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

Optimal. Leaf size=137 \[ -\frac {\sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{d x}-\frac {\left (c d^2-a e^2\right ) \tanh ^{-1}\left (\frac {x \left (a e^2+c d^2\right )+2 a d e}{2 \sqrt {a} \sqrt {d} \sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{2 \sqrt {a} d^{3/2} \sqrt {e}} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {849, 806, 724, 206} \[ -\frac {\sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{d x}-\frac {\left (c d^2-a e^2\right ) \tanh ^{-1}\left (\frac {x \left (a e^2+c d^2\right )+2 a d e}{2 \sqrt {a} \sqrt {d} \sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{2 \sqrt {a} d^{3/2} \sqrt {e}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 724

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 806

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((e*f - d*g)*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 - b*d*e + a*e^2)), x] - Dist[(b
*(e*f + d*g) - 2*(c*d*f + a*e*g))/(2*(c*d^2 - b*d*e + a*e^2)), 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] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[Sim
plify[m + 2*p + 3], 0]

Rule 849

Int[((x_)^(n_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_))/((d_) + (e_.)*(x_)), x_Symbol] :> Int[x^n*(a/d + (c*
x)/e)*(a + b*x + c*x^2)^(p - 1), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b
*d*e + a*e^2, 0] &&  !IntegerQ[p] && ( !IntegerQ[n] ||  !IntegerQ[2*p] || IGtQ[n, 2] || (GtQ[p, 0] && NeQ[n, 2
]))

Rubi steps

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

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Mathematica [A]  time = 0.14, size = 117, normalized size = 0.85 \[ \frac {\sqrt {(d+e x) (a e+c d x)} \left (\frac {\left (a e^2-c d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a e+c d x}}{\sqrt {a} \sqrt {e} \sqrt {d+e x}}\right )}{\sqrt {a} \sqrt {e} \sqrt {d+e x} \sqrt {a e+c d x}}-\frac {\sqrt {d}}{x}\right )}{d^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.08, size = 355, normalized size = 2.59 \[ \left [-\frac {4 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} a d e + {\left (c d^{2} - a e^{2}\right )} \sqrt {a d e} x \log \left (\frac {8 \, a^{2} d^{2} e^{2} + {\left (c^{2} d^{4} + 6 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} x^{2} + 4 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (2 \, a d e + {\left (c d^{2} + a e^{2}\right )} x\right )} \sqrt {a d e} + 8 \, {\left (a c d^{3} e + a^{2} d e^{3}\right )} x}{x^{2}}\right )}{4 \, a d^{2} e x}, -\frac {2 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} a d e - {\left (c d^{2} - a e^{2}\right )} \sqrt {-a d e} x \arctan \left (\frac {\sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (2 \, a d e + {\left (c d^{2} + a e^{2}\right )} x\right )} \sqrt {-a d e}}{2 \, {\left (a c d^{2} e^{2} x^{2} + a^{2} d^{2} e^{2} + {\left (a c d^{3} e + a^{2} d e^{3}\right )} x\right )}}\right )}{2 \, a d^{2} e x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*(4*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*a*d*e + (c*d^2 - a*e^2)*sqrt(a*d*e)*x*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*d^2*e*x), -1/2*(2*sqrt(c*d*e*x^2 + a*d*e + (c*d^
2 + a*e^2)*x)*a*d*e - (c*d^2 - a*e^2)*sqrt(-a*d*e)*x*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*d^2*
e*x)]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: 2*((-2*exp(1)*a*exp(2)+2*exp(1)^3*a)/d/2
/sqrt(-a*d*exp(1)^3+a*d*exp(1)*exp(2))*atan((-d*sqrt(c*d*exp(1))+(sqrt(a*d*exp(1)+a*x*exp(2)+c*d^2*x+c*d*x^2*e
xp(1))-sqrt(c*d*exp(1))*x)*exp(1))/sqrt(-a*d*exp(1)^3+a*d*exp(1)*exp(2)))-(-a*exp(2)+2*exp(1)^2*a-c*d^2)/d/2/s
qrt(-a*d*exp(1))*atan((sqrt(a*d*exp(1)+a*x*exp(2)+c*d^2*x+c*d*x^2*exp(1))-sqrt(c*d*exp(1))*x)/sqrt(-a*d*exp(1)
))-((sqrt(a*d*exp(1)+a*x*exp(2)+c*d^2*x+c*d*x^2*exp(1))-sqrt(c*d*exp(1))*x)*a*exp(2)+c*d^2*(sqrt(a*d*exp(1)+a*
x*exp(2)+c*d^2*x+c*d*x^2*exp(1))-sqrt(c*d*exp(1))*x)-2*d*exp(1)*sqrt(c*d*exp(1))*a)/2/d/((sqrt(a*d*exp(1)+a*x*
exp(2)+c*d^2*x+c*d*x^2*exp(1))-sqrt(c*d*exp(1))*x)^2-d*exp(1)*a))

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maple [B]  time = 0.02, size = 594, normalized size = 4.34 \[ \frac {a \,e^{2} \ln \left (\frac {2 a d e +\left (a \,e^{2}+c \,d^{2}\right ) x +2 \sqrt {a d e}\, \sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}}{x}\right )}{2 \sqrt {a d e}\, d}+\frac {a \,e^{3} \ln \left (\frac {\frac {a \,e^{2}}{2}-\frac {c \,d^{2}}{2}+\left (x +\frac {d}{e}\right ) c d e}{\sqrt {c d e}}+\sqrt {\left (x +\frac {d}{e}\right )^{2} c d e +\left (a \,e^{2}-c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}\right )}{2 \sqrt {c d e}\, d^{2}}-\frac {a \,e^{3} \ln \left (\frac {c d e x +\frac {1}{2} a \,e^{2}+\frac {1}{2} c \,d^{2}}{\sqrt {c d e}}+\sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\right )}{2 \sqrt {c d e}\, d^{2}}-\frac {c d \ln \left (\frac {2 a d e +\left (a \,e^{2}+c \,d^{2}\right ) x +2 \sqrt {a d e}\, \sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}}{x}\right )}{2 \sqrt {a d e}}-\frac {c e \ln \left (\frac {\frac {a \,e^{2}}{2}-\frac {c \,d^{2}}{2}+\left (x +\frac {d}{e}\right ) c d e}{\sqrt {c d e}}+\sqrt {\left (x +\frac {d}{e}\right )^{2} c d e +\left (a \,e^{2}-c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}\right )}{2 \sqrt {c d e}}+\frac {c e \ln \left (\frac {c d e x +\frac {1}{2} a \,e^{2}+\frac {1}{2} c \,d^{2}}{\sqrt {c d e}}+\sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\right )}{2 \sqrt {c d e}}+\frac {\sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\, c x}{a d}+\frac {\sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\, c}{a e}+\frac {\sqrt {\left (x +\frac {d}{e}\right )^{2} c d e +\left (a \,e^{2}-c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}\, e}{d^{2}}-\frac {\left (c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x \right )^{\frac {3}{2}}}{a \,d^{2} e x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x}}{{\left (e x + d\right )} x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}}{x^2\,\left (d+e\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\left (d + e x\right ) \left (a e + c d x\right )}}{x^{2} \left (d + e x\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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