3.368 \(\int \frac{(a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{13/2}} \, dx\)

Optimal. Leaf size=140 \[ \frac{a^3 \cos (e+f x)}{60 c^2 f \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{9/2}}-\frac{a^2 \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{15 c f (c-c \sin (e+f x))^{11/2}}+\frac{a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{6 f (c-c \sin (e+f x))^{13/2}} \]

[Out]

(a*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(6*f*(c - c*Sin[e + f*x])^(13/2)) - (a^2*Cos[e + f*x]*Sqrt[a + a*S
in[e + f*x]])/(15*c*f*(c - c*Sin[e + f*x])^(11/2)) + (a^3*Cos[e + f*x])/(60*c^2*f*Sqrt[a + a*Sin[e + f*x]]*(c
- c*Sin[e + f*x])^(9/2))

________________________________________________________________________________________

Rubi [A]  time = 0.272628, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.067, Rules used = {2739, 2738} \[ \frac{a^3 \cos (e+f x)}{60 c^2 f \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{9/2}}-\frac{a^2 \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{15 c f (c-c \sin (e+f x))^{11/2}}+\frac{a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{6 f (c-c \sin (e+f x))^{13/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^(5/2)/(c - c*Sin[e + f*x])^(13/2),x]

[Out]

(a*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(6*f*(c - c*Sin[e + f*x])^(13/2)) - (a^2*Cos[e + f*x]*Sqrt[a + a*S
in[e + f*x]])/(15*c*f*(c - c*Sin[e + f*x])^(11/2)) + (a^3*Cos[e + f*x])/(60*c^2*f*Sqrt[a + a*Sin[e + f*x]]*(c
- c*Sin[e + f*x])^(9/2))

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))^{5/2}}{(c-c \sin (e+f x))^{13/2}} \, dx &=\frac{a \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{6 f (c-c \sin (e+f x))^{13/2}}-\frac{a \int \frac{(a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{11/2}} \, dx}{3 c}\\ &=\frac{a \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{6 f (c-c \sin (e+f x))^{13/2}}-\frac{a^2 \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{15 c f (c-c \sin (e+f x))^{11/2}}+\frac{a^2 \int \frac{\sqrt{a+a \sin (e+f x)}}{(c-c \sin (e+f x))^{9/2}} \, dx}{15 c^2}\\ &=\frac{a \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{6 f (c-c \sin (e+f x))^{13/2}}-\frac{a^2 \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{15 c f (c-c \sin (e+f x))^{11/2}}+\frac{a^3 \cos (e+f x)}{60 c^2 f \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{9/2}}\\ \end{align*}

Mathematica [A]  time = 4.74052, size = 118, normalized size = 0.84 \[ \frac{a^2 \sqrt{a (\sin (e+f x)+1)} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) (36 \sin (e+f x)-15 \cos (2 (e+f x))+29)}{120 c^6 f (\sin (e+f x)-1)^6 \sqrt{c-c \sin (e+f x)} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])^(5/2)/(c - c*Sin[e + f*x])^(13/2),x]

[Out]

(a^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x])]*(29 - 15*Cos[2*(e + f*x)] + 36*Sin[e + f
*x]))/(120*c^6*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-1 + Sin[e + f*x])^6*Sqrt[c - c*Sin[e + f*x]])

________________________________________________________________________________________

Maple [B]  time = 0.187, size = 252, normalized size = 1.8 \begin{align*}{\frac{ \left ( 7\,\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{5}-7\, \left ( \cos \left ( fx+e \right ) \right ) ^{6}-49\,\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{4}-42\, \left ( \cos \left ( fx+e \right ) \right ) ^{5}-119\,\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{3}+168\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}+343\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) +224\, \left ( \cos \left ( fx+e \right ) \right ) ^{3}+202\,\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) -545\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}-444\,\sin \left ( fx+e \right ) -242\,\cos \left ( fx+e \right ) +444 \right ) \sin \left ( fx+e \right ) }{60\,f \left ( \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) - \left ( \cos \left ( fx+e \right ) \right ) ^{3}+2\,\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) +3\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}-4\,\sin \left ( fx+e \right ) +2\,\cos \left ( fx+e \right ) -4 \right ) } \left ( a \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{{\frac{5}{2}}} \left ( -c \left ( -1+\sin \left ( fx+e \right ) \right ) \right ) ^{-{\frac{13}{2}}}} \end{align*}

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))^(13/2),x)

[Out]

1/60/f*(7*sin(f*x+e)*cos(f*x+e)^5-7*cos(f*x+e)^6-49*sin(f*x+e)*cos(f*x+e)^4-42*cos(f*x+e)^5-119*sin(f*x+e)*cos
(f*x+e)^3+168*cos(f*x+e)^4+343*cos(f*x+e)^2*sin(f*x+e)+224*cos(f*x+e)^3+202*sin(f*x+e)*cos(f*x+e)-545*cos(f*x+
e)^2-444*sin(f*x+e)-242*cos(f*x+e)+444)*sin(f*x+e)*(a*(1+sin(f*x+e)))^(5/2)/(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)/(-c*(-1+sin(f*x+e)))^(13/2)

________________________________________________________________________________________

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))^(5/2)/(c-c*sin(f*x+e))^(13/2),x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

Fricas [A]  time = 1.1942, size = 409, normalized size = 2.92 \begin{align*} \frac{{\left (15 \, a^{2} \cos \left (f x + e\right )^{2} - 18 \, a^{2} \sin \left (f x + e\right ) - 22 \, a^{2}\right )} \sqrt{a \sin \left (f x + e\right ) + a} \sqrt{-c \sin \left (f x + e\right ) + c}}{60 \,{\left (c^{7} f \cos \left (f x + e\right )^{7} - 18 \, c^{7} f \cos \left (f x + e\right )^{5} + 48 \, c^{7} f \cos \left (f x + e\right )^{3} - 32 \, c^{7} f \cos \left (f x + e\right ) + 2 \,{\left (3 \, c^{7} f \cos \left (f x + e\right )^{5} - 16 \, c^{7} f \cos \left (f x + e\right )^{3} + 16 \, c^{7} 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))^(5/2)/(c-c*sin(f*x+e))^(13/2),x, algorithm="fricas")

[Out]

1/60*(15*a^2*cos(f*x + e)^2 - 18*a^2*sin(f*x + e) - 22*a^2)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)
/(c^7*f*cos(f*x + e)^7 - 18*c^7*f*cos(f*x + e)^5 + 48*c^7*f*cos(f*x + e)^3 - 32*c^7*f*cos(f*x + e) + 2*(3*c^7*
f*cos(f*x + e)^5 - 16*c^7*f*cos(f*x + e)^3 + 16*c^7*f*cos(f*x + e))*sin(f*x + e))

________________________________________________________________________________________

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))**(5/2)/(c-c*sin(f*x+e))**(13/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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{13}{2}}}\,{d x} \end{align*}

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))^(13/2),x, algorithm="giac")

[Out]

integrate((a*sin(f*x + e) + a)^(5/2)/(-c*sin(f*x + e) + c)^(13/2), x)