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

Optimal. Leaf size=237 \[ -\frac{2 a (c-3 d) \cos (e+f x)}{3 f (c-d) (c+d)^2 \sqrt{c+d \sin (e+f x)}}-\frac{2 a \cos (e+f x)}{3 f (c+d) (c+d \sin (e+f x))^{3/2}}+\frac{2 a \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{3 d f (c+d) \sqrt{c+d \sin (e+f x)}}-\frac{2 a (c-3 d) \sqrt{c+d \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{3 d f (c-d) (c+d)^2 \sqrt{\frac{c+d \sin (e+f x)}{c+d}}} \]

[Out]

(-2*a*Cos[e + f*x])/(3*(c + d)*f*(c + d*Sin[e + f*x])^(3/2)) - (2*a*(c - 3*d)*Cos[e + f*x])/(3*(c - d)*(c + d)
^2*f*Sqrt[c + d*Sin[e + f*x]]) - (2*a*(c - 3*d)*EllipticE[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[c + d*Sin[e
+ f*x]])/(3*(c - d)*d*(c + d)^2*f*Sqrt[(c + d*Sin[e + f*x])/(c + d)]) + (2*a*EllipticF[(e - Pi/2 + f*x)/2, (2*
d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/(3*d*(c + d)*f*Sqrt[c + d*Sin[e + f*x]])

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Rubi [A]  time = 0.336074, antiderivative size = 237, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {2754, 2752, 2663, 2661, 2655, 2653} \[ -\frac{2 a (c-3 d) \cos (e+f x)}{3 f (c-d) (c+d)^2 \sqrt{c+d \sin (e+f x)}}-\frac{2 a \cos (e+f x)}{3 f (c+d) (c+d \sin (e+f x))^{3/2}}+\frac{2 a \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{3 d f (c+d) \sqrt{c+d \sin (e+f x)}}-\frac{2 a (c-3 d) \sqrt{c+d \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{3 d f (c-d) (c+d)^2 \sqrt{\frac{c+d \sin (e+f x)}{c+d}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*Cos[e + f*x])/(3*(c + d)*f*(c + d*Sin[e + f*x])^(3/2)) - (2*a*(c - 3*d)*Cos[e + f*x])/(3*(c - d)*(c + d)
^2*f*Sqrt[c + d*Sin[e + f*x]]) - (2*a*(c - 3*d)*EllipticE[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[c + d*Sin[e
+ f*x]])/(3*(c - d)*d*(c + d)^2*f*Sqrt[(c + d*Sin[e + f*x])/(c + d)]) + (2*a*EllipticF[(e - Pi/2 + f*x)/2, (2*
d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/(3*d*(c + d)*f*Sqrt[c + d*Sin[e + f*x]])

Rule 2754

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

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 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 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 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 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]

Rubi steps

\begin{align*} \int \frac{a+a \sin (e+f x)}{(c+d \sin (e+f x))^{5/2}} \, dx &=-\frac{2 a \cos (e+f x)}{3 (c+d) f (c+d \sin (e+f x))^{3/2}}-\frac{2 \int \frac{-\frac{3}{2} a (c-d)-\frac{1}{2} a (c-d) \sin (e+f x)}{(c+d \sin (e+f x))^{3/2}} \, dx}{3 \left (c^2-d^2\right )}\\ &=-\frac{2 a \cos (e+f x)}{3 (c+d) f (c+d \sin (e+f x))^{3/2}}-\frac{2 a (c-3 d) \cos (e+f x)}{3 (c-d) (c+d)^2 f \sqrt{c+d \sin (e+f x)}}+\frac{4 \int \frac{\frac{1}{4} a (c-d) (3 c-d)-\frac{1}{4} a (c-3 d) (c-d) \sin (e+f x)}{\sqrt{c+d \sin (e+f x)}} \, dx}{3 \left (c^2-d^2\right )^2}\\ &=-\frac{2 a \cos (e+f x)}{3 (c+d) f (c+d \sin (e+f x))^{3/2}}-\frac{2 a (c-3 d) \cos (e+f x)}{3 (c-d) (c+d)^2 f \sqrt{c+d \sin (e+f x)}}-\frac{(a (c-3 d)) \int \sqrt{c+d \sin (e+f x)} \, dx}{3 (c-d) d (c+d)^2}+\frac{a \int \frac{1}{\sqrt{c+d \sin (e+f x)}} \, dx}{3 d (c+d)}\\ &=-\frac{2 a \cos (e+f x)}{3 (c+d) f (c+d \sin (e+f x))^{3/2}}-\frac{2 a (c-3 d) \cos (e+f x)}{3 (c-d) (c+d)^2 f \sqrt{c+d \sin (e+f x)}}-\frac{\left (a (c-3 d) \sqrt{c+d \sin (e+f x)}\right ) \int \sqrt{\frac{c}{c+d}+\frac{d \sin (e+f x)}{c+d}} \, dx}{3 (c-d) d (c+d)^2 \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}+\frac{\left (a \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}{3 d (c+d) \sqrt{c+d \sin (e+f x)}}\\ &=-\frac{2 a \cos (e+f x)}{3 (c+d) f (c+d \sin (e+f x))^{3/2}}-\frac{2 a (c-3 d) \cos (e+f x)}{3 (c-d) (c+d)^2 f \sqrt{c+d \sin (e+f x)}}-\frac{2 a (c-3 d) 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)}}{3 (c-d) d (c+d)^2 f \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}+\frac{2 a 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}}}{3 d (c+d) f \sqrt{c+d \sin (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 6.69321, size = 1870, normalized size = 7.89 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

a*(((1 + Sin[e + f*x])*Sqrt[c + d*Sin[e + f*x]]*((-2*(c - 3*d)*Csc[e]*Sec[e])/(3*(c - d)*d*(c + d)^2*f) + (2*C
sc[e]*(c*Cos[e] + d*Sin[f*x]))/(3*d*(c + d)*f*(c + d*Sin[e + f*x])^2) - (2*Csc[e]*(3*c*Cos[e] - d*Cos[e] - c*S
in[f*x] + 3*d*Sin[f*x]))/(3*(c - d)*(c + d)^2*f*(c + d*Sin[e + f*x]))))/(Cos[e/2 + (f*x)/2] + Sin[e/2 + (f*x)/
2])^2 - (c*Sec[e]*(1 + Sin[e + f*x])*(-((AppellF1[-1/2, -1/2, -1/2, 1/2, -((Csc[e]*(c + d*Cos[f*x - ArcTan[Cot
[e]]]*Sqrt[1 + Cot[e]^2]*Sin[e]))/(d*Sqrt[1 + Cot[e]^2]*(1 - (c*Csc[e])/(d*Sqrt[1 + Cot[e]^2])))), -((Csc[e]*(
c + d*Cos[f*x - ArcTan[Cot[e]]]*Sqrt[1 + Cot[e]^2]*Sin[e]))/(d*Sqrt[1 + Cot[e]^2]*(-1 - (c*Csc[e])/(d*Sqrt[1 +
Cot[e]^2]))))]*Cot[e]*Sin[f*x - ArcTan[Cot[e]]])/(Sqrt[1 + Cot[e]^2]*Sqrt[(d*Sqrt[1 + Cot[e]^2] + d*Cos[f*x -
ArcTan[Cot[e]]]*Sqrt[1 + Cot[e]^2])/(d*Sqrt[1 + Cot[e]^2] - c*Csc[e])]*Sqrt[(d*Sqrt[1 + Cot[e]^2] - d*Cos[f*x
- ArcTan[Cot[e]]]*Sqrt[1 + Cot[e]^2])/(d*Sqrt[1 + Cot[e]^2] + c*Csc[e])]*Sqrt[c + d*Cos[f*x - ArcTan[Cot[e]]]
*Sqrt[1 + Cot[e]^2]*Sin[e]])) - ((2*d*Sin[e]*(c + d*Cos[f*x - ArcTan[Cot[e]]]*Sqrt[1 + Cot[e]^2]*Sin[e]))/(d^2
*Cos[e]^2 + d^2*Sin[e]^2) - (Cot[e]*Sin[f*x - ArcTan[Cot[e]]])/Sqrt[1 + Cot[e]^2])/Sqrt[c + d*Cos[f*x - ArcTan
[Cot[e]]]*Sqrt[1 + Cot[e]^2]*Sin[e]]))/(3*(c - d)*(c + d)^2*f*(Cos[e/2 + (f*x)/2] + Sin[e/2 + (f*x)/2])^2) + (
d*Sec[e]*(1 + Sin[e + f*x])*(-((AppellF1[-1/2, -1/2, -1/2, 1/2, -((Csc[e]*(c + d*Cos[f*x - ArcTan[Cot[e]]]*Sqr
t[1 + Cot[e]^2]*Sin[e]))/(d*Sqrt[1 + Cot[e]^2]*(1 - (c*Csc[e])/(d*Sqrt[1 + Cot[e]^2])))), -((Csc[e]*(c + d*Cos
[f*x - ArcTan[Cot[e]]]*Sqrt[1 + Cot[e]^2]*Sin[e]))/(d*Sqrt[1 + Cot[e]^2]*(-1 - (c*Csc[e])/(d*Sqrt[1 + Cot[e]^2
]))))]*Cot[e]*Sin[f*x - ArcTan[Cot[e]]])/(Sqrt[1 + Cot[e]^2]*Sqrt[(d*Sqrt[1 + Cot[e]^2] + d*Cos[f*x - ArcTan[C
ot[e]]]*Sqrt[1 + Cot[e]^2])/(d*Sqrt[1 + Cot[e]^2] - c*Csc[e])]*Sqrt[(d*Sqrt[1 + Cot[e]^2] - d*Cos[f*x - ArcTan
[Cot[e]]]*Sqrt[1 + Cot[e]^2])/(d*Sqrt[1 + Cot[e]^2] + c*Csc[e])]*Sqrt[c + d*Cos[f*x - ArcTan[Cot[e]]]*Sqrt[1 +
Cot[e]^2]*Sin[e]])) - ((2*d*Sin[e]*(c + d*Cos[f*x - ArcTan[Cot[e]]]*Sqrt[1 + Cot[e]^2]*Sin[e]))/(d^2*Cos[e]^2
+ d^2*Sin[e]^2) - (Cot[e]*Sin[f*x - ArcTan[Cot[e]]])/Sqrt[1 + Cot[e]^2])/Sqrt[c + d*Cos[f*x - ArcTan[Cot[e]]]
*Sqrt[1 + Cot[e]^2]*Sin[e]]))/((c - d)*(c + d)^2*f*(Cos[e/2 + (f*x)/2] + Sin[e/2 + (f*x)/2])^2) - (2*AppellF1[
1/2, 1/2, 1/2, 3/2, -((Sec[e]*(c + d*Cos[e]*Sin[f*x + ArcTan[Tan[e]]]*Sqrt[1 + Tan[e]^2]))/(d*Sqrt[1 + Tan[e]^
2]*(1 - (c*Sec[e])/(d*Sqrt[1 + Tan[e]^2])))), -((Sec[e]*(c + d*Cos[e]*Sin[f*x + ArcTan[Tan[e]]]*Sqrt[1 + Tan[e
]^2]))/(d*Sqrt[1 + Tan[e]^2]*(-1 - (c*Sec[e])/(d*Sqrt[1 + Tan[e]^2]))))]*Sec[e]*Sec[f*x + ArcTan[Tan[e]]]*(1 +
Sin[e + f*x])*Sqrt[(d*Sqrt[1 + Tan[e]^2] - d*Sin[f*x + ArcTan[Tan[e]]]*Sqrt[1 + Tan[e]^2])/(c*Sec[e] + d*Sqrt
[1 + Tan[e]^2])]*Sqrt[(d*Sqrt[1 + Tan[e]^2] + d*Sin[f*x + ArcTan[Tan[e]]]*Sqrt[1 + Tan[e]^2])/(-(c*Sec[e]) + d
*Sqrt[1 + Tan[e]^2])]*Sqrt[c + d*Cos[e]*Sin[f*x + ArcTan[Tan[e]]]*Sqrt[1 + Tan[e]^2]])/(3*(c - d)*(c + d)^2*f*
(Cos[e/2 + (f*x)/2] + Sin[e/2 + (f*x)/2])^2*Sqrt[1 + Tan[e]^2]) + (2*c*AppellF1[1/2, 1/2, 1/2, 3/2, -((Sec[e]*
(c + d*Cos[e]*Sin[f*x + ArcTan[Tan[e]]]*Sqrt[1 + Tan[e]^2]))/(d*Sqrt[1 + Tan[e]^2]*(1 - (c*Sec[e])/(d*Sqrt[1 +
Tan[e]^2])))), -((Sec[e]*(c + d*Cos[e]*Sin[f*x + ArcTan[Tan[e]]]*Sqrt[1 + Tan[e]^2]))/(d*Sqrt[1 + Tan[e]^2]*(
-1 - (c*Sec[e])/(d*Sqrt[1 + Tan[e]^2]))))]*Sec[e]*Sec[f*x + ArcTan[Tan[e]]]*(1 + Sin[e + f*x])*Sqrt[(d*Sqrt[1
+ Tan[e]^2] - d*Sin[f*x + ArcTan[Tan[e]]]*Sqrt[1 + Tan[e]^2])/(c*Sec[e] + d*Sqrt[1 + Tan[e]^2])]*Sqrt[(d*Sqrt[
1 + Tan[e]^2] + d*Sin[f*x + ArcTan[Tan[e]]]*Sqrt[1 + Tan[e]^2])/(-(c*Sec[e]) + d*Sqrt[1 + Tan[e]^2])]*Sqrt[c +
d*Cos[e]*Sin[f*x + ArcTan[Tan[e]]]*Sqrt[1 + Tan[e]^2]])/((c - d)*d*(c + d)^2*f*(Cos[e/2 + (f*x)/2] + Sin[e/2
+ (f*x)/2])^2*Sqrt[1 + Tan[e]^2]))

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Maple [B]  time = 3.656, size = 884, normalized size = 3.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*a*(1/d*(2*d*cos(f*x+e)^2/(c^2-d^2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(
1/2)+2*c/(c^2-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
))+2/(c^2-d^2)*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))))+(-c+d)/d*(2/3/(c^2-d^2)/d*(-(-d*sin
(f*x+e)-c)*cos(f*x+e)^2)^(1/2)/(sin(f*x+e)+c/d)^2+8/3*d*cos(f*x+e)^2/(c^2-d^2)^2*c/(-(-d*sin(f*x+e)-c)*cos(f*x
+e)^2)^(1/2)+2*(3*c^2+d^2)/(3*c^4-6*c^2*d^2+3*d^4)*(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))+8/3*c*d/(c^2-d^2)^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., size = 0, normalized size = 0. \begin{align*} \int \frac{a \sin \left (f x + e\right ) + a}{{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{{\left (a \sin \left (f x + e\right ) + a\right )} \sqrt{d \sin \left (f x + e\right ) + c}}{3 \, c d^{2} \cos \left (f x + e\right )^{2} - c^{3} - 3 \, c d^{2} +{\left (d^{3} \cos \left (f x + e\right )^{2} - 3 \, c^{2} d - d^{3}\right )} \sin \left (f x + e\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a \sin \left (f x + e\right ) + a}{{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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