3.1073 \(\int \frac{(d+e x)^2}{(c d^2+2 c d e x+c e^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=42 \[ \frac{(d+e x) \log (d+e x)}{c e \sqrt{c d^2+2 c d e x+c e^2 x^2}} \]

[Out]

((d + e*x)*Log[d + e*x])/(c*e*Sqrt[c*d^2 + 2*c*d*e*x + c*e^2*x^2])

________________________________________________________________________________________

Rubi [A]  time = 0.0252252, antiderivative size = 42, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.094, Rules used = {642, 608, 31} \[ \frac{(d+e x) \log (d+e x)}{c e \sqrt{c d^2+2 c d e x+c e^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((d + e*x)*Log[d + e*x])/(c*e*Sqrt[c*d^2 + 2*c*d*e*x + c*e^2*x^2])

Rule 642

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[e^m/c^(m/2), Int[(a +
b*x + c*x^2)^(p + m/2), x], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && EqQ[
2*c*d - b*e, 0] && IntegerQ[m/2]

Rule 608

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[(b/2 + c*x)/Sqrt[a + b*x + c*x^2], Int[1/(b/2
+ c*x), x], x] /; FreeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

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

Mathematica [A]  time = 0.0028975, size = 31, normalized size = 0.74 \[ \frac{(d+e x) \log (d+e x)}{c e \sqrt{c (d+e x)^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((d + e*x)*Log[d + e*x])/(c*e*Sqrt[c*(d + e*x)^2])

________________________________________________________________________________________

Maple [A]  time = 0.042, size = 40, normalized size = 1. \begin{align*}{\frac{ \left ( ex+d \right ) ^{3}\ln \left ( ex+d \right ) }{e} \left ( c{e}^{2}{x}^{2}+2\,cdex+c{d}^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^2/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(3/2),x)

[Out]

1/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(3/2)*(e*x+d)^3*ln(e*x+d)/e

________________________________________________________________________________________

Maxima [B]  time = 1.14424, size = 166, normalized size = 3.95 \begin{align*} \frac{3 \, c^{2} d^{2} e^{4}}{2 \, \left (c e^{2}\right )^{\frac{7}{2}}{\left (x + \frac{d}{e}\right )}^{2}} + \frac{2 \, c d e^{3} x}{\left (c e^{2}\right )^{\frac{5}{2}}{\left (x + \frac{d}{e}\right )}^{2}} + \frac{e^{2} \log \left (x + \frac{d}{e}\right )}{\left (c e^{2}\right )^{\frac{3}{2}}} - \frac{2 \, d}{\sqrt{c e^{2} x^{2} + 2 \, c d e x + c d^{2}} c e} + \frac{d^{2}}{2 \, \left (c e^{2}\right )^{\frac{3}{2}}{\left (x + \frac{d}{e}\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/2*c^2*d^2*e^4/((c*e^2)^(7/2)*(x + d/e)^2) + 2*c*d*e^3*x/((c*e^2)^(5/2)*(x + d/e)^2) + e^2*log(x + d/e)/(c*e^
2)^(3/2) - 2*d/(sqrt(c*e^2*x^2 + 2*c*d*e*x + c*d^2)*c*e) + 1/2*d^2/((c*e^2)^(3/2)*(x + d/e)^2)

________________________________________________________________________________________

Fricas [A]  time = 2.35377, size = 97, normalized size = 2.31 \begin{align*} \frac{\sqrt{c e^{2} x^{2} + 2 \, c d e x + c d^{2}} \log \left (e x + d\right )}{c^{2} e^{2} x + c^{2} d e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(c*e^2*x^2 + 2*c*d*e*x + c*d^2)*log(e*x + d)/(c^2*e^2*x + c^2*d*e)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**2/(c*e**2*x**2+2*c*d*e*x+c*d**2)**(3/2),x)

[Out]

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

________________________________________________________________________________________

Giac [B]  time = 1.41241, size = 119, normalized size = 2.83 \begin{align*} \frac{2 \,{\left (C_{0} d e^{\left (-1\right )} + C_{0} x\right )}}{\sqrt{c x^{2} e^{2} + 2 \, c d x e + c d^{2}}} - \frac{e^{\left (-1\right )} \log \left ({\left | -\sqrt{c} d e^{2} -{\left (\sqrt{c} x e - \sqrt{c x^{2} e^{2} + 2 \, c d x e + c d^{2}}\right )} e^{2} \right |}\right )}{c^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(C_0*d*e^(-1) + C_0*x)/sqrt(c*x^2*e^2 + 2*c*d*x*e + c*d^2) - e^(-1)*log(abs(-sqrt(c)*d*e^2 - (sqrt(c)*x*e -
sqrt(c*x^2*e^2 + 2*c*d*x*e + c*d^2))*e^2))/c^(3/2)