Integrand size = 26, antiderivative size = 286 \[ \int \frac {(e x)^{5/2} \left (A+B x^3\right )}{\left (a+b x^3\right )^{3/2}} \, dx=-\frac {(4 A b-7 a B) e^2 \sqrt {e x}}{6 b^2 \sqrt {a+b x^3}}+\frac {B (e x)^{7/2}}{2 b e \sqrt {a+b x^3}}+\frac {(4 A b-7 a B) e^2 \sqrt {e x} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{a}+\left (1-\sqrt {3}\right ) \sqrt [3]{b} x}{\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{12 \sqrt [4]{3} \sqrt [3]{a} b^2 \sqrt {\frac {\sqrt [3]{b} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}} \]
1/2*B*(e*x)^(7/2)/b/e/(b*x^3+a)^(1/2)-1/6*(4*A*b-7*B*a)*e^2*(e*x)^(1/2)/b^ 2/(b*x^3+a)^(1/2)+1/36*(4*A*b-7*B*a)*e^2*(a^(1/3)+b^(1/3)*x)*((a^(1/3)+b^( 1/3)*x*(1-3^(1/2)))^2/(a^(1/3)+b^(1/3)*x*(1+3^(1/2)))^2)^(1/2)/(a^(1/3)+b^ (1/3)*x*(1-3^(1/2)))*(a^(1/3)+b^(1/3)*x*(1+3^(1/2)))*EllipticF((1-(a^(1/3) +b^(1/3)*x*(1-3^(1/2)))^2/(a^(1/3)+b^(1/3)*x*(1+3^(1/2)))^2)^(1/2),1/4*6^( 1/2)+1/4*2^(1/2))*(e*x)^(1/2)*((a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/(a^ (1/3)+b^(1/3)*x*(1+3^(1/2)))^2)^(1/2)*3^(3/4)/a^(1/3)/b^2/(b*x^3+a)^(1/2)/ (b^(1/3)*x*(a^(1/3)+b^(1/3)*x)/(a^(1/3)+b^(1/3)*x*(1+3^(1/2)))^2)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.13 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.30 \[ \int \frac {(e x)^{5/2} \left (A+B x^3\right )}{\left (a+b x^3\right )^{3/2}} \, dx=\frac {e^2 \sqrt {e x} \left (-4 A b+7 a B+3 b B x^3+(4 A b-7 a B) \sqrt {1+\frac {b x^3}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},-\frac {b x^3}{a}\right )\right )}{6 b^2 \sqrt {a+b x^3}} \]
(e^2*Sqrt[e*x]*(-4*A*b + 7*a*B + 3*b*B*x^3 + (4*A*b - 7*a*B)*Sqrt[1 + (b*x ^3)/a]*Hypergeometric2F1[1/6, 1/2, 7/6, -((b*x^3)/a)]))/(6*b^2*Sqrt[a + b* x^3])
Time = 0.38 (sec) , antiderivative size = 312, normalized size of antiderivative = 1.09, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {959, 817, 851, 766}
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.
| \(\displaystyle \int \frac {(e x)^{5/2} \left (A+B x^3\right )}{\left (a+b x^3\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 959 |
| \(\displaystyle \frac {(4 A b-7 a B) \int \frac {(e x)^{5/2}}{\left (b x^3+a\right )^{3/2}}dx}{4 b}+\frac {B (e x)^{7/2}}{2 b e \sqrt {a+b x^3}}\) |
| \(\Big \downarrow \) 817 |
| \(\displaystyle \frac {(4 A b-7 a B) \left (\frac {e^3 \int \frac {1}{\sqrt {e x} \sqrt {b x^3+a}}dx}{3 b}-\frac {2 e^2 \sqrt {e x}}{3 b \sqrt {a+b x^3}}\right )}{4 b}+\frac {B (e x)^{7/2}}{2 b e \sqrt {a+b x^3}}\) |
| \(\Big \downarrow \) 851 |
| \(\displaystyle \frac {(4 A b-7 a B) \left (\frac {2 e^2 \int \frac {1}{\sqrt {b x^3+a}}d\sqrt {e x}}{3 b}-\frac {2 e^2 \sqrt {e x}}{3 b \sqrt {a+b x^3}}\right )}{4 b}+\frac {B (e x)^{7/2}}{2 b e \sqrt {a+b x^3}}\) |
| \(\Big \downarrow \) 766 |
| \(\displaystyle \frac {(4 A b-7 a B) \left (\frac {e \sqrt {e x} \left (\sqrt [3]{a} e+\sqrt [3]{b} e x\right ) \sqrt {\frac {a^{2/3} e^2-\sqrt [3]{a} \sqrt [3]{b} e^2 x+b^{2/3} e^2 x^2}{\left (\sqrt [3]{a} e+\left (1+\sqrt {3}\right ) \sqrt [3]{b} e x\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{b} x e+\sqrt [3]{a} e}{\left (1+\sqrt {3}\right ) \sqrt [3]{b} x e+\sqrt [3]{a} e}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{3 \sqrt [4]{3} \sqrt [3]{a} b \sqrt {a+b x^3} \sqrt {\frac {\sqrt [3]{b} e x \left (\sqrt [3]{a} e+\sqrt [3]{b} e x\right )}{\left (\sqrt [3]{a} e+\left (1+\sqrt {3}\right ) \sqrt [3]{b} e x\right )^2}}}-\frac {2 e^2 \sqrt {e x}}{3 b \sqrt {a+b x^3}}\right )}{4 b}+\frac {B (e x)^{7/2}}{2 b e \sqrt {a+b x^3}}\) |
(B*(e*x)^(7/2))/(2*b*e*Sqrt[a + b*x^3]) + ((4*A*b - 7*a*B)*((-2*e^2*Sqrt[e *x])/(3*b*Sqrt[a + b*x^3]) + (e*Sqrt[e*x]*(a^(1/3)*e + b^(1/3)*e*x)*Sqrt[( a^(2/3)*e^2 - a^(1/3)*b^(1/3)*e^2*x + b^(2/3)*e^2*x^2)/(a^(1/3)*e + (1 + S qrt[3])*b^(1/3)*e*x)^2]*EllipticF[ArcCos[(a^(1/3)*e + (1 - Sqrt[3])*b^(1/3 )*e*x)/(a^(1/3)*e + (1 + Sqrt[3])*b^(1/3)*e*x)], (2 + Sqrt[3])/4])/(3*3^(1 /4)*a^(1/3)*b*Sqrt[(b^(1/3)*e*x*(a^(1/3)*e + b^(1/3)*e*x))/(a^(1/3)*e + (1 + Sqrt[3])*b^(1/3)*e*x)^2]*Sqrt[a + b*x^3])))/(4*b)
3.6.52.3.1 Defintions of rubi rules used
Int[1/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[x*(s + r*x^2)*(Sqrt[(s^2 - r*s*x^2 + r^2*x^4)/ (s + (1 + Sqrt[3])*r*x^2)^2]/(2*3^(1/4)*s*Sqrt[a + b*x^6]*Sqrt[r*x^2*((s + r*x^2)/(s + (1 + Sqrt[3])*r*x^2)^2)]))*EllipticF[ArcCos[(s + (1 - Sqrt[3])* r*x^2)/(s + (1 + Sqrt[3])*r*x^2)], (2 + Sqrt[3])/4], x]] /; FreeQ[{a, b}, x ]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^( n - 1)*(c*x)^(m - n + 1)*((a + b*x^n)^(p + 1)/(b*n*(p + 1))), x] - Simp[c^n *((m - n + 1)/(b*n*(p + 1))) Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x], x ] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] && ! ILtQ[(m + n*(p + 1) + 1)/n, 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[d*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(b*e*(m + n*(p + 1) + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m + n*(p + 1) + 1)) Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && NeQ[m + n*(p + 1) + 1, 0]
Result contains complex when optimal does not.
Time = 5.91 (sec) , antiderivative size = 787, normalized size of antiderivative = 2.75
| method | result | size |
| elliptic | \(\text {Expression too large to display}\) | \(787\) |
| risch | \(\text {Expression too large to display}\) | \(2115\) |
| default | \(\text {Expression too large to display}\) | \(3760\) |
1/e/x*(e*x)^(1/2)/(b*x^3+a)^(1/2)*((b*x^3+a)*e*x)^(1/2)*(-2/3/b^2*e^3*x*(A *b-B*a)/((x^3+a/b)*b*e*x)^(1/2)+1/2*B*e^2/b^2*(b*e*x^4+a*e*x)^(1/2)+2*(1/3 *(A*b-B*a)*e^3/b^2-1/4*B/b^2*e^3*a)*(1/2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2)/b* (-a*b^2)^(1/3))*((-3/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*x/ (-1/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))/(x-1/b*(-a*b^2)^(1/ 3)))^(1/2)*(x-1/b*(-a*b^2)^(1/3))^2*(1/b*(-a*b^2)^(1/3)*(x+1/2/b*(-a*b^2)^ (1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))/(-1/2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2) /b*(-a*b^2)^(1/3))/(x-1/b*(-a*b^2)^(1/3)))^(1/2)*(1/b*(-a*b^2)^(1/3)*(x+1/ 2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))/(-1/2/b*(-a*b^2)^(1/3)+ 1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))/(x-1/b*(-a*b^2)^(1/3)))^(1/2)/(-3/2/b*(-a* b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*b/(-a*b^2)^(1/3)/(b*e*x*(x-1/b* (-a*b^2)^(1/3))*(x+1/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*(x +1/2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2)/b*(-a*b^2)^(1/3)))^(1/2)*EllipticF(((- 3/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*x/(-1/2/b*(-a*b^2)^(1 /3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))/(x-1/b*(-a*b^2)^(1/3)))^(1/2),((3/2/b* (-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*(1/2/b*(-a*b^2)^(1/3)-1/2*I *3^(1/2)/b*(-a*b^2)^(1/3))/(1/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^ (1/3))/(3/2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2)/b*(-a*b^2)^(1/3)))^(1/2)))
\[ \int \frac {(e x)^{5/2} \left (A+B x^3\right )}{\left (a+b x^3\right )^{3/2}} \, dx=\int { \frac {{\left (B x^{3} + A\right )} \left (e x\right )^{\frac {5}{2}}}{{\left (b x^{3} + a\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {(e x)^{5/2} \left (A+B x^3\right )}{\left (a+b x^3\right )^{3/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(e x)^{5/2} \left (A+B x^3\right )}{\left (a+b x^3\right )^{3/2}} \, dx=\int { \frac {{\left (B x^{3} + A\right )} \left (e x\right )^{\frac {5}{2}}}{{\left (b x^{3} + a\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {(e x)^{5/2} \left (A+B x^3\right )}{\left (a+b x^3\right )^{3/2}} \, dx=\int { \frac {{\left (B x^{3} + A\right )} \left (e x\right )^{\frac {5}{2}}}{{\left (b x^{3} + a\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {(e x)^{5/2} \left (A+B x^3\right )}{\left (a+b x^3\right )^{3/2}} \, dx=\int \frac {\left (B\,x^3+A\right )\,{\left (e\,x\right )}^{5/2}}{{\left (b\,x^3+a\right )}^{3/2}} \,d x \]