\(\int \frac {(a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{11/2}} \, dx\) [170]

Optimal result

Integrand size = 40, antiderivative size = 96 \[ \int \frac {(a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{11/2}} \, dx=\frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{10 f (c-c \sin (e+f x))^{11/2}}+\frac {(A-9 B) \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{80 c f (c-c \sin (e+f x))^{9/2}} \]

[Out]

1/10*(A+B)*cos(f*x+e)*(a+a*sin(f*x+e))^(7/2)/f/(c-c*sin(f*x+e))^(11/2)+1/80*(A-9*B)*cos(f*x+e)*(a+a*sin(f*x+e)
)^(7/2)/c/f/(c-c*sin(f*x+e))^(9/2)

Rubi [A] (verified)

Time = 0.20 (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} \[ \int \frac {(a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{11/2}} \, dx=\frac {(A-9 B) \cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{80 c f (c-c \sin (e+f x))^{9/2}}+\frac {(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{10 f (c-c \sin (e+f x))^{11/2}} \]

[In]

Int[((a + a*Sin[e + f*x])^(7/2)*(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])^(7/2))/(10*f*(c - c*Sin[e + f*x])^(11/2)) + ((A - 9*B)*Cos[e + f*x]
*(a + a*Sin[e + f*x])^(7/2))/(80*c*f*(c - c*Sin[e + f*x])^(9/2))

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))^{7/2}}{10 f (c-c \sin (e+f x))^{11/2}}+\frac {(A-9 B) \int \frac {(a+a \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{9/2}} \, dx}{10 c} \\ & = \frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{10 f (c-c \sin (e+f x))^{11/2}}+\frac {(A-9 B) \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{80 c f (c-c \sin (e+f x))^{9/2}} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(434\) vs. \(2(96)=192\).

Time = 17.07 (sec) , antiderivative size = 434, normalized size of antiderivative = 4.52 \[ \int \frac {(a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{11/2}} \, dx=\frac {8 (A+B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (a (1+\sin (e+f x)))^{7/2}}{5 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^7 (c-c \sin (e+f x))^{11/2}}+\frac {(-3 A-5 B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 (a (1+\sin (e+f x)))^{7/2}}{f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^7 (c-c \sin (e+f x))^{11/2}}+\frac {2 (A+3 B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^5 (a (1+\sin (e+f x)))^{7/2}}{f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^7 (c-c \sin (e+f x))^{11/2}}+\frac {(-A-7 B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^7 (a (1+\sin (e+f x)))^{7/2}}{2 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^7 (c-c \sin (e+f x))^{11/2}}+\frac {B \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^9 (a (1+\sin (e+f x)))^{7/2}}{f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^7 (c-c \sin (e+f x))^{11/2}} \]

[In]

Integrate[((a + a*Sin[e + f*x])^(7/2)*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^(11/2),x]

[Out]

(8*(A + B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(a*(1 + Sin[e + f*x]))^(7/2))/(5*f*(Cos[(e + f*x)/2] + Sin[(e
+ f*x)/2])^7*(c - c*Sin[e + f*x])^(11/2)) + ((-3*A - 5*B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3*(a*(1 + Sin
[e + f*x]))^(7/2))/(f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^7*(c - c*Sin[e + f*x])^(11/2)) + (2*(A + 3*B)*(Cos
[(e + f*x)/2] - Sin[(e + f*x)/2])^5*(a*(1 + Sin[e + f*x]))^(7/2))/(f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^7*(
c - c*Sin[e + f*x])^(11/2)) + ((-A - 7*B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^7*(a*(1 + Sin[e + f*x]))^(7/2)
)/(2*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^7*(c - c*Sin[e + f*x])^(11/2)) + (B*(Cos[(e + f*x)/2] - Sin[(e +
f*x)/2])^9*(a*(1 + Sin[e + f*x]))^(7/2))/(f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^7*(c - c*Sin[e + f*x])^(11/2
))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(189\) vs. \(2(84)=168\).

Time = 4.77 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.98

method	result	size
default	\(\frac {a^{3} \tan \left (f x +e \right ) \left (A \left (\cos ^{4}\left (f x +e \right )\right )-B \left (\sin ^{2}\left (f x +e \right )\right ) \left (\cos ^{2}\left (f x +e \right )\right )+5 A \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right )+5 B \left (\sin ^{3}\left (f x +e \right )\right )-17 A \left (\cos ^{2}\left (f x +e \right )\right )+6 B \left (\sin ^{2}\left (f x +e \right )\right )-10 A \sin \left (f x +e \right )+5 B \sin \left (f x +e \right )+26 A \right ) \sqrt {a \left (1+\sin \left (f x +e \right )\right )}}{10 c^{5} f \left (\cos ^{4}\left (f x +e \right )+4 \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-8 \left (\cos ^{2}\left (f x +e \right )\right )-8 \sin \left (f x +e \right )+8\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}}\)	\(190\)
parts	\(\frac {A \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, a^{3} \left (\left (\cos ^{3}\left (f x +e \right )\right ) \sin \left (f x +e \right )-5 \left (\cos ^{3}\left (f x +e \right )\right )-17 \cos \left (f x +e \right ) \sin \left (f x +e \right )+15 \cos \left (f x +e \right )+26 \tan \left (f x +e \right )-10 \sec \left (f x +e \right )\right )}{10 f \left (\cos ^{4}\left (f x +e \right )+4 \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-8 \left (\cos ^{2}\left (f x +e \right )\right )-8 \sin \left (f x +e \right )+8\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, c^{5}}+\frac {B \sec \left (f x +e \right ) \left (\cos \left (f x +e \right )-1\right ) \left (1+\cos \left (f x +e \right )\right ) \left (\left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+5 \left (\cos ^{2}\left (f x +e \right )\right )-6 \sin \left (f x +e \right )-10\right ) \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, a^{3}}{10 f \left (\cos ^{4}\left (f x +e \right )+4 \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-8 \left (\cos ^{2}\left (f x +e \right )\right )-8 \sin \left (f x +e \right )+8\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, c^{5}}\)	\(289\)

[In]

int((a+a*sin(f*x+e))^(7/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(11/2),x,method=_RETURNVERBOSE)

[Out]

1/10*a^3/c^5/f*tan(f*x+e)*(A*cos(f*x+e)^4-B*sin(f*x+e)^2*cos(f*x+e)^2+5*A*sin(f*x+e)*cos(f*x+e)^2+5*B*sin(f*x+
e)^3-17*A*cos(f*x+e)^2+6*B*sin(f*x+e)^2-10*A*sin(f*x+e)+5*B*sin(f*x+e)+26*A)*(a*(1+sin(f*x+e)))^(1/2)/(cos(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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 199 vs. \(2 (84) = 168\).

Time = 0.30 (sec) , antiderivative size = 199, normalized size of antiderivative = 2.07 \[ \int \frac {(a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{11/2}} \, dx=\frac {{\left (10 \, B a^{3} \cos \left (f x + e\right )^{4} - 5 \, {\left (A + 7 \, B\right )} a^{3} \cos \left (f x + e\right )^{2} + 2 \, {\left (3 \, A + 13 \, B\right )} a^{3} - 5 \, {\left ({\left (A - B\right )} a^{3} \cos \left (f x + e\right )^{2} - 2 \, {\left (A - B\right )} a^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{10 \, {\left (5 \, c^{6} f \cos \left (f x + e\right )^{5} - 20 \, c^{6} f \cos \left (f x + e\right )^{3} + 16 \, c^{6} f \cos \left (f x + e\right ) - {\left (c^{6} f \cos \left (f x + e\right )^{5} - 12 \, c^{6} f \cos \left (f x + e\right )^{3} + 16 \, c^{6} f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )}} \]

[In]

integrate((a+a*sin(f*x+e))^(7/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(11/2),x, algorithm="fricas")

[Out]

1/10*(10*B*a^3*cos(f*x + e)^4 - 5*(A + 7*B)*a^3*cos(f*x + e)^2 + 2*(3*A + 13*B)*a^3 - 5*((A - B)*a^3*cos(f*x +
e)^2 - 2*(A - B)*a^3)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(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. \[ \int \frac {(a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{11/2}} \, dx=\text {Timed out} \]

[In]

integrate((a+a*sin(f*x+e))**(7/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))**(11/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {(a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{11/2}} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {7}{2}}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {11}{2}}} \,d x } \]

[In]

integrate((a+a*sin(f*x+e))^(7/2)*(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)^(7/2)/(-c*sin(f*x + e) + c)^(11/2), x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 341 vs. \(2 (84) = 168\).

Time = 0.42 (sec) , antiderivative size = 341, normalized size of antiderivative = 3.55 \[ \int \frac {(a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{11/2}} \, dx=\frac {{\left (40 \, B a^{3} \sqrt {c} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 10 \, A a^{3} \sqrt {c} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 90 \, B a^{3} \sqrt {c} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 10 \, A a^{3} \sqrt {c} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 90 \, B a^{3} \sqrt {c} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 5 \, A a^{3} \sqrt {c} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 45 \, B a^{3} \sqrt {c} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - A a^{3} \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 9 \, B a^{3} \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}}{80 \, {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{5} c^{6} f \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} \]

[In]

integrate((a+a*sin(f*x+e))^(7/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(11/2),x, algorithm="giac")

[Out]

1/80*(40*B*a^3*sqrt(c)*cos(-1/4*pi + 1/2*f*x + 1/2*e)^8*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 10*A*a^3*sqrt(c)
*cos(-1/4*pi + 1/2*f*x + 1/2*e)^6*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) - 90*B*a^3*sqrt(c)*cos(-1/4*pi + 1/2*f*x
+ 1/2*e)^6*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) - 10*A*a^3*sqrt(c)*cos(-1/4*pi + 1/2*f*x + 1/2*e)^4*sgn(cos(-1
/4*pi + 1/2*f*x + 1/2*e)) + 90*B*a^3*sqrt(c)*cos(-1/4*pi + 1/2*f*x + 1/2*e)^4*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*
e)) + 5*A*a^3*sqrt(c)*cos(-1/4*pi + 1/2*f*x + 1/2*e)^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) - 45*B*a^3*sqrt(c)*
cos(-1/4*pi + 1/2*f*x + 1/2*e)^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) - A*a^3*sqrt(c)*sgn(cos(-1/4*pi + 1/2*f*x
+ 1/2*e)) + 9*B*a^3*sqrt(c)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))*sqrt(a)/((cos(-1/4*pi + 1/2*f*x + 1/2*e)^2 -
1)^5*c^6*f*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)))

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{11/2}} \, dx=\int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{7/2}}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{11/2}} \,d x \]

[In]

int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^(7/2))/(c - c*sin(e + f*x))^(11/2),x)

[Out]

int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^(7/2))/(c - c*sin(e + f*x))^(11/2), x)