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

Optimal. Leaf size=56 \[ \frac {2 \left (a x^3-b\right ) \sqrt {a x^3+b}}{3 x^3}-\frac {4}{3} a \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {a x^3+b}}{\sqrt {b}}\right ) \]

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Rubi [A]  time = 0.05, antiderivative size = 63, normalized size of antiderivative = 1.12, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {446, 78, 50, 63, 208} \begin {gather*} -\frac {2 \left (a x^3+b\right )^{3/2}}{3 x^3}+\frac {4}{3} a \sqrt {a x^3+b}-\frac {4}{3} a \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {a x^3+b}}{\sqrt {b}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Sqrt[b + a*x^3]*(2*b + a*x^3))/x^4,x]

[Out]

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

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 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

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 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 55, normalized size = 0.98 \begin {gather*} -\frac {2 \sqrt {a x^3+b} \left (b-a x^3\right )}{3 x^3}-\frac {4}{3} a \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {a x^3+b}}{\sqrt {b}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[b + a*x^3]*(2*b + a*x^3))/x^4,x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.06, size = 56, normalized size = 1.00 \begin {gather*} \frac {2 \left (-b+a x^3\right ) \sqrt {b+a x^3}}{3 x^3}-\frac {4}{3} a \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b+a x^3}}{\sqrt {b}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(Sqrt[b + a*x^3]*(2*b + a*x^3))/x^4,x]

[Out]

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

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fricas [A]  time = 0.63, size = 117, normalized size = 2.09 \begin {gather*} \left [\frac {2 \, {\left (a \sqrt {b} x^{3} \log \left (\frac {a x^{3} - 2 \, \sqrt {a x^{3} + b} \sqrt {b} + 2 \, b}{x^{3}}\right ) + \sqrt {a x^{3} + b} {\left (a x^{3} - b\right )}\right )}}{3 \, x^{3}}, \frac {2 \, {\left (2 \, a \sqrt {-b} x^{3} \arctan \left (\frac {\sqrt {a x^{3} + b} \sqrt {-b}}{b}\right ) + \sqrt {a x^{3} + b} {\left (a x^{3} - b\right )}\right )}}{3 \, x^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x^3+b)^(1/2)*(a*x^3+2*b)/x^4,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.19, size = 62, normalized size = 1.11 \begin {gather*} \frac {2 \, {\left (\frac {2 \, a^{2} b \arctan \left (\frac {\sqrt {a x^{3} + b}}{\sqrt {-b}}\right )}{\sqrt {-b}} + \sqrt {a x^{3} + b} a^{2} - \frac {\sqrt {a x^{3} + b} a b}{x^{3}}\right )}}{3 \, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x^3+b)^(1/2)*(a*x^3+2*b)/x^4,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.11, size = 49, normalized size = 0.88

method	result	size
elliptic	\(-\frac {2 b \sqrt {a \,x^{3}+b}}{3 x^{3}}+\frac {2 a \sqrt {a \,x^{3}+b}}{3}-\frac {4 a \sqrt {b}\, \arctanh \left (\frac {\sqrt {a \,x^{3}+b}}{\sqrt {b}}\right )}{3}\)	\(49\)
risch	\(-\frac {2 b \sqrt {a \,x^{3}+b}}{3 x^{3}}+a \left (\frac {2 \sqrt {a \,x^{3}+b}}{3}-\frac {4 \sqrt {b}\, \arctanh \left (\frac {\sqrt {a \,x^{3}+b}}{\sqrt {b}}\right )}{3}\right )\)	\(50\)
default	\(2 b \left (-\frac {\sqrt {a \,x^{3}+b}}{3 x^{3}}-\frac {a \arctanh \left (\frac {\sqrt {a \,x^{3}+b}}{\sqrt {b}}\right )}{3 \sqrt {b}}\right )+a \left (\frac {2 \sqrt {a \,x^{3}+b}}{3}-\frac {2 \sqrt {b}\, \arctanh \left (\frac {\sqrt {a \,x^{3}+b}}{\sqrt {b}}\right )}{3}\right )\)	\(73\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^3+b)^(1/2)*(a*x^3+2*b)/x^4,x,method=_RETURNVERBOSE)

[Out]

-2/3*b*(a*x^3+b)^(1/2)/x^3+2/3*a*(a*x^3+b)^(1/2)-4/3*a*b^(1/2)*arctanh((a*x^3+b)^(1/2)/b^(1/2))

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maxima [B]  time = 0.44, size = 107, normalized size = 1.91 \begin {gather*} \frac {1}{3} \, {\left (\sqrt {b} \log \left (\frac {\sqrt {a x^{3} + b} - \sqrt {b}}{\sqrt {a x^{3} + b} + \sqrt {b}}\right ) + 2 \, \sqrt {a x^{3} + b}\right )} a + \frac {1}{3} \, {\left (\frac {a \log \left (\frac {\sqrt {a x^{3} + b} - \sqrt {b}}{\sqrt {a x^{3} + b} + \sqrt {b}}\right )}{\sqrt {b}} - \frac {2 \, \sqrt {a x^{3} + b}}{x^{3}}\right )} b \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.78, size = 69, normalized size = 1.23 \begin {gather*} \frac {2\,a\,\sqrt {a\,x^3+b}}{3}-\frac {2\,b\,\sqrt {a\,x^3+b}}{3\,x^3}+\frac {2\,a\,\sqrt {b}\,\ln \left (\frac {{\left (\sqrt {a\,x^3+b}-\sqrt {b}\right )}^3\,\left (\sqrt {a\,x^3+b}+\sqrt {b}\right )}{x^6}\right )}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((b + a*x^3)^(1/2)*(2*b + a*x^3))/x^4,x)

[Out]

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

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sympy [A]  time = 26.74, size = 105, normalized size = 1.88 \begin {gather*} \frac {2 a^{\frac {3}{2}} x^{\frac {3}{2}}}{3 \sqrt {1 + \frac {b}{a x^{3}}}} - \frac {2 \sqrt {a} b \sqrt {1 + \frac {b}{a x^{3}}}}{3 x^{\frac {3}{2}}} + \frac {2 \sqrt {a} b}{3 x^{\frac {3}{2}} \sqrt {1 + \frac {b}{a x^{3}}}} - \frac {4 a \sqrt {b} \operatorname {asinh}{\left (\frac {\sqrt {b}}{\sqrt {a} x^{\frac {3}{2}}} \right )}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x**3+b)**(1/2)*(a*x**3+2*b)/x**4,x)

[Out]

2*a**(3/2)*x**(3/2)/(3*sqrt(1 + b/(a*x**3))) - 2*sqrt(a)*b*sqrt(1 + b/(a*x**3))/(3*x**(3/2)) + 2*sqrt(a)*b/(3*
x**(3/2)*sqrt(1 + b/(a*x**3))) - 4*a*sqrt(b)*asinh(sqrt(b)/(sqrt(a)*x**(3/2)))/3

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