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

Optimal. Leaf size=99 \[ \frac{\tan ^3(e+f x) \sec ^2(e+f x)^p \left (a+b \sin ^2(e+f x)\right )^p \left (\frac{(a+b) \tan ^2(e+f x)}{a}+1\right )^{-p} F_1\left (\frac{3}{2};p+2,-p;\frac{5}{2};-\tan ^2(e+f x),-\frac{(a+b) \tan ^2(e+f x)}{a}\right )}{3 f} \]

[Out]

(AppellF1[3/2, 2 + p, -p, 5/2, -Tan[e + f*x]^2, -(((a + b)*Tan[e + f*x]^2)/a)]*(Sec[e + f*x]^2)^p*(a + b*Sin[e
+ f*x]^2)^p*Tan[e + f*x]^3)/(3*f*(1 + ((a + b)*Tan[e + f*x]^2)/a)^p)

________________________________________________________________________________________

Rubi [A]  time = 0.165888, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3174, 511, 510} \[ \frac{\tan ^3(e+f x) \sec ^2(e+f x)^p \left (a+b \sin ^2(e+f x)\right )^p \left (\frac{(a+b) \tan ^2(e+f x)}{a}+1\right )^{-p} F_1\left (\frac{3}{2};p+2,-p;\frac{5}{2};-\tan ^2(e+f x),-\frac{(a+b) \tan ^2(e+f x)}{a}\right )}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(AppellF1[3/2, 2 + p, -p, 5/2, -Tan[e + f*x]^2, -(((a + b)*Tan[e + f*x]^2)/a)]*(Sec[e + f*x]^2)^p*(a + b*Sin[e
+ f*x]^2)^p*Tan[e + f*x]^3)/(3*f*(1 + ((a + b)*Tan[e + f*x]^2)/a)^p)

Rule 3174

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Wit
h[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(ff*(a + b*Sin[e + f*x]^2)^p*(Sec[e + f*x]^2)^p)/(f*(a + (a + b)*T
an[e + f*x]^2)^p), Subst[Int[((a + (a + b)*ff^2*x^2)^p*(A + (A + B)*ff^2*x^2))/(1 + ff^2*x^2)^(p + 2), x], x,
Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, A, B}, x] &&  !IntegerQ[p]

Rule 511

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPa
rt[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^FracPart[p], Int[(e*x)^m*(1 + (b*x^n)/a)^p*(c + d*x^n)^q, x], x
] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] &&  !(IntegerQ[
p] || GtQ[a, 0])

Rule 510

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(a^p*c^q
*(e*x)^(m + 1)*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, -((b*x^n)/a), -((d*x^n)/c)])/(e*(m + 1)), x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rubi steps

\begin{align*} \int \sin ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx &=\frac{\left (\sec ^2(e+f x)^p \left (a+b \sin ^2(e+f x)\right )^p \left (a+(a+b) \tan ^2(e+f x)\right )^{-p}\right ) \operatorname{Subst}\left (\int x^2 \left (1+x^2\right )^{-2-p} \left (a+(a+b) x^2\right )^p \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\left (\sec ^2(e+f x)^p \left (a+b \sin ^2(e+f x)\right )^p \left (1+\frac{(a+b) \tan ^2(e+f x)}{a}\right )^{-p}\right ) \operatorname{Subst}\left (\int x^2 \left (1+x^2\right )^{-2-p} \left (1+\frac{(a+b) x^2}{a}\right )^p \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{F_1\left (\frac{3}{2};2+p,-p;\frac{5}{2};-\tan ^2(e+f x),-\frac{(a+b) \tan ^2(e+f x)}{a}\right ) \sec ^2(e+f x)^p \left (a+b \sin ^2(e+f x)\right )^p \tan ^3(e+f x) \left (1+\frac{(a+b) \tan ^2(e+f x)}{a}\right )^{-p}}{3 f}\\ \end{align*}

Mathematica [B]  time = 0.720841, size = 240, normalized size = 2.42 \[ -\frac{2^{-p-2} \csc (2 (e+f x)) \sqrt{-\frac{b \sin ^2(e+f x)}{a}} \sqrt{\frac{b \cos ^2(e+f x)}{a+b}} (2 a-b \cos (2 (e+f x))+b)^{p+1} \left (2 a (p+2) F_1\left (p+1;\frac{1}{2},\frac{1}{2};p+2;\frac{2 a+b-b \cos (2 (e+f x))}{2 (a+b)},\frac{2 a+b-b \cos (2 (e+f x))}{2 a}\right )-(p+1) (2 a-b \cos (2 (e+f x))+b) F_1\left (p+2;\frac{1}{2},\frac{1}{2};p+3;\frac{2 a+b-b \cos (2 (e+f x))}{2 (a+b)},\frac{2 a+b-b \cos (2 (e+f x))}{2 a}\right )\right )}{b^2 f (p+1) (p+2)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [F]  time = 1.733, size = 0, normalized size = 0. \begin{align*} \int \left ( \sin \left ( fx+e \right ) \right ) ^{2} \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)^2*(a+b*sin(f*x+e)^2)^p,x)

[Out]

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

________________________________________________________________________________________

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 )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^2*(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)^2, x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^2*(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, x)

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

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 )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^2*(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)^2, x)