3.1583 \(\int \frac{(b+2 c x) (d+e x)^2}{(a+b x+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=93 \[ \frac{2 e (2 c d-b e) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{c^{3/2}}-\frac{2 (d+e x)^2}{\sqrt{a+b x+c x^2}}+\frac{4 e^2 \sqrt{a+b x+c x^2}}{c} \]

[Out]

(-2*(d + e*x)^2)/Sqrt[a + b*x + c*x^2] + (4*e^2*Sqrt[a + b*x + c*x^2])/c + (2*e*(2*c*d - b*e)*ArcTanh[(b + 2*c
*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/c^(3/2)

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Rubi [A]  time = 0.0518795, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {768, 640, 621, 206} \[ \frac{2 e (2 c d-b e) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{c^{3/2}}-\frac{2 (d+e x)^2}{\sqrt{a+b x+c x^2}}+\frac{4 e^2 \sqrt{a+b x+c x^2}}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(d + e*x)^2)/Sqrt[a + b*x + c*x^2] + (4*e^2*Sqrt[a + b*x + c*x^2])/c + (2*e*(2*c*d - b*e)*ArcTanh[(b + 2*c
*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/c^(3/2)

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 640

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

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

Mathematica [A]  time = 0.172897, size = 95, normalized size = 1.02 \[ \frac{2 e (2 c d-b e) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right )}{c^{3/2}}+\frac{4 e^2 (a+b x)-2 c \left (d^2+2 d e x-e^2 x^2\right )}{c \sqrt{a+x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.009, size = 427, normalized size = 4.6 \begin{align*} -2\,{\frac{{d}^{2}}{\sqrt{c{x}^{2}+bx+a}}}+2\,{\frac{{e}^{2}{x}^{2}}{\sqrt{c{x}^{2}+bx+a}}}+4\,{\frac{de}{\sqrt{c}}\ln \left ({\frac{b/2+cx}{\sqrt{c}}}+\sqrt{c{x}^{2}+bx+a} \right ) }+2\,{\frac{b{e}^{2}x}{c\sqrt{c{x}^{2}+bx+a}}}-{\frac{{e}^{2}{b}^{4}}{{c}^{2} \left ( 4\,ac-{b}^{2} \right ) }{\frac{1}{\sqrt{c{x}^{2}+bx+a}}}}-2\,{\frac{b{e}^{2}}{{c}^{3/2}}\ln \left ({\frac{b/2+cx}{\sqrt{c}}}+\sqrt{c{x}^{2}+bx+a} \right ) }+2\,{\frac{b{d}^{2} \left ( 2\,cx+b \right ) }{ \left ( 4\,ac-{b}^{2} \right ) \sqrt{c{x}^{2}+bx+a}}}-4\,{\frac{bxc{d}^{2}}{ \left ( 4\,ac-{b}^{2} \right ) \sqrt{c{x}^{2}+bx+a}}}-2\,{\frac{{b}^{3}{e}^{2}x}{c \left ( 4\,ac-{b}^{2} \right ) \sqrt{c{x}^{2}+bx+a}}}+8\,{\frac{ab{e}^{2}x}{ \left ( 4\,ac-{b}^{2} \right ) \sqrt{c{x}^{2}+bx+a}}}+4\,{\frac{a{e}^{2}{b}^{2}}{c \left ( 4\,ac-{b}^{2} \right ) \sqrt{c{x}^{2}+bx+a}}}-4\,{\frac{edx}{\sqrt{c{x}^{2}+bx+a}}}-2\,{\frac{{b}^{2}{d}^{2}}{ \left ( 4\,ac-{b}^{2} \right ) \sqrt{c{x}^{2}+bx+a}}}+4\,{\frac{a{e}^{2}}{c\sqrt{c{x}^{2}+bx+a}}}-{\frac{{b}^{2}{e}^{2}}{{c}^{2}}{\frac{1}{\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)^2/(c*x^2+b*x+a)^(3/2),x)

[Out]

-2/(c*x^2+b*x+a)^(1/2)*d^2+2*e^2*x^2/(c*x^2+b*x+a)^(1/2)+4/c^(1/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))
*d*e+2/c*e^2*b*x/(c*x^2+b*x+a)^(1/2)-1/c^2*e^2*b^4/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)-2/c^(3/2)*e^2*b*ln((1/2*b+c
*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+2*b*d^2*(2*c*x+b)/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)-4*b/(4*a*c-b^2)/(c*x^2+b*x+
a)^(1/2)*x*c*d^2-2/c*e^2*b^3/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x+8*e^2*a*b/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x+4/c
*e^2*a*b^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)-4*x/(c*x^2+b*x+a)^(1/2)*d*e-2*b^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*d
^2+4/c*e^2*a/(c*x^2+b*x+a)^(1/2)-1/c^2*e^2*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)^2/(c*x^2+b*x+a)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 3.35496, size = 780, normalized size = 8.39 \begin{align*} \left [-\frac{{\left (2 \, a c d e - a b e^{2} +{\left (2 \, c^{2} d e - b c e^{2}\right )} x^{2} +{\left (2 \, b c d e - b^{2} e^{2}\right )} x\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^{2} e^{2} x^{2} - c^{2} d^{2} + 2 \, a c e^{2} - 2 \,{\left (c^{2} d e - b c e^{2}\right )} x\right )} \sqrt{c x^{2} + b x + a}}{c^{3} x^{2} + b c^{2} x + a c^{2}}, -\frac{2 \,{\left ({\left (2 \, a c d e - a b e^{2} +{\left (2 \, c^{2} d e - b c e^{2}\right )} x^{2} +{\left (2 \, b c d e - b^{2} e^{2}\right )} x\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^{2} e^{2} x^{2} - c^{2} d^{2} + 2 \, a c e^{2} - 2 \,{\left (c^{2} d e - b c e^{2}\right )} x\right )} \sqrt{c x^{2} + b x + a}\right )}}{c^{3} x^{2} + b c^{2} x + a c^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*x+b)*(e*x+d)^2/(c*x^2+b*x+a)^(3/2),x, algorithm="fricas")

[Out]

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

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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 )^{2}}{\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)**2/(c*x**2+b*x+a)**(3/2),x)

[Out]

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

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Giac [B]  time = 1.45249, size = 266, normalized size = 2.86 \begin{align*} \frac{2 \,{\left ({\left (\frac{{\left (b^{2} c e^{2} - 4 \, a c^{2} e^{2}\right )} x}{b^{2} c - 4 \, a c^{2}} - \frac{2 \,{\left (b^{2} c d e - 4 \, a c^{2} d e - b^{3} e^{2} + 4 \, a b c e^{2}\right )}}{b^{2} c - 4 \, a c^{2}}\right )} x - \frac{b^{2} c d^{2} - 4 \, a c^{2} d^{2} - 2 \, a b^{2} e^{2} + 8 \, a^{2} c e^{2}}{b^{2} c - 4 \, a c^{2}}\right )}}{\sqrt{c x^{2} + b x + a}} - \frac{2 \,{\left (2 \, c d e - b e^{2}\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} \sqrt{c} - b \right |}\right )}{c^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*x+b)*(e*x+d)^2/(c*x^2+b*x+a)^(3/2),x, algorithm="giac")

[Out]

2*(((b^2*c*e^2 - 4*a*c^2*e^2)*x/(b^2*c - 4*a*c^2) - 2*(b^2*c*d*e - 4*a*c^2*d*e - b^3*e^2 + 4*a*b*c*e^2)/(b^2*c
- 4*a*c^2))*x - (b^2*c*d^2 - 4*a*c^2*d^2 - 2*a*b^2*e^2 + 8*a^2*c*e^2)/(b^2*c - 4*a*c^2))/sqrt(c*x^2 + b*x + a
) - 2*(2*c*d*e - b*e^2)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) - b))/c^(3/2)