3.5.22 \(\int \frac {\cosh ^5(c+d x)}{(a+b \sinh ^n(c+d x))^2} \, dx\) [422]

Optimal. Leaf size=130 \[ \frac {\, _2F_1\left (2,\frac {1}{n};1+\frac {1}{n};-\frac {b \sinh ^n(c+d x)}{a}\right ) \sinh (c+d x)}{a^2 d}+\frac {2 \, _2F_1\left (2,\frac {3}{n};\frac {3+n}{n};-\frac {b \sinh ^n(c+d x)}{a}\right ) \sinh ^3(c+d x)}{3 a^2 d}+\frac {\, _2F_1\left (2,\frac {5}{n};\frac {5+n}{n};-\frac {b \sinh ^n(c+d x)}{a}\right ) \sinh ^5(c+d x)}{5 a^2 d} \]

[Out]

hypergeom([2, 1/n],[1+1/n],-b*sinh(d*x+c)^n/a)*sinh(d*x+c)/a^2/d+2/3*hypergeom([2, 3/n],[(3+n)/n],-b*sinh(d*x+
c)^n/a)*sinh(d*x+c)^3/a^2/d+1/5*hypergeom([2, 5/n],[(5+n)/n],-b*sinh(d*x+c)^n/a)*sinh(d*x+c)^5/a^2/d

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Rubi [A]
time = 0.09, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3302, 1907, 251, 371} \begin {gather*} \frac {\sinh (c+d x) \, _2F_1\left (2,\frac {1}{n};1+\frac {1}{n};-\frac {b \sinh ^n(c+d x)}{a}\right )}{a^2 d}+\frac {\sinh ^5(c+d x) \, _2F_1\left (2,\frac {5}{n};\frac {n+5}{n};-\frac {b \sinh ^n(c+d x)}{a}\right )}{5 a^2 d}+\frac {2 \sinh ^3(c+d x) \, _2F_1\left (2,\frac {3}{n};\frac {n+3}{n};-\frac {b \sinh ^n(c+d x)}{a}\right )}{3 a^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Hypergeometric2F1[2, n^(-1), 1 + n^(-1), -((b*Sinh[c + d*x]^n)/a)]*Sinh[c + d*x])/(a^2*d) + (2*Hypergeometric
2F1[2, 3/n, (3 + n)/n, -((b*Sinh[c + d*x]^n)/a)]*Sinh[c + d*x]^3)/(3*a^2*d) + (Hypergeometric2F1[2, 5/n, (5 +
n)/n, -((b*Sinh[c + d*x]^n)/a)]*Sinh[c + d*x]^5)/(5*a^2*d)

Rule 251

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[-p, 1/n, 1/n + 1, (-b)*(x^n/a)],
x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p
] || GtQ[a, 0])

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 1907

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, n, p}, x] && (PolyQ[Pq, x] || PolyQ[Pq, x^n])

Rule 3302

Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*((c_.)*sin[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With
[{ff = FreeFactors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*(c*ff*x)^n)^p, x]
, x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[(m - 1)/2] && (EqQ[n, 4] || GtQ[m, 0
] || IGtQ[p, 0] || IntegersQ[m, p])

Rubi steps

\begin {align*} \int \frac {\cosh ^5(c+d x)}{\left (a+b \sinh ^n(c+d x)\right )^2} \, dx &=\frac {\text {Subst}\left (\int \frac {\left (1+x^2\right )^2}{\left (a+b x^n\right )^2} \, dx,x,\sinh (c+d x)\right )}{d}\\ &=\frac {\text {Subst}\left (\int \left (\frac {1}{\left (a+b x^n\right )^2}+\frac {2 x^2}{\left (a+b x^n\right )^2}+\frac {x^4}{\left (a+b x^n\right )^2}\right ) \, dx,x,\sinh (c+d x)\right )}{d}\\ &=\frac {\text {Subst}\left (\int \frac {1}{\left (a+b x^n\right )^2} \, dx,x,\sinh (c+d x)\right )}{d}+\frac {\text {Subst}\left (\int \frac {x^4}{\left (a+b x^n\right )^2} \, dx,x,\sinh (c+d x)\right )}{d}+\frac {2 \text {Subst}\left (\int \frac {x^2}{\left (a+b x^n\right )^2} \, dx,x,\sinh (c+d x)\right )}{d}\\ &=\frac {\, _2F_1\left (2,\frac {1}{n};1+\frac {1}{n};-\frac {b \sinh ^n(c+d x)}{a}\right ) \sinh (c+d x)}{a^2 d}+\frac {2 \, _2F_1\left (2,\frac {3}{n};\frac {3+n}{n};-\frac {b \sinh ^n(c+d x)}{a}\right ) \sinh ^3(c+d x)}{3 a^2 d}+\frac {\, _2F_1\left (2,\frac {5}{n};\frac {5+n}{n};-\frac {b \sinh ^n(c+d x)}{a}\right ) \sinh ^5(c+d x)}{5 a^2 d}\\ \end {align*}

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Mathematica [A]
time = 0.12, size = 119, normalized size = 0.92 \begin {gather*} \frac {15 \, _2F_1\left (2,\frac {1}{n};1+\frac {1}{n};-\frac {b \sinh ^n(c+d x)}{a}\right ) \sinh (c+d x)+10 \, _2F_1\left (2,\frac {3}{n};\frac {3+n}{n};-\frac {b \sinh ^n(c+d x)}{a}\right ) \sinh ^3(c+d x)+3 \, _2F_1\left (2,\frac {5}{n};\frac {5+n}{n};-\frac {b \sinh ^n(c+d x)}{a}\right ) \sinh ^5(c+d x)}{15 a^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(15*Hypergeometric2F1[2, n^(-1), 1 + n^(-1), -((b*Sinh[c + d*x]^n)/a)]*Sinh[c + d*x] + 10*Hypergeometric2F1[2,
3/n, (3 + n)/n, -((b*Sinh[c + d*x]^n)/a)]*Sinh[c + d*x]^3 + 3*Hypergeometric2F1[2, 5/n, (5 + n)/n, -((b*Sinh[
c + d*x]^n)/a)]*Sinh[c + d*x]^5)/(15*a^2*d)

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Maple [F]
time = 2.34, size = 0, normalized size = 0.00 \[\int \frac {\cosh ^{5}\left (d x +c \right )}{\left (a +b \left (\sinh ^{n}\left (d x +c \right )\right )\right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(d*x+c)^5/(a+b*sinh(d*x+c)^n)^2,x)

[Out]

int(cosh(d*x+c)^5/(a+b*sinh(d*x+c)^n)^2,x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(d*x+c)^5/(a+b*sinh(d*x+c)^n)^2,x, algorithm="maxima")

[Out]

1/32*(2^n*e^(c*n + 10*d*x + 10*c) + 3*2^n*e^(c*n + 8*d*x + 8*c) + 2^(n + 1)*e^(c*n + 6*d*x + 6*c) - 2^(n + 1)*
e^(c*n + 4*d*x + 4*c) - 3*2^n*e^(c*n + 2*d*x + 2*c) - 2^n*e^(c*n))*e^(d*n*x)/(2^n*a^2*d*n*e^(d*n*x + c*n + 5*d
*x + 5*c) + a*b*d*n*e^(5*d*x + n*log(e^(d*x + c) + 1) + n*log(e^(d*x + c) - 1) + 5*c)) + 1/32*integrate((2^n*n
*e^(c*n) - 5*2^n*e^(c*n) + (2^n*n*e^(c*n) - 5*2^n*e^(c*n))*e^(10*d*x + 10*c) + (5*2^n*n*e^(c*n) - 9*2^n*e^(c*n
))*e^(8*d*x + 8*c) + (5*2^(n + 1)*n*e^(c*n) - 2^(n + 1)*e^(c*n))*e^(6*d*x + 6*c) + (5*2^(n + 1)*n*e^(c*n) - 2^
(n + 1)*e^(c*n))*e^(4*d*x + 4*c) + (5*2^n*n*e^(c*n) - 9*2^n*e^(c*n))*e^(2*d*x + 2*c))*e^(d*n*x)/(2^n*a^2*n*e^(
d*n*x + c*n + 5*d*x + 5*c) + a*b*n*e^(5*d*x + n*log(e^(d*x + c) + 1) + n*log(e^(d*x + c) - 1) + 5*c)), x)

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Fricas [F]
time = 0.45, size = 43, normalized size = 0.33 \begin {gather*} {\rm integral}\left (\frac {\cosh \left (d x + c\right )^{5}}{b^{2} \sinh \left (d x + c\right )^{2 \, n} + 2 \, a b \sinh \left (d x + c\right )^{n} + a^{2}}, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(d*x+c)^5/(a+b*sinh(d*x+c)^n)^2,x, algorithm="fricas")

[Out]

integral(cosh(d*x + c)^5/(b^2*sinh(d*x + c)^(2*n) + 2*a*b*sinh(d*x + c)^n + a^2), x)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(d*x+c)**5/(a+b*sinh(d*x+c)**n)**2,x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3064 deep

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(d*x+c)^5/(a+b*sinh(d*x+c)^n)^2,x, algorithm="giac")

[Out]

integrate(cosh(d*x + c)^5/(b*sinh(d*x + c)^n + a)^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\mathrm {cosh}\left (c+d\,x\right )}^5}{{\left (a+b\,{\mathrm {sinh}\left (c+d\,x\right )}^n\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(c + d*x)^5/(a + b*sinh(c + d*x)^n)^2,x)

[Out]

int(cosh(c + d*x)^5/(a + b*sinh(c + d*x)^n)^2, x)

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