3.8.28 \(\int \frac {a+b \sin (e+f x)}{(c+d \sin (e+f x))^{5/2}} \, dx\) [728]

Optimal. Leaf size=285 \[ -\frac {2 (b c-a d) \cos (e+f x)}{3 \left (c^2-d^2\right ) f (c+d \sin (e+f x))^{3/2}}+\frac {2 \left (4 a c d-b \left (c^2+3 d^2\right )\right ) \cos (e+f x)}{3 \left (c^2-d^2\right )^2 f \sqrt {c+d \sin (e+f x)}}+\frac {2 \left (4 a c d-b \left (c^2+3 d^2\right )\right ) E\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {c+d \sin (e+f x)}}{3 d \left (c^2-d^2\right )^2 f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {2 (b c-a d) F\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}{3 d \left (c^2-d^2\right ) f \sqrt {c+d \sin (e+f x)}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.26, antiderivative size = 285, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {2833, 2831, 2742, 2740, 2734, 2732} \begin {gather*} \frac {2 \left (4 a c d-b \left (c^2+3 d^2\right )\right ) \cos (e+f x)}{3 f \left (c^2-d^2\right )^2 \sqrt {c+d \sin (e+f x)}}-\frac {2 (b c-a d) \cos (e+f x)}{3 f \left (c^2-d^2\right ) (c+d \sin (e+f x))^{3/2}}+\frac {2 (b c-a d) \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 \left (c^2-d^2\right ) \sqrt {c+d \sin (e+f x)}}+\frac {2 \left (4 a c d-b \left (c^2+3 d^2\right )\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 )}{3 d 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])/(c + d*Sin[e + f*x])^(5/2),x]

[Out]

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]
time = 1.67, size = 199, normalized size = 0.70 \begin {gather*} \frac {2 \left (\frac {\left (\left (-4 a c d+b \left (c^2+3 d^2\right )\right ) E\left (\frac {1}{4} (-2 e+\pi -2 f x)|\frac {2 d}{c+d}\right )-(c-d) (b c-a d) F\left (\frac {1}{4} (-2 e+\pi -2 f x)|\frac {2 d}{c+d}\right )\right ) \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{3/2}}{(c-d)^2 d}-\frac {\cos (e+f x) \left (a d \left (-5 c^2+d^2\right )+2 b c \left (c^2+d^2\right )+d \left (-4 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 f (c+d \sin (e+f x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(886\) vs. \(2(331)=662\).
time = 24.47, size = 887, normalized size = 3.11

method	result	size
default	\(\frac {\sqrt {-\left (-d \sin \left (f x +e \right )-c \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (\frac {b \left (\frac {2 d \left (\cos ^{2}\left (f x +e \right )\right )}{\left (c^{2}-d^{2}\right ) \sqrt {-\left (-d \sin \left (f x +e \right )-c \right ) \left (\cos ^{2}\left (f x +e \right )\right )}}+\frac {2 c \left (\frac {c}{d}-1\right ) \sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}\, \sqrt {\frac {d \left (1-\sin \left (f x +e \right )\right )}{c +d}}\, \sqrt {\frac {\left (-\sin \left (f x +e \right )-1\right ) d}{c -d}}\, \EllipticF \left (\sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}, \sqrt {\frac {c -d}{c +d}}\right )}{\left (c^{2}-d^{2}\right ) \sqrt {-\left (-d \sin \left (f x +e \right )-c \right ) \left (\cos ^{2}\left (f x +e \right )\right )}}+\frac {2 d \left (\frac {c}{d}-1\right ) \sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}\, \sqrt {\frac {d \left (1-\sin \left (f x +e \right )\right )}{c +d}}\, \sqrt {\frac {\left (-\sin \left (f x +e \right )-1\right ) d}{c -d}}\, \left (\left (-\frac {c}{d}-1\right ) \EllipticE \left (\sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}, \sqrt {\frac {c -d}{c +d}}\right )+\EllipticF \left (\sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}, \sqrt {\frac {c -d}{c +d}}\right )\right )}{\left (c^{2}-d^{2}\right ) \sqrt {-\left (-d \sin \left (f x +e \right )-c \right ) \left (\cos ^{2}\left (f x +e \right )\right )}}\right )}{d}+\frac {\left (a d -b c \right ) \left (\frac {2 \sqrt {-\left (-d \sin \left (f x +e \right )-c \right ) \left (\cos ^{2}\left (f x +e \right )\right )}}{3 \left (c^{2}-d^{2}\right ) d \left (\sin \left (f x +e \right )+\frac {c}{d}\right )^{2}}+\frac {8 d \left (\cos ^{2}\left (f x +e \right )\right ) c}{3 \left (c^{2}-d^{2}\right )^{2} \sqrt {-\left (-d \sin \left (f x +e \right )-c \right ) \left (\cos ^{2}\left (f x +e \right )\right )}}+\frac {2 \left (3 c^{2}+d^{2}\right ) \left (\frac {c}{d}-1\right ) \sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}\, \sqrt {\frac {d \left (1-\sin \left (f x +e \right )\right )}{c +d}}\, \sqrt {\frac {\left (-\sin \left (f x +e \right )-1\right ) d}{c -d}}\, \EllipticF \left (\sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}, \sqrt {\frac {c -d}{c +d}}\right )}{\left (3 c^{4}-6 c^{2} d^{2}+3 d^{4}\right ) \sqrt {-\left (-d \sin \left (f x +e \right )-c \right ) \left (\cos ^{2}\left (f x +e \right )\right )}}+\frac {8 d c \left (\frac {c}{d}-1\right ) \sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}\, \sqrt {\frac {d \left (1-\sin \left (f x +e \right )\right )}{c +d}}\, \sqrt {\frac {\left (-\sin \left (f x +e \right )-1\right ) d}{c -d}}\, \left (\left (-\frac {c}{d}-1\right ) \EllipticE \left (\sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}, \sqrt {\frac {c -d}{c +d}}\right )+\EllipticF \left (\sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}, \sqrt {\frac {c -d}{c +d}}\right )\right )}{3 \left (c^{2}-d^{2}\right )^{2} \sqrt {-\left (-d \sin \left (f x +e \right )-c \right ) \left (\cos ^{2}\left (f x +e \right )\right )}}\right )}{d}\right )}{\cos \left (f x +e \right ) \sqrt {c +d \sin \left (f x +e \right )}\, f}\)	\(887\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*sin(f*x + e) + a)/(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 = 0.23, size = 1047, normalized size = 3.67 \begin {gather*} \frac {{\left (\sqrt {2} {\left (2 \, b c^{3} d^{2} + a c^{2} d^{3} - 6 \, b c d^{4} + 3 \, a d^{5}\right )} \cos \left (f x + e\right )^{2} - 2 \, \sqrt {2} {\left (2 \, b c^{4} d + a c^{3} d^{2} - 6 \, b c^{2} d^{3} + 3 \, a c d^{4}\right )} \sin \left (f x + e\right ) - \sqrt {2} {\left (2 \, b c^{5} + a c^{4} d - 4 \, b c^{3} d^{2} + 4 \, a c^{2} d^{3} - 6 \, b c d^{4} + 3 \, a d^{5}\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 (2 \, b c^{3} d^{2} + a c^{2} d^{3} - 6 \, b c d^{4} + 3 \, a d^{5}\right )} \cos \left (f x + e\right )^{2} - 2 \, \sqrt {2} {\left (2 \, b c^{4} d + a c^{3} d^{2} - 6 \, b c^{2} d^{3} + 3 \, a c d^{4}\right )} \sin \left (f x + e\right ) - \sqrt {2} {\left (2 \, b c^{5} + a c^{4} d - 4 \, b c^{3} d^{2} + 4 \, a c^{2} d^{3} - 6 \, b c d^{4} + 3 \, a d^{5}\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 ) + 3 \, {\left (\sqrt {2} {\left (i \, b c^{2} d^{3} - 4 i \, a c d^{4} + 3 i \, b d^{5}\right )} \cos \left (f x + e\right )^{2} + 2 \, \sqrt {2} {\left (-i \, b c^{3} d^{2} + 4 i \, a c^{2} d^{3} - 3 i \, b c d^{4}\right )} \sin \left (f x + e\right ) + \sqrt {2} {\left (-i \, b c^{4} d + 4 i \, a c^{3} d^{2} - 4 i \, b c^{2} d^{3} + 4 i \, a c d^{4} - 3 i \, b 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 ) + 3 \, {\left (\sqrt {2} {\left (-i \, b c^{2} d^{3} + 4 i \, a c d^{4} - 3 i \, b d^{5}\right )} \cos \left (f x + e\right )^{2} + 2 \, \sqrt {2} {\left (i \, b c^{3} d^{2} - 4 i \, a c^{2} d^{3} + 3 i \, b c d^{4}\right )} \sin \left (f x + e\right ) + \sqrt {2} {\left (i \, b c^{4} d - 4 i \, a c^{3} d^{2} + 4 i \, b c^{2} d^{3} - 4 i \, a c d^{4} + 3 i \, b 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 ({\left (b c^{2} d^{3} - 4 \, a c d^{4} + 3 \, b d^{5}\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + {\left (2 \, b c^{3} d^{2} - 5 \, a c^{2} d^{3} + 2 \, b c d^{4} + a d^{5}\right )} \cos \left (f x + e\right )\right )} \sqrt {d \sin \left (f x + e\right ) + c}}{9 \, {\left ({\left (c^{4} d^{4} - 2 \, c^{2} d^{6} + d^{8}\right )} f \cos \left (f x + e\right )^{2} - 2 \, {\left (c^{5} d^{3} - 2 \, c^{3} d^{5} + c d^{7}\right )} f \sin \left (f x + e\right ) - {\left (c^{6} d^{2} - c^{4} d^{4} - c^{2} d^{6} + d^{8}\right )} f\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*((sqrt(2)*(2*b*c^3*d^2 + a*c^2*d^3 - 6*b*c*d^4 + 3*a*d^5)*cos(f*x + e)^2 - 2*sqrt(2)*(2*b*c^4*d + a*c^3*d^
2 - 6*b*c^2*d^3 + 3*a*c*d^4)*sin(f*x + e) - sqrt(2)*(2*b*c^5 + a*c^4*d - 4*b*c^3*d^2 + 4*a*c^2*d^3 - 6*b*c*d^4
+ 3*a*d^5))*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)*(2*b*c^3*d^2 + a*c^2*d^3 - 6*b*c*d^4 + 3*a*d^5)*cos(
f*x + e)^2 - 2*sqrt(2)*(2*b*c^4*d + a*c^3*d^2 - 6*b*c^2*d^3 + 3*a*c*d^4)*sin(f*x + e) - sqrt(2)*(2*b*c^5 + a*c
^4*d - 4*b*c^3*d^2 + 4*a*c^2*d^3 - 6*b*c*d^4 + 3*a*d^5))*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) + 3*(sqrt(2)*(I*b
*c^2*d^3 - 4*I*a*c*d^4 + 3*I*b*d^5)*cos(f*x + e)^2 + 2*sqrt(2)*(-I*b*c^3*d^2 + 4*I*a*c^2*d^3 - 3*I*b*c*d^4)*si
n(f*x + e) + sqrt(2)*(-I*b*c^4*d + 4*I*a*c^3*d^2 - 4*I*b*c^2*d^3 + 4*I*a*c*d^4 - 3*I*b*d^5))*sqrt(I*d)*weierst
rassZeta(-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)) + 3*(sqrt(2)*(-I*
b*c^2*d^3 + 4*I*a*c*d^4 - 3*I*b*d^5)*cos(f*x + e)^2 + 2*sqrt(2)*(I*b*c^3*d^2 - 4*I*a*c^2*d^3 + 3*I*b*c*d^4)*si
n(f*x + e) + sqrt(2)*(I*b*c^4*d - 4*I*a*c^3*d^2 + 4*I*b*c^2*d^3 - 4*I*a*c*d^4 + 3*I*b*d^5))*sqrt(-I*d)*weierst
rassZeta(-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*((b*c^2*d^3
- 4*a*c*d^4 + 3*b*d^5)*cos(f*x + e)*sin(f*x + e) + (2*b*c^3*d^2 - 5*a*c^2*d^3 + 2*b*c*d^4 + a*d^5)*cos(f*x +
e))*sqrt(d*sin(f*x + e) + c))/((c^4*d^4 - 2*c^2*d^6 + d^8)*f*cos(f*x + e)^2 - 2*(c^5*d^3 - 2*c^3*d^5 + c*d^7)*
f*sin(f*x + e) - (c^6*d^2 - c^4*d^4 - c^2*d^6 + d^8)*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))/(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))/(c+d*sin(f*x+e))^(5/2),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________