3.28 \(\int \frac{\cos ^2(e+f x) (a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{11/2}} \, dx\)

Optimal. Leaf size=48 \[ \frac{\cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{8 a c f (c-c \sin (e+f x))^{9/2}} \]

[Out]

(Cos[e + f*x]*(a + a*Sin[e + f*x])^(7/2))/(8*a*c*f*(c - c*Sin[e + f*x])^(9/2))

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Rubi [A]  time = 0.339403, antiderivative size = 48, 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, 2742} \[ \frac{\cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{8 a c f (c-c \sin (e+f x))^{9/2}} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[e + f*x]^2*(a + a*Sin[e + f*x])^(5/2))/(c - c*Sin[e + f*x])^(11/2),x]

[Out]

(Cos[e + f*x]*(a + a*Sin[e + f*x])^(7/2))/(8*a*c*f*(c - c*Sin[e + f*x])^(9/2))

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 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{\cos ^2(e+f x) (a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{11/2}} \, dx &=\frac{\int \frac{(a+a \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{9/2}} \, dx}{a c}\\ &=\frac{\cos (e+f x) (a+a \sin (e+f x))^{7/2}}{8 a c f (c-c \sin (e+f x))^{9/2}}\\ \end{align*}

Mathematica [B]  time = 4.33023, size = 117, normalized size = 2.44 \[ \frac{a^2 (\sin (3 (e+f x))-7 \sin (e+f x)) \sqrt{a (\sin (e+f x)+1)} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^3}{4 c^5 f (\sin (e+f x)-1)^5 \sqrt{c-c \sin (e+f x)} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[e + f*x]^2*(a + a*Sin[e + f*x])^(5/2))/(c - c*Sin[e + f*x])^(11/2),x]

[Out]

(a^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3*Sqrt[a*(1 + Sin[e + f*x])]*(-7*Sin[e + f*x] + Sin[3*(e + f*x)]))/
(4*c^5*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-1 + Sin[e + f*x])^5*Sqrt[c - c*Sin[e + f*x]])

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Maple [B]  time = 0.181, size = 157, normalized size = 3.3 \begin{align*} -{\frac{ \left ( \left ( \cos \left ( fx+e \right ) \right ) ^{2}-2 \right ) \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 ) \sin \left ( fx+e \right ) }{f \left ( \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) - \left ( \cos \left ( fx+e \right ) \right ) ^{3}+2\,\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) +3\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}-4\,\sin \left ( fx+e \right ) +2\,\cos \left ( fx+e \right ) -4 \right ) } \left ( a \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{{\frac{5}{2}}} \left ( -c \left ( -1+\sin \left ( fx+e \right ) \right ) \right ) ^{-{\frac{11}{2}}}} \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))^(11/2),x)

[Out]

-1/f*(cos(f*x+e)^2-2)*(sin(f*x+e)*cos(f*x+e)-cos(f*x+e)^2-2*sin(f*x+e)-cos(f*x+e)+2)*sin(f*x+e)*(a*(1+sin(f*x+
e)))^(5/2)/(cos(f*x+e)^2*sin(f*x+e)-cos(f*x+e)^3+2*sin(f*x+e)*cos(f*x+e)+3*cos(f*x+e)^2-4*sin(f*x+e)+2*cos(f*x
+e)-4)/(-c*(-1+sin(f*x+e)))^(11/2)

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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}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac{11}{2}}}\,{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))^(11/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^(5/2)*cos(f*x + e)^2/(-c*sin(f*x + e) + c)^(11/2), x)

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Fricas [B]  time = 1.88982, size = 309, normalized size = 6.44 \begin{align*} -\frac{{\left (a^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2}\right )} \sqrt{a \sin \left (f x + e\right ) + a} \sqrt{-c \sin \left (f x + e\right ) + c} \sin \left (f x + e\right )}{c^{6} f \cos \left (f x + e\right )^{5} - 8 \, c^{6} f \cos \left (f x + e\right )^{3} + 8 \, c^{6} f \cos \left (f x + e\right ) + 4 \,{\left (c^{6} f \cos \left (f x + e\right )^{3} - 2 \, c^{6} f \cos \left (f x + e\right )\right )} \sin \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))^(11/2),x, algorithm="fricas")

[Out]

-(a^2*cos(f*x + e)^2 - 2*a^2)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)*sin(f*x + e)/(c^6*f*cos(f*x +
e)^5 - 8*c^6*f*cos(f*x + e)^3 + 8*c^6*f*cos(f*x + e) + 4*(c^6*f*cos(f*x + e)^3 - 2*c^6*f*cos(f*x + e))*sin(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))**(11/2),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}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac{11}{2}}}\,{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))^(11/2),x, algorithm="giac")

[Out]

integrate((a*sin(f*x + e) + a)^(5/2)*cos(f*x + e)^2/(-c*sin(f*x + e) + c)^(11/2), x)