\(\int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^{3/2}} \, dx\) [335]

Optimal result

Integrand size = 37, antiderivative size = 213 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^{3/2}} \, dx=-\frac {B \operatorname {AppellF1}\left (\frac {1}{2},-n,1,\frac {3}{2},\frac {d (1-\sin (e+f x))}{c+d},\frac {1}{2} (1-\sin (e+f x))\right ) \cos (e+f x) (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n}}{a f \sqrt {a+a \sin (e+f x)}}-\frac {(A-B) \operatorname {AppellF1}\left (\frac {1}{2},-n,2,\frac {3}{2},\frac {d (1-\sin (e+f x))}{c+d},\frac {1}{2} (1-\sin (e+f x))\right ) \cos (e+f x) (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n}}{2 a f \sqrt {a+a \sin (e+f x)}} \] Output:

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

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(603\) vs. \(2(213)=426\).

Time = 21.51 (sec) , antiderivative size = 603, normalized size of antiderivative = 2.83 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^{3/2}} \, dx=\frac {\sec (e+f x) (c+d \sin (e+f x))^n \left (a B (1+\sin (e+f x)) \left (a \operatorname {AppellF1}\left (1,\frac {1}{2},-n,2,\frac {1}{2} (1+\sin (e+f x)),\frac {d (1+\sin (e+f x))}{-c+d}\right ) \sqrt {2-2 \sin (e+f x)} (1+\sin (e+f x)) \left (\frac {c+d \sin (e+f x)}{c-d}\right )^{-n}-\frac {4 \sqrt {\frac {-1+\sin (e+f x)}{1+\sin (e+f x)}} \left (-2 a (1+2 n) \operatorname {AppellF1}\left (\frac {1}{2}-n,-\frac {1}{2},-n,\frac {3}{2}-n,\frac {2}{1+\sin (e+f x)},\frac {-c+d}{d+d \sin (e+f x)}\right )+a (-1+2 n) \operatorname {AppellF1}\left (-\frac {1}{2}-n,-\frac {1}{2},-n,\frac {1}{2}-n,\frac {2}{1+\sin (e+f x)},\frac {-c+d}{d+d \sin (e+f x)}\right ) (1+\sin (e+f x))\right ) \left (1+\frac {c-d}{d+d \sin (e+f x)}\right )^{-n}}{-1+4 n^2}\right )+a A (1+\sin (e+f x)) \left (a \operatorname {AppellF1}\left (1,\frac {1}{2},-n,2,\frac {1}{2} (1+\sin (e+f x)),\frac {d (1+\sin (e+f x))}{-c+d}\right ) \sqrt {2-2 \sin (e+f x)} (1+\sin (e+f x)) \left (\frac {c+d \sin (e+f x)}{c-d}\right )^{-n}-\frac {4 \sqrt {\frac {-1+\sin (e+f x)}{1+\sin (e+f x)}} \left (2 a (1+2 n) \operatorname {AppellF1}\left (\frac {1}{2}-n,-\frac {1}{2},-n,\frac {3}{2}-n,\frac {2}{1+\sin (e+f x)},\frac {-c+d}{d+d \sin (e+f x)}\right )+a (-1+2 n) \operatorname {AppellF1}\left (-\frac {1}{2}-n,-\frac {1}{2},-n,\frac {1}{2}-n,\frac {2}{1+\sin (e+f x)},\frac {-c+d}{d+d \sin (e+f x)}\right ) (1+\sin (e+f x))\right ) \left (1+\frac {c-d}{d+d \sin (e+f x)}\right )^{-n}}{-1+4 n^2}\right )\right )}{8 a^3 f \sqrt {a (1+\sin (e+f x))}} \] Input:

Integrate[((A + B*Sin[e + f*x])*(c + d*Sin[e + f*x])^n)/(a + a*Sin[e + f*x 
])^(3/2),x]
 

Output:

(Sec[e + f*x]*(c + d*Sin[e + f*x])^n*(a*B*(1 + Sin[e + f*x])*((a*AppellF1[ 
1, 1/2, -n, 2, (1 + Sin[e + f*x])/2, (d*(1 + Sin[e + f*x]))/(-c + d)]*Sqrt 
[2 - 2*Sin[e + f*x]]*(1 + Sin[e + f*x]))/((c + d*Sin[e + f*x])/(c - d))^n 
- (4*Sqrt[(-1 + Sin[e + f*x])/(1 + Sin[e + f*x])]*(-2*a*(1 + 2*n)*AppellF1 
[1/2 - n, -1/2, -n, 3/2 - n, 2/(1 + Sin[e + f*x]), (-c + d)/(d + d*Sin[e + 
 f*x])] + a*(-1 + 2*n)*AppellF1[-1/2 - n, -1/2, -n, 1/2 - n, 2/(1 + Sin[e 
+ f*x]), (-c + d)/(d + d*Sin[e + f*x])]*(1 + Sin[e + f*x])))/((-1 + 4*n^2) 
*(1 + (c - d)/(d + d*Sin[e + f*x]))^n)) + a*A*(1 + Sin[e + f*x])*((a*Appel 
lF1[1, 1/2, -n, 2, (1 + Sin[e + f*x])/2, (d*(1 + Sin[e + f*x]))/(-c + d)]* 
Sqrt[2 - 2*Sin[e + f*x]]*(1 + Sin[e + f*x]))/((c + d*Sin[e + f*x])/(c - d) 
)^n - (4*Sqrt[(-1 + Sin[e + f*x])/(1 + Sin[e + f*x])]*(2*a*(1 + 2*n)*Appel 
lF1[1/2 - n, -1/2, -n, 3/2 - n, 2/(1 + Sin[e + f*x]), (-c + d)/(d + d*Sin[ 
e + f*x])] + a*(-1 + 2*n)*AppellF1[-1/2 - n, -1/2, -n, 1/2 - n, 2/(1 + Sin 
[e + f*x]), (-c + d)/(d + d*Sin[e + f*x])]*(1 + Sin[e + f*x])))/((-1 + 4*n 
^2)*(1 + (c - d)/(d + d*Sin[e + f*x]))^n))))/(8*a^3*f*Sqrt[a*(1 + Sin[e + 
f*x])])
 

Rubi [A] (warning: unable to verify)

Time = 0.70 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.25, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.189, Rules used = {3042, 3466, 3042, 3267, 27, 154, 153}

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.

\(\displaystyle \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^n}{(a \sin (e+f x)+a)^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^n}{(a \sin (e+f x)+a)^{3/2}}dx\)

\(\Big \downarrow \) 3466

\(\displaystyle (A-B) \int \frac {(c+d \sin (e+f x))^n}{(\sin (e+f x) a+a)^{3/2}}dx+\frac {B \int \frac {(c+d \sin (e+f x))^n}{\sqrt {\sin (e+f x) a+a}}dx}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle (A-B) \int \frac {(c+d \sin (e+f x))^n}{(\sin (e+f x) a+a)^{3/2}}dx+\frac {B \int \frac {(c+d \sin (e+f x))^n}{\sqrt {\sin (e+f x) a+a}}dx}{a}\)

\(\Big \downarrow \) 3267

\(\displaystyle \frac {a^2 (A-B) \cos (e+f x) \int \frac {(c+d \sin (e+f x))^n}{a^2 (\sin (e+f x)+1)^2 \sqrt {a-a \sin (e+f x)}}d\sin (e+f x)}{f \sqrt {a-a \sin (e+f x)} \sqrt {a \sin (e+f x)+a}}+\frac {a B \cos (e+f x) \int \frac {(c+d \sin (e+f x))^n}{a (\sin (e+f x)+1) \sqrt {a-a \sin (e+f x)}}d\sin (e+f x)}{f \sqrt {a-a \sin (e+f x)} \sqrt {a \sin (e+f x)+a}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(A-B) \cos (e+f x) \int \frac {(c+d \sin (e+f x))^n}{(\sin (e+f x)+1)^2 \sqrt {a-a \sin (e+f x)}}d\sin (e+f x)}{f \sqrt {a-a \sin (e+f x)} \sqrt {a \sin (e+f x)+a}}+\frac {B \cos (e+f x) \int \frac {(c+d \sin (e+f x))^n}{(\sin (e+f x)+1) \sqrt {a-a \sin (e+f x)}}d\sin (e+f x)}{f \sqrt {a-a \sin (e+f x)} \sqrt {a \sin (e+f x)+a}}\)

\(\Big \downarrow \) 154

\(\displaystyle \frac {(A-B) \cos (e+f x) \sqrt {\frac {d (1-\sin (e+f x))}{c+d}} \int \frac {(c+d \sin (e+f x))^n}{(\sin (e+f x)+1)^2 \sqrt {\frac {d}{c+d}-\frac {d \sin (e+f x)}{c+d}}}d\sin (e+f x)}{f (a-a \sin (e+f x)) \sqrt {a \sin (e+f x)+a}}+\frac {B \cos (e+f x) \sqrt {\frac {d (1-\sin (e+f x))}{c+d}} \int \frac {(c+d \sin (e+f x))^n}{(\sin (e+f x)+1) \sqrt {\frac {d}{c+d}-\frac {d \sin (e+f x)}{c+d}}}d\sin (e+f x)}{f (a-a \sin (e+f x)) \sqrt {a \sin (e+f x)+a}}\)

\(\Big \downarrow \) 153

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

Input:

Int[((A + B*Sin[e + f*x])*(c + d*Sin[e + f*x])^n)/(a + a*Sin[e + f*x])^(3/ 
2),x]
 

Output:

-((B*AppellF1[1 + n, 1/2, 1, 2 + n, (c + d*Sin[e + f*x])/(c + d), (c + d*S 
in[e + f*x])/(c - d)]*Cos[e + f*x]*Sqrt[(d*(1 - Sin[e + f*x]))/(c + d)]*(c 
 + d*Sin[e + f*x])^(1 + n))/((c - d)*f*(1 + n)*(a - a*Sin[e + f*x])*Sqrt[a 
 + a*Sin[e + f*x]])) + ((A - B)*d*AppellF1[1 + n, 1/2, 2, 2 + n, (c + d*Si 
n[e + f*x])/(c + d), (c + d*Sin[e + f*x])/(c - d)]*Cos[e + f*x]*Sqrt[(d*(1 
 - Sin[e + f*x]))/(c + d)]*(c + d*Sin[e + f*x])^(1 + n))/((c - d)^2*f*(1 + 
 n)*(a - a*Sin[e + f*x])*Sqrt[a + a*Sin[e + f*x]])
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 153

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) 
^(p_), x_] :> Simp[(b*e - a*f)^p*((a + b*x)^(m + 1)/(b^(p + 1)*(m + 1)*Simp 
lify[b/(b*c - a*d)]^n))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/(b*c 
 - a*d)), (-f)*((a + b*x)/(b*e - a*f))], x] /; FreeQ[{a, b, c, d, e, f, m, 
n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] && IntegerQ[p] && GtQ[Simplify[b/( 
b*c - a*d)], 0] &&  !(GtQ[Simplify[d/(d*a - c*b)], 0] && SimplerQ[c + d*x, 
a + b*x])
 

rule 154

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) 
^(p_), x_] :> Simp[(c + d*x)^FracPart[n]/(Simplify[b/(b*c - a*d)]^IntPart[n 
]*(b*((c + d*x)/(b*c - a*d)))^FracPart[n])   Int[(a + b*x)^m*Simp[b*(c/(b*c 
 - a*d)) + b*d*(x/(b*c - a*d)), x]^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, 
 d, e, f, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] && IntegerQ[p] &&  !G 
tQ[Simplify[b/(b*c - a*d)], 0] &&  !SimplerQ[c + d*x, a + b*x]
 

rule 3042

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

rule 3267

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

rule 3466

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

Maple [F]

\[\int \frac {\left (A +B \sin \left (f x +e \right )\right ) \left (c +d \sin \left (f x +e \right )\right )^{n}}{\left (a +a \sin \left (f x +e \right )\right )^{\frac {3}{2}}}d x\]

Input:

int((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^n/(a+a*sin(f*x+e))^(3/2),x)
 

Output:

int((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^n/(a+a*sin(f*x+e))^(3/2),x)
 

Fricas [F]

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

integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^n/(a+a*sin(f*x+e))^(3/2),x, al 
gorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^{3/2}} \, dx=\int \frac {\left (A + B \sin {\left (e + f x \right )}\right ) \left (c + d \sin {\left (e + f x \right )}\right )^{n}}{\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \] Input:

integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))**n/(a+a*sin(f*x+e))**(3/2),x)
 

Output:

Integral((A + B*sin(e + f*x))*(c + d*sin(e + f*x))**n/(a*(sin(e + f*x) + 1 
))**(3/2), x)
 

Maxima [F]

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

integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^n/(a+a*sin(f*x+e))^(3/2),x, al 
gorithm="maxima")
 

Output:

integrate((B*sin(f*x + e) + A)*(d*sin(f*x + e) + c)^n/(a*sin(f*x + e) + a) 
^(3/2), x)
 

Giac [F(-1)]

Timed out. \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^{3/2}} \, dx=\text {Timed out} \] Input:

integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^n/(a+a*sin(f*x+e))^(3/2),x, al 
gorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^{3/2}} \, dx=\int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^n}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \] Input:

int(((A + B*sin(e + f*x))*(c + d*sin(e + f*x))^n)/(a + a*sin(e + f*x))^(3/ 
2),x)
 

Output:

int(((A + B*sin(e + f*x))*(c + d*sin(e + f*x))^n)/(a + a*sin(e + f*x))^(3/ 
2), x)
 

Reduce [F]

\[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^{3/2}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\left (\sin \left (f x +e \right ) d +c \right )^{n} \sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right )^{2}+2 \sin \left (f x +e \right )+1}d x \right ) b +\left (\int \frac {\left (\sin \left (f x +e \right ) d +c \right )^{n} \sqrt {\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )^{2}+2 \sin \left (f x +e \right )+1}d x \right ) a \right )}{a^{2}} \] Input:

int((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^n/(a+a*sin(f*x+e))^(3/2),x)
                                                                                    
                                                                                    
 

Output:

(sqrt(a)*(int(((sin(e + f*x)*d + c)**n*sqrt(sin(e + f*x) + 1)*sin(e + f*x) 
)/(sin(e + f*x)**2 + 2*sin(e + f*x) + 1),x)*b + int(((sin(e + f*x)*d + c)* 
*n*sqrt(sin(e + f*x) + 1))/(sin(e + f*x)**2 + 2*sin(e + f*x) + 1),x)*a))/a 
**2