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

Optimal. Leaf size=58 \[ -\frac{\left (a+b x^3\right )^{3/2}}{3 x^3}+b \sqrt{a+b x^3}-\sqrt{a} b \tanh ^{-1}\left (\frac{\sqrt{a+b x^3}}{\sqrt{a}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.0359271, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {266, 47, 50, 63, 208} \[ -\frac{\left (a+b x^3\right )^{3/2}}{3 x^3}+b \sqrt{a+b x^3}-\sqrt{a} b \tanh ^{-1}\left (\frac{\sqrt{a+b x^3}}{\sqrt{a}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3)^(3/2)/x^4,x]

[Out]

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

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 47

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

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

Mathematica [C]  time = 0.0091124, size = 37, normalized size = 0.64 \[ \frac{2 b \left (a+b x^3\right )^{5/2} \, _2F_1\left (2,\frac{5}{2};\frac{7}{2};\frac{b x^3}{a}+1\right )}{15 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^3)^(3/2)/x^4,x]

[Out]

(2*b*(a + b*x^3)^(5/2)*Hypergeometric2F1[2, 5/2, 7/2, 1 + (b*x^3)/a])/(15*a^2)

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Maple [A]  time = 0.017, size = 49, normalized size = 0.8 \begin{align*} -{\frac{a}{3\,{x}^{3}}\sqrt{b{x}^{3}+a}}+{\frac{2\,b}{3}\sqrt{b{x}^{3}+a}}-b{\it Artanh} \left ({\sqrt{b{x}^{3}+a}{\frac{1}{\sqrt{a}}}} \right ) \sqrt{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.41634, size = 285, normalized size = 4.91 \begin{align*} \left [\frac{3 \, \sqrt{a} b x^{3} \log \left (\frac{b x^{3} - 2 \, \sqrt{b x^{3} + a} \sqrt{a} + 2 \, a}{x^{3}}\right ) + 2 \,{\left (2 \, b x^{3} - a\right )} \sqrt{b x^{3} + a}}{6 \, x^{3}}, \frac{3 \, \sqrt{-a} b x^{3} \arctan \left (\frac{\sqrt{b x^{3} + a} \sqrt{-a}}{a}\right ) +{\left (2 \, b x^{3} - a\right )} \sqrt{b x^{3} + a}}{3 \, x^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 2.8285, size = 100, normalized size = 1.72 \begin{align*} - \sqrt{a} b \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x^{\frac{3}{2}}} \right )} - \frac{a^{2}}{3 \sqrt{b} x^{\frac{9}{2}} \sqrt{\frac{a}{b x^{3}} + 1}} + \frac{a \sqrt{b}}{3 x^{\frac{3}{2}} \sqrt{\frac{a}{b x^{3}} + 1}} + \frac{2 b^{\frac{3}{2}} x^{\frac{3}{2}}}{3 \sqrt{\frac{a}{b x^{3}} + 1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(a)*b*asinh(sqrt(a)/(sqrt(b)*x**(3/2))) - a**2/(3*sqrt(b)*x**(9/2)*sqrt(a/(b*x**3) + 1)) + a*sqrt(b)/(3*x
**(3/2)*sqrt(a/(b*x**3) + 1)) + 2*b**(3/2)*x**(3/2)/(3*sqrt(a/(b*x**3) + 1))

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Giac [A]  time = 1.10525, size = 77, normalized size = 1.33 \begin{align*} \frac{1}{3} \,{\left (\frac{3 \, a \arctan \left (\frac{\sqrt{b x^{3} + a}}{\sqrt{-a}}\right )}{\sqrt{-a}} + 2 \, \sqrt{b x^{3} + a} - \frac{\sqrt{b x^{3} + a} a}{b x^{3}}\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*a*arctan(sqrt(b*x^3 + a)/sqrt(-a))/sqrt(-a) + 2*sqrt(b*x^3 + a) - sqrt(b*x^3 + a)*a/(b*x^3))*b