3.4.92 \(\int \frac {(a+b x^3)^{3/2}}{x^4} \, dx\) [392]

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

[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 = 0.03, 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 = {272, 43, 52, 65, 214} \begin {gather*} -\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 ) \end {gather*}

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 43

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, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

Rule 52

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 65

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 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 272

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} \text {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 \text {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) \text {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 \text {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 [A]
time = 0.07, size = 55, normalized size = 0.95 \begin {gather*} \frac {\sqrt {a+b x^3} \left (-a+2 b x^3\right )}{3 x^3}-\sqrt {a} b \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^3)^(3/2)/x^4,x]

[Out]

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

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Maple [A]
time = 0.15, size = 49, normalized size = 0.84

method	result	size
default	\(-\frac {a \sqrt {b \,x^{3}+a}}{3 x^{3}}+\frac {2 b \sqrt {b \,x^{3}+a}}{3}-b \arctanh \left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right ) \sqrt {a}\)	\(49\)
elliptic	\(-\frac {a \sqrt {b \,x^{3}+a}}{3 x^{3}}+\frac {2 b \sqrt {b \,x^{3}+a}}{3}-b \arctanh \left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right ) \sqrt {a}\)	\(49\)
risch	\(-\frac {a \sqrt {b \,x^{3}+a}}{3 x^{3}}+\frac {b \left (\frac {4 \sqrt {b \,x^{3}+a}}{3}-2 \arctanh \left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right ) \sqrt {a}\right )}{2}\)	\(51\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 0.51, size = 66, normalized size = 1.14 \begin {gather*} \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}} \end {gather*}

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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Fricas [A]
time = 0.36, size = 121, normalized size = 2.09 \begin {gather*} \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 {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 100 vs. \(2 (49) = 98\).
time = 1.30, size = 100, normalized size = 1.72 \begin {gather*} - \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 {gather*}

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 = 1.78, size = 63, normalized size = 1.09 \begin {gather*} \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} \end {gather*}

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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Mupad [B]
time = 1.31, size = 69, normalized size = 1.19 \begin {gather*} \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} \end {gather*}

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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