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]

1/6*a*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)/f/(c-c*sin(f*x+e))^(13/2)+1/60*a^3*cos(f*x+e)/c^2/f/(c-c*sin(f*x+e))^(
9/2)/(a+a*sin(f*x+e))^(1/2)-1/15*a^2*cos(f*x+e)*(a+a*sin(f*x+e))^(1/2)/c/f/(c-c*sin(f*x+e))^(11/2)

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Rubi [A]  time = 0.27, antiderivative size = 140, normalized size of antiderivative = 1.00, 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 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)]

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])

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*}

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Mathematica [A]  time = 4.91, 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]])

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fricas [A]  time = 0.48, size = 162, normalized size = 1.16 \[ \frac {{\left (15 \, a^{2} \cos \left (f x + e\right )^{2} - 18 \, a^{2} \sin \left (f x + e\right ) - 22 \, a^{2}\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{60 \, {\left (c^{7} f \cos \left (f x + e\right )^{7} - 18 \, c^{7} f \cos \left (f x + e\right )^{5} + 48 \, c^{7} f \cos \left (f x + e\right )^{3} - 32 \, c^{7} f \cos \left (f x + e\right ) + 2 \, {\left (3 \, c^{7} f \cos \left (f x + e\right )^{5} - 16 \, c^{7} f \cos \left (f x + e\right )^{3} + 16 \, c^{7} f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )}} \]

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))

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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))^(13/2),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.32, size = 252, normalized size = 1.80 \[ \frac {\left (a \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}} \sin \left (f x +e \right ) \left (7 \sin \left (f x +e \right ) \left (\cos ^{5}\left (f x +e \right )\right )-7 \left (\cos ^{6}\left (f x +e \right )\right )-49 \sin \left (f x +e \right ) \left (\cos ^{4}\left (f x +e \right )\right )-42 \left (\cos ^{5}\left (f x +e \right )\right )-119 \sin \left (f x +e \right ) \left (\cos ^{3}\left (f x +e \right )\right )+168 \left (\cos ^{4}\left (f x +e \right )\right )+343 \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+224 \left (\cos ^{3}\left (f x +e \right )\right )+202 \sin \left (f x +e \right ) \cos \left (f x +e \right )-545 \left (\cos ^{2}\left (f x +e \right )\right )-444 \sin \left (f x +e \right )-242 \cos \left (f x +e \right )+444\right )}{60 f \left (-c \left (\sin \left (f x +e \right )-1\right )\right )^{\frac {13}{2}} \left (\left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-\left (\cos ^{3}\left (f x +e \right )\right )+2 \sin \left (f x +e \right ) \cos \left (f x +e \right )+3 \left (\cos ^{2}\left (f x +e \right )\right )-4 \sin \left (f x +e \right )+2 \cos \left (f x +e \right )-4\right )} \]

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))^(13/2),x)

[Out]

1/60/f*(a*(1+sin(f*x+e)))^(5/2)*sin(f*x+e)*(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+20
2*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)/(-c*(sin(f*x+e)-1))^(13/2)/(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 \[ \int \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {13}{2}}}\,{d x} \]

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))^(13/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 12.24, size = 287, normalized size = 2.05 \[ -\frac {\sqrt {c-c\,\sin \left (e+f\,x\right )}\,\left (\frac {464\,a^2\,{\mathrm {e}}^{e\,7{}\mathrm {i}+f\,x\,7{}\mathrm {i}}\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{15\,c^7\,f}+\frac {192\,a^2\,{\mathrm {e}}^{e\,7{}\mathrm {i}+f\,x\,7{}\mathrm {i}}\,\sin \left (e+f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{5\,c^7\,f}-\frac {16\,a^2\,{\mathrm {e}}^{e\,7{}\mathrm {i}+f\,x\,7{}\mathrm {i}}\,\cos \left (2\,e+2\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{c^7\,f}\right )}{-858\,\cos \left (e+f\,x\right )\,{\mathrm {e}}^{e\,7{}\mathrm {i}+f\,x\,7{}\mathrm {i}}+858\,{\mathrm {e}}^{e\,7{}\mathrm {i}+f\,x\,7{}\mathrm {i}}\,\cos \left (3\,e+3\,f\,x\right )-130\,{\mathrm {e}}^{e\,7{}\mathrm {i}+f\,x\,7{}\mathrm {i}}\,\cos \left (5\,e+5\,f\,x\right )+2\,{\mathrm {e}}^{e\,7{}\mathrm {i}+f\,x\,7{}\mathrm {i}}\,\cos \left (7\,e+7\,f\,x\right )+1144\,{\mathrm {e}}^{e\,7{}\mathrm {i}+f\,x\,7{}\mathrm {i}}\,\sin \left (2\,e+2\,f\,x\right )-416\,{\mathrm {e}}^{e\,7{}\mathrm {i}+f\,x\,7{}\mathrm {i}}\,\sin \left (4\,e+4\,f\,x\right )+24\,{\mathrm {e}}^{e\,7{}\mathrm {i}+f\,x\,7{}\mathrm {i}}\,\sin \left (6\,e+6\,f\,x\right )} \]

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))^(13/2),x)

[Out]

-((c - c*sin(e + f*x))^(1/2)*((464*a^2*exp(e*7i + f*x*7i)*(a + a*sin(e + f*x))^(1/2))/(15*c^7*f) + (192*a^2*ex
p(e*7i + f*x*7i)*sin(e + f*x)*(a + a*sin(e + f*x))^(1/2))/(5*c^7*f) - (16*a^2*exp(e*7i + f*x*7i)*cos(2*e + 2*f
*x)*(a + a*sin(e + f*x))^(1/2))/(c^7*f)))/(858*exp(e*7i + f*x*7i)*cos(3*e + 3*f*x) - 858*cos(e + f*x)*exp(e*7i
+ f*x*7i) - 130*exp(e*7i + f*x*7i)*cos(5*e + 5*f*x) + 2*exp(e*7i + f*x*7i)*cos(7*e + 7*f*x) + 1144*exp(e*7i +
f*x*7i)*sin(2*e + 2*f*x) - 416*exp(e*7i + f*x*7i)*sin(4*e + 4*f*x) + 24*exp(e*7i + f*x*7i)*sin(6*e + 6*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))**(13/2),x)

[Out]

Timed out

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