Integrand size = 19, antiderivative size = 292 \[ \int \frac {\sqrt {a x^3}}{\sqrt {1+x^3}} \, dx=\frac {\left (1+\sqrt {3}\right ) \sqrt {a x^3} \sqrt {1+x^3}}{x \left (1+\left (1+\sqrt {3}\right ) x\right )}-\frac {\sqrt [4]{3} \sqrt {a x^3} (1+x) \sqrt {\frac {1-x+x^2}{\left (1+\left (1+\sqrt {3}\right ) x\right )^2}} E\left (\arccos \left (\frac {1+\left (1-\sqrt {3}\right ) x}{1+\left (1+\sqrt {3}\right ) x}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{x \sqrt {\frac {x (1+x)}{\left (1+\left (1+\sqrt {3}\right ) x\right )^2}} \sqrt {1+x^3}}-\frac {\left (1-\sqrt {3}\right ) \sqrt {a x^3} (1+x) \sqrt {\frac {1-x+x^2}{\left (1+\left (1+\sqrt {3}\right ) x\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {1+\left (1-\sqrt {3}\right ) x}{1+\left (1+\sqrt {3}\right ) x}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{2 \sqrt [4]{3} x \sqrt {\frac {x (1+x)}{\left (1+\left (1+\sqrt {3}\right ) x\right )^2}} \sqrt {1+x^3}} \]
(1+3^(1/2))*(a*x^3)^(1/2)*(x^3+1)^(1/2)/x/(1+x*(1+3^(1/2)))-3^(1/4)*(1+x)* ((1+x*(1-3^(1/2)))^2/(1+x*(1+3^(1/2)))^2)^(1/2)/(1+x*(1-3^(1/2)))*(1+x*(1+ 3^(1/2)))*EllipticE((1-(1+x*(1-3^(1/2)))^2/(1+x*(1+3^(1/2)))^2)^(1/2),1/4* 6^(1/2)+1/4*2^(1/2))*(a*x^3)^(1/2)*((x^2-x+1)/(1+x*(1+3^(1/2)))^2)^(1/2)/x /(x^3+1)^(1/2)/(x*(1+x)/(1+x*(1+3^(1/2)))^2)^(1/2)-1/6*(1+x)*((1+x*(1-3^(1 /2)))^2/(1+x*(1+3^(1/2)))^2)^(1/2)/(1+x*(1-3^(1/2)))*(1+x*(1+3^(1/2)))*Ell ipticF((1-(1+x*(1-3^(1/2)))^2/(1+x*(1+3^(1/2)))^2)^(1/2),1/4*6^(1/2)+1/4*2 ^(1/2))*(1-3^(1/2))*(a*x^3)^(1/2)*((x^2-x+1)/(1+x*(1+3^(1/2)))^2)^(1/2)*3^ (3/4)/x/(x^3+1)^(1/2)/(x*(1+x)/(1+x*(1+3^(1/2)))^2)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.02 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.10 \[ \int \frac {\sqrt {a x^3}}{\sqrt {1+x^3}} \, dx=\frac {2}{5} x \sqrt {a x^3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {11}{6},-x^3\right ) \]
Time = 0.39 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {34, 851, 837, 25, 766, 2420}
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.
\(\displaystyle \int \frac {\sqrt {a x^3}}{\sqrt {x^3+1}} \, dx\) |
\(\Big \downarrow \) 34 |
\(\displaystyle \frac {\sqrt {a x^3} \int \frac {x^{3/2}}{\sqrt {x^3+1}}dx}{x^{3/2}}\) |
\(\Big \downarrow \) 851 |
\(\displaystyle \frac {2 \sqrt {a x^3} \int \frac {x^2}{\sqrt {x^3+1}}d\sqrt {x}}{x^{3/2}}\) |
\(\Big \downarrow \) 837 |
\(\displaystyle \frac {2 \sqrt {a x^3} \left (-\frac {1}{2} \left (1-\sqrt {3}\right ) \int \frac {1}{\sqrt {x^3+1}}d\sqrt {x}-\frac {1}{2} \int -\frac {2 x^2-\sqrt {3}+1}{\sqrt {x^3+1}}d\sqrt {x}\right )}{x^{3/2}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {2 \sqrt {a x^3} \left (\frac {1}{2} \int \frac {2 x^2-\sqrt {3}+1}{\sqrt {x^3+1}}d\sqrt {x}-\frac {1}{2} \left (1-\sqrt {3}\right ) \int \frac {1}{\sqrt {x^3+1}}d\sqrt {x}\right )}{x^{3/2}}\) |
\(\Big \downarrow \) 766 |
\(\displaystyle \frac {2 \sqrt {a x^3} \left (\frac {1}{2} \int \frac {2 x^2-\sqrt {3}+1}{\sqrt {x^3+1}}d\sqrt {x}-\frac {\left (1-\sqrt {3}\right ) \sqrt {x} (x+1) \sqrt {\frac {x^2-x+1}{\left (\left (1+\sqrt {3}\right ) x+1\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\left (1-\sqrt {3}\right ) x+1}{\left (1+\sqrt {3}\right ) x+1}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{4 \sqrt [4]{3} \sqrt {\frac {x (x+1)}{\left (\left (1+\sqrt {3}\right ) x+1\right )^2}} \sqrt {x^3+1}}\right )}{x^{3/2}}\) |
\(\Big \downarrow \) 2420 |
\(\displaystyle \frac {2 \sqrt {a x^3} \left (\frac {1}{2} \left (\frac {\left (1+\sqrt {3}\right ) \sqrt {x} \sqrt {x^3+1}}{\left (1+\sqrt {3}\right ) x+1}-\frac {\sqrt [4]{3} \sqrt {x} (x+1) \sqrt {\frac {x^2-x+1}{\left (\left (1+\sqrt {3}\right ) x+1\right )^2}} E\left (\arccos \left (\frac {\left (1-\sqrt {3}\right ) x+1}{\left (1+\sqrt {3}\right ) x+1}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{\sqrt {\frac {x (x+1)}{\left (\left (1+\sqrt {3}\right ) x+1\right )^2}} \sqrt {x^3+1}}\right )-\frac {\left (1-\sqrt {3}\right ) \sqrt {x} (x+1) \sqrt {\frac {x^2-x+1}{\left (\left (1+\sqrt {3}\right ) x+1\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\left (1-\sqrt {3}\right ) x+1}{\left (1+\sqrt {3}\right ) x+1}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{4 \sqrt [4]{3} \sqrt {\frac {x (x+1)}{\left (\left (1+\sqrt {3}\right ) x+1\right )^2}} \sqrt {x^3+1}}\right )}{x^{3/2}}\) |
(2*Sqrt[a*x^3]*((((1 + Sqrt[3])*Sqrt[x]*Sqrt[1 + x^3])/(1 + (1 + Sqrt[3])* x) - (3^(1/4)*Sqrt[x]*(1 + x)*Sqrt[(1 - x + x^2)/(1 + (1 + Sqrt[3])*x)^2]* EllipticE[ArcCos[(1 + (1 - Sqrt[3])*x)/(1 + (1 + Sqrt[3])*x)], (2 + Sqrt[3 ])/4])/(Sqrt[(x*(1 + x))/(1 + (1 + Sqrt[3])*x)^2]*Sqrt[1 + x^3]))/2 - ((1 - Sqrt[3])*Sqrt[x]*(1 + x)*Sqrt[(1 - x + x^2)/(1 + (1 + Sqrt[3])*x)^2]*Ell ipticF[ArcCos[(1 + (1 - Sqrt[3])*x)/(1 + (1 + Sqrt[3])*x)], (2 + Sqrt[3])/ 4])/(4*3^(1/4)*Sqrt[(x*(1 + x))/(1 + (1 + Sqrt[3])*x)^2]*Sqrt[1 + x^3])))/ x^(3/2)
3.4.86.3.1 Defintions of rubi rules used
Int[(u_.)*((a_.)*(x_)^(m_))^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a*x^m)^F racPart[p]/x^(m*FracPart[p])) Int[u*x^(m*p), x], x] /; FreeQ[{a, m, p}, x ] && !IntegerQ[p]
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[(x_)^4/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(Sqrt[3] - 1)*(s^2/(2*r^2)) Int[1/Sqrt[ a + b*x^6], x], x] - Simp[1/(2*r^2) Int[((Sqrt[3] - 1)*s^2 - 2*r^2*x^4)/S qrt[a + b*x^6], x], x]] /; FreeQ[{a, b}, 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[((c_) + (d_.)*(x_)^4)/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(1 + Sqrt[3])*d*s^3*x*(Sqr t[a + b*x^6]/(2*a*r^2*(s + (1 + Sqrt[3])*r*x^2))), x] - Simp[3^(1/4)*d*s*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 *r^2*Sqrt[(r*x^2*(s + r*x^2))/(s + (1 + Sqrt[3])*r*x^2)^2]*Sqrt[a + b*x^6]) )*EllipticE[ArcCos[(s + (1 - Sqrt[3])*r*x^2)/(s + (1 + Sqrt[3])*r*x^2)], (2 + Sqrt[3])/4], x]] /; FreeQ[{a, b, c, d}, x] && EqQ[2*Rt[b/a, 3]^2*c - (1 - Sqrt[3])*d, 0]
Result contains higher order function than in optimal. Order 5 vs. order 4.
Time = 2.30 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.08
method | result | size |
meijerg | \(\frac {2 \sqrt {a \,x^{3}}\, x {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{2},\frac {5}{6};\frac {11}{6};-x^{3}\right )}{5}\) | \(22\) |
default | \(\text {Expression too large to display}\) | \(1521\) |
\[ \int \frac {\sqrt {a x^3}}{\sqrt {1+x^3}} \, dx=\int { \frac {\sqrt {a x^{3}}}{\sqrt {x^{3} + 1}} \,d x } \]
\[ \int \frac {\sqrt {a x^3}}{\sqrt {1+x^3}} \, dx=\int \frac {\sqrt {a x^{3}}}{\sqrt {\left (x + 1\right ) \left (x^{2} - x + 1\right )}}\, dx \]
\[ \int \frac {\sqrt {a x^3}}{\sqrt {1+x^3}} \, dx=\int { \frac {\sqrt {a x^{3}}}{\sqrt {x^{3} + 1}} \,d x } \]
\[ \int \frac {\sqrt {a x^3}}{\sqrt {1+x^3}} \, dx=\int { \frac {\sqrt {a x^{3}}}{\sqrt {x^{3} + 1}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {a x^3}}{\sqrt {1+x^3}} \, dx=\int \frac {\sqrt {a\,x^3}}{\sqrt {x^3+1}} \,d x \]