3.833 \(\int (c (d \sin (e+f x))^p)^n (a+b \sin (e+f x))^2 \, dx\)

Optimal. Leaf size=231 \[ \frac{\left (a^2 (n p+2)+b^2 (n p+1)\right ) \sin (e+f x) \cos (e+f x) \, _2F_1\left (\frac{1}{2},\frac{1}{2} (n p+1);\frac{1}{2} (n p+3);\sin ^2(e+f x)\right ) \left (c (d \sin (e+f x))^p\right )^n}{f (n p+1) (n p+2) \sqrt{\cos ^2(e+f x)}}+\frac{2 a b \sin ^2(e+f x) \cos (e+f x) \, _2F_1\left (\frac{1}{2},\frac{1}{2} (n p+2);\frac{1}{2} (n p+4);\sin ^2(e+f x)\right ) \left (c (d \sin (e+f x))^p\right )^n}{f (n p+2) \sqrt{\cos ^2(e+f x)}}-\frac{b^2 \sin (e+f x) \cos (e+f x) \left (c (d \sin (e+f x))^p\right )^n}{f (n p+2)} \]

[Out]

-((b^2*Cos[e + f*x]*Sin[e + f*x]*(c*(d*Sin[e + f*x])^p)^n)/(f*(2 + n*p))) + ((b^2*(1 + n*p) + a^2*(2 + n*p))*C
os[e + f*x]*Hypergeometric2F1[1/2, (1 + n*p)/2, (3 + n*p)/2, Sin[e + f*x]^2]*Sin[e + f*x]*(c*(d*Sin[e + f*x])^
p)^n)/(f*(1 + n*p)*(2 + n*p)*Sqrt[Cos[e + f*x]^2]) + (2*a*b*Cos[e + f*x]*Hypergeometric2F1[1/2, (2 + n*p)/2, (
4 + n*p)/2, Sin[e + f*x]^2]*Sin[e + f*x]^2*(c*(d*Sin[e + f*x])^p)^n)/(f*(2 + n*p)*Sqrt[Cos[e + f*x]^2])

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Rubi [A]  time = 0.232943, antiderivative size = 221, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {2826, 2789, 2643, 3014} \[ \frac{\left (\frac{a^2}{n p+1}+\frac{b^2}{n p+2}\right ) \sin (e+f x) \cos (e+f x) \, _2F_1\left (\frac{1}{2},\frac{1}{2} (n p+1);\frac{1}{2} (n p+3);\sin ^2(e+f x)\right ) \left (c (d \sin (e+f x))^p\right )^n}{f \sqrt{\cos ^2(e+f x)}}+\frac{2 a b \sin ^2(e+f x) \cos (e+f x) \, _2F_1\left (\frac{1}{2},\frac{1}{2} (n p+2);\frac{1}{2} (n p+4);\sin ^2(e+f x)\right ) \left (c (d \sin (e+f x))^p\right )^n}{f (n p+2) \sqrt{\cos ^2(e+f x)}}-\frac{b^2 \sin (e+f x) \cos (e+f x) \left (c (d \sin (e+f x))^p\right )^n}{f (n p+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b^2*Cos[e + f*x]*Sin[e + f*x]*(c*(d*Sin[e + f*x])^p)^n)/(f*(2 + n*p))) + ((a^2/(1 + n*p) + b^2/(2 + n*p))*C
os[e + f*x]*Hypergeometric2F1[1/2, (1 + n*p)/2, (3 + n*p)/2, Sin[e + f*x]^2]*Sin[e + f*x]*(c*(d*Sin[e + f*x])^
p)^n)/(f*Sqrt[Cos[e + f*x]^2]) + (2*a*b*Cos[e + f*x]*Hypergeometric2F1[1/2, (2 + n*p)/2, (4 + n*p)/2, Sin[e +
f*x]^2]*Sin[e + f*x]^2*(c*(d*Sin[e + f*x])^p)^n)/(f*(2 + n*p)*Sqrt[Cos[e + f*x]^2])

Rule 2826

Int[((c_.)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(p_))^(n_)*((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol]
:> Dist[(c^IntPart[n]*(c*(d*Sin[e + f*x])^p)^FracPart[n])/(d*Sin[e + f*x])^(p*FracPart[n]), Int[(a + b*Sin[e
+ f*x])^m*(d*Sin[e + f*x])^(n*p), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[n]

Rule 2789

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Dist[(2*c*d)/b
, Int[(b*Sin[e + f*x])^(m + 1), x], x] + Int[(b*Sin[e + f*x])^m*(c^2 + d^2*Sin[e + f*x]^2), x] /; FreeQ[{b, c,
d, e, f, m}, x]

Rule 2643

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1)*Hypergeomet
ric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2])/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]), x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 3014

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[
e + f*x]*(b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[(A*(m + 2) + C*(m + 1))/(m + 2), Int[(b*Sin[e + f*
x])^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] &&  !LtQ[m, -1]

Rubi steps

\begin{align*} \int \left (c (d \sin (e+f x))^p\right )^n (a+b \sin (e+f x))^2 \, dx &=\left ((d \sin (e+f x))^{-n p} \left (c (d \sin (e+f x))^p\right )^n\right ) \int (d \sin (e+f x))^{n p} (a+b \sin (e+f x))^2 \, dx\\ &=\left ((d \sin (e+f x))^{-n p} \left (c (d \sin (e+f x))^p\right )^n\right ) \int (d \sin (e+f x))^{n p} \left (a^2+b^2 \sin ^2(e+f x)\right ) \, dx+\frac{\left (2 a b (d \sin (e+f x))^{-n p} \left (c (d \sin (e+f x))^p\right )^n\right ) \int (d \sin (e+f x))^{1+n p} \, dx}{d}\\ &=-\frac{b^2 \cos (e+f x) \sin (e+f x) \left (c (d \sin (e+f x))^p\right )^n}{f (2+n p)}+\frac{2 a b \cos (e+f x) \, _2F_1\left (\frac{1}{2},\frac{1}{2} (2+n p);\frac{1}{2} (4+n p);\sin ^2(e+f x)\right ) \sin ^2(e+f x) \left (c (d \sin (e+f x))^p\right )^n}{f (2+n p) \sqrt{\cos ^2(e+f x)}}+\left (\left (a^2+\frac{b^2 (1+n p)}{2+n p}\right ) (d \sin (e+f x))^{-n p} \left (c (d \sin (e+f x))^p\right )^n\right ) \int (d \sin (e+f x))^{n p} \, dx\\ &=-\frac{b^2 \cos (e+f x) \sin (e+f x) \left (c (d \sin (e+f x))^p\right )^n}{f (2+n p)}+\frac{\left (a^2+\frac{b^2 (1+n p)}{2+n p}\right ) \cos (e+f x) \, _2F_1\left (\frac{1}{2},\frac{1}{2} (1+n p);\frac{1}{2} (3+n p);\sin ^2(e+f x)\right ) \sin (e+f x) \left (c (d \sin (e+f x))^p\right )^n}{f (1+n p) \sqrt{\cos ^2(e+f x)}}+\frac{2 a b \cos (e+f x) \, _2F_1\left (\frac{1}{2},\frac{1}{2} (2+n p);\frac{1}{2} (4+n p);\sin ^2(e+f x)\right ) \sin ^2(e+f x) \left (c (d \sin (e+f x))^p\right )^n}{f (2+n p) \sqrt{\cos ^2(e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.331043, size = 152, normalized size = 0.66 \[ -\frac{\cos (e+f x) \sin ^2(e+f x)^{\frac{1}{2} (-n p-1)} \left (c (d \sin (e+f x))^p\right )^n \left (a \left (a \sin (e+f x) \, _2F_1\left (\frac{1}{2},\frac{1}{2} (1-n p);\frac{3}{2};\cos ^2(e+f x)\right )+2 b \sqrt{\sin ^2(e+f x)} \, _2F_1\left (\frac{1}{2},-\frac{n p}{2};\frac{3}{2};\cos ^2(e+f x)\right )\right )+b^2 \sin (e+f x) \, _2F_1\left (\frac{1}{2},\frac{1}{2} (-n p-1);\frac{3}{2};\cos ^2(e+f x)\right )\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Cos[e + f*x]*(Sin[e + f*x]^2)^((-1 - n*p)/2)*(c*(d*Sin[e + f*x])^p)^n*(b^2*Hypergeometric2F1[1/2, (-1 - n*p
)/2, 3/2, Cos[e + f*x]^2]*Sin[e + f*x] + a*(a*Hypergeometric2F1[1/2, (1 - n*p)/2, 3/2, Cos[e + f*x]^2]*Sin[e +
f*x] + 2*b*Hypergeometric2F1[1/2, -(n*p)/2, 3/2, Cos[e + f*x]^2]*Sqrt[Sin[e + f*x]^2])))/f)

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Maple [F]  time = 0.382, size = 0, normalized size = 0. \begin{align*} \int \left ( c \left ( d\sin \left ( fx+e \right ) \right ) ^{p} \right ) ^{n} \left ( a+b\sin \left ( fx+e \right ) \right ) ^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-{\left (b^{2} \cos \left (f x + e\right )^{2} - 2 \, a b \sin \left (f x + e\right ) - a^{2} - b^{2}\right )} \left (\left (d \sin \left (f x + e\right )\right )^{p} c\right )^{n}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(b^2*cos(f*x + e)^2 - 2*a*b*sin(f*x + e) - a^2 - b^2)*((d*sin(f*x + e))^p*c)^n, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (c \left (d \sin{\left (e + f x \right )}\right )^{p}\right )^{n} \left (a + b \sin{\left (e + f x \right )}\right )^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((c*(d*sin(e + f*x))**p)**n*(a + b*sin(e + f*x))**2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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