3.1410 \(\int \frac {\sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}}{a+b \sin (e+f x)} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.83, 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 = {2909, 2575, 297, 1162, 617, 204, 1165, 628, 2906, 2905, 490, 1218} \[ -\frac {2 \sqrt {2} a d \sqrt {g} \sqrt {\sin (e+f x)} \Pi \left (-\frac {\sqrt {b-a}}{\sqrt {a+b}};\left .\sin ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt {\sin (e+f x)+1}}\right )\right |-1\right )}{b f \sqrt {b-a} \sqrt {a+b} \sqrt {d \sin (e+f x)}}+\frac {2 \sqrt {2} a d \sqrt {g} \sqrt {\sin (e+f x)} \Pi \left (\frac {\sqrt {b-a}}{\sqrt {a+b}};\left .\sin ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt {\sin (e+f x)+1}}\right )\right |-1\right )}{b f \sqrt {b-a} \sqrt {a+b} \sqrt {d \sin (e+f x)}}+\frac {\sqrt {d} \sqrt {g} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt {d \sin (e+f x)}}\right )}{\sqrt {2} b f}-\frac {\sqrt {d} \sqrt {g} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {d} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt {d \sin (e+f x)}}+1\right )}{\sqrt {2} b f}-\frac {\sqrt {d} \sqrt {g} \log \left (-\frac {\sqrt {2} \sqrt {d} \sqrt {g \cos (e+f x)}}{\sqrt {d \sin (e+f x)}}+\sqrt {g} \cot (e+f x)+\sqrt {g}\right )}{2 \sqrt {2} b f}+\frac {\sqrt {d} \sqrt {g} \log \left (\frac {\sqrt {2} \sqrt {d} \sqrt {g \cos (e+f x)}}{\sqrt {d \sin (e+f x)}}+\sqrt {g} \cot (e+f x)+\sqrt {g}\right )}{2 \sqrt {2} b f} \]

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]

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

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 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 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 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 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 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 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 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 2575

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), 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*Cos[e + f*x])^(1/k)/(b*Si
n[e + f*x])^(1/k)], x]] /; FreeQ[{a, b, e, f}, x] && EqQ[m + n, 0] && GtQ[m, 0] && LtQ[m, 1]

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 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 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]

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

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Mathematica [C]  time = 11.03, size = 178, normalized size = 0.35 \[ \frac {2 (d \sin (e+f x))^{3/2} (g \cos (e+f x))^{3/2} \left (a+b \sqrt {\sin ^2(e+f x)}\right ) \left (b F_1\left (\frac {3}{4};-\frac {1}{4},1;\frac {7}{4};\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{b^2-a^2}\right )-a F_1\left (\frac {3}{4};\frac {1}{4},1;\frac {7}{4};\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{b^2-a^2}\right )\right )}{3 d f g \left (a^2-b^2\right ) \sin ^2(e+f x)^{3/4} (a+b \sin (e+f x))} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*(b*AppellF1[3/4, -1/4, 1, 7/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)] - a*AppellF1[3/4, 1/4, 1,
7/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)])*(g*Cos[e + f*x])^(3/2)*(d*Sin[e + f*x])^(3/2)*(a + b
*Sqrt[Sin[e + f*x]^2]))/(3*(a^2 - b^2)*d*f*g*(Sin[e + f*x]^2)^(3/4)*(a + b*Sin[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)*(g*cos(f*x+e))^(1/2)/(a+b*sin(f*x+e)),x, algorithm="fricas")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.86, size = 744, normalized size = 1.46 \[ \frac {\left (a -b \right ) \left (i \EllipticPi \left (\sqrt {-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \sqrt {-a^{2}+b^{2}}-i \EllipticPi \left (\sqrt {-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {1}{2}+\frac {i}{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 {1}{2}-\frac {i}{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 {1}{2}+\frac {i}{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}{a -b +\sqrt {-a^{2}+b^{2}}}, \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 )-\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}{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 )}{\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 )}}\, \sqrt {d \sin \left (f x +e \right )}\, \sqrt {g \cos \left (f x +e \right )}\, \sin \left (f x +e \right ) \sqrt {2}\, a}{f \cos \left (f x +e \right ) \left (-1+\cos \left (f x +e \right )\right ) b \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)*(g*cos(f*x+e))^(1/2)/(a+b*sin(f*x+e)),x)

[Out]

1/f*(a-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^2+b^2)^(1/2)
-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^2+b^2)^(1/2)+EllipticP
i((-(-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^2+b^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^2+b^2)^(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)+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))-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/(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))/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)*(d*sin(f*x+e))^(1/2)*(g*cos(f*x+e))^(1/2)*sin(f*x+e)/
cos(f*x+e)/(-1+cos(f*x+e))*2^(1/2)*a/b/(-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 \[ \int \frac {\sqrt {g \cos \left (f x + e\right )} \sqrt {d \sin \left (f x + e\right )}}{b \sin \left (f x + e\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(g*cos(f*x + e))*sqrt(d*sin(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 \[ \int \frac {\sqrt {g\,\cos \left (e+f\,x\right )}\,\sqrt {d\,\sin \left (e+f\,x\right )}}{a+b\,\sin \left (e+f\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(((g*cos(e + f*x))^(1/2)*(d*sin(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 \[ \int \frac {\sqrt {d \sin {\left (e + f x \right )}} \sqrt {g \cos {\left (e + f x \right )}}}{a + b \sin {\left (e + f x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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