### 3.1911 $$\int \sqrt{a d e+(c d^2+a e^2) x+c d e x^2} \, dx$$

Optimal. Leaf size=159 $\frac{\left (a e^2+c d^2+2 c d e x\right ) \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{4 c d e}-\frac{\left (c d^2-a e^2\right )^2 \tanh ^{-1}\left (\frac{a e^2+c d^2+2 c d e x}{2 \sqrt{c} \sqrt{d} \sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{8 c^{3/2} d^{3/2} e^{3/2}}$

[Out]

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

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Rubi [A]  time = 0.0502793, antiderivative size = 159, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 29, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.103, Rules used = {612, 621, 206} $\frac{\left (a e^2+c d^2+2 c d e x\right ) \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{4 c d e}-\frac{\left (c d^2-a e^2\right )^2 \tanh ^{-1}\left (\frac{a e^2+c d^2+2 c d e x}{2 \sqrt{c} \sqrt{d} \sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{8 c^{3/2} d^{3/2} e^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 612

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

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 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx &=\frac{\left (c d^2+a e^2+2 c d e x\right ) \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 c d e}-\frac{\left (c d^2-a e^2\right )^2 \int \frac{1}{\sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{8 c d e}\\ &=\frac{\left (c d^2+a e^2+2 c d e x\right ) \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 c d e}-\frac{\left (c d^2-a e^2\right )^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 )}{4 c d e}\\ &=\frac{\left (c d^2+a e^2+2 c d e x\right ) \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 c d e}-\frac{\left (c d^2-a e^2\right )^2 \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 )}{8 c^{3/2} d^{3/2} e^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.487997, size = 214, normalized size = 1.35 $\frac{\sqrt{c} \sqrt{d} \left (\sqrt{c} \sqrt{d} \sqrt{e} \sqrt{c d} (d+e x) \left (a^2 e^3+a c d e (d+3 e x)+c^2 d^2 x (d+2 e x)\right )-\left (c d^2-a e^2\right )^{5/2} \sqrt{a e+c d x} \sqrt{\frac{c d (d+e x)}{c d^2-a e^2}} \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{e} \sqrt{a e+c d x}}{\sqrt{c d} \sqrt{c d^2-a e^2}}\right )\right )}{4 e^{3/2} (c d)^{5/2} \sqrt{(d+e x) (a e+c d x)}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(Sqrt[c]*Sqrt[d]*(Sqrt[c]*Sqrt[d]*Sqrt[c*d]*Sqrt[e]*(d + e*x)*(a^2*e^3 + c^2*d^2*x*(d + 2*e*x) + a*c*d*e*(d +
3*e*x)) - (c*d^2 - a*e^2)^(5/2)*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])]))/(4*(c*d)^(5/2)*e^(3/2)*Sqrt[(a*e + c*d*x)*(d
+ e*x)])

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Maple [A]  time = 0.044, size = 265, normalized size = 1.7 \begin{align*}{\frac{2\,cdex+a{e}^{2}+c{d}^{2}}{4\,dec}\sqrt{ade+ \left ( a{e}^{2}+c{d}^{2} \right ) x+cde{x}^{2}}}-{\frac{{a}^{2}{e}^{3}}{8\,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{ade}{4}\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{c{d}^{3}}{8\,e}\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((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2),x)

[Out]

1/4*(2*c*d*e*x+a*e^2+c*d^2)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)/c/d/e-1/8/d*e^3/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^2+1/4*d*e*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-1/8*d^3/e*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)

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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((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 = 1.79373, size = 888, normalized size = 5.58 \begin{align*} \left [\frac{{\left (c^{2} d^{4} - 2 \, a c d^{2} e^{2} + a^{2} e^{4}\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 \,{\left (2 \, c^{2} d^{2} e^{2} x + c^{2} d^{3} e + a c d e^{3}\right )} \sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x}}{16 \, c^{2} d^{2} e^{2}}, \frac{{\left (c^{2} d^{4} - 2 \, a c d^{2} e^{2} + a^{2} e^{4}\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 \,{\left (2 \, c^{2} d^{2} e^{2} x + c^{2} d^{3} e + a c d e^{3}\right )} \sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x}}{8 \, c^{2} d^{2} e^{2}}\right ] \end{align*}

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, algorithm="fricas")

[Out]

[1/16*((c^2*d^4 - 2*a*c*d^2*e^2 + a^2*e^4)*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) + 4*(2*c^2*d^2*e^2*x + c^2*d^3*e + a*c*d*e^3)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))/(c^2*d^
2*e^2), 1/8*((c^2*d^4 - 2*a*c*d^2*e^2 + a^2*e^4)*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)) +
2*(2*c^2*d^2*e^2*x + c^2*d^3*e + a*c*d*e^3)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))/(c^2*d^2*e^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a d e + c d e x^{2} + x \left (a e^{2} + c d^{2}\right )}\, dx \end{align*}

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)

[Out]

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

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Giac [A]  time = 1.35217, size = 219, normalized size = 1.38 \begin{align*} \frac{1}{4} \, \sqrt{c d x^{2} e + c d^{2} x + a x e^{2} + a d e}{\left (2 \, x + \frac{{\left (c d^{2} + a e^{2}\right )} e^{\left (-1\right )}}{c d}\right )} + \frac{{\left (c^{2} d^{4} - 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} \sqrt{c d} e^{\left (-\frac{3}{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 + c d^{2} x + a x e^{2} + a d e}\right )} c d e - \sqrt{c d} a e^{\frac{5}{2}} \right |}\right )}{8 \, c^{2} d^{2}} \end{align*}

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, algorithm="giac")

[Out]

1/4*sqrt(c*d*x^2*e + c*d^2*x + a*x*e^2 + a*d*e)*(2*x + (c*d^2 + a*e^2)*e^(-1)/(c*d)) + 1/8*(c^2*d^4 - 2*a*c*d^
2*e^2 + a^2*e^4)*sqrt(c*d)*e^(-3/2)*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
+ c*d^2*x + a*x*e^2 + a*d*e))*c*d*e - sqrt(c*d)*a*e^(5/2)))/(c^2*d^2)