3.30 \(\int \cos ^2(e+f x) (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{9/2} \, dx\)

Optimal. Leaf size=236 \[ -\frac{a^2 \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{11/2}}{15 c f}-\frac{4 a^3 \cos (e+f x) \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{11/2}}{105 c f}-\frac{4 a^4 \cos (e+f x) (c-c \sin (e+f x))^{11/2}}{315 c f \sqrt{a \sin (e+f x)+a}}-\frac{\cos (e+f x) (a \sin (e+f x)+a)^{7/2} (c-c \sin (e+f x))^{11/2}}{10 c f}-\frac{4 a \cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{11/2}}{45 c f} \]

[Out]

(-4*a^4*Cos[e + f*x]*(c - c*Sin[e + f*x])^(11/2))/(315*c*f*Sqrt[a + a*Sin[e + f*x]]) - (4*a^3*Cos[e + f*x]*Sqr
t[a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(11/2))/(105*c*f) - (a^2*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2)*(c
- c*Sin[e + f*x])^(11/2))/(15*c*f) - (4*a*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2)*(c - c*Sin[e + f*x])^(11/2)
)/(45*c*f) - (Cos[e + f*x]*(a + a*Sin[e + f*x])^(7/2)*(c - c*Sin[e + f*x])^(11/2))/(10*c*f)

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Rubi [A]  time = 0.724664, antiderivative size = 236, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.079, Rules used = {2841, 2740, 2738} \[ -\frac{a^2 \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{11/2}}{15 c f}-\frac{4 a^3 \cos (e+f x) \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{11/2}}{105 c f}-\frac{4 a^4 \cos (e+f x) (c-c \sin (e+f x))^{11/2}}{315 c f \sqrt{a \sin (e+f x)+a}}-\frac{\cos (e+f x) (a \sin (e+f x)+a)^{7/2} (c-c \sin (e+f x))^{11/2}}{10 c f}-\frac{4 a \cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{11/2}}{45 c f} \]

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])^(9/2),x]

[Out]

(-4*a^4*Cos[e + f*x]*(c - c*Sin[e + f*x])^(11/2))/(315*c*f*Sqrt[a + a*Sin[e + f*x]]) - (4*a^3*Cos[e + f*x]*Sqr
t[a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(11/2))/(105*c*f) - (a^2*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2)*(c
- c*Sin[e + f*x])^(11/2))/(15*c*f) - (4*a*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2)*(c - c*Sin[e + f*x])^(11/2)
)/(45*c*f) - (Cos[e + f*x]*(a + a*Sin[e + f*x])^(7/2)*(c - c*Sin[e + f*x])^(11/2))/(10*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))^{7/2} (c-c \sin (e+f x))^{9/2} \, dx &=\frac{\int (a+a \sin (e+f x))^{9/2} (c-c \sin (e+f x))^{11/2} \, dx}{a c}\\ &=-\frac{\cos (e+f x) (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{11/2}}{10 c f}+\frac{4 \int (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{11/2} \, dx}{5 c}\\ &=-\frac{4 a \cos (e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{11/2}}{45 c f}-\frac{\cos (e+f x) (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{11/2}}{10 c f}+\frac{(8 a) \int (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{11/2} \, dx}{15 c}\\ &=-\frac{a^2 \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{11/2}}{15 c f}-\frac{4 a \cos (e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{11/2}}{45 c f}-\frac{\cos (e+f x) (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{11/2}}{10 c f}+\frac{\left (4 a^2\right ) \int (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{11/2} \, dx}{15 c}\\ &=-\frac{4 a^3 \cos (e+f x) \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{11/2}}{105 c f}-\frac{a^2 \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{11/2}}{15 c f}-\frac{4 a \cos (e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{11/2}}{45 c f}-\frac{\cos (e+f x) (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{11/2}}{10 c f}+\frac{\left (8 a^3\right ) \int \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{11/2} \, dx}{105 c}\\ &=-\frac{4 a^4 \cos (e+f x) (c-c \sin (e+f x))^{11/2}}{315 c f \sqrt{a+a \sin (e+f x)}}-\frac{4 a^3 \cos (e+f x) \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{11/2}}{105 c f}-\frac{a^2 \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{11/2}}{15 c f}-\frac{4 a \cos (e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{11/2}}{45 c f}-\frac{\cos (e+f x) (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{11/2}}{10 c f}\\ \end{align*}

Mathematica [A]  time = 5.70095, size = 209, normalized size = 0.89 \[ \frac{a^3 c^4 (\sin (e+f x)-1)^4 (\sin (e+f x)+1)^3 \sqrt{a (\sin (e+f x)+1)} \sqrt{c-c \sin (e+f x)} (158760 \sin (e+f x)+35280 \sin (3 (e+f x))+9072 \sin (5 (e+f x))+1620 \sin (7 (e+f x))+140 \sin (9 (e+f x))+13230 \cos (2 (e+f x))+7560 \cos (4 (e+f x))+2835 \cos (6 (e+f x))+630 \cos (8 (e+f x))+63 \cos (10 (e+f x)))}{322560 f \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^9 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^7} \]

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])^(9/2),x]

[Out]

(a^3*c^4*(-1 + Sin[e + f*x])^4*(1 + Sin[e + f*x])^3*Sqrt[a*(1 + Sin[e + f*x])]*Sqrt[c - c*Sin[e + f*x]]*(13230
*Cos[2*(e + f*x)] + 7560*Cos[4*(e + f*x)] + 2835*Cos[6*(e + f*x)] + 630*Cos[8*(e + f*x)] + 63*Cos[10*(e + f*x)
] + 158760*Sin[e + f*x] + 35280*Sin[3*(e + f*x)] + 9072*Sin[5*(e + f*x)] + 1620*Sin[7*(e + f*x)] + 140*Sin[9*(
e + f*x)]))/(322560*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^9*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^7)

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Maple [A]  time = 0.31, size = 169, normalized size = 0.7 \begin{align*}{\frac{\sin \left ( fx+e \right ) \left ( 63\, \left ( \cos \left ( fx+e \right ) \right ) ^{10}+7\,\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{8}+70\, \left ( \cos \left ( fx+e \right ) \right ) ^{8}+17\, \left ( \cos \left ( fx+e \right ) \right ) ^{6}\sin \left ( fx+e \right ) +80\, \left ( \cos \left ( fx+e \right ) \right ) ^{6}+33\,\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{4}+96\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}+65\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) +128\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}+193\,\sin \left ( fx+e \right ) +193 \right ) }{630\,f \left ( \cos \left ( fx+e \right ) \right ) ^{9}} \left ( -c \left ( -1+\sin \left ( fx+e \right ) \right ) \right ) ^{{\frac{9}{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))^(9/2),x)

[Out]

1/630/f*(-c*(-1+sin(f*x+e)))^(9/2)*sin(f*x+e)*(a*(1+sin(f*x+e)))^(7/2)*(63*cos(f*x+e)^10+7*sin(f*x+e)*cos(f*x+
e)^8+70*cos(f*x+e)^8+17*cos(f*x+e)^6*sin(f*x+e)+80*cos(f*x+e)^6+33*sin(f*x+e)*cos(f*x+e)^4+96*cos(f*x+e)^4+65*
cos(f*x+e)^2*sin(f*x+e)+128*cos(f*x+e)^2+193*sin(f*x+e)+193)/cos(f*x+e)^9

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Maxima [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))^(9/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 2.13492, size = 351, normalized size = 1.49 \begin{align*} \frac{{\left (63 \, a^{3} c^{4} \cos \left (f x + e\right )^{10} - 63 \, a^{3} c^{4} + 2 \,{\left (35 \, a^{3} c^{4} \cos \left (f x + e\right )^{8} + 40 \, a^{3} c^{4} \cos \left (f x + e\right )^{6} + 48 \, a^{3} c^{4} \cos \left (f x + e\right )^{4} + 64 \, a^{3} c^{4} \cos \left (f x + e\right )^{2} + 128 \, a^{3} c^{4}\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}}{630 \, 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))^(9/2),x, algorithm="fricas")

[Out]

1/630*(63*a^3*c^4*cos(f*x + e)^10 - 63*a^3*c^4 + 2*(35*a^3*c^4*cos(f*x + e)^8 + 40*a^3*c^4*cos(f*x + e)^6 + 48
*a^3*c^4*cos(f*x + e)^4 + 64*a^3*c^4*cos(f*x + e)^2 + 128*a^3*c^4)*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))**(9/2),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))^(9/2),x, algorithm="giac")

[Out]

sage2