### 3.2291 $$\int \frac{\sqrt{d+e x}}{a+b x+c x^2} \, dx$$

Optimal. Leaf size=198 $\frac{\sqrt{2} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{c} \sqrt{b^2-4 a c}}-\frac{\sqrt{2} \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}\right )}{\sqrt{c} \sqrt{b^2-4 a c}}$

[Out]

-((Sqrt[2]*Sqrt[2*c*d - (b - Sqrt[b^2 - 4*a*c])*e]*ArcTanh[(Sqrt[2]*Sqrt[c]*Sqrt[d + e*x])/Sqrt[2*c*d - (b - S
qrt[b^2 - 4*a*c])*e]])/(Sqrt[c]*Sqrt[b^2 - 4*a*c])) + (Sqrt[2]*Sqrt[2*c*d - (b + Sqrt[b^2 - 4*a*c])*e]*ArcTanh
[(Sqrt[2]*Sqrt[c]*Sqrt[d + e*x])/Sqrt[2*c*d - (b + Sqrt[b^2 - 4*a*c])*e]])/(Sqrt[c]*Sqrt[b^2 - 4*a*c])

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Rubi [A]  time = 0.288592, antiderivative size = 198, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 22, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.136, Rules used = {699, 1130, 208} $\frac{\sqrt{2} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{c} \sqrt{b^2-4 a c}}-\frac{\sqrt{2} \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}\right )}{\sqrt{c} \sqrt{b^2-4 a c}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-((Sqrt[2]*Sqrt[2*c*d - (b - Sqrt[b^2 - 4*a*c])*e]*ArcTanh[(Sqrt[2]*Sqrt[c]*Sqrt[d + e*x])/Sqrt[2*c*d - (b - S
qrt[b^2 - 4*a*c])*e]])/(Sqrt[c]*Sqrt[b^2 - 4*a*c])) + (Sqrt[2]*Sqrt[2*c*d - (b + Sqrt[b^2 - 4*a*c])*e]*ArcTanh
[(Sqrt[2]*Sqrt[c]*Sqrt[d + e*x])/Sqrt[2*c*d - (b + Sqrt[b^2 - 4*a*c])*e]])/(Sqrt[c]*Sqrt[b^2 - 4*a*c])

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 1130

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

Rule 208

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

Rubi steps

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

Mathematica [A]  time = 0.467059, size = 175, normalized size = 0.88 $\frac{\sqrt{2} \left (\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )-\sqrt{e \sqrt{b^2-4 a c}-b e+2 c d} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{e \sqrt{b^2-4 a c}-b e+2 c d}}\right )\right )}{\sqrt{c} \sqrt{b^2-4 a c}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [B]  time = 0.253, size = 545, normalized size = 2.8 \begin{align*}{{e}^{2}\sqrt{2}b\arctan \left ({c\sqrt{2}\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( be-2\,cd+\sqrt{-{e}^{2} \left ( 4\,ac-{b}^{2} \right ) } \right ) c}}}} \right ){\frac{1}{\sqrt{-{e}^{2} \left ( 4\,ac-{b}^{2} \right ) }}}{\frac{1}{\sqrt{ \left ( be-2\,cd+\sqrt{-{e}^{2} \left ( 4\,ac-{b}^{2} \right ) } \right ) c}}}}-2\,{\frac{ce\sqrt{2}d}{\sqrt{-{e}^{2} \left ( 4\,ac-{b}^{2} \right ) }\sqrt{ \left ( be-2\,cd+\sqrt{-{e}^{2} \left ( 4\,ac-{b}^{2} \right ) } \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c\sqrt{2}}{\sqrt{ \left ( be-2\,cd+\sqrt{-{e}^{2} \left ( 4\,ac-{b}^{2} \right ) } \right ) c}}} \right ) }+{e\sqrt{2}\arctan \left ({c\sqrt{2}\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( be-2\,cd+\sqrt{-{e}^{2} \left ( 4\,ac-{b}^{2} \right ) } \right ) c}}}} \right ){\frac{1}{\sqrt{ \left ( be-2\,cd+\sqrt{-{e}^{2} \left ( 4\,ac-{b}^{2} \right ) } \right ) c}}}}+{{e}^{2}\sqrt{2}b{\it Artanh} \left ({c\sqrt{2}\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( -be+2\,cd+\sqrt{-{e}^{2} \left ( 4\,ac-{b}^{2} \right ) } \right ) c}}}} \right ){\frac{1}{\sqrt{-{e}^{2} \left ( 4\,ac-{b}^{2} \right ) }}}{\frac{1}{\sqrt{ \left ( -be+2\,cd+\sqrt{-{e}^{2} \left ( 4\,ac-{b}^{2} \right ) } \right ) c}}}}-2\,{\frac{ce\sqrt{2}d}{\sqrt{-{e}^{2} \left ( 4\,ac-{b}^{2} \right ) }\sqrt{ \left ( -be+2\,cd+\sqrt{-{e}^{2} \left ( 4\,ac-{b}^{2} \right ) } \right ) c}}{\it Artanh} \left ({\frac{\sqrt{ex+d}c\sqrt{2}}{\sqrt{ \left ( -be+2\,cd+\sqrt{-{e}^{2} \left ( 4\,ac-{b}^{2} \right ) } \right ) c}}} \right ) }-{e\sqrt{2}{\it Artanh} \left ({c\sqrt{2}\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( -be+2\,cd+\sqrt{-{e}^{2} \left ( 4\,ac-{b}^{2} \right ) } \right ) c}}}} \right ){\frac{1}{\sqrt{ \left ( -be+2\,cd+\sqrt{-{e}^{2} \left ( 4\,ac-{b}^{2} \right ) } \right ) c}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^2/(-e^2*(4*a*c-b^2))^(1/2)*2^(1/2)/((b*e-2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)*arctan((e*x+d)^(1/2)*c*2^(
1/2)/((b*e-2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2))*b-2*c*e/(-e^2*(4*a*c-b^2))^(1/2)*2^(1/2)/((b*e-2*c*d+(-e^
2*(4*a*c-b^2))^(1/2))*c)^(1/2)*arctan((e*x+d)^(1/2)*c*2^(1/2)/((b*e-2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2))*
d+e*2^(1/2)/((b*e-2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)*arctan((e*x+d)^(1/2)*c*2^(1/2)/((b*e-2*c*d+(-e^2*(4
*a*c-b^2))^(1/2))*c)^(1/2))+e^2/(-e^2*(4*a*c-b^2))^(1/2)*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/
2)*arctanh((e*x+d)^(1/2)*c*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2))*b-2*c*e/(-e^2*(4*a*c-b^2))
^(1/2)*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)*arctanh((e*x+d)^(1/2)*c*2^(1/2)/((-b*e+2*c*d+(-
e^2*(4*a*c-b^2))^(1/2))*c)^(1/2))*d-e*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)*arctanh((e*x+d)^
(1/2)*c*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(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} + b x + a}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.41623, size = 1466, normalized size = 7.4 \begin{align*} -\frac{1}{2} \, \sqrt{2} \sqrt{\frac{2 \, c d - 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 (\sqrt{2}{\left (b^{2} c - 4 \, a c^{2}\right )} \sqrt{\frac{e^{2}}{b^{2} c^{2} - 4 \, a c^{3}}} \sqrt{\frac{2 \, c d - 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}}} + 2 \, \sqrt{e x + d} e\right ) + \frac{1}{2} \, \sqrt{2} \sqrt{\frac{2 \, c d - 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 (-\sqrt{2}{\left (b^{2} c - 4 \, a c^{2}\right )} \sqrt{\frac{e^{2}}{b^{2} c^{2} - 4 \, a c^{3}}} \sqrt{\frac{2 \, c d - 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}}} + 2 \, \sqrt{e x + d} e\right ) + \frac{1}{2} \, \sqrt{2} \sqrt{\frac{2 \, c d - 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 (\sqrt{2}{\left (b^{2} c - 4 \, a c^{2}\right )} \sqrt{\frac{e^{2}}{b^{2} c^{2} - 4 \, a c^{3}}} \sqrt{\frac{2 \, c d - 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}}} + 2 \, \sqrt{e x + d} e\right ) - \frac{1}{2} \, \sqrt{2} \sqrt{\frac{2 \, c d - 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 (-\sqrt{2}{\left (b^{2} c - 4 \, a c^{2}\right )} \sqrt{\frac{e^{2}}{b^{2} c^{2} - 4 \, a c^{3}}} \sqrt{\frac{2 \, c d - 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}}} + 2 \, \sqrt{e x + d} e\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 10.3997, size = 155, normalized size = 0.78 \begin{align*} 2 e \operatorname{RootSum}{\left (t^{4} \left (256 a^{2} c^{3} e^{4} - 128 a b^{2} c^{2} e^{4} + 16 b^{4} c e^{4}\right ) + t^{2} \left (- 16 a b c e^{3} + 32 a c^{2} d e^{2} + 4 b^{3} e^{3} - 8 b^{2} c d e^{2}\right ) + a e^{2} - b d e + c d^{2}, \left ( t \mapsto t \log{\left (64 t^{3} a c^{2} e^{2} - 16 t^{3} b^{2} c e^{2} - 2 t b e + 4 t c d + \sqrt{d + e x} \right )} \right )\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*e*RootSum(_t**4*(256*a**2*c**3*e**4 - 128*a*b**2*c**2*e**4 + 16*b**4*c*e**4) + _t**2*(-16*a*b*c*e**3 + 32*a*
c**2*d*e**2 + 4*b**3*e**3 - 8*b**2*c*d*e**2) + a*e**2 - b*d*e + c*d**2, Lambda(_t, _t*log(64*_t**3*a*c**2*e**2
- 16*_t**3*b**2*c*e**2 - 2*_t*b*e + 4*_t*c*d + sqrt(d + e*x))))

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out