3.10 \(\int \cos ^2(e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2} \, dx\)

Optimal. Leaf size=140 \[ -\frac{a^2 \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{15 c f \sqrt{a \sin (e+f x)+a}}-\frac{\cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{7/2}}{6 c f}-\frac{2 a \cos (e+f x) \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{7/2}}{15 c f} \]

[Out]

-(a^2*Cos[e + f*x]*(c - c*Sin[e + f*x])^(7/2))/(15*c*f*Sqrt[a + a*Sin[e + f*x]]) - (2*a*Cos[e + f*x]*Sqrt[a +
a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(7/2))/(15*c*f) - (Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2)*(c - c*Sin[e +
f*x])^(7/2))/(6*c*f)

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Rubi [A]  time = 0.529182, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.079, Rules used = {2841, 2740, 2738} \[ -\frac{a^2 \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{15 c f \sqrt{a \sin (e+f x)+a}}-\frac{\cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{7/2}}{6 c f}-\frac{2 a \cos (e+f x) \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{7/2}}{15 c f} \]

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])^(5/2),x]

[Out]

-(a^2*Cos[e + f*x]*(c - c*Sin[e + f*x])^(7/2))/(15*c*f*Sqrt[a + a*Sin[e + f*x]]) - (2*a*Cos[e + f*x]*Sqrt[a +
a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(7/2))/(15*c*f) - (Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2)*(c - c*Sin[e +
f*x])^(7/2))/(6*c*f)

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 2740

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Sim
p[(b*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n)/(f*(m + n)), x] + Dist[(a*(2*m - 1))/(m
+ n), 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, n}, x] && E
qQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[m - 1/2, 0] &&  !LtQ[n, -1] &&  !(IGtQ[n - 1/2, 0] && LtQ[n, m])
&&  !(ILtQ[m + n, 0] && GtQ[2*m + n + 1, 0])

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 \cos ^2(e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2} \, dx &=\frac{\int (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{7/2} \, dx}{a c}\\ &=-\frac{\cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{7/2}}{6 c f}+\frac{2 \int (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{7/2} \, dx}{3 c}\\ &=-\frac{2 a \cos (e+f x) \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{7/2}}{15 c f}-\frac{\cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{7/2}}{6 c f}+\frac{(4 a) \int \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{7/2} \, dx}{15 c}\\ &=-\frac{a^2 \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{15 c f \sqrt{a+a \sin (e+f x)}}-\frac{2 a \cos (e+f x) \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{7/2}}{15 c f}-\frac{\cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{7/2}}{6 c f}\\ \end{align*}

Mathematica [A]  time = 0.97086, size = 156, normalized size = 1.11 \[ \frac{c^2 (\sin (e+f x)-1)^2 (a (\sin (e+f x)+1))^{3/2} \sqrt{c-c \sin (e+f x)} (600 \sin (e+f x)+100 \sin (3 (e+f x))+12 \sin (5 (e+f x))+75 \cos (2 (e+f x))+30 \cos (4 (e+f x))+5 \cos (6 (e+f x)))}{960 f \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^5 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3} \]

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])^(5/2),x]

[Out]

(c^2*(-1 + Sin[e + f*x])^2*(a*(1 + Sin[e + f*x]))^(3/2)*Sqrt[c - c*Sin[e + f*x]]*(75*Cos[2*(e + f*x)] + 30*Cos
[4*(e + f*x)] + 5*Cos[6*(e + f*x)] + 600*Sin[e + f*x] + 100*Sin[3*(e + f*x)] + 12*Sin[5*(e + f*x)]))/(960*f*(C
os[(e + f*x)/2] - Sin[(e + f*x)/2])^5*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3)

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Maple [A]  time = 0.214, size = 116, normalized size = 0.8 \begin{align*}{\frac{\sin \left ( fx+e \right ) \left ( 5\, \left ( \cos \left ( fx+e \right ) \right ) ^{6}+\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{4}+6\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}+3\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) +8\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}+11\,\sin \left ( fx+e \right ) +11 \right ) }{30\,f \left ( \cos \left ( fx+e \right ) \right ) ^{5}} \left ( -c \left ( -1+\sin \left ( fx+e \right ) \right ) \right ) ^{{\frac{5}{2}}} \left ( a \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{{\frac{3}{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))^(5/2),x)

[Out]

1/30/f*(-c*(-1+sin(f*x+e)))^(5/2)*sin(f*x+e)*(a*(1+sin(f*x+e)))^(3/2)*(5*cos(f*x+e)^6+sin(f*x+e)*cos(f*x+e)^4+
6*cos(f*x+e)^4+3*cos(f*x+e)^2*sin(f*x+e)+8*cos(f*x+e)^2+11*sin(f*x+e)+11)/cos(f*x+e)^5

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac{3}{2}}{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac{5}{2}} \cos \left (f x + e\right )^{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))^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.81095, size = 251, normalized size = 1.79 \begin{align*} \frac{{\left (5 \, a c^{2} \cos \left (f x + e\right )^{6} - 5 \, a c^{2} + 2 \,{\left (3 \, a c^{2} \cos \left (f x + e\right )^{4} + 4 \, a c^{2} \cos \left (f x + e\right )^{2} + 8 \, a c^{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}}{30 \, 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))^(3/2)*(c-c*sin(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

1/30*(5*a*c^2*cos(f*x + e)^6 - 5*a*c^2 + 2*(3*a*c^2*cos(f*x + e)^4 + 4*a*c^2*cos(f*x + e)^2 + 8*a*c^2)*sin(f*x
+ e))*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + 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))**(3/2)*(c-c*sin(f*x+e))**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{2} \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))^(5/2),x, algorithm="giac")

[Out]

sage2