3.2.31 \(\int \sinh ^5(e+f x) (a+b \sinh ^2(e+f x))^p \, dx\) [131]

Optimal. Leaf size=226 \[ -\frac {(3 a+2 b (2+p)) \cosh (e+f x) \left (a-b+b \cosh ^2(e+f x)\right )^{1+p}}{b^2 f (3+2 p) (5+2 p)}+\frac {\left (3 a^2+4 a b (1+p)+4 b^2 \left (2+3 p+p^2\right )\right ) \cosh (e+f x) \left (a-b+b \cosh ^2(e+f x)\right )^p \left (1+\frac {b \cosh ^2(e+f x)}{a-b}\right )^{-p} \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b \cosh ^2(e+f x)}{a-b}\right )}{b^2 f (3+2 p) (5+2 p)}+\frac {\cosh (e+f x) \left (a-b+b \cosh ^2(e+f x)\right )^{1+p} \sinh ^2(e+f x)}{b f (5+2 p)} \]

[Out]

-(3*a+2*b*(2+p))*cosh(f*x+e)*(a-b+b*cosh(f*x+e)^2)^(1+p)/b^2/f/(4*p^2+16*p+15)+(3*a^2+4*a*b*(1+p)+4*b^2*(p^2+3
*p+2))*cosh(f*x+e)*(a-b+b*cosh(f*x+e)^2)^p*hypergeom([1/2, -p],[3/2],-b*cosh(f*x+e)^2/(a-b))/b^2/f/(4*p^2+16*p
+15)/((1+b*cosh(f*x+e)^2/(a-b))^p)+cosh(f*x+e)*(a-b+b*cosh(f*x+e)^2)^(1+p)*sinh(f*x+e)^2/b/f/(5+2*p)

________________________________________________________________________________________

Rubi [A]
time = 0.18, antiderivative size = 226, 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 = {3265, 427, 396, 252, 251} \begin {gather*} \frac {\left (3 a^2+4 a b (p+1)+4 b^2 \left (p^2+3 p+2\right )\right ) \cosh (e+f x) \left (a+b \cosh ^2(e+f x)-b\right )^p \left (\frac {b \cosh ^2(e+f x)}{a-b}+1\right )^{-p} \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b \cosh ^2(e+f x)}{a-b}\right )}{b^2 f (2 p+3) (2 p+5)}-\frac {(3 a+2 b (p+2)) \cosh (e+f x) \left (a+b \cosh ^2(e+f x)-b\right )^{p+1}}{b^2 f (2 p+3) (2 p+5)}+\frac {\sinh ^2(e+f x) \cosh (e+f x) \left (a+b \cosh ^2(e+f x)-b\right )^{p+1}}{b f (2 p+5)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(((3*a + 2*b*(2 + p))*Cosh[e + f*x]*(a - b + b*Cosh[e + f*x]^2)^(1 + p))/(b^2*f*(3 + 2*p)*(5 + 2*p))) + ((3*a
^2 + 4*a*b*(1 + p) + 4*b^2*(2 + 3*p + p^2))*Cosh[e + f*x]*(a - b + b*Cosh[e + f*x]^2)^p*Hypergeometric2F1[1/2,
-p, 3/2, -((b*Cosh[e + f*x]^2)/(a - b))])/(b^2*f*(3 + 2*p)*(5 + 2*p)*(1 + (b*Cosh[e + f*x]^2)/(a - b))^p) + (
Cosh[e + f*x]*(a - b + b*Cosh[e + f*x]^2)^(1 + p)*Sinh[e + f*x]^2)/(b*f*(5 + 2*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 3265

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

Rubi steps

\begin {align*} \int \sinh ^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+b x^2\right )^p \, dx,x,\cosh (e+f x)\right )}{f}\\ &=\frac {\cosh (e+f x) \left (a-b+b \cosh ^2(e+f x)\right )^{1+p} \sinh ^2(e+f x)}{b f (5+2 p)}+\frac {\text {Subst}\left (\int \left (a-b+b x^2\right )^p \left (a+2 b (2+p)-(3 a+2 b (2+p)) x^2\right ) \, dx,x,\cosh (e+f x)\right )}{b f (5+2 p)}\\ &=-\frac {(3 a+2 b (2+p)) \cosh (e+f x) \left (a-b+b \cosh ^2(e+f x)\right )^{1+p}}{b^2 f (3+2 p) (5+2 p)}+\frac {\cosh (e+f x) \left (a-b+b \cosh ^2(e+f x)\right )^{1+p} \sinh ^2(e+f x)}{b f (5+2 p)}+\frac {\left (3 a^2+4 a b (1+p)+4 b^2 \left (2+3 p+p^2\right )\right ) \text {Subst}\left (\int \left (a-b+b x^2\right )^p \, dx,x,\cosh (e+f x)\right )}{b^2 f (3+2 p) (5+2 p)}\\ &=-\frac {(3 a+2 b (2+p)) \cosh (e+f x) \left (a-b+b \cosh ^2(e+f x)\right )^{1+p}}{b^2 f (3+2 p) (5+2 p)}+\frac {\cosh (e+f x) \left (a-b+b \cosh ^2(e+f x)\right )^{1+p} \sinh ^2(e+f x)}{b f (5+2 p)}+\frac {\left (\left (3 a^2+4 a b (1+p)+4 b^2 \left (2+3 p+p^2\right )\right ) \left (a-b+b \cosh ^2(e+f x)\right )^p \left (1+\frac {b \cosh ^2(e+f x)}{a-b}\right )^{-p}\right ) \text {Subst}\left (\int \left (1+\frac {b x^2}{a-b}\right )^p \, dx,x,\cosh (e+f x)\right )}{b^2 f (3+2 p) (5+2 p)}\\ &=-\frac {(3 a+2 b (2+p)) \cosh (e+f x) \left (a-b+b \cosh ^2(e+f x)\right )^{1+p}}{b^2 f (3+2 p) (5+2 p)}+\frac {\left (3 a^2+4 a b (1+p)+4 b^2 \left (2+3 p+p^2\right )\right ) \cosh (e+f x) \left (a-b+b \cosh ^2(e+f x)\right )^p \left (1+\frac {b \cosh ^2(e+f x)}{a-b}\right )^{-p} \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b \cosh ^2(e+f x)}{a-b}\right )}{b^2 f (3+2 p) (5+2 p)}+\frac {\cosh (e+f x) \left (a-b+b \cosh ^2(e+f x)\right )^{1+p} \sinh ^2(e+f x)}{b f (5+2 p)}\\ \end {align*}

________________________________________________________________________________________

Mathematica [F]
time = 7.56, size = 0, normalized size = 0.00 \begin {gather*} \int \sinh ^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[Sinh[e + f*x]^5*(a + b*Sinh[e + f*x]^2)^p,x]

[Out]

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

________________________________________________________________________________________

Maple [F]
time = 1.72, size = 0, normalized size = 0.00 \[\int \left (\sinh ^{5}\left (f x +e \right )\right ) \left (a +b \left (\sinh ^{2}\left (f x +e \right )\right )\right )^{p}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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(sinh(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*sinh(f*x + e)^5, x)

________________________________________________________________________________________

Fricas [F]
time = 0.43, size = 25, normalized size = 0.11 \begin {gather*} {\rm integral}\left ({\left (b \sinh \left (f x + e\right )^{2} + a\right )}^{p} \sinh \left (f x + e\right )^{5}, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(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*sinh(f*x + e)^5, x)

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

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(sinh(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*sinh(f*x + e)^5, x)

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________