3.824 \(\int \frac{(c (d \sin (e+f x))^p)^n}{(a+a \sin (e+f x))^2} \, dx\)

Optimal. Leaf size=288 \[ \frac{2 \left (1-n^2 p^2\right ) \sin ^2(e+f x) \cos (e+f x) \, _2F_1\left (\frac{1}{2},\frac{1}{2} (n p+2);\frac{1}{2} (n p+4);\sin ^2(e+f x)\right ) \left (c (d \sin (e+f x))^p\right )^n}{3 a^2 f (n p+2) \sqrt{\cos ^2(e+f x)}}-\frac{n p (1-2 n p) \sin (e+f x) \cos (e+f x) \, _2F_1\left (\frac{1}{2},\frac{1}{2} (n p+1);\frac{1}{2} (n p+3);\sin ^2(e+f x)\right ) \left (c (d \sin (e+f x))^p\right )^n}{3 a^2 f (n p+1) \sqrt{\cos ^2(e+f x)}}+\frac{2 (1-n p) \sin (e+f x) \cos (e+f x) \left (c (d \sin (e+f x))^p\right )^n}{3 a^2 f (\sin (e+f x)+1)}+\frac{\sin (e+f x) \cos (e+f x) \left (c (d \sin (e+f x))^p\right )^n}{3 f (a \sin (e+f x)+a)^2} \]

[Out]

-(n*p*(1 - 2*n*p)*Cos[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)/(3*a^2*f*(1 + n*p)*Sqrt[Cos[e + f*x]^2]) + (2*(1 - n^2*p^2)*Cos[e + f*x]*Hypergeometr
ic2F1[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)/(3*a^2*f*(2 + n*
p)*Sqrt[Cos[e + f*x]^2]) + (2*(1 - n*p)*Cos[e + f*x]*Sin[e + f*x]*(c*(d*Sin[e + f*x])^p)^n)/(3*a^2*f*(1 + Sin[
e + f*x])) + (Cos[e + f*x]*Sin[e + f*x]*(c*(d*Sin[e + f*x])^p)^n)/(3*f*(a + a*Sin[e + f*x])^2)

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Rubi [A]  time = 0.486528, antiderivative size = 288, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {2826, 2766, 2978, 2748, 2643} \[ \frac{2 \left (1-n^2 p^2\right ) \sin ^2(e+f x) \cos (e+f x) \, _2F_1\left (\frac{1}{2},\frac{1}{2} (n p+2);\frac{1}{2} (n p+4);\sin ^2(e+f x)\right ) \left (c (d \sin (e+f x))^p\right )^n}{3 a^2 f (n p+2) \sqrt{\cos ^2(e+f x)}}-\frac{n p (1-2 n p) \sin (e+f x) \cos (e+f x) \, _2F_1\left (\frac{1}{2},\frac{1}{2} (n p+1);\frac{1}{2} (n p+3);\sin ^2(e+f x)\right ) \left (c (d \sin (e+f x))^p\right )^n}{3 a^2 f (n p+1) \sqrt{\cos ^2(e+f x)}}+\frac{2 (1-n p) \sin (e+f x) \cos (e+f x) \left (c (d \sin (e+f x))^p\right )^n}{3 a^2 f (\sin (e+f x)+1)}+\frac{\sin (e+f x) \cos (e+f x) \left (c (d \sin (e+f x))^p\right )^n}{3 f (a \sin (e+f x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(n*p*(1 - 2*n*p)*Cos[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)/(3*a^2*f*(1 + n*p)*Sqrt[Cos[e + f*x]^2]) + (2*(1 - n^2*p^2)*Cos[e + f*x]*Hypergeometr
ic2F1[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)/(3*a^2*f*(2 + n*
p)*Sqrt[Cos[e + f*x]^2]) + (2*(1 - n*p)*Cos[e + f*x]*Sin[e + f*x]*(c*(d*Sin[e + f*x])^p)^n)/(3*a^2*f*(1 + Sin[
e + f*x])) + (Cos[e + f*x]*Sin[e + f*x]*(c*(d*Sin[e + f*x])^p)^n)/(3*f*(a + a*Sin[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 2766

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(a*f*(2*m + 1)*(b*c - a*d)), x] + Dis
t[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[b*c*(m + 1) - a*d*
(2*m + n + 2) + b*d*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d,
0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] &&  !GtQ[n, 0] && (IntegersQ[2*m, 2*n] || (IntegerQ
[m] && EqQ[c, 0]))

Rule 2978

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*
x])^(n + 1))/(a*f*(2*m + 1)*(b*c - a*d)), x] + Dist[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*
(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c,
0])

Rule 2748

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

Rubi steps

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

Mathematica [A]  time = 2.86747, size = 195, normalized size = 0.68 \[ \frac{\sin (e+f x) \cos (e+f x) \left (c (d \sin (e+f x))^p\right )^n \left (-\frac{2 \left (n^2 p^2-1\right ) \sqrt{\cos ^2(e+f x)} \tan (e+f x) \sec (e+f x) \, _2F_1\left (\frac{1}{2},\frac{n p}{2}+1;\frac{n p}{2}+2;\sin ^2(e+f x)\right )}{n p+2}+\frac{n p (2 n p-1) \, _2F_1\left (\frac{1}{2},\frac{1}{2} (n p+1);\frac{1}{2} (n p+3);\sin ^2(e+f x)\right )}{(n p+1) \sqrt{\cos ^2(e+f x)}}+\frac{(2-2 n p) \sin (e+f x)-2 n p+3}{(\sin (e+f x)+1)^2}\right )}{3 a^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.349, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( c \left ( d\sin \left ( fx+e \right ) \right ) ^{p} \right ) ^{n}}{ \left ( a+a\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+a*sin(f*x+e))^2,x)

[Out]

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

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

[Out]

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

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\left (c \left (d \sin{\left (e + f x \right )}\right )^{p}\right )^{n}}{\sin ^{2}{\left (e + f x \right )} + 2 \sin{\left (e + f x \right )} + 1}\, dx}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((c*(d*sin(e + f*x))**p)**n/(sin(e + f*x)**2 + 2*sin(e + f*x) + 1), x)/a**2

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

[Out]

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