\(\int (3+3 \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2} \, dx\) [359]

Optimal result

Integrand size = 30, antiderivative size = 127 \[ \int (3+3 \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2} \, dx=-\frac {18 \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{5 f \sqrt {3+3 \sin (e+f x)}}-\frac {9 \cos (e+f x) \sqrt {3+3 \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}{5 f}-\frac {3 \cos (e+f x) (3+3 \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}}{5 f} \]

[Out]

-1/5*a*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(5/2)/f-2/15*a^3*cos(f*x+e)*(c-c*sin(f*x+e))^(5/2)/f
/(a+a*sin(f*x+e))^(1/2)-1/5*a^2*cos(f*x+e)*(c-c*sin(f*x+e))^(5/2)*(a+a*sin(f*x+e))^(1/2)/f

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.06, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {2819, 2817} \[ \int (3+3 \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2} \, dx=-\frac {2 a^3 \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{15 f \sqrt {a \sin (e+f x)+a}}-\frac {a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}{5 f}-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}{5 f} \]

[In]

Int[(a + a*Sin[e + f*x])^(5/2)*(c - c*Sin[e + f*x])^(5/2),x]

[Out]

(-2*a^3*Cos[e + f*x]*(c - c*Sin[e + f*x])^(5/2))/(15*f*Sqrt[a + a*Sin[e + f*x]]) - (a^2*Cos[e + f*x]*Sqrt[a +
a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(5/2))/(5*f) - (a*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2)*(c - c*Sin[e +
f*x])^(5/2))/(5*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))^{3/2} (c-c \sin (e+f x))^{5/2}}{5 f}+\frac {1}{5} (4 a) \int (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2} \, dx \\ & = -\frac {a^2 \cos (e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}{5 f}-\frac {a \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}}{5 f}+\frac {1}{5} \left (2 a^2\right ) \int \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2} \, dx \\ & = -\frac {2 a^3 \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{15 f \sqrt {a+a \sin (e+f x)}}-\frac {a^2 \cos (e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}{5 f}-\frac {a \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}}{5 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.93 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.69 \[ \int (3+3 \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2} \, dx=\frac {3 \sqrt {3} c^2 \sec ^4(e+f x) (-1+\sin (e+f x))^2 (1+\sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)} \left (15-10 \sin ^2(e+f x)+3 \sin ^4(e+f x)\right ) \tan (e+f x)}{5 f} \]

[In]

Integrate[(3 + 3*Sin[e + f*x])^(5/2)*(c - c*Sin[e + f*x])^(5/2),x]

[Out]

(3*Sqrt[3]*c^2*Sec[e + f*x]^4*(-1 + Sin[e + f*x])^2*(1 + Sin[e + f*x])^(5/2)*Sqrt[c - c*Sin[e + f*x]]*(15 - 10
*Sin[e + f*x]^2 + 3*Sin[e + f*x]^4)*Tan[e + f*x])/(5*f)

Maple [A] (verified)

Time = 4.64 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.51

method	result	size
default	\(\frac {\tan \left (f x +e \right ) a^{2} c^{2} \left (3 \left (\cos ^{4}\left (f x +e \right )\right )+4 \left (\cos ^{2}\left (f x +e \right )\right )+8\right ) \sqrt {a \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}}{15 f}\)	\(65\)

[In]

int((a+a*sin(f*x+e))^(5/2)*(c-c*sin(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/15/f*tan(f*x+e)*a^2*c^2*(3*cos(f*x+e)^4+4*cos(f*x+e)^2+8)*(a*(sin(f*x+e)+1))^(1/2)*(-c*(sin(f*x+e)-1))^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.67 \[ \int (3+3 \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2} \, dx=\frac {{\left (3 \, a^{2} c^{2} \cos \left (f x + e\right )^{4} + 4 \, a^{2} c^{2} \cos \left (f x + e\right )^{2} + 8 \, a^{2} c^{2}\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} \sin \left (f x + e\right )}{15 \, f \cos \left (f x + e\right )} \]

[In]

integrate((a+a*sin(f*x+e))^(5/2)*(c-c*sin(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

1/15*(3*a^2*c^2*cos(f*x + e)^4 + 4*a^2*c^2*cos(f*x + e)^2 + 8*a^2*c^2)*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. \[ \int (3+3 \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2} \, dx=\text {Timed out} \]

[In]

integrate((a+a*sin(f*x+e))**(5/2)*(c-c*sin(f*x+e))**(5/2),x)

[Out]

Timed out

Maxima [F]

\[ \int (3+3 \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2} \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}} \,d x } \]

[In]

integrate((a+a*sin(f*x+e))^(5/2)*(c-c*sin(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^(5/2)*(-c*sin(f*x + e) + c)^(5/2), x)

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.23 \[ \int (3+3 \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2} \, dx=\frac {16 \, {\left (6 \, a^{2} c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{10} - 15 \, a^{2} c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8} + 10 \, a^{2} c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6}\right )} \sqrt {a} \sqrt {c}}{15 \, f} \]

[In]

integrate((a+a*sin(f*x+e))^(5/2)*(c-c*sin(f*x+e))^(5/2),x, algorithm="giac")

[Out]

16/15*(6*a^2*c^2*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^2*c^2*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 + 10*a^2*c^2*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)^6)*sqrt(a)*sqrt(c)/f

Mupad [B] (verification not implemented)

Time = 1.61 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.65 \[ \int (3+3 \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2} \, dx=\frac {a^2\,c^2\,\sqrt {a\,\left (\sin \left (e+f\,x\right )+1\right )}\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (175\,\sin \left (2\,e+2\,f\,x\right )+28\,\sin \left (4\,e+4\,f\,x\right )+3\,\sin \left (6\,e+6\,f\,x\right )\right )}{240\,f\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )} \]

[In]

int((a + a*sin(e + f*x))^(5/2)*(c - c*sin(e + f*x))^(5/2),x)

[Out]

(a^2*c^2*(a*(sin(e + f*x) + 1))^(1/2)*(-c*(sin(e + f*x) - 1))^(1/2)*(175*sin(2*e + 2*f*x) + 28*sin(4*e + 4*f*x
) + 3*sin(6*e + 6*f*x)))/(240*f*(cos(2*e + 2*f*x) + 1))