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

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

[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 = 0.13, 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 = {3223, 1893, 245, 364} \[ \frac {\sinh ^5(c+d x) \, _2F_1\left (2,\frac {5}{n};\frac {n+5}{n};-\frac {b \sinh ^n(c+d x)}{a}\right )}{5 a^2 d}+\frac {2 \sinh ^3(c+d x) \, _2F_1\left (2,\frac {3}{n};\frac {n+3}{n};-\frac {b \sinh ^n(c+d x)}{a}\right )}{3 a^2 d}+\frac {\sinh (c+d x) \, _2F_1\left (2,\frac {1}{n};1+\frac {1}{n};-\frac {b \sinh ^n(c+d x)}{a}\right )}{a^2 d} \]

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 245

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 364

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

Rule 1893

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 3223

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 {\operatorname {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 {\operatorname {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 {\operatorname {Subst}\left (\int \frac {1}{\left (a+b x^n\right )^2} \, dx,x,\sinh (c+d x)\right )}{d}+\frac {\operatorname {Subst}\left (\int \frac {x^4}{\left (a+b x^n\right )^2} \, dx,x,\sinh (c+d x)\right )}{d}+\frac {2 \operatorname {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 \[ \frac {15 \sinh (c+d x) \, _2F_1\left (2,\frac {1}{n};1+\frac {1}{n};-\frac {b \sinh ^n(c+d x)}{a}\right )+3 \sinh ^5(c+d x) \, _2F_1\left (2,\frac {5}{n};\frac {n+5}{n};-\frac {b \sinh ^n(c+d x)}{a}\right )+10 \sinh ^3(c+d x) \, _2F_1\left (2,\frac {3}{n};\frac {n+3}{n};-\frac {b \sinh ^n(c+d x)}{a}\right )}{15 a^2 d} \]

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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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\cosh \left (d x + c\right )^{5}}{b^{2} \sinh \left (d x + c\right )^{2 \, n} + 2 \, a b \sinh \left (d x + c\right )^{n} + a^{2}}, x\right ) \]

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

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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maple [F]  time = 5.82, 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 \[ \frac {{\left (2^{n} e^{\left (c n + 10 \, d x + 10 \, c\right )} + 3 \cdot 2^{n} e^{\left (c n + 8 \, d x + 8 \, c\right )} + 2^{n + 1} e^{\left (c n + 6 \, d x + 6 \, c\right )} - 2^{n + 1} e^{\left (c n + 4 \, d x + 4 \, c\right )} - 3 \cdot 2^{n} e^{\left (c n + 2 \, d x + 2 \, c\right )} - 2^{n} e^{\left (c n\right )}\right )} e^{\left (d n x\right )}}{32 \, {\left (2^{n} a^{2} d n e^{\left (d n x + c n + 5 \, d x + 5 \, c\right )} + a b d n e^{\left (5 \, d x + n \log \left (e^{\left (d x + c\right )} + 1\right ) + n \log \left (e^{\left (d x + c\right )} - 1\right ) + 5 \, c\right )}\right )}} + \frac {1}{32} \, \int \frac {{\left (2^{n} n e^{\left (c n\right )} - 5 \cdot 2^{n} e^{\left (c n\right )} + {\left (2^{n} n e^{\left (c n\right )} - 5 \cdot 2^{n} e^{\left (c n\right )}\right )} e^{\left (10 \, d x + 10 \, c\right )} + {\left (5 \cdot 2^{n} n e^{\left (c n\right )} - 9 \cdot 2^{n} e^{\left (c n\right )}\right )} e^{\left (8 \, d x + 8 \, c\right )} + {\left (5 \cdot 2^{n + 1} n e^{\left (c n\right )} - 2^{n + 1} e^{\left (c n\right )}\right )} e^{\left (6 \, d x + 6 \, c\right )} + {\left (5 \cdot 2^{n + 1} n e^{\left (c n\right )} - 2^{n + 1} e^{\left (c n\right )}\right )} e^{\left (4 \, d x + 4 \, c\right )} + {\left (5 \cdot 2^{n} n e^{\left (c n\right )} - 9 \cdot 2^{n} e^{\left (c n\right )}\right )} e^{\left (2 \, d x + 2 \, c\right )}\right )} e^{\left (d n x\right )}}{2^{n} a^{2} n e^{\left (d n x + c n + 5 \, d x + 5 \, c\right )} + a b n e^{\left (5 \, d x + n \log \left (e^{\left (d x + c\right )} + 1\right ) + n \log \left (e^{\left (d x + c\right )} - 1\right ) + 5 \, c\right )}}\,{d x} \]

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

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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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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]

Timed out

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