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

Optimal. Leaf size=139 \[ \frac {\tanh ^{-1}\left (\frac {a e^2+c d^2+2 c d e x}{2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{\sqrt {c} \sqrt {d} e^{3/2}}-\frac {2 d \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{e (d+e x) \left (c d^2-a e^2\right )} \]

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Rubi [A]  time = 0.11, antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.079, Rules used = {792, 621, 206} \begin {gather*} \frac {\tanh ^{-1}\left (\frac {a e^2+c d^2+2 c d e x}{2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{\sqrt {c} \sqrt {d} e^{3/2}}-\frac {2 d \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{e (d+e x) \left (c d^2-a e^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 206

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

Rule 621

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

Rule 792

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

Rubi steps

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

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Mathematica [A]  time = 0.42, size = 189, normalized size = 1.36 \begin {gather*} \frac {2 \sqrt {c d} \left (c d^2-a e^2\right )^{3/2} \sqrt {a e+c d x} \sqrt {\frac {c d (d+e x)}{c d^2-a e^2}} \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {e} \sqrt {a e+c d x}}{\sqrt {c d} \sqrt {c d^2-a e^2}}\right )-2 c^{3/2} d^{5/2} \sqrt {e} (a e+c d x)}{c^{3/2} d^{3/2} e^{3/2} \left (c d^2-a e^2\right ) \sqrt {(d+e x) (a e+c d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*c^(3/2)*d^(5/2)*Sqrt[e]*(a*e + c*d*x) + 2*Sqrt[c*d]*(c*d^2 - a*e^2)^(3/2)*Sqrt[a*e + c*d*x]*Sqrt[(c*d*(d +
e*x))/(c*d^2 - a*e^2)]*ArcSinh[(Sqrt[c]*Sqrt[d]*Sqrt[e]*Sqrt[a*e + c*d*x])/(Sqrt[c*d]*Sqrt[c*d^2 - a*e^2])])/
(c^(3/2)*d^(3/2)*e^(3/2)*(c*d^2 - a*e^2)*Sqrt[(a*e + c*d*x)*(d + e*x)])

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IntegrateAlgebraic [B]  time = 0.74, size = 296, normalized size = 2.13 \begin {gather*} -\frac {\sqrt {c d e} \log \left (a^2 e^4+8 c d e x \sqrt {c d e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}-2 a c d^2 e^2-4 a c d e^3 x+c^2 d^4-4 c^2 d^3 e x-8 c^2 d^2 e^2 x^2\right )}{2 c d e^2}+\frac {2 d \sqrt {a d e+a e^2 x+c d^2 x+c d e x^2}}{e (d+e x) \left (a e^2-c d^2\right )}-\frac {\tanh ^{-1}\left (\frac {2 \sqrt {c} \sqrt {d} \sqrt {e} x \sqrt {c d e}}{a e^2+c d^2}-\frac {2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{a e^2+c d^2}\right )}{\sqrt {c} \sqrt {d} e^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*d*Sqrt[a*d*e + c*d^2*x + a*e^2*x + c*d*e*x^2])/(e*(-(c*d^2) + a*e^2)*(d + e*x)) - ArcTanh[(2*Sqrt[c]*Sqrt[d
]*Sqrt[e]*Sqrt[c*d*e]*x)/(c*d^2 + a*e^2) - (2*Sqrt[c]*Sqrt[d]*Sqrt[e]*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x
^2])/(c*d^2 + a*e^2)]/(Sqrt[c]*Sqrt[d]*e^(3/2)) - (Sqrt[c*d*e]*Log[c^2*d^4 - 2*a*c*d^2*e^2 + a^2*e^4 - 4*c^2*d
^3*e*x - 4*a*c*d*e^3*x - 8*c^2*d^2*e^2*x^2 + 8*c*d*e*Sqrt[c*d*e]*x*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]
])/(2*c*d*e^2)

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fricas [A]  time = 0.55, size = 443, normalized size = 3.19 \begin {gather*} \left [-\frac {4 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} c d^{2} e - {\left (c d^{3} - a d e^{2} + {\left (c d^{2} e - a e^{3}\right )} x\right )} \sqrt {c d e} \log \left (8 \, c^{2} d^{2} e^{2} x^{2} + c^{2} d^{4} + 6 \, a c d^{2} e^{2} + a^{2} e^{4} + 4 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (2 \, c d e x + c d^{2} + a e^{2}\right )} \sqrt {c d e} + 8 \, {\left (c^{2} d^{3} e + a c d e^{3}\right )} x\right )}{2 \, {\left (c^{2} d^{4} e^{2} - a c d^{2} e^{4} + {\left (c^{2} d^{3} e^{3} - a c d e^{5}\right )} x\right )}}, -\frac {2 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} c d^{2} e + {\left (c d^{3} - a d e^{2} + {\left (c d^{2} e - a e^{3}\right )} x\right )} \sqrt {-c 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 \, c d e x + c d^{2} + a e^{2}\right )} \sqrt {-c d e}}{2 \, {\left (c^{2} d^{2} e^{2} x^{2} + a c d^{2} e^{2} + {\left (c^{2} d^{3} e + a c d e^{3}\right )} x\right )}}\right )}{c^{2} d^{4} e^{2} - a c d^{2} e^{4} + {\left (c^{2} d^{3} e^{3} - a c d e^{5}\right )} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(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,x):;OUTPUT:Erro
r: Bad Argument Type

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maple [A]  time = 0.01, size = 131, normalized size = 0.94 \begin {gather*} \frac {\ln \left (\frac {c d e x +\frac {1}{2} a \,e^{2}+\frac {1}{2} c \,d^{2}}{\sqrt {c d e}}+\sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\right )}{\sqrt {c d e}\, e}+\frac {2 \sqrt {\left (x +\frac {d}{e}\right )^{2} c d e +\left (a \,e^{2}-c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}\, d}{\left (a \,e^{2}-c \,d^{2}\right ) \left (x +\frac {d}{e}\right ) e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(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]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*e^2-c*d^2>0)', see `assume?`
for more details)Is a*e^2-c*d^2 zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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