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

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

[Out]

1/2*arctan(-1+2^(1/2)*d^(1/2)*(g*sin(f*x+e))^(1/2)/g^(1/2)/(d*cos(f*x+e))^(1/2))*d^(1/2)*g^(1/2)/b/f*2^(1/2)+1
/2*arctan(1+2^(1/2)*d^(1/2)*(g*sin(f*x+e))^(1/2)/g^(1/2)/(d*cos(f*x+e))^(1/2))*d^(1/2)*g^(1/2)/b/f*2^(1/2)+1/4
*ln(g^(1/2)-2^(1/2)*d^(1/2)*(g*sin(f*x+e))^(1/2)/(d*cos(f*x+e))^(1/2)+g^(1/2)*tan(f*x+e))*d^(1/2)*g^(1/2)/b/f*
2^(1/2)-1/4*ln(g^(1/2)+2^(1/2)*d^(1/2)*(g*sin(f*x+e))^(1/2)/(d*cos(f*x+e))^(1/2)+g^(1/2)*tan(f*x+e))*d^(1/2)*g
^(1/2)/b/f*2^(1/2)+2*a*d*EllipticPi((g*sin(f*x+e))^(1/2)/g^(1/2)/(1+cos(f*x+e))^(1/2),-(-a+b)^(1/2)/(a+b)^(1/2
),I)*2^(1/2)*g^(1/2)*cos(f*x+e)^(1/2)/b/f/(-a+b)^(1/2)/(a+b)^(1/2)/(d*cos(f*x+e))^(1/2)-2*a*d*EllipticPi((g*si
n(f*x+e))^(1/2)/g^(1/2)/(1+cos(f*x+e))^(1/2),(-a+b)^(1/2)/(a+b)^(1/2),I)*2^(1/2)*g^(1/2)*cos(f*x+e)^(1/2)/b/f/
(-a+b)^(1/2)/(a+b)^(1/2)/(d*cos(f*x+e))^(1/2)

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Rubi [A]
time = 0.55, antiderivative size = 509, normalized size of antiderivative = 1.00, 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 = {2988, 2654, 303, 1176, 631, 210, 1179, 642, 2985, 2984, 504, 1232} \begin {gather*} \frac {2 \sqrt {2} a d \sqrt {g} \sqrt {\cos (e+f x)} \Pi \left (-\frac {\sqrt {b-a}}{\sqrt {a+b}};\left .\text {ArcSin}\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 .\text {ArcSin}\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} \text {ArcTan}\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} \text {ArcTan}\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} \end {gather*}

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 210

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

Rule 303

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 504

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), Int[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 631

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 642

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 1176

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 1179

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 1232

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[-c/a, 4]}, Simp[(1/(d*Sqrt[
a]*q))*EllipticPi[-e/(d*q^2), ArcSin[q*x], -1], x]] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] && GtQ[a, 0]

Rule 2654

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 2984

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 2985

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 2988

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]

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 ) \text {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) \text {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) \text {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 ) \text {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 ) \text {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 ) \text {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 \text {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 \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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*}

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Mathematica [C] Result contains complex when optimal does not.
time = 13.34, size = 264, normalized size = 0.52 \begin {gather*} \frac {2 \sqrt {2} g \sqrt {d \cos (e+f x)} \left (i \sqrt {-a-b} \sqrt {a-b} \Pi \left (-i;\left .\text {ArcSin}\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 .\text {ArcSin}\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 .\text {ArcSin}\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 .\text {ArcSin}\left (\sqrt {\tan \left (\frac {1}{2} (e+f x)\right )}\right )\right |-1\right )\right )\right ) \sqrt {\tan \left (\frac {1}{2} (e+f x)\right )}}{\sqrt {-a-b} \sqrt {a-b} b f \sqrt {\frac {\cos (e+f x)}{1+\cos (e+f x)}} \sqrt {g \sin (e+f x)}} \end {gather*}

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] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 879 vs. \(2 (393 ) = 786\).
time = 1.99, size = 880, normalized size = 1.73

method	result	size
default	\(\text {Expression too large to display}\)	\(880\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \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 )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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