3.6.40 \(\int \frac {\sqrt {d+e x}}{a+c x^2} \, dx\)

Optimal. Leaf size=478 \[ \frac {e \log \left (-\sqrt {2} \sqrt [4]{c} \sqrt {d+e x} \sqrt {\sqrt {a e^2+c d^2}+\sqrt {c} d}+\sqrt {a e^2+c d^2}+\sqrt {c} (d+e x)\right )}{2 \sqrt {2} c^{3/4} \sqrt {\sqrt {a e^2+c d^2}+\sqrt {c} d}}-\frac {e \log \left (\sqrt {2} \sqrt [4]{c} \sqrt {d+e x} \sqrt {\sqrt {a e^2+c d^2}+\sqrt {c} d}+\sqrt {a e^2+c d^2}+\sqrt {c} (d+e x)\right )}{2 \sqrt {2} c^{3/4} \sqrt {\sqrt {a e^2+c d^2}+\sqrt {c} d}}+\frac {e \tanh ^{-1}\left (\frac {\sqrt {\sqrt {a e^2+c d^2}+\sqrt {c} d}-\sqrt {2} \sqrt [4]{c} \sqrt {d+e x}}{\sqrt {\sqrt {c} d-\sqrt {a e^2+c d^2}}}\right )}{\sqrt {2} c^{3/4} \sqrt {\sqrt {c} d-\sqrt {a e^2+c d^2}}}-\frac {e \tanh ^{-1}\left (\frac {\sqrt {\sqrt {a e^2+c d^2}+\sqrt {c} d}+\sqrt {2} \sqrt [4]{c} \sqrt {d+e x}}{\sqrt {\sqrt {c} d-\sqrt {a e^2+c d^2}}}\right )}{\sqrt {2} c^{3/4} \sqrt {\sqrt {c} d-\sqrt {a e^2+c d^2}}} \]

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Rubi [A]  time = 0.40, antiderivative size = 478, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {700, 1129, 634, 618, 206, 628} \begin {gather*} \frac {e \log \left (-\sqrt {2} \sqrt [4]{c} \sqrt {d+e x} \sqrt {\sqrt {a e^2+c d^2}+\sqrt {c} d}+\sqrt {a e^2+c d^2}+\sqrt {c} (d+e x)\right )}{2 \sqrt {2} c^{3/4} \sqrt {\sqrt {a e^2+c d^2}+\sqrt {c} d}}-\frac {e \log \left (\sqrt {2} \sqrt [4]{c} \sqrt {d+e x} \sqrt {\sqrt {a e^2+c d^2}+\sqrt {c} d}+\sqrt {a e^2+c d^2}+\sqrt {c} (d+e x)\right )}{2 \sqrt {2} c^{3/4} \sqrt {\sqrt {a e^2+c d^2}+\sqrt {c} d}}+\frac {e \tanh ^{-1}\left (\frac {\sqrt {\sqrt {a e^2+c d^2}+\sqrt {c} d}-\sqrt {2} \sqrt [4]{c} \sqrt {d+e x}}{\sqrt {\sqrt {c} d-\sqrt {a e^2+c d^2}}}\right )}{\sqrt {2} c^{3/4} \sqrt {\sqrt {c} d-\sqrt {a e^2+c d^2}}}-\frac {e \tanh ^{-1}\left (\frac {\sqrt {\sqrt {a e^2+c d^2}+\sqrt {c} d}+\sqrt {2} \sqrt [4]{c} \sqrt {d+e x}}{\sqrt {\sqrt {c} d-\sqrt {a e^2+c d^2}}}\right )}{\sqrt {2} c^{3/4} \sqrt {\sqrt {c} d-\sqrt {a e^2+c d^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[d + e*x]/(a + c*x^2),x]

[Out]

(e*ArcTanh[(Sqrt[Sqrt[c]*d + Sqrt[c*d^2 + a*e^2]] - Sqrt[2]*c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d - Sqrt[c*d^2
+ a*e^2]]])/(Sqrt[2]*c^(3/4)*Sqrt[Sqrt[c]*d - Sqrt[c*d^2 + a*e^2]]) - (e*ArcTanh[(Sqrt[Sqrt[c]*d + Sqrt[c*d^2
+ a*e^2]] + Sqrt[2]*c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d - Sqrt[c*d^2 + a*e^2]]])/(Sqrt[2]*c^(3/4)*Sqrt[Sqrt
[c]*d - Sqrt[c*d^2 + a*e^2]]) + (e*Log[Sqrt[c*d^2 + a*e^2] - Sqrt[2]*c^(1/4)*Sqrt[Sqrt[c]*d + Sqrt[c*d^2 + a*e
^2]]*Sqrt[d + e*x] + Sqrt[c]*(d + e*x)])/(2*Sqrt[2]*c^(3/4)*Sqrt[Sqrt[c]*d + Sqrt[c*d^2 + a*e^2]]) - (e*Log[Sq
rt[c*d^2 + a*e^2] + Sqrt[2]*c^(1/4)*Sqrt[Sqrt[c]*d + Sqrt[c*d^2 + a*e^2]]*Sqrt[d + e*x] + Sqrt[c]*(d + e*x)])/
(2*Sqrt[2]*c^(3/4)*Sqrt[Sqrt[c]*d + Sqrt[c*d^2 + a*e^2]])

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 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 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 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 700

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

Rubi steps

\begin {align*} \int \frac {\sqrt {d+e x}}{a+c x^2} \, dx &=(2 e) \operatorname {Subst}\left (\int \frac {x^2}{c d^2+a e^2-2 c d 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+a e^2}}{\sqrt {c}}-\frac {\sqrt {2} \sqrt {\sqrt {c} d+\sqrt {c d^2+a e^2}} x}{\sqrt [4]{c}}+x^2} \, dx,x,\sqrt {d+e x}\right )}{\sqrt {2} c^{3/4} \sqrt {\sqrt {c} d+\sqrt {c d^2+a e^2}}}-\frac {e \operatorname {Subst}\left (\int \frac {x}{\frac {\sqrt {c d^2+a e^2}}{\sqrt {c}}+\frac {\sqrt {2} \sqrt {\sqrt {c} d+\sqrt {c d^2+a e^2}} x}{\sqrt [4]{c}}+x^2} \, dx,x,\sqrt {d+e x}\right )}{\sqrt {2} c^{3/4} \sqrt {\sqrt {c} d+\sqrt {c d^2+a e^2}}}\\ &=\frac {e \operatorname {Subst}\left (\int \frac {1}{\frac {\sqrt {c d^2+a e^2}}{\sqrt {c}}-\frac {\sqrt {2} \sqrt {\sqrt {c} d+\sqrt {c d^2+a e^2}} x}{\sqrt [4]{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+a e^2}}{\sqrt {c}}+\frac {\sqrt {2} \sqrt {\sqrt {c} d+\sqrt {c d^2+a e^2}} x}{\sqrt [4]{c}}+x^2} \, dx,x,\sqrt {d+e x}\right )}{2 c}+\frac {e \operatorname {Subst}\left (\int \frac {-\frac {\sqrt {2} \sqrt {\sqrt {c} d+\sqrt {c d^2+a e^2}}}{\sqrt [4]{c}}+2 x}{\frac {\sqrt {c d^2+a e^2}}{\sqrt {c}}-\frac {\sqrt {2} \sqrt {\sqrt {c} d+\sqrt {c d^2+a e^2}} x}{\sqrt [4]{c}}+x^2} \, dx,x,\sqrt {d+e x}\right )}{2 \sqrt {2} c^{3/4} \sqrt {\sqrt {c} d+\sqrt {c d^2+a e^2}}}-\frac {e \operatorname {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt {\sqrt {c} d+\sqrt {c d^2+a e^2}}}{\sqrt [4]{c}}+2 x}{\frac {\sqrt {c d^2+a e^2}}{\sqrt {c}}+\frac {\sqrt {2} \sqrt {\sqrt {c} d+\sqrt {c d^2+a e^2}} x}{\sqrt [4]{c}}+x^2} \, dx,x,\sqrt {d+e x}\right )}{2 \sqrt {2} c^{3/4} \sqrt {\sqrt {c} d+\sqrt {c d^2+a e^2}}}\\ &=\frac {e \log \left (\sqrt {c d^2+a e^2}-\sqrt {2} \sqrt [4]{c} \sqrt {\sqrt {c} d+\sqrt {c d^2+a e^2}} \sqrt {d+e x}+\sqrt {c} (d+e x)\right )}{2 \sqrt {2} c^{3/4} \sqrt {\sqrt {c} d+\sqrt {c d^2+a e^2}}}-\frac {e \log \left (\sqrt {c d^2+a e^2}+\sqrt {2} \sqrt [4]{c} \sqrt {\sqrt {c} d+\sqrt {c d^2+a e^2}} \sqrt {d+e x}+\sqrt {c} (d+e x)\right )}{2 \sqrt {2} c^{3/4} \sqrt {\sqrt {c} d+\sqrt {c d^2+a e^2}}}-\frac {e \operatorname {Subst}\left (\int \frac {1}{2 \left (d-\frac {\sqrt {c d^2+a e^2}}{\sqrt {c}}\right )-x^2} \, dx,x,-\frac {\sqrt {2} \sqrt {\sqrt {c} d+\sqrt {c d^2+a e^2}}}{\sqrt [4]{c}}+2 \sqrt {d+e x}\right )}{c}-\frac {e \operatorname {Subst}\left (\int \frac {1}{2 \left (d-\frac {\sqrt {c d^2+a e^2}}{\sqrt {c}}\right )-x^2} \, dx,x,\frac {\sqrt {2} \sqrt {\sqrt {c} d+\sqrt {c d^2+a e^2}}}{\sqrt [4]{c}}+2 \sqrt {d+e x}\right )}{c}\\ &=\frac {e \tanh ^{-1}\left (\frac {\sqrt [4]{c} \left (\frac {\sqrt {\sqrt {c} d+\sqrt {c d^2+a e^2}}}{\sqrt [4]{c}}-\sqrt {2} \sqrt {d+e x}\right )}{\sqrt {\sqrt {c} d-\sqrt {c d^2+a e^2}}}\right )}{\sqrt {2} c^{3/4} \sqrt {\sqrt {c} d-\sqrt {c d^2+a e^2}}}-\frac {e \tanh ^{-1}\left (\frac {\sqrt [4]{c} \left (\frac {\sqrt {\sqrt {c} d+\sqrt {c d^2+a e^2}}}{\sqrt [4]{c}}+\sqrt {2} \sqrt {d+e x}\right )}{\sqrt {\sqrt {c} d-\sqrt {c d^2+a e^2}}}\right )}{\sqrt {2} c^{3/4} \sqrt {\sqrt {c} d-\sqrt {c d^2+a e^2}}}+\frac {e \log \left (\sqrt {c d^2+a e^2}-\sqrt {2} \sqrt [4]{c} \sqrt {\sqrt {c} d+\sqrt {c d^2+a e^2}} \sqrt {d+e x}+\sqrt {c} (d+e x)\right )}{2 \sqrt {2} c^{3/4} \sqrt {\sqrt {c} d+\sqrt {c d^2+a e^2}}}-\frac {e \log \left (\sqrt {c d^2+a e^2}+\sqrt {2} \sqrt [4]{c} \sqrt {\sqrt {c} d+\sqrt {c d^2+a e^2}} \sqrt {d+e x}+\sqrt {c} (d+e x)\right )}{2 \sqrt {2} c^{3/4} \sqrt {\sqrt {c} d+\sqrt {c d^2+a e^2}}}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 135, normalized size = 0.28 \begin {gather*} \frac {\sqrt {\sqrt {c} d-\sqrt {-a} e} \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d+e x}}{\sqrt {\sqrt {c} d-\sqrt {-a} e}}\right )-\sqrt {\sqrt {-a} e+\sqrt {c} d} \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d+e x}}{\sqrt {\sqrt {-a} e+\sqrt {c} d}}\right )}{\sqrt {-a} c^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Sqrt[c]*d - Sqrt[-a]*e]*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d - Sqrt[-a]*e]] - Sqrt[Sqrt[c]*d +
Sqrt[-a]*e]*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d + Sqrt[-a]*e]])/(Sqrt[-a]*c^(3/4))

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IntegrateAlgebraic [C]  time = 0.40, size = 237, normalized size = 0.50 \begin {gather*} \frac {i \left (\sqrt {c} d+i \sqrt {a} e\right ) \tan ^{-1}\left (\frac {\sqrt {d+e x} \sqrt {-c d-i \sqrt {a} \sqrt {c} e}}{\sqrt {c} d+i \sqrt {a} e}\right )}{\sqrt {a} \sqrt {c} \sqrt {-i \sqrt {c} \left (\sqrt {a} e-i \sqrt {c} d\right )}}-\frac {i \left (\sqrt {c} d-i \sqrt {a} e\right ) \tan ^{-1}\left (\frac {\sqrt {d+e x} \sqrt {-c d+i \sqrt {a} \sqrt {c} e}}{\sqrt {c} d-i \sqrt {a} e}\right )}{\sqrt {a} \sqrt {c} \sqrt {i \sqrt {c} \left (\sqrt {a} e+i \sqrt {c} d\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.42, size = 355, normalized size = 0.74 \begin {gather*} -\frac {1}{2} \, \sqrt {-\frac {a c \sqrt {-\frac {e^{2}}{a c^{3}}} + d}{a c}} \log \left (a c^{2} \sqrt {-\frac {a c \sqrt {-\frac {e^{2}}{a c^{3}}} + d}{a c}} \sqrt {-\frac {e^{2}}{a c^{3}}} + \sqrt {e x + d} e\right ) + \frac {1}{2} \, \sqrt {-\frac {a c \sqrt {-\frac {e^{2}}{a c^{3}}} + d}{a c}} \log \left (-a c^{2} \sqrt {-\frac {a c \sqrt {-\frac {e^{2}}{a c^{3}}} + d}{a c}} \sqrt {-\frac {e^{2}}{a c^{3}}} + \sqrt {e x + d} e\right ) + \frac {1}{2} \, \sqrt {\frac {a c \sqrt {-\frac {e^{2}}{a c^{3}}} - d}{a c}} \log \left (a c^{2} \sqrt {\frac {a c \sqrt {-\frac {e^{2}}{a c^{3}}} - d}{a c}} \sqrt {-\frac {e^{2}}{a c^{3}}} + \sqrt {e x + d} e\right ) - \frac {1}{2} \, \sqrt {\frac {a c \sqrt {-\frac {e^{2}}{a c^{3}}} - d}{a c}} \log \left (-a c^{2} \sqrt {\frac {a c \sqrt {-\frac {e^{2}}{a c^{3}}} - d}{a c}} \sqrt {-\frac {e^{2}}{a c^{3}}} + \sqrt {e x + d} e\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {sage}_{0} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [B]  time = 0.19, size = 1176, normalized size = 2.46 \begin {gather*} -\frac {\sqrt {2 c d +2 \sqrt {a c \,e^{2}+c^{2} d^{2}}}\, \sqrt {2 c d +2 \sqrt {\left (a \,e^{2}+c \,d^{2}\right ) c}}\, d \arctan \left (\frac {2 \sqrt {e x +d}\, \sqrt {c}-\sqrt {2 c d +2 \sqrt {\left (a \,e^{2}+c \,d^{2}\right ) c}}}{\sqrt {-2 c d +4 \sqrt {a \,e^{2}+c \,d^{2}}\, \sqrt {c}-2 \sqrt {\left (a \,e^{2}+c \,d^{2}\right ) c}}}\right )}{2 \sqrt {-2 c d +4 \sqrt {a \,e^{2}+c \,d^{2}}\, \sqrt {c}-2 \sqrt {\left (a \,e^{2}+c \,d^{2}\right ) c}}\, a \sqrt {c}\, e}-\frac {\sqrt {2 c d +2 \sqrt {a c \,e^{2}+c^{2} d^{2}}}\, \sqrt {2 c d +2 \sqrt {\left (a \,e^{2}+c \,d^{2}\right ) c}}\, d \arctan \left (\frac {2 \sqrt {e x +d}\, \sqrt {c}+\sqrt {2 c d +2 \sqrt {\left (a \,e^{2}+c \,d^{2}\right ) c}}}{\sqrt {-2 c d +4 \sqrt {a \,e^{2}+c \,d^{2}}\, \sqrt {c}-2 \sqrt {\left (a \,e^{2}+c \,d^{2}\right ) c}}}\right )}{2 \sqrt {-2 c d +4 \sqrt {a \,e^{2}+c \,d^{2}}\, \sqrt {c}-2 \sqrt {\left (a \,e^{2}+c \,d^{2}\right ) c}}\, a \sqrt {c}\, e}-\frac {\sqrt {2 c d +2 \sqrt {a c \,e^{2}+c^{2} d^{2}}}\, d \ln \left (\left (e x +d \right ) \sqrt {c}-\sqrt {e x +d}\, \sqrt {2 c d +2 \sqrt {\left (a \,e^{2}+c \,d^{2}\right ) c}}+\sqrt {a \,e^{2}+c \,d^{2}}\right )}{4 a \sqrt {c}\, e}+\frac {\sqrt {2 c d +2 \sqrt {a c \,e^{2}+c^{2} d^{2}}}\, d \ln \left (\left (e x +d \right ) \sqrt {c}+\sqrt {e x +d}\, \sqrt {2 c d +2 \sqrt {\left (a \,e^{2}+c \,d^{2}\right ) c}}+\sqrt {a \,e^{2}+c \,d^{2}}\right )}{4 a \sqrt {c}\, e}+\frac {\sqrt {2 c d +2 \sqrt {a c \,e^{2}+c^{2} d^{2}}}\, \sqrt {2 c d +2 \sqrt {\left (a \,e^{2}+c \,d^{2}\right ) c}}\, \sqrt {a c \,e^{2}+c^{2} d^{2}}\, \arctan \left (\frac {2 \sqrt {e x +d}\, \sqrt {c}-\sqrt {2 c d +2 \sqrt {\left (a \,e^{2}+c \,d^{2}\right ) c}}}{\sqrt {-2 c d +4 \sqrt {a \,e^{2}+c \,d^{2}}\, \sqrt {c}-2 \sqrt {\left (a \,e^{2}+c \,d^{2}\right ) c}}}\right )}{2 \sqrt {-2 c d +4 \sqrt {a \,e^{2}+c \,d^{2}}\, \sqrt {c}-2 \sqrt {\left (a \,e^{2}+c \,d^{2}\right ) c}}\, a \,c^{\frac {3}{2}} e}+\frac {\sqrt {2 c d +2 \sqrt {a c \,e^{2}+c^{2} d^{2}}}\, \sqrt {2 c d +2 \sqrt {\left (a \,e^{2}+c \,d^{2}\right ) c}}\, \sqrt {a c \,e^{2}+c^{2} d^{2}}\, \arctan \left (\frac {2 \sqrt {e x +d}\, \sqrt {c}+\sqrt {2 c d +2 \sqrt {\left (a \,e^{2}+c \,d^{2}\right ) c}}}{\sqrt {-2 c d +4 \sqrt {a \,e^{2}+c \,d^{2}}\, \sqrt {c}-2 \sqrt {\left (a \,e^{2}+c \,d^{2}\right ) c}}}\right )}{2 \sqrt {-2 c d +4 \sqrt {a \,e^{2}+c \,d^{2}}\, \sqrt {c}-2 \sqrt {\left (a \,e^{2}+c \,d^{2}\right ) c}}\, a \,c^{\frac {3}{2}} e}+\frac {\sqrt {2 c d +2 \sqrt {a c \,e^{2}+c^{2} d^{2}}}\, \sqrt {a c \,e^{2}+c^{2} d^{2}}\, \ln \left (\left (e x +d \right ) \sqrt {c}-\sqrt {e x +d}\, \sqrt {2 c d +2 \sqrt {\left (a \,e^{2}+c \,d^{2}\right ) c}}+\sqrt {a \,e^{2}+c \,d^{2}}\right )}{4 a \,c^{\frac {3}{2}} e}-\frac {\sqrt {2 c d +2 \sqrt {a c \,e^{2}+c^{2} d^{2}}}\, \sqrt {a c \,e^{2}+c^{2} d^{2}}\, \ln \left (\left (e x +d \right ) \sqrt {c}+\sqrt {e x +d}\, \sqrt {2 c d +2 \sqrt {\left (a \,e^{2}+c \,d^{2}\right ) c}}+\sqrt {a \,e^{2}+c \,d^{2}}\right )}{4 a \,c^{\frac {3}{2}} e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*c*d+2*(a*c*e^2+c^2*d^2)^(1/2))^(1/2)/a/e/c^(1/2)*ln((e*x+d)*c^(1/2)-(e*x+d)^(1/2)*(2*c*d+2*((a*e^2+c*d
^2)*c)^(1/2))^(1/2)+(a*e^2+c*d^2)^(1/2))*d+1/4*(2*c*d+2*(a*c*e^2+c^2*d^2)^(1/2))^(1/2)/a/c^(3/2)/e*ln((e*x+d)*
c^(1/2)-(e*x+d)^(1/2)*(2*c*d+2*((a*e^2+c*d^2)*c)^(1/2))^(1/2)+(a*e^2+c*d^2)^(1/2))*(a*c*e^2+c^2*d^2)^(1/2)-1/2
*(2*c*d+2*(a*c*e^2+c^2*d^2)^(1/2))^(1/2)/a/e/c^(1/2)*(2*c*d+2*((a*e^2+c*d^2)*c)^(1/2))^(1/2)/(-2*c*d+4*(a*e^2+
c*d^2)^(1/2)*c^(1/2)-2*((a*e^2+c*d^2)*c)^(1/2))^(1/2)*arctan((2*(e*x+d)^(1/2)*c^(1/2)-(2*c*d+2*((a*e^2+c*d^2)*
c)^(1/2))^(1/2))/(-2*c*d+4*(a*e^2+c*d^2)^(1/2)*c^(1/2)-2*((a*e^2+c*d^2)*c)^(1/2))^(1/2))*d+1/2*(2*c*d+2*(a*c*e
^2+c^2*d^2)^(1/2))^(1/2)/a/c^(3/2)/e*(2*c*d+2*((a*e^2+c*d^2)*c)^(1/2))^(1/2)/(-2*c*d+4*(a*e^2+c*d^2)^(1/2)*c^(
1/2)-2*((a*e^2+c*d^2)*c)^(1/2))^(1/2)*arctan((2*(e*x+d)^(1/2)*c^(1/2)-(2*c*d+2*((a*e^2+c*d^2)*c)^(1/2))^(1/2))
/(-2*c*d+4*(a*e^2+c*d^2)^(1/2)*c^(1/2)-2*((a*e^2+c*d^2)*c)^(1/2))^(1/2))*(a*c*e^2+c^2*d^2)^(1/2)+1/4*(2*c*d+2*
(a*c*e^2+c^2*d^2)^(1/2))^(1/2)/a/e/c^(1/2)*ln((e*x+d)*c^(1/2)+(e*x+d)^(1/2)*(2*c*d+2*((a*e^2+c*d^2)*c)^(1/2))^
(1/2)+(a*e^2+c*d^2)^(1/2))*d-1/4*(2*c*d+2*(a*c*e^2+c^2*d^2)^(1/2))^(1/2)/a/c^(3/2)/e*ln((e*x+d)*c^(1/2)+(e*x+d
)^(1/2)*(2*c*d+2*((a*e^2+c*d^2)*c)^(1/2))^(1/2)+(a*e^2+c*d^2)^(1/2))*(a*c*e^2+c^2*d^2)^(1/2)-1/2*(2*c*d+2*(a*c
*e^2+c^2*d^2)^(1/2))^(1/2)/a/e/c^(1/2)*(2*c*d+2*((a*e^2+c*d^2)*c)^(1/2))^(1/2)/(-2*c*d+4*(a*e^2+c*d^2)^(1/2)*c
^(1/2)-2*((a*e^2+c*d^2)*c)^(1/2))^(1/2)*arctan((2*(e*x+d)^(1/2)*c^(1/2)+(2*c*d+2*((a*e^2+c*d^2)*c)^(1/2))^(1/2
))/(-2*c*d+4*(a*e^2+c*d^2)^(1/2)*c^(1/2)-2*((a*e^2+c*d^2)*c)^(1/2))^(1/2))*d+1/2*(2*c*d+2*(a*c*e^2+c^2*d^2)^(1
/2))^(1/2)/a/c^(3/2)/e*(2*c*d+2*((a*e^2+c*d^2)*c)^(1/2))^(1/2)/(-2*c*d+4*(a*e^2+c*d^2)^(1/2)*c^(1/2)-2*((a*e^2
+c*d^2)*c)^(1/2))^(1/2)*arctan((2*(e*x+d)^(1/2)*c^(1/2)+(2*c*d+2*((a*e^2+c*d^2)*c)^(1/2))^(1/2))/(-2*c*d+4*(a*
e^2+c*d^2)^(1/2)*c^(1/2)-2*((a*e^2+c*d^2)*c)^(1/2))^(1/2))*(a*c*e^2+c^2*d^2)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {e x + d}}{c x^{2} + a}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.63, size = 308, normalized size = 0.64 \begin {gather*} -2\,\mathrm {atanh}\left (\frac {2\,\left (\left (16\,a\,c^2\,e^4-16\,c^3\,d^2\,e^2\right )\,\sqrt {d+e\,x}+\frac {16\,c\,d\,e^2\,\left (e\,\sqrt {-a^3\,c^3}+a\,c^2\,d\right )\,\sqrt {d+e\,x}}{a}\right )\,\sqrt {-\frac {e\,\sqrt {-a^3\,c^3}+a\,c^2\,d}{4\,a^2\,c^3}}}{16\,c^2\,d^2\,e^3+16\,a\,c\,e^5}\right )\,\sqrt {-\frac {e\,\sqrt {-a^3\,c^3}+a\,c^2\,d}{4\,a^2\,c^3}}-2\,\mathrm {atanh}\left (\frac {2\,\left (\left (16\,a\,c^2\,e^4-16\,c^3\,d^2\,e^2\right )\,\sqrt {d+e\,x}-\frac {16\,c\,d\,e^2\,\left (e\,\sqrt {-a^3\,c^3}-a\,c^2\,d\right )\,\sqrt {d+e\,x}}{a}\right )\,\sqrt {\frac {e\,\sqrt {-a^3\,c^3}-a\,c^2\,d}{4\,a^2\,c^3}}}{16\,c^2\,d^2\,e^3+16\,a\,c\,e^5}\right )\,\sqrt {\frac {e\,\sqrt {-a^3\,c^3}-a\,c^2\,d}{4\,a^2\,c^3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^(1/2)/(a + c*x^2),x)

[Out]

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

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sympy [A]  time = 6.90, size = 75, normalized size = 0.16 \begin {gather*} 2 e \operatorname {RootSum} {\left (256 t^{4} a^{2} c^{3} e^{4} + 32 t^{2} a c^{2} d e^{2} + a e^{2} + c d^{2}, \left (t \mapsto t \log {\left (64 t^{3} a c^{2} e^{2} + 4 t c d + \sqrt {d + e x} \right )} \right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**(1/2)/(c*x**2+a),x)

[Out]

2*e*RootSum(256*_t**4*a**2*c**3*e**4 + 32*_t**2*a*c**2*d*e**2 + a*e**2 + c*d**2, Lambda(_t, _t*log(64*_t**3*a*
c**2*e**2 + 4*_t*c*d + sqrt(d + e*x))))

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