3.18 \(\int \frac{\cos ^2(e+f x) (a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{11/2}} \, dx\)

Optimal. Leaf size=97 \[ \frac{\cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{48 a c^2 f (c-c \sin (e+f x))^{7/2}}+\frac{\cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{8 a c f (c-c \sin (e+f x))^{9/2}} \]

[Out]

(Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(8*a*c*f*(c - c*Sin[e + f*x])^(9/2)) + (Cos[e + f*x]*(a + a*Sin[e +
f*x])^(5/2))/(48*a*c^2*f*(c - c*Sin[e + f*x])^(7/2))

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Rubi [A]  time = 0.435671, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.079, Rules used = {2841, 2743, 2742} \[ \frac{\cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{48 a c^2 f (c-c \sin (e+f x))^{7/2}}+\frac{\cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{8 a c f (c-c \sin (e+f x))^{9/2}} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[e + f*x]^2*(a + a*Sin[e + f*x])^(3/2))/(c - c*Sin[e + f*x])^(11/2),x]

[Out]

(Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(8*a*c*f*(c - c*Sin[e + f*x])^(9/2)) + (Cos[e + f*x]*(a + a*Sin[e +
f*x])^(5/2))/(48*a*c^2*f*(c - c*Sin[e + f*x])^(7/2))

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 2743

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] + Dist[(m + n + 1)/(a*(2*m
+ 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]
&& EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && ILtQ[Simplify[m + n + 1], 0] && NeQ[m, -2^(-1)] && (SumSimplerQ[m
, 1] ||  !SumSimplerQ[n, 1])

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))^{3/2}}{(c-c \sin (e+f x))^{11/2}} \, dx &=\frac{\int \frac{(a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{9/2}} \, dx}{a c}\\ &=\frac{\cos (e+f x) (a+a \sin (e+f x))^{5/2}}{8 a c f (c-c \sin (e+f x))^{9/2}}+\frac{\int \frac{(a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{7/2}} \, dx}{8 a c^2}\\ &=\frac{\cos (e+f x) (a+a \sin (e+f x))^{5/2}}{8 a c f (c-c \sin (e+f x))^{9/2}}+\frac{\cos (e+f x) (a+a \sin (e+f x))^{5/2}}{48 a c^2 f (c-c \sin (e+f x))^{7/2}}\\ \end{align*}

Mathematica [A]  time = 2.00804, size = 118, normalized size = 1.22 \[ -\frac{a \sqrt{a (\sin (e+f x)+1)} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^3 (4 \sin (e+f x)-3 \cos (2 (e+f x))+5)}{12 c^5 f (\sin (e+f x)-1)^5 \sqrt{c-c \sin (e+f x)} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[e + f*x]^2*(a + a*Sin[e + f*x])^(3/2))/(c - c*Sin[e + f*x])^(11/2),x]

[Out]

-(a*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3*Sqrt[a*(1 + Sin[e + f*x])]*(5 - 3*Cos[2*(e + f*x)] + 4*Sin[e + f*x
]))/(12*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 [A]  time = 0.192, size = 152, normalized size = 1.6 \begin{align*}{\frac{ \left ( \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) -4\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}-4\,\sin \left ( fx+e \right ) +10 \right ) \sin \left ( fx+e \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 ) }{6\,f \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 ) } \left ( a \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{{\frac{3}{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))^(3/2)/(c-c*sin(f*x+e))^(11/2),x)

[Out]

1/6/f*(cos(f*x+e)^2*sin(f*x+e)-4*cos(f*x+e)^2-4*sin(f*x+e)+10)*(a*(1+sin(f*x+e)))^(3/2)*sin(f*x+e)*(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)*cos(f*x+e)+cos(f*x+e)^2-2*sin(f*x+e)+cos(f*x+e
)-2)/(-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{3}{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))^(3/2)/(c-c*sin(f*x+e))^(11/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^(3/2)*cos(f*x + e)^2/(-c*sin(f*x + e) + c)^(11/2), x)

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Fricas [A]  time = 1.78047, size = 320, normalized size = 3.3 \begin{align*} -\frac{{\left (3 \, a \cos \left (f x + e\right )^{2} - 2 \, a \sin \left (f x + e\right ) - 4 \, a\right )} \sqrt{a \sin \left (f x + e\right ) + a} \sqrt{-c \sin \left (f x + e\right ) + c}}{6 \,{\left (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 )\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))^(3/2)/(c-c*sin(f*x+e))^(11/2),x, algorithm="fricas")

[Out]

-1/6*(3*a*cos(f*x + e)^2 - 2*a*sin(f*x + e) - 4*a)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(c^6*f*c
os(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))**(3/2)/(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{3}{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))^(3/2)/(c-c*sin(f*x+e))^(11/2),x, algorithm="giac")

[Out]

integrate((a*sin(f*x + e) + a)^(3/2)*cos(f*x + e)^2/(-c*sin(f*x + e) + c)^(11/2), x)