3.2032 \(\int \frac{\sqrt{a d e+(c d^2+a e^2) x+c d e x^2}}{(d+e x)^{3/2}} \, dx\)

Optimal. Leaf size=128 \[ \frac{2 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{e \sqrt{d+e x}}-\frac{2 \sqrt{c d^2-a e^2} \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt{d+e x} \sqrt{c d^2-a e^2}}\right )}{e^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0978779, antiderivative size = 128, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {664, 660, 205} \[ \frac{2 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{e \sqrt{d+e x}}-\frac{2 \sqrt{c d^2-a e^2} \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt{d+e x} \sqrt{c d^2-a e^2}}\right )}{e^{3/2}} \]

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)^(3/2),x]

[Out]

(2*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(e*Sqrt[d + e*x]) - (2*Sqrt[c*d^2 - a*e^2]*ArcTan[(Sqrt[e]*Sqr
t[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(Sqrt[c*d^2 - a*e^2]*Sqrt[d + e*x])])/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 660

Int[1/(Sqrt[(d_.) + (e_.)*(x_)]*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[2*e, Subst[Int[1/(
2*c*d - b*e + e^2*x^2), x], x, Sqrt[a + b*x + c*x^2]/Sqrt[d + e*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]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

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

Mathematica [A]  time = 0.119412, size = 105, normalized size = 0.82 \[ \frac{2 \sqrt{(d+e x) (a e+c d x)} \left (\sqrt{e}-\frac{\sqrt{c d^2-a e^2} \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{a e+c d x}}{\sqrt{c d^2-a e^2}}\right )}{\sqrt{a e+c d x}}\right )}{e^{3/2} \sqrt{d+e x}} \]

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)^(3/2),x]

[Out]

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

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Maple [A]  time = 0.26, size = 163, normalized size = 1.3 \begin{align*} -2\,{\frac{\sqrt{cde{x}^{2}+a{e}^{2}x+c{d}^{2}x+ade}}{\sqrt{ex+d}\sqrt{cdx+ae}e\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}} \left ({\it Artanh} \left ({\frac{e\sqrt{cdx+ae}}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}}} \right ) a{e}^{2}-{\it Artanh} \left ({\frac{e\sqrt{cdx+ae}}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}}} \right ) c{d}^{2}-\sqrt{cdx+ae}\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e} \right ) } \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)^(3/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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 )}^{\frac{3}{2}}}\,{d x} \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)^(3/2),x, algorithm="maxima")

[Out]

integrate(sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)/(e*x + d)^(3/2), x)

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Fricas [A]  time = 1.94932, size = 659, normalized size = 5.15 \begin{align*} \left [\frac{{\left (e x + d\right )} \sqrt{-\frac{c d^{2} - a e^{2}}{e}} \log \left (-\frac{c d e^{2} x^{2} + 2 \, a e^{3} x - c d^{3} + 2 \, a d e^{2} - 2 \, \sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x} \sqrt{e x + d} e \sqrt{-\frac{c d^{2} - a e^{2}}{e}}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) + 2 \, \sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x} \sqrt{e x + d}}{e^{2} x + d e}, \frac{2 \,{\left ({\left (e x + d\right )} \sqrt{\frac{c d^{2} - a e^{2}}{e}} \arctan \left (\frac{\sqrt{e x + d} \sqrt{\frac{c d^{2} - a e^{2}}{e}}}{\sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x}}\right ) + \sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x} \sqrt{e x + d}\right )}}{e^{2} x + d e}\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)^(3/2),x, algorithm="fricas")

[Out]

[((e*x + d)*sqrt(-(c*d^2 - a*e^2)/e)*log(-(c*d*e^2*x^2 + 2*a*e^3*x - c*d^3 + 2*a*d*e^2 - 2*sqrt(c*d*e*x^2 + a*
d*e + (c*d^2 + a*e^2)*x)*sqrt(e*x + d)*e*sqrt(-(c*d^2 - a*e^2)/e))/(e^2*x^2 + 2*d*e*x + d^2)) + 2*sqrt(c*d*e*x
^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(e*x + d))/(e^2*x + d*e), 2*((e*x + d)*sqrt((c*d^2 - a*e^2)/e)*arctan(sqrt
(e*x + d)*sqrt((c*d^2 - a*e^2)/e)/sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)) + sqrt(c*d*e*x^2 + a*d*e + (c*d
^2 + a*e^2)*x)*sqrt(e*x + d))/(e^2*x + d*e)]

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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 )}}{\left (d + e x\right )^{\frac{3}{2}}}\, 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)**(3/2),x)

[Out]

Integral(sqrt((d + e*x)*(a*e + c*d*x))/(d + e*x)**(3/2), x)

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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)^(3/2),x, algorithm="giac")

[Out]

Timed out