3.1584 \(\int \frac{(b+2 c x) (d+e x)}{(a+b x+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=60 \[ \frac{2 e \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{\sqrt{c}}-\frac{2 (d+e x)}{\sqrt{a+b x+c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.0292371, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {768, 621, 206} \[ \frac{2 e \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{\sqrt{c}}-\frac{2 (d+e x)}{\sqrt{a+b x+c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 768

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

Rubi steps

\begin{align*} \int \frac{(b+2 c x) (d+e x)}{\left (a+b x+c x^2\right )^{3/2}} \, dx &=-\frac{2 (d+e x)}{\sqrt{a+b x+c x^2}}+(2 e) \int \frac{1}{\sqrt{a+b x+c x^2}} \, dx\\ &=-\frac{2 (d+e x)}{\sqrt{a+b x+c x^2}}+(4 e) \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 )\\ &=-\frac{2 (d+e x)}{\sqrt{a+b x+c x^2}}+\frac{2 e \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{\sqrt{c}}\\ \end{align*}

Mathematica [A]  time = 0.283108, size = 56, normalized size = 0.93 \[ \frac{2 e \log \left (2 \sqrt{c} \sqrt{a+x (b+c x)}+b+2 c x\right )}{\sqrt{c}}-\frac{2 (d+e x)}{\sqrt{a+x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.006, size = 158, normalized size = 2.6 \begin{align*} -2\,{\frac{ex}{\sqrt{c{x}^{2}+bx+a}}}+2\,{\frac{e}{\sqrt{c}}\ln \left ({\frac{b/2+cx}{\sqrt{c}}}+\sqrt{c{x}^{2}+bx+a} \right ) }-2\,{\frac{d}{\sqrt{c{x}^{2}+bx+a}}}-4\,{\frac{bcdx}{ \left ( 4\,ac-{b}^{2} \right ) \sqrt{c{x}^{2}+bx+a}}}-2\,{\frac{{b}^{2}d}{ \left ( 4\,ac-{b}^{2} \right ) \sqrt{c{x}^{2}+bx+a}}}+2\,{\frac{bd \left ( 2\,cx+b \right ) }{ \left ( 4\,ac-{b}^{2} \right ) \sqrt{c{x}^{2}+bx+a}}} \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)^(3/2),x)

[Out]

-2*e*x/(c*x^2+b*x+a)^(1/2)+2/c^(1/2)*e*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-2/(c*x^2+b*x+a)^(1/2)*d-4*b
/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x*c*d-2*b^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*d+2*b*d*(2*c*x+b)/(4*a*c-b^2)/(c*
x^2+b*x+a)^(1/2)

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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)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.67455, size = 498, normalized size = 8.3 \begin{align*} \left [\frac{{\left (c e x^{2} + b e x + a e\right )} \sqrt{c} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} - 4 \, \sqrt{c x^{2} + b x + a}{\left (2 \, c x + b\right )} \sqrt{c} - 4 \, a c\right ) - 2 \,{\left (c e x + c d\right )} \sqrt{c x^{2} + b x + a}}{c^{2} x^{2} + b c x + a c}, -\frac{2 \,{\left ({\left (c e x^{2} + b e x + a e\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{2} + b x + a}{\left (2 \, c x + b\right )} \sqrt{-c}}{2 \,{\left (c^{2} x^{2} + b c x + a c\right )}}\right ) +{\left (c e x + c d\right )} \sqrt{c x^{2} + b x + a}\right )}}{c^{2} x^{2} + b c x + a c}\right ] \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)^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (b + 2 c x\right ) \left (d + e x\right )}{\left (a + b x + c x^{2}\right )^{\frac{3}{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)**(3/2),x)

[Out]

Integral((b + 2*c*x)*(d + e*x)/(a + b*x + c*x**2)**(3/2), x)

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Giac [A]  time = 1.40621, size = 136, normalized size = 2.27 \begin{align*} -\frac{2 \, e \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} \sqrt{c} - b \right |}\right )}{\sqrt{c}} - \frac{2 \,{\left (\frac{{\left (b^{2} e - 4 \, a c e\right )} x}{b^{2} - 4 \, a c} + \frac{b^{2} d - 4 \, a c d}{b^{2} - 4 \, a c}\right )}}{\sqrt{c x^{2} + b x + a}} \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)^(3/2),x, algorithm="giac")

[Out]

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