\(\int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{7/2}} \, dx\) [146]

Optimal result

Integrand size = 40, antiderivative size = 96 \[ \int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{7/2}} \, dx=\frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{6 f (c-c \sin (e+f x))^{7/2}}+\frac {(A-5 B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{24 c f (c-c \sin (e+f x))^{5/2}} \]

[Out]

1/6*(A+B)*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)/f/(c-c*sin(f*x+e))^(7/2)+1/24*(A-5*B)*cos(f*x+e)*(a+a*sin(f*x+e))^
(3/2)/c/f/(c-c*sin(f*x+e))^(5/2)

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {3051, 2821} \[ \int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{7/2}} \, dx=\frac {(A-5 B) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{24 c f (c-c \sin (e+f x))^{5/2}}+\frac {(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{6 f (c-c \sin (e+f x))^{7/2}} \]

[In]

Int[((a + a*Sin[e + f*x])^(3/2)*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^(7/2),x]

[Out]

((A + B)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(6*f*(c - c*Sin[e + f*x])^(7/2)) + ((A - 5*B)*Cos[e + f*x]*(
a + a*Sin[e + f*x])^(3/2))/(24*c*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 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}}{6 f (c-c \sin (e+f x))^{7/2}}+\frac {(A-5 B) \int \frac {(a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{5/2}} \, dx}{6 c} \\ & = \frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{6 f (c-c \sin (e+f x))^{7/2}}+\frac {(A-5 B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{24 c f (c-c \sin (e+f x))^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 9.78 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.30 \[ \int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{7/2}} \, dx=-\frac {a \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {a (1+\sin (e+f x))} (A+4 B-3 B \cos (2 (e+f x))+3 (A-B) \sin (e+f x))}{6 c^3 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (-1+\sin (e+f x))^3 \sqrt {c-c \sin (e+f x)}} \]

[In]

Integrate[((a + a*Sin[e + f*x])^(3/2)*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^(7/2),x]

[Out]

-1/6*(a*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x])]*(A + 4*B - 3*B*Cos[2*(e + f*x)] + 3*(
A - B)*Sin[e + f*x]))/(c^3*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-1 + Sin[e + f*x])^3*Sqrt[c - c*Sin[e + f*
x]])

Maple [A] (verified)

Time = 3.35 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.08

method	result	size
default	\(\frac {a \tan \left (f x +e \right ) \left (A \left (\cos ^{2}\left (f x +e \right )\right )-B \left (\sin ^{2}\left (f x +e \right )\right )+3 A \sin \left (f x +e \right )-3 B \sin \left (f x +e \right )-7 A \right ) \sqrt {a \left (1+\sin \left (f x +e \right )\right )}}{6 c^{3} f \left (\cos ^{2}\left (f x +e \right )+2 \sin \left (f x +e \right )-2\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}}\)	\(104\)
parts	\(\frac {A \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, a \left (\cos \left (f x +e \right ) \sin \left (f x +e \right )-3 \cos \left (f x +e \right )-7 \tan \left (f x +e \right )+3 \sec \left (f x +e \right )\right )}{6 f \left (\cos ^{2}\left (f x +e \right )+2 \sin \left (f x +e \right )-2\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, c^{3}}+\frac {B \sec \left (f x +e \right ) \left (\cos \left (f x +e \right )-1\right ) \left (1+\cos \left (f x +e \right )\right ) \left (\sin \left (f x +e \right )+3\right ) \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, a}{6 f \left (\cos ^{2}\left (f x +e \right )+2 \sin \left (f x +e \right )-2\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, c^{3}}\)	\(180\)

[In]

int((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(7/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.30 \[ \int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{7/2}} \, dx=\frac {{\left (6 \, B a \cos \left (f x + e\right )^{2} - 3 \, {\left (A - B\right )} a \sin \left (f x + e\right ) - {\left (A + 7 \, B\right )} a\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{6 \, {\left (3 \, c^{4} f \cos \left (f x + e\right )^{3} - 4 \, c^{4} f \cos \left (f x + e\right ) - {\left (c^{4} f \cos \left (f x + e\right )^{3} - 4 \, c^{4} f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )}} \]

[In]

integrate((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(7/2),x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{7/2}} \, dx=\text {Timed out} \]

[In]

integrate((a+a*sin(f*x+e))**(3/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))**(7/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{7/2}} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {7}{2}}} \,d x } \]

[In]

integrate((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(7/2),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)^(7/2), x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 183 vs. \(2 (84) = 168\).

Time = 0.41 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.91 \[ \int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{7/2}} \, dx=-\frac {{\left (12 \, B a \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 3 \, A a \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 9 \, B a \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, A a \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 2 \, B a \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sqrt {a}}{24 \, c^{4} f \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6}} \]

[In]

integrate((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(7/2),x, algorithm="giac")

[Out]

-1/24*(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 - 3*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 - 9*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 + 2*A*a*sqrt(c)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 2*B*a*sqrt(c)*s
gn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))*sqrt(a)/(c^4*f*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x +
1/2*e)^6)

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{7/2}} \, dx=\int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{3/2}}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{7/2}} \,d x \]

[In]

int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^(3/2))/(c - c*sin(e + f*x))^(7/2),x)

[Out]

int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^(3/2))/(c - c*sin(e + f*x))^(7/2), x)