3.402 \(\int \cosh ^3(e+f x) (a+b \sinh ^2(e+f x))^p \, dx\)

Optimal. Leaf size=125 \[ \frac{\sinh (e+f x) \left (a+b \sinh ^2(e+f x)\right )^{p+1}}{b f (2 p+3)}-\frac{(a-b (2 p+3)) \sinh (e+f x) \left (a+b \sinh ^2(e+f x)\right )^p \left (\frac{b \sinh ^2(e+f x)}{a}+1\right )^{-p} \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b \sinh ^2(e+f x)}{a}\right )}{b f (2 p+3)} \]

[Out]

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

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Rubi [A]  time = 0.102418, antiderivative size = 119, normalized size of antiderivative = 0.95, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {3190, 388, 246, 245} \[ \frac{\left (1-\frac{a}{2 b p+3 b}\right ) \sinh (e+f x) \left (a+b \sinh ^2(e+f x)\right )^p \left (\frac{b \sinh ^2(e+f x)}{a}+1\right )^{-p} \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b \sinh ^2(e+f x)}{a}\right )}{f}+\frac{\sinh (e+f x) \left (a+b \sinh ^2(e+f x)\right )^{p+1}}{b f (2 p+3)} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[e + f*x]^3*(a + b*Sinh[e + f*x]^2)^p,x]

[Out]

(Sinh[e + f*x]*(a + b*Sinh[e + f*x]^2)^(1 + p))/(b*f*(3 + 2*p)) + ((1 - a/(3*b + 2*b*p))*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 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

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 ^3(e+f x) \left (a+b \sinh ^2(e+f x)\right )^p \, dx &=\frac{\operatorname{Subst}\left (\int \left (1+x^2\right ) \left (a+b x^2\right )^p \, dx,x,\sinh (e+f x)\right )}{f}\\ &=\frac{\sinh (e+f x) \left (a+b \sinh ^2(e+f x)\right )^{1+p}}{b f (3+2 p)}+\frac{\left (1-\frac{a}{3 b+2 b p}\right ) \operatorname{Subst}\left (\int \left (a+b x^2\right )^p \, dx,x,\sinh (e+f x)\right )}{f}\\ &=\frac{\sinh (e+f x) \left (a+b \sinh ^2(e+f x)\right )^{1+p}}{b f (3+2 p)}+\frac{\left (\left (1-\frac{a}{3 b+2 b p}\right ) \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{\sinh (e+f x) \left (a+b \sinh ^2(e+f x)\right )^{1+p}}{b f (3+2 p)}+\frac{\left (1-\frac{a}{3 b+2 b p}\right ) \, _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 = 0.20196, size = 120, normalized size = 0.96 \[ \frac{\sinh (e+f x) \left (a+b \sinh ^2(e+f x)\right )^p \left (\frac{b \sinh ^2(e+f x)}{a}+1\right )^{-p} \left ((b (2 p+3)-a) \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b \sinh ^2(e+f x)}{a}\right )+\left (a+b \sinh ^2(e+f x)\right ) \left (\frac{b \sinh ^2(e+f x)}{a}+1\right )^p\right )}{b f (2 p+3)} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[e + f*x]^3*(a + b*Sinh[e + f*x]^2)^p,x]

[Out]

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

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Maple [F]  time = 0.453, size = 0, normalized size = 0. \begin{align*} \int \left ( \cosh \left ( fx+e \right ) \right ) ^{3} \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)^3*(a+b*sinh(f*x+e)^2)^p,x)

[Out]

int(cosh(f*x+e)^3*(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 )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(f*x+e)^3*(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)^3, 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 )^{3}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(f*x+e)^3*(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)^3, 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)**3*(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 )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(f*x+e)^3*(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)^3, x)