3.1434 \(\int \frac{1}{\sqrt{g \cos (e+f x)} (d \sin (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx\)
Optimal. Leaf size=320 \[ -\frac{2 \sqrt{2} b^2 \sqrt{\cos (e+f x)} \Pi \left (-\frac{a}{b-\sqrt{b^2-a^2}};\left .\sin ^{-1}\left (\frac{\sqrt{d \sin (e+f x)}}{\sqrt{d} \sqrt{\cos (e+f x)+1}}\right )\right |-1\right )}{a^2 d^{3/2} f \sqrt{b^2-a^2} \sqrt{g \cos (e+f x)}}+\frac{2 \sqrt{2} b^2 \sqrt{\cos (e+f x)} \Pi \left (-\frac{a}{b+\sqrt{b^2-a^2}};\left .\sin ^{-1}\left (\frac{\sqrt{d \sin (e+f x)}}{\sqrt{d} \sqrt{\cos (e+f x)+1}}\right )\right |-1\right )}{a^2 d^{3/2} f \sqrt{b^2-a^2} \sqrt{g \cos (e+f x)}}-\frac{b \sqrt{\sin (2 e+2 f x)} F\left (\left .e+f x-\frac{\pi }{4}\right |2\right )}{a^2 d f \sqrt{d \sin (e+f x)} \sqrt{g \cos (e+f x)}}-\frac{2 \sqrt{g \cos (e+f x)}}{a d f g \sqrt{d \sin (e+f x)}} \]
[Out]
(-2*Sqrt[2]*b^2*Sqrt[Cos[e + f*x]]*EllipticPi[-(a/(b - Sqrt[-a^2 + b^2])), ArcSin[Sqrt[d*Sin[e + f*x]]/(Sqrt[d
]*Sqrt[1 + Cos[e + f*x]])], -1])/(a^2*Sqrt[-a^2 + b^2]*d^(3/2)*f*Sqrt[g*Cos[e + f*x]]) + (2*Sqrt[2]*b^2*Sqrt[C
os[e + f*x]]*EllipticPi[-(a/(b + Sqrt[-a^2 + b^2])), ArcSin[Sqrt[d*Sin[e + f*x]]/(Sqrt[d]*Sqrt[1 + Cos[e + f*x
]])], -1])/(a^2*Sqrt[-a^2 + b^2]*d^(3/2)*f*Sqrt[g*Cos[e + f*x]]) - (2*Sqrt[g*Cos[e + f*x]])/(a*d*f*g*Sqrt[d*Si
n[e + f*x]]) - (b*EllipticF[e - Pi/4 + f*x, 2]*Sqrt[Sin[2*e + 2*f*x]])/(a^2*d*f*Sqrt[g*Cos[e + f*x]]*Sqrt[d*Si
n[e + f*x]])
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Rubi [A] time = 0.83484, antiderivative size = 320, normalized size of antiderivative = 1.,
number of steps used = 9, number of rules used = 7, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.189, Rules used =
{2910, 2563, 2573, 2641, 2908, 2907, 1218} \[ -\frac{2 \sqrt{2} b^2 \sqrt{\cos (e+f x)} \Pi \left (-\frac{a}{b-\sqrt{b^2-a^2}};\left .\sin ^{-1}\left (\frac{\sqrt{d \sin (e+f x)}}{\sqrt{d} \sqrt{\cos (e+f x)+1}}\right )\right |-1\right )}{a^2 d^{3/2} f \sqrt{b^2-a^2} \sqrt{g \cos (e+f x)}}+\frac{2 \sqrt{2} b^2 \sqrt{\cos (e+f x)} \Pi \left (-\frac{a}{b+\sqrt{b^2-a^2}};\left .\sin ^{-1}\left (\frac{\sqrt{d \sin (e+f x)}}{\sqrt{d} \sqrt{\cos (e+f x)+1}}\right )\right |-1\right )}{a^2 d^{3/2} f \sqrt{b^2-a^2} \sqrt{g \cos (e+f x)}}-\frac{b \sqrt{\sin (2 e+2 f x)} F\left (\left .e+f x-\frac{\pi }{4}\right |2\right )}{a^2 d f \sqrt{d \sin (e+f x)} \sqrt{g \cos (e+f x)}}-\frac{2 \sqrt{g \cos (e+f x)}}{a d f g \sqrt{d \sin (e+f x)}} \]
Antiderivative was successfully verified.
[In]
Int[1/(Sqrt[g*Cos[e + f*x]]*(d*Sin[e + f*x])^(3/2)*(a + b*Sin[e + f*x])),x]
[Out]
(-2*Sqrt[2]*b^2*Sqrt[Cos[e + f*x]]*EllipticPi[-(a/(b - Sqrt[-a^2 + b^2])), ArcSin[Sqrt[d*Sin[e + f*x]]/(Sqrt[d
]*Sqrt[1 + Cos[e + f*x]])], -1])/(a^2*Sqrt[-a^2 + b^2]*d^(3/2)*f*Sqrt[g*Cos[e + f*x]]) + (2*Sqrt[2]*b^2*Sqrt[C
os[e + f*x]]*EllipticPi[-(a/(b + Sqrt[-a^2 + b^2])), ArcSin[Sqrt[d*Sin[e + f*x]]/(Sqrt[d]*Sqrt[1 + Cos[e + f*x
]])], -1])/(a^2*Sqrt[-a^2 + b^2]*d^(3/2)*f*Sqrt[g*Cos[e + f*x]]) - (2*Sqrt[g*Cos[e + f*x]])/(a*d*f*g*Sqrt[d*Si
n[e + f*x]]) - (b*EllipticF[e - Pi/4 + f*x, 2]*Sqrt[Sin[2*e + 2*f*x]])/(a^2*d*f*Sqrt[g*Cos[e + f*x]]*Sqrt[d*Si
n[e + f*x]])
Rule 2910
Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_))/((a_) + (b_.)*sin[(e_.) + (f_.
)*(x_)]), x_Symbol] :> Dist[1/a, Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^n, x], x] - Dist[b/(a*d), Int[((g*Cos
[e + f*x])^p*(d*Sin[e + f*x])^(n + 1))/(a + b*Sin[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f, g}, x] && NeQ[a^2
- b^2, 0] && IntegersQ[2*n, 2*p] && LtQ[-1, p, 1] && LtQ[n, 0]
Rule 2563
Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Simp[((a*Sin[e +
f*x])^(m + 1)*(b*Cos[e + f*x])^(n + 1))/(a*b*f*(m + 1)), x] /; FreeQ[{a, b, e, f, m, n}, x] && EqQ[m + n + 2,
0] && NeQ[m, -1]
Rule 2573
Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[Sqrt[Sin[2*
e + 2*f*x]]/(Sqrt[a*Sin[e + f*x]]*Sqrt[b*Cos[e + f*x]]), Int[1/Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b,
e, f}, x]
Rule 2641
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]
Rule 2908
Int[Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(g_.)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(
x_)])), x_Symbol] :> Dist[Sqrt[Cos[e + f*x]]/Sqrt[g*Cos[e + f*x]], Int[Sqrt[d*Sin[e + f*x]]/(Sqrt[Cos[e + f*x]
]*(a + b*Sin[e + f*x])), x], x] /; FreeQ[{a, b, d, e, f, g}, x] && NeQ[a^2 - b^2, 0]
Rule 2907
Int[Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]/(Sqrt[cos[(e_.) + (f_.)*(x_)]]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]))
, x_Symbol] :> With[{q = Rt[-a^2 + b^2, 2]}, Dist[(2*Sqrt[2]*d*(b + q))/(f*q), Subst[Int[1/((d*(b + q) + a*x^2
)*Sqrt[1 - x^4/d^2]), x], x, Sqrt[d*Sin[e + f*x]]/Sqrt[1 + Cos[e + f*x]]], x] - Dist[(2*Sqrt[2]*d*(b - q))/(f*
q), Subst[Int[1/((d*(b - q) + a*x^2)*Sqrt[1 - x^4/d^2]), x], x, Sqrt[d*Sin[e + f*x]]/Sqrt[1 + Cos[e + f*x]]],
x]] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]
Rule 1218
Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[-(c/a), 4]}, Simp[(1*Ellipt
icPi[-(e/(d*q^2)), ArcSin[q*x], -1])/(d*Sqrt[a]*q), x]] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] && GtQ[a, 0]
Rubi steps
\begin{align*} \int \frac{1}{\sqrt{g \cos (e+f x)} (d \sin (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx &=\frac{\int \frac{1}{\sqrt{g \cos (e+f x)} (d \sin (e+f x))^{3/2}} \, dx}{a}-\frac{b \int \frac{1}{\sqrt{g \cos (e+f x)} \sqrt{d \sin (e+f x)} (a+b \sin (e+f x))} \, dx}{a d}\\ &=-\frac{2 \sqrt{g \cos (e+f x)}}{a d f g \sqrt{d \sin (e+f x)}}+\frac{b^2 \int \frac{\sqrt{d \sin (e+f x)}}{\sqrt{g \cos (e+f x)} (a+b \sin (e+f x))} \, dx}{a^2 d^2}-\frac{b \int \frac{1}{\sqrt{g \cos (e+f x)} \sqrt{d \sin (e+f x)}} \, dx}{a^2 d}\\ &=-\frac{2 \sqrt{g \cos (e+f x)}}{a d f g \sqrt{d \sin (e+f x)}}+\frac{\left (b^2 \sqrt{\cos (e+f x)}\right ) \int \frac{\sqrt{d \sin (e+f x)}}{\sqrt{\cos (e+f x)} (a+b \sin (e+f x))} \, dx}{a^2 d^2 \sqrt{g \cos (e+f x)}}-\frac{\left (b \sqrt{\sin (2 e+2 f x)}\right ) \int \frac{1}{\sqrt{\sin (2 e+2 f x)}} \, dx}{a^2 d \sqrt{g \cos (e+f x)} \sqrt{d \sin (e+f x)}}\\ &=-\frac{2 \sqrt{g \cos (e+f x)}}{a d f g \sqrt{d \sin (e+f x)}}-\frac{b F\left (\left .e-\frac{\pi }{4}+f x\right |2\right ) \sqrt{\sin (2 e+2 f x)}}{a^2 d f \sqrt{g \cos (e+f x)} \sqrt{d \sin (e+f x)}}+\frac{\left (2 \sqrt{2} b^2 \left (1-\frac{b}{\sqrt{-a^2+b^2}}\right ) \sqrt{\cos (e+f x)}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (\left (b-\sqrt{-a^2+b^2}\right ) d+a x^2\right ) \sqrt{1-\frac{x^4}{d^2}}} \, dx,x,\frac{\sqrt{d \sin (e+f x)}}{\sqrt{1+\cos (e+f x)}}\right )}{a^2 d f \sqrt{g \cos (e+f x)}}+\frac{\left (2 \sqrt{2} b^2 \left (1+\frac{b}{\sqrt{-a^2+b^2}}\right ) \sqrt{\cos (e+f x)}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (\left (b+\sqrt{-a^2+b^2}\right ) d+a x^2\right ) \sqrt{1-\frac{x^4}{d^2}}} \, dx,x,\frac{\sqrt{d \sin (e+f x)}}{\sqrt{1+\cos (e+f x)}}\right )}{a^2 d f \sqrt{g \cos (e+f x)}}\\ &=-\frac{2 \sqrt{2} b^2 \sqrt{\cos (e+f x)} \Pi \left (-\frac{a}{b-\sqrt{-a^2+b^2}};\left .\sin ^{-1}\left (\frac{\sqrt{d \sin (e+f x)}}{\sqrt{d} \sqrt{1+\cos (e+f x)}}\right )\right |-1\right )}{a^2 \sqrt{-a^2+b^2} d^{3/2} f \sqrt{g \cos (e+f x)}}+\frac{2 \sqrt{2} b^2 \sqrt{\cos (e+f x)} \Pi \left (-\frac{a}{b+\sqrt{-a^2+b^2}};\left .\sin ^{-1}\left (\frac{\sqrt{d \sin (e+f x)}}{\sqrt{d} \sqrt{1+\cos (e+f x)}}\right )\right |-1\right )}{a^2 \sqrt{-a^2+b^2} d^{3/2} f \sqrt{g \cos (e+f x)}}-\frac{2 \sqrt{g \cos (e+f x)}}{a d f g \sqrt{d \sin (e+f x)}}-\frac{b F\left (\left .e-\frac{\pi }{4}+f x\right |2\right ) \sqrt{\sin (2 e+2 f x)}}{a^2 d f \sqrt{g \cos (e+f x)} \sqrt{d \sin (e+f x)}}\\ \end{align*}
Mathematica [A] time = 9.28494, size = 229, normalized size = 0.72 \[ \frac{2 \left (\frac{2 \sqrt{2} b \cos ^2\left (\frac{1}{2} (e+f x)\right ) \sqrt{\frac{\cos (e+f x)}{\cos (e+f x)-1}} \tan ^{\frac{3}{2}}\left (\frac{1}{2} (e+f x)\right ) \left (\sqrt{b^2-a^2} F\left (\left .\sin ^{-1}\left (\frac{1}{\sqrt{\tan \left (\frac{1}{2} (e+f x)\right )}}\right )\right |-1\right )+b \left (\Pi \left (-\frac{a}{b+\sqrt{b^2-a^2}};\left .-\sin ^{-1}\left (\frac{1}{\sqrt{\tan \left (\frac{1}{2} (e+f x)\right )}}\right )\right |-1\right )-\Pi \left (\frac{a}{\sqrt{b^2-a^2}-b};\left .-\sin ^{-1}\left (\frac{1}{\sqrt{\tan \left (\frac{1}{2} (e+f x)\right )}}\right )\right |-1\right )\right )\right )}{\sqrt{b^2-a^2}}-a \cos (e+f x)\right )}{a^2 d f \sqrt{d \sin (e+f x)} \sqrt{g \cos (e+f x)}} \]
Antiderivative was successfully verified.
[In]
Integrate[1/(Sqrt[g*Cos[e + f*x]]*(d*Sin[e + f*x])^(3/2)*(a + b*Sin[e + f*x])),x]
[Out]
(2*(-(a*Cos[e + f*x]) + (2*Sqrt[2]*b*Cos[(e + f*x)/2]^2*Sqrt[Cos[e + f*x]/(-1 + Cos[e + f*x])]*(Sqrt[-a^2 + b^
2]*EllipticF[ArcSin[1/Sqrt[Tan[(e + f*x)/2]]], -1] + b*(-EllipticPi[a/(-b + Sqrt[-a^2 + b^2]), -ArcSin[1/Sqrt[
Tan[(e + f*x)/2]]], -1] + EllipticPi[-(a/(b + Sqrt[-a^2 + b^2])), -ArcSin[1/Sqrt[Tan[(e + f*x)/2]]], -1]))*Tan
[(e + f*x)/2]^(3/2))/Sqrt[-a^2 + b^2]))/(a^2*d*f*Sqrt[g*Cos[e + f*x]]*Sqrt[d*Sin[e + f*x]])
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Maple [B] time = 0.293, size = 2286, normalized size = 7.1 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(d*sin(f*x+e))^(3/2)/(a+b*sin(f*x+e))/(g*cos(f*x+e))^(1/2),x)
[Out]
1/f*2^(1/2)/(-a^2+b^2)^(1/2)/(a-b+(-a^2+b^2)^(1/2))/(b+(-a^2+b^2)^(1/2)-a)/a*((-(-1+cos(f*x+e)-sin(f*x+e))/sin
(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*EllipticPi((-(
-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),a/(a-b+(-a^2+b^2)^(1/2)),1/2*2^(1/2))*cos(f*x+e)*(-a^2+b^2)^(1/2)*
b^2+(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+
e))/sin(f*x+e))^(1/2)*EllipticPi((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),a/(a-b+(-a^2+b^2)^(1/2)),1/2*2
^(1/2))*cos(f*x+e)*a*b^2-(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e)
)^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*EllipticPi((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),a/(a-b+(-
a^2+b^2)^(1/2)),1/2*2^(1/2))*cos(f*x+e)*b^3+EllipticPi((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),-a/(b+(-
a^2+b^2)^(1/2)-a),1/2*2^(1/2))*(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(
f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*cos(f*x+e)*(-a^2+b^2)^(1/2)*b^2-EllipticPi((-(-1+cos(f*x+e)-s
in(f*x+e))/sin(f*x+e))^(1/2),-a/(b+(-a^2+b^2)^(1/2)-a),1/2*2^(1/2))*(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(
1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*cos(f*x+e)*a*b^2+Ellipti
cPi((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),-a/(b+(-a^2+b^2)^(1/2)-a),1/2*2^(1/2))*(-(-1+cos(f*x+e)-sin
(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*co
s(f*x+e)*b^3+2*cos(f*x+e)*(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e
))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*EllipticF((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/
2))*(-a^2+b^2)^(1/2)*a*b-2*cos(f*x+e)*(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e
))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*EllipticF((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/
2),1/2*2^(1/2))*(-a^2+b^2)^(1/2)*b^2+(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e)
)/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*EllipticPi((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/
2),a/(a-b+(-a^2+b^2)^(1/2)),1/2*2^(1/2))*(-a^2+b^2)^(1/2)*b^2+(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*(
(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*EllipticPi((-(-1+cos(f*x+e)-si
n(f*x+e))/sin(f*x+e))^(1/2),a/(a-b+(-a^2+b^2)^(1/2)),1/2*2^(1/2))*a*b^2-(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e
))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*EllipticPi((-(-1+cos
(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),a/(a-b+(-a^2+b^2)^(1/2)),1/2*2^(1/2))*b^3+EllipticPi((-(-1+cos(f*x+e)-si
n(f*x+e))/sin(f*x+e))^(1/2),-a/(b+(-a^2+b^2)^(1/2)-a),1/2*2^(1/2))*(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1
/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*(-a^2+b^2)^(1/2)*b^2-Elli
pticPi((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),-a/(b+(-a^2+b^2)^(1/2)-a),1/2*2^(1/2))*(-(-1+cos(f*x+e)-
sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)
*a*b^2+EllipticPi((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),-a/(b+(-a^2+b^2)^(1/2)-a),1/2*2^(1/2))*(-(-1+
cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*
x+e))^(1/2)*b^3+2*(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)
*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*EllipticF((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))*(-a^
2+b^2)^(1/2)*a*b-2*(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2
)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*EllipticF((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))*(-a
^2+b^2)^(1/2)*b^2+2*(-a^2+b^2)^(1/2)*cos(f*x+e)*2^(1/2)*a^2-2*cos(f*x+e)*2^(1/2)*(-a^2+b^2)^(1/2)*a*b)*sin(f*x
+e)/(d*sin(f*x+e))^(3/2)/(g*cos(f*x+e))^(1/2)
________________________________________________________________________________________
Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{g \cos \left (f x + e\right )}{\left (b \sin \left (f x + e\right ) + a\right )} \left (d \sin \left (f x + e\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(d*sin(f*x+e))^(3/2)/(a+b*sin(f*x+e))/(g*cos(f*x+e))^(1/2),x, algorithm="maxima")
[Out]
integrate(1/(sqrt(g*cos(f*x + e))*(b*sin(f*x + e) + a)*(d*sin(f*x + e))^(3/2)), x)
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Fricas [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(1/(d*sin(f*x+e))^(3/2)/(a+b*sin(f*x+e))/(g*cos(f*x+e))^(1/2),x, algorithm="fricas")
[Out]
Timed out
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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(1/(d*sin(f*x+e))**(3/2)/(a+b*sin(f*x+e))/(g*cos(f*x+e))**(1/2),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{g \cos \left (f x + e\right )}{\left (b \sin \left (f x + e\right ) + a\right )} \left (d \sin \left (f x + e\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(d*sin(f*x+e))^(3/2)/(a+b*sin(f*x+e))/(g*cos(f*x+e))^(1/2),x, algorithm="giac")
[Out]
integrate(1/(sqrt(g*cos(f*x + e))*(b*sin(f*x + e) + a)*(d*sin(f*x + e))^(3/2)), x)