3.9.35 \(\int \frac {(c (d \sin (e+f x))^p)^n}{3+b \sin (e+f x)} \, dx\) [835]

3.9.35.1 Optimal result

Integrand size = 27, antiderivative size = 195 \[ \int \frac {\left (c (d \sin (e+f x))^p\right )^n}{3+b \sin (e+f x)} \, dx=\frac {b \operatorname {AppellF1}\left (\frac {1}{2},-\frac {n p}{2},1,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{9-b^2}\right ) \cos (e+f x) \sin ^2(e+f x)^{-\frac {n p}{2}} \left (c (d \sin (e+f x))^p\right )^n}{\left (9-b^2\right ) f}-\frac {3 \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2} (1-n p),1,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{9-b^2}\right ) \cot (e+f x) \sin ^2(e+f x)^{\frac {1}{2} (1-n p)} \left (c (d \sin (e+f x))^p\right )^n}{\left (9-b^2\right ) f} \]

output

b*AppellF1(1/2,-1/2*n*p,1,3/2,cos(f*x+e)^2,-b^2*cos(f*x+e)^2/(a^2-b^2))*co 
s(f*x+e)*(c*(d*sin(f*x+e))^p)^n/(a^2-b^2)/f/((sin(f*x+e)^2)^(1/2*n*p))-a*A 
ppellF1(1/2,-1/2*n*p+1/2,1,3/2,cos(f*x+e)^2,-b^2*cos(f*x+e)^2/(a^2-b^2))*c 
ot(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)/f
 

3.9.35.2 Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(1730\) vs. \(2(195)=390\).

Time = 16.81 (sec) , antiderivative size = 1730, normalized size of antiderivative = 8.87 \[ \int \frac {\left (c (d \sin (e+f x))^p\right )^n}{3+b \sin (e+f x)} \, dx =\text {Too large to display} \]

input

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

output

-1/9*((Sec[e + f*x]^2)^((n*p)/2)*(c*(d*Sin[e + f*x])^p)^n*Tan[e + f*x]*(Ta 
n[e + f*x]/Sqrt[Sec[e + f*x]^2])^(n*p)*(-3*b*(2 + n*p)*AppellF1[(1 + n*p)/ 
2, (n*p)/2, 1, (3 + n*p)/2, -Tan[e + f*x]^2, ((-9 + b^2)*Tan[e + f*x]^2)/9 
] + (-9 + b^2)*(1 + n*p)*AppellF1[1 + (n*p)/2, (-1 + n*p)/2, 1, 2 + (n*p)/ 
2, -Tan[e + f*x]^2, ((-9 + b^2)*Tan[e + f*x]^2)/9]*Tan[e + f*x] + 9*(1 + n 
*p)*Hypergeometric2F1[1 + (n*p)/2, (1 + n*p)/2, 2 + (n*p)/2, -Tan[e + f*x] 
^2]*Tan[e + f*x]))/(b*f*(1 + n*p)*(2 + n*p)*(3 + b*Sin[e + f*x])*(-1/9*((S 
ec[e + f*x]^2)^(1 + (n*p)/2)*(Tan[e + f*x]/Sqrt[Sec[e + f*x]^2])^(n*p)*(-3 
*b*(2 + n*p)*AppellF1[(1 + n*p)/2, (n*p)/2, 1, (3 + n*p)/2, -Tan[e + f*x]^ 
2, ((-9 + b^2)*Tan[e + f*x]^2)/9] + (-9 + b^2)*(1 + n*p)*AppellF1[1 + (n*p 
)/2, (-1 + n*p)/2, 1, 2 + (n*p)/2, -Tan[e + f*x]^2, ((-9 + b^2)*Tan[e + f* 
x]^2)/9]*Tan[e + f*x] + 9*(1 + n*p)*Hypergeometric2F1[1 + (n*p)/2, (1 + n* 
p)/2, 2 + (n*p)/2, -Tan[e + f*x]^2]*Tan[e + f*x]))/(b*(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*b*(2 + n*p)*AppellF1[(1 + n*p)/2, (n*p)/2, 1, (3 + n 
*p)/2, -Tan[e + f*x]^2, ((-9 + b^2)*Tan[e + f*x]^2)/9] + (-9 + b^2)*(1 + n 
*p)*AppellF1[1 + (n*p)/2, (-1 + n*p)/2, 1, 2 + (n*p)/2, -Tan[e + f*x]^2, ( 
(-9 + b^2)*Tan[e + f*x]^2)/9]*Tan[e + f*x] + 9*(1 + n*p)*Hypergeometric2F1 
[1 + (n*p)/2, (1 + n*p)/2, 2 + (n*p)/2, -Tan[e + f*x]^2]*Tan[e + f*x]))/(9 
*b*(1 + n*p)*(2 + n*p)) - (n*p*(Sec[e + f*x]^2)^((n*p)/2)*Tan[e + f*x]*...
 

3.9.35.3 Rubi [A] (verified)

Time = 0.67 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.18, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {3042, 3305, 3042, 3302, 3042, 3668, 333}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3305

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3302

\(\displaystyle (d \sin (e+f x))^{-n p} \left (c (d \sin (e+f x))^p\right )^n \left (a \int \frac {(d \sin (e+f x))^{n p}}{a^2-b^2 \sin ^2(e+f x)}dx-\frac {b \int \frac {(d \sin (e+f x))^{n p+1}}{a^2-b^2 \sin ^2(e+f x)}dx}{d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle (d \sin (e+f x))^{-n p} \left (c (d \sin (e+f x))^p\right )^n \left (a \int \frac {(d \sin (e+f x))^{n p}}{a^2-b^2 \sin (e+f x)^2}dx-\frac {b \int \frac {(d \sin (e+f x))^{n p+1}}{a^2-b^2 \sin (e+f x)^2}dx}{d}\right )\)

\(\Big \downarrow \) 3668

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

\(\Big \downarrow \) 333

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

input

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

output

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

3.9.35.3.1 Defintions of rubi rules used

rule 333

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[a^p*c^q*x*AppellF1[1/2, -p, -q, 3/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; F 
reeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[p] || GtQ[a, 
0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 3042

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

rule 3302

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

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]
 

rule 3668

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]}, Simp[( 
-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]
 

3.9.35.4 Maple [F]

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

input

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

output

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

3.9.35.5 Fricas [F]

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

input

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

output

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

3.9.35.6 Sympy [F]

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

input

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

output

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

3.9.35.7 Maxima [F]

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

input

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

output

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

3.9.35.8 Giac [F]

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

input

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

output

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

3.9.35.9 Mupad [F(-1)]

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

input

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

output

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