Integrand size = 26, antiderivative size = 75 \[ \int \frac {A+B x^3}{(e x)^{5/2} \sqrt {a+b x^3}} \, dx=-\frac {2 A \sqrt {a+b x^3}}{3 a e (e x)^{3/2}}+\frac {2 B \text {arctanh}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+b x^3}}\right )}{3 \sqrt {b} e^{5/2}} \]
2/3*B*arctanh((e*x)^(3/2)*b^(1/2)/e^(3/2)/(b*x^3+a)^(1/2))/e^(5/2)/b^(1/2) -2/3*A*(b*x^3+a)^(1/2)/a/e/(e*x)^(3/2)
Time = 0.42 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.88 \[ \int \frac {A+B x^3}{(e x)^{5/2} \sqrt {a+b x^3}} \, dx=\frac {2 x \left (-\frac {A \sqrt {a+b x^3}}{a}+\frac {B x^{3/2} \log \left (\sqrt {b} x^{3/2}+\sqrt {a+b x^3}\right )}{\sqrt {b}}\right )}{3 (e x)^{5/2}} \]
(2*x*(-((A*Sqrt[a + b*x^3])/a) + (B*x^(3/2)*Log[Sqrt[b]*x^(3/2) + Sqrt[a + b*x^3]])/Sqrt[b]))/(3*(e*x)^(5/2))
Time = 0.23 (sec) , antiderivative size = 76, 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 = {953, 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.
\(\displaystyle \int \frac {A+B x^3}{(e x)^{5/2} \sqrt {a+b x^3}} \, dx\) |
\(\Big \downarrow \) 953 |
\(\displaystyle \frac {B \int \frac {\sqrt {e x}}{\sqrt {b x^3+a}}dx}{e^3}-\frac {2 A \sqrt {a+b x^3}}{3 a e (e x)^{3/2}}\) |
\(\Big \downarrow \) 851 |
\(\displaystyle \frac {2 B \int \frac {e x}{\sqrt {b x^3+a}}d\sqrt {e x}}{e^4}-\frac {2 A \sqrt {a+b x^3}}{3 a e (e x)^{3/2}}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {2 B \int \frac {1}{\sqrt {a+\frac {b x}{e^2}}}d(e x)^{3/2}}{3 e^4}-\frac {2 A \sqrt {a+b x^3}}{3 a e (e x)^{3/2}}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {2 B \int \frac {1}{1-\frac {b x}{e^2}}d\frac {(e x)^{3/2}}{\sqrt {a+\frac {b x}{e^2}}}}{3 e^4}-\frac {2 A \sqrt {a+b x^3}}{3 a e (e x)^{3/2}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {2 B \text {arctanh}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+\frac {b x}{e^2}}}\right )}{3 \sqrt {b} e^{5/2}}-\frac {2 A \sqrt {a+b x^3}}{3 a e (e x)^{3/2}}\) |
(-2*A*Sqrt[a + b*x^3])/(3*a*e*(e*x)^(3/2)) + (2*B*ArcTanh[(Sqrt[b]*(e*x)^( 3/2))/(e^(3/2)*Sqrt[a + (b*x)/e^2])])/(3*Sqrt[b]*e^(5/2))
3.6.49.3.1 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[c*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] + Simp[d/e^n Int[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n*(p + 1) + 1, 0] && (IntegerQ[n] || GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && G tQ[m + n, -1]))
Time = 4.62 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.16
method | result | size |
risch | \(-\frac {2 A \sqrt {b \,x^{3}+a}}{3 a x \,e^{2} \sqrt {e x}}+\frac {2 B \,\operatorname {arctanh}\left (\frac {\sqrt {\left (b \,x^{3}+a \right ) e x}}{x^{2} \sqrt {b e}}\right ) \sqrt {\left (b \,x^{3}+a \right ) e x}}{3 \sqrt {b e}\, e^{2} \sqrt {e x}\, \sqrt {b \,x^{3}+a}}\) | \(87\) |
default | \(-\frac {2 \sqrt {b \,x^{3}+a}\, \left (-B \,\operatorname {arctanh}\left (\frac {\sqrt {\left (b \,x^{3}+a \right ) e x}}{x^{2} \sqrt {b e}}\right ) a e \,x^{2}+A \sqrt {\left (b \,x^{3}+a \right ) e x}\, \sqrt {b e}\right )}{3 x \,e^{2} \sqrt {e x}\, \sqrt {\left (b \,x^{3}+a \right ) e x}\, a \sqrt {b e}}\) | \(93\) |
elliptic | \(\text {Expression too large to display}\) | \(1037\) |
-2/3/a*A*(b*x^3+a)^(1/2)/x/e^2/(e*x)^(1/2)+2/3*B/(b*e)^(1/2)*arctanh(((b*x ^3+a)*e*x)^(1/2)/x^2/(b*e)^(1/2))/e^2*((b*x^3+a)*e*x)^(1/2)/(e*x)^(1/2)/(b *x^3+a)^(1/2)
Time = 0.38 (sec) , antiderivative size = 183, normalized size of antiderivative = 2.44 \[ \int \frac {A+B x^3}{(e x)^{5/2} \sqrt {a+b x^3}} \, dx=\left [\frac {\sqrt {b e} B a x^{2} \log \left (-8 \, b^{2} e x^{6} - 8 \, a b e x^{3} - a^{2} e - 4 \, {\left (2 \, b x^{4} + a x\right )} \sqrt {b x^{3} + a} \sqrt {b e} \sqrt {e x}\right ) - 4 \, \sqrt {b x^{3} + a} \sqrt {e x} A b}{6 \, a b e^{3} x^{2}}, -\frac {\sqrt {-b e} B a x^{2} \arctan \left (\frac {2 \, \sqrt {b x^{3} + a} \sqrt {-b e} \sqrt {e x} x}{2 \, b e x^{3} + a e}\right ) + 2 \, \sqrt {b x^{3} + a} \sqrt {e x} A b}{3 \, a b e^{3} x^{2}}\right ] \]
[1/6*(sqrt(b*e)*B*a*x^2*log(-8*b^2*e*x^6 - 8*a*b*e*x^3 - a^2*e - 4*(2*b*x^ 4 + a*x)*sqrt(b*x^3 + a)*sqrt(b*e)*sqrt(e*x)) - 4*sqrt(b*x^3 + a)*sqrt(e*x )*A*b)/(a*b*e^3*x^2), -1/3*(sqrt(-b*e)*B*a*x^2*arctan(2*sqrt(b*x^3 + a)*sq rt(-b*e)*sqrt(e*x)*x/(2*b*e*x^3 + a*e)) + 2*sqrt(b*x^3 + a)*sqrt(e*x)*A*b) /(a*b*e^3*x^2)]
Time = 5.12 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.80 \[ \int \frac {A+B x^3}{(e x)^{5/2} \sqrt {a+b x^3}} \, dx=- \frac {2 A \sqrt {b} \sqrt {\frac {a}{b x^{3}} + 1}}{3 a e^{\frac {5}{2}}} + \frac {2 B \operatorname {asinh}{\left (\frac {\sqrt {b} x^{\frac {3}{2}}}{\sqrt {a}} \right )}}{3 \sqrt {b} e^{\frac {5}{2}}} \]
-2*A*sqrt(b)*sqrt(a/(b*x**3) + 1)/(3*a*e**(5/2)) + 2*B*asinh(sqrt(b)*x**(3 /2)/sqrt(a))/(3*sqrt(b)*e**(5/2))
\[ \int \frac {A+B x^3}{(e x)^{5/2} \sqrt {a+b x^3}} \, dx=\int { \frac {B x^{3} + A}{\sqrt {b x^{3} + a} \left (e x\right )^{\frac {5}{2}}} \,d x } \]
B*integrate(sqrt(x)/sqrt(b*x^3 + a), x)/e^(5/2) - 2/3*(b*sqrt(e)*x^4 + a*s qrt(e)*x)*A/(sqrt(b*x^3 + a)*a*e^3*x^(5/2))
Time = 0.31 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.41 \[ \int \frac {A+B x^3}{(e x)^{5/2} \sqrt {a+b x^3}} \, dx=-\frac {2 \, e {\left (\frac {\frac {B \arctan \left (\frac {\sqrt {b e + \frac {a e}{x^{3}}}}{\sqrt {-b e}}\right )}{\sqrt {-b e}} + \frac {\sqrt {b e + \frac {a e}{x^{3}}} A}{a e}}{e} - \frac {B a e \arctan \left (\frac {\sqrt {b e}}{\sqrt {-b e}}\right ) + \sqrt {b e} \sqrt {-b e} A}{\sqrt {-b e} a e^{2}}\right )}}{3 \, {\left | e \right |}^{2}} \]
-2/3*e*((B*arctan(sqrt(b*e + a*e/x^3)/sqrt(-b*e))/sqrt(-b*e) + sqrt(b*e + a*e/x^3)*A/(a*e))/e - (B*a*e*arctan(sqrt(b*e)/sqrt(-b*e)) + sqrt(b*e)*sqrt (-b*e)*A)/(sqrt(-b*e)*a*e^2))/abs(e)^2
Timed out. \[ \int \frac {A+B x^3}{(e x)^{5/2} \sqrt {a+b x^3}} \, dx=\int \frac {B\,x^3+A}{{\left (e\,x\right )}^{5/2}\,\sqrt {b\,x^3+a}} \,d x \]