3.786 \(\int \frac{(a+c x^4)^{3/2}}{x} \, dx\)

Optimal. Leaf size=59 \[ -\frac{1}{2} a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+c x^4}}{\sqrt{a}}\right )+\frac{1}{2} a \sqrt{a+c x^4}+\frac{1}{6} \left (a+c x^4\right )^{3/2} \]

[Out]

(a*Sqrt[a + c*x^4])/2 + (a + c*x^4)^(3/2)/6 - (a^(3/2)*ArcTanh[Sqrt[a + c*x^4]/Sqrt[a]])/2

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Rubi [A]  time = 0.0347514, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {266, 50, 63, 208} \[ -\frac{1}{2} a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+c x^4}}{\sqrt{a}}\right )+\frac{1}{2} a \sqrt{a+c x^4}+\frac{1}{6} \left (a+c x^4\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

Int[(a + c*x^4)^(3/2)/x,x]

[Out]

(a*Sqrt[a + c*x^4])/2 + (a + c*x^4)^(3/2)/6 - (a^(3/2)*ArcTanh[Sqrt[a + c*x^4]/Sqrt[a]])/2

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{\left (a+c x^4\right )^{3/2}}{x} \, dx &=\frac{1}{4} \operatorname{Subst}\left (\int \frac{(a+c x)^{3/2}}{x} \, dx,x,x^4\right )\\ &=\frac{1}{6} \left (a+c x^4\right )^{3/2}+\frac{1}{4} a \operatorname{Subst}\left (\int \frac{\sqrt{a+c x}}{x} \, dx,x,x^4\right )\\ &=\frac{1}{2} a \sqrt{a+c x^4}+\frac{1}{6} \left (a+c x^4\right )^{3/2}+\frac{1}{4} a^2 \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+c x}} \, dx,x,x^4\right )\\ &=\frac{1}{2} a \sqrt{a+c x^4}+\frac{1}{6} \left (a+c x^4\right )^{3/2}+\frac{a^2 \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{c}+\frac{x^2}{c}} \, dx,x,\sqrt{a+c x^4}\right )}{2 c}\\ &=\frac{1}{2} a \sqrt{a+c x^4}+\frac{1}{6} \left (a+c x^4\right )^{3/2}-\frac{1}{2} a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+c x^4}}{\sqrt{a}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0209842, size = 51, normalized size = 0.86 \[ \frac{1}{6} \left (\sqrt{a+c x^4} \left (4 a+c x^4\right )-3 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+c x^4}}{\sqrt{a}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + c*x^4)^(3/2)/x,x]

[Out]

(Sqrt[a + c*x^4]*(4*a + c*x^4) - 3*a^(3/2)*ArcTanh[Sqrt[a + c*x^4]/Sqrt[a]])/6

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Maple [A]  time = 0.014, size = 57, normalized size = 1. \begin{align*}{\frac{c{x}^{4}}{6}\sqrt{c{x}^{4}+a}}+{\frac{2\,a}{3}\sqrt{c{x}^{4}+a}}-{\frac{1}{2}{a}^{{\frac{3}{2}}}\ln \left ({\frac{1}{{x}^{2}} \left ( 2\,a+2\,\sqrt{a}\sqrt{c{x}^{4}+a} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+a)^(3/2)/x,x)

[Out]

1/6*c*x^4*(c*x^4+a)^(1/2)+2/3*a*(c*x^4+a)^(1/2)-1/2*a^(3/2)*ln((2*a+2*a^(1/2)*(c*x^4+a)^(1/2))/x^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((c*x^4+a)^(3/2)/x,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.50878, size = 258, normalized size = 4.37 \begin{align*} \left [\frac{1}{4} \, a^{\frac{3}{2}} \log \left (\frac{c x^{4} - 2 \, \sqrt{c x^{4} + a} \sqrt{a} + 2 \, a}{x^{4}}\right ) + \frac{1}{6} \,{\left (c x^{4} + 4 \, a\right )} \sqrt{c x^{4} + a}, \frac{1}{2} \, \sqrt{-a} a \arctan \left (\frac{\sqrt{c x^{4} + a} \sqrt{-a}}{a}\right ) + \frac{1}{6} \,{\left (c x^{4} + 4 \, a\right )} \sqrt{c x^{4} + a}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+a)^(3/2)/x,x, algorithm="fricas")

[Out]

[1/4*a^(3/2)*log((c*x^4 - 2*sqrt(c*x^4 + a)*sqrt(a) + 2*a)/x^4) + 1/6*(c*x^4 + 4*a)*sqrt(c*x^4 + a), 1/2*sqrt(
-a)*a*arctan(sqrt(c*x^4 + a)*sqrt(-a)/a) + 1/6*(c*x^4 + 4*a)*sqrt(c*x^4 + a)]

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Sympy [A]  time = 2.57206, size = 80, normalized size = 1.36 \begin{align*} \frac{2 a^{\frac{3}{2}} \sqrt{1 + \frac{c x^{4}}{a}}}{3} + \frac{a^{\frac{3}{2}} \log{\left (\frac{c x^{4}}{a} \right )}}{4} - \frac{a^{\frac{3}{2}} \log{\left (\sqrt{1 + \frac{c x^{4}}{a}} + 1 \right )}}{2} + \frac{\sqrt{a} c x^{4} \sqrt{1 + \frac{c x^{4}}{a}}}{6} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+a)**(3/2)/x,x)

[Out]

2*a**(3/2)*sqrt(1 + c*x**4/a)/3 + a**(3/2)*log(c*x**4/a)/4 - a**(3/2)*log(sqrt(1 + c*x**4/a) + 1)/2 + sqrt(a)*
c*x**4*sqrt(1 + c*x**4/a)/6

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Giac [A]  time = 1.11862, size = 68, normalized size = 1.15 \begin{align*} \frac{a^{2} \arctan \left (\frac{\sqrt{c x^{4} + a}}{\sqrt{-a}}\right )}{2 \, \sqrt{-a}} + \frac{1}{6} \,{\left (c x^{4} + a\right )}^{\frac{3}{2}} + \frac{1}{2} \, \sqrt{c x^{4} + a} a \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+a)^(3/2)/x,x, algorithm="giac")

[Out]

1/2*a^2*arctan(sqrt(c*x^4 + a)/sqrt(-a))/sqrt(-a) + 1/6*(c*x^4 + a)^(3/2) + 1/2*sqrt(c*x^4 + a)*a