3.5.2 \(\int \cosh ^3(e+f x) (a+b \sinh ^2(e+f x))^p \, dx\) [402]

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

[Out]

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

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Rubi [A]
time = 0.07, 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 = {3269, 396, 252, 251} \begin {gather*} \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)} \end {gather*}

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 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 252

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^Fra
cPart[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 396

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 3269

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]

Rubi steps

\begin {align*} \int \cosh ^3(e+f x) \left (a+b \sinh ^2(e+f x)\right )^p \, dx &=\frac {\text {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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.23, size = 120, normalized size = 0.96 \begin {gather*} \frac {\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} \left ((-a+b (3+2 p)) \, _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 (1+\frac {b \sinh ^2(e+f x)}{a}\right )^p\right )}{b f (3+2 p)} \end {gather*}

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 = 2.17, size = 0, normalized size = 0.00 \[\int \left (\cosh ^{3}\left (f x +e \right )\right ) \left (a +b \left (\sinh ^{2}\left (f x +e \right )\right )\right )^{p}\, dx\]

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 = 0.00, 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(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 = 0.40, size = 25, normalized size = 0.20 \begin {gather*} {\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 {gather*}

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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, 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(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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\mathrm {cosh}\left (e+f\,x\right )}^3\,{\left (b\,{\mathrm {sinh}\left (e+f\,x\right )}^2+a\right )}^p \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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