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

Optimal. Leaf size=209 \[ \frac{2 \sqrt{2} \sqrt{d} \sqrt{\sin (e+f x)} \Pi \left (-\frac{a}{b-\sqrt{b^2-a^2}};\left .\sin ^{-1}\left (\frac{\sqrt{d \cos (e+f x)}}{\sqrt{d} \sqrt{\sin (e+f x)+1}}\right )\right |-1\right )}{f \sqrt{b^2-a^2} \sqrt{g \sin (e+f x)}}-\frac{2 \sqrt{2} \sqrt{d} \sqrt{\sin (e+f x)} \Pi \left (-\frac{a}{b+\sqrt{b^2-a^2}};\left .\sin ^{-1}\left (\frac{\sqrt{d \cos (e+f x)}}{\sqrt{d} \sqrt{\sin (e+f x)+1}}\right )\right |-1\right )}{f \sqrt{b^2-a^2} \sqrt{g \sin (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.409849, antiderivative size = 209, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.081, Rules used = {2908, 2907, 1218} \[ \frac{2 \sqrt{2} \sqrt{d} \sqrt{\sin (e+f x)} \Pi \left (-\frac{a}{b-\sqrt{b^2-a^2}};\left .\sin ^{-1}\left (\frac{\sqrt{d \cos (e+f x)}}{\sqrt{d} \sqrt{\sin (e+f x)+1}}\right )\right |-1\right )}{f \sqrt{b^2-a^2} \sqrt{g \sin (e+f x)}}-\frac{2 \sqrt{2} \sqrt{d} \sqrt{\sin (e+f x)} \Pi \left (-\frac{a}{b+\sqrt{b^2-a^2}};\left .\sin ^{-1}\left (\frac{\sqrt{d \cos (e+f x)}}{\sqrt{d} \sqrt{\sin (e+f x)+1}}\right )\right |-1\right )}{f \sqrt{b^2-a^2} \sqrt{g \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [C]  time = 9.23293, size = 594, normalized size = 2.84 \[ \frac{2 \sqrt{\tan (e+f x)} \sec ^2(e+f x) \sqrt{d \cos (e+f x)} \left (a \sqrt{\tan ^2(e+f x)+1}+b\right ) \left (\frac{5 b \left (a^2-b^2\right ) \sqrt{\tan (e+f x)} F_1\left (\frac{1}{4};\frac{1}{2},1;\frac{5}{4};-\tan ^2(e+f x),-\frac{a^2 \tan ^2(e+f x)}{a^2-b^2}\right )}{\sqrt{\tan ^2(e+f x)+1} \left (a^2 \left (\tan ^2(e+f x)+1\right )-b^2\right ) \left (2 \tan ^2(e+f x) \left (2 a^2 F_1\left (\frac{5}{4};\frac{1}{2},2;\frac{9}{4};-\tan ^2(e+f x),-\frac{a^2 \tan ^2(e+f x)}{a^2-b^2}\right )+\left (a^2-b^2\right ) F_1\left (\frac{5}{4};\frac{3}{2},1;\frac{9}{4};-\tan ^2(e+f x),-\frac{a^2 \tan ^2(e+f x)}{a^2-b^2}\right )\right )-5 \left (a^2-b^2\right ) F_1\left (\frac{1}{4};\frac{1}{2},1;\frac{5}{4};-\tan ^2(e+f x),-\frac{a^2 \tan ^2(e+f x)}{a^2-b^2}\right )\right )}+\frac{\sqrt{a} \left (-2 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{a} \sqrt{\tan (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )+2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{\tan (e+f x)}}{\sqrt [4]{a^2-b^2}}+1\right )-\log \left (-\sqrt{2} \sqrt{a} \sqrt [4]{a^2-b^2} \sqrt{\tan (e+f x)}+\sqrt{a^2-b^2}+a \tan (e+f x)\right )+\log \left (\sqrt{2} \sqrt{a} \sqrt [4]{a^2-b^2} \sqrt{\tan (e+f x)}+\sqrt{a^2-b^2}+a \tan (e+f x)\right )\right )}{4 \sqrt{2} \left (a^2-b^2\right )^{3/4}}\right )}{f \left (\tan ^2(e+f x)+1\right )^{3/2} \sqrt{g \sin (e+f x)} (a+b \cos (e+f x))} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*Sqrt[d*Cos[e + f*x]]*Sec[e + f*x]^2*Sqrt[Tan[e + f*x]]*(b + a*Sqrt[1 + Tan[e + f*x]^2])*((Sqrt[a]*(-2*ArcTa
n[1 - (Sqrt[2]*Sqrt[a]*Sqrt[Tan[e + f*x]])/(a^2 - b^2)^(1/4)] + 2*ArcTan[1 + (Sqrt[2]*Sqrt[a]*Sqrt[Tan[e + f*x
]])/(a^2 - b^2)^(1/4)] - Log[Sqrt[a^2 - b^2] - Sqrt[2]*Sqrt[a]*(a^2 - b^2)^(1/4)*Sqrt[Tan[e + f*x]] + a*Tan[e
+ f*x]] + Log[Sqrt[a^2 - b^2] + Sqrt[2]*Sqrt[a]*(a^2 - b^2)^(1/4)*Sqrt[Tan[e + f*x]] + a*Tan[e + f*x]]))/(4*Sq
rt[2]*(a^2 - b^2)^(3/4)) + (5*b*(a^2 - b^2)*AppellF1[1/4, 1/2, 1, 5/4, -Tan[e + f*x]^2, -((a^2*Tan[e + f*x]^2)
/(a^2 - b^2))]*Sqrt[Tan[e + f*x]])/(Sqrt[1 + Tan[e + f*x]^2]*(-5*(a^2 - b^2)*AppellF1[1/4, 1/2, 1, 5/4, -Tan[e
+ f*x]^2, -((a^2*Tan[e + f*x]^2)/(a^2 - b^2))] + 2*(2*a^2*AppellF1[5/4, 1/2, 2, 9/4, -Tan[e + f*x]^2, -((a^2*
Tan[e + f*x]^2)/(a^2 - b^2))] + (a^2 - b^2)*AppellF1[5/4, 3/2, 1, 9/4, -Tan[e + f*x]^2, -((a^2*Tan[e + f*x]^2)
/(a^2 - b^2))])*Tan[e + f*x]^2)*(-b^2 + a^2*(1 + Tan[e + f*x]^2)))))/(f*(a + b*Cos[e + f*x])*Sqrt[g*Sin[e + f*
x]]*(1 + Tan[e + f*x]^2)^(3/2))

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Maple [B]  time = 0.548, size = 608, normalized size = 2.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/f*2^(1/2)*a/(-a^2+b^2)^(1/2)/(a-b+(-a^2+b^2)^(1/2))/(b+(-a^2+b^2)^(1/2)-a)*(d*cos(f*x+e))^(1/2)*(2*(-a^2+b^
2)^(1/2)*EllipticF(((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))+a*EllipticPi(((1-cos(f*x+e)+sin(f
*x+e))/sin(f*x+e))^(1/2),(a-b)/(a-b+(-(a-b)*(a+b))^(1/2)),1/2*2^(1/2))-EllipticPi(((1-cos(f*x+e)+sin(f*x+e))/s
in(f*x+e))^(1/2),(a-b)/(a-b+(-(a-b)*(a+b))^(1/2)),1/2*2^(1/2))*b-(-a^2+b^2)^(1/2)*EllipticPi(((1-cos(f*x+e)+si
n(f*x+e))/sin(f*x+e))^(1/2),(a-b)/(a-b+(-(a-b)*(a+b))^(1/2)),1/2*2^(1/2))-a*EllipticPi(((1-cos(f*x+e)+sin(f*x+
e))/sin(f*x+e))^(1/2),-(a-b)/(-a+b+(-(a-b)*(a+b))^(1/2)),1/2*2^(1/2))+EllipticPi(((1-cos(f*x+e)+sin(f*x+e))/si
n(f*x+e))^(1/2),-(a-b)/(-a+b+(-(a-b)*(a+b))^(1/2)),1/2*2^(1/2))*b-(-a^2+b^2)^(1/2)*EllipticPi(((1-cos(f*x+e)+s
in(f*x+e))/sin(f*x+e))^(1/2),-(a-b)/(-a+b+(-(a-b)*(a+b))^(1/2)),1/2*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/(g*
sin(f*x+e))^(1/2)/cos(f*x+e)/(-1+cos(f*x+e))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{d \cos \left (f x + e\right )}}{{\left (b \cos \left (f x + e\right ) + a\right )} \sqrt{g \sin \left (f x + e\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cos(f*x+e))^(1/2)/(a+b*cos(f*x+e))/(g*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(d*cos(f*x + e))/((b*cos(f*x + e) + a)*sqrt(g*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((d*cos(f*x+e))^(1/2)/(a+b*cos(f*x+e))/(g*sin(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{\sqrt{d \cos{\left (e + f x \right )}}}{\sqrt{g \sin{\left (e + f x \right )}} \left (a + b \cos{\left (e + f x \right )}\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{d \cos \left (f x + e\right )}}{{\left (b \cos \left (f x + e\right ) + a\right )} \sqrt{g \sin \left (f x + e\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cos(f*x+e))^(1/2)/(a+b*cos(f*x+e))/(g*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

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