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

Optimal. Leaf size=39 \[ \frac {(d+e x) \log (d+e x)}{e \sqrt {c d^2+2 c d e x+c e^2 x^2}} \]

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Rubi [A]  time = 0.01, antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {608, 31} \begin {gather*} \frac {(d+e x) \log (d+e x)}{e \sqrt {c d^2+2 c d e x+c e^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

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]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {c d^2+2 c d e x+c e^2 x^2}} \, dx &=\frac {\left (c d e+c e^2 x\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 {(d+e x) \log (d+e x)}{e \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 = 28, normalized size = 0.72 \begin {gather*} \frac {(d+e x) \log (d+e x)}{e \sqrt {c (d+e x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [B]  time = 0.23, size = 136, normalized size = 3.49 \begin {gather*} -\frac {\sqrt {c e^2} \log \left (x \left (c d e+c e^2 x\right )-x \sqrt {c e^2} \sqrt {c d^2+2 c d e x+c e^2 x^2}\right )}{2 c e^2}-\frac {\tanh ^{-1}\left (\frac {x \sqrt {c e^2}}{\sqrt {c} d}-\frac {\sqrt {c d^2+2 c d e x+c e^2 x^2}}{\sqrt {c} d}\right )}{\sqrt {c} e} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/Sqrt[c*d^2 + 2*c*d*e*x + c*e^2*x^2],x]

[Out]

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

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fricas [A]  time = 0.40, size = 42, normalized size = 1.08 \begin {gather*} \frac {\sqrt {c e^{2} x^{2} + 2 \, c d e x + c d^{2}} \log \left (e x + d\right )}{c e^{2} x + c d e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.33, size = 54, normalized size = 1.38 \begin {gather*} -\frac {e^{\left (-1\right )} \log \left ({\left | -\sqrt {c} d e^{2} - {\left (\sqrt {c} x e - \sqrt {c x^{2} e^{2} + 2 \, c d x e + c d^{2}}\right )} e^{2} \right |}\right )}{\sqrt {c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 38, normalized size = 0.97 \begin {gather*} \frac {\left (e x +d \right ) \ln \left (e x +d \right )}{\sqrt {c \,e^{2} x^{2}+2 c d e x +c \,d^{2}}\, e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.35, size = 15, normalized size = 0.38 \begin {gather*} \frac {\log \left (x + \frac {d}{e}\right )}{\sqrt {c} e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x + d/e)/(sqrt(c)*e)

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mupad [B]  time = 0.47, size = 29, normalized size = 0.74 \begin {gather*} \frac {\ln \left (c\,x\,e^2+c\,d\,e\right )\,\mathrm {sign}\left (c\,e\,\left (d+e\,x\right )\right )}{\sqrt {c\,e^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/sqrt(c*d**2 + 2*c*d*e*x + c*e**2*x**2), x)

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