\(\int (3+3 \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2} \, dx\) [349]

Optimal result

Integrand size = 30, antiderivative size = 85 \[ \int (3+3 \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2} \, dx=-\frac {3 \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{2 f \sqrt {3+3 \sin (e+f x)}}-\frac {3 \cos (e+f x) \sqrt {3+3 \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}{4 f} \]

[Out]

-1/6*a^2*cos(f*x+e)*(c-c*sin(f*x+e))^(5/2)/f/(a+a*sin(f*x+e))^(1/2)-1/4*a*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 = 0.13 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.05, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {2819, 2817} \[ \int (3+3 \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2} \, dx=-\frac {a^2 \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{6 f \sqrt {a \sin (e+f x)+a}}-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}{4 f} \]

[In]

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

[Out]

-1/6*(a^2*Cos[e + f*x]*(c - c*Sin[e + f*x])^(5/2))/(f*Sqrt[a + a*Sin[e + f*x]]) - (a*Cos[e + f*x]*Sqrt[a + a*S
in[e + f*x]]*(c - c*Sin[e + f*x])^(5/2))/(4*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) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}{4 f}+\frac {1}{2} a \int \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2} \, dx \\ & = -\frac {a^2 \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{6 f \sqrt {a+a \sin (e+f x)}}-\frac {a \cos (e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}{4 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.95 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.65 \[ \int (3+3 \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2} \, dx=\frac {\sqrt {3} c^2 (-1+\sin (e+f x))^2 (1+\sin (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)} (12 \cos (2 (e+f x))+3 \cos (4 (e+f x))+8 (9 \sin (e+f x)+\sin (3 (e+f x))))}{32 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^5 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3} \]

[In]

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

[Out]

(Sqrt[3]*c^2*(-1 + Sin[e + f*x])^2*(1 + Sin[e + f*x])^(3/2)*Sqrt[c - c*Sin[e + f*x]]*(12*Cos[2*(e + f*x)] + 3*
Cos[4*(e + f*x)] + 8*(9*Sin[e + f*x] + Sin[3*(e + f*x)])))/(32*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5*(Cos[
(e + f*x)/2] + Sin[(e + f*x)/2])^3)

Maple [A] (verified)

Time = 2.93 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.89

method	result	size
default	\(\frac {\sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {a \left (\sin \left (f x +e \right )+1\right )}\, c^{2} a \left (3 \left (\cos ^{3}\left (f x +e \right )\right )+4 \sin \left (f x +e \right ) \cos \left (f x +e \right )+8 \tan \left (f x +e \right )-3 \sec \left (f x +e \right )\right )}{12 f}\)	\(76\)

[In]

int((a+a*sin(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/12/f*(-c*(sin(f*x+e)-1))^(1/2)*(a*(sin(f*x+e)+1))^(1/2)*c^2*a*(3*cos(f*x+e)^3+4*sin(f*x+e)*cos(f*x+e)+8*tan(
f*x+e)-3*sec(f*x+e))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.02 \[ \int (3+3 \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2} \, dx=\frac {{\left (3 \, a c^{2} \cos \left (f x + e\right )^{4} - 3 \, a c^{2} + 4 \, {\left (a c^{2} \cos \left (f x + e\right )^{2} + 2 \, a c^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{12 \, f \cos \left (f x + e\right )} \]

[In]

integrate((a+a*sin(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

1/12*(3*a*c^2*cos(f*x + e)^4 - 3*a*c^2 + 4*(a*c^2*cos(f*x + e)^2 + 2*a*c^2)*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. \[ \int (3+3 \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2} \, dx=\text {Timed out} \]

[In]

integrate((a+a*sin(f*x+e))**(3/2)*(c-c*sin(f*x+e))**(5/2),x)

[Out]

Timed out

Maxima [F]

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

[In]

integrate((a+a*sin(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^(3/2)*(-c*sin(f*x + e) + c)^(5/2), x)

Giac [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.22 \[ \int (3+3 \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2} \, dx=-\frac {4 \, {\left (3 \, a c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8} - 4 \, a c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6}\right )} \sqrt {a} \sqrt {c}}{3 \, f} \]

[In]

integrate((a+a*sin(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(5/2),x, algorithm="giac")

[Out]

-4/3*(3*a*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 - 4*a*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 = 8.42 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.18 \[ \int (3+3 \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2} \, dx=\frac {a\,c^2\,\sqrt {a\,\left (\sin \left (e+f\,x\right )+1\right )}\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (12\,\cos \left (e+f\,x\right )+15\,\cos \left (3\,e+3\,f\,x\right )+3\,\cos \left (5\,e+5\,f\,x\right )+80\,\sin \left (2\,e+2\,f\,x\right )+8\,\sin \left (4\,e+4\,f\,x\right )\right )}{96\,f\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )} \]

[In]

int((a + a*sin(e + f*x))^(3/2)*(c - c*sin(e + f*x))^(5/2),x)

[Out]

(a*c^2*(a*(sin(e + f*x) + 1))^(1/2)*(-c*(sin(e + f*x) - 1))^(1/2)*(12*cos(e + f*x) + 15*cos(3*e + 3*f*x) + 3*c
os(5*e + 5*f*x) + 80*sin(2*e + 2*f*x) + 8*sin(4*e + 4*f*x)))/(96*f*(cos(2*e + 2*f*x) + 1))