3.2005 \(\int \frac{1}{\sqrt{d+e x} (a d e+(c d^2+a e^2) x+c d e x^2)} \, dx\)

Optimal. Leaf size=91 \[ \frac{2}{\sqrt{d+e x} \left (c d^2-a e^2\right )}-\frac{2 \sqrt{c} \sqrt{d} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{d+e x}}{\sqrt{c d^2-a e^2}}\right )}{\left (c d^2-a e^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0830422, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.108, Rules used = {626, 51, 63, 208} \[ \frac{2}{\sqrt{d+e x} \left (c d^2-a e^2\right )}-\frac{2 \sqrt{c} \sqrt{d} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{d+e x}}{\sqrt{c d^2-a e^2}}\right )}{\left (c d^2-a e^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 626

Int[((d_) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a
/d + (c*x)/e)^p, x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&
IntegerQ[p]

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rubi steps

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

Mathematica [C]  time = 0.0151336, size = 55, normalized size = 0.6 \[ -\frac{2 \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{c d (d+e x)}{c d^2-a e^2}\right )}{\sqrt{d+e x} \left (a e^2-c d^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.195, size = 88, normalized size = 1. \begin{align*} -2\,{\frac{cd}{ \left ( a{e}^{2}-c{d}^{2} \right ) \sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) cd}}\arctan \left ({\frac{\sqrt{ex+d}cd}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) cd}}} \right ) }-2\,{\frac{1}{ \left ( a{e}^{2}-c{d}^{2} \right ) \sqrt{ex+d}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.85558, size = 529, normalized size = 5.81 \begin{align*} \left [-\frac{{\left (e x + d\right )} \sqrt{\frac{c d}{c d^{2} - a e^{2}}} \log \left (\frac{c d e x + 2 \, c d^{2} - a e^{2} + 2 \,{\left (c d^{2} - a e^{2}\right )} \sqrt{e x + d} \sqrt{\frac{c d}{c d^{2} - a e^{2}}}}{c d x + a e}\right ) - 2 \, \sqrt{e x + d}}{c d^{3} - a d e^{2} +{\left (c d^{2} e - a e^{3}\right )} x}, -\frac{2 \,{\left ({\left (e x + d\right )} \sqrt{-\frac{c d}{c d^{2} - a e^{2}}} \arctan \left (-\frac{{\left (c d^{2} - a e^{2}\right )} \sqrt{e x + d} \sqrt{-\frac{c d}{c d^{2} - a e^{2}}}}{c d e x + c d^{2}}\right ) - \sqrt{e x + d}\right )}}{c d^{3} - a d e^{2} +{\left (c d^{2} e - a e^{3}\right )} x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 7.51262, size = 82, normalized size = 0.9 \begin{align*} \frac{2 c d \operatorname{atan}{\left (\frac{1}{\sqrt{\frac{c d}{a e^{2} - c d^{2}}} \sqrt{d + e x}} \right )}}{\sqrt{\frac{c d}{a e^{2} - c d^{2}}} \left (a e^{2} - c d^{2}\right )^{2}} - \frac{2}{\sqrt{d + e x} \left (a e^{2} - c d^{2}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*c*d*atan(1/(sqrt(c*d/(a*e**2 - c*d**2))*sqrt(d + e*x)))/(sqrt(c*d/(a*e**2 - c*d**2))*(a*e**2 - c*d**2)**2) -
2/(sqrt(d + e*x)*(a*e**2 - c*d**2))

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

[Out]

Timed out