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]

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

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Rubi [A]  time = 0.09, antiderivative size = 42, normalized size of antiderivative = 1.00, 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*}

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Mathematica [B]  time = 0.46, 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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fricas [B]  time = 0.44, size = 80, normalized size = 1.90 \[ -\frac {\sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} c \sin \left (f x + e\right )}{a^{3} f \cos \left (f x + e\right )^{3} - 2 \, a^{3} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a^{3} f \cos \left (f x + e\right )} \]

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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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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]

Timed out

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maple [B]  time = 0.29, size = 93, normalized size = 2.21 \[ -\frac {\sin \left (f x +e \right ) \left (-c \left (\sin \left (f x +e \right )-1\right )\right )^{\frac {3}{2}} \left (-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )\right )}{f \left (\cos ^{2}\left (f x +e \right )-\sin \left (f x +e \right ) \cos \left (f x +e \right )+\cos \left (f x +e \right )+2 \sin \left (f x +e \right )-2\right ) \left (a \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}}} \]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

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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mupad [B]  time = 8.21, size = 118, normalized size = 2.81 \[ \frac {2\,c\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (-2\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+2\,{\sin \left (\frac {3\,e}{2}+\frac {3\,f\,x}{2}\right )}^2+2\,\sin \left (2\,e+2\,f\,x\right )\right )}{a^2\,f\,\sqrt {a\,\left (\sin \left (e+f\,x\right )+1\right )}\,\left (-8\,{\sin \left (e+f\,x\right )}^2+4\,\sin \left (e+f\,x\right )+2\,{\sin \left (2\,e+2\,f\,x\right )}^2+4\,\sin \left (3\,e+3\,f\,x\right )+8\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (- c \left (\sin {\left (e + f x \right )} - 1\right )\right )^{\frac {3}{2}}}{\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \]

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]

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

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