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

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

[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)

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Rubi [A]
time = 0.12, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3253, 525, 524} \begin {gather*} \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} \end {gather*}

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 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*} \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 ) \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 {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*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(240\) vs. \(2(99)=198\).
time = 0.50, size = 240, normalized size = 2.42 \begin {gather*} -\frac {2^{-2-p} \sqrt {\frac {b \cos ^2(e+f x)}{a+b}} (2 a+b-b \cos (2 (e+f x)))^{1+p} \left (2 a (2+p) F_1\left (1+p;\frac {1}{2},\frac {1}{2};2+p;\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 )-(1+p) F_1\left (2+p;\frac {1}{2},\frac {1}{2};3+p;\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 ) (2 a+b-b \cos (2 (e+f x)))\right ) \csc (2 (e+f x)) \sqrt {-\frac {b \sin ^2(e+f x)}{a}}}{b^2 f (1+p) (2+p)} \end {gather*}

Antiderivative was successfully verified.

[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
)))

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Maple [F]
time = 1.03, size = 0, normalized size = 0.00 \[\int \left (\sin ^{2}\left (f x +e \right )\right ) \left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )^{p}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)^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)

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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)^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)

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Fricas [F]
time = 0.40, size = 30, normalized size = 0.30 \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}, x\right ) \end {gather*}

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)

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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)**2*(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)^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)

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

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