### 3.1046 $$\int \frac{(c d^2+2 c d e x+c e^2 x^2)^{3/2}}{(d+e x)^4} \, dx$$

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

[Out]

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

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Rubi [A]  time = 0.0246343, 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} $\frac{c^2 (d+e x) \log (d+e x)}{e \sqrt{c d^2+2 c d e x+c e^2 x^2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^(3/2)/(d + e*x)^4,x]

[Out]

(c^2*(d + e*x)*Log[d + e*x])/(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{\left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}}{(d+e x)^4} \, dx &=c^2 \int \frac{1}{\sqrt{c d^2+2 c d e x+c e^2 x^2}} \, dx\\ &=\frac{\left (c^2 \left (c d e+c e^2 x\right )\right ) \int \frac{1}{c d e+c e^2 x} \, dx}{\sqrt{c d^2+2 c d e x+c e^2 x^2}}\\ &=\frac{c^2 (d+e x) \log (d+e x)}{e \sqrt{c d^2+2 c d e x+c e^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0038717, size = 31, normalized size = 0.74 $\frac{c^2 (d+e x) \log (d+e x)}{e \sqrt{c (d+e x)^2}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^(3/2)/(d + e*x)^4,x]

[Out]

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

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Maple [A]  time = 0.043, size = 40, normalized size = 1. \begin{align*}{\frac{\ln \left ( ex+d \right ) }{ \left ( ex+d \right ) ^{3}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((c*e^2*x^2+2*c*d*e*x+c*d^2)^(3/2)/(e*x+d)^4,x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*e^2*x^2+2*c*d*e*x+c*d^2)^(3/2)/(e*x+d)^4,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 2.2917, size = 89, normalized size = 2.12 \begin{align*} \frac{\sqrt{c e^{2} x^{2} + 2 \, c d e x + c d^{2}} c \log \left (e x + d\right )}{e^{2} x + d e} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*e^2*x^2+2*c*d*e*x+c*d^2)^(3/2)/(e*x+d)^4,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (c \left (d + e x\right )^{2}\right )^{\frac{3}{2}}}{\left (d + e x\right )^{4}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*e**2*x**2+2*c*d*e*x+c*d**2)**(3/2)/(e*x+d)**4,x)

[Out]

Integral((c*(d + e*x)**2)**(3/2)/(d + e*x)**4, x)

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*e^2*x^2+2*c*d*e*x+c*d^2)^(3/2)/(e*x+d)^4,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError