3.2962 \(\int \frac{\sqrt{a+b \sqrt{c x^3}}}{x} \, dx\)

Optimal. Leaf size=55 \[ \frac{4}{3} \sqrt{a+b \sqrt{c x^3}}-\frac{4}{3} \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+b \sqrt{c x^3}}}{\sqrt{a}}\right ) \]

[Out]

(4*Sqrt[a + b*Sqrt[c*x^3]])/3 - (4*Sqrt[a]*ArcTanh[Sqrt[a + b*Sqrt[c*x^3]]/Sqrt[a]])/3

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Rubi [A]  time = 0.0347186, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {369, 266, 50, 63, 208} \[ \frac{4}{3} \sqrt{a+b \sqrt{c x^3}}-\frac{4}{3} \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+b \sqrt{c x^3}}}{\sqrt{a}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*Sqrt[c*x^3]]/x,x]

[Out]

(4*Sqrt[a + b*Sqrt[c*x^3]])/3 - (4*Sqrt[a]*ArcTanh[Sqrt[a + b*Sqrt[c*x^3]]/Sqrt[a]])/3

Rule 369

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*((c_.)*(x_)^(q_))^(n_))^(p_.), x_Symbol] :> With[{k = Denominator[n]}, Su
bst[Int[(d*x)^m*(a + b*c^n*x^(n*q))^p, x], x^(1/k), (c*x^q)^(1/k)/(c^(1/k)*(x^(1/k))^(q - 1))]] /; FreeQ[{a, b
, c, d, m, p, q}, x] && FractionQ[n]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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 \sqrt{c x^3}}}{x} \, dx &=\operatorname{Subst}\left (\int \frac{\sqrt{a+b \sqrt{c} x^{3/2}}}{x} \, dx,\sqrt{x},\frac{\sqrt{c x^3}}{\sqrt{c} x}\right )\\ &=\operatorname{Subst}\left (\frac{2}{3} \operatorname{Subst}\left (\int \frac{\sqrt{a+b \sqrt{c} x}}{x} \, dx,x,x^{3/2}\right ),\sqrt{x},\frac{\sqrt{c x^3}}{\sqrt{c} x}\right )\\ &=\frac{4}{3} \sqrt{a+b \sqrt{c x^3}}+\operatorname{Subst}\left (\frac{1}{3} (2 a) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b \sqrt{c} x}} \, dx,x,x^{3/2}\right ),\sqrt{x},\frac{\sqrt{c x^3}}{\sqrt{c} x}\right )\\ &=\frac{4}{3} \sqrt{a+b \sqrt{c x^3}}+\operatorname{Subst}\left (\frac{(4 a) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b \sqrt{c}}+\frac{x^2}{b \sqrt{c}}} \, dx,x,\sqrt{a+b \sqrt{c} x^{3/2}}\right )}{3 b \sqrt{c}},\sqrt{x},\frac{\sqrt{c x^3}}{\sqrt{c} x}\right )\\ &=\frac{4}{3} \sqrt{a+b \sqrt{c x^3}}-\frac{4}{3} \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+b \sqrt{c x^3}}}{\sqrt{a}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0432338, size = 55, normalized size = 1. \[ \frac{4}{3} \sqrt{a+b \sqrt{c x^3}}-\frac{4}{3} \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+b \sqrt{c x^3}}}{\sqrt{a}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*Sqrt[c*x^3]]/x,x]

[Out]

(4*Sqrt[a + b*Sqrt[c*x^3]])/3 - (4*Sqrt[a]*ArcTanh[Sqrt[a + b*Sqrt[c*x^3]]/Sqrt[a]])/3

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Maple [A]  time = 0.181, size = 40, normalized size = 0.7 \begin{align*} -{\frac{4}{3}{\it Artanh} \left ({\sqrt{a+b\sqrt{c{x}^{3}}}{\frac{1}{\sqrt{a}}}} \right ) \sqrt{a}}+{\frac{4}{3}\sqrt{a+b\sqrt{c{x}^{3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*(c*x^3)^(1/2))^(1/2)/x,x)

[Out]

-4/3*arctanh((a+b*(c*x^3)^(1/2))^(1/2)/a^(1/2))*a^(1/2)+4/3*(a+b*(c*x^3)^(1/2))^(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((a+b*(c*x^3)^(1/2))^(1/2)/x,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [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+b*(c*x^3)^(1/2))^(1/2)/x,x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{a + b \sqrt{c x^{3}}}}{x}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(a + b*sqrt(c*x**3))/x, x)

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Giac [B]  time = 1.13353, size = 124, normalized size = 2.25 \begin{align*} \frac{4 \,{\left (\frac{a c \arctan \left (\frac{\sqrt{\sqrt{c x} b c x + a c}}{\sqrt{-a c}}\right )}{\sqrt{-a c}} + \sqrt{\sqrt{c x} b c x + a c} - \frac{a c \arctan \left (\frac{\sqrt{a c}}{\sqrt{-a c}}\right ) + \sqrt{a c} \sqrt{-a c}}{\sqrt{-a c}}\right )}{\left | c \right |}}{3 \, c^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/3*(a*c*arctan(sqrt(sqrt(c*x)*b*c*x + a*c)/sqrt(-a*c))/sqrt(-a*c) + sqrt(sqrt(c*x)*b*c*x + a*c) - (a*c*arctan
(sqrt(a*c)/sqrt(-a*c)) + sqrt(a*c)*sqrt(-a*c))/sqrt(-a*c))*abs(c)/c^(3/2)