3.24.7 \(\int \frac {\sqrt {d+e x}}{a+i b x+c x^2} \, dx\) [2307]

Optimal. Leaf size=629 \[ \frac {e \tanh ^{-1}\left (\frac {\sqrt {2 c d-i b e+2 \sqrt {c} \sqrt {c d^2-e (i b d-a e)}}-2 \sqrt {c} \sqrt {d+e x}}{\sqrt {2 c d-i b e-2 \sqrt {c} \sqrt {c d^2-e (i b d-a e)}}}\right )}{\sqrt {c} \sqrt {2 c d-i b e-2 \sqrt {c} \sqrt {c d^2-e (i b d-a e)}}}-\frac {e \tanh ^{-1}\left (\frac {\sqrt {2 c d-i b e+2 \sqrt {c} \sqrt {c d^2-e (i b d-a e)}}+2 \sqrt {c} \sqrt {d+e x}}{\sqrt {2 c d-i b e-2 \sqrt {c} \sqrt {c d^2-e (i b d-a e)}}}\right )}{\sqrt {c} \sqrt {2 c d-i b e-2 \sqrt {c} \sqrt {c d^2-e (i b d-a e)}}}+\frac {e \log \left (\sqrt {c d^2-e (i b d-a e)}-\sqrt {2 c d-i b e+2 \sqrt {c} \sqrt {c d^2-e (i b d-a e)}} \sqrt {d+e x}+\sqrt {c} (d+e x)\right )}{2 \sqrt {c} \sqrt {2 c d-i b e+2 \sqrt {c} \sqrt {c d^2-e (i b d-a e)}}}-\frac {e \log \left (\sqrt {c d^2-e (i b d-a e)}+\sqrt {2 c d-i b e+2 \sqrt {c} \sqrt {c d^2-e (i b d-a e)}} \sqrt {d+e x}+\sqrt {c} (d+e x)\right )}{2 \sqrt {c} \sqrt {2 c d-i b e+2 \sqrt {c} \sqrt {c d^2-e (i b d-a e)}}} \]

[Out]

e*arctanh((-2*c^(1/2)*(e*x+d)^(1/2)+(2*c*d-I*b*e+2*c^(1/2)*(c*d^2-e*(I*b*d-a*e))^(1/2))^(1/2))/(2*c*d-I*b*e-2*
c^(1/2)*(c*d^2-e*(I*b*d-a*e))^(1/2))^(1/2))/c^(1/2)/(2*c*d-I*b*e-2*c^(1/2)*(c*d^2-e*(I*b*d-a*e))^(1/2))^(1/2)-
e*arctanh((2*c^(1/2)*(e*x+d)^(1/2)+(2*c*d-I*b*e+2*c^(1/2)*(c*d^2-e*(I*b*d-a*e))^(1/2))^(1/2))/(2*c*d-I*b*e-2*c
^(1/2)*(c*d^2-e*(I*b*d-a*e))^(1/2))^(1/2))/c^(1/2)/(2*c*d-I*b*e-2*c^(1/2)*(c*d^2-e*(I*b*d-a*e))^(1/2))^(1/2)+1
/2*e*ln((e*x+d)*c^(1/2)+(c*d^2-e*(I*b*d-a*e))^(1/2)-(e*x+d)^(1/2)*(2*c*d-I*b*e+2*c^(1/2)*(c*d^2-e*(I*b*d-a*e))
^(1/2))^(1/2))/c^(1/2)/(2*c*d-I*b*e+2*c^(1/2)*(c*d^2-e*(I*b*d-a*e))^(1/2))^(1/2)-1/2*e*ln((e*x+d)*c^(1/2)+(c*d
^2-e*(I*b*d-a*e))^(1/2)+(e*x+d)^(1/2)*(2*c*d-I*b*e+2*c^(1/2)*(c*d^2-e*(I*b*d-a*e))^(1/2))^(1/2))/c^(1/2)/(2*c*
d-I*b*e+2*c^(1/2)*(c*d^2-e*(I*b*d-a*e))^(1/2))^(1/2)

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Rubi [A]
time = 0.83, antiderivative size = 629, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {713, 1143, 648, 632, 212, 642} \begin {gather*} \frac {e \log \left (-\sqrt {d+e x} \sqrt {2 \sqrt {c} \sqrt {c d^2-e (-a e+i b d)}-i b e+2 c d}+\sqrt {c d^2-e (-a e+i b d)}+\sqrt {c} (d+e x)\right )}{2 \sqrt {c} \sqrt {2 \sqrt {c} \sqrt {c d^2-e (-a e+i b d)}-i b e+2 c d}}-\frac {e \log \left (\sqrt {d+e x} \sqrt {2 \sqrt {c} \sqrt {c d^2-e (-a e+i b d)}-i b e+2 c d}+\sqrt {c d^2-e (-a e+i b d)}+\sqrt {c} (d+e x)\right )}{2 \sqrt {c} \sqrt {2 \sqrt {c} \sqrt {c d^2-e (-a e+i b d)}-i b e+2 c d}}+\frac {e \tanh ^{-1}\left (\frac {-2 \sqrt {c} \sqrt {d+e x}+\sqrt {2 \sqrt {c} \sqrt {c d^2-e (-a e+i b d)}-i b e+2 c d}}{\sqrt {-2 \sqrt {c} \sqrt {c d^2-e (-a e+i b d)}-i b e+2 c d}}\right )}{\sqrt {c} \sqrt {-2 \sqrt {c} \sqrt {c d^2-e (-a e+i b d)}-i b e+2 c d}}-\frac {e \tanh ^{-1}\left (\frac {2 \sqrt {c} \sqrt {d+e x}+\sqrt {2 \sqrt {c} \sqrt {c d^2-e (-a e+i b d)}-i b e+2 c d}}{\sqrt {-2 \sqrt {c} \sqrt {c d^2-e (-a e+i b d)}-i b e+2 c d}}\right )}{\sqrt {c} \sqrt {-2 \sqrt {c} \sqrt {c d^2-e (-a e+i b d)}-i b e+2 c d}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[d + e*x]/(a + I*b*x + c*x^2),x]

[Out]

(e*ArcTanh[(Sqrt[2*c*d - I*b*e + 2*Sqrt[c]*Sqrt[c*d^2 - e*(I*b*d - a*e)]] - 2*Sqrt[c]*Sqrt[d + e*x])/Sqrt[2*c*
d - I*b*e - 2*Sqrt[c]*Sqrt[c*d^2 - e*(I*b*d - a*e)]]])/(Sqrt[c]*Sqrt[2*c*d - I*b*e - 2*Sqrt[c]*Sqrt[c*d^2 - e*
(I*b*d - a*e)]]) - (e*ArcTanh[(Sqrt[2*c*d - I*b*e + 2*Sqrt[c]*Sqrt[c*d^2 - e*(I*b*d - a*e)]] + 2*Sqrt[c]*Sqrt[
d + e*x])/Sqrt[2*c*d - I*b*e - 2*Sqrt[c]*Sqrt[c*d^2 - e*(I*b*d - a*e)]]])/(Sqrt[c]*Sqrt[2*c*d - I*b*e - 2*Sqrt
[c]*Sqrt[c*d^2 - e*(I*b*d - a*e)]]) + (e*Log[Sqrt[c*d^2 - e*(I*b*d - a*e)] - Sqrt[2*c*d - I*b*e + 2*Sqrt[c]*Sq
rt[c*d^2 - e*(I*b*d - a*e)]]*Sqrt[d + e*x] + Sqrt[c]*(d + e*x)])/(2*Sqrt[c]*Sqrt[2*c*d - I*b*e + 2*Sqrt[c]*Sqr
t[c*d^2 - e*(I*b*d - a*e)]]) - (e*Log[Sqrt[c*d^2 - e*(I*b*d - a*e)] + Sqrt[2*c*d - I*b*e + 2*Sqrt[c]*Sqrt[c*d^
2 - e*(I*b*d - a*e)]]*Sqrt[d + e*x] + Sqrt[c]*(d + e*x)])/(2*Sqrt[c]*Sqrt[2*c*d - I*b*e + 2*Sqrt[c]*Sqrt[c*d^2
- e*(I*b*d - a*e)]])

Rule 212

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

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; 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 648

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 713

Int[Sqrt[(d_.) + (e_.)*(x_)]/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[2*e, Subst[Int[x^2/(c*d^2
- b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^
2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0]

Rule 1143

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a/c, 2]}, With[{r = Rt[2*q - b/
c, 2]}, Dist[1/(2*c*r), Int[x^(m - 1)/(q - r*x + x^2), x], x] - Dist[1/(2*c*r), Int[x^(m - 1)/(q + r*x + x^2),
x], x]]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && GeQ[m, 1] && LtQ[m, 3] && NegQ[b^2 - 4*a*c]

Rubi steps

\begin {align*} \int \frac {\sqrt {d+e x}}{a+i b x+c x^2} \, dx &=(2 e) \text {Subst}\left (\int \frac {x^2}{c d^2-i b d e+a e^2+(-2 c d+i b e) x^2+c x^4} \, dx,x,\sqrt {d+e x}\right )\\ &=\frac {e \text {Subst}\left (\int \frac {x}{\frac {\sqrt {c d^2-i b d e+a e^2}}{\sqrt {c}}-\frac {\sqrt {2 c d-i b e+2 \sqrt {c} \sqrt {c d^2-i b d e+a e^2}} x}{\sqrt {c}}+x^2} \, dx,x,\sqrt {d+e x}\right )}{\sqrt {c} \sqrt {2 c d-i b e+2 \sqrt {c} \sqrt {c d^2-e (i b d-a e)}}}-\frac {e \text {Subst}\left (\int \frac {x}{\frac {\sqrt {c d^2-i b d e+a e^2}}{\sqrt {c}}+\frac {\sqrt {2 c d-i b e+2 \sqrt {c} \sqrt {c d^2-i b d e+a e^2}} x}{\sqrt {c}}+x^2} \, dx,x,\sqrt {d+e x}\right )}{\sqrt {c} \sqrt {2 c d-i b e+2 \sqrt {c} \sqrt {c d^2-e (i b d-a e)}}}\\ &=\frac {e \text {Subst}\left (\int \frac {1}{\frac {\sqrt {c d^2-i b d e+a e^2}}{\sqrt {c}}-\frac {\sqrt {2 c d-i b e+2 \sqrt {c} \sqrt {c d^2-i b d e+a e^2}} x}{\sqrt {c}}+x^2} \, dx,x,\sqrt {d+e x}\right )}{2 c}+\frac {e \text {Subst}\left (\int \frac {1}{\frac {\sqrt {c d^2-i b d e+a e^2}}{\sqrt {c}}+\frac {\sqrt {2 c d-i b e+2 \sqrt {c} \sqrt {c d^2-i b d e+a e^2}} x}{\sqrt {c}}+x^2} \, dx,x,\sqrt {d+e x}\right )}{2 c}+\frac {e \text {Subst}\left (\int \frac {-\frac {\sqrt {2 c d-i b e+2 \sqrt {c} \sqrt {c d^2-i b d e+a e^2}}}{\sqrt {c}}+2 x}{\frac {\sqrt {c d^2-i b d e+a e^2}}{\sqrt {c}}-\frac {\sqrt {2 c d-i b e+2 \sqrt {c} \sqrt {c d^2-i b d e+a e^2}} x}{\sqrt {c}}+x^2} \, dx,x,\sqrt {d+e x}\right )}{2 \sqrt {c} \sqrt {2 c d-i b e+2 \sqrt {c} \sqrt {c d^2-e (i b d-a e)}}}-\frac {e \text {Subst}\left (\int \frac {\frac {\sqrt {2 c d-i b e+2 \sqrt {c} \sqrt {c d^2-i b d e+a e^2}}}{\sqrt {c}}+2 x}{\frac {\sqrt {c d^2-i b d e+a e^2}}{\sqrt {c}}+\frac {\sqrt {2 c d-i b e+2 \sqrt {c} \sqrt {c d^2-i b d e+a e^2}} x}{\sqrt {c}}+x^2} \, dx,x,\sqrt {d+e x}\right )}{2 \sqrt {c} \sqrt {2 c d-i b e+2 \sqrt {c} \sqrt {c d^2-e (i b d-a e)}}}\\ &=\frac {e \log \left (\sqrt {c d^2-e (i b d-a e)}-\sqrt {2 c d-i b e+2 \sqrt {c} \sqrt {c d^2-e (i b d-a e)}} \sqrt {d+e x}+\sqrt {c} (d+e x)\right )}{2 \sqrt {c} \sqrt {2 c d-i b e+2 \sqrt {c} \sqrt {c d^2-e (i b d-a e)}}}-\frac {e \log \left (\sqrt {c d^2-e (i b d-a e)}+\sqrt {2 c d-i b e+2 \sqrt {c} \sqrt {c d^2-e (i b d-a e)}} \sqrt {d+e x}+\sqrt {c} (d+e x)\right )}{2 \sqrt {c} \sqrt {2 c d-i b e+2 \sqrt {c} \sqrt {c d^2-e (i b d-a e)}}}-\frac {e \text {Subst}\left (\int \frac {1}{2 d-\frac {i b e}{c}-\frac {2 \sqrt {c d^2-e (i b d-a e)}}{\sqrt {c}}-x^2} \, dx,x,-\frac {\sqrt {2 c d-i b e+2 \sqrt {c} \sqrt {c d^2-i b d e+a e^2}}}{\sqrt {c}}+2 \sqrt {d+e x}\right )}{c}-\frac {e \text {Subst}\left (\int \frac {1}{2 d-\frac {i b e}{c}-\frac {2 \sqrt {c d^2-e (i b d-a e)}}{\sqrt {c}}-x^2} \, dx,x,\frac {\sqrt {2 c d-i b e+2 \sqrt {c} \sqrt {c d^2-i b d e+a e^2}}}{\sqrt {c}}+2 \sqrt {d+e x}\right )}{c}\\ &=\frac {e \tanh ^{-1}\left (\frac {\sqrt {c} \left (\frac {\sqrt {2 c d-i b e+2 \sqrt {c} \sqrt {c d^2-e (i b d-a e)}}}{\sqrt {c}}-2 \sqrt {d+e x}\right )}{\sqrt {2 c d-i b e-2 \sqrt {c} \sqrt {c d^2-e (i b d-a e)}}}\right )}{\sqrt {c} \sqrt {2 c d-i b e-2 \sqrt {c} \sqrt {c d^2-e (i b d-a e)}}}-\frac {e \tanh ^{-1}\left (\frac {\sqrt {c} \left (\frac {\sqrt {2 c d-i b e+2 \sqrt {c} \sqrt {c d^2-e (i b d-a e)}}}{\sqrt {c}}+2 \sqrt {d+e x}\right )}{\sqrt {2 c d-i b e-2 \sqrt {c} \sqrt {c d^2-e (i b d-a e)}}}\right )}{\sqrt {c} \sqrt {2 c d-i b e-2 \sqrt {c} \sqrt {c d^2-e (i b d-a e)}}}+\frac {e \log \left (\sqrt {c d^2-e (i b d-a e)}-\sqrt {2 c d-i b e+2 \sqrt {c} \sqrt {c d^2-e (i b d-a e)}} \sqrt {d+e x}+\sqrt {c} (d+e x)\right )}{2 \sqrt {c} \sqrt {2 c d-i b e+2 \sqrt {c} \sqrt {c d^2-e (i b d-a e)}}}-\frac {e \log \left (\sqrt {c d^2-e (i b d-a e)}+\sqrt {2 c d-i b e+2 \sqrt {c} \sqrt {c d^2-e (i b d-a e)}} \sqrt {d+e x}+\sqrt {c} (d+e x)\right )}{2 \sqrt {c} \sqrt {2 c d-i b e+2 \sqrt {c} \sqrt {c d^2-e (i b d-a e)}}}\\ \end {align*}

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Mathematica [A]
time = 0.64, size = 232, normalized size = 0.37 \begin {gather*} \frac {\sqrt {2} \left (\frac {\left (-2 i c d+\left (-b+\sqrt {b^2+4 a c}\right ) e\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {d+e x}}{\sqrt {-2 c d-i \left (-b+\sqrt {b^2+4 a c}\right ) e}}\right )}{\sqrt {-2 c d-i \left (-b+\sqrt {b^2+4 a c}\right ) e}}+\frac {\left (2 i c d+\left (b+\sqrt {b^2+4 a c}\right ) e\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {d+e x}}{\sqrt {-2 c d+i \left (b+\sqrt {b^2+4 a c}\right ) e}}\right )}{\sqrt {-2 c d+i \left (b+\sqrt {b^2+4 a c}\right ) e}}\right )}{\sqrt {c} \sqrt {b^2+4 a c}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[d + e*x]/(a + I*b*x + c*x^2),x]

[Out]

(Sqrt[2]*((((-2*I)*c*d + (-b + Sqrt[b^2 + 4*a*c])*e)*ArcTan[(Sqrt[2]*Sqrt[c]*Sqrt[d + e*x])/Sqrt[-2*c*d - I*(-
b + Sqrt[b^2 + 4*a*c])*e]])/Sqrt[-2*c*d - I*(-b + Sqrt[b^2 + 4*a*c])*e] + (((2*I)*c*d + (b + Sqrt[b^2 + 4*a*c]
)*e)*ArcTan[(Sqrt[2]*Sqrt[c]*Sqrt[d + e*x])/Sqrt[-2*c*d + I*(b + Sqrt[b^2 + 4*a*c])*e]])/Sqrt[-2*c*d + I*(b +
Sqrt[b^2 + 4*a*c])*e]))/(Sqrt[c]*Sqrt[b^2 + 4*a*c])

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Maple [A]
time = 1.66, size = 691, normalized size = 1.10

method	result	size
derivativedivides	\(2 e \left (\frac {\frac {\ln \left (\left (e x +d \right ) \sqrt {c}-\sqrt {e x +d}\, \sqrt {2 \sqrt {-c \left (i b d e -e^{2} a -c \,d^{2}\right )}-i b e +2 c d}+\sqrt {-i b d e +e^{2} a +c \,d^{2}}\right )}{2 \sqrt {c}}+\frac {\sqrt {2 \sqrt {-c \left (i b d e -e^{2} a -c \,d^{2}\right )}-i b e +2 c d}\, \arctan \left (\frac {2 \sqrt {c}\, \sqrt {e x +d}-\sqrt {2 \sqrt {-c \left (i b d e -e^{2} a -c \,d^{2}\right )}-i b e +2 c d}}{\sqrt {4 \sqrt {-i b d e +e^{2} a +c \,d^{2}}\, \sqrt {c}-2 \sqrt {-c \left (i b d e -e^{2} a -c \,d^{2}\right )}+i b e -2 c d}}\right )}{\sqrt {c}\, \sqrt {4 \sqrt {-i b d e +e^{2} a +c \,d^{2}}\, \sqrt {c}-2 \sqrt {-c \left (i b d e -e^{2} a -c \,d^{2}\right )}+i b e -2 c d}}}{2 \sqrt {2 \sqrt {-c \left (i b d e -e^{2} a -c \,d^{2}\right )}-i b e +2 c d}}-\frac {\frac {\ln \left (\left (e x +d \right ) \sqrt {c}+\sqrt {e x +d}\, \sqrt {2 \sqrt {-c \left (i b d e -e^{2} a -c \,d^{2}\right )}-i b e +2 c d}+\sqrt {-i b d e +e^{2} a +c \,d^{2}}\right )}{2 \sqrt {c}}-\frac {\sqrt {2 \sqrt {-c \left (i b d e -e^{2} a -c \,d^{2}\right )}-i b e +2 c d}\, \arctan \left (\frac {2 \sqrt {c}\, \sqrt {e x +d}+\sqrt {2 \sqrt {-c \left (i b d e -e^{2} a -c \,d^{2}\right )}-i b e +2 c d}}{\sqrt {4 \sqrt {-i b d e +e^{2} a +c \,d^{2}}\, \sqrt {c}-2 \sqrt {-c \left (i b d e -e^{2} a -c \,d^{2}\right )}+i b e -2 c d}}\right )}{\sqrt {c}\, \sqrt {4 \sqrt {-i b d e +e^{2} a +c \,d^{2}}\, \sqrt {c}-2 \sqrt {-c \left (i b d e -e^{2} a -c \,d^{2}\right )}+i b e -2 c d}}}{2 \sqrt {2 \sqrt {-c \left (i b d e -e^{2} a -c \,d^{2}\right )}-i b e +2 c d}}\right )\)	\(691\)
default	\(2 e \left (\frac {\frac {\ln \left (\left (e x +d \right ) \sqrt {c}-\sqrt {e x +d}\, \sqrt {2 \sqrt {-c \left (i b d e -e^{2} a -c \,d^{2}\right )}-i b e +2 c d}+\sqrt {-i b d e +e^{2} a +c \,d^{2}}\right )}{2 \sqrt {c}}+\frac {\sqrt {2 \sqrt {-c \left (i b d e -e^{2} a -c \,d^{2}\right )}-i b e +2 c d}\, \arctan \left (\frac {2 \sqrt {c}\, \sqrt {e x +d}-\sqrt {2 \sqrt {-c \left (i b d e -e^{2} a -c \,d^{2}\right )}-i b e +2 c d}}{\sqrt {4 \sqrt {-i b d e +e^{2} a +c \,d^{2}}\, \sqrt {c}-2 \sqrt {-c \left (i b d e -e^{2} a -c \,d^{2}\right )}+i b e -2 c d}}\right )}{\sqrt {c}\, \sqrt {4 \sqrt {-i b d e +e^{2} a +c \,d^{2}}\, \sqrt {c}-2 \sqrt {-c \left (i b d e -e^{2} a -c \,d^{2}\right )}+i b e -2 c d}}}{2 \sqrt {2 \sqrt {-c \left (i b d e -e^{2} a -c \,d^{2}\right )}-i b e +2 c d}}-\frac {\frac {\ln \left (\left (e x +d \right ) \sqrt {c}+\sqrt {e x +d}\, \sqrt {2 \sqrt {-c \left (i b d e -e^{2} a -c \,d^{2}\right )}-i b e +2 c d}+\sqrt {-i b d e +e^{2} a +c \,d^{2}}\right )}{2 \sqrt {c}}-\frac {\sqrt {2 \sqrt {-c \left (i b d e -e^{2} a -c \,d^{2}\right )}-i b e +2 c d}\, \arctan \left (\frac {2 \sqrt {c}\, \sqrt {e x +d}+\sqrt {2 \sqrt {-c \left (i b d e -e^{2} a -c \,d^{2}\right )}-i b e +2 c d}}{\sqrt {4 \sqrt {-i b d e +e^{2} a +c \,d^{2}}\, \sqrt {c}-2 \sqrt {-c \left (i b d e -e^{2} a -c \,d^{2}\right )}+i b e -2 c d}}\right )}{\sqrt {c}\, \sqrt {4 \sqrt {-i b d e +e^{2} a +c \,d^{2}}\, \sqrt {c}-2 \sqrt {-c \left (i b d e -e^{2} a -c \,d^{2}\right )}+i b e -2 c d}}}{2 \sqrt {2 \sqrt {-c \left (i b d e -e^{2} a -c \,d^{2}\right )}-i b e +2 c d}}\right )\)	\(691\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(1/2)/(a+I*b*x+c*x^2),x,method=_RETURNVERBOSE)

[Out]

2*e*(1/2/(2*(-c*(I*b*d*e-e^2*a-c*d^2))^(1/2)-I*b*e+2*c*d)^(1/2)*(1/2/c^(1/2)*ln((e*x+d)*c^(1/2)-(e*x+d)^(1/2)*
(2*(-c*(I*b*d*e-e^2*a-c*d^2))^(1/2)-I*b*e+2*c*d)^(1/2)+(-I*b*d*e+e^2*a+c*d^2)^(1/2))+1/c^(1/2)*(2*(-c*(I*b*d*e
-e^2*a-c*d^2))^(1/2)-I*b*e+2*c*d)^(1/2)/(4*(-I*b*d*e+e^2*a+c*d^2)^(1/2)*c^(1/2)-2*(-c*(I*b*d*e-e^2*a-c*d^2))^(
1/2)+I*e*b-2*c*d)^(1/2)*arctan((2*c^(1/2)*(e*x+d)^(1/2)-(2*(-c*(I*b*d*e-e^2*a-c*d^2))^(1/2)-I*b*e+2*c*d)^(1/2)
)/(4*(-I*b*d*e+e^2*a+c*d^2)^(1/2)*c^(1/2)-2*(-c*(I*b*d*e-e^2*a-c*d^2))^(1/2)+I*e*b-2*c*d)^(1/2)))-1/2/(2*(-c*(
I*b*d*e-e^2*a-c*d^2))^(1/2)-I*b*e+2*c*d)^(1/2)*(1/2/c^(1/2)*ln((e*x+d)*c^(1/2)+(e*x+d)^(1/2)*(2*(-c*(I*b*d*e-e
^2*a-c*d^2))^(1/2)-I*b*e+2*c*d)^(1/2)+(-I*b*d*e+e^2*a+c*d^2)^(1/2))-1/c^(1/2)*(2*(-c*(I*b*d*e-e^2*a-c*d^2))^(1
/2)-I*b*e+2*c*d)^(1/2)/(4*(-I*b*d*e+e^2*a+c*d^2)^(1/2)*c^(1/2)-2*(-c*(I*b*d*e-e^2*a-c*d^2))^(1/2)+I*e*b-2*c*d)
^(1/2)*arctan((2*c^(1/2)*(e*x+d)^(1/2)+(2*(-c*(I*b*d*e-e^2*a-c*d^2))^(1/2)-I*b*e+2*c*d)^(1/2))/(4*(-I*b*d*e+e^
2*a+c*d^2)^(1/2)*c^(1/2)-2*(-c*(I*b*d*e-e^2*a-c*d^2))^(1/2)+I*e*b-2*c*d)^(1/2))))

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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((e*x+d)^(1/2)/(a+I*b*x+c*x^2),x, algorithm="maxima")

[Out]

integrate(sqrt(x*e + d)/(c*x^2 + I*b*x + a), x)

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Fricas [A]
time = 4.88, size = 753, normalized size = 1.20 \begin {gather*} -\frac {1}{2} \, \sqrt {2} \sqrt {-\frac {2 \, c d - i \, b e + {\left (b^{2} c + 4 \, a c^{2}\right )} \sqrt {-\frac {e^{2}}{b^{2} c^{2} + 4 \, a c^{3}}}}{b^{2} c + 4 \, a c^{2}}} \log \left (\frac {1}{2} \, {\left (\sqrt {2} {\left (b^{2} c + 4 \, a c^{2}\right )} \sqrt {-\frac {2 \, c d - i \, b e + {\left (b^{2} c + 4 \, a c^{2}\right )} \sqrt {-\frac {e^{2}}{b^{2} c^{2} + 4 \, a c^{3}}}}{b^{2} c + 4 \, a c^{2}}} \sqrt {-\frac {e^{2}}{b^{2} c^{2} + 4 \, a c^{3}}} + 2 \, \sqrt {x e + d} e\right )} e^{\left (-1\right )}\right ) + \frac {1}{2} \, \sqrt {2} \sqrt {-\frac {2 \, c d - i \, b e + {\left (b^{2} c + 4 \, a c^{2}\right )} \sqrt {-\frac {e^{2}}{b^{2} c^{2} + 4 \, a c^{3}}}}{b^{2} c + 4 \, a c^{2}}} \log \left (-\frac {1}{2} \, {\left (\sqrt {2} {\left (b^{2} c + 4 \, a c^{2}\right )} \sqrt {-\frac {2 \, c d - i \, b e + {\left (b^{2} c + 4 \, a c^{2}\right )} \sqrt {-\frac {e^{2}}{b^{2} c^{2} + 4 \, a c^{3}}}}{b^{2} c + 4 \, a c^{2}}} \sqrt {-\frac {e^{2}}{b^{2} c^{2} + 4 \, a c^{3}}} - 2 \, \sqrt {x e + d} e\right )} e^{\left (-1\right )}\right ) + \frac {1}{2} \, \sqrt {2} \sqrt {-\frac {2 \, c d - i \, b e - {\left (b^{2} c + 4 \, a c^{2}\right )} \sqrt {-\frac {e^{2}}{b^{2} c^{2} + 4 \, a c^{3}}}}{b^{2} c + 4 \, a c^{2}}} \log \left (\frac {1}{2} \, {\left (\sqrt {2} {\left (b^{2} c + 4 \, a c^{2}\right )} \sqrt {-\frac {2 \, c d - i \, b e - {\left (b^{2} c + 4 \, a c^{2}\right )} \sqrt {-\frac {e^{2}}{b^{2} c^{2} + 4 \, a c^{3}}}}{b^{2} c + 4 \, a c^{2}}} \sqrt {-\frac {e^{2}}{b^{2} c^{2} + 4 \, a c^{3}}} + 2 \, \sqrt {x e + d} e\right )} e^{\left (-1\right )}\right ) - \frac {1}{2} \, \sqrt {2} \sqrt {-\frac {2 \, c d - i \, b e - {\left (b^{2} c + 4 \, a c^{2}\right )} \sqrt {-\frac {e^{2}}{b^{2} c^{2} + 4 \, a c^{3}}}}{b^{2} c + 4 \, a c^{2}}} \log \left (-\frac {1}{2} \, {\left (\sqrt {2} {\left (b^{2} c + 4 \, a c^{2}\right )} \sqrt {-\frac {2 \, c d - i \, b e - {\left (b^{2} c + 4 \, a c^{2}\right )} \sqrt {-\frac {e^{2}}{b^{2} c^{2} + 4 \, a c^{3}}}}{b^{2} c + 4 \, a c^{2}}} \sqrt {-\frac {e^{2}}{b^{2} c^{2} + 4 \, a c^{3}}} - 2 \, \sqrt {x e + d} e\right )} e^{\left (-1\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(1/2)/(a+I*b*x+c*x^2),x, algorithm="fricas")

[Out]

-1/2*sqrt(2)*sqrt(-(2*c*d - I*b*e + (b^2*c + 4*a*c^2)*sqrt(-e^2/(b^2*c^2 + 4*a*c^3)))/(b^2*c + 4*a*c^2))*log(1
/2*(sqrt(2)*(b^2*c + 4*a*c^2)*sqrt(-(2*c*d - I*b*e + (b^2*c + 4*a*c^2)*sqrt(-e^2/(b^2*c^2 + 4*a*c^3)))/(b^2*c
+ 4*a*c^2))*sqrt(-e^2/(b^2*c^2 + 4*a*c^3)) + 2*sqrt(x*e + d)*e)*e^(-1)) + 1/2*sqrt(2)*sqrt(-(2*c*d - I*b*e + (
b^2*c + 4*a*c^2)*sqrt(-e^2/(b^2*c^2 + 4*a*c^3)))/(b^2*c + 4*a*c^2))*log(-1/2*(sqrt(2)*(b^2*c + 4*a*c^2)*sqrt(-
(2*c*d - I*b*e + (b^2*c + 4*a*c^2)*sqrt(-e^2/(b^2*c^2 + 4*a*c^3)))/(b^2*c + 4*a*c^2))*sqrt(-e^2/(b^2*c^2 + 4*a
*c^3)) - 2*sqrt(x*e + d)*e)*e^(-1)) + 1/2*sqrt(2)*sqrt(-(2*c*d - I*b*e - (b^2*c + 4*a*c^2)*sqrt(-e^2/(b^2*c^2
+ 4*a*c^3)))/(b^2*c + 4*a*c^2))*log(1/2*(sqrt(2)*(b^2*c + 4*a*c^2)*sqrt(-(2*c*d - I*b*e - (b^2*c + 4*a*c^2)*sq
rt(-e^2/(b^2*c^2 + 4*a*c^3)))/(b^2*c + 4*a*c^2))*sqrt(-e^2/(b^2*c^2 + 4*a*c^3)) + 2*sqrt(x*e + d)*e)*e^(-1)) -
1/2*sqrt(2)*sqrt(-(2*c*d - I*b*e - (b^2*c + 4*a*c^2)*sqrt(-e^2/(b^2*c^2 + 4*a*c^3)))/(b^2*c + 4*a*c^2))*log(-
1/2*(sqrt(2)*(b^2*c + 4*a*c^2)*sqrt(-(2*c*d - I*b*e - (b^2*c + 4*a*c^2)*sqrt(-e^2/(b^2*c^2 + 4*a*c^3)))/(b^2*c
+ 4*a*c^2))*sqrt(-e^2/(b^2*c^2 + 4*a*c^3)) - 2*sqrt(x*e + d)*e)*e^(-1))

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Sympy [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((e*x+d)**(1/2)/(a+I*b*x+c*x**2),x)

[Out]

Timed out

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(1/2)/(a+I*b*x+c*x^2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, need to choose a branch for the root of a polynomial with parameters. This might be wrong.Non regu
lar value [

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Mupad [B]
time = 1.82, size = 711, normalized size = 1.13 \begin {gather*} -2\,\mathrm {atanh}\left (\frac {\left (8\,c^2\,\sqrt {d+e\,x}\,\left (b^2\,e^4+b\,c\,d\,e^3\,2{}\mathrm {i}-2\,c^2\,d^2\,e^2+2\,a\,c\,e^4\right )-\frac {4\,c^2\,\sqrt {d+e\,x}\,\left (b^3\,c\,e^3\,1{}\mathrm {i}-2\,d\,b^2\,c^2\,e^2+4{}\mathrm {i}\,a\,b\,c^2\,e^3-8\,a\,d\,c^3\,e^2\right )\,\left (e\,\sqrt {-{\left (b^2+4\,a\,c\right )}^3}-b^3\,e\,1{}\mathrm {i}+8\,a\,c^2\,d+2\,b^2\,c\,d-a\,b\,c\,e\,4{}\mathrm {i}\right )}{16\,a^2\,c^3+8\,a\,b^2\,c^2+b^4\,c}\right )\,\sqrt {-\frac {e\,\sqrt {-{\left (b^2+4\,a\,c\right )}^3}-b^3\,e\,1{}\mathrm {i}+8\,a\,c^2\,d+2\,b^2\,c\,d-a\,b\,c\,e\,4{}\mathrm {i}}{2\,\left (16\,a^2\,c^3+8\,a\,b^2\,c^2+b^4\,c\right )}}}{8\,c^2\,\left (c\,d^2\,e^3-1{}\mathrm {i}\,b\,d\,e^4+a\,e^5\right )}\right )\,\sqrt {-\frac {e\,\sqrt {-{\left (b^2+4\,a\,c\right )}^3}-b^3\,e\,1{}\mathrm {i}+8\,a\,c^2\,d+2\,b^2\,c\,d-a\,b\,c\,e\,4{}\mathrm {i}}{2\,\left (16\,a^2\,c^3+8\,a\,b^2\,c^2+b^4\,c\right )}}-2\,\mathrm {atanh}\left (\frac {\left (8\,c^2\,\sqrt {d+e\,x}\,\left (b^2\,e^4+b\,c\,d\,e^3\,2{}\mathrm {i}-2\,c^2\,d^2\,e^2+2\,a\,c\,e^4\right )+\frac {4\,c^2\,\sqrt {d+e\,x}\,\left (b^3\,c\,e^3\,1{}\mathrm {i}-2\,d\,b^2\,c^2\,e^2+4{}\mathrm {i}\,a\,b\,c^2\,e^3-8\,a\,d\,c^3\,e^2\right )\,\left (e\,\sqrt {-{\left (b^2+4\,a\,c\right )}^3}+b^3\,e\,1{}\mathrm {i}-8\,a\,c^2\,d-2\,b^2\,c\,d+a\,b\,c\,e\,4{}\mathrm {i}\right )}{16\,a^2\,c^3+8\,a\,b^2\,c^2+b^4\,c}\right )\,\sqrt {\frac {e\,\sqrt {-{\left (b^2+4\,a\,c\right )}^3}+b^3\,e\,1{}\mathrm {i}-8\,a\,c^2\,d-2\,b^2\,c\,d+a\,b\,c\,e\,4{}\mathrm {i}}{2\,\left (16\,a^2\,c^3+8\,a\,b^2\,c^2+b^4\,c\right )}}}{8\,c^2\,\left (c\,d^2\,e^3-1{}\mathrm {i}\,b\,d\,e^4+a\,e^5\right )}\right )\,\sqrt {\frac {e\,\sqrt {-{\left (b^2+4\,a\,c\right )}^3}+b^3\,e\,1{}\mathrm {i}-8\,a\,c^2\,d-2\,b^2\,c\,d+a\,b\,c\,e\,4{}\mathrm {i}}{2\,\left (16\,a^2\,c^3+8\,a\,b^2\,c^2+b^4\,c\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^(1/2)/(a + b*x*1i + c*x^2),x)

[Out]

- 2*atanh(((8*c^2*(d + e*x)^(1/2)*(b^2*e^4 - 2*c^2*d^2*e^2 + 2*a*c*e^4 + b*c*d*e^3*2i) - (4*c^2*(d + e*x)^(1/2
)*(b^3*c*e^3*1i - 2*b^2*c^2*d*e^2 + a*b*c^2*e^3*4i - 8*a*c^3*d*e^2)*(e*(-(4*a*c + b^2)^3)^(1/2) - b^3*e*1i + 8
*a*c^2*d + 2*b^2*c*d - a*b*c*e*4i))/(b^4*c + 16*a^2*c^3 + 8*a*b^2*c^2))*(-(e*(-(4*a*c + b^2)^3)^(1/2) - b^3*e*
1i + 8*a*c^2*d + 2*b^2*c*d - a*b*c*e*4i)/(2*(b^4*c + 16*a^2*c^3 + 8*a*b^2*c^2)))^(1/2))/(8*c^2*(a*e^5 + c*d^2*
e^3 - b*d*e^4*1i)))*(-(e*(-(4*a*c + b^2)^3)^(1/2) - b^3*e*1i + 8*a*c^2*d + 2*b^2*c*d - a*b*c*e*4i)/(2*(b^4*c +
16*a^2*c^3 + 8*a*b^2*c^2)))^(1/2) - 2*atanh(((8*c^2*(d + e*x)^(1/2)*(b^2*e^4 - 2*c^2*d^2*e^2 + 2*a*c*e^4 + b*
c*d*e^3*2i) + (4*c^2*(d + e*x)^(1/2)*(b^3*c*e^3*1i - 2*b^2*c^2*d*e^2 + a*b*c^2*e^3*4i - 8*a*c^3*d*e^2)*(e*(-(4
*a*c + b^2)^3)^(1/2) + b^3*e*1i - 8*a*c^2*d - 2*b^2*c*d + a*b*c*e*4i))/(b^4*c + 16*a^2*c^3 + 8*a*b^2*c^2))*((e
*(-(4*a*c + b^2)^3)^(1/2) + b^3*e*1i - 8*a*c^2*d - 2*b^2*c*d + a*b*c*e*4i)/(2*(b^4*c + 16*a^2*c^3 + 8*a*b^2*c^
2)))^(1/2))/(8*c^2*(a*e^5 + c*d^2*e^3 - b*d*e^4*1i)))*((e*(-(4*a*c + b^2)^3)^(1/2) + b^3*e*1i - 8*a*c^2*d - 2*
b^2*c*d + a*b*c*e*4i)/(2*(b^4*c + 16*a^2*c^3 + 8*a*b^2*c^2)))^(1/2)

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