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

Optimal. Leaf size=182 \[ \frac{e \tanh ^{-1}\left (\frac{-2 a e+x (2 c d-b e)+b d}{2 \sqrt{a+b x+c x^2} \sqrt{a e^2-b d e+c d^2}}\right )}{(e f-d g) \sqrt{a e^2-b d e+c d^2}}-\frac{g \tanh ^{-1}\left (\frac{-2 a g+x (2 c f-b g)+b f}{2 \sqrt{a+b x+c x^2} \sqrt{a g^2-b f g+c f^2}}\right )}{(e f-d g) \sqrt{a g^2-b f g+c f^2}} \]

[Out]

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

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Rubi [A]  time = 0.218878, antiderivative size = 182, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {960, 724, 206} \[ \frac{e \tanh ^{-1}\left (\frac{-2 a e+x (2 c d-b e)+b d}{2 \sqrt{a+b x+c x^2} \sqrt{a e^2-b d e+c d^2}}\right )}{(e f-d g) \sqrt{a e^2-b d e+c d^2}}-\frac{g \tanh ^{-1}\left (\frac{-2 a g+x (2 c f-b g)+b f}{2 \sqrt{a+b x+c x^2} \sqrt{a g^2-b f g+c f^2}}\right )}{(e f-d g) \sqrt{a g^2-b f g+c f^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 960

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && (IntegerQ[p] || (ILtQ[m, 0] &&
ILtQ[n, 0])) &&  !(IGtQ[m, 0] || IGtQ[n, 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) (f+g x) \sqrt{a+b x+c x^2}} \, dx &=\int \left (\frac{e}{(e f-d g) (d+e x) \sqrt{a+b x+c x^2}}-\frac{g}{(e f-d g) (f+g x) \sqrt{a+b x+c x^2}}\right ) \, dx\\ &=\frac{e \int \frac{1}{(d+e x) \sqrt{a+b x+c x^2}} \, dx}{e f-d g}-\frac{g \int \frac{1}{(f+g x) \sqrt{a+b x+c x^2}} \, dx}{e f-d g}\\ &=-\frac{(2 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 )}{e f-d g}+\frac{(2 g) \operatorname{Subst}\left (\int \frac{1}{4 c f^2-4 b f g+4 a g^2-x^2} \, dx,x,\frac{-b f+2 a g-(2 c f-b g) x}{\sqrt{a+b x+c x^2}}\right )}{e f-d g}\\ &=\frac{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 )}{\sqrt{c d^2-b d e+a e^2} (e f-d g)}-\frac{g \tanh ^{-1}\left (\frac{b f-2 a g+(2 c f-b g) x}{2 \sqrt{c f^2-b f g+a g^2} \sqrt{a+b x+c x^2}}\right )}{(e f-d g) \sqrt{c f^2-b f g+a g^2}}\\ \end{align*}

Mathematica [A]  time = 0.260909, size = 169, normalized size = 0.93 \[ \frac{\frac{g \tanh ^{-1}\left (\frac{-2 a g+b (f-g x)+2 c f x}{2 \sqrt{a+x (b+c x)} \sqrt{g (a g-b f)+c f^2}}\right )}{\sqrt{g (a g-b f)+c f^2}}-\frac{e \tanh ^{-1}\left (\frac{-2 a e+b (d-e x)+2 c d x}{2 \sqrt{a+x (b+c x)} \sqrt{e (a e-b d)+c d^2}}\right )}{\sqrt{e (a e-b d)+c d^2}}}{d g-e f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.387, size = 327, normalized size = 1.8 \begin{align*}{\frac{1}{dg-ef}\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{1}{dg-ef}\ln \left ({ \left ( 2\,{\frac{a{g}^{2}-bfg+c{f}^{2}}{{g}^{2}}}+{\frac{bg-2\,cf}{g} \left ( x+{\frac{f}{g}} \right ) }+2\,\sqrt{{\frac{a{g}^{2}-bfg+c{f}^{2}}{{g}^{2}}}}\sqrt{ \left ( x+{\frac{f}{g}} \right ) ^{2}c+{\frac{bg-2\,cf}{g} \left ( x+{\frac{f}{g}} \right ) }+{\frac{a{g}^{2}-bfg+c{f}^{2}}{{g}^{2}}}} \right ) \left ( x+{\frac{f}{g}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{{\frac{a{g}^{2}-bfg+c{f}^{2}}{{g}^{2}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(sqrt(c*x^2 + b*x + a)*(e*x + d)*(g*x + f)), x)

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Fricas [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(1/(e*x+d)/(g*x+f)/(c*x^2+b*x+a)^(1/2),x, algorithm="fricas")

[Out]

Timed out

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

[Out]

Integral(1/((d + e*x)*(f + g*x)*sqrt(a + 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(1/(e*x+d)/(g*x+f)/(c*x^2+b*x+a)^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError