\(\int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{9/2}} \, dx\) [147]

Optimal result

Integrand size = 40, antiderivative size = 146 \[ \int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{9/2}} \, dx=\frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{8 f (c-c \sin (e+f x))^{9/2}}+\frac {(A-3 B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{24 c f (c-c \sin (e+f x))^{7/2}}+\frac {(A-3 B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{96 c^2 f (c-c \sin (e+f x))^{5/2}} \]

[Out]

1/8*(A+B)*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)/f/(c-c*sin(f*x+e))^(9/2)+1/24*(A-3*B)*cos(f*x+e)*(a+a*sin(f*x+e))^
(3/2)/c/f/(c-c*sin(f*x+e))^(7/2)+1/96*(A-3*B)*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)/c^2/f/(c-c*sin(f*x+e))^(5/2)

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.075, Rules used = {3051, 2822, 2821} \[ \int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{9/2}} \, dx=\frac {(A-3 B) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{96 c^2 f (c-c \sin (e+f x))^{5/2}}+\frac {(A-3 B) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{24 c f (c-c \sin (e+f x))^{7/2}}+\frac {(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{8 f (c-c \sin (e+f x))^{9/2}} \]

[In]

Int[((a + a*Sin[e + f*x])^(3/2)*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^(9/2),x]

[Out]

((A + B)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(8*f*(c - c*Sin[e + f*x])^(9/2)) + ((A - 3*B)*Cos[e + f*x]*(
a + a*Sin[e + f*x])^(3/2))/(24*c*f*(c - c*Sin[e + f*x])^(7/2)) + ((A - 3*B)*Cos[e + f*x]*(a + a*Sin[e + f*x])^
(3/2))/(96*c^2*f*(c - c*Sin[e + f*x])^(5/2))

Rule 2821

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

Rule 2822

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

Rule 3051

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x]
)^n/(a*f*(2*m + 1))), x] + Dist[(a*B*(m - n) + A*b*(m + n + 1))/(a*b*(2*m + 1)), 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] && EqQ[b*c + a*d, 0] && EqQ[a^2
- b^2, 0] && (LtQ[m, -2^(-1)] || (ILtQ[m + n, 0] &&  !SumSimplerQ[n, 1])) && NeQ[2*m + 1, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{8 f (c-c \sin (e+f x))^{9/2}}+\frac {(A-3 B) \int \frac {(a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{7/2}} \, dx}{4 c} \\ & = \frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{8 f (c-c \sin (e+f x))^{9/2}}+\frac {(A-3 B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{24 c f (c-c \sin (e+f x))^{7/2}}+\frac {(A-3 B) \int \frac {(a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{5/2}} \, dx}{24 c^2} \\ & = \frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{8 f (c-c \sin (e+f x))^{9/2}}+\frac {(A-3 B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{24 c f (c-c \sin (e+f x))^{7/2}}+\frac {(A-3 B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{96 c^2 f (c-c \sin (e+f x))^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 11.60 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.84 \[ \int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{9/2}} \, dx=\frac {a \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {a (1+\sin (e+f x))} (2 A+3 B-3 B \cos (2 (e+f x))+4 A \sin (e+f x))}{12 c^4 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (-1+\sin (e+f x))^4 \sqrt {c-c \sin (e+f x)}} \]

[In]

Integrate[((a + a*Sin[e + f*x])^(3/2)*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^(9/2),x]

[Out]

(a*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x])]*(2*A + 3*B - 3*B*Cos[2*(e + f*x)] + 4*A*Si
n[e + f*x]))/(12*c^4*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-1 + Sin[e + f*x])^4*Sqrt[c - c*Sin[e + f*x]])

Maple [A] (verified)

Time = 3.97 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.87

method	result	size
default	\(\frac {a \tan \left (f x +e \right ) \left (A \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right )-4 A \left (\cos ^{2}\left (f x +e \right )\right )-7 A \sin \left (f x +e \right )+3 B \sin \left (f x +e \right )+10 A \right ) \sqrt {a \left (1+\sin \left (f x +e \right )\right )}}{6 c^{4} f \left (\left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-3 \left (\cos ^{2}\left (f x +e \right )\right )-4 \sin \left (f x +e \right )+4\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}}\)	\(127\)
parts	\(-\frac {A \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, a \left (\cos ^{3}\left (f x +e \right )+4 \cos \left (f x +e \right ) \sin \left (f x +e \right )-8 \cos \left (f x +e \right )-10 \tan \left (f x +e \right )+7 \sec \left (f x +e \right )\right )}{6 f \left (\left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-3 \left (\cos ^{2}\left (f x +e \right )\right )-4 \sin \left (f x +e \right )+4\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, c^{4}}-\frac {B a \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, \left (\cos \left (f x +e \right )-\sec \left (f x +e \right )\right )}{2 f \left (\left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-3 \left (\cos ^{2}\left (f x +e \right )\right )-4 \sin \left (f x +e \right )+4\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, c^{4}}\)	\(208\)

[In]

int((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(9/2),x,method=_RETURNVERBOSE)

[Out]

1/6*a/c^4/f*tan(f*x+e)*(A*sin(f*x+e)*cos(f*x+e)^2-4*A*cos(f*x+e)^2-7*A*sin(f*x+e)+3*B*sin(f*x+e)+10*A)*(a*(1+s
in(f*x+e)))^(1/2)/(cos(f*x+e)^2*sin(f*x+e)-3*cos(f*x+e)^2-4*sin(f*x+e)+4)/(-c*(sin(f*x+e)-1))^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.92 \[ \int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{9/2}} \, dx=-\frac {{\left (3 \, B a \cos \left (f x + e\right )^{2} - 2 \, A a \sin \left (f x + e\right ) - {\left (A + 3 \, B\right )} a\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{6 \, {\left (c^{5} f \cos \left (f x + e\right )^{5} - 8 \, c^{5} f \cos \left (f x + e\right )^{3} + 8 \, c^{5} f \cos \left (f x + e\right ) + 4 \, {\left (c^{5} f \cos \left (f x + e\right )^{3} - 2 \, c^{5} f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )}} \]

[In]

integrate((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(9/2),x, algorithm="fricas")

[Out]

-1/6*(3*B*a*cos(f*x + e)^2 - 2*A*a*sin(f*x + e) - (A + 3*B)*a)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) +
c)/(c^5*f*cos(f*x + e)^5 - 8*c^5*f*cos(f*x + e)^3 + 8*c^5*f*cos(f*x + e) + 4*(c^5*f*cos(f*x + e)^3 - 2*c^5*f*
cos(f*x + e))*sin(f*x + e))

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{9/2}} \, dx=\text {Timed out} \]

[In]

integrate((a+a*sin(f*x+e))**(3/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))**(9/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{9/2}} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {9}{2}}} \,d x } \]

[In]

integrate((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(9/2),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^(3/2)/(-c*sin(f*x + e) + c)^(9/2), x)

Giac [A] (verification not implemented)

none

Time = 0.42 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.25 \[ \int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{9/2}} \, dx=-\frac {{\left (12 \, B a \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 4 \, A a \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 12 \, B a \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 3 \, A a \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 3 \, B a \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sqrt {a}}{96 \, c^{5} f \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8}} \]

[In]

integrate((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(9/2),x, algorithm="giac")

[Out]

-1/96*(12*B*a*sqrt(c)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^4 - 4*A*a*sqrt(c)*sgn
(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^2 - 12*B*a*sqrt(c)*sgn(cos(-1/4*pi + 1/2*f*x +
1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^2 + 3*A*a*sqrt(c)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 3*B*a*sqrt(c)*
sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))*sqrt(a)/(c^5*f*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x
+ 1/2*e)^8)

Mupad [B] (verification not implemented)

Time = 19.27 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.68 \[ \int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{9/2}} \, dx=\frac {\sqrt {c-c\,\sin \left (e+f\,x\right )}\,\left (\frac {8\,a\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,\left (2\,A+3\,B\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{3\,c^5\,f}+\frac {32\,A\,a\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,\sin \left (e+f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{3\,c^5\,f}-\frac {8\,B\,a\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,\cos \left (2\,e+2\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{c^5\,f}\right )}{84\,\cos \left (e+f\,x\right )\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}-54\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,\cos \left (3\,e+3\,f\,x\right )+2\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,\cos \left (5\,e+5\,f\,x\right )-96\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,\sin \left (2\,e+2\,f\,x\right )+16\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,\sin \left (4\,e+4\,f\,x\right )} \]

[In]

int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^(3/2))/(c - c*sin(e + f*x))^(9/2),x)

[Out]

((c - c*sin(e + f*x))^(1/2)*((8*a*exp(e*5i + f*x*5i)*(2*A + 3*B)*(a + a*sin(e + f*x))^(1/2))/(3*c^5*f) + (32*A
*a*exp(e*5i + f*x*5i)*sin(e + f*x)*(a + a*sin(e + f*x))^(1/2))/(3*c^5*f) - (8*B*a*exp(e*5i + f*x*5i)*cos(2*e +
2*f*x)*(a + a*sin(e + f*x))^(1/2))/(c^5*f)))/(84*cos(e + f*x)*exp(e*5i + f*x*5i) - 54*exp(e*5i + f*x*5i)*cos(
3*e + 3*f*x) + 2*exp(e*5i + f*x*5i)*cos(5*e + 5*f*x) - 96*exp(e*5i + f*x*5i)*sin(2*e + 2*f*x) + 16*exp(e*5i +
f*x*5i)*sin(4*e + 4*f*x))