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

Optimal. Leaf size=45 \[ \frac{\cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{4 a f \sqrt{c-c \sin (e+f x)}} \]

[Out]

(Cos[e + f*x]*(a + a*Sin[e + f*x])^(7/2))/(4*a*f*Sqrt[c - c*Sin[e + f*x]])

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Rubi [A]  time = 0.306869, antiderivative size = 45, normalized size of antiderivative = 1., 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} \[ \frac{\cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{4 a f \sqrt{c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[e + f*x]*(a + a*Sin[e + f*x])^(7/2))/(4*a*f*Sqrt[c - c*Sin[e + f*x]])

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]

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)]

Rubi steps

\begin{align*} \int \frac{\cos ^2(e+f x) (a+a \sin (e+f x))^{5/2}}{\sqrt{c-c \sin (e+f x)}} \, dx &=\frac{\int (a+a \sin (e+f x))^{7/2} \sqrt{c-c \sin (e+f x)} \, dx}{a c}\\ &=\frac{\cos (e+f x) (a+a \sin (e+f x))^{7/2}}{4 a f \sqrt{c-c \sin (e+f x)}}\\ \end{align*}

Mathematica [B]  time = 0.920495, size = 119, normalized size = 2.64 \[ \frac{(a (\sin (e+f x)+1))^{5/2} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) (56 \sin (e+f x)-8 \sin (3 (e+f x))-28 \cos (2 (e+f x))+\cos (4 (e+f x)))}{32 f \sqrt{c-c \sin (e+f x)} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(a*(1 + Sin[e + f*x]))^(5/2)*(-28*Cos[2*(e + f*x)] + Cos[4*(e + f*x)] +
56*Sin[e + f*x] - 8*Sin[3*(e + f*x)]))/(32*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5*Sqrt[c - c*Sin[e + f*x]]
)

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Maple [B]  time = 0.227, size = 195, normalized size = 4.3 \begin{align*}{\frac{ \left ( \left ( \cos \left ( fx+e \right ) \right ) ^{4}-\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{3}-4\, \left ( \cos \left ( fx+e \right ) \right ) ^{3}-3\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) -4\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}+7\,\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) +8\,\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) -1 \right ) \sin \left ( fx+e \right ) }{4\,f \left ( \left ( \cos \left ( fx+e \right ) \right ) ^{3}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) -3\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}-2\,\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) -2\,\cos \left ( fx+e \right ) +4\,\sin \left ( fx+e \right ) +4 \right ) } \left ( a \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{{\frac{5}{2}}}{\frac{1}{\sqrt{-c \left ( -1+\sin \left ( fx+e \right ) \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/f*(cos(f*x+e)^4-sin(f*x+e)*cos(f*x+e)^3-4*cos(f*x+e)^3-3*cos(f*x+e)^2*sin(f*x+e)-4*cos(f*x+e)^2+7*sin(f*x+
e)*cos(f*x+e)+8*cos(f*x+e)+sin(f*x+e)-1)*sin(f*x+e)*(a*(1+sin(f*x+e)))^(5/2)/(cos(f*x+e)^3-cos(f*x+e)^2*sin(f*
x+e)-3*cos(f*x+e)^2-2*sin(f*x+e)*cos(f*x+e)-2*cos(f*x+e)+4*sin(f*x+e)+4)/(-c*(-1+sin(f*x+e)))^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.70971, size = 234, normalized size = 5.2 \begin{align*} \frac{{\left (a^{2} \cos \left (f x + e\right )^{4} - 8 \, a^{2} \cos \left (f x + e\right )^{2} + 7 \, a^{2} - 4 \,{\left (a^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt{a \sin \left (f x + e\right ) + a} \sqrt{-c \sin \left (f x + e\right ) + c}}{4 \, c f \cos \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(a^2*cos(f*x + e)^4 - 8*a^2*cos(f*x + e)^2 + 7*a^2 - 4*(a^2*cos(f*x + e)^2 - 2*a^2)*sin(f*x + e))*sqrt(a*s
in(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(c*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(cos(f*x+e)**2*(a+a*sin(f*x+e))**(5/2)/(c-c*sin(f*x+e))**(1/2),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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