3.1577 \(\int \frac{b+2 c x}{(d+e x) \sqrt{a+b x+c x^2}} \, dx\)

Optimal. Leaf size=133 \[ \frac{2 \sqrt{c} \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{e}-\frac{(2 c d-b e) \tanh ^{-1}\left (\frac{-2 a e+x (2 c d-b e)+b d}{2 \sqrt{a+b x+c x^2} \sqrt{a e^2-b d e+c d^2}}\right )}{e \sqrt{a e^2-b d e+c d^2}} \]

[Out]

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

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Rubi [A]  time = 0.0907349, antiderivative size = 133, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {843, 621, 206, 724} \[ \frac{2 \sqrt{c} \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{e}-\frac{(2 c d-b e) \tanh ^{-1}\left (\frac{-2 a e+x (2 c d-b e)+b d}{2 \sqrt{a+b x+c x^2} \sqrt{a e^2-b d e+c d^2}}\right )}{e \sqrt{a e^2-b d e+c d^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 843

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + b*x + c*x^
2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 0]

Rule 621

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

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]

Rubi steps

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

Mathematica [A]  time = 0.14292, size = 127, normalized size = 0.95 \[ \frac{\frac{(2 c d-b e) \tanh ^{-1}\left (\frac{2 a e-b d+b e x-2 c d x}{2 \sqrt{a+x (b+c x)} \sqrt{e (a e-b d)+c d^2}}\right )}{\sqrt{e (a e-b d)+c d^2}}+2 \sqrt{c} \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right )}{e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.01, size = 350, normalized size = 2.6 \begin{align*} 2\,{\frac{\sqrt{c}}{e}\ln \left ({\frac{b/2+cx}{\sqrt{c}}}+\sqrt{c{x}^{2}+bx+a} \right ) }-{\frac{b}{e}\ln \left ({ \left ( 2\,{\frac{a{e}^{2}-bde+c{d}^{2}}{{e}^{2}}}+{\frac{be-2\,cd}{e} \left ( x+{\frac{d}{e}} \right ) }+2\,\sqrt{{\frac{a{e}^{2}-bde+c{d}^{2}}{{e}^{2}}}}\sqrt{ \left ( x+{\frac{d}{e}} \right ) ^{2}c+{\frac{be-2\,cd}{e} \left ( x+{\frac{d}{e}} \right ) }+{\frac{a{e}^{2}-bde+c{d}^{2}}{{e}^{2}}}} \right ) \left ( x+{\frac{d}{e}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{{\frac{a{e}^{2}-bde+c{d}^{2}}{{e}^{2}}}}}}}+2\,{\frac{cd}{{e}^{2}}\ln \left ({ \left ( 2\,{\frac{a{e}^{2}-bde+c{d}^{2}}{{e}^{2}}}+{\frac{be-2\,cd}{e} \left ( x+{\frac{d}{e}} \right ) }+2\,\sqrt{{\frac{a{e}^{2}-bde+c{d}^{2}}{{e}^{2}}}}\sqrt{ \left ( x+{\frac{d}{e}} \right ) ^{2}c+{\frac{be-2\,cd}{e} \left ( x+{\frac{d}{e}} \right ) }+{\frac{a{e}^{2}-bde+c{d}^{2}}{{e}^{2}}}} \right ) \left ( x+{\frac{d}{e}} \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((2*c*x+b)/(e*x+d)/(c*x^2+b*x+a)^(1/2),x)

[Out]

2*c^(1/2)/e*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-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*(x+d/e)+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*((x+d/e)^2*c+(b*e-2*c*d)/e*(x+d/e)+(a*e^2-
b*d*e+c*d^2)/e^2)^(1/2))/(x+d/e))*b+2/e^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*(x+d/e)+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*((x+d/e)^2*c+(b*e-2*c*d)/e*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^
(1/2))/(x+d/e))*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((2*c*x+b)/(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 = 9.77439, size = 2302, normalized size = 17.31 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

[Out]

Exception raised: TypeError