Integrand size = 15, antiderivative size = 93 \[ \int x^2 \left (3+4 x^4\right )^{5/4} \, dx=\frac {15}{32} x^3 \sqrt [4]{3+4 x^4}+\frac {1}{8} x^3 \left (3+4 x^4\right )^{5/4}-\frac {45 \arctan \left (\frac {\sqrt {2} x}{\sqrt [4]{3+4 x^4}}\right )}{128 \sqrt {2}}+\frac {45 \text {arctanh}\left (\frac {\sqrt {2} x}{\sqrt [4]{3+4 x^4}}\right )}{128 \sqrt {2}} \]
15/32*x^3*(4*x^4+3)^(1/4)+1/8*x^3*(4*x^4+3)^(5/4)-45/256*arctan(x*2^(1/2)/ (4*x^4+3)^(1/4))*2^(1/2)+45/256*arctanh(x*2^(1/2)/(4*x^4+3)^(1/4))*2^(1/2)
Time = 0.39 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.88 \[ \int x^2 \left (3+4 x^4\right )^{5/4} \, dx=\frac {1}{32} x^3 \sqrt [4]{3+4 x^4} \left (27+16 x^4\right )-\frac {45 \arctan \left (\frac {\sqrt {2} x}{\sqrt [4]{3+4 x^4}}\right )}{128 \sqrt {2}}+\frac {45 \text {arctanh}\left (\frac {\sqrt {2} x}{\sqrt [4]{3+4 x^4}}\right )}{128 \sqrt {2}} \]
(x^3*(3 + 4*x^4)^(1/4)*(27 + 16*x^4))/32 - (45*ArcTan[(Sqrt[2]*x)/(3 + 4*x ^4)^(1/4)])/(128*Sqrt[2]) + (45*ArcTanh[(Sqrt[2]*x)/(3 + 4*x^4)^(1/4)])/(1 28*Sqrt[2])
Time = 0.20 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.11, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {811, 811, 854, 827, 216, 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.
| \(\displaystyle \int x^2 \left (4 x^4+3\right )^{5/4} \, dx\) |
| \(\Big \downarrow \) 811 |
| \(\displaystyle \frac {15}{8} \int x^2 \sqrt [4]{4 x^4+3}dx+\frac {1}{8} \left (4 x^4+3\right )^{5/4} x^3\) |
| \(\Big \downarrow \) 811 |
| \(\displaystyle \frac {15}{8} \left (\frac {3}{4} \int \frac {x^2}{\left (4 x^4+3\right )^{3/4}}dx+\frac {1}{4} \sqrt [4]{4 x^4+3} x^3\right )+\frac {1}{8} \left (4 x^4+3\right )^{5/4} x^3\) |
| \(\Big \downarrow \) 854 |
| \(\displaystyle \frac {15}{8} \left (\frac {3}{4} \int \frac {x^2}{\sqrt {4 x^4+3} \left (1-\frac {4 x^4}{4 x^4+3}\right )}d\frac {x}{\sqrt [4]{4 x^4+3}}+\frac {1}{4} \sqrt [4]{4 x^4+3} x^3\right )+\frac {1}{8} \left (4 x^4+3\right )^{5/4} x^3\) |
| \(\Big \downarrow \) 827 |
| \(\displaystyle \frac {15}{8} \left (\frac {3}{4} \left (\frac {1}{4} \int \frac {1}{1-\frac {2 x^2}{\sqrt {4 x^4+3}}}d\frac {x}{\sqrt [4]{4 x^4+3}}-\frac {1}{4} \int \frac {1}{\frac {2 x^2}{\sqrt {4 x^4+3}}+1}d\frac {x}{\sqrt [4]{4 x^4+3}}\right )+\frac {1}{4} \sqrt [4]{4 x^4+3} x^3\right )+\frac {1}{8} \left (4 x^4+3\right )^{5/4} x^3\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {15}{8} \left (\frac {3}{4} \left (\frac {1}{4} \int \frac {1}{1-\frac {2 x^2}{\sqrt {4 x^4+3}}}d\frac {x}{\sqrt [4]{4 x^4+3}}-\frac {\arctan \left (\frac {\sqrt {2} x}{\sqrt [4]{4 x^4+3}}\right )}{4 \sqrt {2}}\right )+\frac {1}{4} \sqrt [4]{4 x^4+3} x^3\right )+\frac {1}{8} \left (4 x^4+3\right )^{5/4} x^3\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {15}{8} \left (\frac {3}{4} \left (\frac {\text {arctanh}\left (\frac {\sqrt {2} x}{\sqrt [4]{4 x^4+3}}\right )}{4 \sqrt {2}}-\frac {\arctan \left (\frac {\sqrt {2} x}{\sqrt [4]{4 x^4+3}}\right )}{4 \sqrt {2}}\right )+\frac {1}{4} \sqrt [4]{4 x^4+3} x^3\right )+\frac {1}{8} \left (4 x^4+3\right )^{5/4} x^3\) |
(x^3*(3 + 4*x^4)^(5/4))/8 + (15*((x^3*(3 + 4*x^4)^(1/4))/4 + (3*(-1/4*ArcT an[(Sqrt[2]*x)/(3 + 4*x^4)^(1/4)]/Sqrt[2] + ArcTanh[(Sqrt[2]*x)/(3 + 4*x^4 )^(1/4)]/(4*Sqrt[2])))/4))/8
3.4.4.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
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[((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[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b) Int[1/(r + s*x^2), x], x] - Simp[s/(2*b) Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ [a/b, 0]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^(p + (m + 1)/n) Subst[Int[x^m/(1 - b*x^n)^(p + (m + 1)/n + 1), x], x, x/(a + b*x^n )^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, - 2^(-1)] && IntegersQ[m, p + (m + 1)/n]
Result contains higher order function than in optimal. Order 5 vs. order 3.
Time = 2.23 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.20
| method | result | size |
| meijerg | \(3^{\frac {1}{4}} x^{3} {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (-\frac {5}{4},\frac {3}{4};\frac {7}{4};-\frac {4 x^{4}}{3}\right )\) | \(19\) |
| risch | \(\frac {x^{3} \left (16 x^{4}+27\right ) \left (4 x^{4}+3\right )^{\frac {1}{4}}}{32}+\frac {5 \,3^{\frac {1}{4}} x^{3} {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {3}{4},\frac {3}{4};\frac {7}{4};-\frac {4 x^{4}}{3}\right )}{32}\) | \(42\) |
| pseudoelliptic | \(\frac {\frac {9 x^{7} \left (4 x^{4}+3\right )^{\frac {1}{4}}}{2}+\frac {243 x^{3} \left (4 x^{4}+3\right )^{\frac {1}{4}}}{32}+\frac {405 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\left (4 x^{4}+3\right )^{\frac {1}{4}} \sqrt {2}}{2 x}\right )}{256}+\frac {405 \sqrt {2}\, \arctan \left (\frac {\left (4 x^{4}+3\right )^{\frac {1}{4}} \sqrt {2}}{2 x}\right )}{256}}{\left (-2 x^{2}+\sqrt {4 x^{4}+3}\right )^{2} \left (2 x^{2}+\sqrt {4 x^{4}+3}\right )^{2}}\) | \(112\) |
| trager | \(\frac {x^{3} \left (16 x^{4}+27\right ) \left (4 x^{4}+3\right )^{\frac {1}{4}}}{32}-\frac {45 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) \ln \left (-4 \sqrt {4 x^{4}+3}\, \operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) x^{2}-8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) x^{4}+4 \left (4 x^{4}+3\right )^{\frac {3}{4}} x +8 x^{3} \left (4 x^{4}+3\right )^{\frac {1}{4}}-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right )\right )}{512}-\frac {45 \operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) \ln \left (-4 \operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) \sqrt {4 x^{4}+3}\, x^{2}+8 \operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) x^{4}+4 \left (4 x^{4}+3\right )^{\frac {3}{4}} x -8 x^{3} \left (4 x^{4}+3\right )^{\frac {1}{4}}+3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right )\right )}{512}\) | \(173\) |
Time = 0.26 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.13 \[ \int x^2 \left (3+4 x^4\right )^{5/4} \, dx=\frac {45}{256} \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (4 \, x^{4} + 3\right )}^{\frac {1}{4}}}{2 \, x}\right ) + \frac {45}{512} \, \sqrt {2} \log \left (8 \, x^{4} + 4 \, \sqrt {2} {\left (4 \, x^{4} + 3\right )}^{\frac {1}{4}} x^{3} + 4 \, \sqrt {4 \, x^{4} + 3} x^{2} + 2 \, \sqrt {2} {\left (4 \, x^{4} + 3\right )}^{\frac {3}{4}} x + 3\right ) + \frac {1}{32} \, {\left (16 \, x^{7} + 27 \, x^{3}\right )} {\left (4 \, x^{4} + 3\right )}^{\frac {1}{4}} \]
45/256*sqrt(2)*arctan(1/2*sqrt(2)*(4*x^4 + 3)^(1/4)/x) + 45/512*sqrt(2)*lo g(8*x^4 + 4*sqrt(2)*(4*x^4 + 3)^(1/4)*x^3 + 4*sqrt(4*x^4 + 3)*x^2 + 2*sqrt (2)*(4*x^4 + 3)^(3/4)*x + 3) + 1/32*(16*x^7 + 27*x^3)*(4*x^4 + 3)^(1/4)
Result contains complex when optimal does not.
Time = 0.92 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.44 \[ \int x^2 \left (3+4 x^4\right )^{5/4} \, dx=\frac {3 \cdot \sqrt [4]{3} x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {4 x^{4} e^{i \pi }}{3}} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} \]
3*3**(1/4)*x**3*gamma(3/4)*hyper((-5/4, 3/4), (7/4,), 4*x**4*exp_polar(I*p i)/3)/(4*gamma(7/4))
Time = 0.27 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.40 \[ \int x^2 \left (3+4 x^4\right )^{5/4} \, dx=\frac {45}{256} \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (4 \, x^{4} + 3\right )}^{\frac {1}{4}}}{2 \, x}\right ) - \frac {45}{512} \, \sqrt {2} \log \left (-\frac {\sqrt {2} - \frac {{\left (4 \, x^{4} + 3\right )}^{\frac {1}{4}}}{x}}{\sqrt {2} + \frac {{\left (4 \, x^{4} + 3\right )}^{\frac {1}{4}}}{x}}\right ) + \frac {9 \, {\left (\frac {20 \, {\left (4 \, x^{4} + 3\right )}^{\frac {1}{4}}}{x} - \frac {9 \, {\left (4 \, x^{4} + 3\right )}^{\frac {5}{4}}}{x^{5}}\right )}}{32 \, {\left (\frac {8 \, {\left (4 \, x^{4} + 3\right )}}{x^{4}} - \frac {{\left (4 \, x^{4} + 3\right )}^{2}}{x^{8}} - 16\right )}} \]
45/256*sqrt(2)*arctan(1/2*sqrt(2)*(4*x^4 + 3)^(1/4)/x) - 45/512*sqrt(2)*lo g(-(sqrt(2) - (4*x^4 + 3)^(1/4)/x)/(sqrt(2) + (4*x^4 + 3)^(1/4)/x)) + 9/32 *(20*(4*x^4 + 3)^(1/4)/x - 9*(4*x^4 + 3)^(5/4)/x^5)/(8*(4*x^4 + 3)/x^4 - ( 4*x^4 + 3)^2/x^8 - 16)
Time = 0.31 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.18 \[ \int x^2 \left (3+4 x^4\right )^{5/4} \, dx=\frac {1}{32} \, x^{8} {\left (\frac {9 \, {\left (4 \, x^{4} + 3\right )}^{\frac {1}{4}} {\left (\frac {3}{x^{4}} + 4\right )}}{x} - \frac {20 \, {\left (4 \, x^{4} + 3\right )}^{\frac {1}{4}}}{x}\right )} + \frac {45}{256} \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (4 \, x^{4} + 3\right )}^{\frac {1}{4}}}{2 \, x}\right ) - \frac {45}{512} \, \sqrt {2} \log \left (-\frac {\sqrt {2} - \frac {{\left (4 \, x^{4} + 3\right )}^{\frac {1}{4}}}{x}}{\sqrt {2} + \frac {{\left (4 \, x^{4} + 3\right )}^{\frac {1}{4}}}{x}}\right ) \]
1/32*x^8*(9*(4*x^4 + 3)^(1/4)*(3/x^4 + 4)/x - 20*(4*x^4 + 3)^(1/4)/x) + 45 /256*sqrt(2)*arctan(1/2*sqrt(2)*(4*x^4 + 3)^(1/4)/x) - 45/512*sqrt(2)*log( -(sqrt(2) - (4*x^4 + 3)^(1/4)/x)/(sqrt(2) + (4*x^4 + 3)^(1/4)/x))
Timed out. \[ \int x^2 \left (3+4 x^4\right )^{5/4} \, dx=\int x^2\,{\left (4\,x^4+3\right )}^{5/4} \,d x \]