3.1069 \(\int \frac{1}{(d+e x)^3 \sqrt{c d^2+2 c d e x+c e^2 x^2}} \, dx\)

Optimal. Leaf size=32 \[ -\frac{c}{3 e \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0235819, antiderivative size = 32, 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} \[ -\frac{c}{3 e \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((d + e*x)^3*Sqrt[c*d^2 + 2*c*d*e*x + c*e^2*x^2]),x]

[Out]

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

Mathematica [A]  time = 0.0147183, size = 21, normalized size = 0.66 \[ -\frac{c}{3 e \left (c (d+e x)^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((d + e*x)^3*Sqrt[c*d^2 + 2*c*d*e*x + c*e^2*x^2]),x]

[Out]

-c/(3*e*(c*(d + e*x)^2)^(3/2))

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Maple [A]  time = 0.041, size = 35, normalized size = 1.1 \begin{align*} -{\frac{1}{3\, \left ( ex+d \right ) ^{2}e}{\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(1/(e*x+d)^3/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(1/2),x)

[Out]

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

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Maxima [A]  time = 1.06041, size = 63, normalized size = 1.97 \begin{align*} -\frac{1}{3 \,{\left (\sqrt{c} e^{4} x^{3} + 3 \, \sqrt{c} d e^{3} x^{2} + 3 \, \sqrt{c} d^{2} e^{2} x + \sqrt{c} d^{3} e\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^3/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.38018, size = 154, normalized size = 4.81 \begin{align*} -\frac{\sqrt{c e^{2} x^{2} + 2 \, c d e x + c d^{2}}}{3 \,{\left (c e^{5} x^{4} + 4 \, c d e^{4} x^{3} + 6 \, c d^{2} e^{3} x^{2} + 4 \, c d^{3} e^{2} x + c d^{4} e\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^3/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{c \left (d + e x\right )^{2}} \left (d + e x\right )^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)**3/(c*e**2*x**2+2*c*d*e*x+c*d**2)**(1/2),x)

[Out]

Integral(1/(sqrt(c*(d + e*x)**2)*(d + e*x)**3), 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(1/(e*x+d)^3/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(1/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError