### 3.2307 $$\int \frac{\sqrt{d+e x}}{a+i b x+c x^2} \, dx$$

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

[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)]])

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Rubi [A]  time = 1.22707, antiderivative size = 629, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 25, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.24, Rules used = {699, 1129, 634, 618, 206, 628} $\frac{e \log \left (-\sqrt{d+e x} \sqrt{2 \sqrt{c} \sqrt{c d^2-e (-a e+i b d)}-i b e+2 c d}+\sqrt{c d^2-e (-a e+i b d)}+\sqrt{c} (d+e x)\right )}{2 \sqrt{c} \sqrt{2 \sqrt{c} \sqrt{c d^2-e (-a e+i b d)}-i b e+2 c d}}-\frac{e \log \left (\sqrt{d+e x} \sqrt{2 \sqrt{c} \sqrt{c d^2-e (-a e+i b d)}-i b e+2 c d}+\sqrt{c d^2-e (-a e+i b d)}+\sqrt{c} (d+e x)\right )}{2 \sqrt{c} \sqrt{2 \sqrt{c} \sqrt{c d^2-e (-a e+i b d)}-i b e+2 c d}}+\frac{e \tanh ^{-1}\left (\frac{-2 \sqrt{c} \sqrt{d+e x}+\sqrt{2 \sqrt{c} \sqrt{c d^2-e (-a e+i b d)}-i b e+2 c d}}{\sqrt{-2 \sqrt{c} \sqrt{c d^2-e (-a e+i b d)}-i b e+2 c d}}\right )}{\sqrt{c} \sqrt{-2 \sqrt{c} \sqrt{c d^2-e (-a e+i b d)}-i b e+2 c d}}-\frac{e \tanh ^{-1}\left (\frac{2 \sqrt{c} \sqrt{d+e x}+\sqrt{2 \sqrt{c} \sqrt{c d^2-e (-a e+i b d)}-i b e+2 c d}}{\sqrt{-2 \sqrt{c} \sqrt{c d^2-e (-a e+i b d)}-i b e+2 c d}}\right )}{\sqrt{c} \sqrt{-2 \sqrt{c} \sqrt{c d^2-e (-a e+i b d)}-i b e+2 c d}}$

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 699

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 1129

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]

Rule 634

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 618

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 206

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

Rubi steps

\begin{align*} \int \frac{\sqrt{d+e x}}{a+i b x+c x^2} \, dx &=(2 e) \operatorname{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 \operatorname{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 \operatorname{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 \operatorname{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 \operatorname{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 \operatorname{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 \operatorname{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 \operatorname{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 \operatorname{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*}

Mathematica [A]  time = 0.498221, size = 197, normalized size = 0.31 $\frac{\sqrt{2} \left (\sqrt{2 c d-e \left (\sqrt{-4 a c-b^2}+i b\right )} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-e \left (\sqrt{-4 a c-b^2}+i b\right )}}\right )-\sqrt{e \sqrt{-4 a c-b^2}-i b e+2 c d} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{e \sqrt{-4 a c-b^2}-i b e+2 c d}}\right )\right )}{\sqrt{c} \sqrt{-4 a c-b^2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(Sqrt[2]*(-(Sqrt[2*c*d - I*b*e + Sqrt[-b^2 - 4*a*c]*e]*ArcTanh[(Sqrt[2]*Sqrt[c]*Sqrt[d + e*x])/Sqrt[2*c*d - I*
b*e + Sqrt[-b^2 - 4*a*c]*e]]) + Sqrt[2*c*d - (I*b + Sqrt[-b^2 - 4*a*c])*e]*ArcTanh[(Sqrt[2]*Sqrt[c]*Sqrt[d + e
*x])/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 = 0.379, size = 609, normalized size = 1. \begin{align*}{\frac{e}{2}\ln \left ( \left ( ex+d \right ) \sqrt{c}-\sqrt{ex+d}\sqrt{2\,\sqrt{-c \left ( ibde-a{e}^{2}-c{d}^{2} \right ) }-ibe+2\,cd}+\sqrt{-ibde+a{e}^{2}+c{d}^{2}} \right ){\frac{1}{\sqrt{2\,\sqrt{-c \left ( ibde-a{e}^{2}-c{d}^{2} \right ) }-ibe+2\,cd}}}{\frac{1}{\sqrt{c}}}}+{e\arctan \left ({ \left ( 2\,\sqrt{c}\sqrt{ex+d}-\sqrt{2\,\sqrt{-c \left ( ibde-a{e}^{2}-c{d}^{2} \right ) }-ibe+2\,cd} \right ){\frac{1}{\sqrt{4\,\sqrt{c}\sqrt{-ibde+a{e}^{2}+c{d}^{2}}-2\,\sqrt{-c \left ( ibde-a{e}^{2}-c{d}^{2} \right ) }+ibe-2\,cd}}}} \right ){\frac{1}{\sqrt{c}}}{\frac{1}{\sqrt{4\,\sqrt{c}\sqrt{-ibde+a{e}^{2}+c{d}^{2}}-2\,\sqrt{-c \left ( ibde-a{e}^{2}-c{d}^{2} \right ) }+ibe-2\,cd}}}}-{\frac{e}{2}\ln \left ( \left ( ex+d \right ) \sqrt{c}+\sqrt{ex+d}\sqrt{2\,\sqrt{-c \left ( ibde-a{e}^{2}-c{d}^{2} \right ) }-ibe+2\,cd}+\sqrt{-ibde+a{e}^{2}+c{d}^{2}} \right ){\frac{1}{\sqrt{2\,\sqrt{-c \left ( ibde-a{e}^{2}-c{d}^{2} \right ) }-ibe+2\,cd}}}{\frac{1}{\sqrt{c}}}}+{e\arctan \left ({ \left ( 2\,\sqrt{c}\sqrt{ex+d}+\sqrt{2\,\sqrt{-c \left ( ibde-a{e}^{2}-c{d}^{2} \right ) }-ibe+2\,cd} \right ){\frac{1}{\sqrt{4\,\sqrt{c}\sqrt{-ibde+a{e}^{2}+c{d}^{2}}-2\,\sqrt{-c \left ( ibde-a{e}^{2}-c{d}^{2} \right ) }+ibe-2\,cd}}}} \right ){\frac{1}{\sqrt{c}}}{\frac{1}{\sqrt{4\,\sqrt{c}\sqrt{-ibde+a{e}^{2}+c{d}^{2}}-2\,\sqrt{-c \left ( ibde-a{e}^{2}-c{d}^{2} \right ) }+ibe-2\,cd}}}} \end{align*}

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)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{e x + d}}{c x^{2} + i \, b x + a}\,{d x} \end{align*}

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(e*x + d)/(c*x^2 + I*b*x + a), x)

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Fricas [A]  time = 2.79583, size = 1515, normalized size = 2.41 \begin{align*} -\frac{1}{2} \, \sqrt{-\frac{4 \, c d - 2 i \, b e + 2 \,{\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{{\left (b^{2} c + 4 \, a c^{2}\right )} \sqrt{-\frac{e^{2}}{b^{2} c^{2} + 4 \, a c^{3}}} \sqrt{-\frac{4 \, c d - 2 i \, b e + 2 \,{\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}}} + 2 \, \sqrt{e x + d} e}{2 \, e}\right ) + \frac{1}{2} \, \sqrt{-\frac{4 \, c d - 2 i \, b e + 2 \,{\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{{\left (b^{2} c + 4 \, a c^{2}\right )} \sqrt{-\frac{e^{2}}{b^{2} c^{2} + 4 \, a c^{3}}} \sqrt{-\frac{4 \, c d - 2 i \, b e + 2 \,{\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}}} - 2 \, \sqrt{e x + d} e}{2 \, e}\right ) + \frac{1}{2} \, \sqrt{-\frac{4 \, c d - 2 i \, b e - 2 \,{\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{{\left (b^{2} c + 4 \, a c^{2}\right )} \sqrt{-\frac{e^{2}}{b^{2} c^{2} + 4 \, a c^{3}}} \sqrt{-\frac{4 \, c d - 2 i \, b e - 2 \,{\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}}} + 2 \, \sqrt{e x + d} e}{2 \, e}\right ) - \frac{1}{2} \, \sqrt{-\frac{4 \, c d - 2 i \, b e - 2 \,{\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{{\left (b^{2} c + 4 \, a c^{2}\right )} \sqrt{-\frac{e^{2}}{b^{2} c^{2} + 4 \, a c^{3}}} \sqrt{-\frac{4 \, c d - 2 i \, b e - 2 \,{\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}}} - 2 \, \sqrt{e x + d} e}{2 \, e}\right ) \end{align*}

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{d + e x}}{a + i b x + c x^{2}}\, dx \end{align*}

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]

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

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

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