\(\int \frac {1}{x (d+e x) \sqrt {a d e+(c d^2+a e^2) x+c d e x^2}} \, dx\) [474]

Optimal result

Integrand size = 40, antiderivative size = 143 \[ \int \frac {1}{x (d+e x) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=-\frac {2 e (a e+c d x)}{d \left (c d^2-a e^2\right ) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}-\frac {\text {arctanh}\left (\frac {2 a d e+\left (c d^2+a e^2\right ) x}{2 \sqrt {a} \sqrt {d} \sqrt {e} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\right )}{\sqrt {a} d^{3/2} \sqrt {e}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {865, 836, 12, 738, 212} \[ \int \frac {1}{x (d+e x) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=-\frac {\text {arctanh}\left (\frac {x \left (a e^2+c d^2\right )+2 a d e}{2 \sqrt {a} \sqrt {d} \sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{\sqrt {a} d^{3/2} \sqrt {e}}-\frac {2 e (a e+c d x)}{d \left (c d^2-a e^2\right ) \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}} \]

[In]

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

[Out]

(-2*e*(a*e + c*d*x))/(d*(c*d^2 - a*e^2)*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) - ArcTanh[(2*a*d*e + (c*d
^2 + a*e^2)*x)/(2*Sqrt[a]*Sqrt[d]*Sqrt[e]*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])]/(Sqrt[a]*d^(3/2)*Sqrt[
e])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 738

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rule 836

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[(d + e*x)^(m + 1)*(f*(b*c*d - b^2*e + 2*a*c*e) - a*g*(2*c*d - b*e) + c*(f*(2*c*d - b*e) - g*(b*d - 2*a*e))*x)
*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2))), x] + Dist[1/((p + 1)*(b^2 - 4*a*
c)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^m*(a + b*x + c*x^2)^(p + 1)*Simp[f*(b*c*d*e*(2*p - m + 2) + b^2*e^2
*(p + m + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3)) - g*(a*e*(b*e - 2*c*d*m + b*e*m) - b*d*(3*c*d -
b*e + 2*c*d*p - b*e*p)) + c*e*(g*(b*d - 2*a*e) - f*(2*c*d - b*e))*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, b,
c, d, e, f, g, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] ||
IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 865

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :>
Int[((f + g*x)^n*(a + b*x + c*x^2)^(m + p))/(a/d + c*(x/e))^m, x] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] &&
NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && ILtQ[m, 0] && In
tegerQ[n] && (LtQ[n, 0] || GtQ[p, 0])

Rubi steps \begin{align*} \text {integral}& = \int \frac {a e+c d x}{x \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx \\ & = -\frac {2 e (a e+c d x)}{d \left (c d^2-a e^2\right ) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}-\frac {2 \int -\frac {a e \left (c d^2-a e^2\right )^2}{2 x \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{a d e \left (c d^2-a e^2\right )^2} \\ & = -\frac {2 e (a e+c d x)}{d \left (c d^2-a e^2\right ) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {\int \frac {1}{x \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{d} \\ & = -\frac {2 e (a e+c d x)}{d \left (c d^2-a e^2\right ) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}-\frac {2 \text {Subst}\left (\int \frac {1}{4 a d e-x^2} \, dx,x,\frac {2 a d e-\left (-c d^2-a e^2\right ) x}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\right )}{d} \\ & = -\frac {2 e (a e+c d x)}{d \left (c d^2-a e^2\right ) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}-\frac {\tanh ^{-1}\left (\frac {2 a d e+\left (c d^2+a e^2\right ) x}{2 \sqrt {a} \sqrt {d} \sqrt {e} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\right )}{\sqrt {a} d^{3/2} \sqrt {e}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.92 \[ \int \frac {1}{x (d+e x) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=\frac {2 \left (-\frac {\sqrt {d} e^{3/2} (a e+c d x)}{c d^2-a e^2}-\frac {\sqrt {a e+c d x} \sqrt {d+e x} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a e+c d x}}{\sqrt {a} \sqrt {e} \sqrt {d+e x}}\right )}{\sqrt {a}}\right )}{d^{3/2} \sqrt {e} \sqrt {(a e+c d x) (d+e x)}} \]

[In]

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

[Out]

(2*(-((Sqrt[d]*e^(3/2)*(a*e + c*d*x))/(c*d^2 - a*e^2)) - (Sqrt[a*e + c*d*x]*Sqrt[d + e*x]*ArcTanh[(Sqrt[d]*Sqr
t[a*e + c*d*x])/(Sqrt[a]*Sqrt[e]*Sqrt[d + e*x])])/Sqrt[a]))/(d^(3/2)*Sqrt[e]*Sqrt[(a*e + c*d*x)*(d + e*x)])

Maple [A] (verified)

Time = 0.59 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.95

method	result	size
default	\(-\frac {\ln \left (\frac {2 a d e +\left (e^{2} a +c \,d^{2}\right ) x +2 \sqrt {a d e}\, \sqrt {a d e +\left (e^{2} a +c \,d^{2}\right ) x +c d e \,x^{2}}}{x}\right )}{d \sqrt {a d e}}+\frac {2 \sqrt {c d e \left (x +\frac {d}{e}\right )^{2}+\left (e^{2} a -c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}}{d \left (e^{2} a -c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}\)	\(136\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.51 (sec) , antiderivative size = 454, normalized size of antiderivative = 3.17 \[ \int \frac {1}{x (d+e x) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=\left [-\frac {4 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} a d e^{2} - {\left (c d^{3} - a d e^{2} + {\left (c d^{2} e - a e^{3}\right )} x\right )} \sqrt {a d e} \log \left (\frac {8 \, a^{2} d^{2} e^{2} + {\left (c^{2} d^{4} + 6 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} x^{2} - 4 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (2 \, a d e + {\left (c d^{2} + a e^{2}\right )} x\right )} \sqrt {a d e} + 8 \, {\left (a c d^{3} e + a^{2} d e^{3}\right )} x}{x^{2}}\right )}{2 \, {\left (a c d^{5} e - a^{2} d^{3} e^{3} + {\left (a c d^{4} e^{2} - a^{2} d^{2} e^{4}\right )} x\right )}}, -\frac {2 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} a d e^{2} - {\left (c d^{3} - a d e^{2} + {\left (c d^{2} e - a e^{3}\right )} x\right )} \sqrt {-a d e} \arctan \left (\frac {\sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (2 \, a d e + {\left (c d^{2} + a e^{2}\right )} x\right )} \sqrt {-a d e}}{2 \, {\left (a c d^{2} e^{2} x^{2} + a^{2} d^{2} e^{2} + {\left (a c d^{3} e + a^{2} d e^{3}\right )} x\right )}}\right )}{a c d^{5} e - a^{2} d^{3} e^{3} + {\left (a c d^{4} e^{2} - a^{2} d^{2} e^{4}\right )} x}\right ] \]

[In]

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

[Out]

[-1/2*(4*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*a*d*e^2 - (c*d^3 - a*d*e^2 + (c*d^2*e - a*e^3)*x)*sqrt(a*
d*e)*log((8*a^2*d^2*e^2 + (c^2*d^4 + 6*a*c*d^2*e^2 + a^2*e^4)*x^2 - 4*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)
*x)*(2*a*d*e + (c*d^2 + a*e^2)*x)*sqrt(a*d*e) + 8*(a*c*d^3*e + a^2*d*e^3)*x)/x^2))/(a*c*d^5*e - a^2*d^3*e^3 +
(a*c*d^4*e^2 - a^2*d^2*e^4)*x), -(2*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*a*d*e^2 - (c*d^3 - a*d*e^2 + (
c*d^2*e - a*e^3)*x)*sqrt(-a*d*e)*arctan(1/2*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*(2*a*d*e + (c*d^2 + a*
e^2)*x)*sqrt(-a*d*e)/(a*c*d^2*e^2*x^2 + a^2*d^2*e^2 + (a*c*d^3*e + a^2*d*e^3)*x)))/(a*c*d^5*e - a^2*d^3*e^3 +
(a*c*d^4*e^2 - a^2*d^2*e^4)*x)]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F(-2)]

Exception generated. \[ \int \frac {1}{x (d+e x) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Unable to divide, perhaps due to rounding error%%%{%%%{1,[0,1,5]%%%},[2,2]%%%}+%%%{%%%{-2,[1,3,3]%%%},[2,1]
%%%}+%%%{%%

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x (d+e x) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=\int \frac {1}{x\,\left (d+e\,x\right )\,\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}} \,d x \]

[In]

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

[Out]

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