\(\int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{11/2}} \, dx\) [148]

Optimal result

Integrand size = 40, antiderivative size = 154 \[ \int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{11/2}} \, dx=\frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{10 f (c-c \sin (e+f x))^{11/2}}+\frac {a (3 A-7 B) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{40 c f (c-c \sin (e+f x))^{9/2}}-\frac {a^2 (3 A-7 B) \cos (e+f x)}{120 c^2 f \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{7/2}} \]

[Out]

1/10*(A+B)*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)/f/(c-c*sin(f*x+e))^(11/2)-1/120*a^2*(3*A-7*B)*cos(f*x+e)/c^2/f/(c
-c*sin(f*x+e))^(7/2)/(a+a*sin(f*x+e))^(1/2)+1/40*a*(3*A-7*B)*cos(f*x+e)*(a+a*sin(f*x+e))^(1/2)/c/f/(c-c*sin(f*
x+e))^(9/2)

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 154, 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, 2818, 2817} \[ \int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{11/2}} \, dx=-\frac {a^2 (3 A-7 B) \cos (e+f x)}{120 c^2 f \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{7/2}}+\frac {a (3 A-7 B) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{40 c f (c-c \sin (e+f x))^{9/2}}+\frac {(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{10 f (c-c \sin (e+f x))^{11/2}} \]

[In]

Int[((a + a*Sin[e + f*x])^(3/2)*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^(11/2),x]

[Out]

((A + B)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(10*f*(c - c*Sin[e + f*x])^(11/2)) + (a*(3*A - 7*B)*Cos[e +
f*x]*Sqrt[a + a*Sin[e + f*x]])/(40*c*f*(c - c*Sin[e + f*x])^(9/2)) - (a^2*(3*A - 7*B)*Cos[e + f*x])/(120*c^2*f
*Sqrt[a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(7/2))

Rule 2817

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

Rule 2818

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

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}}{10 f (c-c \sin (e+f x))^{11/2}}+\frac {(3 A-7 B) \int \frac {(a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{9/2}} \, dx}{10 c} \\ & = \frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{10 f (c-c \sin (e+f x))^{11/2}}+\frac {a (3 A-7 B) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{40 c f (c-c \sin (e+f x))^{9/2}}-\frac {(a (3 A-7 B)) \int \frac {\sqrt {a+a \sin (e+f x)}}{(c-c \sin (e+f x))^{7/2}} \, dx}{40 c^2} \\ & = \frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{10 f (c-c \sin (e+f x))^{11/2}}+\frac {a (3 A-7 B) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{40 c f (c-c \sin (e+f x))^{9/2}}-\frac {a^2 (3 A-7 B) \cos (e+f x)}{120 c^2 f \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 11.91 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.82 \[ \int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{11/2}} \, dx=-\frac {a \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {a (1+\sin (e+f x))} (9 (A+B)-10 B \cos (2 (e+f x))+5 (3 A+B) \sin (e+f x))}{60 c^5 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (-1+\sin (e+f x))^5 \sqrt {c-c \sin (e+f x)}} \]

[In]

Integrate[((a + a*Sin[e + f*x])^(3/2)*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^(11/2),x]

[Out]

-1/60*(a*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x])]*(9*(A + B) - 10*B*Cos[2*(e + f*x)] +
5*(3*A + B)*Sin[e + f*x]))/(c^5*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-1 + Sin[e + f*x])^5*Sqrt[c - c*Sin[
e + f*x]])

Maple [A] (verified)

Time = 4.24 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.22

method	result	size
default	\(\frac {a \tan \left (f x +e \right ) \left (9 A \left (\cos ^{4}\left (f x +e \right )\right )+B \left (\sin ^{2}\left (f x +e \right )\right ) \left (\cos ^{2}\left (f x +e \right )\right )+45 A \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right )+5 B \left (\sin ^{3}\left (f x +e \right )\right )-108 A \left (\cos ^{2}\left (f x +e \right )\right )-11 B \left (\sin ^{2}\left (f x +e \right )\right )-135 A \sin \left (f x +e \right )+30 B \sin \left (f x +e \right )+159 A \right ) \sqrt {a \left (1+\sin \left (f x +e \right )\right )}}{60 c^{5} f \left (\cos ^{4}\left (f x +e \right )+4 \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-8 \left (\cos ^{2}\left (f x +e \right )\right )-8 \sin \left (f x +e \right )+8\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}}\)	\(188\)
parts	\(\frac {A \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, a \left (3 \left (\cos ^{3}\left (f x +e \right )\right ) \sin \left (f x +e \right )-15 \left (\cos ^{3}\left (f x +e \right )\right )-36 \cos \left (f x +e \right ) \sin \left (f x +e \right )+60 \cos \left (f x +e \right )+53 \tan \left (f x +e \right )-45 \sec \left (f x +e \right )\right )}{20 f \left (\cos ^{4}\left (f x +e \right )+4 \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-8 \left (\cos ^{2}\left (f x +e \right )\right )-8 \sin \left (f x +e \right )+8\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, c^{5}}-\frac {B \sec \left (f x +e \right ) \left (\cos \left (f x +e \right )-1\right ) \left (1+\cos \left (f x +e \right )\right ) \left (\left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-5 \left (\cos ^{2}\left (f x +e \right )\right )-11 \sin \left (f x +e \right )+35\right ) \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, a}{60 f \left (\cos ^{4}\left (f x +e \right )+4 \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-8 \left (\cos ^{2}\left (f x +e \right )\right )-8 \sin \left (f x +e \right )+8\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, c^{5}}\)	\(286\)

[In]

int((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(11/2),x,method=_RETURNVERBOSE)

[Out]

1/60*a/c^5/f*tan(f*x+e)*(9*A*cos(f*x+e)^4+B*sin(f*x+e)^2*cos(f*x+e)^2+45*A*sin(f*x+e)*cos(f*x+e)^2+5*B*sin(f*x
+e)^3-108*A*cos(f*x+e)^2-11*B*sin(f*x+e)^2-135*A*sin(f*x+e)+30*B*sin(f*x+e)+159*A)*(a*(1+sin(f*x+e)))^(1/2)/(c
os(f*x+e)^4+4*cos(f*x+e)^2*sin(f*x+e)-8*cos(f*x+e)^2-8*sin(f*x+e)+8)/(-c*(sin(f*x+e)-1))^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.01 \[ \int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{11/2}} \, dx=-\frac {{\left (20 \, B a \cos \left (f x + e\right )^{2} - 5 \, {\left (3 \, A + B\right )} a \sin \left (f x + e\right ) - {\left (9 \, A + 19 \, B\right )} a\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{60 \, {\left (5 \, c^{6} f \cos \left (f x + e\right )^{5} - 20 \, c^{6} f \cos \left (f x + e\right )^{3} + 16 \, c^{6} f \cos \left (f x + e\right ) - {\left (c^{6} f \cos \left (f x + e\right )^{5} - 12 \, c^{6} f \cos \left (f x + e\right )^{3} + 16 \, c^{6} f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )}} \]

[In]

integrate((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(11/2),x, algorithm="fricas")

[Out]

-1/60*(20*B*a*cos(f*x + e)^2 - 5*(3*A + B)*a*sin(f*x + e) - (9*A + 19*B)*a)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*s
in(f*x + e) + c)/(5*c^6*f*cos(f*x + e)^5 - 20*c^6*f*cos(f*x + e)^3 + 16*c^6*f*cos(f*x + e) - (c^6*f*cos(f*x +
e)^5 - 12*c^6*f*cos(f*x + e)^3 + 16*c^6*f*cos(f*x + e))*sin(f*x + e))

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{11/2}} \, dx=\text {Timed out} \]

[In]

integrate((a+a*sin(f*x+e))**(3/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))**(11/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{11/2}} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {11}{2}}} \,d x } \]

[In]

integrate((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(11/2),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)^(11/2), x)

Giac [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.19 \[ \int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{11/2}} \, dx=-\frac {{\left (40 \, B a \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 15 \, A a \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 45 \, B a \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 12 \, A a \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 12 \, B a \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sqrt {a}}{960 \, c^{6} f \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{10}} \]

[In]

integrate((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(11/2),x, algorithm="giac")

[Out]

-1/960*(40*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 - 15*A*a*sqrt(c)*s
gn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^2 - 45*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 + 12*A*a*sqrt(c)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 12*B*a*sqrt
(c)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))*sqrt(a)/(c^6*f*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*
f*x + 1/2*e)^10)

Mupad [B] (verification not implemented)

Time = 21.01 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.81 \[ \int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{11/2}} \, dx=\frac {\sqrt {c-c\,\sin \left (e+f\,x\right )}\,\left (\frac {a\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\left (A+B\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,48{}\mathrm {i}}{5\,c^6\,f}-\frac {B\,a\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\cos \left (2\,e+2\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,32{}\mathrm {i}}{3\,c^6\,f}+\frac {16\,a\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\sin \left (e+f\,x\right )\,\left (A\,3{}\mathrm {i}+B\,1{}\mathrm {i}\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{3\,c^6\,f}\right )}{\cos \left (e+f\,x\right )\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,264{}\mathrm {i}-{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\cos \left (3\,e+3\,f\,x\right )\,220{}\mathrm {i}+{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\cos \left (5\,e+5\,f\,x\right )\,20{}\mathrm {i}-{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\sin \left (2\,e+2\,f\,x\right )\,330{}\mathrm {i}+{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\sin \left (4\,e+4\,f\,x\right )\,88{}\mathrm {i}-{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\sin \left (6\,e+6\,f\,x\right )\,2{}\mathrm {i}} \]

[In]

int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^(3/2))/(c - c*sin(e + f*x))^(11/2),x)

[Out]

((c - c*sin(e + f*x))^(1/2)*((a*exp(e*6i + f*x*6i)*(A + B)*(a + a*sin(e + f*x))^(1/2)*48i)/(5*c^6*f) - (B*a*ex
p(e*6i + f*x*6i)*cos(2*e + 2*f*x)*(a + a*sin(e + f*x))^(1/2)*32i)/(3*c^6*f) + (16*a*exp(e*6i + f*x*6i)*sin(e +
f*x)*(A*3i + B*1i)*(a + a*sin(e + f*x))^(1/2))/(3*c^6*f)))/(cos(e + f*x)*exp(e*6i + f*x*6i)*264i - exp(e*6i +
f*x*6i)*cos(3*e + 3*f*x)*220i + exp(e*6i + f*x*6i)*cos(5*e + 5*f*x)*20i - exp(e*6i + f*x*6i)*sin(2*e + 2*f*x)
*330i + exp(e*6i + f*x*6i)*sin(4*e + 4*f*x)*88i - exp(e*6i + f*x*6i)*sin(6*e + 6*f*x)*2i)