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

Optimal. Leaf size=322 \[ \frac {2 a b \cos (e+f x) \sin ^2(e+f x)^{-\frac {n p}{2}} \left (c (d \sin (e+f x))^p\right )^n F_1\left (\frac {1}{2};-\frac {n p}{2},2;\frac {3}{2};\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^2}-\frac {b^2 \sin (e+f x) \cos (e+f x) \sin ^2(e+f x)^{\frac {1}{2} (-n p-1)} \left (c (d \sin (e+f x))^p\right )^n F_1\left (\frac {1}{2};\frac {1}{2} (-n p-1),2;\frac {3}{2};\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^2}-\frac {a^2 \cot (e+f x) \sin ^2(e+f x)^{\frac {1}{2} (1-n p)} \left (c (d \sin (e+f x))^p\right )^n F_1\left (\frac {1}{2};\frac {1}{2} (1-n p),2;\frac {3}{2};\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^2} \]

[Out]

2*a*b*AppellF1(1/2,-1/2*n*p,2,3/2,cos(f*x+e)^2,-b^2*cos(f*x+e)^2/(a^2-b^2))*cos(f*x+e)*(c*(d*sin(f*x+e))^p)^n/
(a^2-b^2)^2/f/((sin(f*x+e)^2)^(1/2*n*p))-b^2*AppellF1(1/2,-1/2*n*p-1/2,2,3/2,cos(f*x+e)^2,-b^2*cos(f*x+e)^2/(a
^2-b^2))*cos(f*x+e)*sin(f*x+e)*(sin(f*x+e)^2)^(-1/2*n*p-1/2)*(c*(d*sin(f*x+e))^p)^n/(a^2-b^2)^2/f-a^2*AppellF1
(1/2,-1/2*n*p+1/2,2,3/2,cos(f*x+e)^2,-b^2*cos(f*x+e)^2/(a^2-b^2))*cot(f*x+e)*(sin(f*x+e)^2)^(-1/2*n*p+1/2)*(c*
(d*sin(f*x+e))^p)^n/(a^2-b^2)^2/f

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Rubi [A]  time = 0.51, antiderivative size = 322, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2826, 2824, 3189, 429, 16} \[ \frac {2 a b \cos (e+f x) \sin ^2(e+f x)^{-\frac {n p}{2}} \left (c (d \sin (e+f x))^p\right )^n F_1\left (\frac {1}{2};-\frac {n p}{2},2;\frac {3}{2};\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^2}-\frac {b^2 \sin (e+f x) \cos (e+f x) \sin ^2(e+f x)^{\frac {1}{2} (-n p-1)} \left (c (d \sin (e+f x))^p\right )^n F_1\left (\frac {1}{2};\frac {1}{2} (-n p-1),2;\frac {3}{2};\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^2}-\frac {a^2 \cot (e+f x) \sin ^2(e+f x)^{\frac {1}{2} (1-n p)} \left (c (d \sin (e+f x))^p\right )^n F_1\left (\frac {1}{2};\frac {1}{2} (1-n p),2;\frac {3}{2};\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*b*AppellF1[1/2, -(n*p)/2, 2, 3/2, Cos[e + f*x]^2, -((b^2*Cos[e + f*x]^2)/(a^2 - b^2))]*Cos[e + f*x]*(c*(d
*Sin[e + f*x])^p)^n)/((a^2 - b^2)^2*f*(Sin[e + f*x]^2)^((n*p)/2)) - (b^2*AppellF1[1/2, (-1 - n*p)/2, 2, 3/2, C
os[e + f*x]^2, -((b^2*Cos[e + f*x]^2)/(a^2 - b^2))]*Cos[e + f*x]*Sin[e + f*x]*(Sin[e + f*x]^2)^((-1 - n*p)/2)*
(c*(d*Sin[e + f*x])^p)^n)/((a^2 - b^2)^2*f) - (a^2*AppellF1[1/2, (1 - n*p)/2, 2, 3/2, Cos[e + f*x]^2, -((b^2*C
os[e + f*x]^2)/(a^2 - b^2))]*Cot[e + f*x]*(Sin[e + f*x]^2)^((1 - n*p)/2)*(c*(d*Sin[e + f*x])^p)^n)/((a^2 - b^2
)^2*f)

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

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])

Rule 2824

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Int[Expan
dTrig[(d*sin[e + f*x])^n/((a - b*sin[e + f*x])^m/(a^2 - b^2*sin[e + f*x]^2)^m), x], x] /; FreeQ[{a, b, d, e, f
, n}, x] && NeQ[a^2 - b^2, 0] && ILtQ[m, -1]

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 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]

Rubi steps

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

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Mathematica [B]  time = 19.02, size = 1970, normalized size = 6.12 \[ \text {result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(((Sec[e + f*x]^2)^((n*p)/2)*(c*(d*Sin[e + f*x])^p)^n*Tan[e + f*x]*(Tan[e + f*x]/Sqrt[Sec[e + f*x]^2])^(n*p)*
(-(a*(2 + n*p)*((a^2 + b^2)*AppellF1[(1 + n*p)/2, (n*p)/2, 1, (3 + n*p)/2, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan
[e + f*x]^2] - 2*b^2*AppellF1[(1 + n*p)/2, (n*p)/2, 2, (3 + n*p)/2, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*
x]^2])) + 2*b*(a^2 - b^2)*(1 + n*p)*AppellF1[1 + (n*p)/2, (-1 + n*p)/2, 2, 2 + (n*p)/2, -Tan[e + f*x]^2, (-1 +
b^2/a^2)*Tan[e + f*x]^2]*Tan[e + f*x]))/(a^3*(a^2 - b^2)*f*(1 + n*p)*(2 + n*p)*(a + b*Sin[e + f*x])^2*(-(((Se
c[e + f*x]^2)^(1 + (n*p)/2)*(Tan[e + f*x]/Sqrt[Sec[e + f*x]^2])^(n*p)*(-(a*(2 + n*p)*((a^2 + b^2)*AppellF1[(1
+ n*p)/2, (n*p)/2, 1, (3 + n*p)/2, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^2] - 2*b^2*AppellF1[(1 + n*p)/
2, (n*p)/2, 2, (3 + n*p)/2, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^2])) + 2*b*(a^2 - b^2)*(1 + n*p)*Appe
llF1[1 + (n*p)/2, (-1 + n*p)/2, 2, 2 + (n*p)/2, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^2]*Tan[e + f*x]))
/(a^3*(a^2 - b^2)*(1 + n*p)*(2 + n*p))) - (n*p*(Sec[e + f*x]^2)^((n*p)/2)*Tan[e + f*x]^2*(Tan[e + f*x]/Sqrt[Se
c[e + f*x]^2])^(n*p)*(-(a*(2 + n*p)*((a^2 + b^2)*AppellF1[(1 + n*p)/2, (n*p)/2, 1, (3 + n*p)/2, -Tan[e + f*x]^
2, (-1 + b^2/a^2)*Tan[e + f*x]^2] - 2*b^2*AppellF1[(1 + n*p)/2, (n*p)/2, 2, (3 + n*p)/2, -Tan[e + f*x]^2, (-1
+ b^2/a^2)*Tan[e + f*x]^2])) + 2*b*(a^2 - b^2)*(1 + n*p)*AppellF1[1 + (n*p)/2, (-1 + n*p)/2, 2, 2 + (n*p)/2, -
Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^2]*Tan[e + f*x]))/(a^3*(a^2 - b^2)*(1 + n*p)*(2 + n*p)) - (n*p*(Se
c[e + f*x]^2)^((n*p)/2)*Tan[e + f*x]*(Tan[e + f*x]/Sqrt[Sec[e + f*x]^2])^(-1 + n*p)*(-(a*(2 + n*p)*((a^2 + b^2
)*AppellF1[(1 + n*p)/2, (n*p)/2, 1, (3 + n*p)/2, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^2] - 2*b^2*Appel
lF1[(1 + n*p)/2, (n*p)/2, 2, (3 + n*p)/2, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^2])) + 2*b*(a^2 - b^2)*
(1 + n*p)*AppellF1[1 + (n*p)/2, (-1 + n*p)/2, 2, 2 + (n*p)/2, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^2]*
Tan[e + f*x])*(Sqrt[Sec[e + f*x]^2] - Tan[e + f*x]^2/Sqrt[Sec[e + f*x]^2]))/(a^3*(a^2 - b^2)*(1 + n*p)*(2 + n*
p)) - ((Sec[e + f*x]^2)^((n*p)/2)*Tan[e + f*x]*(Tan[e + f*x]/Sqrt[Sec[e + f*x]^2])^(n*p)*(2*b*(a^2 - b^2)*(1 +
n*p)*AppellF1[1 + (n*p)/2, (-1 + n*p)/2, 2, 2 + (n*p)/2, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^2]*Sec[
e + f*x]^2 + 2*b*(a^2 - b^2)*(1 + n*p)*Tan[e + f*x]*((4*(-1 + b^2/a^2)*(1 + (n*p)/2)*AppellF1[2 + (n*p)/2, (-1
+ n*p)/2, 3, 3 + (n*p)/2, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^2]*Sec[e + f*x]^2*Tan[e + f*x])/(2 + (
n*p)/2) - ((1 + (n*p)/2)*(-1 + n*p)*AppellF1[2 + (n*p)/2, 1 + (-1 + n*p)/2, 2, 3 + (n*p)/2, -Tan[e + f*x]^2, (
-1 + b^2/a^2)*Tan[e + f*x]^2]*Sec[e + f*x]^2*Tan[e + f*x])/(2 + (n*p)/2)) - a*(2 + n*p)*((a^2 + b^2)*((2*(-1 +
b^2/a^2)*(1 + n*p)*AppellF1[1 + (1 + n*p)/2, (n*p)/2, 2, 1 + (3 + n*p)/2, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan
[e + f*x]^2]*Sec[e + f*x]^2*Tan[e + f*x])/(3 + n*p) - (n*p*(1 + n*p)*AppellF1[1 + (1 + n*p)/2, 1 + (n*p)/2, 1,
1 + (3 + n*p)/2, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^2]*Sec[e + f*x]^2*Tan[e + f*x])/(3 + n*p)) - 2*
b^2*((4*(-1 + b^2/a^2)*(1 + n*p)*AppellF1[1 + (1 + n*p)/2, (n*p)/2, 3, 1 + (3 + n*p)/2, -Tan[e + f*x]^2, (-1 +
b^2/a^2)*Tan[e + f*x]^2]*Sec[e + f*x]^2*Tan[e + f*x])/(3 + n*p) - (n*p*(1 + n*p)*AppellF1[1 + (1 + n*p)/2, 1
+ (n*p)/2, 2, 1 + (3 + n*p)/2, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^2]*Sec[e + f*x]^2*Tan[e + f*x])/(3
+ n*p)))))/(a^3*(a^2 - b^2)*(1 + n*p)*(2 + n*p)))))

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\left (\left (d \sin \left (f x + e\right )\right )^{p} c\right )^{n}}{b^{2} \cos \left (f x + e\right )^{2} - 2 \, a b \sin \left (f x + e\right ) - a^{2} - b^{2}}, x\right ) \]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (\left (d \sin \left (f x + e\right )\right )^{p} c\right )^{n}}{{\left (b \sin \left (f x + e\right ) + a\right )}^{2}}\,{d x} \]

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

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maple [F]  time = 1.14, size = 0, normalized size = 0.00 \[ \int \frac {\left (c \left (d \sin \left (f x +e \right )\right )^{p}\right )^{n}}{\left (a +b \sin \left (f x +e \right )\right )^{2}}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (\left (d \sin \left (f x + e\right )\right )^{p} c\right )^{n}}{{\left (b \sin \left (f x + e\right ) + a\right )}^{2}}\,{d x} \]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (c\,{\left (d\,\sin \left (e+f\,x\right )\right )}^p\right )}^n}{{\left (a+b\,\sin \left (e+f\,x\right )\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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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