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

Optimal. Leaf size=134 \[ \frac {\left (c d^2-a e^2\right ) \tanh ^{-1}\left (\frac {a e^2+c d^2+2 c d e x}{2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{2 c^{3/2} d^{3/2} \sqrt {e}}+\frac {\sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{c d} \]

________________________________________________________________________________________

Rubi [A]  time = 0.06, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {640, 621, 206} \begin {gather*} \frac {\left (c d^2-a e^2\right ) \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 )}{2 c^{3/2} d^{3/2} \sqrt {e}}+\frac {\sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{c d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]/(c*d) + ((c*d^2 - a*e^2)*ArcTanh[(c*d^2 + a*e^2 + 2*c*d*e*x)/(2*Sq
rt[c]*Sqrt[d]*Sqrt[e]*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])])/(2*c^(3/2)*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 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 640

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.21, size = 203, normalized size = 1.51 \begin {gather*} \frac {\sqrt {(d+e x) (a e+c d x)} \left (\sqrt {c} \sqrt {d} \sqrt {e} \sqrt {a e+c d x} \sqrt {\frac {c d (d+e x)}{c d^2-a e^2}}+\sqrt {c d} \sqrt {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 )\right )}{c^{3/2} d^{3/2} \sqrt {e} \sqrt {a e+c d x} \sqrt {\frac {c d (d+e x)}{c d^2-a e^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

IntegrateAlgebraic [B]  time = 0.97, size = 275, normalized size = 2.05 \begin {gather*} -\frac {\sqrt {c d e} \left (c d^2-a e^2\right ) \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 )}{4 c^2 d^2 e}+\frac {\left (c d^2-a e^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {e} \left (2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}-2 x \sqrt {c d e}\right )}{a e^2+c d^2}\right )}{2 c^{3/2} d^{3/2} \sqrt {e}}+\frac {\sqrt {a d e+a e^2 x+c d^2 x+c d e x^2}}{c d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.43, size = 338, normalized size = 2.52 \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 e - {\left (c d^{2} - a e^{2}\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 )}{4 \, c^{2} d^{2} e}, \frac {2 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} c d e - {\left (c d^{2} - a e^{2}\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 )}{2 \, c^{2} d^{2} e}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.83, size = 134, normalized size = 1.00 \begin {gather*} -\frac {{\left (c d^{2} - a e^{2}\right )} \sqrt {c d} e^{\left (-\frac {1}{2}\right )} \log \left ({\left | -\sqrt {c d} c d^{2} e^{\frac {1}{2}} - 2 \, {\left (\sqrt {c d} x e^{\frac {1}{2}} - \sqrt {c d x^{2} e + a d e + {\left (c d^{2} + a e^{2}\right )} x}\right )} c d e - \sqrt {c d} a e^{\frac {5}{2}} \right |}\right )}{2 \, c^{2} d^{2}} + \frac {\sqrt {c d x^{2} e + a d e + {\left (c d^{2} + a e^{2}\right )} x}}{c d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(c*d^2 - a*e^2)*sqrt(c*d)*e^(-1/2)*log(abs(-sqrt(c*d)*c*d^2*e^(1/2) - 2*(sqrt(c*d)*x*e^(1/2) - sqrt(c*d*x
^2*e + a*d*e + (c*d^2 + a*e^2)*x))*c*d*e - sqrt(c*d)*a*e^(5/2)))/(c^2*d^2) + sqrt(c*d*x^2*e + a*d*e + (c*d^2 +
a*e^2)*x)/(c*d)

________________________________________________________________________________________

maple [A]  time = 0.06, size = 171, normalized size = 1.28 \begin {gather*} -\frac {a \,e^{2} \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}\, c d}+\frac {d \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 d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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((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?

________________________________________________________________________________________

mupad [B]  time = 1.07, size = 144, normalized size = 1.07 \begin {gather*} \frac {\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}}{c\,d}-\frac {a\,e^3\,\ln \left (2\,\sqrt {\left (a\,e+c\,d\,x\right )\,\left (d+e\,x\right )}\,\sqrt {c\,d\,e}+a\,e^2+c\,d^2+2\,c\,d\,e\,x\right )}{2\,{\left (c\,d\,e\right )}^{3/2}}+\frac {c\,d^2\,e\,\ln \left (2\,\sqrt {\left (a\,e+c\,d\,x\right )\,\left (d+e\,x\right )}\,\sqrt {c\,d\,e}+a\,e^2+c\,d^2+2\,c\,d\,e\,x\right )}{2\,{\left (c\,d\,e\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________