3.172 \(\int \sin ^5(e+f x) (a+b \sin ^2(e+f x))^p \, dx\)

Optimal. Leaf size=220 \[ -\frac{\left (3 a^2-4 a b (p+1)+4 b^2 \left (p^2+3 p+2\right )\right ) \cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^p \left (1-\frac{b \cos ^2(e+f x)}{a+b}\right )^{-p} \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};\frac{b \cos ^2(e+f x)}{a+b}\right )}{b^2 f (2 p+3) (2 p+5)}+\frac{(3 a-2 b (p+2)) \cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^{p+1}}{b^2 f (2 p+3) (2 p+5)}-\frac{\sin ^2(e+f x) \cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^{p+1}}{b f (2 p+5)} \]

[Out]

((3*a - 2*b*(2 + p))*Cos[e + f*x]*(a + b - b*Cos[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))*Cos[e + f*x]*(a + b - b*Cos[e + f*x]^2)^p*Hypergeometric2F1[1/2, -p, 3/
2, (b*Cos[e + f*x]^2)/(a + b)])/(b^2*f*(3 + 2*p)*(5 + 2*p)*(1 - (b*Cos[e + f*x]^2)/(a + b))^p) - (Cos[e + f*x]
*(a + b - b*Cos[e + f*x]^2)^(1 + p)*Sin[e + f*x]^2)/(b*f*(5 + 2*p))

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Rubi [A]  time = 0.224846, antiderivative size = 220, normalized size of antiderivative = 1., 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 = {3186, 416, 388, 246, 245} \[ -\frac{\left (3 a^2-4 a b (p+1)+4 b^2 \left (p^2+3 p+2\right )\right ) \cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^p \left (1-\frac{b \cos ^2(e+f x)}{a+b}\right )^{-p} \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};\frac{b \cos ^2(e+f x)}{a+b}\right )}{b^2 f (2 p+3) (2 p+5)}+\frac{(3 a-2 b (p+2)) \cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^{p+1}}{b^2 f (2 p+3) (2 p+5)}-\frac{\sin ^2(e+f x) \cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^{p+1}}{b f (2 p+5)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*a - 2*b*(2 + p))*Cos[e + f*x]*(a + b - b*Cos[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))*Cos[e + f*x]*(a + b - b*Cos[e + f*x]^2)^p*Hypergeometric2F1[1/2, -p, 3/
2, (b*Cos[e + f*x]^2)/(a + b)])/(b^2*f*(3 + 2*p)*(5 + 2*p)*(1 - (b*Cos[e + f*x]^2)/(a + b))^p) - (Cos[e + f*x]
*(a + b - b*Cos[e + f*x]^2)^(1 + p)*Sin[e + f*x]^2)/(b*f*(5 + 2*p))

Rule 3186

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]

Rule 416

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

Mathematica [C]  time = 0.546825, size = 98, normalized size = 0.45 \[ \frac{\sin ^5(e+f x) \sqrt{\cos ^2(e+f x)} \tan (e+f x) \left (a+b \sin ^2(e+f x)\right )^p \left (\frac{a+b \sin ^2(e+f x)}{a}\right )^{-p} F_1\left (3;\frac{1}{2},-p;4;\sin ^2(e+f x),-\frac{b \sin ^2(e+f x)}{a}\right )}{6 f} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(AppellF1[3, 1/2, -p, 4, Sin[e + f*x]^2, -((b*Sin[e + f*x]^2)/a)]*Sqrt[Cos[e + f*x]^2]*Sin[e + f*x]^5*(a + b*S
in[e + f*x]^2)^p*Tan[e + f*x])/(6*f*((a + b*Sin[e + f*x]^2)/a)^p)

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Maple [F]  time = 1.431, size = 0, normalized size = 0. \begin{align*} \int \left ( \sin \left ( fx+e \right ) \right ) ^{5} \left ( a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) ^{p}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{p} \sin \left (f x + e\right )^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^5*(a+b*sin(f*x+e)^2)^p,x, algorithm="maxima")

[Out]

integrate((b*sin(f*x + e)^2 + a)^p*sin(f*x + e)^5, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (\cos \left (f x + e\right )^{4} - 2 \, \cos \left (f x + e\right )^{2} + 1\right )}{\left (-b \cos \left (f x + e\right )^{2} + a + b\right )}^{p} \sin \left (f x + e\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^5*(a+b*sin(f*x+e)^2)^p,x, algorithm="fricas")

[Out]

integral((cos(f*x + e)^4 - 2*cos(f*x + e)^2 + 1)*(-b*cos(f*x + e)^2 + a + b)^p*sin(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(sin(f*x+e)**5*(a+b*sin(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 \sin \left (f x + e\right )^{2} + a\right )}^{p} \sin \left (f x + e\right )^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^5*(a+b*sin(f*x+e)^2)^p,x, algorithm="giac")

[Out]

integrate((b*sin(f*x + e)^2 + a)^p*sin(f*x + e)^5, x)