### 3.2378 $$\int \frac{1}{(d+e x) \sqrt{a+b x+c x^2}} \, dx$$

Optimal. Leaf size=79 $\frac{\tanh ^{-1}\left (\frac{-2 a e+x (2 c d-b e)+b d}{2 \sqrt{a+b x+c x^2} \sqrt{a e^2-b d e+c d^2}}\right )}{\sqrt{a e^2-b d e+c d^2}}$

[Out]

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

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Rubi [A]  time = 0.0386044, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 22, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.091, Rules used = {724, 206} $\frac{\tanh ^{-1}\left (\frac{-2 a e+x (2 c d-b e)+b d}{2 \sqrt{a+b x+c x^2} \sqrt{a e^2-b d e+c d^2}}\right )}{\sqrt{a e^2-b d e+c d^2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[1/((d + e*x)*Sqrt[a + b*x + c*x^2]),x]

[Out]

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

Rule 724

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

Rubi steps

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

Mathematica [A]  time = 0.0365629, size = 78, normalized size = 0.99 $-\frac{\tanh ^{-1}\left (\frac{2 a e-b d+b e x-2 c d x}{2 \sqrt{a+x (b+c x)} \sqrt{e (a e-b d)+c d^2}}\right )}{\sqrt{e (a e-b d)+c d^2}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[1/((d + e*x)*Sqrt[a + b*x + c*x^2]),x]

[Out]

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

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Maple [B]  time = 0.228, size = 157, normalized size = 2. \begin{align*} -{\frac{1}{e}\ln \left ({ \left ( 2\,{\frac{a{e}^{2}-bde+c{d}^{2}}{{e}^{2}}}+{\frac{be-2\,cd}{e} \left ({\frac{d}{e}}+x \right ) }+2\,\sqrt{{\frac{a{e}^{2}-bde+c{d}^{2}}{{e}^{2}}}}\sqrt{ \left ({\frac{d}{e}}+x \right ) ^{2}c+{\frac{be-2\,cd}{e} \left ({\frac{d}{e}}+x \right ) }+{\frac{a{e}^{2}-bde+c{d}^{2}}{{e}^{2}}}} \right ) \left ({\frac{d}{e}}+x \right ) ^{-1}} \right ){\frac{1}{\sqrt{{\frac{a{e}^{2}-bde+c{d}^{2}}{{e}^{2}}}}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 3.07362, size = 743, normalized size = 9.41 \begin{align*} \left [\frac{\log \left (\frac{8 \, a b d e - 8 \, a^{2} e^{2} -{\left (b^{2} + 4 \, a c\right )} d^{2} -{\left (8 \, c^{2} d^{2} - 8 \, b c d e +{\left (b^{2} + 4 \, a c\right )} e^{2}\right )} x^{2} - 4 \, \sqrt{c d^{2} - b d e + a e^{2}} \sqrt{c x^{2} + b x + a}{\left (b d - 2 \, a e +{\left (2 \, c d - b e\right )} x\right )} - 2 \,{\left (4 \, b c d^{2} + 4 \, a b e^{2} -{\left (3 \, b^{2} + 4 \, a c\right )} d e\right )} x}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right )}{2 \, \sqrt{c d^{2} - b d e + a e^{2}}}, \frac{\sqrt{-c d^{2} + b d e - a e^{2}} \arctan \left (-\frac{\sqrt{-c d^{2} + b d e - a e^{2}} \sqrt{c x^{2} + b x + a}{\left (b d - 2 \, a e +{\left (2 \, c d - b e\right )} x\right )}}{2 \,{\left (a c d^{2} - a b d e + a^{2} e^{2} +{\left (c^{2} d^{2} - b c d e + a c e^{2}\right )} x^{2} +{\left (b c d^{2} - b^{2} d e + a b e^{2}\right )} x\right )}}\right )}{c d^{2} - b d e + a e^{2}}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*log((8*a*b*d*e - 8*a^2*e^2 - (b^2 + 4*a*c)*d^2 - (8*c^2*d^2 - 8*b*c*d*e + (b^2 + 4*a*c)*e^2)*x^2 - 4*sqrt
(c*d^2 - b*d*e + a*e^2)*sqrt(c*x^2 + b*x + a)*(b*d - 2*a*e + (2*c*d - b*e)*x) - 2*(4*b*c*d^2 + 4*a*b*e^2 - (3*
b^2 + 4*a*c)*d*e)*x)/(e^2*x^2 + 2*d*e*x + d^2))/sqrt(c*d^2 - b*d*e + a*e^2), sqrt(-c*d^2 + b*d*e - a*e^2)*arct
an(-1/2*sqrt(-c*d^2 + b*d*e - a*e^2)*sqrt(c*x^2 + b*x + a)*(b*d - 2*a*e + (2*c*d - b*e)*x)/(a*c*d^2 - a*b*d*e
+ a^2*e^2 + (c^2*d^2 - b*c*d*e + a*c*e^2)*x^2 + (b*c*d^2 - b^2*d*e + a*b*e^2)*x))/(c*d^2 - b*d*e + a*e^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (d + e x\right ) \sqrt{a + b x + c x^{2}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.10382, size = 97, normalized size = 1.23 \begin{align*} \frac{2 \, \arctan \left (-\frac{{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} e + \sqrt{c} d}{\sqrt{-c d^{2} + b d e - a e^{2}}}\right )}{\sqrt{-c d^{2} + b d e - a e^{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*arctan(-((sqrt(c)*x - sqrt(c*x^2 + b*x + a))*e + sqrt(c)*d)/sqrt(-c*d^2 + b*d*e - a*e^2))/sqrt(-c*d^2 + b*d*
e - a*e^2)