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]

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

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Rubi [A]  time = 0.04, antiderivative size = 58, normalized size of antiderivative = 1.00, 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 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]

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]]

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*}

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Mathematica [C]  time = 0.01, 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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fricas [A]  time = 0.82, size = 121, normalized size = 2.09 \[ \left [\frac {3 \, \sqrt {a} b x^{3} \log \left (\frac {b x^{3} - 2 \, \sqrt {b x^{3} + a} \sqrt {a} + 2 \, a}{x^{3}}\right ) + 2 \, {\left (2 \, b x^{3} - a\right )} \sqrt {b x^{3} + a}}{6 \, x^{3}}, \frac {3 \, \sqrt {-a} b x^{3} \arctan \left (\frac {\sqrt {b x^{3} + a} \sqrt {-a}}{a}\right ) + {\left (2 \, b x^{3} - a\right )} \sqrt {b x^{3} + a}}{3 \, x^{3}}\right ] \]

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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giac [A]  time = 0.18, size = 63, normalized size = 1.09 \[ \frac {\frac {3 \, a b^{2} \arctan \left (\frac {\sqrt {b x^{3} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + 2 \, \sqrt {b x^{3} + a} b^{2} - \frac {\sqrt {b x^{3} + a} a b}{x^{3}}}{3 \, b} \]

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

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maple [A]  time = 0.03, size = 49, normalized size = 0.84 \[ -\sqrt {a}\, b \arctanh \left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )+\frac {2 \sqrt {b \,x^{3}+a}\, b}{3}-\frac {\sqrt {b \,x^{3}+a}\, a}{3 x^{3}} \]

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 [A]  time = 2.99, size = 66, normalized size = 1.14 \[ \frac {1}{2} \, \sqrt {a} b \log \left (\frac {\sqrt {b x^{3} + a} - \sqrt {a}}{\sqrt {b x^{3} + a} + \sqrt {a}}\right ) + \frac {2}{3} \, \sqrt {b x^{3} + a} b - \frac {\sqrt {b x^{3} + a} a}{3 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.31, size = 69, normalized size = 1.19 \[ \frac {2\,b\,\sqrt {b\,x^3+a}}{3}-\frac {a\,\sqrt {b\,x^3+a}}{3\,x^3}+\frac {\sqrt {a}\,b\,\ln \left (\frac {{\left (\sqrt {b\,x^3+a}-\sqrt {a}\right )}^3\,\left (\sqrt {b\,x^3+a}+\sqrt {a}\right )}{x^6}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^3)^(3/2)/x^4,x)

[Out]

(2*b*(a + b*x^3)^(1/2))/3 - (a*(a + b*x^3)^(1/2))/(3*x^3) + (a^(1/2)*b*log((((a + b*x^3)^(1/2) - a^(1/2))^3*((
a + b*x^3)^(1/2) + a^(1/2)))/x^6))/2

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sympy [B]  time = 5.25, size = 100, normalized size = 1.72 \[ - \sqrt {a} b \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x^{\frac {3}{2}}} \right )} - \frac {a^{2}}{3 \sqrt {b} x^{\frac {9}{2}} \sqrt {\frac {a}{b x^{3}} + 1}} + \frac {a \sqrt {b}}{3 x^{\frac {3}{2}} \sqrt {\frac {a}{b x^{3}} + 1}} + \frac {2 b^{\frac {3}{2}} x^{\frac {3}{2}}}{3 \sqrt {\frac {a}{b x^{3}} + 1}} \]

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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