3.2379 \(\int \frac{1}{(d+e x)^2 \sqrt{a+b x+c x^2}} \, dx\)

Optimal. Leaf size=134 \[ \frac{(2 c d-b e) \tanh ^{-1}\left (\frac{-2 a e+x (2 c d-b e)+b d}{2 \sqrt{a+b x+c x^2} \sqrt{a e^2-b d e+c d^2}}\right )}{2 \left (a e^2-b d e+c d^2\right )^{3/2}}-\frac{e \sqrt{a+b x+c x^2}}{(d+e x) \left (a e^2-b d e+c d^2\right )} \]

[Out]

-((e*Sqrt[a + b*x + c*x^2])/((c*d^2 - b*d*e + a*e^2)*(d + e*x))) + ((2*c*d - b*e)*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])])/(2*(c*d^2 - b*d*e + a*e^2)^(3/2))

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Rubi [A]  time = 0.0784632, antiderivative size = 134, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {730, 724, 206} \[ \frac{(2 c d-b e) \tanh ^{-1}\left (\frac{-2 a e+x (2 c d-b e)+b d}{2 \sqrt{a+b x+c x^2} \sqrt{a e^2-b d e+c d^2}}\right )}{2 \left (a e^2-b d e+c d^2\right )^{3/2}}-\frac{e \sqrt{a+b x+c x^2}}{(d+e x) \left (a e^2-b d e+c d^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((e*Sqrt[a + b*x + c*x^2])/((c*d^2 - b*d*e + a*e^2)*(d + e*x))) + ((2*c*d - b*e)*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])])/(2*(c*d^2 - b*d*e + a*e^2)^(3/2))

Rule 730

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m + 1)
*(a + b*x + c*x^2)^(p + 1))/((m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[(2*c*d - b*e)/(2*(c*d^2 - b*d*e + a*e
^2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, 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] && EqQ[m + 2*p + 3, 0]

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)^2 \sqrt{a+b x+c x^2}} \, dx &=-\frac{e \sqrt{a+b x+c x^2}}{\left (c d^2-b d e+a e^2\right ) (d+e x)}+\frac{(2 c d-b e) \int \frac{1}{(d+e x) \sqrt{a+b x+c x^2}} \, dx}{2 \left (c d^2-b d e+a e^2\right )}\\ &=-\frac{e \sqrt{a+b x+c x^2}}{\left (c d^2-b d e+a e^2\right ) (d+e x)}-\frac{(2 c d-b e) \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 )}{c d^2-b d e+a e^2}\\ &=-\frac{e \sqrt{a+b x+c x^2}}{\left (c d^2-b d e+a e^2\right ) (d+e x)}+\frac{(2 c d-b e) \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 )}{2 \left (c d^2-b d e+a e^2\right )^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.108025, size = 131, normalized size = 0.98 \[ -\frac{e \sqrt{a+x (b+c x)}}{(d+e x) \left (e (a e-b d)+c d^2\right )}-\frac{(2 c d-b e) \tanh ^{-1}\left (\frac{2 a e-b d+b e x-2 c d x}{2 \sqrt{a+x (b+c x)} \sqrt{e (a e-b d)+c d^2}}\right )}{2 \left (e (a e-b d)+c d^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((e*Sqrt[a + x*(b + c*x)])/((c*d^2 + e*(-(b*d) + a*e))*(d + e*x))) - ((2*c*d - b*e)*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)])])/(2*(c*d^2 + e*(-(b*d) + a*e))^(3/
2))

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Maple [B]  time = 0.228, size = 432, normalized size = 3.2 \begin{align*} -{\frac{1}{a{e}^{2}-bde+c{d}^{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}}}} \left ({\frac{d}{e}}+x \right ) ^{-1}}+{\frac{b}{2\,a{e}^{2}-2\,bde+2\,c{d}^{2}}\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}}}}}}}-{\frac{cd}{e \left ( a{e}^{2}-bde+c{d}^{2} \right ) }\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)^2/(c*x^2+b*x+a)^(1/2),x)

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 6.00199, size = 1397, normalized size = 10.43 \begin{align*} \left [-\frac{{\left (2 \, c d^{2} - b d e +{\left (2 \, c d e - b e^{2}\right )} x\right )} \sqrt{c d^{2} - b d e + a e^{2}} \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 ) + 4 \,{\left (c d^{2} e - b d e^{2} + a e^{3}\right )} \sqrt{c x^{2} + b x + a}}{4 \,{\left (c^{2} d^{5} - 2 \, b c d^{4} e - 2 \, a b d^{2} e^{3} + a^{2} d e^{4} +{\left (b^{2} + 2 \, a c\right )} d^{3} e^{2} +{\left (c^{2} d^{4} e - 2 \, b c d^{3} e^{2} - 2 \, a b d e^{4} + a^{2} e^{5} +{\left (b^{2} + 2 \, a c\right )} d^{2} e^{3}\right )} x\right )}}, \frac{{\left (2 \, c d^{2} - b d e +{\left (2 \, c d e - b e^{2}\right )} x\right )} \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 ) - 2 \,{\left (c d^{2} e - b d e^{2} + a e^{3}\right )} \sqrt{c x^{2} + b x + a}}{2 \,{\left (c^{2} d^{5} - 2 \, b c d^{4} e - 2 \, a b d^{2} e^{3} + a^{2} d e^{4} +{\left (b^{2} + 2 \, a c\right )} d^{3} e^{2} +{\left (c^{2} d^{4} e - 2 \, b c d^{3} e^{2} - 2 \, a b d e^{4} + a^{2} e^{5} +{\left (b^{2} + 2 \, a c\right )} d^{2} e^{3}\right )} x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*((2*c*d^2 - b*d*e + (2*c*d*e - b*e^2)*x)*sqrt(c*d^2 - b*d*e + a*e^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)) + 4*(c*d^2*e - b*d*e^2 + a*e^3)*sqrt(c*x^2 + b*x + a))/(c^2*d^5 - 2*b*c*d^4*e - 2*a*b*d^2*e^3 + a^2*d*
e^4 + (b^2 + 2*a*c)*d^3*e^2 + (c^2*d^4*e - 2*b*c*d^3*e^2 - 2*a*b*d*e^4 + a^2*e^5 + (b^2 + 2*a*c)*d^2*e^3)*x),
1/2*((2*c*d^2 - b*d*e + (2*c*d*e - b*e^2)*x)*sqrt(-c*d^2 + b*d*e - a*e^2)*arctan(-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)) - 2*(c*d^2*e - b*d*e^2 + a*e^3)*sqrt(c*x^2 + b*x + a))/(c^2*
d^5 - 2*b*c*d^4*e - 2*a*b*d^2*e^3 + a^2*d*e^4 + (b^2 + 2*a*c)*d^3*e^2 + (c^2*d^4*e - 2*b*c*d^3*e^2 - 2*a*b*d*e
^4 + a^2*e^5 + (b^2 + 2*a*c)*d^2*e^3)*x)]

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

[Out]

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

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Giac [B]  time = 18.7616, size = 859, normalized size = 6.41 \begin{align*} \frac{{\left (2 \, c^{\frac{3}{2}} d^{2} - 2 \, b \sqrt{c} d e + 2 \, \sqrt{c d^{2} - b d e + a e^{2}} c d \log \left ({\left | 2 \, c^{\frac{3}{2}} d^{2} - 2 \, b \sqrt{c} d e - 2 \, \sqrt{c d^{2} - b d e + a e^{2}} c d + 2 \, a \sqrt{c} e^{2} + \sqrt{c d^{2} - b d e + a e^{2}} b e \right |}\right ) - \sqrt{c d^{2} - b d e + a e^{2}} b e \log \left ({\left | 2 \, c^{\frac{3}{2}} d^{2} - 2 \, b \sqrt{c} d e - 2 \, \sqrt{c d^{2} - b d e + a e^{2}} c d + 2 \, a \sqrt{c} e^{2} + \sqrt{c d^{2} - b d e + a e^{2}} b e \right |}\right ) + 2 \, a \sqrt{c} e^{2}\right )} \mathrm{sgn}\left (\frac{1}{x e + d}\right )}{2 \,{\left (c^{2} d^{4} - 2 \, b c d^{3} e + b^{2} d^{2} e^{2} + 2 \, a c d^{2} e^{2} - 2 \, a b d e^{3} + a^{2} e^{4}\right )}} - \frac{\sqrt{c d^{2} - b d e + a e^{2}}{\left (2 \, c d e - b e^{2}\right )} \log \left ({\left | 2 \,{\left (c d^{2} - b d e + a e^{2}\right )}{\left (\sqrt{c - \frac{2 \, c d}{x e + d} + \frac{c d^{2}}{{\left (x e + d\right )}^{2}} + \frac{b e}{x e + d} - \frac{b d e}{{\left (x e + d\right )}^{2}} + \frac{a e^{2}}{{\left (x e + d\right )}^{2}}} + \frac{\sqrt{c d^{2} e^{2} - b d e^{3} + a e^{4}} e^{\left (-1\right )}}{x e + d}\right )} - \sqrt{c d^{2} - b d e + a e^{2}}{\left (2 \, c d - b e\right )} \right |}\right )}{2 \,{\left (c^{2} d^{4} e - 2 \, b c d^{3} e^{2} + b^{2} d^{2} e^{3} + 2 \, a c d^{2} e^{3} - 2 \, a b d e^{4} + a^{2} e^{5}\right )} \mathrm{sgn}\left (\frac{1}{x e + d}\right )} - \frac{\sqrt{c - \frac{2 \, c d}{x e + d} + \frac{c d^{2}}{{\left (x e + d\right )}^{2}} + \frac{b e}{x e + d} - \frac{b d e}{{\left (x e + d\right )}^{2}} + \frac{a e^{2}}{{\left (x e + d\right )}^{2}}}}{c d^{2} \mathrm{sgn}\left (\frac{1}{x e + d}\right ) - b d e \mathrm{sgn}\left (\frac{1}{x e + d}\right ) + a e^{2} \mathrm{sgn}\left (\frac{1}{x e + d}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*c^(3/2)*d^2 - 2*b*sqrt(c)*d*e + 2*sqrt(c*d^2 - b*d*e + a*e^2)*c*d*log(abs(2*c^(3/2)*d^2 - 2*b*sqrt(c)*d
*e - 2*sqrt(c*d^2 - b*d*e + a*e^2)*c*d + 2*a*sqrt(c)*e^2 + sqrt(c*d^2 - b*d*e + a*e^2)*b*e)) - sqrt(c*d^2 - b*
d*e + a*e^2)*b*e*log(abs(2*c^(3/2)*d^2 - 2*b*sqrt(c)*d*e - 2*sqrt(c*d^2 - b*d*e + a*e^2)*c*d + 2*a*sqrt(c)*e^2
+ sqrt(c*d^2 - b*d*e + a*e^2)*b*e)) + 2*a*sqrt(c)*e^2)*sgn(1/(x*e + d))/(c^2*d^4 - 2*b*c*d^3*e + b^2*d^2*e^2
+ 2*a*c*d^2*e^2 - 2*a*b*d*e^3 + a^2*e^4) - 1/2*sqrt(c*d^2 - b*d*e + a*e^2)*(2*c*d*e - b*e^2)*log(abs(2*(c*d^2
- b*d*e + a*e^2)*(sqrt(c - 2*c*d/(x*e + d) + c*d^2/(x*e + d)^2 + b*e/(x*e + d) - b*d*e/(x*e + d)^2 + a*e^2/(x*
e + d)^2) + sqrt(c*d^2*e^2 - b*d*e^3 + a*e^4)*e^(-1)/(x*e + d)) - sqrt(c*d^2 - b*d*e + a*e^2)*(2*c*d - b*e)))/
((c^2*d^4*e - 2*b*c*d^3*e^2 + b^2*d^2*e^3 + 2*a*c*d^2*e^3 - 2*a*b*d*e^4 + a^2*e^5)*sgn(1/(x*e + d))) - sqrt(c
- 2*c*d/(x*e + d) + c*d^2/(x*e + d)^2 + b*e/(x*e + d) - b*d*e/(x*e + d)^2 + a*e^2/(x*e + d)^2)/(c*d^2*sgn(1/(x
*e + d)) - b*d*e*sgn(1/(x*e + d)) + a*e^2*sgn(1/(x*e + d)))