\(\int \frac {1}{\sqrt {c d^2+2 c d e x+c e^2 x^2}} \, dx\) [1066]

Optimal result

Integrand size = 24, antiderivative size = 39 \[ \int \frac {1}{\sqrt {c d^2+2 c d e x+c e^2 x^2}} \, dx=\frac {(d+e x) \log (d+e x)}{e \sqrt {c d^2+2 c d e x+c e^2 x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {622, 31} \[ \int \frac {1}{\sqrt {c d^2+2 c d e x+c e^2 x^2}} \, dx=\frac {(d+e x) \log (d+e x)}{e \sqrt {c d^2+2 c d e x+c e^2 x^2}} \]

[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 622

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*} \text {integral}& = \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*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.72 \[ \int \frac {1}{\sqrt {c d^2+2 c d e x+c e^2 x^2}} \, dx=\frac {(d+e x) \log (d+e x)}{e \sqrt {c (d+e x)^2}} \]

[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])

Maple [A] (verified)

Time = 2.31 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.69

method	result	size
risch	\(\frac {\left (e x +d \right ) \ln \left (e x +d \right )}{\sqrt {c \left (e x +d \right )^{2}}\, e}\)	\(27\)
default	\(\frac {\left (e x +d \right ) \ln \left (e x +d \right )}{e \sqrt {c \,x^{2} e^{2}+2 x c d e +c \,d^{2}}}\)	\(38\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.08 \[ \int \frac {1}{\sqrt {c d^2+2 c d e x+c e^2 x^2}} \, dx=\frac {\sqrt {c e^{2} x^{2} + 2 \, c d e x + c d^{2}} \log \left (e x + d\right )}{c e^{2} x + c d e} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.67 \[ \int \frac {1}{\sqrt {c d^2+2 c d e x+c e^2 x^2}} \, dx=\frac {\left (\frac {d}{e} + x\right ) \log {\left (\frac {d}{e} + x \right )}}{\sqrt {c e^{2} \left (\frac {d}{e} + x\right )^{2}}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.46 \[ \int \frac {1}{\sqrt {c d^2+2 c d e x+c e^2 x^2}} \, dx=\sqrt {\frac {1}{c e^{2}}} \log \left (x + \frac {d}{e}\right ) \]

[In]

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

[Out]

sqrt(1/(c*e^2))*log(x + d/e)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.87 \[ \int \frac {1}{\sqrt {c d^2+2 c d e x+c e^2 x^2}} \, dx=\frac {\log \left ({\left | e x + d \right |} \sqrt {{\left | c \right |}} {\left | \mathrm {sgn}\left (e x + d\right ) \right |}\right )}{\sqrt {c} e \mathrm {sgn}\left (e x + d\right )} \]

[In]

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

[Out]

log(abs(e*x + d)*sqrt(abs(c))*abs(sgn(e*x + d)))/(sqrt(c)*e*sgn(e*x + d))

Mupad [B] (verification not implemented)

Time = 9.68 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.74 \[ \int \frac {1}{\sqrt {c d^2+2 c d e x+c e^2 x^2}} \, dx=\frac {\ln \left (c\,x\,e^2+c\,d\,e\right )\,\mathrm {sign}\left (c\,e\,\left (d+e\,x\right )\right )}{\sqrt {c\,e^2}} \]

[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)