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

Optimal. Leaf size=208 \[ \frac{2 \sqrt{2} \sqrt{g} \sqrt{\cos (e+f x)} \Pi \left (\frac{\sqrt{b-a}}{\sqrt{a+b}};\left .\sin ^{-1}\left (\frac{\sqrt{g \sin (e+f x)}}{\sqrt{g} \sqrt{\cos (e+f x)+1}}\right )\right |-1\right )}{f \sqrt{b-a} \sqrt{a+b} \sqrt{d \cos (e+f x)}}-\frac{2 \sqrt{2} \sqrt{g} \sqrt{\cos (e+f x)} \Pi \left (-\frac{\sqrt{b-a}}{\sqrt{a+b}};\left .\sin ^{-1}\left (\frac{\sqrt{g \sin (e+f x)}}{\sqrt{g} \sqrt{\cos (e+f x)+1}}\right )\right |-1\right )}{f \sqrt{b-a} \sqrt{a+b} \sqrt{d \cos (e+f x)}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.390143, antiderivative size = 208, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.108, Rules used = {2906, 2905, 490, 1218} \[ \frac{2 \sqrt{2} \sqrt{g} \sqrt{\cos (e+f x)} \Pi \left (\frac{\sqrt{b-a}}{\sqrt{a+b}};\left .\sin ^{-1}\left (\frac{\sqrt{g \sin (e+f x)}}{\sqrt{g} \sqrt{\cos (e+f x)+1}}\right )\right |-1\right )}{f \sqrt{b-a} \sqrt{a+b} \sqrt{d \cos (e+f x)}}-\frac{2 \sqrt{2} \sqrt{g} \sqrt{\cos (e+f x)} \Pi \left (-\frac{\sqrt{b-a}}{\sqrt{a+b}};\left .\sin ^{-1}\left (\frac{\sqrt{g \sin (e+f x)}}{\sqrt{g} \sqrt{\cos (e+f x)+1}}\right )\right |-1\right )}{f \sqrt{b-a} \sqrt{a+b} \sqrt{d \cos (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2906

Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(g_.)]/(Sqrt[(d_)*sin[(e_.) + (f_.)*(x_)]]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x
_)])), x_Symbol] :> Dist[Sqrt[Sin[e + f*x]]/Sqrt[d*Sin[e + f*x]], Int[Sqrt[g*Cos[e + f*x]]/(Sqrt[Sin[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 2905

Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(g_.)]/(Sqrt[sin[(e_.) + (f_.)*(x_)]]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]))
, x_Symbol] :> Dist[(-4*Sqrt[2]*g)/f, Subst[Int[x^2/(((a + b)*g^2 + (a - b)*x^4)*Sqrt[1 - x^4/g^2]), x], x, Sq
rt[g*Cos[e + f*x]]/Sqrt[1 + Sin[e + f*x]]], x] /; FreeQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0]

Rule 490

Int[(x_)^2/(((a_) + (b_.)*(x_)^4)*Sqrt[(c_) + (d_.)*(x_)^4]), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]],
s = Denominator[Rt[-(a/b), 2]]}, Dist[s/(2*b), Int[1/((r + s*x^2)*Sqrt[c + d*x^4]), x], x] - Dist[s/(2*b), In
t[1/((r - s*x^2)*Sqrt[c + d*x^4]), x], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 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{g \sin (e+f x)}}{\sqrt{d \cos (e+f x)} (a+b \cos (e+f x))} \, dx &=\frac{\sqrt{\cos (e+f x)} \int \frac{\sqrt{g \sin (e+f x)}}{\sqrt{\cos (e+f x)} (a+b \cos (e+f x))} \, dx}{\sqrt{d \cos (e+f x)}}\\ &=\frac{\left (4 \sqrt{2} g \sqrt{\cos (e+f x)}\right ) \operatorname{Subst}\left (\int \frac{x^2}{\left ((a+b) g^2+(a-b) x^4\right ) \sqrt{1-\frac{x^4}{g^2}}} \, dx,x,\frac{\sqrt{g \sin (e+f x)}}{\sqrt{1+\cos (e+f x)}}\right )}{f \sqrt{d \cos (e+f x)}}\\ &=\frac{\left (2 \sqrt{2} g \sqrt{\cos (e+f x)}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (\sqrt{a+b} g-\sqrt{-a+b} x^2\right ) \sqrt{1-\frac{x^4}{g^2}}} \, dx,x,\frac{\sqrt{g \sin (e+f x)}}{\sqrt{1+\cos (e+f x)}}\right )}{\sqrt{-a+b} f \sqrt{d \cos (e+f x)}}-\frac{\left (2 \sqrt{2} g \sqrt{\cos (e+f x)}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (\sqrt{a+b} g+\sqrt{-a+b} x^2\right ) \sqrt{1-\frac{x^4}{g^2}}} \, dx,x,\frac{\sqrt{g \sin (e+f x)}}{\sqrt{1+\cos (e+f x)}}\right )}{\sqrt{-a+b} f \sqrt{d \cos (e+f x)}}\\ &=-\frac{2 \sqrt{2} \sqrt{g} \sqrt{\cos (e+f x)} \Pi \left (-\frac{\sqrt{-a+b}}{\sqrt{a+b}};\left .\sin ^{-1}\left (\frac{\sqrt{g \sin (e+f x)}}{\sqrt{g} \sqrt{1+\cos (e+f x)}}\right )\right |-1\right )}{\sqrt{-a+b} \sqrt{a+b} f \sqrt{d \cos (e+f x)}}+\frac{2 \sqrt{2} \sqrt{g} \sqrt{\cos (e+f x)} \Pi \left (\frac{\sqrt{-a+b}}{\sqrt{a+b}};\left .\sin ^{-1}\left (\frac{\sqrt{g \sin (e+f x)}}{\sqrt{g} \sqrt{1+\cos (e+f x)}}\right )\right |-1\right )}{\sqrt{-a+b} \sqrt{a+b} f \sqrt{d \cos (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 2.73567, size = 375, normalized size = 1.8 \[ \frac{2 \sqrt{g \sin (e+f x)} \left (a \sqrt{\sec ^2(e+f x)}+b\right ) \left (\frac{b \tan ^{\frac{3}{2}}(e+f x) F_1\left (\frac{3}{4};\frac{1}{2},1;\frac{7}{4};-\tan ^2(e+f x),-\frac{a^2 \tan ^2(e+f x)}{a^2-b^2}\right )}{3 \left (b^2-a^2\right )}+\frac{-2 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{a} \sqrt{\tan (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )+2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{\tan (e+f x)}}{\sqrt [4]{a^2-b^2}}+1\right )+\log \left (-\sqrt{2} \sqrt{a} \sqrt [4]{a^2-b^2} \sqrt{\tan (e+f x)}+\sqrt{a^2-b^2}+a \tan (e+f x)\right )-\log \left (\sqrt{2} \sqrt{a} \sqrt [4]{a^2-b^2} \sqrt{\tan (e+f x)}+\sqrt{a^2-b^2}+a \tan (e+f x)\right )}{4 \sqrt{2} \sqrt{a} \sqrt [4]{a^2-b^2}}\right )}{f \sqrt{\tan (e+f x)} \sqrt{\sec ^2(e+f x)} \sqrt{d \cos (e+f x)} (a+b \cos (e+f x))} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*(b + a*Sqrt[Sec[e + f*x]^2])*Sqrt[g*Sin[e + f*x]]*((-2*ArcTan[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*Sqrt[2]*Sqrt[a]*(a^2 - b^2)^(1/4)) + (b*AppellF1[3
/4, 1/2, 1, 7/4, -Tan[e + f*x]^2, -((a^2*Tan[e + f*x]^2)/(a^2 - b^2))]*Tan[e + f*x]^(3/2))/(3*(-a^2 + b^2))))/
(f*Sqrt[d*Cos[e + f*x]]*(a + b*Cos[e + f*x])*Sqrt[Sec[e + f*x]^2]*Sqrt[Tan[e + f*x]])

________________________________________________________________________________________

Maple [B]  time = 0.463, size = 538, normalized size = 2.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*2^(1/2)/(b+(-a^2+b^2)^(1/2)-a)/(a-b+(-a^2+b^2)^(1/2))*((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+co
s(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)*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))-(-a^2+b^2)^(1/2)*EllipticP
i(((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))-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))/sin(f*x+e))^(1/2),(a-b)/(a-b+(-(a-b)*(a+b))^(1/2)),1/2*2^(1/2))*b-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))/sin(f*x+e))^(1/2),-(a-b)/(-a+b+(-(a-b)*(a+b))^(1/2)),1/2*2^(1/2))*b)*(g*sin(f*x+e))^(1/2)*sin(f*x+e)/(-
1+cos(f*x+e))/(d*cos(f*x+e))^(1/2)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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