3.18 \(\int \frac{\sqrt{a+b x} (e+f x)}{x} \, dx\)

Optimal. Leaf size=54 \[ 2 e \sqrt{a+b x}-2 \sqrt{a} e \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )+\frac{2 f (a+b x)^{3/2}}{3 b} \]

[Out]

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

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Rubi [A]  time = 0.0158209, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {80, 50, 63, 208} \[ 2 e \sqrt{a+b x}-2 \sqrt{a} e \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )+\frac{2 f (a+b x)^{3/2}}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[a + b*x]*(e + f*x))/x,x]

[Out]

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

Rule 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

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

Mathematica [A]  time = 0.0462837, size = 55, normalized size = 1.02 \[ e \left (2 \sqrt{a+b x}-2 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )\right )+\frac{2 f (a+b x)^{3/2}}{3 b} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[a + b*x]*(e + f*x))/x,x]

[Out]

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

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Maple [A]  time = 0.006, size = 46, normalized size = 0.9 \begin{align*} 2\,{\frac{1}{b} \left ( 1/3\,f \left ( bx+a \right ) ^{3/2}+\sqrt{bx+a}be-\sqrt{a}eb{\it Artanh} \left ({\frac{\sqrt{bx+a}}{\sqrt{a}}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)*(b*x+a)^(1/2)/x,x)

[Out]

2/b*(1/3*f*(b*x+a)^(3/2)+(b*x+a)^(1/2)*b*e-a^(1/2)*e*b*arctanh((b*x+a)^(1/2)/a^(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((f*x+e)*(b*x+a)^(1/2)/x,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.29334, size = 279, normalized size = 5.17 \begin{align*} \left [\frac{3 \, \sqrt{a} b e \log \left (\frac{b x - 2 \, \sqrt{b x + a} \sqrt{a} + 2 \, a}{x}\right ) + 2 \,{\left (b f x + 3 \, b e + a f\right )} \sqrt{b x + a}}{3 \, b}, \frac{2 \,{\left (3 \, \sqrt{-a} b e \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) +{\left (b f x + 3 \, b e + a f\right )} \sqrt{b x + a}\right )}}{3 \, b}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/3*(3*sqrt(a)*b*e*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(b*f*x + 3*b*e + a*f)*sqrt(b*x + a))/b, 2
/3*(3*sqrt(-a)*b*e*arctan(sqrt(b*x + a)*sqrt(-a)/a) + (b*f*x + 3*b*e + a*f)*sqrt(b*x + a))/b]

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Sympy [A]  time = 3.90109, size = 54, normalized size = 1. \begin{align*} \frac{2 a e \operatorname{atan}{\left (\frac{\sqrt{a + b x}}{\sqrt{- a}} \right )}}{\sqrt{- a}} + 2 e \sqrt{a + b x} + \frac{2 f \left (a + b x\right )^{\frac{3}{2}}}{3 b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*(b*x+a)**(1/2)/x,x)

[Out]

2*a*e*atan(sqrt(a + b*x)/sqrt(-a))/sqrt(-a) + 2*e*sqrt(a + b*x) + 2*f*(a + b*x)**(3/2)/(3*b)

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Giac [A]  time = 2.64331, size = 77, normalized size = 1.43 \begin{align*} \frac{2 \, a \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right ) e}{\sqrt{-a}} + \frac{2 \,{\left ({\left (b x + a\right )}^{\frac{3}{2}} b^{2} f + 3 \, \sqrt{b x + a} b^{3} e\right )}}{3 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*(b*x+a)^(1/2)/x,x, algorithm="giac")

[Out]

2*a*arctan(sqrt(b*x + a)/sqrt(-a))*e/sqrt(-a) + 2/3*((b*x + a)^(3/2)*b^2*f + 3*sqrt(b*x + a)*b^3*e)/b^3