*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^(5/3)*d) + (2*a*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^(5/3)*d) + (2*a*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^(5/3)*d) + (Cosh[c + d*x]*Sinh[c + d*x])/(2*b*d)

________________________________________________________________________________________

Rubi [A]  time = 0.558601, antiderivative size = 295, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {3220, 2635, 8, 2660, 618, 206, 204} \[ \frac{2 a \tanh ^{-1}\left (\frac{\sqrt [3]{b}-\sqrt [3]{a} \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^{2/3}+b^{2/3}}}\right )}{3 b^{5/3} d \sqrt{a^{2/3}+b^{2/3}}}+\frac{2 a \tan ^{-1}\left (\frac{(-1)^{5/6} \left (\sqrt [6]{-1} \sqrt [3]{b}+i \sqrt [3]{a} \tanh \left (\frac{1}{2} (c+d x)\right )\right )}{\sqrt{-(-1)^{2/3} a^{2/3}-b^{2/3}}}\right )}{3 b^{5/3} d \sqrt{-(-1)^{2/3} a^{2/3}-b^{2/3}}}+\frac{2 a \tan ^{-1}\left (\frac{\sqrt [6]{-1} \left ((-1)^{5/6} \sqrt [3]{b}+i \sqrt [3]{a} \tanh \left (\frac{1}{2} (c+d x)\right )\right )}{\sqrt{\sqrt [3]{-1} a^{2/3}-b^{2/3}}}\right )}{3 b^{5/3} d \sqrt{\sqrt [3]{-1} a^{2/3}-b^{2/3}}}+\frac{\sinh (c+d x) \cosh (c+d x)}{2 b d}-\frac{x}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[c + d*x]^5/(a + b*Sinh[c + d*x]^3),x]

[Out]

-x/(2*b) + (2*a*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^(5/3)*d) + (2*a*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^(5/3)*d) + (2*a*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^(5/3)*d) + (Cosh[c + d*x]*Sinh[c + d*x])/(2*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 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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

Mathematica [C]  time = 0.316158, size = 299, normalized size = 1.01 \[ \frac{-2 a \text{RootSum}\left [8 \text{$\#$1}^3 a+\text{$\#$1}^6 b-3 \text{$\#$1}^4 b+3 \text{$\#$1}^2 b-b\& ,\frac{2 \text{$\#$1}^4 \log \left (-\text{$\#$1} \sinh \left (\frac{1}{2} (c+d x)\right )+\text{$\#$1} \cosh \left (\frac{1}{2} (c+d x)\right )-\sinh \left (\frac{1}{2} (c+d x)\right )-\cosh \left (\frac{1}{2} (c+d x)\right )\right )-4 \text{$\#$1}^2 \log \left (-\text{$\#$1} \sinh \left (\frac{1}{2} (c+d x)\right )+\text{$\#$1} \cosh \left (\frac{1}{2} (c+d x)\right )-\sinh \left (\frac{1}{2} (c+d x)\right )-\cosh \left (\frac{1}{2} (c+d x)\right )\right )+\text{$\#$1}^4 c-2 \text{$\#$1}^2 c+\text{$\#$1}^4 d x-2 \text{$\#$1}^2 d x+2 \log \left (-\text{$\#$1} \sinh \left (\frac{1}{2} (c+d x)\right )+\text{$\#$1} \cosh \left (\frac{1}{2} (c+d x)\right )-\sinh \left (\frac{1}{2} (c+d x)\right )-\cosh \left (\frac{1}{2} (c+d x)\right )\right )+c+d x}{4 \text{$\#$1}^2 a+\text{$\#$1}^5 b-2 \text{$\#$1}^3 b+\text{$\#$1} b}\& \right ]-6 (c+d x)+3 \sinh (2 (c+d x))}{12 b d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[c + d*x]^5/(a + b*Sinh[c + d*x]^3),x]

[Out]

(-6*(c + d*x) - 2*a*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) & ] + 3*Sinh[2*(c + d*x)])/(12*b*d)

________________________________________________________________________________________

Maple [C]  time = 0.06, size = 207, normalized size = 0.7 \begin{align*} -{\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{1}{2\,bd}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }+{\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{1}{2\,bd}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+{\frac{4\,a}{3\,bd}\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)^5/(a+b*sinh(d*x+c)^3),x)

[Out]

-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)-1/2/d/b*ln(tanh(1/2*d*x+1/2*c)+1)+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)+1/2/d/b*ln(tanh(1/2*d*x+1/2*c)-1)+4/3/d*a/b*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*} -\frac{{\left (4 \, d x e^{\left (2 \, d x + 2 \, c\right )} - e^{\left (4 \, d x + 4 \, c\right )} + 1\right )} e^{\left (-2 \, d x - 2 \, c\right )}}{8 \, b d} - \frac{1}{32} \, \int \frac{64 \,{\left (a e^{\left (5 \, d x + 5 \, c\right )} - 2 \, a e^{\left (3 \, d x + 3 \, c\right )} + a e^{\left (d x + c\right )}\right )}}{b^{2} e^{\left (6 \, d x + 6 \, c\right )} - 3 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 8 \, a b e^{\left (3 \, d x + 3 \, c\right )} + 3 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} - b^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)^5/(a+b*sinh(d*x+c)^3),x, algorithm="maxima")

[Out]

-1/8*(4*d*x*e^(2*d*x + 2*c) - e^(4*d*x + 4*c) + 1)*e^(-2*d*x - 2*c)/(b*d) - 1/32*integrate(64*(a*e^(5*d*x + 5*
c) - 2*a*e^(3*d*x + 3*c) + a*e^(d*x + c))/(b^2*e^(6*d*x + 6*c) - 3*b^2*e^(4*d*x + 4*c) + 8*a*b*e^(3*d*x + 3*c)
+ 3*b^2*e^(2*d*x + 2*c) - b^2), 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)^5/(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)**5/(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 )^{5}}{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)^5/(a+b*sinh(d*x+c)^3),x, algorithm="giac")

[Out]

integrate(sinh(d*x + c)^5/(b*sinh(d*x + c)^3 + a), x)