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

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

[Out]

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

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Rubi [A]  time = 0.618151, 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} \[ -\frac{2 \sqrt{2} b \sqrt{\sin (e+f x)} \Pi \left (-\frac{a}{b-\sqrt{b^2-a^2}};\left .\sin ^{-1}\left (\frac{\sqrt{d \cos (e+f x)}}{\sqrt{d} \sqrt{\sin (e+f x)+1}}\right )\right |-1\right )}{a \sqrt{d} f \sqrt{b^2-a^2} \sqrt{g \sin (e+f x)}}+\frac{2 \sqrt{2} b \sqrt{\sin (e+f x)} \Pi \left (-\frac{a}{b+\sqrt{b^2-a^2}};\left .\sin ^{-1}\left (\frac{\sqrt{d \cos (e+f x)}}{\sqrt{d} \sqrt{\sin (e+f x)+1}}\right )\right |-1\right )}{a \sqrt{d} f \sqrt{b^2-a^2} \sqrt{g \sin (e+f x)}}+\frac{\sqrt{\sin (2 e+2 f x)} F\left (\left .e+f x-\frac{\pi }{4}\right |2\right )}{a f \sqrt{d \cos (e+f x)} \sqrt{g \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [C]  time = 8.20153, size = 504, normalized size = 1.85 \[ \frac{18 (a+b) \sqrt{g \sin (e+f x)} \left (\tan ^2\left (\frac{1}{2} (e+f x)\right ) F_1\left (\frac{5}{4};\frac{1}{2},1;\frac{9}{4};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\frac{(a-b) \tan ^2\left (\frac{1}{2} (e+f x)\right )}{a+b}\right )+5 F_1\left (\frac{1}{4};\frac{1}{2},1;\frac{5}{4};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\frac{(a-b) \tan ^2\left (\frac{1}{2} (e+f x)\right )}{a+b}\right )\right )}{f g \sqrt{d \cos (e+f x)} (a+b \cos (e+f x)) \left (\tan ^2\left (\frac{1}{2} (e+f x)\right ) \left (10 \tan ^2\left (\frac{1}{2} (e+f x)\right ) \left ((a+b) F_1\left (\frac{9}{4};\frac{3}{2},1;\frac{13}{4};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\frac{(a-b) \tan ^2\left (\frac{1}{2} (e+f x)\right )}{a+b}\right )-2 (a-b) F_1\left (\frac{9}{4};\frac{1}{2},2;\frac{13}{4};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\frac{(a-b) \tan ^2\left (\frac{1}{2} (e+f x)\right )}{a+b}\right )\right )+45 (a+b) F_1\left (\frac{5}{4};\frac{1}{2},1;\frac{9}{4};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\frac{(a-b) \tan ^2\left (\frac{1}{2} (e+f x)\right )}{a+b}\right )-36 (a-b) F_1\left (\frac{5}{4};\frac{1}{2},2;\frac{9}{4};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\frac{(a-b) \tan ^2\left (\frac{1}{2} (e+f x)\right )}{a+b}\right )+18 (a+b) F_1\left (\frac{5}{4};\frac{3}{2},1;\frac{9}{4};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\frac{(a-b) \tan ^2\left (\frac{1}{2} (e+f x)\right )}{a+b}\right )\right )+45 (a+b) F_1\left (\frac{1}{4};\frac{1}{2},1;\frac{5}{4};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\frac{(a-b) \tan ^2\left (\frac{1}{2} (e+f x)\right )}{a+b}\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(18*(a + b)*Sqrt[g*Sin[e + f*x]]*(5*AppellF1[1/4, 1/2, 1, 5/4, Tan[(e + f*x)/2]^2, -(((a - b)*Tan[(e + f*x)/2]
^2)/(a + b))] + AppellF1[5/4, 1/2, 1, 9/4, Tan[(e + f*x)/2]^2, -(((a - b)*Tan[(e + f*x)/2]^2)/(a + b))]*Tan[(e
+ f*x)/2]^2))/(f*g*Sqrt[d*Cos[e + f*x]]*(a + b*Cos[e + f*x])*(45*(a + b)*AppellF1[1/4, 1/2, 1, 5/4, Tan[(e +
f*x)/2]^2, -(((a - b)*Tan[(e + f*x)/2]^2)/(a + b))] + Tan[(e + f*x)/2]^2*(45*(a + b)*AppellF1[5/4, 1/2, 1, 9/4
, Tan[(e + f*x)/2]^2, -(((a - b)*Tan[(e + f*x)/2]^2)/(a + b))] - 36*(a - b)*AppellF1[5/4, 1/2, 2, 9/4, Tan[(e
+ f*x)/2]^2, -(((a - b)*Tan[(e + f*x)/2]^2)/(a + b))] + 18*(a + b)*AppellF1[5/4, 3/2, 1, 9/4, Tan[(e + f*x)/2]
^2, -(((a - b)*Tan[(e + f*x)/2]^2)/(a + b))] + 10*(-2*(a - b)*AppellF1[9/4, 1/2, 2, 13/4, Tan[(e + f*x)/2]^2,
-(((a - b)*Tan[(e + f*x)/2]^2)/(a + b))] + (a + b)*AppellF1[9/4, 3/2, 1, 13/4, Tan[(e + f*x)/2]^2, -(((a - b)*
Tan[(e + f*x)/2]^2)/(a + b))])*Tan[(e + f*x)/2]^2)))

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Maple [B]  time = 0.533, size = 607, normalized size = 2.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

[Out]

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

[Out]

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

[Out]

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