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

Optimal result

Integrand size = 37, antiderivative size = 269 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^{3/2}} \, dx=-\frac {B \operatorname {AppellF1}\left (1+n,\frac {1}{2},1,2+n,\frac {c+d \sin (e+f x)}{c+d},\frac {c+d \sin (e+f x)}{c-d}\right ) \cos (e+f x) \sqrt {\frac {d (1-\sin (e+f x))}{c+d}} (c+d \sin (e+f x))^{1+n}}{a (c-d) f (1+n) (1-\sin (e+f x)) \sqrt {a+a \sin (e+f x)}}+\frac {(A-B) d \operatorname {AppellF1}\left (1+n,\frac {1}{2},2,2+n,\frac {c+d \sin (e+f x)}{c+d},\frac {c+d \sin (e+f x)}{c-d}\right ) \cos (e+f x) \sqrt {\frac {d (1-\sin (e+f x))}{c+d}} (c+d \sin (e+f x))^{1+n}}{(c-d)^2 f (1+n) (a-a \sin (e+f x)) \sqrt {a+a \sin (e+f x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.108, Rules used = {3066, 2867, 142, 141} \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^{3/2}} \, dx=\frac {d (A-B) \cos (e+f x) \sqrt {\frac {d (1-\sin (e+f x))}{c+d}} (c+d \sin (e+f x))^{n+1} \operatorname {AppellF1}\left (n+1,\frac {1}{2},2,n+2,\frac {c+d \sin (e+f x)}{c+d},\frac {c+d \sin (e+f x)}{c-d}\right )}{f (n+1) (c-d)^2 (a-a \sin (e+f x)) \sqrt {a \sin (e+f x)+a}}-\frac {B \cos (e+f x) \sqrt {\frac {d (1-\sin (e+f x))}{c+d}} (c+d \sin (e+f x))^{n+1} \operatorname {AppellF1}\left (n+1,\frac {1}{2},1,n+2,\frac {c+d \sin (e+f x)}{c+d},\frac {c+d \sin (e+f x)}{c-d}\right )}{a f (n+1) (c-d) (1-\sin (e+f x)) \sqrt {a \sin (e+f x)+a}} \]

[In]

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

[Out]

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

Rule 141

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

Rule 142

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

Rule 2867

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dis
t[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 3066

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(A*b - a*B)/b, Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n, x
], x] + Dist[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]

Rubi steps \begin{align*} \text {integral}& = (A-B) \int \frac {(c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^{3/2}} \, dx+\frac {B \int \frac {(c+d \sin (e+f x))^n}{\sqrt {a+a \sin (e+f x)}} \, dx}{a} \\ & = \frac {\left (a^2 (A-B) \cos (e+f x)\right ) \text {Subst}\left (\int \frac {(c+d x)^n}{\sqrt {a-a x} (a+a x)^2} \, dx,x,\sin (e+f x)\right )}{f \sqrt {a-a \sin (e+f x)} \sqrt {a+a \sin (e+f x)}}+\frac {(a B \cos (e+f x)) \text {Subst}\left (\int \frac {(c+d x)^n}{\sqrt {a-a x} (a+a x)} \, dx,x,\sin (e+f x)\right )}{f \sqrt {a-a \sin (e+f x)} \sqrt {a+a \sin (e+f x)}} \\ & = \frac {\left (a^2 (A-B) \cos (e+f x) \sqrt {\frac {d (a-a \sin (e+f x))}{a c+a d}}\right ) \text {Subst}\left (\int \frac {(c+d x)^n}{(a+a x)^2 \sqrt {\frac {a d}{a c+a d}-\frac {a d x}{a c+a d}}} \, dx,x,\sin (e+f x)\right )}{f (a-a \sin (e+f x)) \sqrt {a+a \sin (e+f x)}}+\frac {\left (a B \cos (e+f x) \sqrt {\frac {d (a-a \sin (e+f x))}{a c+a d}}\right ) \text {Subst}\left (\int \frac {(c+d x)^n}{(a+a x) \sqrt {\frac {a d}{a c+a d}-\frac {a d x}{a c+a d}}} \, dx,x,\sin (e+f x)\right )}{f (a-a \sin (e+f x)) \sqrt {a+a \sin (e+f x)}} \\ & = -\frac {B \operatorname {AppellF1}\left (1+n,\frac {1}{2},1,2+n,\frac {c+d \sin (e+f x)}{c+d},\frac {c+d \sin (e+f x)}{c-d}\right ) \cos (e+f x) \sqrt {\frac {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)}}+\frac {(A-B) d \operatorname {AppellF1}\left (1+n,\frac {1}{2},2,2+n,\frac {c+d \sin (e+f x)}{c+d},\frac {c+d \sin (e+f x)}{c-d}\right ) \cos (e+f x) \sqrt {\frac {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)}} \\ \end{align*}

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(603\) vs. \(2(269)=538\).

Time = 16.63 (sec) , antiderivative size = 603, normalized size of antiderivative = 2.24 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^{3/2}} \, dx=\frac {\sec (e+f x) (c+d \sin (e+f x))^n \left (a B (1+\sin (e+f x)) \left (a \operatorname {AppellF1}\left (1,\frac {1}{2},-n,2,\frac {1}{2} (1+\sin (e+f x)),\frac {d (1+\sin (e+f x))}{-c+d}\right ) \sqrt {2-2 \sin (e+f x)} (1+\sin (e+f x)) \left (\frac {c+d \sin (e+f x)}{c-d}\right )^{-n}-\frac {4 \sqrt {\frac {-1+\sin (e+f x)}{1+\sin (e+f x)}} \left (-2 a (1+2 n) \operatorname {AppellF1}\left (\frac {1}{2}-n,-\frac {1}{2},-n,\frac {3}{2}-n,\frac {2}{1+\sin (e+f x)},\frac {-c+d}{d+d \sin (e+f x)}\right )+a (-1+2 n) \operatorname {AppellF1}\left (-\frac {1}{2}-n,-\frac {1}{2},-n,\frac {1}{2}-n,\frac {2}{1+\sin (e+f x)},\frac {-c+d}{d+d \sin (e+f x)}\right ) (1+\sin (e+f x))\right ) \left (1+\frac {c-d}{d+d \sin (e+f x)}\right )^{-n}}{-1+4 n^2}\right )+a A (1+\sin (e+f x)) \left (a \operatorname {AppellF1}\left (1,\frac {1}{2},-n,2,\frac {1}{2} (1+\sin (e+f x)),\frac {d (1+\sin (e+f x))}{-c+d}\right ) \sqrt {2-2 \sin (e+f x)} (1+\sin (e+f x)) \left (\frac {c+d \sin (e+f x)}{c-d}\right )^{-n}-\frac {4 \sqrt {\frac {-1+\sin (e+f x)}{1+\sin (e+f x)}} \left (2 a (1+2 n) \operatorname {AppellF1}\left (\frac {1}{2}-n,-\frac {1}{2},-n,\frac {3}{2}-n,\frac {2}{1+\sin (e+f x)},\frac {-c+d}{d+d \sin (e+f x)}\right )+a (-1+2 n) \operatorname {AppellF1}\left (-\frac {1}{2}-n,-\frac {1}{2},-n,\frac {1}{2}-n,\frac {2}{1+\sin (e+f x)},\frac {-c+d}{d+d \sin (e+f x)}\right ) (1+\sin (e+f x))\right ) \left (1+\frac {c-d}{d+d \sin (e+f x)}\right )^{-n}}{-1+4 n^2}\right )\right )}{8 a^3 f \sqrt {a (1+\sin (e+f x))}} \]

[In]

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

[Out]

(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 + Si
n[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*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))))/(8*a^3*f*Sqrt[a*(1
+ Sin[e + f*x])])

Maple [F]

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

[In]

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

[Out]

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]

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

[In]

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

[Out]

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]

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

[In]

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

[Out]

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]

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

[In]

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

[Out]

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

[In]

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

[Out]

Timed out

Mupad [F(-1)]

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

[In]

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

[Out]

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