3.2003 \(\int \frac{(d+e x)^{3/2}}{a d e+(c d^2+a e^2) x+c d e x^2} \, dx\)

Optimal. Leaf size=83 \[ \frac{2 \sqrt{d+e x}}{c d}-\frac{2 \sqrt{c d^2-a e^2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{d+e x}}{\sqrt{c d^2-a e^2}}\right )}{c^{3/2} d^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0546732, antiderivative size = 83, 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, 50, 63, 208} \[ \frac{2 \sqrt{d+e x}}{c d}-\frac{2 \sqrt{c d^2-a e^2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{d+e x}}{\sqrt{c d^2-a e^2}}\right )}{c^{3/2} d^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[d + e*x])/(c*d) - (2*Sqrt[c*d^2 - a*e^2]*ArcTanh[(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[c*d^2 - a*e^2]])
/(c^(3/2)*d^(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 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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{(d+e x)^{3/2}}{a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx &=\int \frac{\sqrt{d+e x}}{a e+c d x} \, dx\\ &=\frac{2 \sqrt{d+e x}}{c d}+\frac{\left (c d^2-a e^2\right ) \int \frac{1}{(a e+c d x) \sqrt{d+e x}} \, dx}{c d}\\ &=\frac{2 \sqrt{d+e x}}{c d}+\left (2 \left (\frac{d}{e}-\frac{a e}{c d}\right )\right ) \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 )\\ &=\frac{2 \sqrt{d+e x}}{c d}-\frac{2 \sqrt{c d^2-a e^2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{d+e x}}{\sqrt{c d^2-a e^2}}\right )}{c^{3/2} d^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0340531, size = 83, normalized size = 1. \[ \frac{2 \sqrt{d+e x}}{c d}-\frac{2 \sqrt{c d^2-a e^2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{d+e x}}{\sqrt{c d^2-a e^2}}\right )}{c^{3/2} d^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.192, size = 122, normalized size = 1.5 \begin{align*} 2\,{\frac{\sqrt{ex+d}}{cd}}-2\,{\frac{a{e}^{2}}{cd\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{d}{\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 ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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((e*x+d)^(3/2)/(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 = 1.97021, size = 385, normalized size = 4.64 \begin{align*} \left [\frac{\sqrt{\frac{c d^{2} - a e^{2}}{c d}} \log \left (\frac{c d e x + 2 \, c d^{2} - a e^{2} - 2 \, \sqrt{e x + d} c d \sqrt{\frac{c d^{2} - a e^{2}}{c d}}}{c d x + a e}\right ) + 2 \, \sqrt{e x + d}}{c d}, -\frac{2 \,{\left (\sqrt{-\frac{c d^{2} - a e^{2}}{c d}} \arctan \left (-\frac{\sqrt{e x + d} c d \sqrt{-\frac{c d^{2} - a e^{2}}{c d}}}{c d^{2} - a e^{2}}\right ) - \sqrt{e x + d}\right )}}{c d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 12.6772, size = 80, normalized size = 0.96 \begin{align*} \frac{2 \left (\frac{e \sqrt{d + e x}}{c d} - \frac{e \left (a e^{2} - c d^{2}\right ) \operatorname{atan}{\left (\frac{\sqrt{d + e x}}{\sqrt{\frac{a e^{2} - c d^{2}}{c d}}} \right )}}{c^{2} d^{2} \sqrt{\frac{a e^{2} - c d^{2}}{c d}}}\right )}{e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out