\(\int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx\) [139]

Optimal result

Integrand size = 40, antiderivative size = 146 \[ \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx=-\frac {a^2 (3 A-B) \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{30 f \sqrt {a+a \sin (e+f x)}}-\frac {a (3 A-B) \cos (e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{7/2}}{15 f}-\frac {B \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{7/2}}{6 f} \]

[Out]

-1/6*B*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(7/2)/f-1/30*a^2*(3*A-B)*cos(f*x+e)*(c-c*sin(f*x+e))
^(7/2)/f/(a+a*sin(f*x+e))^(1/2)-1/15*a*(3*A-B)*cos(f*x+e)*(c-c*sin(f*x+e))^(7/2)*(a+a*sin(f*x+e))^(1/2)/f

Rubi [A] (verified)

Time = 0.26 (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 = {3052, 2819, 2817} \[ \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx=-\frac {a^2 (3 A-B) \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{30 f \sqrt {a \sin (e+f x)+a}}-\frac {a (3 A-B) \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{7/2}}{15 f}-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{7/2}}{6 f} \]

[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]

-1/30*(a^2*(3*A - B)*Cos[e + f*x]*(c - c*Sin[e + f*x])^(7/2))/(f*Sqrt[a + a*Sin[e + f*x]]) - (a*(3*A - B)*Cos[
e + f*x]*Sqrt[a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(7/2))/(15*f) - (B*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3
/2)*(c - c*Sin[e + f*x])^(7/2))/(6*f)

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 2819

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 - 1)*((c + d*Sin[e + f*x])^n/(f*(m + n))), x] + Dist[a*((2*m - 1)/(
m + n)), 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, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[m - 1/2, 0] &&  !LtQ[n, -1] &&  !(IGtQ[n - 1/2, 0] && LtQ[n, m
]) &&  !(ILtQ[m + n, 0] && GtQ[2*m + n + 1, 0])

Rule 3052

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((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/(f*
(m + n + 1))), x] - Dist[(B*c*(m - n) - A*d*(m + n + 1))/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^m*(c + d*Si
n[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)] && NeQ[m + n + 1, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {B \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{7/2}}{6 f}+\frac {1}{3} (3 A-B) \int (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{7/2} \, dx \\ & = -\frac {a (3 A-B) \cos (e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{7/2}}{15 f}-\frac {B \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{7/2}}{6 f}+\frac {1}{15} (2 a (3 A-B)) \int \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{7/2} \, dx \\ & = -\frac {a^2 (3 A-B) \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{30 f \sqrt {a+a \sin (e+f x)}}-\frac {a (3 A-B) \cos (e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{7/2}}{15 f}-\frac {B \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{7/2}}{6 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 4.91 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.40 \[ \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx=-\frac {c^3 (-1+\sin (e+f x))^3 (a (1+\sin (e+f x)))^{3/2} \sqrt {c-c \sin (e+f x)} (15 (16 A-11 B) \cos (2 (e+f x))+30 (2 A-B) \cos (4 (e+f x))+5 B \cos (6 (e+f x))+840 A \sin (e+f x)-240 B \sin (e+f x)+60 A \sin (3 (e+f x))+40 B \sin (3 (e+f x))-12 A \sin (5 (e+f x))+24 B \sin (5 (e+f x)))}{960 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^7 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3} \]

[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]

-1/960*(c^3*(-1 + Sin[e + f*x])^3*(a*(1 + Sin[e + f*x]))^(3/2)*Sqrt[c - c*Sin[e + f*x]]*(15*(16*A - 11*B)*Cos[
2*(e + f*x)] + 30*(2*A - B)*Cos[4*(e + f*x)] + 5*B*Cos[6*(e + f*x)] + 840*A*Sin[e + f*x] - 240*B*Sin[e + f*x]
+ 60*A*Sin[3*(e + f*x)] + 40*B*Sin[3*(e + f*x)] - 12*A*Sin[5*(e + f*x)] + 24*B*Sin[5*(e + f*x)]))/(f*(Cos[(e +
f*x)/2] - Sin[(e + f*x)/2])^7*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3)

Maple [A] (verified)

Time = 3.31 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.92

method	result	size
default	\(-\frac {a \,c^{3} \tan \left (f x +e \right ) \left (5 B \left (\sin ^{5}\left (f x +e \right )\right )+6 A \left (\cos ^{4}\left (f x +e \right )\right )-12 \left (\sin ^{4}\left (f x +e \right )\right ) B +15 A \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right )-12 A \left (\cos ^{2}\left (f x +e \right )\right )+20 B \left (\sin ^{2}\left (f x +e \right )\right )+15 A \sin \left (f x +e \right )-15 B \sin \left (f x +e \right )-24 A \right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+\sin \left (f x +e \right )\right )}}{30 f}\)	\(135\)
parts	\(-\frac {A \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, c^{3} a \left (2 \left (\cos ^{3}\left (f x +e \right )\right ) \sin \left (f x +e \right )-5 \left (\cos ^{3}\left (f x +e \right )\right )-4 \cos \left (f x +e \right ) \sin \left (f x +e \right )-8 \tan \left (f x +e \right )+5 \sec \left (f x +e \right )\right )}{10 f}+\frac {B \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, c^{3} a \left (5 \left (\cos ^{5}\left (f x +e \right )\right )+12 \left (\cos ^{3}\left (f x +e \right )\right ) \sin \left (f x +e \right )-15 \left (\cos ^{3}\left (f x +e \right )\right )-4 \cos \left (f x +e \right ) \sin \left (f x +e \right )-8 \tan \left (f x +e \right )+10 \sec \left (f x +e \right )\right )}{30 f}\)	\(196\)

[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]

-1/30*a*c^3/f*tan(f*x+e)*(5*B*sin(f*x+e)^5+6*A*cos(f*x+e)^4-12*sin(f*x+e)^4*B+15*A*sin(f*x+e)*cos(f*x+e)^2-12*
A*cos(f*x+e)^2+20*B*sin(f*x+e)^2+15*A*sin(f*x+e)-15*B*sin(f*x+e)-24*A)*(-c*(sin(f*x+e)-1))^(1/2)*(a*(1+sin(f*x
+e)))^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.01 \[ \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx=\frac {{\left (5 \, B a c^{3} \cos \left (f x + e\right )^{6} + 15 \, {\left (A - B\right )} a c^{3} \cos \left (f x + e\right )^{4} - 5 \, {\left (3 \, A - 2 \, B\right )} a c^{3} - 2 \, {\left (3 \, {\left (A - 2 \, B\right )} a c^{3} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, A - B\right )} a c^{3} \cos \left (f x + e\right )^{2} - 4 \, {\left (3 \, A - B\right )} a c^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{30 \, f \cos \left (f x + e\right )} \]

[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]

1/30*(5*B*a*c^3*cos(f*x + e)^6 + 15*(A - B)*a*c^3*cos(f*x + e)^4 - 5*(3*A - 2*B)*a*c^3 - 2*(3*(A - 2*B)*a*c^3*
cos(f*x + e)^4 - 2*(3*A - B)*a*c^3*cos(f*x + e)^2 - 4*(3*A - B)*a*c^3)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*
sqrt(-c*sin(f*x + e) + c)/(f*cos(f*x + e))

Sympy [F(-1)]

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

[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]

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

[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 [A] (verification not implemented)

none

Time = 0.50 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.69 \[ \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx=\frac {8 \, {\left (20 \, B a c^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{12} - 12 \, A a c^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{10} - 36 \, B a c^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{10} + 15 \, A a c^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8} + 15 \, B a c^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8}\right )} \sqrt {a} \sqrt {c}}{15 \, f} \]

[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]

8/15*(20*B*a*c^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x
+ 1/2*e)^12 - 12*A*a*c^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi
+ 1/2*f*x + 1/2*e)^10 - 36*B*a*c^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin
(-1/4*pi + 1/2*f*x + 1/2*e)^10 + 15*A*a*c^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/
2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^8 + 15*B*a*c^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*
f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^8)*sqrt(a)*sqrt(c)/f

Mupad [B] (verification not implemented)

Time = 17.67 (sec) , antiderivative size = 323, normalized size of antiderivative = 2.21 \[ \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx=\frac {{\mathrm {e}}^{-e\,6{}\mathrm {i}-f\,x\,6{}\mathrm {i}}\,\sqrt {c-c\,\sin \left (e+f\,x\right )}\,\left (\frac {B\,a\,c^3\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\cos \left (6\,e+6\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{96\,f}-\frac {a\,c^3\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\sin \left (e+f\,x\right )\,\left (A\,7{}\mathrm {i}-B\,2{}\mathrm {i}\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,1{}\mathrm {i}}{4\,f}+\frac {a\,c^3\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\cos \left (4\,e+4\,f\,x\right )\,\left (2\,A-B\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{16\,f}+\frac {a\,c^3\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\cos \left (2\,e+2\,f\,x\right )\,\left (16\,A-11\,B\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{32\,f}-\frac {a\,c^3\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\sin \left (3\,e+3\,f\,x\right )\,\left (A\,3{}\mathrm {i}+B\,2{}\mathrm {i}\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,1{}\mathrm {i}}{24\,f}+\frac {a\,c^3\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\sin \left (5\,e+5\,f\,x\right )\,\left (A\,1{}\mathrm {i}-B\,2{}\mathrm {i}\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,1{}\mathrm {i}}{40\,f}\right )}{2\,\cos \left (e+f\,x\right )} \]

[In]

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

[Out]

(exp(- e*6i - f*x*6i)*(c - c*sin(e + f*x))^(1/2)*((B*a*c^3*exp(e*6i + f*x*6i)*cos(6*e + 6*f*x)*(a + a*sin(e +
f*x))^(1/2))/(96*f) - (a*c^3*exp(e*6i + f*x*6i)*sin(e + f*x)*(A*7i - B*2i)*(a + a*sin(e + f*x))^(1/2)*1i)/(4*f
) + (a*c^3*exp(e*6i + f*x*6i)*cos(4*e + 4*f*x)*(2*A - B)*(a + a*sin(e + f*x))^(1/2))/(16*f) + (a*c^3*exp(e*6i
+ f*x*6i)*cos(2*e + 2*f*x)*(16*A - 11*B)*(a + a*sin(e + f*x))^(1/2))/(32*f) - (a*c^3*exp(e*6i + f*x*6i)*sin(3*
e + 3*f*x)*(A*3i + B*2i)*(a + a*sin(e + f*x))^(1/2)*1i)/(24*f) + (a*c^3*exp(e*6i + f*x*6i)*sin(5*e + 5*f*x)*(A
*1i - B*2i)*(a + a*sin(e + f*x))^(1/2)*1i)/(40*f)))/(2*cos(e + f*x))