3.2.73 \(\int \sin ^3(e+f x) (a+b \sin ^2(e+f x))^p \, dx\) [173]

Optimal. Leaf size=131 \[ -\frac {\cos (e+f x) \left (a+b-b \cos ^2(e+f x)\right )^{1+p}}{b f (3+2 p)}+\frac {(a-2 b (1+p)) \cos (e+f x) \left (a+b-b \cos ^2(e+f x)\right )^p \left (1-\frac {b \cos ^2(e+f x)}{a+b}\right )^{-p} \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};\frac {b \cos ^2(e+f x)}{a+b}\right )}{b f (3+2 p)} \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 131, normalized size of antiderivative = 1.00, 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 = {3265, 396, 252, 251} \begin {gather*} \frac {(a-2 b (p+1)) \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 f (2 p+3)}-\frac {\cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^{p+1}}{b f (2 p+3)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((Cos[e + f*x]*(a + b - b*Cos[e + f*x]^2)^(1 + p))/(b*f*(3 + 2*p))) + ((a - 2*b*(1 + p))*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*f*(3 + 2*p)*(1 - (b*Cos[
e + f*x]^2)/(a + b))^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 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 \sin ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx &=-\frac {\text {Subst}\left (\int \left (1-x^2\right ) \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}}{b f (3+2 p)}+\frac {(a-2 b (1+p)) \text {Subst}\left (\int \left (a+b-b x^2\right )^p \, dx,x,\cos (e+f x)\right )}{b f (3+2 p)}\\ &=-\frac {\cos (e+f x) \left (a+b-b \cos ^2(e+f x)\right )^{1+p}}{b f (3+2 p)}+\frac {\left ((a-2 b (1+p)) \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 ) \text {Subst}\left (\int \left (1-\frac {b x^2}{a+b}\right )^p \, dx,x,\cos (e+f x)\right )}{b f (3+2 p)}\\ &=-\frac {\cos (e+f x) \left (a+b-b \cos ^2(e+f x)\right )^{1+p}}{b f (3+2 p)}+\frac {(a-2 b (1+p)) \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 f (3+2 p)}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
time = 0.27, size = 98, normalized size = 0.75 \begin {gather*} \frac {F_1\left (2;\frac {1}{2},-p;3;\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right ) \sqrt {\cos ^2(e+f x)} \sin ^3(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} \tan (e+f x)}{4 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.65, size = 0, normalized size = 0.00 \[\int \left (\sin ^{3}\left (f x +e \right )\right ) \left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )^{p}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(sin(f*x+e)^3*(a+b*sin(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(sin(f*x+e)^3*(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)^3, x)

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Fricas [F]
time = 0.43, size = 36, normalized size = 0.27 \begin {gather*} {\rm integral}\left (-{\left (\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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^3*(a+b*sin(f*x+e)^2)^p,x, algorithm="fricas")

[Out]

integral(-(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)] 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(sin(f*x+e)**3*(a+b*sin(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(sin(f*x+e)^3*(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)^3, x)

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

[Out]

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

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