3.382 \(\int \frac{(a+a \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{17/2}} \, dx\)

Optimal. Leaf size=188 \[ -\frac{a^4 \cos (e+f x)}{280 c^3 f \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{11/2}}+\frac{a^3 \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{56 c^2 f (c-c \sin (e+f x))^{13/2}}-\frac{3 a^2 \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{56 c f (c-c \sin (e+f x))^{15/2}}+\frac{a \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{8 f (c-c \sin (e+f x))^{17/2}} \]

[Out]

(a*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(8*f*(c - c*Sin[e + f*x])^(17/2)) - (3*a^2*Cos[e + f*x]*(a + a*Sin
[e + f*x])^(3/2))/(56*c*f*(c - c*Sin[e + f*x])^(15/2)) + (a^3*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(56*c^2*f
*(c - c*Sin[e + f*x])^(13/2)) - (a^4*Cos[e + f*x])/(280*c^3*f*Sqrt[a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(1
1/2))

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Rubi [A]  time = 0.384105, antiderivative size = 188, 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 = {2739, 2738} \[ -\frac{a^4 \cos (e+f x)}{280 c^3 f \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{11/2}}+\frac{a^3 \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{56 c^2 f (c-c \sin (e+f x))^{13/2}}-\frac{3 a^2 \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{56 c f (c-c \sin (e+f x))^{15/2}}+\frac{a \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{8 f (c-c \sin (e+f x))^{17/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^(7/2)/(c - c*Sin[e + f*x])^(17/2),x]

[Out]

(a*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(8*f*(c - c*Sin[e + f*x])^(17/2)) - (3*a^2*Cos[e + f*x]*(a + a*Sin
[e + f*x])^(3/2))/(56*c*f*(c - c*Sin[e + f*x])^(15/2)) + (a^3*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(56*c^2*f
*(c - c*Sin[e + f*x])^(13/2)) - (a^4*Cos[e + f*x])/(280*c^3*f*Sqrt[a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(1
1/2))

Rule 2739

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[(-2*b*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n)/(f*(2*n + 1)), x] - Dist[(b*(2*m - 1)
)/(d*(2*n + 1)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e
, f}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[m - 1/2, 0] && LtQ[n, -1] &&  !(ILtQ[m + n, 0] && G
tQ[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 \frac{(a+a \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{17/2}} \, dx &=\frac{a \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{8 f (c-c \sin (e+f x))^{17/2}}-\frac{(3 a) \int \frac{(a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{15/2}} \, dx}{8 c}\\ &=\frac{a \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{8 f (c-c \sin (e+f x))^{17/2}}-\frac{3 a^2 \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{56 c f (c-c \sin (e+f x))^{15/2}}+\frac{\left (3 a^2\right ) \int \frac{(a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{13/2}} \, dx}{28 c^2}\\ &=\frac{a \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{8 f (c-c \sin (e+f x))^{17/2}}-\frac{3 a^2 \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{56 c f (c-c \sin (e+f x))^{15/2}}+\frac{a^3 \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{56 c^2 f (c-c \sin (e+f x))^{13/2}}-\frac{a^3 \int \frac{\sqrt{a+a \sin (e+f x)}}{(c-c \sin (e+f x))^{11/2}} \, dx}{56 c^3}\\ &=\frac{a \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{8 f (c-c \sin (e+f x))^{17/2}}-\frac{3 a^2 \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{56 c f (c-c \sin (e+f x))^{15/2}}+\frac{a^3 \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{56 c^2 f (c-c \sin (e+f x))^{13/2}}-\frac{a^4 \cos (e+f x)}{280 c^3 f \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{11/2}}\\ \end{align*}

Mathematica [A]  time = 6.02393, size = 128, normalized size = 0.68 \[ \frac{a^3 \sqrt{a (\sin (e+f x)+1)} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) (65 \sin (e+f x)-7 \sin (3 (e+f x))-28 \cos (2 (e+f x))+40)}{140 c^8 f (\sin (e+f x)-1)^8 \sqrt{c-c \sin (e+f x)} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])^(7/2)/(c - c*Sin[e + f*x])^(17/2),x]

[Out]

(a^3*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x])]*(40 - 28*Cos[2*(e + f*x)] + 65*Sin[e + f
*x] - 7*Sin[3*(e + f*x)]))/(140*c^8*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-1 + Sin[e + f*x])^8*Sqrt[c - c*S
in[e + f*x]])

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Maple [A]  time = 0.228, size = 328, normalized size = 1.7 \begin{align*}{\frac{ \left ( 3\,\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{7}-3\, \left ( \cos \left ( fx+e \right ) \right ) ^{8}-27\, \left ( \cos \left ( fx+e \right ) \right ) ^{6}\sin \left ( fx+e \right ) -24\, \left ( \cos \left ( fx+e \right ) \right ) ^{7}-93\,\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{5}+120\, \left ( \cos \left ( fx+e \right ) \right ) ^{6}+333\,\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{4}+240\, \left ( \cos \left ( fx+e \right ) \right ) ^{5}+387\,\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{3}-720\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}-970\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) -583\, \left ( \cos \left ( fx+e \right ) \right ) ^{3}-367\,\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) +1337\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}+769\,\sin \left ( fx+e \right ) +402\,\cos \left ( fx+e \right ) -769 \right ) \sin \left ( fx+e \right ) }{35\,f \left ( \sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{3}+ \left ( \cos \left ( fx+e \right ) \right ) ^{4}-4\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) +3\, \left ( \cos \left ( fx+e \right ) \right ) ^{3}-4\,\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) -8\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}+8\,\sin \left ( fx+e \right ) -4\,\cos \left ( fx+e \right ) +8 \right ) } \left ( a \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{{\frac{7}{2}}} \left ( -c \left ( -1+\sin \left ( fx+e \right ) \right ) \right ) ^{-{\frac{17}{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))^(17/2),x)

[Out]

1/35/f*(3*sin(f*x+e)*cos(f*x+e)^7-3*cos(f*x+e)^8-27*cos(f*x+e)^6*sin(f*x+e)-24*cos(f*x+e)^7-93*sin(f*x+e)*cos(
f*x+e)^5+120*cos(f*x+e)^6+333*sin(f*x+e)*cos(f*x+e)^4+240*cos(f*x+e)^5+387*sin(f*x+e)*cos(f*x+e)^3-720*cos(f*x
+e)^4-970*cos(f*x+e)^2*sin(f*x+e)-583*cos(f*x+e)^3-367*sin(f*x+e)*cos(f*x+e)+1337*cos(f*x+e)^2+769*sin(f*x+e)+
402*cos(f*x+e)-769)*sin(f*x+e)*(a*(1+sin(f*x+e)))^(7/2)/(sin(f*x+e)*cos(f*x+e)^3+cos(f*x+e)^4-4*cos(f*x+e)^2*s
in(f*x+e)+3*cos(f*x+e)^3-4*sin(f*x+e)*cos(f*x+e)-8*cos(f*x+e)^2+8*sin(f*x+e)-4*cos(f*x+e)+8)/(-c*(-1+sin(f*x+e
)))^(17/2)

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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((a+a*sin(f*x+e))^(7/2)/(c-c*sin(f*x+e))^(17/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 1.34539, size = 516, normalized size = 2.74 \begin{align*} -\frac{{\left (14 \, a^{3} \cos \left (f x + e\right )^{2} - 17 \, a^{3} +{\left (7 \, a^{3} \cos \left (f x + e\right )^{2} - 18 \, a^{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}}{35 \,{\left (c^{9} f \cos \left (f x + e\right )^{9} - 32 \, c^{9} f \cos \left (f x + e\right )^{7} + 160 \, c^{9} f \cos \left (f x + e\right )^{5} - 256 \, c^{9} f \cos \left (f x + e\right )^{3} + 128 \, c^{9} f \cos \left (f x + e\right ) + 8 \,{\left (c^{9} f \cos \left (f x + e\right )^{7} - 10 \, c^{9} f \cos \left (f x + e\right )^{5} + 24 \, c^{9} f \cos \left (f x + e\right )^{3} - 16 \, c^{9} f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\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))^(17/2),x, algorithm="fricas")

[Out]

-1/35*(14*a^3*cos(f*x + e)^2 - 17*a^3 + (7*a^3*cos(f*x + e)^2 - 18*a^3)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)
*sqrt(-c*sin(f*x + e) + c)/(c^9*f*cos(f*x + e)^9 - 32*c^9*f*cos(f*x + e)^7 + 160*c^9*f*cos(f*x + e)^5 - 256*c^
9*f*cos(f*x + e)^3 + 128*c^9*f*cos(f*x + e) + 8*(c^9*f*cos(f*x + e)^7 - 10*c^9*f*cos(f*x + e)^5 + 24*c^9*f*cos
(f*x + e)^3 - 16*c^9*f*cos(f*x + e))*sin(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))**(17/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac{7}{2}}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac{17}{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))^(17/2),x, algorithm="giac")

[Out]

integrate((a*sin(f*x + e) + a)^(7/2)/(-c*sin(f*x + e) + c)^(17/2), x)