3.403 \(\int \cosh (e+f x) (a+b \sinh ^2(e+f x))^p \, dx\)

Optimal. Leaf size=67 \[ \frac{\sinh (e+f x) \left (a+b \sinh ^2(e+f x)\right )^p \left (\frac{b \sinh ^2(e+f x)}{a}+1\right )^{-p} \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b \sinh ^2(e+f x)}{a}\right )}{f} \]

[Out]

(Hypergeometric2F1[1/2, -p, 3/2, -((b*Sinh[e + f*x]^2)/a)]*Sinh[e + f*x]*(a + b*Sinh[e + f*x]^2)^p)/(f*(1 + (b
*Sinh[e + f*x]^2)/a)^p)

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Rubi [A]  time = 0.0462517, antiderivative size = 67, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3190, 246, 245} \[ \frac{\sinh (e+f x) \left (a+b \sinh ^2(e+f x)\right )^p \left (\frac{b \sinh ^2(e+f x)}{a}+1\right )^{-p} \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b \sinh ^2(e+f x)}{a}\right )}{f} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[e + f*x]*(a + b*Sinh[e + f*x]^2)^p,x]

[Out]

(Hypergeometric2F1[1/2, -p, 3/2, -((b*Sinh[e + f*x]^2)/a)]*Sinh[e + f*x]*(a + b*Sinh[e + f*x]^2)^p)/(f*(1 + (b
*Sinh[e + f*x]^2)/a)^p)

Rule 3190

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

Rule 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^Fr
acPart[p], Int[(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILt
Q[Simplify[1/n + p], 0] &&  !(IntegerQ[p] || GtQ[a, 0])

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])

Rubi steps

\begin{align*} \int \cosh (e+f x) \left (a+b \sinh ^2(e+f x)\right )^p \, dx &=\frac{\operatorname{Subst}\left (\int \left (a+b x^2\right )^p \, dx,x,\sinh (e+f x)\right )}{f}\\ &=\frac{\left (\left (a+b \sinh ^2(e+f x)\right )^p \left (1+\frac{b \sinh ^2(e+f x)}{a}\right )^{-p}\right ) \operatorname{Subst}\left (\int \left (1+\frac{b x^2}{a}\right )^p \, dx,x,\sinh (e+f x)\right )}{f}\\ &=\frac{\, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b \sinh ^2(e+f x)}{a}\right ) \sinh (e+f x) \left (a+b \sinh ^2(e+f x)\right )^p \left (1+\frac{b \sinh ^2(e+f x)}{a}\right )^{-p}}{f}\\ \end{align*}

Mathematica [A]  time = 0.0255693, size = 67, normalized size = 1. \[ \frac{\sinh (e+f x) \left (a+b \sinh ^2(e+f x)\right )^p \left (\frac{b \sinh ^2(e+f x)}{a}+1\right )^{-p} \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b \sinh ^2(e+f x)}{a}\right )}{f} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[e + f*x]*(a + b*Sinh[e + f*x]^2)^p,x]

[Out]

(Hypergeometric2F1[1/2, -p, 3/2, -((b*Sinh[e + f*x]^2)/a)]*Sinh[e + f*x]*(a + b*Sinh[e + f*x]^2)^p)/(f*(1 + (b
*Sinh[e + f*x]^2)/a)^p)

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Maple [F]  time = 0.341, size = 0, normalized size = 0. \begin{align*} \int \cosh \left ( fx+e \right ) \left ( a+b \left ( \sinh \left ( fx+e \right ) \right ) ^{2} \right ) ^{p}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(f*x+e)*(a+b*sinh(f*x+e)^2)^p,x)

[Out]

int(cosh(f*x+e)*(a+b*sinh(f*x+e)^2)^p,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sinh \left (f x + e\right )^{2} + a\right )}^{p} \cosh \left (f x + e\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(f*x+e)*(a+b*sinh(f*x+e)^2)^p,x, algorithm="maxima")

[Out]

integrate((b*sinh(f*x + e)^2 + a)^p*cosh(f*x + e), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b \sinh \left (f x + e\right )^{2} + a\right )}^{p} \cosh \left (f x + e\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(f*x+e)*(a+b*sinh(f*x+e)^2)^p,x, algorithm="fricas")

[Out]

integral((b*sinh(f*x + e)^2 + a)^p*cosh(f*x + e), x)

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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(cosh(f*x+e)*(a+b*sinh(f*x+e)**2)**p,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sinh \left (f x + e\right )^{2} + a\right )}^{p} \cosh \left (f x + e\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(f*x+e)*(a+b*sinh(f*x+e)^2)^p,x, algorithm="giac")

[Out]

integrate((b*sinh(f*x + e)^2 + a)^p*cosh(f*x + e), x)