Integrand size = 26, antiderivative size = 118 \[ \int \frac {\sqrt {a+b x^3} \left (A+B x^3\right )}{(e x)^{5/2}} \, dx=\frac {(2 A b+a B) (e x)^{3/2} \sqrt {a+b x^3}}{3 a e^4}-\frac {2 A \left (a+b x^3\right )^{3/2}}{3 a e (e x)^{3/2}}+\frac {(2 A b+a B) \text {arctanh}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+b x^3}}\right )}{3 \sqrt {b} e^{5/2}} \]
-2/3*A*(b*x^3+a)^(3/2)/a/e/(e*x)^(3/2)+1/3*(2*A*b+B*a)*arctanh((e*x)^(3/2) *b^(1/2)/e^(3/2)/(b*x^3+a)^(1/2))/e^(5/2)/b^(1/2)+1/3*(2*A*b+B*a)*(e*x)^(3 /2)*(b*x^3+a)^(1/2)/a/e^4
Time = 0.39 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.69 \[ \int \frac {\sqrt {a+b x^3} \left (A+B x^3\right )}{(e x)^{5/2}} \, dx=\frac {x \left (\sqrt {b} \sqrt {a+b x^3} \left (-2 A+B x^3\right )+(2 A b+a B) x^{3/2} \log \left (\sqrt {b} x^{3/2}+\sqrt {a+b x^3}\right )\right )}{3 \sqrt {b} (e x)^{5/2}} \]
(x*(Sqrt[b]*Sqrt[a + b*x^3]*(-2*A + B*x^3) + (2*A*b + a*B)*x^(3/2)*Log[Sqr t[b]*x^(3/2) + Sqrt[a + b*x^3]]))/(3*Sqrt[b]*(e*x)^(5/2))
Time = 0.27 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.99, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {955, 811, 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.
\(\displaystyle \int \frac {\sqrt {a+b x^3} \left (A+B x^3\right )}{(e x)^{5/2}} \, dx\) |
\(\Big \downarrow \) 955 |
\(\displaystyle \frac {(a B+2 A b) \int \sqrt {e x} \sqrt {b x^3+a}dx}{a e^3}-\frac {2 A \left (a+b x^3\right )^{3/2}}{3 a e (e x)^{3/2}}\) |
\(\Big \downarrow \) 811 |
\(\displaystyle \frac {(a B+2 A b) \left (\frac {1}{2} a \int \frac {\sqrt {e x}}{\sqrt {b x^3+a}}dx+\frac {(e x)^{3/2} \sqrt {a+b x^3}}{3 e}\right )}{a e^3}-\frac {2 A \left (a+b x^3\right )^{3/2}}{3 a e (e x)^{3/2}}\) |
\(\Big \downarrow \) 851 |
\(\displaystyle \frac {(a B+2 A b) \left (\frac {a \int \frac {e x}{\sqrt {b x^3+a}}d\sqrt {e x}}{e}+\frac {(e x)^{3/2} \sqrt {a+b x^3}}{3 e}\right )}{a e^3}-\frac {2 A \left (a+b x^3\right )^{3/2}}{3 a e (e x)^{3/2}}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {(a B+2 A b) \left (\frac {a \int \frac {1}{\sqrt {a+\frac {b x}{e^2}}}d(e x)^{3/2}}{3 e}+\frac {(e x)^{3/2} \sqrt {a+b x^3}}{3 e}\right )}{a e^3}-\frac {2 A \left (a+b x^3\right )^{3/2}}{3 a e (e x)^{3/2}}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {(a B+2 A b) \left (\frac {a \int \frac {1}{1-\frac {b x}{e^2}}d\frac {(e x)^{3/2}}{\sqrt {a+\frac {b x}{e^2}}}}{3 e}+\frac {(e x)^{3/2} \sqrt {a+b x^3}}{3 e}\right )}{a e^3}-\frac {2 A \left (a+b x^3\right )^{3/2}}{3 a e (e x)^{3/2}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {(a B+2 A b) \left (\frac {a \sqrt {e} \text {arctanh}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+\frac {b x}{e^2}}}\right )}{3 \sqrt {b}}+\frac {(e x)^{3/2} \sqrt {a+b x^3}}{3 e}\right )}{a e^3}-\frac {2 A \left (a+b x^3\right )^{3/2}}{3 a e (e x)^{3/2}}\) |
(-2*A*(a + b*x^3)^(3/2))/(3*a*e*(e*x)^(3/2)) + ((2*A*b + a*B)*(((e*x)^(3/2 )*Sqrt[a + b*x^3])/(3*e) + (a*Sqrt[e]*ArcTanh[(Sqrt[b]*(e*x)^(3/2))/(e^(3/ 2)*Sqrt[a + (b*x)/e^2])])/(3*Sqrt[b])))/(a*e^3)
3.6.22.3.1 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] :> Simp[(c* x)^(m + 1)*((a + b*x^n)^p/(c*(m + n*p + 1))), x] + Simp[a*n*(p/(m + n*p + 1 )) Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x] && I GtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b, c, n, m , p, x]
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[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)) Int[(e *x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b* c - a*d, 0] && (IntegerQ[n] || GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) && !ILtQ[p, -1]
Time = 4.60 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.85
method | result | size |
risch | \(-\frac {\sqrt {b \,x^{3}+a}\, \left (-x^{3} B +2 A \right )}{3 x \,e^{2} \sqrt {e x}}+\frac {2 \left (A b +\frac {B a}{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {\left (b \,x^{3}+a \right ) e x}}{x^{2} \sqrt {b e}}\right ) \sqrt {\left (b \,x^{3}+a \right ) e x}}{3 \sqrt {b e}\, e^{2} \sqrt {e x}\, \sqrt {b \,x^{3}+a}}\) | \(100\) |
default | \(\frac {\sqrt {b \,x^{3}+a}\, \left (2 A \,\operatorname {arctanh}\left (\frac {\sqrt {\left (b \,x^{3}+a \right ) e x}}{x^{2} \sqrt {b e}}\right ) b e \,x^{2}+B \,\operatorname {arctanh}\left (\frac {\sqrt {\left (b \,x^{3}+a \right ) e x}}{x^{2} \sqrt {b e}}\right ) a e \,x^{2}+B \sqrt {\left (b \,x^{3}+a \right ) e x}\, \sqrt {b e}\, x^{3}-2 A \sqrt {\left (b \,x^{3}+a \right ) e x}\, \sqrt {b e}\right )}{3 x \,e^{2} \sqrt {e x}\, \sqrt {\left (b \,x^{3}+a \right ) e x}\, \sqrt {b e}}\) | \(142\) |
elliptic | \(\text {Expression too large to display}\) | \(1069\) |
-1/3*(b*x^3+a)^(1/2)*(-B*x^3+2*A)/x/e^2/(e*x)^(1/2)+2/3*(A*b+1/2*B*a)/(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 = 0.57 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.75 \[ \int \frac {\sqrt {a+b x^3} \left (A+B x^3\right )}{(e x)^{5/2}} \, dx=\left [\frac {{\left (B a + 2 \, A b\right )} \sqrt {b e} x^{2} \log \left (-8 \, b^{2} e x^{6} - 8 \, a b e x^{3} - a^{2} e - 4 \, {\left (2 \, b x^{4} + a x\right )} \sqrt {b x^{3} + a} \sqrt {b e} \sqrt {e x}\right ) + 4 \, {\left (B b x^{3} - 2 \, A b\right )} \sqrt {b x^{3} + a} \sqrt {e x}}{12 \, b e^{3} x^{2}}, -\frac {{\left (B a + 2 \, A b\right )} \sqrt {-b e} x^{2} \arctan \left (\frac {2 \, \sqrt {b x^{3} + a} \sqrt {-b e} \sqrt {e x} x}{2 \, b e x^{3} + a e}\right ) - 2 \, {\left (B b x^{3} - 2 \, A b\right )} \sqrt {b x^{3} + a} \sqrt {e x}}{6 \, b e^{3} x^{2}}\right ] \]
[1/12*((B*a + 2*A*b)*sqrt(b*e)*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*(B*b*x^3 - 2* A*b)*sqrt(b*x^3 + a)*sqrt(e*x))/(b*e^3*x^2), -1/6*((B*a + 2*A*b)*sqrt(-b*e )*x^2*arctan(2*sqrt(b*x^3 + a)*sqrt(-b*e)*sqrt(e*x)*x/(2*b*e*x^3 + a*e)) - 2*(B*b*x^3 - 2*A*b)*sqrt(b*x^3 + a)*sqrt(e*x))/(b*e^3*x^2)]
Time = 6.39 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.36 \[ \int \frac {\sqrt {a+b x^3} \left (A+B x^3\right )}{(e x)^{5/2}} \, dx=- \frac {2 A \sqrt {a}}{3 e^{\frac {5}{2}} x^{\frac {3}{2}} \sqrt {1 + \frac {b x^{3}}{a}}} + \frac {2 A \sqrt {b} \operatorname {asinh}{\left (\frac {\sqrt {b} x^{\frac {3}{2}}}{\sqrt {a}} \right )}}{3 e^{\frac {5}{2}}} - \frac {2 A b x^{\frac {3}{2}}}{3 \sqrt {a} e^{\frac {5}{2}} \sqrt {1 + \frac {b x^{3}}{a}}} + \frac {B \sqrt {a} x^{\frac {3}{2}} \sqrt {1 + \frac {b x^{3}}{a}}}{3 e^{\frac {5}{2}}} + \frac {B a \operatorname {asinh}{\left (\frac {\sqrt {b} x^{\frac {3}{2}}}{\sqrt {a}} \right )}}{3 \sqrt {b} e^{\frac {5}{2}}} \]
-2*A*sqrt(a)/(3*e**(5/2)*x**(3/2)*sqrt(1 + b*x**3/a)) + 2*A*sqrt(b)*asinh( sqrt(b)*x**(3/2)/sqrt(a))/(3*e**(5/2)) - 2*A*b*x**(3/2)/(3*sqrt(a)*e**(5/2 )*sqrt(1 + b*x**3/a)) + B*sqrt(a)*x**(3/2)*sqrt(1 + b*x**3/a)/(3*e**(5/2)) + B*a*asinh(sqrt(b)*x**(3/2)/sqrt(a))/(3*sqrt(b)*e**(5/2))
\[ \int \frac {\sqrt {a+b x^3} \left (A+B x^3\right )}{(e x)^{5/2}} \, dx=\int { \frac {{\left (B x^{3} + A\right )} \sqrt {b x^{3} + a}}{\left (e x\right )^{\frac {5}{2}}} \,d x } \]
Exception generated. \[ \int \frac {\sqrt {a+b x^3} \left (A+B x^3\right )}{(e x)^{5/2}} \, dx=\text {Exception raised: NotImplementedError} \]
Exception raised: NotImplementedError >> unable to parse Giac output: Recu rsive assumption sageVARa>=(-sageVARb*sageVARe/(sageVARe^4*t_nostep^6)) ig nored2/sageVARe^3*sageVARB/6/sageVARe^3*sqrt(sageVARe*sageVARx)*sqrt(sageV ARe*sageVARx)*sqrt(
Timed out. \[ \int \frac {\sqrt {a+b x^3} \left (A+B x^3\right )}{(e x)^{5/2}} \, dx=\int \frac {\left (B\,x^3+A\right )\,\sqrt {b\,x^3+a}}{{\left (e\,x\right )}^{5/2}} \,d x \]