3.1433 \(\int \frac{1}{\sqrt{g \cos (e+f x)} \sqrt{d \sin (e+f x)} (a+b \sin (e+f x))} \, dx\)

Optimal. Leaf size=273 \[ \frac{2 \sqrt{2} b \sqrt{\cos (e+f x)} \Pi \left (-\frac{a}{b-\sqrt{b^2-a^2}};\left .\sin ^{-1}\left (\frac{\sqrt{d \sin (e+f x)}}{\sqrt{d} \sqrt{\cos (e+f x)+1}}\right )\right |-1\right )}{a \sqrt{d} f \sqrt{b^2-a^2} \sqrt{g \cos (e+f x)}}-\frac{2 \sqrt{2} b \sqrt{\cos (e+f x)} \Pi \left (-\frac{a}{b+\sqrt{b^2-a^2}};\left .\sin ^{-1}\left (\frac{\sqrt{d \sin (e+f x)}}{\sqrt{d} \sqrt{\cos (e+f x)+1}}\right )\right |-1\right )}{a \sqrt{d} f \sqrt{b^2-a^2} \sqrt{g \cos (e+f x)}}+\frac{\sqrt{\sin (2 e+2 f x)} F\left (\left .e+f x-\frac{\pi }{4}\right |2\right )}{a f \sqrt{d \sin (e+f x)} \sqrt{g \cos (e+f x)}} \]

[Out]

(2*Sqrt[2]*b*Sqrt[Cos[e + f*x]]*EllipticPi[-(a/(b - Sqrt[-a^2 + b^2])), ArcSin[Sqrt[d*Sin[e + f*x]]/(Sqrt[d]*S
qrt[1 + Cos[e + f*x]])], -1])/(a*Sqrt[-a^2 + b^2]*Sqrt[d]*f*Sqrt[g*Cos[e + f*x]]) - (2*Sqrt[2]*b*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*Sqrt[-a^2 + b^2]*Sqrt[d]*f*Sqrt[g*Cos[e + f*x]]) + (EllipticF[e - Pi/4 + f*x, 2]*Sqrt[Sin[2*e + 2*f*x]]
)/(a*f*Sqrt[g*Cos[e + f*x]]*Sqrt[d*Sin[e + f*x]])

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Rubi [A]  time = 0.586007, antiderivative size = 273, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.162, Rules used = {2910, 2573, 2641, 2908, 2907, 1218} \[ \frac{2 \sqrt{2} b \sqrt{\cos (e+f x)} \Pi \left (-\frac{a}{b-\sqrt{b^2-a^2}};\left .\sin ^{-1}\left (\frac{\sqrt{d \sin (e+f x)}}{\sqrt{d} \sqrt{\cos (e+f x)+1}}\right )\right |-1\right )}{a \sqrt{d} f \sqrt{b^2-a^2} \sqrt{g \cos (e+f x)}}-\frac{2 \sqrt{2} b \sqrt{\cos (e+f x)} \Pi \left (-\frac{a}{b+\sqrt{b^2-a^2}};\left .\sin ^{-1}\left (\frac{\sqrt{d \sin (e+f x)}}{\sqrt{d} \sqrt{\cos (e+f x)+1}}\right )\right |-1\right )}{a \sqrt{d} f \sqrt{b^2-a^2} \sqrt{g \cos (e+f x)}}+\frac{\sqrt{\sin (2 e+2 f x)} F\left (\left .e+f x-\frac{\pi }{4}\right |2\right )}{a f \sqrt{d \sin (e+f x)} \sqrt{g \cos (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[g*Cos[e + f*x]]*Sqrt[d*Sin[e + f*x]]*(a + b*Sin[e + f*x])),x]

[Out]

(2*Sqrt[2]*b*Sqrt[Cos[e + f*x]]*EllipticPi[-(a/(b - Sqrt[-a^2 + b^2])), ArcSin[Sqrt[d*Sin[e + f*x]]/(Sqrt[d]*S
qrt[1 + Cos[e + f*x]])], -1])/(a*Sqrt[-a^2 + b^2]*Sqrt[d]*f*Sqrt[g*Cos[e + f*x]]) - (2*Sqrt[2]*b*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*Sqrt[-a^2 + b^2]*Sqrt[d]*f*Sqrt[g*Cos[e + f*x]]) + (EllipticF[e - Pi/4 + f*x, 2]*Sqrt[Sin[2*e + 2*f*x]]
)/(a*f*Sqrt[g*Cos[e + f*x]]*Sqrt[d*Sin[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 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)} \sqrt{d \sin (e+f x)} (a+b \sin (e+f x))} \, dx &=\frac{\int \frac{1}{\sqrt{g \cos (e+f x)} \sqrt{d \sin (e+f x)}} \, dx}{a}-\frac{b \int \frac{\sqrt{d \sin (e+f x)}}{\sqrt{g \cos (e+f x)} (a+b \sin (e+f x))} \, dx}{a d}\\ &=-\frac{\left (b \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 d \sqrt{g \cos (e+f x)}}+\frac{\sqrt{\sin (2 e+2 f x)} \int \frac{1}{\sqrt{\sin (2 e+2 f x)}} \, dx}{a \sqrt{g \cos (e+f x)} \sqrt{d \sin (e+f x)}}\\ &=\frac{F\left (\left .e-\frac{\pi }{4}+f x\right |2\right ) \sqrt{\sin (2 e+2 f x)}}{a f \sqrt{g \cos (e+f x)} \sqrt{d \sin (e+f x)}}-\frac{\left (2 \sqrt{2} b \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 f \sqrt{g \cos (e+f x)}}-\frac{\left (2 \sqrt{2} b \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 f \sqrt{g \cos (e+f x)}}\\ &=\frac{2 \sqrt{2} b \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 \sqrt{-a^2+b^2} \sqrt{d} f \sqrt{g \cos (e+f x)}}-\frac{2 \sqrt{2} b \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 \sqrt{-a^2+b^2} \sqrt{d} f \sqrt{g \cos (e+f x)}}+\frac{F\left (\left .e-\frac{\pi }{4}+f x\right |2\right ) \sqrt{\sin (2 e+2 f x)}}{a f \sqrt{g \cos (e+f x)} \sqrt{d \sin (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 8.51983, size = 213, normalized size = 0.78 \[ -\frac{4 \sqrt{2} \cos ^2\left (\frac{1}{2} (e+f x)\right ) \sqrt{\frac{\cos (e+f x)}{\cos (e+f x)-1}} \tan ^{\frac{3}{2}}\left (\frac{1}{2} (e+f x)\right ) \left (\sqrt{b^2-a^2} F\left (\left .\sin ^{-1}\left (\frac{1}{\sqrt{\tan \left (\frac{1}{2} (e+f x)\right )}}\right )\right |-1\right )+b \left (\Pi \left (-\frac{a}{b+\sqrt{b^2-a^2}};\left .-\sin ^{-1}\left (\frac{1}{\sqrt{\tan \left (\frac{1}{2} (e+f x)\right )}}\right )\right |-1\right )-\Pi \left (\frac{a}{\sqrt{b^2-a^2}-b};\left .-\sin ^{-1}\left (\frac{1}{\sqrt{\tan \left (\frac{1}{2} (e+f x)\right )}}\right )\right |-1\right )\right )\right )}{a f \sqrt{b^2-a^2} \sqrt{d \sin (e+f x)} \sqrt{g \cos (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[g*Cos[e + f*x]]*Sqrt[d*Sin[e + f*x]]*(a + b*Sin[e + f*x])),x]

[Out]

(-4*Sqrt[2]*Cos[(e + f*x)/2]^2*Sqrt[Cos[e + f*x]/(-1 + Cos[e + f*x])]*(Sqrt[-a^2 + b^2]*EllipticF[ArcSin[1/Sqr
t[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))/(a*S
qrt[-a^2 + b^2]*f*Sqrt[g*Cos[e + f*x]]*Sqrt[d*Sin[e + f*x]])

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Maple [B]  time = 0.269, size = 631, normalized size = 2.3 \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))^(1/2)/(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)*(2*EllipticF((-(-1+cos(f*x+e)-sin(f
*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))*a*(-a^2+b^2)^(1/2)-2*EllipticF((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^
(1/2),1/2*2^(1/2))*b*(-a^2+b^2)^(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*b-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^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*(-a^2+b^2)^(1/2)-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))*a*b+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))
*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))*b*(-a^2+
b^2)^(1/2))*((-1+cos(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))/sin(f*x+e))^(1/2)*sin(f*x+e)^2/(d*sin(f*x+e))^(1/2)/(-1+cos(f*x+e))/(g*cos(f*x+e))^(1/2)

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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 )} \sqrt{d \sin \left (f x + e\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*sin(f*x+e))^(1/2)/(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)*sqrt(d*sin(f*x + e))), 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))^(1/2)/(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]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{d \sin{\left (e + f x \right )}} \sqrt{g \cos{\left (e + f x \right )}} \left (a + b \sin{\left (e + f x \right )}\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*sin(f*x+e))**(1/2)/(a+b*sin(f*x+e))/(g*cos(f*x+e))**(1/2),x)

[Out]

Integral(1/(sqrt(d*sin(e + f*x))*sqrt(g*cos(e + f*x))*(a + b*sin(e + f*x))), x)

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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 )} \sqrt{d \sin \left (f x + e\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*sin(f*x+e))^(1/2)/(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)*sqrt(d*sin(f*x + e))), x)