3.1068 \(\int \frac {1}{(d+e x)^2 \sqrt {c d^2+2 c d e x+c e^2 x^2}} \, dx\)

Optimal. Leaf size=38 \[ -\frac {1}{2 e (d+e x) \sqrt {c d^2+2 c d e x+c e^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 38, normalized size of antiderivative = 1.00, 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, 607} \[ -\frac {1}{2 e (d+e x) \sqrt {c d^2+2 c d e x+c e^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(2*e*(d + e*x)*Sqrt[c*d^2 + 2*c*d*e*x + c*e^2*x^2])

Rule 607

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 26, normalized size = 0.68 \[ -\frac {c (d+e x)}{2 e \left (c (d+e x)^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.62, size = 61, normalized size = 1.61 \[ -\frac {\sqrt {c e^{2} x^{2} + 2 \, c d e x + c d^{2}}}{2 \, {\left (c e^{4} x^{3} + 3 \, c d e^{3} x^{2} + 3 \, c d^{2} e^{2} x + c d^{3} e\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [A]  time = 0.05, size = 35, normalized size = 0.92 \[ -\frac {1}{2 \left (e x +d \right ) \sqrt {c \,e^{2} x^{2}+2 c d e x +c \,d^{2}}\, e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.45, size = 33, normalized size = 0.87 \[ -\frac {1}{2 \, {\left (\sqrt {c} e^{3} x^{2} + 2 \, \sqrt {c} d e^{2} x + \sqrt {c} d^{2} e\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.45, size = 37, normalized size = 0.97 \[ -\frac {\sqrt {c\,d^2+2\,c\,d\,e\,x+c\,e^2\,x^2}}{2\,c\,e\,{\left (d+e\,x\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {c \left (d + e x\right )^{2}} \left (d + e x\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)**2/(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)**2), x)

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