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

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

[Out]

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

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Rubi [A]  time = 0.524338, 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} \[ -\frac{4 a^2 \cos (e+f x) (c-c \sin (e+f x))^{9/2}}{105 c f \sqrt{a \sin (e+f x)+a}}-\frac{\cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{9/2}}{7 c f}-\frac{2 a \cos (e+f x) \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{9/2}}{21 c f} \]

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])^(7/2),x]

[Out]

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

Mathematica [A]  time = 1.24586, size = 166, normalized size = 1.19 \[ -\frac{c^3 (\sin (e+f x)-1)^3 (a (\sin (e+f x)+1))^{3/2} \sqrt{c-c \sin (e+f x)} (4725 \sin (e+f x)+665 \sin (3 (e+f x))+21 \sin (5 (e+f x))-15 \sin (7 (e+f x))+1050 \cos (2 (e+f x))+420 \cos (4 (e+f x))+70 \cos (6 (e+f x)))}{6720 f \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^7 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3} \]

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])^(7/2),x]

[Out]

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

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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{7}{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))^(7/2),x)

[Out]

1/105/f*(-c*(-1+sin(f*x+e)))^(7/2)*sin(f*x+e)*(a*(1+sin(f*x+e)))^(3/2)*(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)/co
s(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{3}{2}}{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac{7}{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))^(7/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.8217, size = 292, normalized size = 2.09 \begin{align*} \frac{{\left (35 \, a c^{3} \cos \left (f x + e\right )^{6} - 35 \, a c^{3} -{\left (15 \, a c^{3} \cos \left (f x + e\right )^{6} - 24 \, a c^{3} \cos \left (f x + e\right )^{4} - 32 \, a c^{3} \cos \left (f x + e\right )^{2} - 64 \, a c^{3}\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))^(3/2)*(c-c*sin(f*x+e))^(7/2),x, algorithm="fricas")

[Out]

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

[Out]

sage2