3.33 \(\int \cos ^2(e+f x) (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{3/2} \, dx\)

Optimal. Leaf size=140 \[ \frac{4 c^2 \cos (e+f x) (a \sin (e+f x)+a)^{9/2}}{105 a f \sqrt{c-c \sin (e+f x)}}+\frac{\cos (e+f x) (a \sin (e+f x)+a)^{9/2} (c-c \sin (e+f x))^{3/2}}{7 a f}+\frac{2 c \cos (e+f x) (a \sin (e+f x)+a)^{9/2} \sqrt{c-c \sin (e+f x)}}{21 a f} \]

[Out]

(4*c^2*Cos[e + f*x]*(a + a*Sin[e + f*x])^(9/2))/(105*a*f*Sqrt[c - c*Sin[e + f*x]]) + (2*c*Cos[e + f*x]*(a + a*
Sin[e + f*x])^(9/2)*Sqrt[c - c*Sin[e + f*x]])/(21*a*f) + (Cos[e + f*x]*(a + a*Sin[e + f*x])^(9/2)*(c - c*Sin[e
+ f*x])^(3/2))/(7*a*f)

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Rubi [A]  time = 0.515013, 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} \[ \frac{4 c^2 \cos (e+f x) (a \sin (e+f x)+a)^{9/2}}{105 a f \sqrt{c-c \sin (e+f x)}}+\frac{\cos (e+f x) (a \sin (e+f x)+a)^{9/2} (c-c \sin (e+f x))^{3/2}}{7 a f}+\frac{2 c \cos (e+f x) (a \sin (e+f x)+a)^{9/2} \sqrt{c-c \sin (e+f x)}}{21 a f} \]

Antiderivative was successfully verified.

[In]

Int[Cos[e + f*x]^2*(a + a*Sin[e + f*x])^(7/2)*(c - c*Sin[e + f*x])^(3/2),x]

[Out]

(4*c^2*Cos[e + f*x]*(a + a*Sin[e + f*x])^(9/2))/(105*a*f*Sqrt[c - c*Sin[e + f*x]]) + (2*c*Cos[e + f*x]*(a + a*
Sin[e + f*x])^(9/2)*Sqrt[c - c*Sin[e + f*x]])/(21*a*f) + (Cos[e + f*x]*(a + a*Sin[e + f*x])^(9/2)*(c - c*Sin[e
+ f*x])^(3/2))/(7*a*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))^{7/2} (c-c \sin (e+f x))^{3/2} \, dx &=\frac{\int (a+a \sin (e+f x))^{9/2} (c-c \sin (e+f x))^{5/2} \, dx}{a c}\\ &=\frac{\cos (e+f x) (a+a \sin (e+f x))^{9/2} (c-c \sin (e+f x))^{3/2}}{7 a f}+\frac{4 \int (a+a \sin (e+f x))^{9/2} (c-c \sin (e+f x))^{3/2} \, dx}{7 a}\\ &=\frac{2 c \cos (e+f x) (a+a \sin (e+f x))^{9/2} \sqrt{c-c \sin (e+f x)}}{21 a f}+\frac{\cos (e+f x) (a+a \sin (e+f x))^{9/2} (c-c \sin (e+f x))^{3/2}}{7 a f}+\frac{(4 c) \int (a+a \sin (e+f x))^{9/2} \sqrt{c-c \sin (e+f x)} \, dx}{21 a}\\ &=\frac{4 c^2 \cos (e+f x) (a+a \sin (e+f x))^{9/2}}{105 a f \sqrt{c-c \sin (e+f x)}}+\frac{2 c \cos (e+f x) (a+a \sin (e+f x))^{9/2} \sqrt{c-c \sin (e+f x)}}{21 a f}+\frac{\cos (e+f x) (a+a \sin (e+f x))^{9/2} (c-c \sin (e+f x))^{3/2}}{7 a f}\\ \end{align*}

Mathematica [A]  time = 1.60899, size = 115, normalized size = 0.82 \[ \frac{a^3 c \sec (e+f x) \sqrt{a (\sin (e+f x)+1)} \sqrt{c-c \sin (e+f x)} (4725 \sin (e+f x)+665 \sin (3 (e+f x))+21 \sin (5 (e+f x))-15 \sin (7 (e+f x))-1050 \cos (2 (e+f x))-420 \cos (4 (e+f x))-70 \cos (6 (e+f x)))}{6720 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[e + f*x]^2*(a + a*Sin[e + f*x])^(7/2)*(c - c*Sin[e + f*x])^(3/2),x]

[Out]

(a^3*c*Sec[e + f*x]*Sqrt[a*(1 + Sin[e + f*x])]*Sqrt[c - c*Sin[e + f*x]]*(-1050*Cos[2*(e + f*x)] - 420*Cos[4*(e
+ f*x)] - 70*Cos[6*(e + f*x)] + 4725*Sin[e + f*x] + 665*Sin[3*(e + f*x)] + 21*Sin[5*(e + f*x)] - 15*Sin[7*(e
+ f*x)]))/(6720*f)

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Maple [A]  time = 0.228, size = 133, normalized size = 1. \begin{align*} -{\frac{\sin \left ( fx+e \right ) \left ( -15\, \left ( \cos \left ( fx+e \right ) \right ) ^{8}+5\, \left ( \cos \left ( fx+e \right ) \right ) ^{6}\sin \left ( fx+e \right ) -16\, \left ( \cos \left ( fx+e \right ) \right ) ^{6}+13\,\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{4}-16\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}+29\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) +58\,\sin \left ( fx+e \right ) -58 \right ) }{105\,f \left ( \cos \left ( fx+e \right ) \right ) ^{7}} \left ( -c \left ( -1+\sin \left ( fx+e \right ) \right ) \right ) ^{{\frac{3}{2}}} \left ( a \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{{\frac{7}{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))^(7/2)*(c-c*sin(f*x+e))^(3/2),x)

[Out]

-1/105/f*(-c*(-1+sin(f*x+e)))^(3/2)*sin(f*x+e)*(a*(1+sin(f*x+e)))^(7/2)*(-15*cos(f*x+e)^8+5*cos(f*x+e)^6*sin(f
*x+e)-16*cos(f*x+e)^6+13*sin(f*x+e)*cos(f*x+e)^4-16*cos(f*x+e)^4+29*cos(f*x+e)^2*sin(f*x+e)+58*sin(f*x+e)-58)/
cos(f*x+e)^7

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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{7}{2}}{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac{3}{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))^(7/2)*(c-c*sin(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^(7/2)*(-c*sin(f*x + e) + c)^(3/2)*cos(f*x + e)^2, x)

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Fricas [A]  time = 1.85931, size = 293, normalized size = 2.09 \begin{align*} -\frac{{\left (35 \, a^{3} c \cos \left (f x + e\right )^{6} - 35 \, a^{3} c +{\left (15 \, a^{3} c \cos \left (f x + e\right )^{6} - 24 \, a^{3} c \cos \left (f x + e\right )^{4} - 32 \, a^{3} c \cos \left (f x + e\right )^{2} - 64 \, a^{3} c\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}}{105 \, 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))^(7/2)*(c-c*sin(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

-1/105*(35*a^3*c*cos(f*x + e)^6 - 35*a^3*c + (15*a^3*c*cos(f*x + e)^6 - 24*a^3*c*cos(f*x + e)^4 - 32*a^3*c*cos
(f*x + e)^2 - 64*a^3*c)*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))**(7/2)*(c-c*sin(f*x+e))**(3/2),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))^(7/2)*(c-c*sin(f*x+e))^(3/2),x, algorithm="giac")

[Out]

sage2