Integrand size = 19, antiderivative size = 82 \[ \int \frac {\sqrt [4]{b x^3+a x^4}}{x} \, dx=\sqrt [4]{b x^3+a x^4}-\frac {b \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^3+a x^4}}\right )}{2 a^{3/4}}+\frac {b \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^3+a x^4}}\right )}{2 a^{3/4}} \]
(a*x^4+b*x^3)^(1/4)-1/2*b*arctan(a^(1/4)*x/(a*x^4+b*x^3)^(1/4))/a^(3/4)+1/ 2*b*arctanh(a^(1/4)*x/(a*x^4+b*x^3)^(1/4))/a^(3/4)
Time = 0.27 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.28 \[ \int \frac {\sqrt [4]{b x^3+a x^4}}{x} \, dx=\frac {x^{9/4} (b+a x)^{3/4} \left (2 a^{3/4} x^{3/4} \sqrt [4]{b+a x}-b \arctan \left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{b+a x}}\right )+b \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{b+a x}}\right )\right )}{2 a^{3/4} \left (x^3 (b+a x)\right )^{3/4}} \]
(x^(9/4)*(b + a*x)^(3/4)*(2*a^(3/4)*x^(3/4)*(b + a*x)^(1/4) - b*ArcTan[(a^ (1/4)*x^(1/4))/(b + a*x)^(1/4)] + b*ArcTanh[(a^(1/4)*x^(1/4))/(b + a*x)^(1 /4)]))/(2*a^(3/4)*(x^3*(b + a*x))^(3/4))
Time = 0.25 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.32, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {1927, 1938, 73, 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.
\(\displaystyle \int \frac {\sqrt [4]{a x^4+b x^3}}{x} \, dx\) |
\(\Big \downarrow \) 1927 |
\(\displaystyle \frac {1}{4} b \int \frac {x^2}{\left (a x^4+b x^3\right )^{3/4}}dx+\sqrt [4]{a x^4+b x^3}\) |
\(\Big \downarrow \) 1938 |
\(\displaystyle \frac {b x^{9/4} (a x+b)^{3/4} \int \frac {1}{\sqrt [4]{x} (b+a x)^{3/4}}dx}{4 \left (a x^4+b x^3\right )^{3/4}}+\sqrt [4]{a x^4+b x^3}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {b x^{9/4} (a x+b)^{3/4} \int \frac {\sqrt {x}}{(b+a x)^{3/4}}d\sqrt [4]{x}}{\left (a x^4+b x^3\right )^{3/4}}+\sqrt [4]{a x^4+b x^3}\) |
\(\Big \downarrow \) 854 |
\(\displaystyle \frac {b x^{9/4} (a x+b)^{3/4} \int \frac {\sqrt {x}}{1-a x}d\frac {\sqrt [4]{x}}{\sqrt [4]{b+a x}}}{\left (a x^4+b x^3\right )^{3/4}}+\sqrt [4]{a x^4+b x^3}\) |
\(\Big \downarrow \) 827 |
\(\displaystyle \frac {b x^{9/4} (a x+b)^{3/4} \left (\frac {\int \frac {1}{1-\sqrt {a} \sqrt {x}}d\frac {\sqrt [4]{x}}{\sqrt [4]{b+a x}}}{2 \sqrt {a}}-\frac {\int \frac {1}{\sqrt {a} \sqrt {x}+1}d\frac {\sqrt [4]{x}}{\sqrt [4]{b+a x}}}{2 \sqrt {a}}\right )}{\left (a x^4+b x^3\right )^{3/4}}+\sqrt [4]{a x^4+b x^3}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {b x^{9/4} (a x+b)^{3/4} \left (\frac {\int \frac {1}{1-\sqrt {a} \sqrt {x}}d\frac {\sqrt [4]{x}}{\sqrt [4]{b+a x}}}{2 \sqrt {a}}-\frac {\arctan \left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{a x+b}}\right )}{2 a^{3/4}}\right )}{\left (a x^4+b x^3\right )^{3/4}}+\sqrt [4]{a x^4+b x^3}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {b x^{9/4} (a x+b)^{3/4} \left (\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{a x+b}}\right )}{2 a^{3/4}}-\frac {\arctan \left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{a x+b}}\right )}{2 a^{3/4}}\right )}{\left (a x^4+b x^3\right )^{3/4}}+\sqrt [4]{a x^4+b x^3}\) |
(b*x^3 + a*x^4)^(1/4) + (b*x^(9/4)*(b + a*x)^(3/4)*(-1/2*ArcTan[(a^(1/4)*x ^(1/4))/(b + a*x)^(1/4)]/a^(3/4) + ArcTanh[(a^(1/4)*x^(1/4))/(b + a*x)^(1/ 4)]/(2*a^(3/4))))/(b*x^3 + a*x^4)^(3/4)
3.12.2.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
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[(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]
Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol ] :> Simp[(c*x)^(m + 1)*((a*x^j + b*x^n)^p/(c*(m + n*p + 1))), x] + Simp[a* (n - j)*(p/(c^j*(m + n*p + 1))) Int[(c*x)^(m + j)*(a*x^j + b*x^n)^(p - 1) , x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[p] && LtQ[0, j, n] && (Int egersQ[j, n] || GtQ[c, 0]) && GtQ[p, 0] && NeQ[m + n*p + 1, 0]
Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol ] :> Simp[c^IntPart[m]*(c*x)^FracPart[m]*((a*x^j + b*x^n)^FracPart[p]/(x^(F racPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p])) Int[x^(m + j* p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] && !Inte gerQ[p] && NeQ[n, j] && PosQ[n - j]
Time = 1.19 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.05
method | result | size |
pseudoelliptic | \(\frac {4 \left (x^{3} \left (a x +b \right )\right )^{\frac {1}{4}} a^{\frac {3}{4}}+\ln \left (\frac {a^{\frac {1}{4}} x +\left (x^{3} \left (a x +b \right )\right )^{\frac {1}{4}}}{-a^{\frac {1}{4}} x +\left (x^{3} \left (a x +b \right )\right )^{\frac {1}{4}}}\right ) b +2 \arctan \left (\frac {\left (x^{3} \left (a x +b \right )\right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right ) b}{4 a^{\frac {3}{4}}}\) | \(86\) |
1/4*(4*(x^3*(a*x+b))^(1/4)*a^(3/4)+ln((a^(1/4)*x+(x^3*(a*x+b))^(1/4))/(-a^ (1/4)*x+(x^3*(a*x+b))^(1/4)))*b+2*arctan(1/a^(1/4)/x*(x^3*(a*x+b))^(1/4))* b)/a^(3/4)
Result contains complex when optimal does not.
Time = 0.26 (sec) , antiderivative size = 194, normalized size of antiderivative = 2.37 \[ \int \frac {\sqrt [4]{b x^3+a x^4}}{x} \, dx=\frac {1}{4} \, \left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}} \log \left (\frac {a \left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}} x + {\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}} b}{x}\right ) - \frac {1}{4} \, \left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}} \log \left (-\frac {a \left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}} x - {\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}} b}{x}\right ) + \frac {1}{4} i \, \left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}} \log \left (\frac {i \, a \left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}} x + {\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}} b}{x}\right ) - \frac {1}{4} i \, \left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}} \log \left (\frac {-i \, a \left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}} x + {\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}} b}{x}\right ) + {\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}} \]
1/4*(b^4/a^3)^(1/4)*log((a*(b^4/a^3)^(1/4)*x + (a*x^4 + b*x^3)^(1/4)*b)/x) - 1/4*(b^4/a^3)^(1/4)*log(-(a*(b^4/a^3)^(1/4)*x - (a*x^4 + b*x^3)^(1/4)*b )/x) + 1/4*I*(b^4/a^3)^(1/4)*log((I*a*(b^4/a^3)^(1/4)*x + (a*x^4 + b*x^3)^ (1/4)*b)/x) - 1/4*I*(b^4/a^3)^(1/4)*log((-I*a*(b^4/a^3)^(1/4)*x + (a*x^4 + b*x^3)^(1/4)*b)/x) + (a*x^4 + b*x^3)^(1/4)
\[ \int \frac {\sqrt [4]{b x^3+a x^4}}{x} \, dx=\int \frac {\sqrt [4]{x^{3} \left (a x + b\right )}}{x}\, dx \]
\[ \int \frac {\sqrt [4]{b x^3+a x^4}}{x} \, dx=\int { \frac {{\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}}}{x} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 211 vs. \(2 (64) = 128\).
Time = 0.28 (sec) , antiderivative size = 211, normalized size of antiderivative = 2.57 \[ \int \frac {\sqrt [4]{b x^3+a x^4}}{x} \, dx=\frac {\frac {2 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (a + \frac {b}{x}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{a} + \frac {2 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (a + \frac {b}{x}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{a} + \frac {\sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{2} \log \left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a + \frac {b}{x}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a + \frac {b}{x}}\right )}{a} + \frac {\sqrt {2} b^{2} \log \left (-\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a + \frac {b}{x}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a + \frac {b}{x}}\right )}{\left (-a\right )^{\frac {3}{4}}} + 8 \, {\left (a + \frac {b}{x}\right )}^{\frac {1}{4}} b x}{8 \, b} \]
1/8*(2*sqrt(2)*(-a)^(1/4)*b^2*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*( a + b/x)^(1/4))/(-a)^(1/4))/a + 2*sqrt(2)*(-a)^(1/4)*b^2*arctan(-1/2*sqrt( 2)*(sqrt(2)*(-a)^(1/4) - 2*(a + b/x)^(1/4))/(-a)^(1/4))/a + sqrt(2)*(-a)^( 1/4)*b^2*log(sqrt(2)*(-a)^(1/4)*(a + b/x)^(1/4) + sqrt(-a) + sqrt(a + b/x) )/a + sqrt(2)*b^2*log(-sqrt(2)*(-a)^(1/4)*(a + b/x)^(1/4) + sqrt(-a) + sqr t(a + b/x))/(-a)^(3/4) + 8*(a + b/x)^(1/4)*b*x)/b
Timed out. \[ \int \frac {\sqrt [4]{b x^3+a x^4}}{x} \, dx=\int \frac {{\left (a\,x^4+b\,x^3\right )}^{1/4}}{x} \,d x \]