3.30 \(\int \frac{\sinh ^5(c+d x)}{a+b \sinh ^2(c+d x)} \, dx\)

Optimal. Leaf size=79 \[ \frac{a^2 \tan ^{-1}\left (\frac{\sqrt{b} \cosh (c+d x)}{\sqrt{a-b}}\right )}{b^{5/2} d \sqrt{a-b}}-\frac{(a+b) \cosh (c+d x)}{b^2 d}+\frac{\cosh ^3(c+d x)}{3 b d} \]

[Out]

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

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Rubi [A]  time = 0.108028, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3186, 390, 205} \[ \frac{a^2 \tan ^{-1}\left (\frac{\sqrt{b} \cosh (c+d x)}{\sqrt{a-b}}\right )}{b^{5/2} d \sqrt{a-b}}-\frac{(a+b) \cosh (c+d x)}{b^2 d}+\frac{\cosh ^3(c+d x)}{3 b d} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[c + d*x]^5/(a + b*Sinh[c + d*x]^2),x]

[Out]

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

Rule 3186

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(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 - b*ff^2*x^2)^p, x], x, Cos
[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rule 390

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]

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]

Rubi steps

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

Mathematica [C]  time = 0.434318, size = 134, normalized size = 1.7 \[ \frac{\frac{12 a^2 \left (\tan ^{-1}\left (\frac{\sqrt{b}-i \sqrt{a} \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a-b}}\right )+\tan ^{-1}\left (\frac{\sqrt{b}+i \sqrt{a} \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a-b}}\right )\right )}{\sqrt{a-b}}-3 \sqrt{b} (4 a+3 b) \cosh (c+d x)+b^{3/2} \cosh (3 (c+d x))}{12 b^{5/2} d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[c + d*x]^5/(a + b*Sinh[c + d*x]^2),x]

[Out]

((12*a^2*(ArcTan[(Sqrt[b] - I*Sqrt[a]*Tanh[(c + d*x)/2])/Sqrt[a - b]] + ArcTan[(Sqrt[b] + I*Sqrt[a]*Tanh[(c +
d*x)/2])/Sqrt[a - b]]))/Sqrt[a - b] - 3*Sqrt[b]*(4*a + 3*b)*Cosh[c + d*x] + b^(3/2)*Cosh[3*(c + d*x)])/(12*b^(
5/2)*d)

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Maple [B]  time = 0.033, size = 227, normalized size = 2.9 \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{a}{d{b}^{2}} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{1}{2\,bd} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\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{a}{d{b}^{2}} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+{\frac{1}{2\,bd} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+{\frac{{a}^{2}}{d{b}^{2}}\arctan \left ({\frac{1}{4} \left ( 2\, \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a-2\,a+4\,b \right ){\frac{1}{\sqrt{ab-{b}^{2}}}}} \right ){\frac{1}{\sqrt{ab-{b}^{2}}}}} \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)^2),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/d/b^2/(tanh(1/2*d*x+1/2*c)+1)*a-1/2/d/b/
(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/d/b^2/(tanh(1/2*
d*x+1/2*c)-1)*a+1/2/d/b/(tanh(1/2*d*x+1/2*c)-1)+1/d*a^2/b^2/(a*b-b^2)^(1/2)*arctan(1/4*(2*tanh(1/2*d*x+1/2*c)^
2*a-2*a+4*b)/(a*b-b^2)^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{{\left (3 \,{\left (4 \, a e^{\left (4 \, c\right )} + 3 \, b e^{\left (4 \, c\right )}\right )} e^{\left (4 \, d x\right )} + 3 \,{\left (4 \, a e^{\left (2 \, c\right )} + 3 \, b e^{\left (2 \, c\right )}\right )} e^{\left (2 \, d x\right )} - b e^{\left (6 \, d x + 6 \, c\right )} - b\right )} e^{\left (-3 \, d x - 3 \, c\right )}}{24 \, b^{2} d} + \frac{1}{32} \, \int \frac{64 \,{\left (a^{2} e^{\left (3 \, d x + 3 \, c\right )} - a^{2} e^{\left (d x + c\right )}\right )}}{b^{3} e^{\left (4 \, d x + 4 \, c\right )} + b^{3} + 2 \,{\left (2 \, a b^{2} e^{\left (2 \, c\right )} - b^{3} e^{\left (2 \, c\right )}\right )} e^{\left (2 \, 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)^2),x, algorithm="maxima")

[Out]

-1/24*(3*(4*a*e^(4*c) + 3*b*e^(4*c))*e^(4*d*x) + 3*(4*a*e^(2*c) + 3*b*e^(2*c))*e^(2*d*x) - b*e^(6*d*x + 6*c) -
b)*e^(-3*d*x - 3*c)/(b^2*d) + 1/32*integrate(64*(a^2*e^(3*d*x + 3*c) - a^2*e^(d*x + c))/(b^3*e^(4*d*x + 4*c)
+ b^3 + 2*(2*a*b^2*e^(2*c) - b^3*e^(2*c))*e^(2*d*x)), x)

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Fricas [B]  time = 1.96975, size = 3993, normalized size = 50.54 \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)^2),x, algorithm="fricas")

[Out]

[1/24*((a*b^2 - b^3)*cosh(d*x + c)^6 + 6*(a*b^2 - b^3)*cosh(d*x + c)*sinh(d*x + c)^5 + (a*b^2 - b^3)*sinh(d*x
+ c)^6 - 3*(4*a^2*b - a*b^2 - 3*b^3)*cosh(d*x + c)^4 - 3*(4*a^2*b - a*b^2 - 3*b^3 - 5*(a*b^2 - b^3)*cosh(d*x +
c)^2)*sinh(d*x + c)^4 + 4*(5*(a*b^2 - b^3)*cosh(d*x + c)^3 - 3*(4*a^2*b - a*b^2 - 3*b^3)*cosh(d*x + c))*sinh(
d*x + c)^3 + a*b^2 - b^3 - 3*(4*a^2*b - a*b^2 - 3*b^3)*cosh(d*x + c)^2 + 3*(5*(a*b^2 - b^3)*cosh(d*x + c)^4 -
4*a^2*b + a*b^2 + 3*b^3 - 6*(4*a^2*b - a*b^2 - 3*b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^2 - 12*(a^2*cosh(d*x + c)
^3 + 3*a^2*cosh(d*x + c)^2*sinh(d*x + c) + 3*a^2*cosh(d*x + c)*sinh(d*x + c)^2 + a^2*sinh(d*x + c)^3)*sqrt(-a*
b + b^2)*log((b*cosh(d*x + c)^4 + 4*b*cosh(d*x + c)*sinh(d*x + c)^3 + b*sinh(d*x + c)^4 - 2*(2*a - 3*b)*cosh(d
*x + c)^2 + 2*(3*b*cosh(d*x + c)^2 - 2*a + 3*b)*sinh(d*x + c)^2 + 4*(b*cosh(d*x + c)^3 - (2*a - 3*b)*cosh(d*x
+ c))*sinh(d*x + c) - 4*(cosh(d*x + c)^3 + 3*cosh(d*x + c)*sinh(d*x + c)^2 + sinh(d*x + c)^3 + (3*cosh(d*x + c
)^2 + 1)*sinh(d*x + c) + cosh(d*x + c))*sqrt(-a*b + b^2) + b)/(b*cosh(d*x + c)^4 + 4*b*cosh(d*x + c)*sinh(d*x
+ c)^3 + b*sinh(d*x + c)^4 + 2*(2*a - b)*cosh(d*x + c)^2 + 2*(3*b*cosh(d*x + c)^2 + 2*a - b)*sinh(d*x + c)^2 +
4*(b*cosh(d*x + c)^3 + (2*a - b)*cosh(d*x + c))*sinh(d*x + c) + b)) + 6*((a*b^2 - b^3)*cosh(d*x + c)^5 - 2*(4
*a^2*b - a*b^2 - 3*b^3)*cosh(d*x + c)^3 - (4*a^2*b - a*b^2 - 3*b^3)*cosh(d*x + c))*sinh(d*x + c))/((a*b^3 - b^
4)*d*cosh(d*x + c)^3 + 3*(a*b^3 - b^4)*d*cosh(d*x + c)^2*sinh(d*x + c) + 3*(a*b^3 - b^4)*d*cosh(d*x + c)*sinh(
d*x + c)^2 + (a*b^3 - b^4)*d*sinh(d*x + c)^3), 1/24*((a*b^2 - b^3)*cosh(d*x + c)^6 + 6*(a*b^2 - b^3)*cosh(d*x
+ c)*sinh(d*x + c)^5 + (a*b^2 - b^3)*sinh(d*x + c)^6 - 3*(4*a^2*b - a*b^2 - 3*b^3)*cosh(d*x + c)^4 - 3*(4*a^2*
b - a*b^2 - 3*b^3 - 5*(a*b^2 - b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^4 + 4*(5*(a*b^2 - b^3)*cosh(d*x + c)^3 - 3*
(4*a^2*b - a*b^2 - 3*b^3)*cosh(d*x + c))*sinh(d*x + c)^3 + a*b^2 - b^3 - 3*(4*a^2*b - a*b^2 - 3*b^3)*cosh(d*x
+ c)^2 + 3*(5*(a*b^2 - b^3)*cosh(d*x + c)^4 - 4*a^2*b + a*b^2 + 3*b^3 - 6*(4*a^2*b - a*b^2 - 3*b^3)*cosh(d*x +
c)^2)*sinh(d*x + c)^2 + 24*(a^2*cosh(d*x + c)^3 + 3*a^2*cosh(d*x + c)^2*sinh(d*x + c) + 3*a^2*cosh(d*x + c)*s
inh(d*x + c)^2 + a^2*sinh(d*x + c)^3)*sqrt(a*b - b^2)*arctan(-1/2*(b*cosh(d*x + c)^3 + 3*b*cosh(d*x + c)*sinh(
d*x + c)^2 + b*sinh(d*x + c)^3 + (4*a - 3*b)*cosh(d*x + c) + (3*b*cosh(d*x + c)^2 + 4*a - 3*b)*sinh(d*x + c))/
sqrt(a*b - b^2)) - 24*(a^2*cosh(d*x + c)^3 + 3*a^2*cosh(d*x + c)^2*sinh(d*x + c) + 3*a^2*cosh(d*x + c)*sinh(d*
x + c)^2 + a^2*sinh(d*x + c)^3)*sqrt(a*b - b^2)*arctan(-1/2*sqrt(a*b - b^2)*(cosh(d*x + c) + sinh(d*x + c))/(a
- b)) + 6*((a*b^2 - b^3)*cosh(d*x + c)^5 - 2*(4*a^2*b - a*b^2 - 3*b^3)*cosh(d*x + c)^3 - (4*a^2*b - a*b^2 - 3
*b^3)*cosh(d*x + c))*sinh(d*x + c))/((a*b^3 - b^4)*d*cosh(d*x + c)^3 + 3*(a*b^3 - b^4)*d*cosh(d*x + c)^2*sinh(
d*x + c) + 3*(a*b^3 - b^4)*d*cosh(d*x + c)*sinh(d*x + c)^2 + (a*b^3 - b^4)*d*sinh(d*x + c)^3)]

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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)**2),x)

[Out]

Timed out

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

[Out]

Exception raised: TypeError