\(\int \sin ^2(e+f x) (a+b \sin ^2(e+f x))^p \, dx\) [179]

Optimal result

Integrand size = 23, antiderivative size = 99 \[ \int \sin ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\frac {\operatorname {AppellF1}\left (\frac {3}{2},2+p,-p,\frac {5}{2},-\tan ^2(e+f x),-\frac {(a+b) \tan ^2(e+f x)}{a}\right ) \sec ^2(e+f x)^p \left (a+b \sin ^2(e+f x)\right )^p \tan ^3(e+f x) \left (1+\frac {(a+b) \tan ^2(e+f x)}{a}\right )^{-p}}{3 f} \]

[Out]

1/3*AppellF1(3/2,2+p,-p,5/2,-tan(f*x+e)^2,-(a+b)*tan(f*x+e)^2/a)*(sec(f*x+e)^2)^p*(a+b*sin(f*x+e)^2)^p*tan(f*x
+e)^3/f/((1+(a+b)*tan(f*x+e)^2/a)^p)

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3253, 525, 524} \[ \int \sin ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\frac {\tan ^3(e+f x) \sec ^2(e+f x)^p \left (a+b \sin ^2(e+f x)\right )^p \left (\frac {(a+b) \tan ^2(e+f x)}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {3}{2},p+2,-p,\frac {5}{2},-\tan ^2(e+f x),-\frac {(a+b) \tan ^2(e+f x)}{a}\right )}{3 f} \]

[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 524

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)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], 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])

Rule 525

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[a^IntPar
t[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 3253

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)*Ta
n[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]

Rubi steps \begin{align*} \text {integral}& = \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 ) \text {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 ) \text {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 {\operatorname {AppellF1}\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 [A] (verified)

Time = 0.32 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.03 \[ \int \sin ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\frac {\operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{2},-p,\frac {5}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right ) \sqrt {\cos ^2(e+f x)} \sin ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \left (\frac {a+b \sin ^2(e+f x)}{a}\right )^{-p} \tan (e+f x)}{3 f} \]

[In]

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

[Out]

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

Maple [F]

\[\int \left (\sin ^{2}\left (f x +e \right )\right ) {\left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )}^{p}d x\]

[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)

Fricas [F]

\[ \int \sin ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{p} \sin \left (f x + e\right )^{2} \,d x } \]

[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)]

Timed out. \[ \int \sin ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \sin ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{p} \sin \left (f x + e\right )^{2} \,d x } \]

[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)

Giac [F]

\[ \int \sin ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{p} \sin \left (f x + e\right )^{2} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \sin ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int {\sin \left (e+f\,x\right )}^2\,{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^p \,d x \]

[In]

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

[Out]

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