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

Optimal. Leaf size=122 \[ -\frac{d \cos (e+f x) \sin ^2(e+f x)^{\frac{1-m}{2}} (d \sin (e+f x))^{m-1} \left (a-b \cos ^2(e+f x)+b\right )^p \left (1-\frac{b \cos ^2(e+f x)}{a+b}\right )^{-p} F_1\left (\frac{1}{2};\frac{1-m}{2},-p;\frac{3}{2};\cos ^2(e+f x),\frac{b \cos ^2(e+f x)}{a+b}\right )}{f} \]

[Out]

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

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Rubi [A]  time = 0.115666, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {3189, 430, 429} \[ -\frac{d \cos (e+f x) \sin ^2(e+f x)^{\frac{1-m}{2}} (d \sin (e+f x))^{m-1} \left (a-b \cos ^2(e+f x)+b\right )^p \left (1-\frac{b \cos ^2(e+f x)}{a+b}\right )^{-p} F_1\left (\frac{1}{2};\frac{1-m}{2},-p;\frac{3}{2};\cos ^2(e+f x),\frac{b \cos ^2(e+f x)}{a+b}\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3189

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff
= FreeFactors[Cos[e + f*x], x]}, -Dist[(ff*d^(2*IntPart[(m - 1)/2] + 1)*(d*Sin[e + f*x])^(2*FracPart[(m - 1)/
2]))/(f*(Sin[e + f*x]^2)^FracPart[(m - 1)/2]), 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, d, e, f, m, p}, x] &&  !IntegerQ[m]

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

Mathematica [A]  time = 0.44214, size = 113, normalized size = 0.93 \[ \frac{\sqrt{\cos ^2(e+f x)} \tan (e+f x) (d \sin (e+f x))^m \left (a+b \sin ^2(e+f x)\right )^p \left (\frac{b \sin ^2(e+f x)}{a}+1\right )^{-p} F_1\left (\frac{m+1}{2};\frac{1}{2},-p;\frac{m+3}{2};\sin ^2(e+f x),-\frac{b \sin ^2(e+f x)}{a}\right )}{f (m+1)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [F]  time = 1.32, size = 0, normalized size = 0. \begin{align*} \int \left ( d\sin \left ( fx+e \right ) \right ) ^{m} \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((d*sin(f*x+e))^m*(a+b*sin(f*x+e)^2)^p,x)

[Out]

int((d*sin(f*x+e))^m*(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} \left (d \sin \left (f x + e\right )\right )^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*sin(f*x + e)^2 + a)^p*(d*sin(f*x + e))^m, 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} \left (d \sin \left (f x + e\right )\right )^{m}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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