3.4.100 \(\int \frac {\sqrt {c-c \sin (e+f x)}}{(a+a \sin (e+f x))^{5/2}} \, dx\) [400]

Optimal. Leaf size=43 \[ -\frac {c \cos (e+f x)}{2 f (a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 43, 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 = {2817} \begin {gather*} -\frac {c \cos (e+f x)}{2 f (a \sin (e+f x)+a)^{5/2} \sqrt {c-c \sin (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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)]

Rubi steps

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

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(87\) vs. \(2(43)=86\).
time = 0.15, size = 87, normalized size = 2.02 \begin {gather*} -\frac {\sqrt {a (1+\sin (e+f x))} \sqrt {c-c \sin (e+f x)}}{2 a^3 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(Sqrt[a*(1 + Sin[e + f*x])]*Sqrt[c - c*Sin[e + f*x]])/(a^3*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(
e + f*x)/2] + Sin[(e + f*x)/2])^5)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(91\) vs. \(2(37)=74\).
time = 19.16, size = 92, normalized size = 2.14

method	result	size
default	\(-\frac {\sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )+\cos \left (f x +e \right ) \sin \left (f x +e \right )+2 \cos \left (f x +e \right )-3 \sin \left (f x +e \right )-3\right )}{2 f \left (a \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}} \left (-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )\right )}\)	\(92\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c-c*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.34, size = 79, normalized size = 1.84 \begin {gather*} \frac {\sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{2 \, {\left (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 )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

1/2*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {- c \left (\sin {\left (e + f x \right )} - 1\right )}}{\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.50, size = 56, normalized size = 1.30 \begin {gather*} \frac {\sqrt {c} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{8 \, a^{\frac {5}{2}} f \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e))^(5/2),x, algorithm="giac")

[Out]

1/8*sqrt(c)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(a^(5/2)*f*cos(-1/4*pi + 1/2*f*x + 1/2*e)^4*sgn(cos(-1/4*pi +
1/2*f*x + 1/2*e)))

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Mupad [B]
time = 7.70, size = 103, normalized size = 2.40 \begin {gather*} -\frac {2\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (-4\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+\sin \left (2\,e+2\,f\,x\right )+2\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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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