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

Optimal. Leaf size=37 \[ \frac {\sinh (c+d x) \, _2F_1\left (2,\frac {1}{n};1+\frac {1}{n};-\frac {b \sinh ^n(c+d x)}{a}\right )}{a^2 d} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {3223, 245} \[ \frac {\sinh (c+d x) \, _2F_1\left (2,\frac {1}{n};1+\frac {1}{n};-\frac {b \sinh ^n(c+d x)}{a}\right )}{a^2 d} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[c + d*x]/(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)

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 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 (c+d x)}{\left (a+b \sinh ^n(c+d x)\right )^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{\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}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 37, normalized size = 1.00 \[ \frac {\sinh (c+d x) \, _2F_1\left (2,\frac {1}{n};1+\frac {1}{n};-\frac {b \sinh ^n(c+d x)}{a}\right )}{a^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[c + d*x]/(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)

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fricas [F]  time = 2.03, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\cosh \left (d x + c\right )}{b^{2} \sinh \left (d x + c\right )^{2 \, n} + 2 \, a b \sinh \left (d x + c\right )^{n} + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(cosh(d*x + c)/(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 \[ \int \frac {\cosh \left (d x + c\right )}{{\left (b \sinh \left (d x + c\right )^{n} + a\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 6.03, size = 0, normalized size = 0.00 \[ \int \frac {\cosh \left (d x +c \right )}{\left (a +b \left (\sinh ^{n}\left (d x +c \right )\right )\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {{\left (2^{n} e^{\left (c n + 2 \, d x + 2 \, c\right )} - 2^{n} e^{\left (c n\right )}\right )} e^{\left (d n x\right )}}{2 \, {\left (2^{n} a^{2} d n e^{\left (d n x + c n + d x + c\right )} + a b d n e^{\left (d x + n \log \left (e^{\left (d x + c\right )} + 1\right ) + n \log \left (e^{\left (d x + c\right )} - 1\right ) + c\right )}\right )}} + \frac {1}{2} \, \int \frac {{\left (2^{n} n e^{\left (c n\right )} - 2^{n} e^{\left (c n\right )} + {\left (2^{n} n e^{\left (c n\right )} - 2^{n} e^{\left (c n\right )}\right )} e^{\left (2 \, d x + 2 \, c\right )}\right )} e^{\left (d n x\right )}}{2^{n} a^{2} n e^{\left (d n x + c n + d x + c\right )} + a b n e^{\left (d x + n \log \left (e^{\left (d x + c\right )} + 1\right ) + n \log \left (e^{\left (d x + c\right )} - 1\right ) + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(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 + d*x + c) + a*b*d*n*e^(d*
x + n*log(e^(d*x + c) + 1) + n*log(e^(d*x + c) - 1) + c)) + 1/2*integrate((2^n*n*e^(c*n) - 2^n*e^(c*n) + (2^n*
n*e^(c*n) - 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 + d*x + c) + a*b*n*e^(d*x + n*lo
g(e^(d*x + c) + 1) + n*log(e^(d*x + c) - 1) + c)), x)

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mupad [B]  time = 0.95, size = 38, normalized size = 1.03 \[ \frac {\mathrm {sinh}\left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (2,\frac {1}{n};\ \frac {1}{n}+1;\ -\frac {b\,{\mathrm {sinh}\left (c+d\,x\right )}^n}{a}\right )}{a^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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