3.7.66 \(\int \frac {(-b+a x^3) \sqrt {b+a x^3}}{x} \, dx\)

Optimal. Leaf size=53 \[ \frac {2}{3} b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a x^3+b}}{\sqrt {b}}\right )-\frac {2}{9} \left (2 b-a x^3\right ) \sqrt {a x^3+b} \]

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Rubi [A]  time = 0.04, antiderivative size = 59, normalized size of antiderivative = 1.11, 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, 80, 50, 63, 208} \begin {gather*} \frac {2}{3} b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a x^3+b}}{\sqrt {b}}\right )-\frac {2}{3} b \sqrt {a x^3+b}+\frac {2}{9} \left (a x^3+b\right )^{3/2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b*Sqrt[b + a*x^3])/3 + (2*(b + a*x^3)^(3/2))/9 + (2*b^(3/2)*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 80

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

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 {\left (-b+a x^3\right ) \sqrt {b+a x^3}}{x} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {(-b+a x) \sqrt {b+a x}}{x} \, dx,x,x^3\right )\\ &=\frac {2}{9} \left (b+a x^3\right )^{3/2}-\frac {1}{3} b \operatorname {Subst}\left (\int \frac {\sqrt {b+a x}}{x} \, dx,x,x^3\right )\\ &=-\frac {2}{3} b \sqrt {b+a x^3}+\frac {2}{9} \left (b+a x^3\right )^{3/2}-\frac {1}{3} b^2 \operatorname {Subst}\left (\int \frac {1}{x \sqrt {b+a x}} \, dx,x,x^3\right )\\ &=-\frac {2}{3} b \sqrt {b+a x^3}+\frac {2}{9} \left (b+a x^3\right )^{3/2}-\frac {\left (2 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {b}{a}+\frac {x^2}{a}} \, dx,x,\sqrt {b+a x^3}\right )}{3 a}\\ &=-\frac {2}{3} b \sqrt {b+a x^3}+\frac {2}{9} \left (b+a x^3\right )^{3/2}+\frac {2}{3} b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b+a x^3}}{\sqrt {b}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 52, normalized size = 0.98 \begin {gather*} \frac {1}{9} \left (6 b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a x^3+b}}{\sqrt {b}}\right )+2 \left (a x^3-2 b\right ) \sqrt {a x^3+b}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-2*b + a*x^3)*Sqrt[b + a*x^3] + 6*b^(3/2)*ArcTanh[Sqrt[b + a*x^3]/Sqrt[b]])/9

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IntegrateAlgebraic [A]  time = 0.04, size = 53, normalized size = 1.00 \begin {gather*} -\frac {2}{9} \left (2 b-a x^3\right ) \sqrt {b+a x^3}+\frac {2}{3} b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b+a x^3}}{\sqrt {b}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(2*b - a*x^3)*Sqrt[b + a*x^3])/9 + (2*b^(3/2)*ArcTanh[Sqrt[b + a*x^3]/Sqrt[b]])/3

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fricas [A]  time = 0.48, size = 103, normalized size = 1.94 \begin {gather*} \left [\frac {1}{3} \, b^{\frac {3}{2}} \log \left (\frac {a x^{3} + 2 \, \sqrt {a x^{3} + b} \sqrt {b} + 2 \, b}{x^{3}}\right ) + \frac {2}{9} \, \sqrt {a x^{3} + b} {\left (a x^{3} - 2 \, b\right )}, -\frac {2}{3} \, \sqrt {-b} b \arctan \left (\frac {\sqrt {a x^{3} + b} \sqrt {-b}}{b}\right ) + \frac {2}{9} \, \sqrt {a x^{3} + b} {\left (a x^{3} - 2 \, b\right )}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/3*b^(3/2)*log((a*x^3 + 2*sqrt(a*x^3 + b)*sqrt(b) + 2*b)/x^3) + 2/9*sqrt(a*x^3 + b)*(a*x^3 - 2*b), -2/3*sqrt
(-b)*b*arctan(sqrt(a*x^3 + b)*sqrt(-b)/b) + 2/9*sqrt(a*x^3 + b)*(a*x^3 - 2*b)]

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giac [A]  time = 0.19, size = 50, normalized size = 0.94 \begin {gather*} -\frac {2 \, b^{2} \arctan \left (\frac {\sqrt {a x^{3} + b}}{\sqrt {-b}}\right )}{3 \, \sqrt {-b}} + \frac {2}{9} \, {\left (a x^{3} + b\right )}^{\frac {3}{2}} - \frac {2}{3} \, \sqrt {a x^{3} + b} b \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*b^2*arctan(sqrt(a*x^3 + b)/sqrt(-b))/sqrt(-b) + 2/9*(a*x^3 + b)^(3/2) - 2/3*sqrt(a*x^3 + b)*b

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maple [A]  time = 0.09, size = 47, normalized size = 0.89

method	result	size
default	\(\frac {2 \left (a \,x^{3}+b \right )^{\frac {3}{2}}}{9}-b \left (\frac {2 \sqrt {a \,x^{3}+b}}{3}-\frac {2 \sqrt {b}\, \arctanh \left (\frac {\sqrt {a \,x^{3}+b}}{\sqrt {b}}\right )}{3}\right )\)	\(47\)
elliptic	\(\frac {2 a \,x^{3} \sqrt {a \,x^{3}+b}}{9}-\frac {4 b \sqrt {a \,x^{3}+b}}{9}+\frac {2 b^{\frac {3}{2}} \arctanh \left (\frac {\sqrt {a \,x^{3}+b}}{\sqrt {b}}\right )}{3}\)	\(48\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/9*(a*x^3+b)^(3/2)-b*(2/3*(a*x^3+b)^(1/2)-2/3*b^(1/2)*arctanh((a*x^3+b)^(1/2)/b^(1/2)))

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maxima [A]  time = 0.67, size = 63, normalized size = 1.19 \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 )} b + \frac {2}{9} \, {\left (a x^{3} + b\right )}^{\frac {3}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x^3-b)*(a*x^3+b)^(1/2)/x,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))*b + 2/9*(a*x^3
+ b)^(3/2)

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mupad [B]  time = 0.71, size = 68, normalized size = 1.28 \begin {gather*} \frac {b^{3/2}\,\ln \left (\frac {\left (\sqrt {a\,x^3+b}-\sqrt {b}\right )\,{\left (\sqrt {a\,x^3+b}+\sqrt {b}\right )}^3}{x^6}\right )}{3}-\frac {4\,b\,\sqrt {a\,x^3+b}}{9}+\frac {2\,a\,x^3\,\sqrt {a\,x^3+b}}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b^(3/2)*log((((b + a*x^3)^(1/2) - b^(1/2))*((b + a*x^3)^(1/2) + b^(1/2))^3)/x^6))/3 - (4*b*(b + a*x^3)^(1/2))
/9 + (2*a*x^3*(b + a*x^3)^(1/2))/9

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sympy [A]  time = 21.16, size = 75, normalized size = 1.42 \begin {gather*} - \frac {a \left (\begin {cases} - \sqrt {b} x^{3} & \text {for}\: a = 0 \\- \frac {2 \left (a x^{3} + b\right )^{\frac {3}{2}}}{3 a} & \text {otherwise} \end {cases}\right )}{3} + \frac {b \left (- \frac {2 b \operatorname {atan}{\left (\frac {\sqrt {a x^{3} + b}}{\sqrt {- b}} \right )}}{\sqrt {- b}} - 2 \sqrt {a x^{3} + b}\right )}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*Piecewise((-sqrt(b)*x**3, Eq(a, 0)), (-2*(a*x**3 + b)**(3/2)/(3*a), True))/3 + b*(-2*b*atan(sqrt(a*x**3 + b
)/sqrt(-b))/sqrt(-b) - 2*sqrt(a*x**3 + b))/3

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