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

Optimal. Leaf size=509 \[ \frac{2 \sqrt{2} a d \sqrt{g} \sqrt{\cos (e+f x)} \Pi \left (-\frac{\sqrt{b-a}}{\sqrt{a+b}};\left .\sin ^{-1}\left (\frac{\sqrt{g \sin (e+f x)}}{\sqrt{g} \sqrt{\cos (e+f x)+1}}\right )\right |-1\right )}{b f \sqrt{b-a} \sqrt{a+b} \sqrt{d \cos (e+f x)}}-\frac{2 \sqrt{2} a d \sqrt{g} \sqrt{\cos (e+f x)} \Pi \left (\frac{\sqrt{b-a}}{\sqrt{a+b}};\left .\sin ^{-1}\left (\frac{\sqrt{g \sin (e+f x)}}{\sqrt{g} \sqrt{\cos (e+f x)+1}}\right )\right |-1\right )}{b f \sqrt{b-a} \sqrt{a+b} \sqrt{d \cos (e+f x)}}-\frac{\sqrt{d} \sqrt{g} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d} \sqrt{g \sin (e+f x)}}{\sqrt{g} \sqrt{d \cos (e+f x)}}\right )}{\sqrt{2} b f}+\frac{\sqrt{d} \sqrt{g} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d} \sqrt{g \sin (e+f x)}}{\sqrt{g} \sqrt{d \cos (e+f x)}}+1\right )}{\sqrt{2} b f}+\frac{\sqrt{d} \sqrt{g} \log \left (-\frac{\sqrt{2} \sqrt{d} \sqrt{g \sin (e+f x)}}{\sqrt{d \cos (e+f x)}}+\sqrt{g} \tan (e+f x)+\sqrt{g}\right )}{2 \sqrt{2} b f}-\frac{\sqrt{d} \sqrt{g} \log \left (\frac{\sqrt{2} \sqrt{d} \sqrt{g \sin (e+f x)}}{\sqrt{d \cos (e+f x)}}+\sqrt{g} \tan (e+f x)+\sqrt{g}\right )}{2 \sqrt{2} b f} \]

[Out]

-((Sqrt[d]*Sqrt[g]*ArcTan[1 - (Sqrt[2]*Sqrt[d]*Sqrt[g*Sin[e + f*x]])/(Sqrt[g]*Sqrt[d*Cos[e + f*x]])])/(Sqrt[2]
*b*f)) + (Sqrt[d]*Sqrt[g]*ArcTan[1 + (Sqrt[2]*Sqrt[d]*Sqrt[g*Sin[e + f*x]])/(Sqrt[g]*Sqrt[d*Cos[e + f*x]])])/(
Sqrt[2]*b*f) + (2*Sqrt[2]*a*d*Sqrt[g]*Sqrt[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])/(b*Sqrt[-a + b]*Sqrt[a + b]*f*Sqrt[d*Cos[e + f*x]]) - (
2*Sqrt[2]*a*d*Sqrt[g]*Sqrt[Cos[e + f*x]]*EllipticPi[Sqrt[-a + b]/Sqrt[a + b], ArcSin[Sqrt[g*Sin[e + f*x]]/(Sqr
t[g]*Sqrt[1 + Cos[e + f*x]])], -1])/(b*Sqrt[-a + b]*Sqrt[a + b]*f*Sqrt[d*Cos[e + f*x]]) + (Sqrt[d]*Sqrt[g]*Log
[Sqrt[g] - (Sqrt[2]*Sqrt[d]*Sqrt[g*Sin[e + f*x]])/Sqrt[d*Cos[e + f*x]] + Sqrt[g]*Tan[e + f*x]])/(2*Sqrt[2]*b*f
) - (Sqrt[d]*Sqrt[g]*Log[Sqrt[g] + (Sqrt[2]*Sqrt[d]*Sqrt[g*Sin[e + f*x]])/Sqrt[d*Cos[e + f*x]] + Sqrt[g]*Tan[e
+ f*x]])/(2*Sqrt[2]*b*f)

________________________________________________________________________________________

Rubi [A]  time = 0.81537, antiderivative size = 509, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 12, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.324, Rules used = {2909, 2574, 297, 1162, 617, 204, 1165, 628, 2906, 2905, 490, 1218} \[ \frac{2 \sqrt{2} a d \sqrt{g} \sqrt{\cos (e+f x)} \Pi \left (-\frac{\sqrt{b-a}}{\sqrt{a+b}};\left .\sin ^{-1}\left (\frac{\sqrt{g \sin (e+f x)}}{\sqrt{g} \sqrt{\cos (e+f x)+1}}\right )\right |-1\right )}{b f \sqrt{b-a} \sqrt{a+b} \sqrt{d \cos (e+f x)}}-\frac{2 \sqrt{2} a d \sqrt{g} \sqrt{\cos (e+f x)} \Pi \left (\frac{\sqrt{b-a}}{\sqrt{a+b}};\left .\sin ^{-1}\left (\frac{\sqrt{g \sin (e+f x)}}{\sqrt{g} \sqrt{\cos (e+f x)+1}}\right )\right |-1\right )}{b f \sqrt{b-a} \sqrt{a+b} \sqrt{d \cos (e+f x)}}-\frac{\sqrt{d} \sqrt{g} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d} \sqrt{g \sin (e+f x)}}{\sqrt{g} \sqrt{d \cos (e+f x)}}\right )}{\sqrt{2} b f}+\frac{\sqrt{d} \sqrt{g} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d} \sqrt{g \sin (e+f x)}}{\sqrt{g} \sqrt{d \cos (e+f x)}}+1\right )}{\sqrt{2} b f}+\frac{\sqrt{d} \sqrt{g} \log \left (-\frac{\sqrt{2} \sqrt{d} \sqrt{g \sin (e+f x)}}{\sqrt{d \cos (e+f x)}}+\sqrt{g} \tan (e+f x)+\sqrt{g}\right )}{2 \sqrt{2} b f}-\frac{\sqrt{d} \sqrt{g} \log \left (\frac{\sqrt{2} \sqrt{d} \sqrt{g \sin (e+f x)}}{\sqrt{d \cos (e+f x)}}+\sqrt{g} \tan (e+f x)+\sqrt{g}\right )}{2 \sqrt{2} b f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[d]*Sqrt[g]*ArcTan[1 - (Sqrt[2]*Sqrt[d]*Sqrt[g*Sin[e + f*x]])/(Sqrt[g]*Sqrt[d*Cos[e + f*x]])])/(Sqrt[2]
*b*f)) + (Sqrt[d]*Sqrt[g]*ArcTan[1 + (Sqrt[2]*Sqrt[d]*Sqrt[g*Sin[e + f*x]])/(Sqrt[g]*Sqrt[d*Cos[e + f*x]])])/(
Sqrt[2]*b*f) + (2*Sqrt[2]*a*d*Sqrt[g]*Sqrt[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])/(b*Sqrt[-a + b]*Sqrt[a + b]*f*Sqrt[d*Cos[e + f*x]]) - (
2*Sqrt[2]*a*d*Sqrt[g]*Sqrt[Cos[e + f*x]]*EllipticPi[Sqrt[-a + b]/Sqrt[a + b], ArcSin[Sqrt[g*Sin[e + f*x]]/(Sqr
t[g]*Sqrt[1 + Cos[e + f*x]])], -1])/(b*Sqrt[-a + b]*Sqrt[a + b]*f*Sqrt[d*Cos[e + f*x]]) + (Sqrt[d]*Sqrt[g]*Log
[Sqrt[g] - (Sqrt[2]*Sqrt[d]*Sqrt[g*Sin[e + f*x]])/Sqrt[d*Cos[e + f*x]] + Sqrt[g]*Tan[e + f*x]])/(2*Sqrt[2]*b*f
) - (Sqrt[d]*Sqrt[g]*Log[Sqrt[g] + (Sqrt[2]*Sqrt[d]*Sqrt[g*Sin[e + f*x]])/Sqrt[d*Cos[e + f*x]] + Sqrt[g]*Tan[e
+ f*x]])/(2*Sqrt[2]*b*f)

Rule 2909

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_))/((a_) + (b_.)*sin[(e_.) + (f_.
)*(x_)]), x_Symbol] :> Dist[d/b, Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n - 1), x], x] - Dist[(a*d)/b, 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] && N
eQ[a^2 - b^2, 0] && IntegersQ[2*n, 2*p] && LtQ[-1, p, 1] && GtQ[n, 0]

Rule 2574

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> With[{k = Denomina
tor[m]}, Dist[(k*a*b)/f, Subst[Int[x^(k*(m + 1) - 1)/(a^2 + b^2*x^(2*k)), x], x, (a*Sin[e + f*x])^(1/k)/(b*Cos
[e + f*x])^(1/k)], x]] /; FreeQ[{a, b, e, f}, x] && EqQ[m + n, 0] && GtQ[m, 0] && LtQ[m, 1]

Rule 297

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

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{d \cos (e+f x)} \sqrt{g \sin (e+f x)}}{a+b \cos (e+f x)} \, dx &=\frac{d \int \frac{\sqrt{g \sin (e+f x)}}{\sqrt{d \cos (e+f x)}} \, dx}{b}-\frac{(a d) \int \frac{\sqrt{g \sin (e+f x)}}{\sqrt{d \cos (e+f x)} (a+b \cos (e+f x))} \, dx}{b}\\ &=\frac{\left (2 d^2 g\right ) \operatorname{Subst}\left (\int \frac{x^2}{g^2+d^2 x^4} \, dx,x,\frac{\sqrt{g \sin (e+f x)}}{\sqrt{d \cos (e+f x)}}\right )}{b f}-\frac{\left (a d \sqrt{\cos (e+f x)}\right ) \int \frac{\sqrt{g \sin (e+f x)}}{\sqrt{\cos (e+f x)} (a+b \cos (e+f x))} \, dx}{b \sqrt{d \cos (e+f x)}}\\ &=-\frac{(d g) \operatorname{Subst}\left (\int \frac{g-d x^2}{g^2+d^2 x^4} \, dx,x,\frac{\sqrt{g \sin (e+f x)}}{\sqrt{d \cos (e+f x)}}\right )}{b f}+\frac{(d g) \operatorname{Subst}\left (\int \frac{g+d x^2}{g^2+d^2 x^4} \, dx,x,\frac{\sqrt{g \sin (e+f x)}}{\sqrt{d \cos (e+f x)}}\right )}{b f}-\frac{\left (4 \sqrt{2} a d 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 )}{b f \sqrt{d \cos (e+f x)}}\\ &=\frac{\left (\sqrt{d} \sqrt{g}\right ) \operatorname{Subst}\left (\int \frac{\frac{\sqrt{2} \sqrt{g}}{\sqrt{d}}+2 x}{-\frac{g}{d}-\frac{\sqrt{2} \sqrt{g} x}{\sqrt{d}}-x^2} \, dx,x,\frac{\sqrt{g \sin (e+f x)}}{\sqrt{d \cos (e+f x)}}\right )}{2 \sqrt{2} b f}+\frac{\left (\sqrt{d} \sqrt{g}\right ) \operatorname{Subst}\left (\int \frac{\frac{\sqrt{2} \sqrt{g}}{\sqrt{d}}-2 x}{-\frac{g}{d}+\frac{\sqrt{2} \sqrt{g} x}{\sqrt{d}}-x^2} \, dx,x,\frac{\sqrt{g \sin (e+f x)}}{\sqrt{d \cos (e+f x)}}\right )}{2 \sqrt{2} b f}+\frac{g \operatorname{Subst}\left (\int \frac{1}{\frac{g}{d}-\frac{\sqrt{2} \sqrt{g} x}{\sqrt{d}}+x^2} \, dx,x,\frac{\sqrt{g \sin (e+f x)}}{\sqrt{d \cos (e+f x)}}\right )}{2 b f}+\frac{g \operatorname{Subst}\left (\int \frac{1}{\frac{g}{d}+\frac{\sqrt{2} \sqrt{g} x}{\sqrt{d}}+x^2} \, dx,x,\frac{\sqrt{g \sin (e+f x)}}{\sqrt{d \cos (e+f x)}}\right )}{2 b f}-\frac{\left (2 \sqrt{2} a d 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 )}{b \sqrt{-a+b} f \sqrt{d \cos (e+f x)}}+\frac{\left (2 \sqrt{2} a d 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 )}{b \sqrt{-a+b} f \sqrt{d \cos (e+f x)}}\\ &=\frac{2 \sqrt{2} a d \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 )}{b \sqrt{-a+b} \sqrt{a+b} f \sqrt{d \cos (e+f x)}}-\frac{2 \sqrt{2} a d \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 )}{b \sqrt{-a+b} \sqrt{a+b} f \sqrt{d \cos (e+f x)}}+\frac{\sqrt{d} \sqrt{g} \log \left (\sqrt{g}-\frac{\sqrt{2} \sqrt{d} \sqrt{g \sin (e+f x)}}{\sqrt{d \cos (e+f x)}}+\sqrt{g} \tan (e+f x)\right )}{2 \sqrt{2} b f}-\frac{\sqrt{d} \sqrt{g} \log \left (\sqrt{g}+\frac{\sqrt{2} \sqrt{d} \sqrt{g \sin (e+f x)}}{\sqrt{d \cos (e+f x)}}+\sqrt{g} \tan (e+f x)\right )}{2 \sqrt{2} b f}+\frac{\left (\sqrt{d} \sqrt{g}\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\frac{\sqrt{2} \sqrt{d} \sqrt{g \sin (e+f x)}}{\sqrt{g} \sqrt{d \cos (e+f x)}}\right )}{\sqrt{2} b f}-\frac{\left (\sqrt{d} \sqrt{g}\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\frac{\sqrt{2} \sqrt{d} \sqrt{g \sin (e+f x)}}{\sqrt{g} \sqrt{d \cos (e+f x)}}\right )}{\sqrt{2} b f}\\ &=-\frac{\sqrt{d} \sqrt{g} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d} \sqrt{g \sin (e+f x)}}{\sqrt{g} \sqrt{d \cos (e+f x)}}\right )}{\sqrt{2} b f}+\frac{\sqrt{d} \sqrt{g} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{d} \sqrt{g \sin (e+f x)}}{\sqrt{g} \sqrt{d \cos (e+f x)}}\right )}{\sqrt{2} b f}+\frac{2 \sqrt{2} a d \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 )}{b \sqrt{-a+b} \sqrt{a+b} f \sqrt{d \cos (e+f x)}}-\frac{2 \sqrt{2} a d \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 )}{b \sqrt{-a+b} \sqrt{a+b} f \sqrt{d \cos (e+f x)}}+\frac{\sqrt{d} \sqrt{g} \log \left (\sqrt{g}-\frac{\sqrt{2} \sqrt{d} \sqrt{g \sin (e+f x)}}{\sqrt{d \cos (e+f x)}}+\sqrt{g} \tan (e+f x)\right )}{2 \sqrt{2} b f}-\frac{\sqrt{d} \sqrt{g} \log \left (\sqrt{g}+\frac{\sqrt{2} \sqrt{d} \sqrt{g \sin (e+f x)}}{\sqrt{d \cos (e+f x)}}+\sqrt{g} \tan (e+f x)\right )}{2 \sqrt{2} b f}\\ \end{align*}

Mathematica [C]  time = 4.88328, size = 272, normalized size = 0.53 \[ \frac{2 \sqrt{2} g \sqrt{\tan \left (\frac{1}{2} (e+f x)\right )} \sqrt{d \cos (e+f x)} \left (-i \sqrt{-a-b} \sqrt{a-b} \Pi \left (-i;\left .-\sin ^{-1}\left (\sqrt{\tan \left (\frac{1}{2} (e+f x)\right )}\right )\right |-1\right )+i \sqrt{-a-b} \sqrt{a-b} \Pi \left (i;\left .-\sin ^{-1}\left (\sqrt{\tan \left (\frac{1}{2} (e+f x)\right )}\right )\right |-1\right )+a \left (\Pi \left (-\frac{\sqrt{a-b}}{\sqrt{-a-b}};\left .-\sin ^{-1}\left (\sqrt{\tan \left (\frac{1}{2} (e+f x)\right )}\right )\right |-1\right )-\Pi \left (\frac{\sqrt{a-b}}{\sqrt{-a-b}};\left .-\sin ^{-1}\left (\sqrt{\tan \left (\frac{1}{2} (e+f x)\right )}\right )\right |-1\right )\right )\right )}{b f \sqrt{-a-b} \sqrt{a-b} \sqrt{\frac{\cos (e+f x)}{\cos (e+f x)+1}} \sqrt{g \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.662, size = 864, normalized size = 1.7 \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)*(g*sin(f*x+e))^(1/2)/(a+b*cos(f*x+e)),x)

[Out]

1/f*2^(1/2)*a/b/(a-b+(-a^2+b^2)^(1/2))/(b+(-a^2+b^2)^(1/2)-a)*(I*EllipticPi(((1-cos(f*x+e)+sin(f*x+e))/sin(f*x
+e))^(1/2),1/2-1/2*I,1/2*2^(1/2))*a-I*EllipticPi(((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2),1/2-1/2*I,1/2*2^
(1/2))*b-I*EllipticPi(((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2),1/2+1/2*I,1/2*2^(1/2))*a+I*EllipticPi(((1-c
os(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2),1/2+1/2*I,1/2*2^(1/2))*b-EllipticPi(((1-cos(f*x+e)+sin(f*x+e))/sin(f*x
+e))^(1/2),1/2-1/2*I,1/2*2^(1/2))*a+EllipticPi(((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2),1/2-1/2*I,1/2*2^(1
/2))*b-EllipticPi(((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2),1/2+1/2*I,1/2*2^(1/2))*a+EllipticPi(((1-cos(f*x
+e)+sin(f*x+e))/sin(f*x+e))^(1/2),1/2+1/2*I,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^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*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)*sin(f*x+e)*(d*cos(f*x+e))^(1/2)*(g*sin(f*x+e))^(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)/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 )} \sqrt{g \sin \left (f x + e\right )}}{b \cos \left (f x + e\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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