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

Optimal. Leaf size=179 \[ -\frac{3 a^2 \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{9/2}}{28 f}-\frac{a^3 \cos (e+f x) \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{9/2}}{14 f}-\frac{a^4 \cos (e+f x) (c-c \sin (e+f x))^{9/2}}{35 f \sqrt{a \sin (e+f x)+a}}-\frac{a \cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{9/2}}{8 f} \]

[Out]

-(a^4*Cos[e + f*x]*(c - c*Sin[e + f*x])^(9/2))/(35*f*Sqrt[a + a*Sin[e + f*x]]) - (a^3*Cos[e + f*x]*Sqrt[a + a*
Sin[e + f*x]]*(c - c*Sin[e + f*x])^(9/2))/(14*f) - (3*a^2*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2)*(c - c*Sin[e
+ f*x])^(9/2))/(28*f) - (a*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2)*(c - c*Sin[e + f*x])^(9/2))/(8*f)

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Rubi [A]  time = 0.36556, antiderivative size = 179, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.067, Rules used = {2740, 2738} \[ -\frac{3 a^2 \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{9/2}}{28 f}-\frac{a^3 \cos (e+f x) \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{9/2}}{14 f}-\frac{a^4 \cos (e+f x) (c-c \sin (e+f x))^{9/2}}{35 f \sqrt{a \sin (e+f x)+a}}-\frac{a \cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{9/2}}{8 f} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^(7/2)*(c - c*Sin[e + f*x])^(9/2),x]

[Out]

-(a^4*Cos[e + f*x]*(c - c*Sin[e + f*x])^(9/2))/(35*f*Sqrt[a + a*Sin[e + f*x]]) - (a^3*Cos[e + f*x]*Sqrt[a + a*
Sin[e + f*x]]*(c - c*Sin[e + f*x])^(9/2))/(14*f) - (3*a^2*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2)*(c - c*Sin[e
+ f*x])^(9/2))/(28*f) - (a*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2)*(c - c*Sin[e + f*x])^(9/2))/(8*f)

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 (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{9/2} \, dx &=-\frac{a \cos (e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{9/2}}{8 f}+\frac{1}{4} (3 a) \int (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{9/2} \, dx\\ &=-\frac{3 a^2 \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{9/2}}{28 f}-\frac{a \cos (e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{9/2}}{8 f}+\frac{1}{7} \left (3 a^2\right ) \int (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{9/2} \, dx\\ &=-\frac{a^3 \cos (e+f x) \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{9/2}}{14 f}-\frac{3 a^2 \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{9/2}}{28 f}-\frac{a \cos (e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{9/2}}{8 f}+\frac{1}{7} a^3 \int \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{9/2} \, dx\\ &=-\frac{a^4 \cos (e+f x) (c-c \sin (e+f x))^{9/2}}{35 f \sqrt{a+a \sin (e+f x)}}-\frac{a^3 \cos (e+f x) \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{9/2}}{14 f}-\frac{3 a^2 \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{9/2}}{28 f}-\frac{a \cos (e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{9/2}}{8 f}\\ \end{align*}

Mathematica [A]  time = 5.62273, size = 127, normalized size = 0.71 \[ \frac{a^3 c^4 \sec (e+f x) \sqrt{a (\sin (e+f x)+1)} \sqrt{c-c \sin (e+f x)} (19600 \sin (e+f x)+3920 \sin (3 (e+f x))+784 \sin (5 (e+f x))+80 \sin (7 (e+f x))+1960 \cos (2 (e+f x))+980 \cos (4 (e+f x))+280 \cos (6 (e+f x))+35 \cos (8 (e+f x)))}{35840 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])^(7/2)*(c - c*Sin[e + f*x])^(9/2),x]

[Out]

(a^3*c^4*Sec[e + f*x]*Sqrt[a*(1 + Sin[e + f*x])]*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)

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Maple [A]  time = 0.203, 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 ) ^{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((a+a*sin(f*x+e))^(7/2)*(c-c*sin(f*x+e))^(9/2),x)

[Out]

1/280/f*(-c*(-1+sin(f*x+e)))^(9/2)*sin(f*x+e)*(a*(1+sin(f*x+e)))^(7/2)*(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)^9

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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{9}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(7/2)*(c-c*sin(f*x+e))^(9/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^(7/2)*(-c*sin(f*x + e) + c)^(9/2), x)

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Fricas [A]  time = 1.20983, size = 306, normalized size = 1.71 \begin{align*} \frac{{\left (35 \, a^{3} c^{4} \cos \left (f x + e\right )^{8} - 35 \, a^{3} c^{4} + 8 \,{\left (5 \, a^{3} c^{4} \cos \left (f x + e\right )^{6} + 6 \, a^{3} c^{4} \cos \left (f x + e\right )^{4} + 8 \, a^{3} c^{4} \cos \left (f x + e\right )^{2} + 16 \, 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}}{280 \, f \cos \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(7/2)*(c-c*sin(f*x+e))^(9/2),x, algorithm="fricas")

[Out]

1/280*(35*a^3*c^4*cos(f*x + e)^8 - 35*a^3*c^4 + 8*(5*a^3*c^4*cos(f*x + e)^6 + 6*a^3*c^4*cos(f*x + e)^4 + 8*a^3
*c^4*cos(f*x + e)^2 + 16*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((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((a+a*sin(f*x+e))^(7/2)*(c-c*sin(f*x+e))^(9/2),x, algorithm="giac")

[Out]

sage2