3.230 \(\int \frac{\sinh ^5(c+d x)}{a-b \sinh ^4(c+d x)} \, dx\)

Optimal. Leaf size=139 \[ \frac{\sqrt{a} \tan ^{-1}\left (\frac{\sqrt [4]{b} \cosh (c+d x)}{\sqrt{\sqrt{a}-\sqrt{b}}}\right )}{2 b^{5/4} d \sqrt{\sqrt{a}-\sqrt{b}}}+\frac{\sqrt{a} \tanh ^{-1}\left (\frac{\sqrt [4]{b} \cosh (c+d x)}{\sqrt{\sqrt{a}+\sqrt{b}}}\right )}{2 b^{5/4} d \sqrt{\sqrt{a}+\sqrt{b}}}-\frac{\cosh (c+d x)}{b d} \]

[Out]

(Sqrt[a]*ArcTan[(b^(1/4)*Cosh[c + d*x])/Sqrt[Sqrt[a] - Sqrt[b]]])/(2*Sqrt[Sqrt[a] - Sqrt[b]]*b^(5/4)*d) + (Sqr
t[a]*ArcTanh[(b^(1/4)*Cosh[c + d*x])/Sqrt[Sqrt[a] + Sqrt[b]]])/(2*Sqrt[Sqrt[a] + Sqrt[b]]*b^(5/4)*d) - Cosh[c
+ d*x]/(b*d)

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Rubi [A]  time = 0.197599, antiderivative size = 139, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {3215, 1170, 1093, 205, 208} \[ \frac{\sqrt{a} \tan ^{-1}\left (\frac{\sqrt [4]{b} \cosh (c+d x)}{\sqrt{\sqrt{a}-\sqrt{b}}}\right )}{2 b^{5/4} d \sqrt{\sqrt{a}-\sqrt{b}}}+\frac{\sqrt{a} \tanh ^{-1}\left (\frac{\sqrt [4]{b} \cosh (c+d x)}{\sqrt{\sqrt{a}+\sqrt{b}}}\right )}{2 b^{5/4} d \sqrt{\sqrt{a}+\sqrt{b}}}-\frac{\cosh (c+d x)}{b d} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[c + d*x]^5/(a - b*Sinh[c + d*x]^4),x]

[Out]

(Sqrt[a]*ArcTan[(b^(1/4)*Cosh[c + d*x])/Sqrt[Sqrt[a] - Sqrt[b]]])/(2*Sqrt[Sqrt[a] - Sqrt[b]]*b^(5/4)*d) + (Sqr
t[a]*ArcTanh[(b^(1/4)*Cosh[c + d*x])/Sqrt[Sqrt[a] + Sqrt[b]]])/(2*Sqrt[Sqrt[a] + Sqrt[b]]*b^(5/4)*d) - Cosh[c
+ d*x]/(b*d)

Rule 3215

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Cos[e + f*x], x]}, -Dist[ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - 2*b*ff^2*x^2 + b*ff^4*x^4
)^p, x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rule 1170

Int[((d_) + (e_.)*(x_)^2)^(q_)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> Int[ExpandIntegrand[(d + e*x
^2)^q/(a + b*x^2 + c*x^4), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a
*e^2, 0] && IntegerQ[q]

Rule 1093

Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(-1), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[c/q, Int[1/(b/
2 - q/2 + c*x^2), x], x] - Dist[c/q, Int[1/(b/2 + q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*
a*c, 0] && PosQ[b^2 - 4*a*c]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

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

Mathematica [C]  time = 0.259954, size = 235, normalized size = 1.69 \[ -\frac{a \text{RootSum}\left [-16 \text{$\#$1}^4 a+\text{$\#$1}^8 b-4 \text{$\#$1}^6 b+6 \text{$\#$1}^4 b-4 \text{$\#$1}^2 b+b\& ,\frac{2 \text{$\#$1}^3 \log \left (-\text{$\#$1} \sinh \left (\frac{1}{2} (c+d x)\right )+\text{$\#$1} \cosh \left (\frac{1}{2} (c+d x)\right )-\sinh \left (\frac{1}{2} (c+d x)\right )-\cosh \left (\frac{1}{2} (c+d x)\right )\right )+\text{$\#$1}^3 c+\text{$\#$1}^3 d x-2 \text{$\#$1} \log \left (-\text{$\#$1} \sinh \left (\frac{1}{2} (c+d x)\right )+\text{$\#$1} \cosh \left (\frac{1}{2} (c+d x)\right )-\sinh \left (\frac{1}{2} (c+d x)\right )-\cosh \left (\frac{1}{2} (c+d x)\right )\right )-\text{$\#$1} c-\text{$\#$1} d x}{-8 \text{$\#$1}^2 a+\text{$\#$1}^6 b-3 \text{$\#$1}^4 b+3 \text{$\#$1}^2 b-b}\& \right ]+2 \cosh (c+d x)}{2 b d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[c + d*x]^5/(a - b*Sinh[c + d*x]^4),x]

[Out]

-(2*Cosh[c + d*x] + a*RootSum[b - 4*b*#1^2 - 16*a*#1^4 + 6*b*#1^4 - 4*b*#1^6 + b*#1^8 & , (-(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 + c*#1^3 + d*x*
#1^3 + 2*Log[-Cosh[(c + d*x)/2] - Sinh[(c + d*x)/2] + Cosh[(c + d*x)/2]*#1 - Sinh[(c + d*x)/2]*#1]*#1^3)/(-b -
8*a*#1^2 + 3*b*#1^2 - 3*b*#1^4 + b*#1^6) & ])/(2*b*d)

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Maple [A]  time = 0.048, size = 175, normalized size = 1.3 \begin{align*} -{\frac{1}{bd} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{a}{2\,bd}\arctan \left ({\frac{1}{4} \left ( -2\, \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a+4\,\sqrt{ab}+2\,a \right ){\frac{1}{\sqrt{-ab-\sqrt{ab}a}}}} \right ){\frac{1}{\sqrt{-ab-\sqrt{ab}a}}}}+{\frac{a}{2\,bd}\arctan \left ({\frac{1}{4} \left ( 2\, \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a+4\,\sqrt{ab}-2\,a \right ){\frac{1}{\sqrt{-ab+\sqrt{ab}a}}}} \right ){\frac{1}{\sqrt{-ab+\sqrt{ab}a}}}}+{\frac{1}{bd} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}} \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)^4),x)

[Out]

-1/d/b/(tanh(1/2*d*x+1/2*c)+1)-1/2/d*a/b/(-a*b-(a*b)^(1/2)*a)^(1/2)*arctan(1/4*(-2*tanh(1/2*d*x+1/2*c)^2*a+4*(
a*b)^(1/2)+2*a)/(-a*b-(a*b)^(1/2)*a)^(1/2))+1/2/d*a/b/(-a*b+(a*b)^(1/2)*a)^(1/2)*arctan(1/4*(2*tanh(1/2*d*x+1/
2*c)^2*a+4*(a*b)^(1/2)-2*a)/(-a*b+(a*b)^(1/2)*a)^(1/2))+1/d/b/(tanh(1/2*d*x+1/2*c)-1)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )} e^{\left (-d x - c\right )}}{2 \, b d} - \frac{1}{32} \, \int \frac{256 \,{\left (a e^{\left (5 \, d x + 5 \, c\right )} - a e^{\left (3 \, d x + 3 \, c\right )}\right )}}{b^{2} e^{\left (8 \, d x + 8 \, c\right )} - 4 \, b^{2} e^{\left (6 \, d x + 6 \, c\right )} - 4 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + b^{2} - 2 \,{\left (8 \, a b e^{\left (4 \, c\right )} - 3 \, b^{2} e^{\left (4 \, c\right )}\right )} e^{\left (4 \, d x\right )}}\,{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)^4),x, algorithm="maxima")

[Out]

-1/2*(e^(2*d*x + 2*c) + 1)*e^(-d*x - c)/(b*d) - 1/32*integrate(256*(a*e^(5*d*x + 5*c) - a*e^(3*d*x + 3*c))/(b^
2*e^(8*d*x + 8*c) - 4*b^2*e^(6*d*x + 6*c) - 4*b^2*e^(2*d*x + 2*c) + b^2 - 2*(8*a*b*e^(4*c) - 3*b^2*e^(4*c))*e^
(4*d*x)), x)

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Fricas [B]  time = 2.11141, size = 2755, normalized size = 19.82 \begin{align*} \text{result too large to display} \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)^4),x, algorithm="fricas")

[Out]

1/4*((b*d*cosh(d*x + c) + b*d*sinh(d*x + c))*sqrt(-((a*b^2 - b^3)*d^2*sqrt(a^3/((a^2*b^5 - 2*a*b^6 + b^7)*d^4)
) + a)/((a*b^2 - b^3)*d^2))*log(a^2*cosh(d*x + c)^2 + 2*a^2*cosh(d*x + c)*sinh(d*x + c) + a^2*sinh(d*x + c)^2
+ a^2 + 2*(a^2*b*d*cosh(d*x + c) + a^2*b*d*sinh(d*x + c) - ((a*b^4 - b^5)*d^3*cosh(d*x + c) + (a*b^4 - b^5)*d^
3*sinh(d*x + c))*sqrt(a^3/((a^2*b^5 - 2*a*b^6 + b^7)*d^4)))*sqrt(-((a*b^2 - b^3)*d^2*sqrt(a^3/((a^2*b^5 - 2*a*
b^6 + b^7)*d^4)) + a)/((a*b^2 - b^3)*d^2))) - (b*d*cosh(d*x + c) + b*d*sinh(d*x + c))*sqrt(-((a*b^2 - b^3)*d^2
*sqrt(a^3/((a^2*b^5 - 2*a*b^6 + b^7)*d^4)) + a)/((a*b^2 - b^3)*d^2))*log(a^2*cosh(d*x + c)^2 + 2*a^2*cosh(d*x
+ c)*sinh(d*x + c) + a^2*sinh(d*x + c)^2 + a^2 - 2*(a^2*b*d*cosh(d*x + c) + a^2*b*d*sinh(d*x + c) - ((a*b^4 -
b^5)*d^3*cosh(d*x + c) + (a*b^4 - b^5)*d^3*sinh(d*x + c))*sqrt(a^3/((a^2*b^5 - 2*a*b^6 + b^7)*d^4)))*sqrt(-((a
*b^2 - b^3)*d^2*sqrt(a^3/((a^2*b^5 - 2*a*b^6 + b^7)*d^4)) + a)/((a*b^2 - b^3)*d^2))) + (b*d*cosh(d*x + c) + b*
d*sinh(d*x + c))*sqrt(((a*b^2 - b^3)*d^2*sqrt(a^3/((a^2*b^5 - 2*a*b^6 + b^7)*d^4)) - a)/((a*b^2 - b^3)*d^2))*l
og(a^2*cosh(d*x + c)^2 + 2*a^2*cosh(d*x + c)*sinh(d*x + c) + a^2*sinh(d*x + c)^2 + a^2 + 2*(a^2*b*d*cosh(d*x +
c) + a^2*b*d*sinh(d*x + c) + ((a*b^4 - b^5)*d^3*cosh(d*x + c) + (a*b^4 - b^5)*d^3*sinh(d*x + c))*sqrt(a^3/((a
^2*b^5 - 2*a*b^6 + b^7)*d^4)))*sqrt(((a*b^2 - b^3)*d^2*sqrt(a^3/((a^2*b^5 - 2*a*b^6 + b^7)*d^4)) - a)/((a*b^2
- b^3)*d^2))) - (b*d*cosh(d*x + c) + b*d*sinh(d*x + c))*sqrt(((a*b^2 - b^3)*d^2*sqrt(a^3/((a^2*b^5 - 2*a*b^6 +
b^7)*d^4)) - a)/((a*b^2 - b^3)*d^2))*log(a^2*cosh(d*x + c)^2 + 2*a^2*cosh(d*x + c)*sinh(d*x + c) + a^2*sinh(d
*x + c)^2 + a^2 - 2*(a^2*b*d*cosh(d*x + c) + a^2*b*d*sinh(d*x + c) + ((a*b^4 - b^5)*d^3*cosh(d*x + c) + (a*b^4
- b^5)*d^3*sinh(d*x + c))*sqrt(a^3/((a^2*b^5 - 2*a*b^6 + b^7)*d^4)))*sqrt(((a*b^2 - b^3)*d^2*sqrt(a^3/((a^2*b
^5 - 2*a*b^6 + b^7)*d^4)) - a)/((a*b^2 - b^3)*d^2))) - 2*cosh(d*x + c)^2 - 4*cosh(d*x + c)*sinh(d*x + c) - 2*s
inh(d*x + c)^2 - 2)/(b*d*cosh(d*x + c) + b*d*sinh(d*x + 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(d*x+c)**5/(a-b*sinh(d*x+c)**4),x)

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \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)^4),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError