### 3.1037 $$\int \frac{\sqrt{c d^2+2 c d e x+c e^2 x^2}}{(d+e x)^5} \, dx$$

Optimal. Leaf size=34 $-\frac{c^2}{3 e \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}}$

[Out]

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

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Rubi [A]  time = 0.0229168, antiderivative size = 34, 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 = {643, 629} $-\frac{c^2}{3 e \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[Sqrt[c*d^2 + 2*c*d*e*x + c*e^2*x^2]/(d + e*x)^5,x]

[Out]

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

Rule 643

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[e^(m - 1)/c^((m - 1)/2
), Int[(d + e*x)*(a + b*x + c*x^2)^(p + (m - 1)/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 - 1)/2]

Rule 629

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

Rubi steps

\begin{align*} \int \frac{\sqrt{c d^2+2 c d e x+c e^2 x^2}}{(d+e x)^5} \, dx &=c^3 \int \frac{d+e x}{\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}} \, dx\\ &=-\frac{c^2}{3 e \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0073914, size = 27, normalized size = 0.79 $-\frac{\sqrt{c (d+e x)^2}}{3 e (d+e x)^4}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Sqrt[c*d^2 + 2*c*d*e*x + c*e^2*x^2]/(d + e*x)^5,x]

[Out]

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

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Maple [A]  time = 0.041, size = 35, normalized size = 1. \begin{align*} -{\frac{1}{3\, \left ( ex+d \right ) ^{4}e}\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((c*e^2*x^2+2*c*d*e*x+c*d^2)^(1/2)/(e*x+d)^5,x)

[Out]

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

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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)^(1/2)/(e*x+d)^5,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [B]  time = 2.31593, size = 140, normalized size = 4.12 \begin{align*} -\frac{\sqrt{c e^{2} x^{2} + 2 \, c d e x + c d^{2}}}{3 \,{\left (e^{5} x^{4} + 4 \, d e^{4} x^{3} + 6 \, d^{2} e^{3} x^{2} + 4 \, d^{3} e^{2} x + d^{4} e\right )}} \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)^(1/2)/(e*x+d)^5,x, algorithm="fricas")

[Out]

-1/3*sqrt(c*e^2*x^2 + 2*c*d*e*x + c*d^2)/(e^5*x^4 + 4*d*e^4*x^3 + 6*d^2*e^3*x^2 + 4*d^3*e^2*x + d^4*e)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{c \left (d + e x\right )^{2}}}{\left (d + e x\right )^{5}}\, 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)**(1/2)/(e*x+d)**5,x)

[Out]

Integral(sqrt(c*(d + e*x)**2)/(d + e*x)**5, 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)^(1/2)/(e*x+d)^5,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError