### 3.1064 $$\int \frac{(d+e x)^2}{\sqrt{c d^2+2 c d e x+c e^2 x^2}} \, dx$$

Optimal. Leaf size=39 $\frac{(d+e x) \sqrt{c d^2+2 c d e x+c e^2 x^2}}{2 c e}$

[Out]

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

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Rubi [A]  time = 0.0215158, antiderivative size = 39, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 32, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.062, Rules used = {642, 609} $\frac{(d+e x) \sqrt{c d^2+2 c d e x+c e^2 x^2}}{2 c e}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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 609

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

Rubi steps

\begin{align*} \int \frac{(d+e x)^2}{\sqrt{c d^2+2 c d e x+c e^2 x^2}} \, dx &=\frac{\int \sqrt{c d^2+2 c d e x+c e^2 x^2} \, dx}{c}\\ &=\frac{(d+e x) \sqrt{c d^2+2 c d e x+c e^2 x^2}}{2 c e}\\ \end{align*}

Mathematica [A]  time = 0.003141, size = 30, normalized size = 0.77 $\frac{x (d+e x) (2 d+e x)}{2 \sqrt{c (d+e x)^2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.04, size = 38, normalized size = 1. \begin{align*}{\frac{x \left ( ex+2\,d \right ) \left ( ex+d \right ) }{2}{\frac{1}{\sqrt{c{e}^{2}{x}^{2}+2\,cdex+c{d}^{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)^(1/2),x)

[Out]

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

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Maxima [B]  time = 1.16027, size = 147, normalized size = 3.77 \begin{align*} \frac{c^{2} d^{2} e^{4} \log \left (x + \frac{d}{e}\right )}{\left (c e^{2}\right )^{\frac{5}{2}}} - \frac{c d e^{3} x}{\left (c e^{2}\right )^{\frac{3}{2}}} + \frac{e^{2} x^{2}}{2 \, \sqrt{c e^{2}}} - d^{2} \sqrt{\frac{1}{c e^{2}}} \log \left (x + \frac{d}{e}\right ) + \frac{2 \, \sqrt{c e^{2} x^{2} + 2 \, c d e x + c d^{2}} d}{c 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)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.38543, size = 96, normalized size = 2.46 \begin{align*} \frac{\sqrt{c e^{2} x^{2} + 2 \, c d e x + c d^{2}}{\left (e x^{2} + 2 \, d x\right )}}{2 \,{\left (c e x + c d\right )}} \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)^(1/2),x, algorithm="fricas")

[Out]

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

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

[Out]

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

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Giac [A]  time = 1.39398, size = 50, normalized size = 1.28 \begin{align*} \frac{1}{2} \, \sqrt{c x^{2} e^{2} + 2 \, c d x e + c d^{2}}{\left (\frac{d e^{\left (-1\right )}}{c} + \frac{x}{c}\right )} \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)^(1/2),x, algorithm="giac")

[Out]

1/2*sqrt(c*x^2*e^2 + 2*c*d*x*e + c*d^2)*(d*e^(-1)/c + x/c)