### 3.1912 $$\int \frac{\sqrt{a d e+(c d^2+a e^2) x+c d e x^2}}{d+e x} \, dx$$

Optimal. Leaf size=131 $\frac{\sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{e}-\frac{\left (c d^2-a e^2\right ) \tanh ^{-1}\left (\frac{a e^2+c d^2+2 c d e x}{2 \sqrt{c} \sqrt{d} \sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{2 \sqrt{c} \sqrt{d} e^{3/2}}$

[Out]

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

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Rubi [A]  time = 0.0672677, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 37, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.081, Rules used = {664, 621, 206} $\frac{\sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{e}-\frac{\left (c d^2-a e^2\right ) \tanh ^{-1}\left (\frac{a e^2+c d^2+2 c d e x}{2 \sqrt{c} \sqrt{d} \sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{2 \sqrt{c} \sqrt{d} e^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 664

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + 2*p + 1)), x] - Dist[(p*(2*c*d - b*e))/(e^2*(m + 2*p + 1)), Int[(d + e*x)^(m + 1)*
(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a
*e^2, 0] && GtQ[p, 0] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && NeQ[m + 2*p + 1, 0] && IntegerQ[2*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 \frac{\sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{d+e x} \, dx &=\frac{\sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{e}-\frac{\left (2 c d^2 e-e \left (c d^2+a e^2\right )\right ) \int \frac{1}{\sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{2 e^2}\\ &=\frac{\sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{e}-\frac{\left (2 c d^2 e-e \left (c d^2+a e^2\right )\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 )}{e^2}\\ &=\frac{\sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{e}-\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 \sqrt{c} \sqrt{d} e^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.786715, size = 155, normalized size = 1.18 $\frac{\sqrt{(d+e x) (a e+c d x)} \left (\sqrt{e}-\frac{c^{3/2} d^{3/2} \sqrt{c d^2-a e^2} \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{e} \sqrt{a e+c d x}}{\sqrt{c d} \sqrt{c d^2-a e^2}}\right )}{(c d)^{3/2} \sqrt{a e+c d x} \sqrt{\frac{c d (d+e x)}{c d^2-a e^2}}}\right )}{e^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.045, size = 205, normalized size = 1.6 \begin{align*}{\frac{1}{e}\sqrt{cde \left ({\frac{d}{e}}+x \right ) ^{2}+ \left ( a{e}^{2}-c{d}^{2} \right ) \left ({\frac{d}{e}}+x \right ) }}+{\frac{ae}{2}\ln \left ({ \left ({\frac{a{e}^{2}}{2}}-{\frac{c{d}^{2}}{2}}+ \left ({\frac{d}{e}}+x \right ) cde \right ){\frac{1}{\sqrt{dec}}}}+\sqrt{cde \left ({\frac{d}{e}}+x \right ) ^{2}+ \left ( a{e}^{2}-c{d}^{2} \right ) \left ({\frac{d}{e}}+x \right ) } \right ){\frac{1}{\sqrt{dec}}}}-{\frac{c{d}^{2}}{2\,e}\ln \left ({ \left ({\frac{a{e}^{2}}{2}}-{\frac{c{d}^{2}}{2}}+ \left ({\frac{d}{e}}+x \right ) cde \right ){\frac{1}{\sqrt{dec}}}}+\sqrt{cde \left ({\frac{d}{e}}+x \right ) ^{2}+ \left ( a{e}^{2}-c{d}^{2} \right ) \left ({\frac{d}{e}}+x \right ) } \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)/(e*x+d),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.70254, size = 725, normalized size = 5.53 \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 d e^{2}}, \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 d 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)/(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)*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*d*e^2), 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*d*e^2)]

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

[Out]

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

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \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)/(e*x+d),x, algorithm="giac")

[Out]

Exception raised: TypeError