Integrand size = 15, antiderivative size = 100 \[ \int x^2 \left (a+b x^4\right )^{5/4} \, dx=\frac {5}{32} a x^3 \sqrt [4]{a+b x^4}+\frac {1}{8} x^3 \left (a+b x^4\right )^{5/4}-\frac {5 a^2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 b^{3/4}}+\frac {5 a^2 \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 b^{3/4}} \]
5/32*a*x^3*(b*x^4+a)^(1/4)+1/8*x^3*(b*x^4+a)^(5/4)-5/64*a^2*arctan(b^(1/4) *x/(b*x^4+a)^(1/4))/b^(3/4)+5/64*a^2*arctanh(b^(1/4)*x/(b*x^4+a)^(1/4))/b^ (3/4)
Time = 0.48 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.91 \[ \int x^2 \left (a+b x^4\right )^{5/4} \, dx=\frac {1}{32} x^3 \sqrt [4]{a+b x^4} \left (9 a+4 b x^4\right )-\frac {5 a^2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 b^{3/4}}+\frac {5 a^2 \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 b^{3/4}} \]
(x^3*(a + b*x^4)^(1/4)*(9*a + 4*b*x^4))/32 - (5*a^2*ArcTan[(b^(1/4)*x)/(a + b*x^4)^(1/4)])/(64*b^(3/4)) + (5*a^2*ArcTanh[(b^(1/4)*x)/(a + b*x^4)^(1/ 4)])/(64*b^(3/4))
Time = 0.22 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.05, 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.
\(\displaystyle \int x^2 \left (a+b x^4\right )^{5/4} \, dx\) |
\(\Big \downarrow \) 811 |
\(\displaystyle \frac {5}{8} a \int x^2 \sqrt [4]{b x^4+a}dx+\frac {1}{8} x^3 \left (a+b x^4\right )^{5/4}\) |
\(\Big \downarrow \) 811 |
\(\displaystyle \frac {5}{8} a \left (\frac {1}{4} a \int \frac {x^2}{\left (b x^4+a\right )^{3/4}}dx+\frac {1}{4} x^3 \sqrt [4]{a+b x^4}\right )+\frac {1}{8} x^3 \left (a+b x^4\right )^{5/4}\) |
\(\Big \downarrow \) 854 |
\(\displaystyle \frac {5}{8} a \left (\frac {1}{4} a \int \frac {x^2}{\sqrt {b x^4+a} \left (1-\frac {b x^4}{b x^4+a}\right )}d\frac {x}{\sqrt [4]{b x^4+a}}+\frac {1}{4} x^3 \sqrt [4]{a+b x^4}\right )+\frac {1}{8} x^3 \left (a+b x^4\right )^{5/4}\) |
\(\Big \downarrow \) 827 |
\(\displaystyle \frac {5}{8} a \left (\frac {1}{4} a \left (\frac {\int \frac {1}{1-\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}}d\frac {x}{\sqrt [4]{b x^4+a}}}{2 \sqrt {b}}-\frac {\int \frac {1}{\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}+1}d\frac {x}{\sqrt [4]{b x^4+a}}}{2 \sqrt {b}}\right )+\frac {1}{4} x^3 \sqrt [4]{a+b x^4}\right )+\frac {1}{8} x^3 \left (a+b x^4\right )^{5/4}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {5}{8} a \left (\frac {1}{4} a \left (\frac {\int \frac {1}{1-\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}}d\frac {x}{\sqrt [4]{b x^4+a}}}{2 \sqrt {b}}-\frac {\arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 b^{3/4}}\right )+\frac {1}{4} x^3 \sqrt [4]{a+b x^4}\right )+\frac {1}{8} x^3 \left (a+b x^4\right )^{5/4}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {5}{8} a \left (\frac {1}{4} a \left (\frac {\text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 b^{3/4}}-\frac {\arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 b^{3/4}}\right )+\frac {1}{4} x^3 \sqrt [4]{a+b x^4}\right )+\frac {1}{8} x^3 \left (a+b x^4\right )^{5/4}\) |
(x^3*(a + b*x^4)^(5/4))/8 + (5*a*((x^3*(a + b*x^4)^(1/4))/4 + (a*(-1/2*Arc Tan[(b^(1/4)*x)/(a + b*x^4)^(1/4)]/b^(3/4) + ArcTanh[(b^(1/4)*x)/(a + b*x^ 4)^(1/4)]/(2*b^(3/4))))/4))/8
3.11.64.3.1 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]
Time = 4.38 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.08
method | result | size |
pseudoelliptic | \(\frac {16 b^{\frac {7}{4}} \left (b \,x^{4}+a \right )^{\frac {1}{4}} x^{7}+36 a \,x^{3} b^{\frac {3}{4}} \left (b \,x^{4}+a \right )^{\frac {1}{4}}+5 \ln \left (\frac {-b^{\frac {1}{4}} x -\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{b^{\frac {1}{4}} x -\left (b \,x^{4}+a \right )^{\frac {1}{4}}}\right ) a^{2}+10 \arctan \left (\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right ) a^{2}}{128 b^{\frac {3}{4}}}\) | \(108\) |
1/128*(16*b^(7/4)*(b*x^4+a)^(1/4)*x^7+36*a*x^3*b^(3/4)*(b*x^4+a)^(1/4)+5*l n((-b^(1/4)*x-(b*x^4+a)^(1/4))/(b^(1/4)*x-(b*x^4+a)^(1/4)))*a^2+10*arctan( 1/b^(1/4)/x*(b*x^4+a)^(1/4))*a^2)/b^(3/4)
Result contains complex when optimal does not.
Time = 0.28 (sec) , antiderivative size = 200, normalized size of antiderivative = 2.00 \[ \int x^2 \left (a+b x^4\right )^{5/4} \, dx=\frac {1}{32} \, {\left (4 \, b x^{7} + 9 \, a x^{3}\right )} {\left (b x^{4} + a\right )}^{\frac {1}{4}} + \frac {5}{128} \, \left (\frac {a^{8}}{b^{3}}\right )^{\frac {1}{4}} \log \left (\frac {5 \, {\left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{2} + \left (\frac {a^{8}}{b^{3}}\right )^{\frac {1}{4}} b x\right )}}{x}\right ) + \frac {5}{128} i \, \left (\frac {a^{8}}{b^{3}}\right )^{\frac {1}{4}} \log \left (\frac {5 \, {\left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{2} + i \, \left (\frac {a^{8}}{b^{3}}\right )^{\frac {1}{4}} b x\right )}}{x}\right ) - \frac {5}{128} i \, \left (\frac {a^{8}}{b^{3}}\right )^{\frac {1}{4}} \log \left (\frac {5 \, {\left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{2} - i \, \left (\frac {a^{8}}{b^{3}}\right )^{\frac {1}{4}} b x\right )}}{x}\right ) - \frac {5}{128} \, \left (\frac {a^{8}}{b^{3}}\right )^{\frac {1}{4}} \log \left (\frac {5 \, {\left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{2} - \left (\frac {a^{8}}{b^{3}}\right )^{\frac {1}{4}} b x\right )}}{x}\right ) \]
1/32*(4*b*x^7 + 9*a*x^3)*(b*x^4 + a)^(1/4) + 5/128*(a^8/b^3)^(1/4)*log(5*( (b*x^4 + a)^(1/4)*a^2 + (a^8/b^3)^(1/4)*b*x)/x) + 5/128*I*(a^8/b^3)^(1/4)* log(5*((b*x^4 + a)^(1/4)*a^2 + I*(a^8/b^3)^(1/4)*b*x)/x) - 5/128*I*(a^8/b^ 3)^(1/4)*log(5*((b*x^4 + a)^(1/4)*a^2 - I*(a^8/b^3)^(1/4)*b*x)/x) - 5/128* (a^8/b^3)^(1/4)*log(5*((b*x^4 + a)^(1/4)*a^2 - (a^8/b^3)^(1/4)*b*x)/x)
Result contains complex when optimal does not.
Time = 1.04 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.39 \[ \int x^2 \left (a+b x^4\right )^{5/4} \, dx=\frac {a^{\frac {5}{4}} x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} \]
a**(5/4)*x**3*gamma(3/4)*hyper((-5/4, 3/4), (7/4,), b*x**4*exp_polar(I*pi) /a)/(4*gamma(7/4))
Time = 0.29 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.44 \[ \int x^2 \left (a+b x^4\right )^{5/4} \, dx=\frac {5 \, a^{2} \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right )}{64 \, b^{\frac {3}{4}}} - \frac {5 \, a^{2} \log \left (-\frac {b^{\frac {1}{4}} - \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}{b^{\frac {1}{4}} + \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}\right )}{128 \, b^{\frac {3}{4}}} - \frac {\frac {5 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{2} b}{x} - \frac {9 \, {\left (b x^{4} + a\right )}^{\frac {5}{4}} a^{2}}{x^{5}}}{32 \, {\left (b^{2} - \frac {2 \, {\left (b x^{4} + a\right )} b}{x^{4}} + \frac {{\left (b x^{4} + a\right )}^{2}}{x^{8}}\right )}} \]
5/64*a^2*arctan((b*x^4 + a)^(1/4)/(b^(1/4)*x))/b^(3/4) - 5/128*a^2*log(-(b ^(1/4) - (b*x^4 + a)^(1/4)/x)/(b^(1/4) + (b*x^4 + a)^(1/4)/x))/b^(3/4) - 1 /32*(5*(b*x^4 + a)^(1/4)*a^2*b/x - 9*(b*x^4 + a)^(5/4)*a^2/x^5)/(b^2 - 2*( b*x^4 + a)*b/x^4 + (b*x^4 + a)^2/x^8)
\[ \int x^2 \left (a+b x^4\right )^{5/4} \, dx=\int { {\left (b x^{4} + a\right )}^{\frac {5}{4}} x^{2} \,d x } \]
Timed out. \[ \int x^2 \left (a+b x^4\right )^{5/4} \, dx=\int x^2\,{\left (b\,x^4+a\right )}^{5/4} \,d x \]