3.1432 \(\int \frac {\sqrt {d \sin (e+f x)}}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.35, antiderivative size = 209, normalized size of antiderivative = 1.00, 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} \[ \frac {2 \sqrt {2} \sqrt {d} \sqrt {\cos (e+f x)} \Pi \left (-\frac {a}{b+\sqrt {b^2-a^2}};\left .\sin ^{-1}\left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {\cos (e+f x)+1}}\right )\right |-1\right )}{f \sqrt {b^2-a^2} \sqrt {g \cos (e+f x)}}-\frac {2 \sqrt {2} \sqrt {d} \sqrt {\cos (e+f x)} \Pi \left (-\frac {a}{b-\sqrt {b^2-a^2}};\left .\sin ^{-1}\left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {\cos (e+f x)+1}}\right )\right |-1\right )}{f \sqrt {b^2-a^2} \sqrt {g \cos (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 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]

Rubi steps

\begin {align*} \int \frac {\sqrt {d \sin (e+f x)}}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx &=\frac {\sqrt {\cos (e+f x)} \int \frac {\sqrt {d \sin (e+f x)}}{\sqrt {\cos (e+f x)} (a+b \sin (e+f x))} \, dx}{\sqrt {g \cos (e+f x)}}\\ &=\frac {\left (2 \sqrt {2} \left (1-\frac {b}{\sqrt {-a^2+b^2}}\right ) d \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 )}{f \sqrt {g \cos (e+f x)}}+\frac {\left (2 \sqrt {2} \left (1+\frac {b}{\sqrt {-a^2+b^2}}\right ) d \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 )}{f \sqrt {g \cos (e+f x)}}\\ &=-\frac {2 \sqrt {2} \sqrt {d} \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 )}{\sqrt {-a^2+b^2} f \sqrt {g \cos (e+f x)}}+\frac {2 \sqrt {2} \sqrt {d} \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 )}{\sqrt {-a^2+b^2} f \sqrt {g \cos (e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 4.44, size = 166, normalized size = 0.79 \[ -\frac {2 \sqrt {2} \sqrt {\tan \left (\frac {1}{2} (e+f x)\right )} \cot (e+f x) \sqrt {d \sin (e+f x)} \left (\Pi \left (\frac {a}{\sqrt {b^2-a^2}-b};\left .\sin ^{-1}\left (\sqrt {\tan \left (\frac {1}{2} (e+f x)\right )}\right )\right |-1\right )-\Pi \left (-\frac {a}{b+\sqrt {b^2-a^2}};\left .\sin ^{-1}\left (\sqrt {\tan \left (\frac {1}{2} (e+f x)\right )}\right )\right |-1\right )\right )}{f \sqrt {b^2-a^2} \sqrt {\frac {\cos (e+f x)}{\cos (e+f x)+1}} \sqrt {g \cos (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {d \sin \left (f x + e\right )}}{\sqrt {g \cos \left (f x + e\right )} {\left (b \sin \left (f x + e\right ) + a\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [B]  time = 0.80, size = 527, normalized size = 2.52 \[ -\frac {\sqrt {d \sin \left (f x +e \right )}\, \left (\EllipticPi \left (\sqrt {-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {a}{a -b +\sqrt {-a^{2}+b^{2}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-a^{2}+b^{2}}+\EllipticPi \left (\sqrt {-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, -\frac {a}{b +\sqrt {-a^{2}+b^{2}}-a}, \frac {\sqrt {2}}{2}\right ) \sqrt {-a^{2}+b^{2}}+a \EllipticPi \left (\sqrt {-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {a}{a -b +\sqrt {-a^{2}+b^{2}}}, \frac {\sqrt {2}}{2}\right )-b \EllipticPi \left (\sqrt {-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {a}{a -b +\sqrt {-a^{2}+b^{2}}}, \frac {\sqrt {2}}{2}\right )-a \EllipticPi \left (\sqrt {-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, -\frac {a}{b +\sqrt {-a^{2}+b^{2}}-a}, \frac {\sqrt {2}}{2}\right )+b \EllipticPi \left (\sqrt {-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, -\frac {a}{b +\sqrt {-a^{2}+b^{2}}-a}, \frac {\sqrt {2}}{2}\right )\right ) \sqrt {\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sin \left (f x +e \right ) \sqrt {2}\, a}{f \left (-1+\cos \left (f x +e \right )\right ) \sqrt {g \cos \left (f x +e \right )}\, \sqrt {-a^{2}+b^{2}}\, \left (a -b +\sqrt {-a^{2}+b^{2}}\right ) \left (b +\sqrt {-a^{2}+b^{2}}-a \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/f*(d*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)+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^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)))*((-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)/(-1+cos(f*x+e))/(g*cos(f*x+e))^(1/2)*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)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {d \sin \left (f x + e\right )}}{\sqrt {g \cos \left (f x + e\right )} {\left (b \sin \left (f x + e\right ) + a\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\sqrt {d\,\sin \left (e+f\,x\right )}}{\sqrt {g\,\cos \left (e+f\,x\right )}\,\left (a+b\,\sin \left (e+f\,x\right )\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {d \sin {\left (e + f x \right )}}}{\sqrt {g \cos {\left (e + f x \right )}} \left (a + b \sin {\left (e + f x \right )}\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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