\(\int \frac {(3+3 \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{5/2}} \, dx\) [354]

Optimal result

Integrand size = 30, antiderivative size = 42 \[ \int \frac {(3+3 \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {\cos (e+f x) (3+3 \sin (e+f x))^{3/2}}{4 f (c-c \sin (e+f x))^{5/2}} \]

[Out]

1/4*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)/f/(c-c*sin(f*x+e))^(5/2)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.033, Rules used = {2821} \[ \int \frac {(3+3 \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {\cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{4 f (c-c \sin (e+f x))^{5/2}} \]

[In]

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

[Out]

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

Rule 2821

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*((c + d*Sin[e + f*x])^n/(a*f*(2*m + 1))), x] /; FreeQ[{a, b, c, d, e, f
, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && EqQ[m + n + 1, 0] && NeQ[m, -2^(-1)]

Rubi steps \begin{align*} \text {integral}& = \frac {\cos (e+f x) (a+a \sin (e+f x))^{3/2}}{4 f (c-c \sin (e+f x))^{5/2}} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(102\) vs. \(2(42)=84\).

Time = 2.43 (sec) , antiderivative size = 102, normalized size of antiderivative = 2.43 \[ \int \frac {(3+3 \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {3 \sqrt {3} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sin (e+f x) (1+\sin (e+f x))^{3/2}}{c^2 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 (-1+\sin (e+f x))^2 \sqrt {c-c \sin (e+f x)}} \]

[In]

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

[Out]

(3*Sqrt[3]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sin[e + f*x]*(1 + Sin[e + f*x])^(3/2))/(c^2*f*(Cos[(e + f*x)/
2] + Sin[(e + f*x)/2])^3*(-1 + Sin[e + f*x])^2*Sqrt[c - c*Sin[e + f*x]])

Maple [A] (verified)

Time = 3.07 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.21

method	result	size
default	\(-\frac {\tan \left (f x +e \right ) \sqrt {a \left (\sin \left (f x +e \right )+1\right )}\, a}{f \left (\sin \left (f x +e \right )-1\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, c^{2}}\)	\(51\)

[In]

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

[Out]

-1/f*tan(f*x+e)*(a*(sin(f*x+e)+1))^(1/2)*a/(sin(f*x+e)-1)/(-c*(sin(f*x+e)-1))^(1/2)/c^2

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 80 vs. \(2 (36) = 72\).

Time = 0.28 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.90 \[ \int \frac {(3+3 \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{5/2}} \, dx=-\frac {\sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} a \sin \left (f x + e\right )}{c^{3} f \cos \left (f x + e\right )^{3} + 2 \, c^{3} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, c^{3} f \cos \left (f x + e\right )} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {(3+3 \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{5/2}} \, dx=\int \frac {\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}}}{\left (- c \left (\sin {\left (e + f x \right )} - 1\right )\right )^{\frac {5}{2}}}\, dx \]

[In]

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

[Out]

Integral((a*(sin(e + f*x) + 1))**(3/2)/(-c*(sin(e + f*x) - 1))**(5/2), x)

Maxima [F]

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

[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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 87 vs. \(2 (36) = 72\).

Time = 0.34 (sec) , antiderivative size = 87, normalized size of antiderivative = 2.07 \[ \int \frac {(3+3 \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {{\left (2 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sqrt {a}}{4 \, c^{\frac {5}{2}} f \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4}} \]

[In]

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

[Out]

1/4*(2*a*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^2 - a*sgn(cos(-1/4*pi + 1/2*f*x +
1/2*e)))*sqrt(a)/(c^(5/2)*f*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^4)

Mupad [F(-1)]

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

[In]

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

[Out]

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