3.2.78 \(\int \frac {\sqrt {a+b x^3} (A+B x^3)}{x} \, dx\)

Optimal. Leaf size=64 \[ \frac {2}{3} A \sqrt {a+b x^3}-\frac {2}{3} \sqrt {a} A \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )+\frac {2 B \left (a+b x^3\right )^{3/2}}{9 b} \]

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Rubi [A]  time = 0.04, antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {446, 80, 50, 63, 208} \begin {gather*} \frac {2}{3} A \sqrt {a+b x^3}-\frac {2}{3} \sqrt {a} A \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )+\frac {2 B \left (a+b x^3\right )^{3/2}}{9 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.05, size = 60, normalized size = 0.94 \begin {gather*} \frac {2}{9} \left (\frac {\sqrt {a+b x^3} \left (B \left (a+b x^3\right )+3 A b\right )}{b}-3 \sqrt {a} A \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.05, size = 61, normalized size = 0.95 \begin {gather*} \frac {2 \sqrt {a+b x^3} \left (a B+3 A b+b B x^3\right )}{9 b}-\frac {2}{3} \sqrt {a} A \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.77, size = 125, normalized size = 1.95 \begin {gather*} \left [\frac {3 \, A \sqrt {a} b \log \left (\frac {b x^{3} - 2 \, \sqrt {b x^{3} + a} \sqrt {a} + 2 \, a}{x^{3}}\right ) + 2 \, {\left (B b x^{3} + B a + 3 \, A b\right )} \sqrt {b x^{3} + a}}{9 \, b}, \frac {2 \, {\left (3 \, A \sqrt {-a} b \arctan \left (\frac {\sqrt {b x^{3} + a} \sqrt {-a}}{a}\right ) + {\left (B b x^{3} + B a + 3 \, A b\right )} \sqrt {b x^{3} + a}\right )}}{9 \, b}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.15, size = 61, normalized size = 0.95 \begin {gather*} \frac {2 \, A a \arctan \left (\frac {\sqrt {b x^{3} + a}}{\sqrt {-a}}\right )}{3 \, \sqrt {-a}} + \frac {2 \, {\left ({\left (b x^{3} + a\right )}^{\frac {3}{2}} B b^{2} + 3 \, \sqrt {b x^{3} + a} A b^{3}\right )}}{9 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 50, normalized size = 0.78 \begin {gather*} \left (-\frac {2 \sqrt {a}\, \arctanh \left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{3}+\frac {2 \sqrt {b \,x^{3}+a}}{3}\right ) A +\frac {2 \left (b \,x^{3}+a \right )^{\frac {3}{2}} B}{9 b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.20, size = 67, normalized size = 1.05 \begin {gather*} \frac {1}{3} \, {\left (\sqrt {a} \log \left (\frac {\sqrt {b x^{3} + a} - \sqrt {a}}{\sqrt {b x^{3} + a} + \sqrt {a}}\right ) + 2 \, \sqrt {b x^{3} + a}\right )} A + \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} B}{9 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(sqrt(a)*log((sqrt(b*x^3 + a) - sqrt(a))/(sqrt(b*x^3 + a) + sqrt(a))) + 2*sqrt(b*x^3 + a))*A + 2/9*(b*x^3
+ a)^(3/2)*B/b

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mupad [B]  time = 2.71, size = 80, normalized size = 1.25 \begin {gather*} \frac {2\,B\,x^3\,\sqrt {b\,x^3+a}}{9}+\frac {\sqrt {b\,x^3+a}\,\left (2\,A\,b+\frac {2\,B\,a}{3}\right )}{3\,b}+\frac {A\,\sqrt {a}\,\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 )}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 25.83, size = 76, normalized size = 1.19 \begin {gather*} - \frac {A \left (- \frac {2 a \operatorname {atan}{\left (\frac {\sqrt {a + b x^{3}}}{\sqrt {- a}} \right )}}{\sqrt {- a}} - 2 \sqrt {a + b x^{3}}\right )}{3} - \frac {B \left (\begin {cases} - \sqrt {a} x^{3} & \text {for}\: b = 0 \\- \frac {2 \left (a + b x^{3}\right )^{\frac {3}{2}}}{3 b} & \text {otherwise} \end {cases}\right )}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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