3.1411 \(\int \frac{\sqrt{g \cos (e+f x)}}{\sqrt{d \sin (e+f x)} (a+b \sin (e+f x))} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.414891, 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} \[ \frac{2 \sqrt{2} \sqrt{g} \sqrt{\sin (e+f x)} \Pi \left (-\frac{\sqrt{b-a}}{\sqrt{a+b}};\left .\sin ^{-1}\left (\frac{\sqrt{g \cos (e+f x)}}{\sqrt{g} \sqrt{\sin (e+f x)+1}}\right )\right |-1\right )}{f \sqrt{b-a} \sqrt{a+b} \sqrt{d \sin (e+f x)}}-\frac{2 \sqrt{2} \sqrt{g} \sqrt{\sin (e+f x)} \Pi \left (\frac{\sqrt{b-a}}{\sqrt{a+b}};\left .\sin ^{-1}\left (\frac{\sqrt{g \cos (e+f x)}}{\sqrt{g} \sqrt{\sin (e+f x)+1}}\right )\right |-1\right )}{f \sqrt{b-a} \sqrt{a+b} \sqrt{d \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 7.18232, size = 182, normalized size = 0.88 \[ -\frac{4 \sqrt{2} g \cos ^2\left (\frac{1}{2} (e+f x)\right ) \sqrt{\frac{\cos (e+f x)}{\cos (e+f x)-1}} \tan ^{\frac{3}{2}}\left (\frac{1}{2} (e+f x)\right ) \left (\Pi \left (\frac{a}{\sqrt{b^2-a^2}-b};\left .-\sin ^{-1}\left (\frac{1}{\sqrt{\tan \left (\frac{1}{2} (e+f x)\right )}}\right )\right |-1\right )+\Pi \left (-\frac{a}{b+\sqrt{b^2-a^2}};\left .-\sin ^{-1}\left (\frac{1}{\sqrt{\tan \left (\frac{1}{2} (e+f x)\right )}}\right )\right |-1\right )+F\left (\left .\sin ^{-1}\left (\frac{1}{\sqrt{\tan \left (\frac{1}{2} (e+f x)\right )}}\right )\right |-1\right )\right )}{a f \sqrt{d \sin (e+f x)} \sqrt{g \cos (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*Sqrt[2]*g*Cos[(e + f*x)/2]^2*Sqrt[Cos[e + f*x]/(-1 + Cos[e + f*x])]*(EllipticF[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] + 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*f*Sqrt[g*Cos[e + f*x]
]*Sqrt[d*Sin[e + f*x]])

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Maple [B]  time = 0.303, size = 590, normalized size = 2.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*cos(f*x+e))^(1/2)/(d*sin(f*x+e))^(1/2)/(a+b*sin(f*x+e)),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)*(g*cos(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)*((-1+cos(f*x+e))/sin(f*x+e))^(
1/2)*(a-b)*(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))-EllipticP
i((-(-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)-Ellipt
icPi((-(-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^2+b^2)^(1/2)+a*
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*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*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*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)))*sin(f*x+e)^2/(d*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{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((g*cos(f*x+e))^(1/2)/(d*sin(f*x+e))^(1/2)/(a+b*sin(f*x+e)),x, algorithm="maxima")

[Out]

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

[Out]

Timed out

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

[Out]

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

[Out]

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