Integrand size = 26, antiderivative size = 85 \[ \int \frac {\sqrt {e x} \left (A+B x^3\right )}{\left (a+b x^3\right )^{3/2}} \, dx=\frac {2 (A b-a B) (e x)^{3/2}}{3 a b e \sqrt {a+b x^3}}+\frac {2 B \sqrt {e} \text {arctanh}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+b x^3}}\right )}{3 b^{3/2}} \]
2/3*B*arctanh((e*x)^(3/2)*b^(1/2)/e^(3/2)/(b*x^3+a)^(1/2))*e^(1/2)/b^(3/2) +2/3*(A*b-B*a)*(e*x)^(3/2)/a/b/e/(b*x^3+a)^(1/2)
Time = 0.74 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.95 \[ \int \frac {\sqrt {e x} \left (A+B x^3\right )}{\left (a+b x^3\right )^{3/2}} \, dx=\frac {2 \sqrt {e x} \left (\frac {\sqrt {b} (A b-a B) x^{3/2}}{a \sqrt {a+b x^3}}+B \log \left (\sqrt {b} x^{3/2}+\sqrt {a+b x^3}\right )\right )}{3 b^{3/2} \sqrt {x}} \]
(2*Sqrt[e*x]*((Sqrt[b]*(A*b - a*B)*x^(3/2))/(a*Sqrt[a + b*x^3]) + B*Log[Sq rt[b]*x^(3/2) + Sqrt[a + b*x^3]]))/(3*b^(3/2)*Sqrt[x])
Time = 0.24 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.01, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {954, 851, 807, 224, 219}
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.
\(\displaystyle \int \frac {\sqrt {e x} \left (A+B x^3\right )}{\left (a+b x^3\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 954 |
\(\displaystyle \frac {B \int \frac {\sqrt {e x}}{\sqrt {b x^3+a}}dx}{b}+\frac {2 (e x)^{3/2} (A b-a B)}{3 a b e \sqrt {a+b x^3}}\) |
\(\Big \downarrow \) 851 |
\(\displaystyle \frac {2 B \int \frac {e x}{\sqrt {b x^3+a}}d\sqrt {e x}}{b e}+\frac {2 (e x)^{3/2} (A b-a B)}{3 a b e \sqrt {a+b x^3}}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {2 B \int \frac {1}{\sqrt {a+\frac {b x}{e^2}}}d(e x)^{3/2}}{3 b e}+\frac {2 (e x)^{3/2} (A b-a B)}{3 a b e \sqrt {a+b x^3}}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {2 B \int \frac {1}{1-\frac {b x}{e^2}}d\frac {(e x)^{3/2}}{\sqrt {a+\frac {b x}{e^2}}}}{3 b e}+\frac {2 (e x)^{3/2} (A b-a B)}{3 a b e \sqrt {a+b x^3}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {2 (e x)^{3/2} (A b-a B)}{3 a b e \sqrt {a+b x^3}}+\frac {2 B \sqrt {e} \text {arctanh}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+\frac {b x}{e^2}}}\right )}{3 b^{3/2}}\) |
(2*(A*b - a*B)*(e*x)^(3/2))/(3*a*b*e*Sqrt[a + b*x^3]) + (2*B*Sqrt[e]*ArcTa nh[(Sqrt[b]*(e*x)^(3/2))/(e^(3/2)*Sqrt[a + (b*x)/e^2])])/(3*b^(3/2))
3.6.54.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[1/k Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^ n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n _)), x_Symbol] :> Simp[(b*c - a*d)*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*b* e*(m + 1))), x] + Simp[d/b Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; Fre eQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n*(p + 1) + 1, 0] && NeQ[m, -1]
Time = 4.28 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.08
method | result | size |
default | \(\frac {2 \sqrt {e x}\, \left (A \sqrt {b e}\, b \,x^{2}-B \sqrt {b e}\, a \,x^{2}+B \sqrt {\left (b \,x^{3}+a \right ) e x}\, \operatorname {arctanh}\left (\frac {\sqrt {\left (b \,x^{3}+a \right ) e x}}{x^{2} \sqrt {b e}}\right ) a \right )}{3 \sqrt {b \,x^{3}+a}\, x b \sqrt {b e}\, a}\) | \(92\) |
elliptic | \(\text {Expression too large to display}\) | \(1050\) |
2/3*(e*x)^(1/2)/(b*x^3+a)^(1/2)*(A*(b*e)^(1/2)*b*x^2-B*(b*e)^(1/2)*a*x^2+B *((b*x^3+a)*e*x)^(1/2)*arctanh(((b*x^3+a)*e*x)^(1/2)/x^2/(b*e)^(1/2))*a)/x /b/(b*e)^(1/2)/a
Time = 0.39 (sec) , antiderivative size = 234, normalized size of antiderivative = 2.75 \[ \int \frac {\sqrt {e x} \left (A+B x^3\right )}{\left (a+b x^3\right )^{3/2}} \, dx=\left [-\frac {4 \, \sqrt {b x^{3} + a} {\left (B a - A b\right )} \sqrt {e x} x - {\left (B a b x^{3} + B a^{2}\right )} \sqrt {\frac {e}{b}} \log \left (-8 \, b^{2} e x^{6} - 8 \, a b e x^{3} - a^{2} e - 4 \, {\left (2 \, b^{2} x^{4} + a b x\right )} \sqrt {b x^{3} + a} \sqrt {e x} \sqrt {\frac {e}{b}}\right )}{6 \, {\left (a b^{2} x^{3} + a^{2} b\right )}}, -\frac {2 \, \sqrt {b x^{3} + a} {\left (B a - A b\right )} \sqrt {e x} x + {\left (B a b x^{3} + B a^{2}\right )} \sqrt {-\frac {e}{b}} \arctan \left (\frac {2 \, \sqrt {b x^{3} + a} \sqrt {e x} b x \sqrt {-\frac {e}{b}}}{2 \, b e x^{3} + a e}\right )}{3 \, {\left (a b^{2} x^{3} + a^{2} b\right )}}\right ] \]
[-1/6*(4*sqrt(b*x^3 + a)*(B*a - A*b)*sqrt(e*x)*x - (B*a*b*x^3 + B*a^2)*sqr t(e/b)*log(-8*b^2*e*x^6 - 8*a*b*e*x^3 - a^2*e - 4*(2*b^2*x^4 + a*b*x)*sqrt (b*x^3 + a)*sqrt(e*x)*sqrt(e/b)))/(a*b^2*x^3 + a^2*b), -1/3*(2*sqrt(b*x^3 + a)*(B*a - A*b)*sqrt(e*x)*x + (B*a*b*x^3 + B*a^2)*sqrt(-e/b)*arctan(2*sqr t(b*x^3 + a)*sqrt(e*x)*b*x*sqrt(-e/b)/(2*b*e*x^3 + a*e)))/(a*b^2*x^3 + a^2 *b)]
Time = 19.57 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.12 \[ \int \frac {\sqrt {e x} \left (A+B x^3\right )}{\left (a+b x^3\right )^{3/2}} \, dx=\frac {2 A \sqrt {e} x^{\frac {3}{2}}}{3 a^{\frac {3}{2}} \sqrt {1 + \frac {b x^{3}}{a}}} + B \left (\frac {2 \sqrt {e} \operatorname {asinh}{\left (\frac {\sqrt {b} x^{\frac {3}{2}}}{\sqrt {a}} \right )}}{3 b^{\frac {3}{2}}} - \frac {2 \sqrt {e} x^{\frac {3}{2}}}{3 \sqrt {a} b \sqrt {1 + \frac {b x^{3}}{a}}}\right ) \]
2*A*sqrt(e)*x**(3/2)/(3*a**(3/2)*sqrt(1 + b*x**3/a)) + B*(2*sqrt(e)*asinh( sqrt(b)*x**(3/2)/sqrt(a))/(3*b**(3/2)) - 2*sqrt(e)*x**(3/2)/(3*sqrt(a)*b*s qrt(1 + b*x**3/a)))
\[ \int \frac {\sqrt {e x} \left (A+B x^3\right )}{\left (a+b x^3\right )^{3/2}} \, dx=\int { \frac {{\left (B x^{3} + A\right )} \sqrt {e x}}{{\left (b x^{3} + a\right )}^{\frac {3}{2}}} \,d x } \]
Time = 0.32 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.09 \[ \int \frac {\sqrt {e x} \left (A+B x^3\right )}{\left (a+b x^3\right )^{3/2}} \, dx=-\frac {2 \, B e^{3} \log \left ({\left | -\sqrt {b e} \sqrt {e x} e x + \sqrt {b e^{4} x^{3} + a e^{4}} \right |}\right )}{3 \, \sqrt {b e} b {\left | e \right |}^{2}} - \frac {2 \, {\left (B a e - A b e\right )} \sqrt {e x} e x}{3 \, \sqrt {b e^{4} x^{3} + a e^{4}} a b} \]
-2/3*B*e^3*log(abs(-sqrt(b*e)*sqrt(e*x)*e*x + sqrt(b*e^4*x^3 + a*e^4)))/(s qrt(b*e)*b*abs(e)^2) - 2/3*(B*a*e - A*b*e)*sqrt(e*x)*e*x/(sqrt(b*e^4*x^3 + a*e^4)*a*b)
Timed out. \[ \int \frac {\sqrt {e x} \left (A+B x^3\right )}{\left (a+b x^3\right )^{3/2}} \, dx=\int \frac {\left (B\,x^3+A\right )\,\sqrt {e\,x}}{{\left (b\,x^3+a\right )}^{3/2}} \,d x \]