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

Optimal. Leaf size=90 \[ \frac{\sqrt{\cos ^2(e+f x)} \tan (e+f x) \left (a+b \sin ^2(e+f x)\right )^p \left (\frac{b \sin ^2(e+f x)}{a}+1\right )^{-p} F_1\left (\frac{1}{2};\frac{1}{2},-p;\frac{3}{2};\sin ^2(e+f x),-\frac{b \sin ^2(e+f x)}{a}\right )}{f} \]

[Out]

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

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Rubi [A]  time = 0.052151, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {3185, 430, 429} \[ \frac{\sqrt{\cos ^2(e+f x)} \tan (e+f x) \left (a+b \sin ^2(e+f x)\right )^p \left (\frac{b \sin ^2(e+f x)}{a}+1\right )^{-p} F_1\left (\frac{1}{2};\frac{1}{2},-p;\frac{3}{2};\sin ^2(e+f x),-\frac{b \sin ^2(e+f x)}{a}\right )}{f} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[e + f*x]^2)^p,x]

[Out]

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

Rule 3185

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Dis
t[(ff*Sqrt[Cos[e + f*x]^2])/(f*Cos[e + f*x]), Subst[Int[(a + b*ff^2*x^2)^p/Sqrt[1 - ff^2*x^2], x], x, Sin[e +
f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] &&  !IntegerQ[p]

Rule 430

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

Rule 429

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

Rubi steps

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

Mathematica [A]  time = 0.522443, size = 145, normalized size = 1.61 \[ \frac{2^{-p-1} \csc (2 (e+f x)) \sqrt{-\frac{b \sin ^2(e+f x)}{a}} \sqrt{\frac{b \cos ^2(e+f x)}{a+b}} (2 a-b \cos (2 (e+f x))+b)^{p+1} F_1\left (p+1;\frac{1}{2},\frac{1}{2};p+2;\frac{2 a+b-b \cos (2 (e+f x))}{2 (a+b)},\frac{2 a+b-b \cos (2 (e+f x))}{2 a}\right )}{b f (p+1)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*Sin[e + f*x]^2)^p,x]

[Out]

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

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Maple [F]  time = 0.546, size = 0, normalized size = 0. \begin{align*} \int \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((a+b*sin(f*x+e)^2)^p,x)

[Out]

int((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}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\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((a+b*sin(f*x+e)^2)^p,x, algorithm="fricas")

[Out]

integral((-b*cos(f*x + e)^2 + a + b)^p, 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((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}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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