Integrand size = 22, antiderivative size = 81 \[ \int \frac {\left (a+b x^3\right )^{3/2} \left (A+B x^3\right )}{x} \, dx=\frac {2}{3} a A \sqrt {a+b x^3}+\frac {2}{9} A \left (a+b x^3\right )^{3/2}+\frac {2 B \left (a+b x^3\right )^{5/2}}{15 b}-\frac {2}{3} a^{3/2} A \text {arctanh}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right ) \]
2/9*A*(b*x^3+a)^(3/2)+2/15*B*(b*x^3+a)^(5/2)/b-2/3*a^(3/2)*A*arctanh((b*x^ 3+a)^(1/2)/a^(1/2))+2/3*a*A*(b*x^3+a)^(1/2)
Time = 0.15 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.05 \[ \int \frac {\left (a+b x^3\right )^{3/2} \left (A+B x^3\right )}{x} \, dx=\frac {2 \sqrt {a+b x^3} \left (20 a A b+3 a^2 B+5 A b^2 x^3+6 a b B x^3+3 b^2 B x^6\right )}{45 b}-\frac {2}{3} a^{3/2} A \text {arctanh}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right ) \]
(2*Sqrt[a + b*x^3]*(20*a*A*b + 3*a^2*B + 5*A*b^2*x^3 + 6*a*b*B*x^3 + 3*b^2 *B*x^6))/(45*b) - (2*a^(3/2)*A*ArcTanh[Sqrt[a + b*x^3]/Sqrt[a]])/3
Time = 0.20 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {948, 90, 60, 60, 73, 221}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\left (a+b x^3\right )^{3/2} \left (A+B x^3\right )}{x} \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {1}{3} \int \frac {\left (b x^3+a\right )^{3/2} \left (B x^3+A\right )}{x^3}dx^3\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {1}{3} \left (A \int \frac {\left (b x^3+a\right )^{3/2}}{x^3}dx^3+\frac {2 B \left (a+b x^3\right )^{5/2}}{5 b}\right )\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{3} \left (A \left (a \int \frac {\sqrt {b x^3+a}}{x^3}dx^3+\frac {2}{3} \left (a+b x^3\right )^{3/2}\right )+\frac {2 B \left (a+b x^3\right )^{5/2}}{5 b}\right )\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{3} \left (A \left (a \left (a \int \frac {1}{x^3 \sqrt {b x^3+a}}dx^3+2 \sqrt {a+b x^3}\right )+\frac {2}{3} \left (a+b x^3\right )^{3/2}\right )+\frac {2 B \left (a+b x^3\right )^{5/2}}{5 b}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{3} \left (A \left (a \left (\frac {2 a \int \frac {1}{\frac {x^6}{b}-\frac {a}{b}}d\sqrt {b x^3+a}}{b}+2 \sqrt {a+b x^3}\right )+\frac {2}{3} \left (a+b x^3\right )^{3/2}\right )+\frac {2 B \left (a+b x^3\right )^{5/2}}{5 b}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{3} \left (A \left (a \left (2 \sqrt {a+b x^3}-2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )\right )+\frac {2}{3} \left (a+b x^3\right )^{3/2}\right )+\frac {2 B \left (a+b x^3\right )^{5/2}}{5 b}\right )\) |
((2*B*(a + b*x^3)^(5/2))/(5*b) + A*((2*(a + b*x^3)^(3/2))/3 + a*(2*Sqrt[a + b*x^3] - 2*Sqrt[a]*ArcTanh[Sqrt[a + b*x^3]/Sqrt[a]])))/3
3.2.99.3.1 Defintions of rubi rules used
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] + Simp[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] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[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] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(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]
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]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. ), x_Symbol] :> Simp[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]]
Time = 4.20 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.81
method | result | size |
default | \(\frac {2 B \left (b \,x^{3}+a \right )^{\frac {5}{2}}}{15 b}+A \left (\frac {2 b \,x^{3} \sqrt {b \,x^{3}+a}}{9}+\frac {8 a \sqrt {b \,x^{3}+a}}{9}-\frac {2 a^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{3}\right )\) | \(66\) |
pseudoelliptic | \(\frac {-\frac {2 a^{\frac {3}{2}} b A \,\operatorname {arctanh}\left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{3}+\frac {8 \left (\frac {x^{3} \left (\frac {3 x^{3} B}{5}+A \right ) b^{2}}{4}+a \left (\frac {3 x^{3} B}{10}+A \right ) b +\frac {3 a^{2} B}{20}\right ) \sqrt {b \,x^{3}+a}}{9}}{b}\) | \(73\) |
elliptic | \(\frac {2 B b \,x^{6} \sqrt {b \,x^{3}+a}}{15}+\frac {2 \left (b^{2} A +\frac {6}{5} a b B \right ) x^{3} \sqrt {b \,x^{3}+a}}{9 b}+\frac {2 \left (2 a b A +a^{2} B -\frac {2 \left (b^{2} A +\frac {6}{5} a b B \right ) a}{3 b}\right ) \sqrt {b \,x^{3}+a}}{3 b}-\frac {2 a^{\frac {3}{2}} A \,\operatorname {arctanh}\left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{3}\) | \(108\) |
2/15*B*(b*x^3+a)^(5/2)/b+A*(2/9*b*x^3*(b*x^3+a)^(1/2)+8/9*a*(b*x^3+a)^(1/2 )-2/3*a^(3/2)*arctanh((b*x^3+a)^(1/2)/a^(1/2)))
Time = 0.26 (sec) , antiderivative size = 172, normalized size of antiderivative = 2.12 \[ \int \frac {\left (a+b x^3\right )^{3/2} \left (A+B x^3\right )}{x} \, dx=\left [\frac {15 \, A a^{\frac {3}{2}} b \log \left (\frac {b x^{3} - 2 \, \sqrt {b x^{3} + a} \sqrt {a} + 2 \, a}{x^{3}}\right ) + 2 \, {\left (3 \, B b^{2} x^{6} + {\left (6 \, B a b + 5 \, A b^{2}\right )} x^{3} + 3 \, B a^{2} + 20 \, A a b\right )} \sqrt {b x^{3} + a}}{45 \, b}, \frac {2 \, {\left (15 \, A \sqrt {-a} a b \arctan \left (\frac {\sqrt {b x^{3} + a} \sqrt {-a}}{a}\right ) + {\left (3 \, B b^{2} x^{6} + {\left (6 \, B a b + 5 \, A b^{2}\right )} x^{3} + 3 \, B a^{2} + 20 \, A a b\right )} \sqrt {b x^{3} + a}\right )}}{45 \, b}\right ] \]
[1/45*(15*A*a^(3/2)*b*log((b*x^3 - 2*sqrt(b*x^3 + a)*sqrt(a) + 2*a)/x^3) + 2*(3*B*b^2*x^6 + (6*B*a*b + 5*A*b^2)*x^3 + 3*B*a^2 + 20*A*a*b)*sqrt(b*x^3 + a))/b, 2/45*(15*A*sqrt(-a)*a*b*arctan(sqrt(b*x^3 + a)*sqrt(-a)/a) + (3* B*b^2*x^6 + (6*B*a*b + 5*A*b^2)*x^3 + 3*B*a^2 + 20*A*a*b)*sqrt(b*x^3 + a)) /b]
Time = 10.22 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.35 \[ \int \frac {\left (a+b x^3\right )^{3/2} \left (A+B x^3\right )}{x} \, dx=\frac {\begin {cases} \frac {2 A a^{2} \operatorname {atan}{\left (\frac {\sqrt {a + b x^{3}}}{\sqrt {- a}} \right )}}{\sqrt {- a}} + 2 A a \sqrt {a + b x^{3}} + \frac {2 A \left (a + b x^{3}\right )^{\frac {3}{2}}}{3} + \frac {2 B \left (a + b x^{3}\right )^{\frac {5}{2}}}{5 b} & \text {for}\: b \neq 0 \\A a^{\frac {3}{2}} \log {\left (B a^{\frac {3}{2}} x^{3} \right )} + B a^{\frac {3}{2}} x^{3} & \text {otherwise} \end {cases}}{3} \]
Piecewise((2*A*a**2*atan(sqrt(a + b*x**3)/sqrt(-a))/sqrt(-a) + 2*A*a*sqrt( a + b*x**3) + 2*A*(a + b*x**3)**(3/2)/3 + 2*B*(a + b*x**3)**(5/2)/(5*b), N e(b, 0)), (A*a**(3/2)*log(B*a**(3/2)*x**3) + B*a**(3/2)*x**3, True))/3
Time = 0.28 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.99 \[ \int \frac {\left (a+b x^3\right )^{3/2} \left (A+B x^3\right )}{x} \, dx=\frac {2 \, {\left (b x^{3} + a\right )}^{\frac {5}{2}} B}{15 \, b} + \frac {1}{9} \, {\left (3 \, a^{\frac {3}{2}} \log \left (\frac {\sqrt {b x^{3} + a} - \sqrt {a}}{\sqrt {b x^{3} + a} + \sqrt {a}}\right ) + 2 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} + 6 \, \sqrt {b x^{3} + a} a\right )} A \]
2/15*(b*x^3 + a)^(5/2)*B/b + 1/9*(3*a^(3/2)*log((sqrt(b*x^3 + a) - sqrt(a) )/(sqrt(b*x^3 + a) + sqrt(a))) + 2*(b*x^3 + a)^(3/2) + 6*sqrt(b*x^3 + a)*a )*A
Time = 0.27 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.99 \[ \int \frac {\left (a+b x^3\right )^{3/2} \left (A+B x^3\right )}{x} \, dx=\frac {2 \, A a^{2} \arctan \left (\frac {\sqrt {b x^{3} + a}}{\sqrt {-a}}\right )}{3 \, \sqrt {-a}} + \frac {2 \, {\left (3 \, {\left (b x^{3} + a\right )}^{\frac {5}{2}} B b^{4} + 5 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} A b^{5} + 15 \, \sqrt {b x^{3} + a} A a b^{5}\right )}}{45 \, b^{5}} \]
2/3*A*a^2*arctan(sqrt(b*x^3 + a)/sqrt(-a))/sqrt(-a) + 2/45*(3*(b*x^3 + a)^ (5/2)*B*b^4 + 5*(b*x^3 + a)^(3/2)*A*b^5 + 15*sqrt(b*x^3 + a)*A*a*b^5)/b^5
Time = 6.91 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.62 \[ \int \frac {\left (a+b x^3\right )^{3/2} \left (A+B x^3\right )}{x} \, dx=\frac {A\,a^{3/2}\,\ln \left (\frac {{\left (\sqrt {b\,x^3+a}-\sqrt {a}\right )}^3\,\left (\sqrt {b\,x^3+a}+\sqrt {a}\right )}{x^6}\right )}{3}+\frac {\sqrt {b\,x^3+a}\,\left (2\,B\,a^2+4\,A\,a\,b-\frac {2\,a\,\left (2\,A\,b^2+\frac {12\,B\,a\,b}{5}\right )}{3\,b}\right )}{3\,b}+\frac {2\,B\,b\,x^6\,\sqrt {b\,x^3+a}}{15}+\frac {x^3\,\left (2\,A\,b^2+\frac {12\,B\,a\,b}{5}\right )\,\sqrt {b\,x^3+a}}{9\,b} \]