\(\int (3+3 \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{7/2} \, dx\) [370]

Optimal result

Integrand size = 30, antiderivative size = 169 \[ \int (3+3 \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{7/2} \, dx=-\frac {162 \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{35 f \sqrt {3+3 \sin (e+f x)}}-\frac {108 \cos (e+f x) \sqrt {3+3 \sin (e+f x)} (c-c \sin (e+f x))^{7/2}}{35 f}-\frac {9 \cos (e+f x) (3+3 \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{7/2}}{7 f}-\frac {3 \cos (e+f x) (3+3 \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{7/2}}{7 f} \]

[Out]

-1/7*a^2*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(7/2)/f-1/7*a*cos(f*x+e)*(a+a*sin(f*x+e))^(5/2)*(c
-c*sin(f*x+e))^(7/2)/f-2/35*a^4*cos(f*x+e)*(c-c*sin(f*x+e))^(7/2)/f/(a+a*sin(f*x+e))^(1/2)-4/35*a^3*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 = 179, normalized size of antiderivative = 1.06, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {2819, 2817} \[ \int (3+3 \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{7/2} \, dx=-\frac {2 a^4 \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{35 f \sqrt {a \sin (e+f x)+a}}-\frac {4 a^3 \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{7/2}}{35 f}-\frac {a^2 \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{7/2}}{7 f}-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{7/2}}{7 f} \]

[In]

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

[Out]

(-2*a^4*Cos[e + f*x]*(c - c*Sin[e + f*x])^(7/2))/(35*f*Sqrt[a + a*Sin[e + f*x]]) - (4*a^3*Cos[e + f*x]*Sqrt[a
+ a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(7/2))/(35*f) - (a^2*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2)*(c - c*Sin
[e + f*x])^(7/2))/(7*f) - (a*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2)*(c - c*Sin[e + f*x])^(7/2))/(7*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])

Rubi steps \begin{align*} \text {integral}& = -\frac {a \cos (e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{7/2}}{7 f}+\frac {1}{7} (6 a) \int (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{7/2} \, dx \\ & = -\frac {a^2 \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{7/2}}{7 f}-\frac {a \cos (e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{7/2}}{7 f}+\frac {1}{7} \left (4 a^2\right ) \int (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{7/2} \, dx \\ & = -\frac {4 a^3 \cos (e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{7/2}}{35 f}-\frac {a^2 \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{7/2}}{7 f}-\frac {a \cos (e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{7/2}}{7 f}+\frac {1}{35} \left (8 a^3\right ) \int \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{7/2} \, dx \\ & = -\frac {2 a^4 \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{35 f \sqrt {a+a \sin (e+f x)}}-\frac {4 a^3 \cos (e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{7/2}}{35 f}-\frac {a^2 \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{7/2}}{7 f}-\frac {a \cos (e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{7/2}}{7 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.48 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.58 \[ \int (3+3 \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{7/2} \, dx=\frac {27 \sqrt {3} c^3 \sec ^6(e+f x) (-1+\sin (e+f x))^3 (1+\sin (e+f x))^{7/2} \sqrt {c-c \sin (e+f x)} \left (-35+35 \sin ^2(e+f x)-21 \sin ^4(e+f x)+5 \sin ^6(e+f x)\right ) \tan (e+f x)}{35 f} \]

[In]

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

[Out]

(27*Sqrt[3]*c^3*Sec[e + f*x]^6*(-1 + Sin[e + f*x])^3*(1 + Sin[e + f*x])^(7/2)*Sqrt[c - c*Sin[e + f*x]]*(-35 +
35*Sin[e + f*x]^2 - 21*Sin[e + f*x]^4 + 5*Sin[e + f*x]^6)*Tan[e + f*x])/(35*f)

Maple [A] (verified)

Time = 4.84 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.44

method	result	size
default	\(\frac {\tan \left (f x +e \right ) a^{3} c^{3} \left (5 \left (\cos ^{6}\left (f x +e \right )\right )+6 \left (\cos ^{4}\left (f x +e \right )\right )+8 \left (\cos ^{2}\left (f x +e \right )\right )+16\right ) \sqrt {a \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}}{35 f}\)	\(75\)

[In]

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

[Out]

1/35/f*tan(f*x+e)*a^3*c^3*(5*cos(f*x+e)^6+6*cos(f*x+e)^4+8*cos(f*x+e)^2+16)*(a*(sin(f*x+e)+1))^(1/2)*(-c*(sin(
f*x+e)-1))^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.60 \[ \int (3+3 \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{7/2} \, dx=\frac {{\left (5 \, a^{3} c^{3} \cos \left (f x + e\right )^{6} + 6 \, a^{3} c^{3} \cos \left (f x + e\right )^{4} + 8 \, a^{3} c^{3} \cos \left (f x + e\right )^{2} + 16 \, a^{3} c^{3}\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} \sin \left (f x + e\right )}{35 \, f \cos \left (f x + e\right )} \]

[In]

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

[Out]

1/35*(5*a^3*c^3*cos(f*x + e)^6 + 6*a^3*c^3*cos(f*x + e)^4 + 8*a^3*c^3*cos(f*x + e)^2 + 16*a^3*c^3)*sqrt(a*sin(
f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)*sin(f*x + e)/(f*cos(f*x + e))

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.21 \[ \int (3+3 \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{7/2} \, dx=-\frac {32 \, {\left (20 \, a^{3} c^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{14} - 70 \, a^{3} c^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{12} + 84 \, a^{3} c^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{10} - 35 \, a^{3} c^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8}\right )} \sqrt {a} \sqrt {c}}{35 \, f} \]

[In]

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

[Out]

-32/35*(20*a^3*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)^14 - 70*a^3*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*p
i + 1/2*f*x + 1/2*e)^12 + 84*a^3*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))*s
in(-1/4*pi + 1/2*f*x + 1/2*e)^10 - 35*a^3*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 = 11.11 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.06 \[ \int (3+3 \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{7/2} \, dx=\frac {\frac {1225\,a^3\,c^3\,\sin \left (e+f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,\sqrt {c-c\,\sin \left (e+f\,x\right )}}{32}+\frac {245\,a^3\,c^3\,\sin \left (3\,e+3\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,\sqrt {c-c\,\sin \left (e+f\,x\right )}}{32}+\frac {49\,a^3\,c^3\,\sin \left (5\,e+5\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,\sqrt {c-c\,\sin \left (e+f\,x\right )}}{32}+\frac {5\,a^3\,c^3\,\sin \left (7\,e+7\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,\sqrt {c-c\,\sin \left (e+f\,x\right )}}{32}}{70\,f\,\cos \left (e+f\,x\right )} \]

[In]

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

[Out]

((1225*a^3*c^3*sin(e + f*x)*(a + a*sin(e + f*x))^(1/2)*(c - c*sin(e + f*x))^(1/2))/32 + (245*a^3*c^3*sin(3*e +
3*f*x)*(a + a*sin(e + f*x))^(1/2)*(c - c*sin(e + f*x))^(1/2))/32 + (49*a^3*c^3*sin(5*e + 5*f*x)*(a + a*sin(e
+ f*x))^(1/2)*(c - c*sin(e + f*x))^(1/2))/32 + (5*a^3*c^3*sin(7*e + 7*f*x)*(a + a*sin(e + f*x))^(1/2)*(c - c*s
in(e + f*x))^(1/2))/32)/(70*f*cos(e + f*x))