3.366 \(\int \frac {(a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{9/2}} \, dx\)

Optimal. Leaf size=88 \[ \frac {\cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{48 c f (c-c \sin (e+f x))^{7/2}}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{8 f (c-c \sin (e+f x))^{9/2}} \]

[Out]

1/8*cos(f*x+e)*(a+a*sin(f*x+e))^(5/2)/f/(c-c*sin(f*x+e))^(9/2)+1/48*cos(f*x+e)*(a+a*sin(f*x+e))^(5/2)/c/f/(c-c
*sin(f*x+e))^(7/2)

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Rubi [A]  time = 0.19, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {2743, 2742} \[ \frac {\cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{48 c f (c-c \sin (e+f x))^{7/2}}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{8 f (c-c \sin (e+f x))^{9/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^(5/2)/(c - c*Sin[e + f*x])^(9/2),x]

[Out]

(Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(8*f*(c - c*Sin[e + f*x])^(9/2)) + (Cos[e + f*x]*(a + a*Sin[e + f*x]
)^(5/2))/(48*c*f*(c - c*Sin[e + f*x])^(7/2))

Rule 2742

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[(b*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n)/(a*f*(2*m + 1)), x] /; FreeQ[{a, b, c, d, e, f
, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && EqQ[m + n + 1, 0] && NeQ[m, -2^(-1)]

Rule 2743

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[(b*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n)/(a*f*(2*m + 1)), x] + Dist[(m + n + 1)/(a*(2*m
+ 1)), 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, m, n}, x]
&& EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && ILtQ[Simplify[m + n + 1], 0] && NeQ[m, -2^(-1)] && (SumSimplerQ[m
, 1] ||  !SumSimplerQ[n, 1])

Rubi steps

\begin {align*} \int \frac {(a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{9/2}} \, dx &=\frac {\cos (e+f x) (a+a \sin (e+f x))^{5/2}}{8 f (c-c \sin (e+f x))^{9/2}}+\frac {\int \frac {(a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{7/2}} \, dx}{8 c}\\ &=\frac {\cos (e+f x) (a+a \sin (e+f x))^{5/2}}{8 f (c-c \sin (e+f x))^{9/2}}+\frac {\cos (e+f x) (a+a \sin (e+f x))^{5/2}}{48 c f (c-c \sin (e+f x))^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 2.25, size = 118, normalized size = 1.34 \[ \frac {a^2 \sqrt {a (\sin (e+f x)+1)} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (4 \sin (e+f x)-3 \cos (2 (e+f x))+5)}{12 c^4 f (\sin (e+f x)-1)^4 \sqrt {c-c \sin (e+f x)} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])^(5/2)/(c - c*Sin[e + f*x])^(9/2),x]

[Out]

(a^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x])]*(5 - 3*Cos[2*(e + f*x)] + 4*Sin[e + f*x]
))/(12*c^4*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-1 + Sin[e + f*x])^4*Sqrt[c - c*Sin[e + f*x]])

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fricas [A]  time = 0.45, size = 133, normalized size = 1.51 \[ -\frac {{\left (3 \, a^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2} \sin \left (f x + e\right ) - 4 \, a^{2}\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{6 \, {\left (c^{5} f \cos \left (f x + e\right )^{5} - 8 \, c^{5} f \cos \left (f x + e\right )^{3} + 8 \, c^{5} f \cos \left (f x + e\right ) + 4 \, {\left (c^{5} f \cos \left (f x + e\right )^{3} - 2 \, c^{5} f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(9/2),x, algorithm="fricas")

[Out]

-1/6*(3*a^2*cos(f*x + e)^2 - 2*a^2*sin(f*x + e) - 4*a^2)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(c
^5*f*cos(f*x + e)^5 - 8*c^5*f*cos(f*x + e)^3 + 8*c^5*f*cos(f*x + e) + 4*(c^5*f*cos(f*x + e)^3 - 2*c^5*f*cos(f*
x + e))*sin(f*x + e))

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(9/2),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.28, size = 199, normalized size = 2.26 \[ -\frac {\left (a \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}} \sin \left (f x +e \right ) \left (\sin \left (f x +e \right ) \left (\cos ^{3}\left (f x +e \right )\right )-\left (\cos ^{4}\left (f x +e \right )\right )-5 \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-4 \left (\cos ^{3}\left (f x +e \right )\right )-4 \sin \left (f x +e \right ) \cos \left (f x +e \right )+9 \left (\cos ^{2}\left (f x +e \right )\right )+14 \sin \left (f x +e \right )+10 \cos \left (f x +e \right )-14\right )}{6 f \left (-c \left (\sin \left (f x +e \right )-1\right )\right )^{\frac {9}{2}} \left (\left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-\left (\cos ^{3}\left (f x +e \right )\right )+2 \sin \left (f x +e \right ) \cos \left (f x +e \right )+3 \left (\cos ^{2}\left (f x +e \right )\right )-4 \sin \left (f x +e \right )+2 \cos \left (f x +e \right )-4\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(9/2),x)

[Out]

-1/6/f*(a*(1+sin(f*x+e)))^(5/2)*sin(f*x+e)*(sin(f*x+e)*cos(f*x+e)^3-cos(f*x+e)^4-5*cos(f*x+e)^2*sin(f*x+e)-4*c
os(f*x+e)^3-4*sin(f*x+e)*cos(f*x+e)+9*cos(f*x+e)^2+14*sin(f*x+e)+10*cos(f*x+e)-14)/(-c*(sin(f*x+e)-1))^(9/2)/(
cos(f*x+e)^2*sin(f*x+e)-cos(f*x+e)^3+2*sin(f*x+e)*cos(f*x+e)+3*cos(f*x+e)^2-4*sin(f*x+e)+2*cos(f*x+e)-4)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {9}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(9/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^(5/2)/(-c*sin(f*x + e) + c)^(9/2), x)

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mupad [B]  time = 11.57, size = 242, normalized size = 2.75 \[ \frac {\sqrt {c-c\,\sin \left (e+f\,x\right )}\,\left (\frac {40\,a^2\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{3\,c^5\,f}+\frac {32\,a^2\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,\sin \left (e+f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{3\,c^5\,f}-\frac {8\,a^2\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,\cos \left (2\,e+2\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{c^5\,f}\right )}{84\,\cos \left (e+f\,x\right )\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}-54\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,\cos \left (3\,e+3\,f\,x\right )+2\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,\cos \left (5\,e+5\,f\,x\right )-96\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,\sin \left (2\,e+2\,f\,x\right )+16\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,\sin \left (4\,e+4\,f\,x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(e + f*x))^(5/2)/(c - c*sin(e + f*x))^(9/2),x)

[Out]

((c - c*sin(e + f*x))^(1/2)*((40*a^2*exp(e*5i + f*x*5i)*(a + a*sin(e + f*x))^(1/2))/(3*c^5*f) + (32*a^2*exp(e*
5i + f*x*5i)*sin(e + f*x)*(a + a*sin(e + f*x))^(1/2))/(3*c^5*f) - (8*a^2*exp(e*5i + f*x*5i)*cos(2*e + 2*f*x)*(
a + a*sin(e + f*x))^(1/2))/(c^5*f)))/(84*cos(e + f*x)*exp(e*5i + f*x*5i) - 54*exp(e*5i + f*x*5i)*cos(3*e + 3*f
*x) + 2*exp(e*5i + f*x*5i)*cos(5*e + 5*f*x) - 96*exp(e*5i + f*x*5i)*sin(2*e + 2*f*x) + 16*exp(e*5i + f*x*5i)*s
in(4*e + 4*f*x))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))**(5/2)/(c-c*sin(f*x+e))**(9/2),x)

[Out]

Timed out

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