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

Optimal. Leaf size=42 \[ -\frac{\cos (e+f x) (c-c \sin (e+f x))^{3/2}}{4 f (a \sin (e+f x)+a)^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.0924086, antiderivative size = 42, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.033, Rules used = {2742} \[ -\frac{\cos (e+f x) (c-c \sin (e+f x))^{3/2}}{4 f (a \sin (e+f x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2742

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*} \int \frac{(c-c \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^{5/2}} \, dx &=-\frac{\cos (e+f x) (c-c \sin (e+f x))^{3/2}}{4 f (a+a \sin (e+f x))^{5/2}}\\ \end{align*}

Mathematica [B]  time = 0.467379, size = 86, normalized size = 2.05 \[ \frac{c \sin (e+f x) \sqrt{c-c \sin (e+f x)} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )}{f (a (\sin (e+f x)+1))^{5/2} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sin[e + f*x]*Sqrt[c - c*Sin[e + f*x]])/(f*(Cos[(e + f*x)/2] - Sin[(e
+ f*x)/2])*(a*(1 + Sin[e + f*x]))^(5/2))

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Maple [B]  time = 0.132, size = 96, normalized size = 2.3 \begin{align*} -{\frac{\sin \left ( fx+e \right ) \left ( 1-\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) \right ) }{f \left ( \sin \left ( fx+e \right ) \cos \left ( fx+e \right ) - \left ( \cos \left ( fx+e \right ) \right ) ^{2}-2\,\sin \left ( fx+e \right ) -\cos \left ( fx+e \right ) +2 \right ) } \left ( -c \left ( -1+\sin \left ( fx+e \right ) \right ) \right ) ^{{\frac{3}{2}}} \left ( a \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac{3}{2}}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.39044, size = 203, normalized size = 4.83 \begin{align*} -\frac{\sqrt{a \sin \left (f x + e\right ) + a} \sqrt{-c \sin \left (f x + e\right ) + c} c \sin \left (f x + e\right )}{a^{3} f \cos \left (f x + e\right )^{3} - 2 \, a^{3} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a^{3} f \cos \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac{3}{2}}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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