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

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

[Out]

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

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Rubi [A]  time = 0.32, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {2841, 2738} \[ \frac {\cos (e+f x)}{c f \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2738

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 2841

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*
(x_)])^(n_.), x_Symbol] :> Dist[1/(a^(p/2)*c^(p/2)), Int[(a + b*Sin[e + f*x])^(m + p/2)*(c + d*Sin[e + f*x])^(
n + p/2), x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[p
/2]

Rubi steps

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

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Mathematica [A]  time = 0.49, size = 79, normalized size = 1.88 \[ \frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )}{f \sqrt {a (\sin (e+f x)+1)} (c-c \sin (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.45, size = 61, normalized size = 1.45 \[ -\frac {\sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{a c^{3} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - a c^{3} f \cos \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.36, size = 51, normalized size = 1.21 \[ -\frac {\left (\sin \left (f x +e \right )-1\right ) \cos \left (f x +e \right ) \sin \left (f x +e \right )}{f \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, \left (-c \left (\sin \left (f x +e \right )-1\right )\right )^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 9.77, size = 88, normalized size = 2.10 \[ -\frac {2\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (4\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+\sin \left (2\,e+2\,f\,x\right )-2\right )}{c^3\,f\,\sqrt {a\,\left (\sin \left (e+f\,x\right )+1\right )}\,\left (12\,{\sin \left (e+f\,x\right )}^2-15\,\sin \left (e+f\,x\right )+\sin \left (3\,e+3\,f\,x\right )+4\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(e + f*x)^2/((a + a*sin(e + f*x))^(1/2)*(c - c*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))/(c^3*f*(a*(sin(e + f*x) + 1
))^(1/2)*(sin(3*e + 3*f*x) - 15*sin(e + f*x) + 12*sin(e + f*x)^2 + 4))

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

[Out]

Timed out

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