3.19 \(\int \cos ^2(e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{7/2} \, dx\)

Optimal. Leaf size=188 \[ -\frac{a^2 \cos (e+f x) \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{9/2}}{14 c f}-\frac{a^3 \cos (e+f x) (c-c \sin (e+f x))^{9/2}}{35 c f \sqrt{a \sin (e+f x)+a}}-\frac{\cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{9/2}}{8 c f}-\frac{3 a \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{9/2}}{28 c f} \]

[Out]

-(a^3*Cos[e + f*x]*(c - c*Sin[e + f*x])^(9/2))/(35*c*f*Sqrt[a + a*Sin[e + f*x]]) - (a^2*Cos[e + f*x]*Sqrt[a +
a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(9/2))/(14*c*f) - (3*a*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2)*(c - c*Sin
[e + f*x])^(9/2))/(28*c*f) - (Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2)*(c - c*Sin[e + f*x])^(9/2))/(8*c*f)

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Rubi [A]  time = 0.624161, antiderivative size = 188, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.079, Rules used = {2841, 2740, 2738} \[ -\frac{a^2 \cos (e+f x) \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{9/2}}{14 c f}-\frac{a^3 \cos (e+f x) (c-c \sin (e+f x))^{9/2}}{35 c f \sqrt{a \sin (e+f x)+a}}-\frac{\cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{9/2}}{8 c f}-\frac{3 a \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{9/2}}{28 c f} \]

Antiderivative was successfully verified.

[In]

Int[Cos[e + f*x]^2*(a + a*Sin[e + f*x])^(5/2)*(c - c*Sin[e + f*x])^(7/2),x]

[Out]

-(a^3*Cos[e + f*x]*(c - c*Sin[e + f*x])^(9/2))/(35*c*f*Sqrt[a + a*Sin[e + f*x]]) - (a^2*Cos[e + f*x]*Sqrt[a +
a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(9/2))/(14*c*f) - (3*a*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2)*(c - c*Sin
[e + f*x])^(9/2))/(28*c*f) - (Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2)*(c - c*Sin[e + f*x])^(9/2))/(8*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))^{5/2} (c-c \sin (e+f x))^{7/2} \, dx &=\frac{\int (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{9/2} \, dx}{a c}\\ &=-\frac{\cos (e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{9/2}}{8 c f}+\frac{3 \int (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{9/2} \, dx}{4 c}\\ &=-\frac{3 a \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{9/2}}{28 c f}-\frac{\cos (e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{9/2}}{8 c f}+\frac{(3 a) \int (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{9/2} \, dx}{7 c}\\ &=-\frac{a^2 \cos (e+f x) \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{9/2}}{14 c f}-\frac{3 a \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{9/2}}{28 c f}-\frac{\cos (e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{9/2}}{8 c f}+\frac{a^2 \int \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{9/2} \, dx}{7 c}\\ &=-\frac{a^3 \cos (e+f x) (c-c \sin (e+f x))^{9/2}}{35 c f \sqrt{a+a \sin (e+f x)}}-\frac{a^2 \cos (e+f x) \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{9/2}}{14 c f}-\frac{3 a \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{9/2}}{28 c f}-\frac{\cos (e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{9/2}}{8 c f}\\ \end{align*}

Mathematica [A]  time = 3.21113, size = 176, normalized size = 0.94 \[ -\frac{c^3 (\sin (e+f x)-1)^3 (a (\sin (e+f x)+1))^{5/2} \sqrt{c-c \sin (e+f x)} (19600 \sin (e+f x)+3920 \sin (3 (e+f x))+784 \sin (5 (e+f x))+80 \sin (7 (e+f x))+1960 \cos (2 (e+f x))+980 \cos (4 (e+f x))+280 \cos (6 (e+f x))+35 \cos (8 (e+f x)))}{35840 f \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^7 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^5} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[e + f*x]^2*(a + a*Sin[e + f*x])^(5/2)*(c - c*Sin[e + f*x])^(7/2),x]

[Out]

-(c^3*(-1 + Sin[e + f*x])^3*(a*(1 + Sin[e + f*x]))^(5/2)*Sqrt[c - c*Sin[e + f*x]]*(1960*Cos[2*(e + f*x)] + 980
*Cos[4*(e + f*x)] + 280*Cos[6*(e + f*x)] + 35*Cos[8*(e + f*x)] + 19600*Sin[e + f*x] + 3920*Sin[3*(e + f*x)] +
784*Sin[5*(e + f*x)] + 80*Sin[7*(e + f*x)]))/(35840*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^7*(Cos[(e + f*x)/2
] + Sin[(e + f*x)/2])^5)

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Maple [A]  time = 0.24, size = 143, normalized size = 0.8 \begin{align*}{\frac{\sin \left ( fx+e \right ) \left ( 35\, \left ( \cos \left ( fx+e \right ) \right ) ^{8}+5\, \left ( \cos \left ( fx+e \right ) \right ) ^{6}\sin \left ( fx+e \right ) +40\, \left ( \cos \left ( fx+e \right ) \right ) ^{6}+13\,\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{4}+48\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}+29\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) +64\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}+93\,\sin \left ( fx+e \right ) +93 \right ) }{280\,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{5}{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))^(5/2)*(c-c*sin(f*x+e))^(7/2),x)

[Out]

1/280/f*(-c*(-1+sin(f*x+e)))^(7/2)*sin(f*x+e)*(a*(1+sin(f*x+e)))^(5/2)*(35*cos(f*x+e)^8+5*cos(f*x+e)^6*sin(f*x
+e)+40*cos(f*x+e)^6+13*sin(f*x+e)*cos(f*x+e)^4+48*cos(f*x+e)^4+29*cos(f*x+e)^2*sin(f*x+e)+64*cos(f*x+e)^2+93*s
in(f*x+e)+93)/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{5}{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))^(5/2)*(c-c*sin(f*x+e))^(7/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^(5/2)*(-c*sin(f*x + e) + c)^(7/2)*cos(f*x + e)^2, x)

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Fricas [A]  time = 2.00811, size = 306, normalized size = 1.63 \begin{align*} \frac{{\left (35 \, a^{2} c^{3} \cos \left (f x + e\right )^{8} - 35 \, a^{2} c^{3} + 8 \,{\left (5 \, a^{2} c^{3} \cos \left (f x + e\right )^{6} + 6 \, a^{2} c^{3} \cos \left (f x + e\right )^{4} + 8 \, a^{2} c^{3} \cos \left (f x + e\right )^{2} + 16 \, a^{2} 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}}{280 \, 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))^(5/2)*(c-c*sin(f*x+e))^(7/2),x, algorithm="fricas")

[Out]

1/280*(35*a^2*c^3*cos(f*x + e)^8 - 35*a^2*c^3 + 8*(5*a^2*c^3*cos(f*x + e)^6 + 6*a^2*c^3*cos(f*x + e)^4 + 8*a^2
*c^3*cos(f*x + e)^2 + 16*a^2*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))**(5/2)*(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))^(5/2)*(c-c*sin(f*x+e))^(7/2),x, algorithm="giac")

[Out]

sage2