\(\int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx\) [81]

Optimal result

Integrand size = 36, antiderivative size = 198 \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx=\frac {256 a (11 A-5 B) c^5 \cos ^3(e+f x)}{3465 f (c-c \sin (e+f x))^{3/2}}+\frac {64 a (11 A-5 B) c^4 \cos ^3(e+f x)}{1155 f \sqrt {c-c \sin (e+f x)}}+\frac {8 a (11 A-5 B) c^3 \cos ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{231 f}+\frac {2 a (11 A-5 B) c^2 \cos ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{99 f}-\frac {2 a B c \cos ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{11 f} \]

[Out]

256/3465*a*(11*A-5*B)*c^5*cos(f*x+e)^3/f/(c-c*sin(f*x+e))^(3/2)+2/99*a*(11*A-5*B)*c^2*cos(f*x+e)^3*(c-c*sin(f*
x+e))^(3/2)/f-2/11*a*B*c*cos(f*x+e)^3*(c-c*sin(f*x+e))^(5/2)/f+64/1155*a*(11*A-5*B)*c^4*cos(f*x+e)^3/f/(c-c*si
n(f*x+e))^(1/2)+8/231*a*(11*A-5*B)*c^3*cos(f*x+e)^3*(c-c*sin(f*x+e))^(1/2)/f

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {3046, 2935, 2753, 2752} \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx=\frac {256 a c^5 (11 A-5 B) \cos ^3(e+f x)}{3465 f (c-c \sin (e+f x))^{3/2}}+\frac {64 a c^4 (11 A-5 B) \cos ^3(e+f x)}{1155 f \sqrt {c-c \sin (e+f x)}}+\frac {8 a c^3 (11 A-5 B) \cos ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{231 f}+\frac {2 a c^2 (11 A-5 B) \cos ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{99 f}-\frac {2 a B c \cos ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{11 f} \]

[In]

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

[Out]

(256*a*(11*A - 5*B)*c^5*Cos[e + f*x]^3)/(3465*f*(c - c*Sin[e + f*x])^(3/2)) + (64*a*(11*A - 5*B)*c^4*Cos[e + f
*x]^3)/(1155*f*Sqrt[c - c*Sin[e + f*x]]) + (8*a*(11*A - 5*B)*c^3*Cos[e + f*x]^3*Sqrt[c - c*Sin[e + f*x]])/(231
*f) + (2*a*(11*A - 5*B)*c^2*Cos[e + f*x]^3*(c - c*Sin[e + f*x])^(3/2))/(99*f) - (2*a*B*c*Cos[e + f*x]^3*(c - c
*Sin[e + f*x])^(5/2))/(11*f)

Rule 2752

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b*(g*C
os[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^(m - 1)/(f*g*(m - 1))), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && Eq
Q[a^2 - b^2, 0] && EqQ[2*m + p - 1, 0] && NeQ[m, 1]

Rule 2753

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(-b)*(
g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^(m - 1)/(f*g*(m + p))), x] + Dist[a*((2*m + p - 1)/(m + p)), Int
[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2,
0] && IGtQ[Simplify[(2*m + p - 1)/2], 0] && NeQ[m + p, 0]

Rule 2935

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[(-d)*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(f*g*(m + p + 1))), x
] + Dist[(a*d*m + b*c*(m + p + 1))/(b*(m + p + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^m, x], x] /; F
reeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[Simplify[(2*m + p + 1)/2], 0] && NeQ[m + p +
1, 0]

Rule 3046

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a^m*c^m, Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m)*(A + B
*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && I
ntegerQ[m] &&  !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))

Rubi steps \begin{align*} \text {integral}& = (a c) \int \cos ^2(e+f x) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx \\ & = -\frac {2 a B c \cos ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{11 f}+\frac {1}{11} (a (11 A-5 B) c) \int \cos ^2(e+f x) (c-c \sin (e+f x))^{5/2} \, dx \\ & = \frac {2 a (11 A-5 B) c^2 \cos ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{99 f}-\frac {2 a B c \cos ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{11 f}+\frac {1}{33} \left (4 a (11 A-5 B) c^2\right ) \int \cos ^2(e+f x) (c-c \sin (e+f x))^{3/2} \, dx \\ & = \frac {8 a (11 A-5 B) c^3 \cos ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{231 f}+\frac {2 a (11 A-5 B) c^2 \cos ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{99 f}-\frac {2 a B c \cos ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{11 f}+\frac {1}{231} \left (32 a (11 A-5 B) c^3\right ) \int \cos ^2(e+f x) \sqrt {c-c \sin (e+f x)} \, dx \\ & = \frac {64 a (11 A-5 B) c^4 \cos ^3(e+f x)}{1155 f \sqrt {c-c \sin (e+f x)}}+\frac {8 a (11 A-5 B) c^3 \cos ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{231 f}+\frac {2 a (11 A-5 B) c^2 \cos ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{99 f}-\frac {2 a B c \cos ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{11 f}+\frac {\left (128 a (11 A-5 B) c^4\right ) \int \frac {\cos ^2(e+f x)}{\sqrt {c-c \sin (e+f x)}} \, dx}{1155} \\ & = \frac {256 a (11 A-5 B) c^5 \cos ^3(e+f x)}{3465 f (c-c \sin (e+f x))^{3/2}}+\frac {64 a (11 A-5 B) c^4 \cos ^3(e+f x)}{1155 f \sqrt {c-c \sin (e+f x)}}+\frac {8 a (11 A-5 B) c^3 \cos ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{231 f}+\frac {2 a (11 A-5 B) c^2 \cos ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{99 f}-\frac {2 a B c \cos ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{11 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 4.16 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.75 \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx=-\frac {a c^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 \sqrt {c-c \sin (e+f x)} (-35332 A+27085 B+60 (121 A-202 B) \cos (2 (e+f x))+315 B \cos (4 (e+f x))+30558 A \sin (e+f x)-31530 B \sin (e+f x)-770 A \sin (3 (e+f x))+2870 B \sin (3 (e+f x)))}{13860 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )} \]

[In]

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

[Out]

-1/13860*(a*c^3*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3*Sqrt[c - c*Sin[e + f*x]]*(-35332*A + 27085*B + 60*(121
*A - 202*B)*Cos[2*(e + f*x)] + 315*B*Cos[4*(e + f*x)] + 30558*A*Sin[e + f*x] - 31530*B*Sin[e + f*x] - 770*A*Si
n[3*(e + f*x)] + 2870*B*Sin[3*(e + f*x)]))/(f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]))

Maple [A] (verified)

Time = 7.46 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.60

method	result	size
default	\(\frac {2 \left (\sin \left (f x +e \right )-1\right ) c^{4} \left (1+\sin \left (f x +e \right )\right )^{2} a \left (315 B \left (\cos ^{4}\left (f x +e \right )\right )+\left (-385 A +1435 B \right ) \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+\left (1815 A -3345 B \right ) \left (\cos ^{2}\left (f x +e \right )\right )+\left (3916 A -4300 B \right ) \sin \left (f x +e \right )-5324 A +4940 B \right )}{3465 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(119\)
parts	\(\frac {2 a A \left (\sin \left (f x +e \right )-1\right ) c^{4} \left (1+\sin \left (f x +e \right )\right ) \left (5 \left (\sin ^{3}\left (f x +e \right )\right )-27 \left (\sin ^{2}\left (f x +e \right )\right )+71 \sin \left (f x +e \right )-177\right )}{35 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}+\frac {2 B a \left (\sin \left (f x +e \right )-1\right ) c^{4} \left (1+\sin \left (f x +e \right )\right ) \left (315 \left (\sin ^{5}\left (f x +e \right )\right )-1505 \left (\sin ^{4}\left (f x +e \right )\right )+3205 \left (\sin ^{3}\left (f x +e \right )\right )-4539 \left (\sin ^{2}\left (f x +e \right )\right )+6052 \sin \left (f x +e \right )-12104\right )}{3465 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}+\frac {2 a \left (A +B \right ) \left (\sin \left (f x +e \right )-1\right ) c^{4} \left (1+\sin \left (f x +e \right )\right ) \left (5 \left (\sin ^{4}\left (f x +e \right )\right )-25 \left (\sin ^{3}\left (f x +e \right )\right )+57 \left (\sin ^{2}\left (f x +e \right )\right )-91 \sin \left (f x +e \right )+182\right )}{45 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(265\)

[In]

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

[Out]

2/3465*(sin(f*x+e)-1)*c^4*(1+sin(f*x+e))^2*a*(315*B*cos(f*x+e)^4+(-385*A+1435*B)*cos(f*x+e)^2*sin(f*x+e)+(1815
*A-3345*B)*cos(f*x+e)^2+(3916*A-4300*B)*sin(f*x+e)-5324*A+4940*B)/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.45 \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx=\frac {2 \, {\left (315 \, B a c^{3} \cos \left (f x + e\right )^{6} - 35 \, {\left (11 \, A - 32 \, B\right )} a c^{3} \cos \left (f x + e\right )^{5} + 5 \, {\left (209 \, A - 221 \, B\right )} a c^{3} \cos \left (f x + e\right )^{4} + 2 \, {\left (1243 \, A - 1195 \, B\right )} a c^{3} \cos \left (f x + e\right )^{3} - 32 \, {\left (11 \, A - 5 \, B\right )} a c^{3} \cos \left (f x + e\right )^{2} + 128 \, {\left (11 \, A - 5 \, B\right )} a c^{3} \cos \left (f x + e\right ) + 256 \, {\left (11 \, A - 5 \, B\right )} a c^{3} - {\left (315 \, B a c^{3} \cos \left (f x + e\right )^{5} + 35 \, {\left (11 \, A - 23 \, B\right )} a c^{3} \cos \left (f x + e\right )^{4} + 10 \, {\left (143 \, A - 191 \, B\right )} a c^{3} \cos \left (f x + e\right )^{3} - 96 \, {\left (11 \, A - 5 \, B\right )} a c^{3} \cos \left (f x + e\right )^{2} - 128 \, {\left (11 \, A - 5 \, B\right )} a c^{3} \cos \left (f x + e\right ) - 256 \, {\left (11 \, A - 5 \, B\right )} a c^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{3465 \, {\left (f \cos \left (f x + e\right ) - f \sin \left (f x + e\right ) + f\right )}} \]

[In]

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

[Out]

2/3465*(315*B*a*c^3*cos(f*x + e)^6 - 35*(11*A - 32*B)*a*c^3*cos(f*x + e)^5 + 5*(209*A - 221*B)*a*c^3*cos(f*x +
e)^4 + 2*(1243*A - 1195*B)*a*c^3*cos(f*x + e)^3 - 32*(11*A - 5*B)*a*c^3*cos(f*x + e)^2 + 128*(11*A - 5*B)*a*c
^3*cos(f*x + e) + 256*(11*A - 5*B)*a*c^3 - (315*B*a*c^3*cos(f*x + e)^5 + 35*(11*A - 23*B)*a*c^3*cos(f*x + e)^4
+ 10*(143*A - 191*B)*a*c^3*cos(f*x + e)^3 - 96*(11*A - 5*B)*a*c^3*cos(f*x + e)^2 - 128*(11*A - 5*B)*a*c^3*cos
(f*x + e) - 256*(11*A - 5*B)*a*c^3)*sin(f*x + e))*sqrt(-c*sin(f*x + e) + c)/(f*cos(f*x + e) - f*sin(f*x + e) +
f)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

Giac [A] (verification not implemented)

none

Time = 0.46 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.49 \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx=-\frac {\sqrt {2} {\left (6930 \, B a c^{3} \cos \left (-\frac {3}{4} \, \pi + \frac {3}{2} \, f x + \frac {3}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 315 \, B a c^{3} \cos \left (-\frac {11}{4} \, \pi + \frac {11}{2} \, f x + \frac {11}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 48510 \, {\left (2 \, A a c^{3} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - B a c^{3} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 693 \, {\left (16 \, A a c^{3} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 5 \, B a c^{3} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \cos \left (-\frac {5}{4} \, \pi + \frac {5}{2} \, f x + \frac {5}{2} \, e\right ) + 495 \, {\left (10 \, A a c^{3} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 9 \, B a c^{3} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \cos \left (-\frac {7}{4} \, \pi + \frac {7}{2} \, f x + \frac {7}{2} \, e\right ) - 385 \, {\left (2 \, A a c^{3} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 5 \, B a c^{3} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \cos \left (-\frac {9}{4} \, \pi + \frac {9}{2} \, f x + \frac {9}{2} \, e\right )\right )} \sqrt {c}}{55440 \, f} \]

[In]

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

[Out]

-1/55440*sqrt(2)*(6930*B*a*c^3*cos(-3/4*pi + 3/2*f*x + 3/2*e)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 315*B*a*c^
3*cos(-11/4*pi + 11/2*f*x + 11/2*e)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) + 48510*(2*A*a*c^3*sgn(sin(-1/4*pi + 1
/2*f*x + 1/2*e)) - B*a*c^3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)))*cos(-1/4*pi + 1/2*f*x + 1/2*e) - 693*(16*A*a*c
^3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 5*B*a*c^3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)))*cos(-5/4*pi + 5/2*f*x
+ 5/2*e) + 495*(10*A*a*c^3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 9*B*a*c^3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))
)*cos(-7/4*pi + 7/2*f*x + 7/2*e) - 385*(2*A*a*c^3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 5*B*a*c^3*sgn(sin(-1/4
*pi + 1/2*f*x + 1/2*e)))*cos(-9/4*pi + 9/2*f*x + 9/2*e))*sqrt(c)/f

Mupad [F(-1)]

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

[In]

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

[Out]

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