3.11.68 \(\int \frac {1}{(d+e x)^2 \sqrt {c d^2+2 c d e x+c e^2 x^2}} \, dx\) [1068]

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 = 0.01, 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 = {656, 621} \begin {gather*} -\frac {1}{2 e (d+e x) \sqrt {c d^2+2 c d e x+c e^2 x^2}} \end {gather*}

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

Rule 621

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 656

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 \begin {gather*} -\frac {c (d+e x)}{2 e \left (c (d+e x)^2\right )^{3/2}} \end {gather*}

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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Maple [A]
time = 0.59, size = 35, normalized size = 0.92

method	result	size
risch	\(-\frac {1}{2 \left (e x +d \right ) \sqrt {\left (e x +d \right )^{2} c}\, e}\)	\(24\)
gosper	\(-\frac {1}{2 e \left (e x +d \right ) \sqrt {x^{2} c \,e^{2}+2 c d e x +c \,d^{2}}}\)	\(35\)
default	\(-\frac {1}{2 e \left (e x +d \right ) \sqrt {x^{2} c \,e^{2}+2 c d e x +c \,d^{2}}}\)	\(35\)
trager	\(\frac {\left (e x +2 d \right ) x \sqrt {x^{2} c \,e^{2}+2 c d e x +c \,d^{2}}}{2 d^{2} c \left (e x +d \right )^{3}}\)	\(46\)

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,method=_RETURNVERBOSE)

[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 = 0.27, size = 32, normalized size = 0.84 \begin {gather*} -\frac {1}{2 \, {\left (\sqrt {c} x^{2} e^{3} + 2 \, \sqrt {c} d x e^{2} + \sqrt {c} d^{2} e\right )}} \end {gather*}

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

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Fricas [A]
time = 2.84, size = 59, normalized size = 1.55 \begin {gather*} -\frac {\sqrt {c x^{2} e^{2} + 2 \, c d x e + c d^{2}}}{2 \, {\left (c x^{3} e^{4} + 3 \, c d x^{2} e^{3} + 3 \, c d^{2} x e^{2} + c d^{3} e\right )}} \end {gather*}

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

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

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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Giac [A]
time = 1.82, size = 24, normalized size = 0.63 \begin {gather*} -\frac {e^{\left (-1\right )}}{2 \, {\left (x e + d\right )}^{2} \sqrt {c} \mathrm {sgn}\left (x e + d\right )} \end {gather*}

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]

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

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

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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