3.5.1 \(\int \cosh ^5(e+f x) (a+b \sinh ^2(e+f x))^p \, dx\) [401]

Optimal. Leaf size=214 \[ -\frac {(3 a-b (7+2 p)) \sinh (e+f x) \left (a+b \sinh ^2(e+f x)\right )^{1+p}}{b^2 f (3+2 p) (5+2 p)}+\frac {\cosh ^2(e+f x) \sinh (e+f x) \left (a+b \sinh ^2(e+f x)\right )^{1+p}}{b f (5+2 p)}+\frac {\left (3 a^2-2 a b (5+2 p)+b^2 \left (15+16 p+4 p^2\right )\right ) \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b \sinh ^2(e+f x)}{a}\right ) \sinh (e+f x) \left (a+b \sinh ^2(e+f x)\right )^p \left (1+\frac {b \sinh ^2(e+f x)}{a}\right )^{-p}}{b^2 f (3+2 p) (5+2 p)} \]

[Out]

-(3*a-b*(7+2*p))*sinh(f*x+e)*(a+b*sinh(f*x+e)^2)^(1+p)/b^2/f/(4*p^2+16*p+15)+cosh(f*x+e)^2*sinh(f*x+e)*(a+b*si
nh(f*x+e)^2)^(1+p)/b/f/(5+2*p)+(3*a^2-2*a*b*(5+2*p)+b^2*(4*p^2+16*p+15))*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^2/f/(4*p^2+16*p+15)/((1+b*sinh(f*x+e)^2/a)^p)

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Rubi [A]
time = 0.15, antiderivative size = 214, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3269, 427, 396, 252, 251} \begin {gather*} \frac {\left (3 a^2-2 a b (2 p+5)+b^2 \left (4 p^2+16 p+15\right )\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 )}{b^2 f (2 p+3) (2 p+5)}-\frac {(3 a-b (2 p+7)) \sinh (e+f x) \left (a+b \sinh ^2(e+f x)\right )^{p+1}}{b^2 f (2 p+3) (2 p+5)}+\frac {\sinh (e+f x) \cosh ^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )^{p+1}}{b f (2 p+5)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cosh[e + f*x]^5*(a + b*Sinh[e + f*x]^2)^p,x]

[Out]

-(((3*a - b*(7 + 2*p))*Sinh[e + f*x]*(a + b*Sinh[e + f*x]^2)^(1 + p))/(b^2*f*(3 + 2*p)*(5 + 2*p))) + (Cosh[e +
f*x]^2*Sinh[e + f*x]*(a + b*Sinh[e + f*x]^2)^(1 + p))/(b*f*(5 + 2*p)) + ((3*a^2 - 2*a*b*(5 + 2*p) + b^2*(15 +
16*p + 4*p^2))*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^2*f*(3 + 2*p)*(5 + 2*p)*(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 427

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[d*x*(a + b*x^n)^(p + 1)*((c
+ d*x^n)^(q - 1)/(b*(n*(p + q) + 1))), x] + Dist[1/(b*(n*(p + q) + 1)), Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)
*Simp[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q - 1) + 1))*x^n, x], x], x] /; F
reeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntB
inomialQ[a, b, c, d, n, p, q, x]

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 ^5(e+f x) \left (a+b \sinh ^2(e+f x)\right )^p \, dx &=\frac {\text {Subst}\left (\int \left (1+x^2\right )^2 \left (a+b x^2\right )^p \, dx,x,\sinh (e+f x)\right )}{f}\\ &=\frac {\cosh ^2(e+f x) \sinh (e+f x) \left (a+b \sinh ^2(e+f x)\right )^{1+p}}{b f (5+2 p)}+\frac {\text {Subst}\left (\int \left (a+b x^2\right )^p \left (-a+b (5+2 p)-(3 a-b (7+2 p)) x^2\right ) \, dx,x,\sinh (e+f x)\right )}{b f (5+2 p)}\\ &=-\frac {(3 a-b (7+2 p)) \sinh (e+f x) \left (a+b \sinh ^2(e+f x)\right )^{1+p}}{b^2 f (3+2 p) (5+2 p)}+\frac {\cosh ^2(e+f x) \sinh (e+f x) \left (a+b \sinh ^2(e+f x)\right )^{1+p}}{b f (5+2 p)}+\frac {\left (3 a^2-2 a b (5+2 p)+b^2 \left (15+16 p+4 p^2\right )\right ) \text {Subst}\left (\int \left (a+b x^2\right )^p \, dx,x,\sinh (e+f x)\right )}{b^2 f (3+2 p) (5+2 p)}\\ &=-\frac {(3 a-b (7+2 p)) \sinh (e+f x) \left (a+b \sinh ^2(e+f x)\right )^{1+p}}{b^2 f (3+2 p) (5+2 p)}+\frac {\cosh ^2(e+f x) \sinh (e+f x) \left (a+b \sinh ^2(e+f x)\right )^{1+p}}{b f (5+2 p)}+\frac {\left (\left (3 a^2-2 a b (5+2 p)+b^2 \left (15+16 p+4 p^2\right )\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 )}{b^2 f (3+2 p) (5+2 p)}\\ &=-\frac {(3 a-b (7+2 p)) \sinh (e+f x) \left (a+b \sinh ^2(e+f x)\right )^{1+p}}{b^2 f (3+2 p) (5+2 p)}+\frac {\cosh ^2(e+f x) \sinh (e+f x) \left (a+b \sinh ^2(e+f x)\right )^{1+p}}{b f (5+2 p)}+\frac {\left (3 a^2-2 a b (5+2 p)+b^2 \left (15+16 p+4 p^2\right )\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}}{b^2 f (3+2 p) (5+2 p)}\\ \end {align*}

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Mathematica [F]
time = 8.00, size = 0, normalized size = 0.00 \begin {gather*} \int \cosh ^5(e+f x) \left (a+b \sinh ^2(e+f x)\right )^p \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[Cosh[e + f*x]^5*(a + b*Sinh[e + f*x]^2)^p,x]

[Out]

Integrate[Cosh[e + f*x]^5*(a + b*Sinh[e + f*x]^2)^p, x]

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Maple [F]
time = 2.03, size = 0, normalized size = 0.00 \[\int \left (\cosh ^{5}\left (f x +e \right )\right ) \left (a +b \left (\sinh ^{2}\left (f x +e \right )\right )\right )^{p}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(f*x+e)^5*(a+b*sinh(f*x+e)^2)^p,x)

[Out]

int(cosh(f*x+e)^5*(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)^5*(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)^5, x)

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Fricas [F]
time = 0.41, size = 25, normalized size = 0.12 \begin {gather*} {\rm integral}\left ({\left (b \sinh \left (f x + e\right )^{2} + a\right )}^{p} \cosh \left (f x + e\right )^{5}, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(f*x+e)^5*(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)^5, 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)**5*(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)^5*(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)^5, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\mathrm {cosh}\left (e+f\,x\right )}^5\,{\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)^5*(a + b*sinh(e + f*x)^2)^p,x)

[Out]

int(cosh(e + f*x)^5*(a + b*sinh(e + f*x)^2)^p, x)

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