3.516 \(\int \frac {(c+d \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^3} \, dx\)

Optimal. Leaf size=322 \[ -\frac {\left (4 c^2+15 c d+27 d^2\right ) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{30 f \left (a^3 \sin (e+f x)+a^3\right )}+\frac {(c+d) \left (4 c^2+11 c d+15 d^2\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} F\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{30 a^3 f \sqrt {c+d \sin (e+f x)}}-\frac {\left (4 c^2+15 c d+27 d^2\right ) \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{30 a^3 f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 f (a \sin (e+f x)+a)^3}-\frac {2 (c-d) (c+3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{15 a f (a \sin (e+f x)+a)^2} \]

[Out]

-1/5*(c-d)*cos(f*x+e)*(c+d*sin(f*x+e))^(3/2)/f/(a+a*sin(f*x+e))^3-2/15*(c-d)*(c+3*d)*cos(f*x+e)*(c+d*sin(f*x+e
))^(1/2)/a/f/(a+a*sin(f*x+e))^2-1/30*(4*c^2+15*c*d+27*d^2)*cos(f*x+e)*(c+d*sin(f*x+e))^(1/2)/f/(a^3+a^3*sin(f*
x+e))+1/30*(4*c^2+15*c*d+27*d^2)*(sin(1/2*e+1/4*Pi+1/2*f*x)^2)^(1/2)/sin(1/2*e+1/4*Pi+1/2*f*x)*EllipticE(cos(1
/2*e+1/4*Pi+1/2*f*x),2^(1/2)*(d/(c+d))^(1/2))*(c+d*sin(f*x+e))^(1/2)/a^3/f/((c+d*sin(f*x+e))/(c+d))^(1/2)-1/30
*(c+d)*(4*c^2+11*c*d+15*d^2)*(sin(1/2*e+1/4*Pi+1/2*f*x)^2)^(1/2)/sin(1/2*e+1/4*Pi+1/2*f*x)*EllipticF(cos(1/2*e
+1/4*Pi+1/2*f*x),2^(1/2)*(d/(c+d))^(1/2))*((c+d*sin(f*x+e))/(c+d))^(1/2)/a^3/f/(c+d*sin(f*x+e))^(1/2)

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Rubi [A]  time = 0.83, antiderivative size = 322, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {2765, 2977, 2978, 2752, 2663, 2661, 2655, 2653} \[ -\frac {\left (4 c^2+15 c d+27 d^2\right ) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{30 f \left (a^3 \sin (e+f x)+a^3\right )}+\frac {(c+d) \left (4 c^2+11 c d+15 d^2\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} F\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{30 a^3 f \sqrt {c+d \sin (e+f x)}}-\frac {\left (4 c^2+15 c d+27 d^2\right ) \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{30 a^3 f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 f (a \sin (e+f x)+a)^3}-\frac {2 (c-d) (c+3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{15 a f (a \sin (e+f x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*Sin[e + f*x])^(5/2)/(a + a*Sin[e + f*x])^3,x]

[Out]

(-2*(c - d)*(c + 3*d)*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(15*a*f*(a + a*Sin[e + f*x])^2) - ((4*c^2 + 15*c*
d + 27*d^2)*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(30*f*(a^3 + a^3*Sin[e + f*x])) - ((c - d)*Cos[e + f*x]*(c
+ d*Sin[e + f*x])^(3/2))/(5*f*(a + a*Sin[e + f*x])^3) - ((4*c^2 + 15*c*d + 27*d^2)*EllipticE[(e - Pi/2 + f*x)/
2, (2*d)/(c + d)]*Sqrt[c + d*Sin[e + f*x]])/(30*a^3*f*Sqrt[(c + d*Sin[e + f*x])/(c + d)]) + ((c + d)*(4*c^2 +
11*c*d + 15*d^2)*EllipticF[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/(30*a^3*f*Sq
rt[c + d*Sin[e + f*x]])

Rule 2653

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*Sqrt[a + b]*EllipticE[(1*(c - Pi/2 + d*x)
)/2, (2*b)/(a + b)])/d, x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2655

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2661

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, (2*b)
/(a + b)])/(d*Sqrt[a + b]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2663

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2752

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

Rule 2765

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

Rule 2977

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

Rule 2978

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

Rubi steps

\begin {align*} \int \frac {(c+d \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^3} \, dx &=-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 f (a+a \sin (e+f x))^3}-\frac {\int \frac {\sqrt {c+d \sin (e+f x)} \left (-\frac {1}{2} a \left (4 c^2+9 c d-3 d^2\right )-\frac {1}{2} a d (c+9 d) \sin (e+f x)\right )}{(a+a \sin (e+f x))^2} \, dx}{5 a^2}\\ &=-\frac {2 (c-d) (c+3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{15 a f (a+a \sin (e+f x))^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 f (a+a \sin (e+f x))^3}-\frac {\int \frac {-\frac {1}{2} a^2 \left (4 c^3+13 c^2 d+19 c d^2-6 d^3\right )-\frac {1}{2} a^2 d \left (2 c^2+7 c d+21 d^2\right ) \sin (e+f x)}{(a+a \sin (e+f x)) \sqrt {c+d \sin (e+f x)}} \, dx}{15 a^4}\\ &=-\frac {2 (c-d) (c+3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{15 a f (a+a \sin (e+f x))^2}-\frac {\left (4 c^2+15 c d+27 d^2\right ) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{30 f \left (a^3+a^3 \sin (e+f x)\right )}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 f (a+a \sin (e+f x))^3}+\frac {\int \frac {-\frac {1}{4} a^3 (c-15 d) (c-d) d^2-\frac {1}{4} a^3 (c-d) d \left (4 c^2+15 c d+27 d^2\right ) \sin (e+f x)}{\sqrt {c+d \sin (e+f x)}} \, dx}{15 a^6 (c-d)}\\ &=-\frac {2 (c-d) (c+3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{15 a f (a+a \sin (e+f x))^2}-\frac {\left (4 c^2+15 c d+27 d^2\right ) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{30 f \left (a^3+a^3 \sin (e+f x)\right )}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 f (a+a \sin (e+f x))^3}+\frac {\left ((c+d) \left (4 c^2+11 c d+15 d^2\right )\right ) \int \frac {1}{\sqrt {c+d \sin (e+f x)}} \, dx}{60 a^3}-\frac {\left (4 c^2+15 c d+27 d^2\right ) \int \sqrt {c+d \sin (e+f x)} \, dx}{60 a^3}\\ &=-\frac {2 (c-d) (c+3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{15 a f (a+a \sin (e+f x))^2}-\frac {\left (4 c^2+15 c d+27 d^2\right ) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{30 f \left (a^3+a^3 \sin (e+f x)\right )}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 f (a+a \sin (e+f x))^3}-\frac {\left (\left (4 c^2+15 c d+27 d^2\right ) \sqrt {c+d \sin (e+f x)}\right ) \int \sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}} \, dx}{60 a^3 \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {\left ((c+d) \left (4 c^2+11 c d+15 d^2\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}\right ) \int \frac {1}{\sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}}} \, dx}{60 a^3 \sqrt {c+d \sin (e+f x)}}\\ &=-\frac {2 (c-d) (c+3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{15 a f (a+a \sin (e+f x))^2}-\frac {\left (4 c^2+15 c d+27 d^2\right ) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{30 f \left (a^3+a^3 \sin (e+f x)\right )}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 f (a+a \sin (e+f x))^3}-\frac {\left (4 c^2+15 c d+27 d^2\right ) E\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {c+d \sin (e+f x)}}{30 a^3 f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {(c+d) \left (4 c^2+11 c d+15 d^2\right ) F\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}{30 a^3 f \sqrt {c+d \sin (e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 5.81, size = 385, normalized size = 1.20 \[ \frac {\left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^6 \left (-\left (4 c^2+15 c d+27 d^2\right ) (c+d \sin (e+f x))-\frac {(c+d \sin (e+f x)) \left (\left (20 c^2+74 c d+90 d^2\right ) \cos \left (\frac {3}{2} (e+f x)\right )+2 \sin \left (\frac {1}{2} (e+f x)\right ) \left (2 \left (2 c^2+7 c d-9 d^2\right ) \cos (e+f x)+\left (4 c^2+15 c d+27 d^2\right ) \cos (2 (e+f x))-3 \left (6 c^2+11 c d+29 d^2\right )\right )-2 d (35 c+57 d) \cos \left (\frac {1}{2} (e+f x)\right )\right )}{2 \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^5}+\left (4 c^2+15 c d+27 d^2\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} \left ((c+d) E\left (\frac {1}{4} (-2 e-2 f x+\pi )|\frac {2 d}{c+d}\right )-c F\left (\frac {1}{4} (-2 e-2 f x+\pi )|\frac {2 d}{c+d}\right )\right )+d^2 (c-15 d) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} F\left (\frac {1}{4} (-2 e-2 f x+\pi )|\frac {2 d}{c+d}\right )\right )}{30 a^3 f (\sin (e+f x)+1)^3 \sqrt {c+d \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*Sin[e + f*x])^(5/2)/(a + a*Sin[e + f*x])^3,x]

[Out]

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^6*(-((4*c^2 + 15*c*d + 27*d^2)*(c + d*Sin[e + f*x])) - ((-2*d*(35*c + 5
7*d)*Cos[(e + f*x)/2] + (20*c^2 + 74*c*d + 90*d^2)*Cos[(3*(e + f*x))/2] + 2*(-3*(6*c^2 + 11*c*d + 29*d^2) + 2*
(2*c^2 + 7*c*d - 9*d^2)*Cos[e + f*x] + (4*c^2 + 15*c*d + 27*d^2)*Cos[2*(e + f*x)])*Sin[(e + f*x)/2])*(c + d*Si
n[e + f*x]))/(2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5) + (c - 15*d)*d^2*EllipticF[(-2*e + Pi - 2*f*x)/4, (2*
d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)] + (4*c^2 + 15*c*d + 27*d^2)*((c + d)*EllipticE[(-2*e + Pi - 2*f
*x)/4, (2*d)/(c + d)] - c*EllipticF[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)])*Sqrt[(c + d*Sin[e + f*x])/(c + d)])
)/(30*a^3*f*(1 + Sin[e + f*x])^3*Sqrt[c + d*Sin[e + f*x]])

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fricas [F]  time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (d^{2} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2}\right )} \sqrt {d \sin \left (f x + e\right ) + c}}{3 \, a^{3} \cos \left (f x + e\right )^{2} - 4 \, a^{3} + {\left (a^{3} \cos \left (f x + e\right )^{2} - 4 \, a^{3}\right )} \sin \left (f x + e\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^3,x, algorithm="fricas")

[Out]

integral((d^2*cos(f*x + e)^2 - 2*c*d*sin(f*x + e) - c^2 - d^2)*sqrt(d*sin(f*x + e) + c)/(3*a^3*cos(f*x + e)^2
- 4*a^3 + (a^3*cos(f*x + e)^2 - 4*a^3)*sin(f*x + e)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^3,x, algorithm="giac")

[Out]

integrate((d*sin(f*x + e) + c)^(5/2)/(a*sin(f*x + e) + a)^3, x)

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maple [B]  time = 7.63, size = 1615, normalized size = 5.02 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c+d*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^3,x)

[Out]

(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)/a^3*(2*d^3*(c/d-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*x+e))/(c
+d))^(1/2)*((-sin(f*x+e)-1)*d/(c-d))^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*EllipticF(((c+d*sin(f*x+e))
/(c-d))^(1/2),((c-d)/(c+d))^(1/2))+(c^3-3*c^2*d+3*c*d^2-d^3)*(-1/5/(c-d)*(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/
2)/(1+sin(f*x+e))^3-2/15*(c-3*d)/(c-d)^2*(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)/(1+sin(f*x+e))^2-1/30*(-sin(f
*x+e)^2*d-c*sin(f*x+e)+d*sin(f*x+e)+c)/(c-d)^3*(4*c^2-15*c*d+27*d^2)/((-d*sin(f*x+e)-c)*(sin(f*x+e)-1)*(1+sin(
f*x+e)))^(1/2)+2*(-c*d^2-15*d^3)/(60*c^3-180*c^2*d+180*c*d^2-60*d^3)*(c/d-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d
*(1-sin(f*x+e))/(c+d))^(1/2)*((-sin(f*x+e)-1)*d/(c-d))^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*EllipticF
(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))-1/30*d*(4*c^2-15*c*d+27*d^2)/(c-d)^3*(c/d-1)*((c+d*sin(f*
x+e))/(c-d))^(1/2)*(d*(1-sin(f*x+e))/(c+d))^(1/2)*((-sin(f*x+e)-1)*d/(c-d))^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+
e)^2)^(1/2)*((-c/d-1)*EllipticE(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))+EllipticF(((c+d*sin(f*x+e)
)/(c-d))^(1/2),((c-d)/(c+d))^(1/2))))+3*d^2*(c-d)*(-(-sin(f*x+e)^2*d-c*sin(f*x+e)+d*sin(f*x+e)+c)/(c-d)/((-d*s
in(f*x+e)-c)*(sin(f*x+e)-1)*(1+sin(f*x+e)))^(1/2)-2*d/(2*c-2*d)*(c/d-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-s
in(f*x+e))/(c+d))^(1/2)*((-sin(f*x+e)-1)*d/(c-d))^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*EllipticF(((c+
d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))-d/(c-d)*(c/d-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*x+e
))/(c+d))^(1/2)*((-sin(f*x+e)-1)*d/(c-d))^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*((-c/d-1)*EllipticE(((
c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))+EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2)
)))+3*d*(c^2-2*c*d+d^2)*(-1/3/(c-d)*(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)/(1+sin(f*x+e))^2-1/3*(-sin(f*x+e)^
2*d-c*sin(f*x+e)+d*sin(f*x+e)+c)/(c-d)^2*(c-3*d)/((-d*sin(f*x+e)-c)*(sin(f*x+e)-1)*(1+sin(f*x+e)))^(1/2)+2*d^2
/(3*c^2-6*c*d+3*d^2)*(c/d-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*x+e))/(c+d))^(1/2)*((-sin(f*x+e)-1)*d/
(c-d))^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1
/2))-1/3*d*(c-3*d)/(c-d)^2*(c/d-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*x+e))/(c+d))^(1/2)*((-sin(f*x+e)
-1)*d/(c-d))^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*((-c/d-1)*EllipticE(((c+d*sin(f*x+e))/(c-d))^(1/2),
((c-d)/(c+d))^(1/2))+EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2)))))/cos(f*x+e)/(c+d*sin(f*x+
e))^(1/2)/f

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^3,x, algorithm="maxima")

[Out]

integrate((d*sin(f*x + e) + c)^(5/2)/(a*sin(f*x + e) + a)^3, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{5/2}}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*sin(e + f*x))^(5/2)/(a + a*sin(e + f*x))^3,x)

[Out]

int((c + d*sin(e + f*x))^(5/2)/(a + a*sin(e + f*x))^3, 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((c+d*sin(f*x+e))**(5/2)/(a+a*sin(f*x+e))**3,x)

[Out]

Timed out

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