Integrand size = 26, antiderivative size = 299 \[ \int \frac {(e x)^{5/2} \left (A+B x^3\right )}{\left (a+b x^3\right )^{5/2}} \, dx=\frac {2 (A b-a B) (e x)^{7/2}}{9 a b e \left (a+b x^3\right )^{3/2}}-\frac {2 (2 A b+7 a B) e^2 \sqrt {e x}}{27 a b^2 \sqrt {a+b x^3}}+\frac {(2 A b+7 a B) e^2 \sqrt {e x} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{a}+\left (1-\sqrt {3}\right ) \sqrt [3]{b} x}{\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{27 \sqrt [4]{3} a^{4/3} b^2 \sqrt {\frac {\sqrt [3]{b} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}} \]
2/9*(A*b-B*a)*(e*x)^(7/2)/a/b/e/(b*x^3+a)^(3/2)-2/27*(2*A*b+7*B*a)*e^2*(e* x)^(1/2)/a/b^2/(b*x^3+a)^(1/2)+1/81*(2*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)))*Elliptic F((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^(4/3)/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 = 10.16 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.36 \[ \int \frac {(e x)^{5/2} \left (A+B x^3\right )}{\left (a+b x^3\right )^{5/2}} \, dx=\frac {2 e^2 \sqrt {e x} \left (-7 a^2 B+A b^2 x^3-2 a b \left (A+5 B x^3\right )+(2 A b+7 a B) \left (a+b x^3\right ) \sqrt {1+\frac {b x^3}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},-\frac {b x^3}{a}\right )\right )}{27 a b^2 \left (a+b x^3\right )^{3/2}} \]
(2*e^2*Sqrt[e*x]*(-7*a^2*B + A*b^2*x^3 - 2*a*b*(A + 5*B*x^3) + (2*A*b + 7* a*B)*(a + b*x^3)*Sqrt[1 + (b*x^3)/a]*Hypergeometric2F1[1/6, 1/2, 7/6, -((b *x^3)/a)]))/(27*a*b^2*(a + b*x^3)^(3/2))
Time = 0.40 (sec) , antiderivative size = 325, 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 = {957, 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.
| \(\displaystyle \int \frac {(e x)^{5/2} \left (A+B x^3\right )}{\left (a+b x^3\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 957 |
| \(\displaystyle \frac {(7 a B+2 A b) \int \frac {(e x)^{5/2}}{\left (b x^3+a\right )^{3/2}}dx}{9 a b}+\frac {2 (e x)^{7/2} (A b-a B)}{9 a b e \left (a+b x^3\right )^{3/2}}\) |
| \(\Big \downarrow \) 817 |
| \(\displaystyle \frac {(7 a B+2 A b) \left (\frac {e^3 \int \frac {1}{\sqrt {e x} \sqrt {b x^3+a}}dx}{3 b}-\frac {2 e^2 \sqrt {e x}}{3 b \sqrt {a+b x^3}}\right )}{9 a b}+\frac {2 (e x)^{7/2} (A b-a B)}{9 a b e \left (a+b x^3\right )^{3/2}}\) |
| \(\Big \downarrow \) 851 |
| \(\displaystyle \frac {(7 a B+2 A b) \left (\frac {2 e^2 \int \frac {1}{\sqrt {b x^3+a}}d\sqrt {e x}}{3 b}-\frac {2 e^2 \sqrt {e x}}{3 b \sqrt {a+b x^3}}\right )}{9 a b}+\frac {2 (e x)^{7/2} (A b-a B)}{9 a b e \left (a+b x^3\right )^{3/2}}\) |
| \(\Big \downarrow \) 766 |
| \(\displaystyle \frac {(7 a B+2 A b) \left (\frac {e \sqrt {e x} \left (\sqrt [3]{a} e+\sqrt [3]{b} e x\right ) \sqrt {\frac {a^{2/3} e^2-\sqrt [3]{a} \sqrt [3]{b} e^2 x+b^{2/3} e^2 x^2}{\left (\sqrt [3]{a} e+\left (1+\sqrt {3}\right ) \sqrt [3]{b} e x\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{b} x e+\sqrt [3]{a} e}{\left (1+\sqrt {3}\right ) \sqrt [3]{b} x e+\sqrt [3]{a} e}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{3 \sqrt [4]{3} \sqrt [3]{a} b \sqrt {a+b x^3} \sqrt {\frac {\sqrt [3]{b} e x \left (\sqrt [3]{a} e+\sqrt [3]{b} e x\right )}{\left (\sqrt [3]{a} e+\left (1+\sqrt {3}\right ) \sqrt [3]{b} e x\right )^2}}}-\frac {2 e^2 \sqrt {e x}}{3 b \sqrt {a+b x^3}}\right )}{9 a b}+\frac {2 (e x)^{7/2} (A b-a B)}{9 a b e \left (a+b x^3\right )^{3/2}}\) |
(2*(A*b - a*B)*(e*x)^(7/2))/(9*a*b*e*(a + b*x^3)^(3/2)) + ((2*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 + Sqrt[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])))/(9*a*b)
3.6.60.3.1 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[(-(b*c - a*d))*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a *b*e*n*(p + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b*n* (p + 1)) Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && (( !IntegerQ[p + 1/2] && N eQ[p, -5/4]) || !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] && LeQ[-1 , m, (-n)*(p + 1)]))
Result contains complex when optimal does not.
Time = 4.60 (sec) , antiderivative size = 809, normalized size of antiderivative = 2.71
| method | result | size |
| elliptic | \(\text {Expression too large to display}\) | \(809\) |
| default | \(\text {Expression too large to display}\) | \(7083\) |
1/e/x*(e*x)^(1/2)/(b*x^3+a)^(1/2)*((b*x^3+a)*e*x)^(1/2)*(-2/9*e^2/b^4*(A*b -B*a)*(b*e*x^4+a*e*x)^(1/2)/(x^3+a/b)^2+2/27/b^2*e^3*x/a*(A*b-10*B*a)/((x^ 3+a/b)*b*e*x)^(1/2)+2*(B*e^3/b^2+2/27/b^2/a*e^3*(A*b-10*B*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)))
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.09 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.57 \[ \int \frac {(e x)^{5/2} \left (A+B x^3\right )}{\left (a+b x^3\right )^{5/2}} \, dx=-\frac {2 \, {\left ({\left ({\left (7 \, B a b^{2} + 2 \, A b^{3}\right )} e^{2} x^{6} + 2 \, {\left (7 \, B a^{2} b + 2 \, A a b^{2}\right )} e^{2} x^{3} + {\left (7 \, B a^{3} + 2 \, A a^{2} b\right )} e^{2}\right )} \sqrt {a e} {\rm weierstrassPInverse}\left (0, -\frac {4 \, b}{a}, \frac {1}{x}\right ) + {\left ({\left (10 \, B a^{2} b - A a b^{2}\right )} e^{2} x^{3} + {\left (7 \, B a^{3} + 2 \, A a^{2} b\right )} e^{2}\right )} \sqrt {b x^{3} + a} \sqrt {e x}\right )}}{27 \, {\left (a^{2} b^{4} x^{6} + 2 \, a^{3} b^{3} x^{3} + a^{4} b^{2}\right )}} \]
-2/27*(((7*B*a*b^2 + 2*A*b^3)*e^2*x^6 + 2*(7*B*a^2*b + 2*A*a*b^2)*e^2*x^3 + (7*B*a^3 + 2*A*a^2*b)*e^2)*sqrt(a*e)*weierstrassPInverse(0, -4*b/a, 1/x) + ((10*B*a^2*b - A*a*b^2)*e^2*x^3 + (7*B*a^3 + 2*A*a^2*b)*e^2)*sqrt(b*x^3 + a)*sqrt(e*x))/(a^2*b^4*x^6 + 2*a^3*b^3*x^3 + a^4*b^2)
Timed out. \[ \int \frac {(e x)^{5/2} \left (A+B x^3\right )}{\left (a+b x^3\right )^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(e x)^{5/2} \left (A+B x^3\right )}{\left (a+b x^3\right )^{5/2}} \, dx=\int { \frac {{\left (B x^{3} + A\right )} \left (e x\right )^{\frac {5}{2}}}{{\left (b x^{3} + a\right )}^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {(e x)^{5/2} \left (A+B x^3\right )}{\left (a+b x^3\right )^{5/2}} \, dx=\int { \frac {{\left (B x^{3} + A\right )} \left (e x\right )^{\frac {5}{2}}}{{\left (b x^{3} + a\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {(e x)^{5/2} \left (A+B x^3\right )}{\left (a+b x^3\right )^{5/2}} \, dx=\int \frac {\left (B\,x^3+A\right )\,{\left (e\,x\right )}^{5/2}}{{\left (b\,x^3+a\right )}^{5/2}} \,d x \]