\(\int \frac {(c (d \sin (e+f x))^p)^n}{(a+b \sin (e+f x))^3} \, dx\) [838]

Optimal result

Integrand size = 27, antiderivative size = 428 \[ \int \frac {\left (c (d \sin (e+f x))^p\right )^n}{(a+b \sin (e+f x))^3} \, dx=\frac {3 a^2 b \operatorname {AppellF1}\left (\frac {1}{2},-\frac {n p}{2},3,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right ) \cos (e+f x) \sin ^2(e+f x)^{-\frac {n p}{2}} \left (c (d \sin (e+f x))^p\right )^n}{\left (a^2-b^2\right )^3 f}+\frac {b^3 \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2} (-2-n p),3,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right ) \cos (e+f x) \sin ^2(e+f x)^{-\frac {n p}{2}} \left (c (d \sin (e+f x))^p\right )^n}{\left (a^2-b^2\right )^3 f}-\frac {3 a b^2 \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2} (-1-n p),3,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right ) \cos (e+f x) \sin (e+f x) \sin ^2(e+f x)^{\frac {1}{2} (-1-n p)} \left (c (d \sin (e+f x))^p\right )^n}{\left (a^2-b^2\right )^3 f}-\frac {a^3 \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2} (1-n p),3,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right ) \cot (e+f x) \sin ^2(e+f x)^{\frac {1}{2} (1-n p)} \left (c (d \sin (e+f x))^p\right )^n}{\left (a^2-b^2\right )^3 f} \] Output:

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

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(2570\) vs. \(2(428)=856\).

Time = 20.31 (sec) , antiderivative size = 2570, normalized size of antiderivative = 6.00 \[ \int \frac {\left (c (d \sin (e+f x))^p\right )^n}{(a+b \sin (e+f x))^3} \, dx=\text {Result too large to show} \] Input:

Integrate[(c*(d*Sin[e + f*x])^p)^n/(a + b*Sin[e + f*x])^3,x]
 

Output:

-(((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 + 3*b^2)*AppellF1 
[(1 + n*p)/2, -1 + (n*p)/2, 2, (3 + n*p)/2, -Tan[e + f*x]^2, (-1 + b^2/a^2 
)*Tan[e + f*x]^2] - 4*b^2*AppellF1[(1 + 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])) + b*(3*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] - 4*b^3*(1 + n*p)*AppellF1[ 
1 + (n*p)/2, (-1 + n*p)/2, 3, 2 + (n*p)/2, -Tan[e + f*x]^2, (-1 + b^2/a^2) 
*Tan[e + f*x]^2]*Tan[e + f*x]))/(a^4*(a^2 - b^2)*f*(1 + n*p)*(2 + n*p)*(a 
+ b*Sin[e + f*x])^3*(-(((Sec[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 + 3*b^2)*AppellF1[(1 + n*p)/2, 
 -1 + (n*p)/2, 2, (3 + n*p)/2, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x 
]^2] - 4*b^2*AppellF1[(1 + 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])) + b*(3*a^2 + b^2)*(1 + n*p)*Appel 
lF1[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] - 4*b^3*(1 + n*p)*AppellF1[1 + (n*p)/2, 
(-1 + n*p)/2, 3, 2 + (n*p)/2, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x] 
^2]*Tan[e + f*x]))/(a^4*(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[Sec[e + f*x]^2])^(n*p) 
*(-(a*(2 + n*p)*((a^2 + 3*b^2)*AppellF1[(1 + n*p)/2, -1 + (n*p)/2, 2, (...
 

Rubi [A] (verified)

Time = 1.14 (sec) , antiderivative size = 450, normalized size of antiderivative = 1.05, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {3042, 3305, 3042, 3303, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\left (c (d \sin (e+f x))^p\right )^n}{(a+b \sin (e+f x))^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (c (d \sin (e+f x))^p\right )^n}{(a+b \sin (e+f x))^3}dx\)

\(\Big \downarrow \) 3305

\(\displaystyle (d \sin (e+f x))^{-n p} \left (c (d \sin (e+f x))^p\right )^n \int \frac {(d \sin (e+f x))^{n p}}{(a+b \sin (e+f x))^3}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle (d \sin (e+f x))^{-n p} \left (c (d \sin (e+f x))^p\right )^n \int \frac {(d \sin (e+f x))^{n p}}{(a+b \sin (e+f x))^3}dx\)

\(\Big \downarrow \) 3303

\(\displaystyle (d \sin (e+f x))^{-n p} \left (c (d \sin (e+f x))^p\right )^n \int \left (\frac {a^3 (d \sin (e+f x))^{n p}}{\left (a^2-b^2 \sin ^2(e+f x)\right )^3}+\frac {3 a b^2 \sin ^2(e+f x) (d \sin (e+f x))^{n p}}{\left (a^2-b^2 \sin ^2(e+f x)\right )^3}-\frac {3 a^2 b \sin (e+f x) (d \sin (e+f x))^{n p}}{\left (a^2-b^2 \sin ^2(e+f x)\right )^3}+\frac {b^3 \sin ^3(e+f x) (d \sin (e+f x))^{n p}}{\left (b^2 \sin ^2(e+f x)-a^2\right )^3}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle (d \sin (e+f x))^{-n p} \left (c (d \sin (e+f x))^p\right )^n \left (\frac {3 a^2 b \cos (e+f x) \sin ^2(e+f x)^{-\frac {n p}{2}} (d \sin (e+f x))^{n p} \operatorname {AppellF1}\left (\frac {1}{2},-\frac {n p}{2},3,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^3}-\frac {3 a b^2 \cos (e+f x) \sin ^2(e+f x)^{\frac {1}{2} (-n p-1)} (d \sin (e+f x))^{n p+1} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2} (-n p-1),3,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{d f \left (a^2-b^2\right )^3}+\frac {b^3 \cos (e+f x) \sin ^2(e+f x)^{-\frac {n p}{2}} (d \sin (e+f x))^{n p} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2} (-n p-2),3,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^3}-\frac {a^3 d \cos (e+f x) \sin ^2(e+f x)^{\frac {1}{2} (1-n p)} (d \sin (e+f x))^{n p-1} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2} (1-n p),3,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^3}\right )\)

Input:

Int[(c*(d*Sin[e + f*x])^p)^n/(a + b*Sin[e + f*x])^3,x]
 

Output:

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

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3303

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*( 
x_)])^(m_.), x_Symbol] :> Int[ExpandTrig[(d*sin[e + f*x])^n*(1/((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 3305

Int[((c_.)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(p_))^(n_)*((a_.) + (b_.)*sin[(e 
_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Simp[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]
 

Maple [F]

\[\int \frac {\left (c \left (d \sin \left (f x +e \right )\right )^{p}\right )^{n}}{\left (a +b \sin \left (f x +e \right )\right )^{3}}d x\]

Input:

int((c*(d*sin(f*x+e))^p)^n/(a+b*sin(f*x+e))^3,x)
 

Output:

int((c*(d*sin(f*x+e))^p)^n/(a+b*sin(f*x+e))^3,x)
 

Fricas [F]

\[ \int \frac {\left (c (d \sin (e+f x))^p\right )^n}{(a+b \sin (e+f x))^3} \, dx=\int { \frac {\left (\left (d \sin \left (f x + e\right )\right )^{p} c\right )^{n}}{{\left (b \sin \left (f x + e\right ) + a\right )}^{3}} \,d x } \] Input:

integrate((c*(d*sin(f*x+e))^p)^n/(a+b*sin(f*x+e))^3,x, algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int \frac {\left (c (d \sin (e+f x))^p\right )^n}{(a+b \sin (e+f x))^3} \, dx=\int \frac {\left (c \left (d \sin {\left (e + f x \right )}\right )^{p}\right )^{n}}{\left (a + b \sin {\left (e + f x \right )}\right )^{3}}\, dx \] Input:

integrate((c*(d*sin(f*x+e))**p)**n/(a+b*sin(f*x+e))**3,x)
 

Output:

Integral((c*(d*sin(e + f*x))**p)**n/(a + b*sin(e + f*x))**3, x)
 

Maxima [F]

\[ \int \frac {\left (c (d \sin (e+f x))^p\right )^n}{(a+b \sin (e+f x))^3} \, dx=\int { \frac {\left (\left (d \sin \left (f x + e\right )\right )^{p} c\right )^{n}}{{\left (b \sin \left (f x + e\right ) + a\right )}^{3}} \,d x } \] Input:

integrate((c*(d*sin(f*x+e))^p)^n/(a+b*sin(f*x+e))^3,x, algorithm="maxima")
 

Output:

integrate(((d*sin(f*x + e))^p*c)^n/(b*sin(f*x + e) + a)^3, x)
 

Giac [F]

\[ \int \frac {\left (c (d \sin (e+f x))^p\right )^n}{(a+b \sin (e+f x))^3} \, dx=\int { \frac {\left (\left (d \sin \left (f x + e\right )\right )^{p} c\right )^{n}}{{\left (b \sin \left (f x + e\right ) + a\right )}^{3}} \,d x } \] Input:

integrate((c*(d*sin(f*x+e))^p)^n/(a+b*sin(f*x+e))^3,x, algorithm="giac")
 

Output:

integrate(((d*sin(f*x + e))^p*c)^n/(b*sin(f*x + e) + a)^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (c (d \sin (e+f x))^p\right )^n}{(a+b \sin (e+f x))^3} \, dx=\int \frac {{\left (c\,{\left (d\,\sin \left (e+f\,x\right )\right )}^p\right )}^n}{{\left (a+b\,\sin \left (e+f\,x\right )\right )}^3} \,d x \] Input:

int((c*(d*sin(e + f*x))^p)^n/(a + b*sin(e + f*x))^3,x)
 

Output:

int((c*(d*sin(e + f*x))^p)^n/(a + b*sin(e + f*x))^3, x)
 

Reduce [F]

\[ \int \frac {\left (c (d \sin (e+f x))^p\right )^n}{(a+b \sin (e+f x))^3} \, dx=d^{n p} c^{n} \left (\int \frac {\sin \left (f x +e \right )^{n p}}{\sin \left (f x +e \right )^{3} b^{3}+3 \sin \left (f x +e \right )^{2} a \,b^{2}+3 \sin \left (f x +e \right ) a^{2} b +a^{3}}d x \right ) \] Input:

int((c*(d*sin(f*x+e))^p)^n/(a+b*sin(f*x+e))^3,x)
 

Output:

d**(n*p)*c**n*int(sin(e + f*x)**(n*p)/(sin(e + f*x)**3*b**3 + 3*sin(e + f* 
x)**2*a*b**2 + 3*sin(e + f*x)*a**2*b + a**3),x)