3.1948 \(\int \frac{d+e x}{\sqrt{a d e+(c d^2+a e^2) x+c d e x^2}} \, dx\)

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

[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])

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Rubi [A]  time = 0.055865, antiderivative size = 134, normalized size of antiderivative = 1., 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} \[ \frac{\left (c d^2-a e^2\right ) \tanh ^{-1}\left (\frac{a e^2+c d^2+2 c d e x}{2 \sqrt{c} \sqrt{d} \sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{2 c^{3/2} d^{3/2} \sqrt{e}}+\frac{\sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{c d} \]

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 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]

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 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])

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 = 0.203789, size = 203, normalized size = 1.51 \[ \frac{\sqrt{(d+e x) (a e+c d x)} \left (\sqrt{c} \sqrt{d} \sqrt{e} \sqrt{a e+c d x} \sqrt{\frac{c d (d+e x)}{c d^2-a e^2}}+\sqrt{c d} \sqrt{c d^2-a e^2} \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{e} \sqrt{a e+c d x}}{\sqrt{c d} \sqrt{c d^2-a e^2}}\right )\right )}{c^{3/2} d^{3/2} \sqrt{e} \sqrt{a e+c d x} \sqrt{\frac{c d (d+e x)}{c d^2-a e^2}}} \]

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)])

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Maple [A]  time = 0.049, size = 171, normalized size = 1.3 \begin{align*}{\frac{1}{cd}\sqrt{ade+ \left ( a{e}^{2}+c{d}^{2} \right ) x+cde{x}^{2}}}-{\frac{a{e}^{2}}{2\,cd}\ln \left ({ \left ({\frac{a{e}^{2}}{2}}+{\frac{c{d}^{2}}{2}}+cdex \right ){\frac{1}{\sqrt{dec}}}}+\sqrt{ade+ \left ( a{e}^{2}+c{d}^{2} \right ) x+cde{x}^{2}} \right ){\frac{1}{\sqrt{dec}}}}+{\frac{d}{2}\ln \left ({ \left ({\frac{a{e}^{2}}{2}}+{\frac{c{d}^{2}}{2}}+cdex \right ){\frac{1}{\sqrt{dec}}}}+\sqrt{ade+ \left ( a{e}^{2}+c{d}^{2} \right ) x+cde{x}^{2}} \right ){\frac{1}{\sqrt{dec}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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

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Fricas [A]  time = 1.96632, size = 730, normalized size = 5.45 \begin{align*} \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{align*}

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)]

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

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)

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Giac [A]  time = 1.30856, size = 181, normalized size = 1.35 \begin{align*} -\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{align*}

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)