Integrand size = 27, antiderivative size = 405 \[ \int \frac {\left (c (d \sin (e+f x))^p\right )^n}{(3+b \sin (e+f x))^3} \, dx=\frac {27 b \operatorname {AppellF1}\left (\frac {1}{2},-\frac {n p}{2},3,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{9-b^2}\right ) \cos (e+f x) \sin ^2(e+f x)^{-\frac {n p}{2}} \left (c (d \sin (e+f x))^p\right )^n}{\left (9-b^2\right )^3 f}+\frac {b^3 \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2} (-2-n p),3,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{9-b^2}\right ) \cos (e+f x) \sin ^2(e+f x)^{-\frac {n p}{2}} \left (c (d \sin (e+f x))^p\right )^n}{\left (9-b^2\right )^3 f}-\frac {9 b^2 \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2} (-1-n p),3,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{9-b^2}\right ) \cos (e+f x) \sin (e+f x) \sin ^2(e+f x)^{\frac {1}{2} (-1-n p)} \left (c (d \sin (e+f x))^p\right )^n}{\left (9-b^2\right )^3 f}-\frac {27 \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2} (1-n p),3,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{9-b^2}\right ) \cot (e+f x) \sin ^2(e+f x)^{\frac {1}{2} (1-n p)} \left (c (d \sin (e+f x))^p\right )^n}{\left (9-b^2\right )^3 f} \]
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
Leaf count is larger than twice the leaf count of optimal. \(2467\) vs. \(2(405)=810\).
Time = 19.28 (sec) , antiderivative size = 2467, normalized size of antiderivative = 6.09 \[ \int \frac {\left (c (d \sin (e+f x))^p\right )^n}{(3+b \sin (e+f x))^3} \, dx=\text {Result too large to show} \]
((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)*(-3*(2 + n*p)*(3*(3 + b^2)*AppellF1[(1 + n*p)/2, -1 + (n*p)/2, 2, (3 + n*p)/2, -Tan[e + f*x]^2, ((-9 + b^2)*Tan[e + f*x]^2)/9] - 4*b^2*AppellF1[(1 + n*p)/2, -1 + (n*p)/2, 3, (3 + n*p)/2, - Tan[e + f*x]^2, ((-9 + b^2)*Tan[e + f*x]^2)/9]) + b*(27 + b^2)*(1 + n*p)*A ppellF1[1 + (n*p)/2, (-1 + n*p)/2, 2, 2 + (n*p)/2, -Tan[e + f*x]^2, ((-9 + b^2)*Tan[e + f*x]^2)/9]*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, ((-9 + b^2)*Tan[e + f*x ]^2)/9]*Tan[e + f*x]))/(81*(-9 + b^2)*f*(1 + n*p)*(2 + n*p)*(3 + 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)*(-3*(2 + n*p)*(3*(3 + b^2)*AppellF1[(1 + n*p)/2, -1 + (n*p)/2, 2 , (3 + n*p)/2, -Tan[e + f*x]^2, ((-9 + b^2)*Tan[e + f*x]^2)/9] - 4*b^2*App ellF1[(1 + n*p)/2, -1 + (n*p)/2, 3, (3 + n*p)/2, -Tan[e + f*x]^2, ((-9 + b ^2)*Tan[e + f*x]^2)/9]) + b*(27 + b^2)*(1 + n*p)*AppellF1[1 + (n*p)/2, (-1 + n*p)/2, 2, 2 + (n*p)/2, -Tan[e + f*x]^2, ((-9 + b^2)*Tan[e + f*x]^2)/9] *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, ((-9 + b^2)*Tan[e + f*x]^2)/9]*Tan[e + f*x]))/( 81*(-9 + 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)*(-3*(2 + n*p)*(3*(3 + b^2)*AppellF1[(1 + n*p)/2, -1 + (n*p)/2, 2, (3 + n*p)/2, -Tan[e + f*x]^...
Time = 0.99 (sec) , antiderivative size = 450, normalized size of antiderivative = 1.11, 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 |
| \(\displaystyle (d \sin (e+f x))^{-n p} \left (c (d \sin (e+f x))^p\right )^n \left (\frac {3 a^2 b \cos (e+f x) \sin ^2(e+f x)^{-\frac {n p}{2}} (d \sin (e+f x))^{n p} \operatorname {AppellF1}\left (\frac {1}{2},-\frac {n p}{2},3,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^3}-\frac {3 a b^2 \cos (e+f x) \sin ^2(e+f x)^{\frac {1}{2} (-n p-1)} (d \sin (e+f x))^{n p+1} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2} (-n p-1),3,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{d f \left (a^2-b^2\right )^3}+\frac {b^3 \cos (e+f x) \sin ^2(e+f x)^{-\frac {n p}{2}} (d \sin (e+f x))^{n p} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2} (-n p-2),3,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^3}-\frac {a^3 d \cos (e+f x) \sin ^2(e+f x)^{\frac {1}{2} (1-n p)} (d \sin (e+f x))^{n p-1} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2} (1-n p),3,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^3}\right )\) |
((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)
3.9.37.3.1 Defintions of rubi rules used
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]
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]
\[\int \frac {\left (c \left (d \sin \left (f x +e \right )\right )^{p}\right )^{n}}{\left (a +b \sin \left (f x +e \right )\right )^{3}}d x\]
\[ \int \frac {\left (c (d \sin (e+f x))^p\right )^n}{(3+b \sin (e+f x))^3} \, dx=\int { \frac {\left (\left (d \sin \left (f x + e\right )\right )^{p} c\right )^{n}}{{\left (b \sin \left (f x + e\right ) + a\right )}^{3}} \,d x } \]
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)
\[ \int \frac {\left (c (d \sin (e+f x))^p\right )^n}{(3+b \sin (e+f x))^3} \, dx=\int \frac {\left (c \left (d \sin {\left (e + f x \right )}\right )^{p}\right )^{n}}{\left (a + b \sin {\left (e + f x \right )}\right )^{3}}\, dx \]
\[ \int \frac {\left (c (d \sin (e+f x))^p\right )^n}{(3+b \sin (e+f x))^3} \, dx=\int { \frac {\left (\left (d \sin \left (f x + e\right )\right )^{p} c\right )^{n}}{{\left (b \sin \left (f x + e\right ) + a\right )}^{3}} \,d x } \]
\[ \int \frac {\left (c (d \sin (e+f x))^p\right )^n}{(3+b \sin (e+f x))^3} \, dx=\int { \frac {\left (\left (d \sin \left (f x + e\right )\right )^{p} c\right )^{n}}{{\left (b \sin \left (f x + e\right ) + a\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {\left (c (d \sin (e+f x))^p\right )^n}{(3+b \sin (e+f x))^3} \, dx=\int \frac {{\left (c\,{\left (d\,\sin \left (e+f\,x\right )\right )}^p\right )}^n}{{\left (a+b\,\sin \left (e+f\,x\right )\right )}^3} \,d x \]