3.8.35 \(\int \frac {(a+b \sin (e+f x))^2}{(c+d \sin (e+f x))^{5/2}} \, dx\) [735]

Optimal. Leaf size=329 \[ \frac {2 (b c-a d)^2 \cos (e+f x)}{3 d \left (c^2-d^2\right ) f (c+d \sin (e+f x))^{3/2}}-\frac {4 (b c-a d) \left (2 a c d+b \left (c^2-3 d^2\right )\right ) \cos (e+f x)}{3 d \left (c^2-d^2\right )^2 f \sqrt {c+d \sin (e+f x)}}-\frac {4 (b c-a d) \left (2 a c d+b \left (c^2-3 d^2\right )\right ) E\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {c+d \sin (e+f x)}}{3 d^2 \left (c^2-d^2\right )^2 f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {2 \left (2 a b c d-a^2 d^2+b^2 \left (2 c^2-3 d^2\right )\right ) F\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}{3 d^2 \left (c^2-d^2\right ) f \sqrt {c+d \sin (e+f x)}} \]

[Out]

2/3*(-a*d+b*c)^2*cos(f*x+e)/d/(c^2-d^2)/f/(c+d*sin(f*x+e))^(3/2)-4/3*(-a*d+b*c)*(2*a*c*d+b*(c^2-3*d^2))*cos(f*
x+e)/d/(c^2-d^2)^2/f/(c+d*sin(f*x+e))^(1/2)+4/3*(-a*d+b*c)*(2*a*c*d+b*(c^2-3*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)/d^2/(c^2-d^2)^2/f/((c+d*sin(f*x+e))/(c+d))^(1/2)-2/3*(2*a*b*c*d-a^2*d^2+b^2*(2*c^2-3*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)/d^2/(c^2-d^2)/f/(c+d*sin(f*x+e))^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.32, antiderivative size = 329, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {2869, 2833, 2831, 2742, 2740, 2734, 2732} \begin {gather*} \frac {2 \left (-a^2 d^2+2 a b c d+b^2 \left (2 c^2-3 d^2\right )\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 )}{3 d^2 f \left (c^2-d^2\right ) \sqrt {c+d \sin (e+f x)}}+\frac {2 (b c-a d)^2 \cos (e+f x)}{3 d f \left (c^2-d^2\right ) (c+d \sin (e+f x))^{3/2}}-\frac {4 \left (2 a c d+b \left (c^2-3 d^2\right )\right ) (b c-a d) \cos (e+f x)}{3 d f \left (c^2-d^2\right )^2 \sqrt {c+d \sin (e+f x)}}-\frac {4 \left (2 a c d+b \left (c^2-3 d^2\right )\right ) (b c-a 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^2 f \left (c^2-d^2\right )^2 \sqrt {\frac {c+d \sin (e+f x)}{c+d}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(b*c - a*d)^2*Cos[e + f*x])/(3*d*(c^2 - d^2)*f*(c + d*Sin[e + f*x])^(3/2)) - (4*(b*c - a*d)*(2*a*c*d + b*(c
^2 - 3*d^2))*Cos[e + f*x])/(3*d*(c^2 - d^2)^2*f*Sqrt[c + d*Sin[e + f*x]]) - (4*(b*c - a*d)*(2*a*c*d + b*(c^2 -
3*d^2))*EllipticE[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[c + d*Sin[e + f*x]])/(3*d^2*(c^2 - d^2)^2*f*Sqrt[(c
+ d*Sin[e + f*x])/(c + d)]) + (2*(2*a*b*c*d - a^2*d^2 + b^2*(2*c^2 - 3*d^2))*EllipticF[(e - Pi/2 + f*x)/2, (2
*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/(3*d^2*(c^2 - d^2)*f*Sqrt[c + d*Sin[e + f*x]])

Rule 2732

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

Rule 2734

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/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2740

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

Rule 2742

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/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2831

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 2833

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 2869

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

Rubi steps

\begin {align*} \int \frac {(a+b \sin (e+f x))^2}{(c+d \sin (e+f x))^{5/2}} \, dx &=\frac {2 (b c-a d)^2 \cos (e+f x)}{3 d \left (c^2-d^2\right ) f (c+d \sin (e+f x))^{3/2}}+\frac {2 \int \frac {\frac {3}{2} d \left (a^2 c+b^2 c-2 a b d\right )+\frac {1}{2} \left (2 b^2 c^2+2 a b c d-\left (a^2+3 b^2\right ) d^2\right ) \sin (e+f x)}{(c+d \sin (e+f x))^{3/2}} \, dx}{3 d \left (c^2-d^2\right )}\\ &=\frac {2 (b c-a d)^2 \cos (e+f x)}{3 d \left (c^2-d^2\right ) f (c+d \sin (e+f x))^{3/2}}-\frac {4 (b c-a d) \left (2 a c d+b \left (c^2-3 d^2\right )\right ) \cos (e+f x)}{3 d \left (c^2-d^2\right )^2 f \sqrt {c+d \sin (e+f x)}}-\frac {4 \int \frac {\frac {1}{4} d \left (8 a b c d-a^2 \left (3 c^2+d^2\right )-b^2 \left (c^2+3 d^2\right )\right )+\frac {1}{2} (b c-a d) \left (b c^2+2 a c d-3 b d^2\right ) \sin (e+f x)}{\sqrt {c+d \sin (e+f x)}} \, dx}{3 d \left (c^2-d^2\right )^2}\\ &=\frac {2 (b c-a d)^2 \cos (e+f x)}{3 d \left (c^2-d^2\right ) f (c+d \sin (e+f x))^{3/2}}-\frac {4 (b c-a d) \left (2 a c d+b \left (c^2-3 d^2\right )\right ) \cos (e+f x)}{3 d \left (c^2-d^2\right )^2 f \sqrt {c+d \sin (e+f x)}}-\frac {\left (2 (b c-a d) \left (2 a c d+b \left (c^2-3 d^2\right )\right )\right ) \int \sqrt {c+d \sin (e+f x)} \, dx}{3 d^2 \left (c^2-d^2\right )^2}+\frac {\left (2 a b c d-a^2 d^2+b^2 \left (2 c^2-3 d^2\right )\right ) \int \frac {1}{\sqrt {c+d \sin (e+f x)}} \, dx}{3 d^2 \left (c^2-d^2\right )}\\ &=\frac {2 (b c-a d)^2 \cos (e+f x)}{3 d \left (c^2-d^2\right ) f (c+d \sin (e+f x))^{3/2}}-\frac {4 (b c-a d) \left (2 a c d+b \left (c^2-3 d^2\right )\right ) \cos (e+f x)}{3 d \left (c^2-d^2\right )^2 f \sqrt {c+d \sin (e+f x)}}-\frac {\left (2 (b c-a d) \left (2 a c d+b \left (c^2-3 d^2\right )\right ) \sqrt {c+d \sin (e+f x)}\right ) \int \sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}} \, dx}{3 d^2 \left (c^2-d^2\right )^2 \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {\left (\left (2 a b c d-a^2 d^2+b^2 \left (2 c^2-3 d^2\right )\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}{3 d^2 \left (c^2-d^2\right ) \sqrt {c+d \sin (e+f x)}}\\ &=\frac {2 (b c-a d)^2 \cos (e+f x)}{3 d \left (c^2-d^2\right ) f (c+d \sin (e+f x))^{3/2}}-\frac {4 (b c-a d) \left (2 a c d+b \left (c^2-3 d^2\right )\right ) \cos (e+f x)}{3 d \left (c^2-d^2\right )^2 f \sqrt {c+d \sin (e+f x)}}-\frac {4 (b c-a d) \left (2 a c d+b \left (c^2-3 d^2\right )\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)}}{3 d^2 \left (c^2-d^2\right )^2 f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {2 \left (2 a b c d-a^2 d^2+b^2 \left (2 c^2-3 d^2\right )\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}}}{3 d^2 \left (c^2-d^2\right ) f \sqrt {c+d \sin (e+f x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 2.66, size = 302, normalized size = 0.92 \begin {gather*} \frac {2 \left (\frac {\left (d^2 \left (-8 a b c d+a^2 \left (3 c^2+d^2\right )+b^2 \left (c^2+3 d^2\right )\right ) F\left (\frac {1}{4} (-2 e+\pi -2 f x)|\frac {2 d}{c+d}\right )-2 \left (-2 a^2 c d^2+a b d \left (c^2+3 d^2\right )+b^2 \left (c^3-3 c d^2\right )\right ) \left ((c+d) E\left (\frac {1}{4} (-2 e+\pi -2 f x)|\frac {2 d}{c+d}\right )-c F\left (\frac {1}{4} (-2 e+\pi -2 f x)|\frac {2 d}{c+d}\right )\right )\right ) (-c-d \sin (e+f x)) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}{(c-d)^2 (c+d)^2}-\frac {d (-b c+a d) \cos (e+f x) \left (-b c^3-5 a c^2 d+5 b c d^2+a d^3-2 d \left (2 a c d+b \left (c^2-3 d^2\right )\right ) \sin (e+f x)\right )}{\left (c^2-d^2\right )^2}\right )}{3 d^2 f (c+d \sin (e+f x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(((d^2*(-8*a*b*c*d + a^2*(3*c^2 + d^2) + b^2*(c^2 + 3*d^2))*EllipticF[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)]
- 2*(-2*a^2*c*d^2 + a*b*d*(c^2 + 3*d^2) + b^2*(c^3 - 3*c*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)]))*(-c - d*Sin[e + f*x])*Sqrt[(c + d*Sin[e + f*x
])/(c + d)])/((c - d)^2*(c + d)^2) - (d*(-(b*c) + a*d)*Cos[e + f*x]*(-(b*c^3) - 5*a*c^2*d + 5*b*c*d^2 + a*d^3
- 2*d*(2*a*c*d + b*(c^2 - 3*d^2))*Sin[e + f*x]))/(c^2 - d^2)^2))/(3*d^2*f*(c + d*Sin[e + f*x])^(3/2))

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1042\) vs. \(2(375)=750\).
time = 26.65, size = 1043, normalized size = 3.17

method	result	size
default	\(\text {Expression too large to display}\)	\(1043\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*(2*b^2/d^2*(1/d*c-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*b/d^2*(a*d-b*c)*(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)*(1/d*c-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*(1/d*c-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)*((-1/d*c-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))))+1/d^2*(a^2*d^2-2*a*b*
c*d+b^2*c^2)*(2/3/(c^2-d^2)/d*(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)/(sin(f*x+e)+1/d*c)^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)*(1/d*c-1)*((c+d*si
n(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*d*c/(c^2-d^2)^2*(1/d*c-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)*((-1/d*c-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

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.25, size = 1422, normalized size = 4.32 \begin {gather*} \frac {{\left (\sqrt {2} {\left (4 \, b^{2} c^{4} d^{2} + 4 \, a b c^{3} d^{3} - 12 \, a b c d^{5} + {\left (a^{2} - 9 \, b^{2}\right )} c^{2} d^{4} + 3 \, {\left (a^{2} + 3 \, b^{2}\right )} d^{6}\right )} \cos \left (f x + e\right )^{2} - 2 \, \sqrt {2} {\left (4 \, b^{2} c^{5} d + 4 \, a b c^{4} d^{2} - 12 \, a b c^{2} d^{4} + {\left (a^{2} - 9 \, b^{2}\right )} c^{3} d^{3} + 3 \, {\left (a^{2} + 3 \, b^{2}\right )} c d^{5}\right )} \sin \left (f x + e\right ) - \sqrt {2} {\left (4 \, b^{2} c^{6} + 4 \, a b c^{5} d - 8 \, a b c^{3} d^{3} + 4 \, a^{2} c^{2} d^{4} - 12 \, a b c d^{5} + {\left (a^{2} - 5 \, b^{2}\right )} c^{4} d^{2} + 3 \, {\left (a^{2} + 3 \, b^{2}\right )} d^{6}\right )}\right )} \sqrt {i \, d} {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (4 \, c^{2} - 3 \, d^{2}\right )}}{3 \, d^{2}}, -\frac {8 \, {\left (8 i \, c^{3} - 9 i \, c d^{2}\right )}}{27 \, d^{3}}, \frac {3 \, d \cos \left (f x + e\right ) - 3 i \, d \sin \left (f x + e\right ) - 2 i \, c}{3 \, d}\right ) + {\left (\sqrt {2} {\left (4 \, b^{2} c^{4} d^{2} + 4 \, a b c^{3} d^{3} - 12 \, a b c d^{5} + {\left (a^{2} - 9 \, b^{2}\right )} c^{2} d^{4} + 3 \, {\left (a^{2} + 3 \, b^{2}\right )} d^{6}\right )} \cos \left (f x + e\right )^{2} - 2 \, \sqrt {2} {\left (4 \, b^{2} c^{5} d + 4 \, a b c^{4} d^{2} - 12 \, a b c^{2} d^{4} + {\left (a^{2} - 9 \, b^{2}\right )} c^{3} d^{3} + 3 \, {\left (a^{2} + 3 \, b^{2}\right )} c d^{5}\right )} \sin \left (f x + e\right ) - \sqrt {2} {\left (4 \, b^{2} c^{6} + 4 \, a b c^{5} d - 8 \, a b c^{3} d^{3} + 4 \, a^{2} c^{2} d^{4} - 12 \, a b c d^{5} + {\left (a^{2} - 5 \, b^{2}\right )} c^{4} d^{2} + 3 \, {\left (a^{2} + 3 \, b^{2}\right )} d^{6}\right )}\right )} \sqrt {-i \, d} {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (4 \, c^{2} - 3 \, d^{2}\right )}}{3 \, d^{2}}, -\frac {8 \, {\left (-8 i \, c^{3} + 9 i \, c d^{2}\right )}}{27 \, d^{3}}, \frac {3 \, d \cos \left (f x + e\right ) + 3 i \, d \sin \left (f x + e\right ) + 2 i \, c}{3 \, d}\right ) + 6 \, {\left (\sqrt {2} {\left (i \, b^{2} c^{3} d^{3} + i \, a b c^{2} d^{4} + 3 i \, a b d^{6} - i \, {\left (2 \, a^{2} + 3 \, b^{2}\right )} c d^{5}\right )} \cos \left (f x + e\right )^{2} + 2 \, \sqrt {2} {\left (-i \, b^{2} c^{4} d^{2} - i \, a b c^{3} d^{3} - 3 i \, a b c d^{5} + i \, {\left (2 \, a^{2} + 3 \, b^{2}\right )} c^{2} d^{4}\right )} \sin \left (f x + e\right ) + \sqrt {2} {\left (-i \, b^{2} c^{5} d - i \, a b c^{4} d^{2} - 4 i \, a b c^{2} d^{4} - 3 i \, a b d^{6} + 2 i \, {\left (a^{2} + b^{2}\right )} c^{3} d^{3} + i \, {\left (2 \, a^{2} + 3 \, b^{2}\right )} c d^{5}\right )}\right )} \sqrt {i \, d} {\rm weierstrassZeta}\left (-\frac {4 \, {\left (4 \, c^{2} - 3 \, d^{2}\right )}}{3 \, d^{2}}, -\frac {8 \, {\left (8 i \, c^{3} - 9 i \, c d^{2}\right )}}{27 \, d^{3}}, {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (4 \, c^{2} - 3 \, d^{2}\right )}}{3 \, d^{2}}, -\frac {8 \, {\left (8 i \, c^{3} - 9 i \, c d^{2}\right )}}{27 \, d^{3}}, \frac {3 \, d \cos \left (f x + e\right ) - 3 i \, d \sin \left (f x + e\right ) - 2 i \, c}{3 \, d}\right )\right ) + 6 \, {\left (\sqrt {2} {\left (-i \, b^{2} c^{3} d^{3} - i \, a b c^{2} d^{4} - 3 i \, a b d^{6} + i \, {\left (2 \, a^{2} + 3 \, b^{2}\right )} c d^{5}\right )} \cos \left (f x + e\right )^{2} + 2 \, \sqrt {2} {\left (i \, b^{2} c^{4} d^{2} + i \, a b c^{3} d^{3} + 3 i \, a b c d^{5} - i \, {\left (2 \, a^{2} + 3 \, b^{2}\right )} c^{2} d^{4}\right )} \sin \left (f x + e\right ) + \sqrt {2} {\left (i \, b^{2} c^{5} d + i \, a b c^{4} d^{2} + 4 i \, a b c^{2} d^{4} + 3 i \, a b d^{6} - 2 i \, {\left (a^{2} + b^{2}\right )} c^{3} d^{3} - i \, {\left (2 \, a^{2} + 3 \, b^{2}\right )} c d^{5}\right )}\right )} \sqrt {-i \, d} {\rm weierstrassZeta}\left (-\frac {4 \, {\left (4 \, c^{2} - 3 \, d^{2}\right )}}{3 \, d^{2}}, -\frac {8 \, {\left (-8 i \, c^{3} + 9 i \, c d^{2}\right )}}{27 \, d^{3}}, {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (4 \, c^{2} - 3 \, d^{2}\right )}}{3 \, d^{2}}, -\frac {8 \, {\left (-8 i \, c^{3} + 9 i \, c d^{2}\right )}}{27 \, d^{3}}, \frac {3 \, d \cos \left (f x + e\right ) + 3 i \, d \sin \left (f x + e\right ) + 2 i \, c}{3 \, d}\right )\right ) + 6 \, {\left (2 \, {\left (b^{2} c^{3} d^{3} + a b c^{2} d^{4} + 3 \, a b d^{6} - {\left (2 \, a^{2} + 3 \, b^{2}\right )} c d^{5}\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + {\left (b^{2} c^{4} d^{2} + 4 \, a b c^{3} d^{3} + 4 \, a b c d^{5} + a^{2} d^{6} - 5 \, {\left (a^{2} + b^{2}\right )} c^{2} d^{4}\right )} \cos \left (f x + e\right )\right )} \sqrt {d \sin \left (f x + e\right ) + c}}{9 \, {\left ({\left (c^{4} d^{5} - 2 \, c^{2} d^{7} + d^{9}\right )} f \cos \left (f x + e\right )^{2} - 2 \, {\left (c^{5} d^{4} - 2 \, c^{3} d^{6} + c d^{8}\right )} f \sin \left (f x + e\right ) - {\left (c^{6} d^{3} - c^{4} d^{5} - c^{2} d^{7} + d^{9}\right )} f\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*((sqrt(2)*(4*b^2*c^4*d^2 + 4*a*b*c^3*d^3 - 12*a*b*c*d^5 + (a^2 - 9*b^2)*c^2*d^4 + 3*(a^2 + 3*b^2)*d^6)*cos
(f*x + e)^2 - 2*sqrt(2)*(4*b^2*c^5*d + 4*a*b*c^4*d^2 - 12*a*b*c^2*d^4 + (a^2 - 9*b^2)*c^3*d^3 + 3*(a^2 + 3*b^2
)*c*d^5)*sin(f*x + e) - sqrt(2)*(4*b^2*c^6 + 4*a*b*c^5*d - 8*a*b*c^3*d^3 + 4*a^2*c^2*d^4 - 12*a*b*c*d^5 + (a^2
- 5*b^2)*c^4*d^2 + 3*(a^2 + 3*b^2)*d^6))*sqrt(I*d)*weierstrassPInverse(-4/3*(4*c^2 - 3*d^2)/d^2, -8/27*(8*I*c
^3 - 9*I*c*d^2)/d^3, 1/3*(3*d*cos(f*x + e) - 3*I*d*sin(f*x + e) - 2*I*c)/d) + (sqrt(2)*(4*b^2*c^4*d^2 + 4*a*b*
c^3*d^3 - 12*a*b*c*d^5 + (a^2 - 9*b^2)*c^2*d^4 + 3*(a^2 + 3*b^2)*d^6)*cos(f*x + e)^2 - 2*sqrt(2)*(4*b^2*c^5*d
+ 4*a*b*c^4*d^2 - 12*a*b*c^2*d^4 + (a^2 - 9*b^2)*c^3*d^3 + 3*(a^2 + 3*b^2)*c*d^5)*sin(f*x + e) - sqrt(2)*(4*b^
2*c^6 + 4*a*b*c^5*d - 8*a*b*c^3*d^3 + 4*a^2*c^2*d^4 - 12*a*b*c*d^5 + (a^2 - 5*b^2)*c^4*d^2 + 3*(a^2 + 3*b^2)*d
^6))*sqrt(-I*d)*weierstrassPInverse(-4/3*(4*c^2 - 3*d^2)/d^2, -8/27*(-8*I*c^3 + 9*I*c*d^2)/d^3, 1/3*(3*d*cos(f
*x + e) + 3*I*d*sin(f*x + e) + 2*I*c)/d) + 6*(sqrt(2)*(I*b^2*c^3*d^3 + I*a*b*c^2*d^4 + 3*I*a*b*d^6 - I*(2*a^2
+ 3*b^2)*c*d^5)*cos(f*x + e)^2 + 2*sqrt(2)*(-I*b^2*c^4*d^2 - I*a*b*c^3*d^3 - 3*I*a*b*c*d^5 + I*(2*a^2 + 3*b^2)
*c^2*d^4)*sin(f*x + e) + sqrt(2)*(-I*b^2*c^5*d - I*a*b*c^4*d^2 - 4*I*a*b*c^2*d^4 - 3*I*a*b*d^6 + 2*I*(a^2 + b^
2)*c^3*d^3 + I*(2*a^2 + 3*b^2)*c*d^5))*sqrt(I*d)*weierstrassZeta(-4/3*(4*c^2 - 3*d^2)/d^2, -8/27*(8*I*c^3 - 9*
I*c*d^2)/d^3, weierstrassPInverse(-4/3*(4*c^2 - 3*d^2)/d^2, -8/27*(8*I*c^3 - 9*I*c*d^2)/d^3, 1/3*(3*d*cos(f*x
+ e) - 3*I*d*sin(f*x + e) - 2*I*c)/d)) + 6*(sqrt(2)*(-I*b^2*c^3*d^3 - I*a*b*c^2*d^4 - 3*I*a*b*d^6 + I*(2*a^2 +
3*b^2)*c*d^5)*cos(f*x + e)^2 + 2*sqrt(2)*(I*b^2*c^4*d^2 + I*a*b*c^3*d^3 + 3*I*a*b*c*d^5 - I*(2*a^2 + 3*b^2)*c
^2*d^4)*sin(f*x + e) + sqrt(2)*(I*b^2*c^5*d + I*a*b*c^4*d^2 + 4*I*a*b*c^2*d^4 + 3*I*a*b*d^6 - 2*I*(a^2 + b^2)*
c^3*d^3 - I*(2*a^2 + 3*b^2)*c*d^5))*sqrt(-I*d)*weierstrassZeta(-4/3*(4*c^2 - 3*d^2)/d^2, -8/27*(-8*I*c^3 + 9*I
*c*d^2)/d^3, weierstrassPInverse(-4/3*(4*c^2 - 3*d^2)/d^2, -8/27*(-8*I*c^3 + 9*I*c*d^2)/d^3, 1/3*(3*d*cos(f*x
+ e) + 3*I*d*sin(f*x + e) + 2*I*c)/d)) + 6*(2*(b^2*c^3*d^3 + a*b*c^2*d^4 + 3*a*b*d^6 - (2*a^2 + 3*b^2)*c*d^5)*
cos(f*x + e)*sin(f*x + e) + (b^2*c^4*d^2 + 4*a*b*c^3*d^3 + 4*a*b*c*d^5 + a^2*d^6 - 5*(a^2 + b^2)*c^2*d^4)*cos(
f*x + e))*sqrt(d*sin(f*x + e) + c))/((c^4*d^5 - 2*c^2*d^7 + d^9)*f*cos(f*x + e)^2 - 2*(c^5*d^4 - 2*c^3*d^6 + c
*d^8)*f*sin(f*x + e) - (c^6*d^3 - c^4*d^5 - c^2*d^7 + d^9)*f)

________________________________________________________________________________________

Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________