### 3.2068 $$\int \frac{\sqrt{d+e x}}{(a d e+(c d^2+a e^2) x+c d e x^2)^{3/2}} \, dx$$

Optimal. Leaf size=139 $-\frac{2 \sqrt{d+e x}}{\left (c d^2-a e^2\right ) \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}-\frac{2 \sqrt{e} \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt{d+e x} \sqrt{c d^2-a e^2}}\right )}{\left (c d^2-a e^2\right )^{3/2}}$

[Out]

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

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Rubi [A]  time = 0.0941894, antiderivative size = 139, 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 = {666, 660, 205} $-\frac{2 \sqrt{d+e x}}{\left (c d^2-a e^2\right ) \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}-\frac{2 \sqrt{e} \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt{d+e x} \sqrt{c d^2-a e^2}}\right )}{\left (c d^2-a e^2\right )^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 666

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

Mathematica [C]  time = 0.0214777, size = 75, normalized size = 0.54 $-\frac{2 \sqrt{d+e x} \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{e (a e+c d x)}{a e^2-c d^2}\right )}{\left (c d^2-a e^2\right ) \sqrt{(d+e x) (a e+c d x)}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-2*Sqrt[d + e*x]*Hypergeometric2F1[-1/2, 1, 1/2, (e*(a*e + c*d*x))/(-(c*d^2) + a*e^2)])/((c*d^2 - a*e^2)*Sqrt
[(a*e + c*d*x)*(d + e*x)])

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Maple [A]  time = 0.207, size = 136, normalized size = 1. \begin{align*} -2\,{\frac{\sqrt{cde{x}^{2}+a{e}^{2}x+c{d}^{2}x+ade}}{\sqrt{ex+d} \left ( cdx+ae \right ) \left ( a{e}^{2}-c{d}^{2} \right ) \sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}} \left ( e{\it Artanh} \left ({\frac{e\sqrt{cdx+ae}}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}}} \right ) \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((e*x+d)^(1/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x)

[Out]

-2*(c*d*e*x^2+a*e^2*x+c*d^2*x+a*d*e)^(1/2)*(e*arctanh(e*(c*d*x+a*e)^(1/2)/((a*e^2-c*d^2)*e)^(1/2))*(c*d*x+a*e)
^(1/2)-((a*e^2-c*d^2)*e)^(1/2))/(e*x+d)^(1/2)/(c*d*x+a*e)/(a*e^2-c*d^2)/((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{e x + d}}{{\left (c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(1/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.15886, size = 1019, normalized size = 7.33 \begin{align*} \left [-\frac{{\left (c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x\right )} \sqrt{-\frac{e}{c d^{2} - a e^{2}}} \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}{\left (c d^{2} - a e^{2}\right )} \sqrt{e x + d} \sqrt{-\frac{e}{c d^{2} - a e^{2}}}}{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}}{a c d^{3} e - a^{2} d e^{3} +{\left (c^{2} d^{3} e - a c d e^{3}\right )} x^{2} +{\left (c^{2} d^{4} - a^{2} e^{4}\right )} x}, -\frac{2 \,{\left ({\left (c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x\right )} \sqrt{\frac{e}{c d^{2} - a e^{2}}} \arctan \left (-\frac{\sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x}{\left (c d^{2} - a e^{2}\right )} \sqrt{e x + d} \sqrt{\frac{e}{c d^{2} - a e^{2}}}}{c d e^{2} x^{2} + a d e^{2} +{\left (c d^{2} e + a e^{3}\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 )}}{a c d^{3} e - a^{2} d e^{3} +{\left (c^{2} d^{3} e - a c d e^{3}\right )} x^{2} +{\left (c^{2} d^{4} - a^{2} e^{4}\right )} x}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(1/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x, algorithm="fricas")

[Out]

[-((c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(-e/(c*d^2 - a*e^2))*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)*(c*d^2 - a*e^2)*sqrt(e*x + d)*sqrt(-e/(c*d^2 - a*e^2))
)/(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))/(a*c*d^3*e - a^2*d
*e^3 + (c^2*d^3*e - a*c*d*e^3)*x^2 + (c^2*d^4 - a^2*e^4)*x), -2*((c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(
e/(c*d^2 - a*e^2))*arctan(-sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*(c*d^2 - a*e^2)*sqrt(e*x + d)*sqrt(e/(c
*d^2 - a*e^2))/(c*d*e^2*x^2 + a*d*e^2 + (c*d^2*e + a*e^3)*x)) + sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sq
rt(e*x + d))/(a*c*d^3*e - a^2*d*e^3 + (c^2*d^3*e - a*c*d*e^3)*x^2 + (c^2*d^4 - a^2*e^4)*x)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{d + e x}}{\left (\left (d + e x\right ) \left (a e + c d x\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**(1/2)/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(3/2),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(1/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x, algorithm="giac")

[Out]

sage0*x