Integrand size = 15, antiderivative size = 91 \[ \int \sqrt [4]{b x^5+a x^8} \, dx=\frac {1}{3} x \sqrt [4]{b x^5+a x^8}+\frac {b \arctan \left (\frac {\sqrt [4]{b x^5+a x^8}}{\sqrt [4]{a} x^2}\right )}{6 a^{3/4}}+\frac {b \text {arctanh}\left (\frac {\sqrt [4]{b x^5+a x^8}}{\sqrt [4]{a} x^2}\right )}{6 a^{3/4}} \]
1/3*x*(a*x^8+b*x^5)^(1/4)+1/6*b*arctan((a*x^8+b*x^5)^(1/4)/a^(1/4)/x^2)/a^ (3/4)+1/6*b*arctanh((a*x^8+b*x^5)^(1/4)/a^(1/4)/x^2)/a^(3/4)
Time = 6.70 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.26 \[ \int \sqrt [4]{b x^5+a x^8} \, dx=\frac {\sqrt [4]{x^5 \left (b+a x^3\right )} \left (2 a^{3/4} x^{9/4} \sqrt [4]{b+a x^3}-b \arctan \left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{b+a x^3}}\right )+b \text {arctanh}\left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{b+a x^3}}\right )\right )}{6 a^{3/4} x^{5/4} \sqrt [4]{b+a x^3}} \]
((x^5*(b + a*x^3))^(1/4)*(2*a^(3/4)*x^(9/4)*(b + a*x^3)^(1/4) - b*ArcTan[( a^(1/4)*x^(3/4))/(b + a*x^3)^(1/4)] + b*ArcTanh[(a^(1/4)*x^(3/4))/(b + a*x ^3)^(1/4)]))/(6*a^(3/4)*x^(5/4)*(b + a*x^3)^(1/4))
Time = 0.28 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.30, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {1910, 1938, 851, 807, 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.
\(\displaystyle \int \sqrt [4]{a x^8+b x^5} \, dx\) |
\(\Big \downarrow \) 1910 |
\(\displaystyle \frac {1}{4} b \int \frac {x^5}{\left (a x^8+b x^5\right )^{3/4}}dx+\frac {1}{3} x \sqrt [4]{a x^8+b x^5}\) |
\(\Big \downarrow \) 1938 |
\(\displaystyle \frac {b x^{15/4} \left (a x^3+b\right )^{3/4} \int \frac {x^{5/4}}{\left (a x^3+b\right )^{3/4}}dx}{4 \left (a x^8+b x^5\right )^{3/4}}+\frac {1}{3} x \sqrt [4]{a x^8+b x^5}\) |
\(\Big \downarrow \) 851 |
\(\displaystyle \frac {b x^{15/4} \left (a x^3+b\right )^{3/4} \int \frac {x^2}{\left (a x^3+b\right )^{3/4}}d\sqrt [4]{x}}{\left (a x^8+b x^5\right )^{3/4}}+\frac {1}{3} x \sqrt [4]{a x^8+b x^5}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {b x^{15/4} \left (a x^3+b\right )^{3/4} \int \frac {\sqrt {x}}{(b+a x)^{3/4}}dx^{3/4}}{3 \left (a x^8+b x^5\right )^{3/4}}+\frac {1}{3} x \sqrt [4]{a x^8+b x^5}\) |
\(\Big \downarrow \) 854 |
\(\displaystyle \frac {b x^{15/4} \left (a x^3+b\right )^{3/4} \int \frac {\sqrt {x}}{1-a x}d\frac {x^{3/4}}{\sqrt [4]{b+a x}}}{3 \left (a x^8+b x^5\right )^{3/4}}+\frac {1}{3} x \sqrt [4]{a x^8+b x^5}\) |
\(\Big \downarrow \) 827 |
\(\displaystyle \frac {b x^{15/4} \left (a x^3+b\right )^{3/4} \left (\frac {\int \frac {1}{1-\sqrt {a} \sqrt {x}}d\frac {x^{3/4}}{\sqrt [4]{b+a x}}}{2 \sqrt {a}}-\frac {\int \frac {1}{\sqrt {a} \sqrt {x}+1}d\frac {x^{3/4}}{\sqrt [4]{b+a x}}}{2 \sqrt {a}}\right )}{3 \left (a x^8+b x^5\right )^{3/4}}+\frac {1}{3} x \sqrt [4]{a x^8+b x^5}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {b x^{15/4} \left (a x^3+b\right )^{3/4} \left (\frac {\int \frac {1}{1-\sqrt {a} \sqrt {x}}d\frac {x^{3/4}}{\sqrt [4]{b+a x}}}{2 \sqrt {a}}-\frac {\arctan \left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{a x+b}}\right )}{2 a^{3/4}}\right )}{3 \left (a x^8+b x^5\right )^{3/4}}+\frac {1}{3} x \sqrt [4]{a x^8+b x^5}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {b x^{15/4} \left (a x^3+b\right )^{3/4} \left (\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{a x+b}}\right )}{2 a^{3/4}}-\frac {\arctan \left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{a x+b}}\right )}{2 a^{3/4}}\right )}{3 \left (a x^8+b x^5\right )^{3/4}}+\frac {1}{3} x \sqrt [4]{a x^8+b x^5}\) |
(x*(b*x^5 + a*x^8)^(1/4))/3 + (b*x^(15/4)*(b + a*x^3)^(3/4)*(-1/2*ArcTan[( a^(1/4)*x^(3/4))/(b + a*x)^(1/4)]/a^(3/4) + ArcTanh[(a^(1/4)*x^(3/4))/(b + a*x)^(1/4)]/(2*a^(3/4))))/(3*(b*x^5 + a*x^8)^(3/4))
3.13.58.3.1 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[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[1/k Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
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[((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[(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[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[x*((a*x^j + b*x^n)^p/(n*p + 1)), x] + Simp[a*(n - j)*(p/(n*p + 1)) Int[x^j*(a*x^j + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && !IntegerQ[p] && LtQ[0, j, n] && GtQ[p, 0] && NeQ[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 = 3.52 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.13
method | result | size |
pseudoelliptic | \(\frac {4 \left (x^{5} \left (a \,x^{3}+b \right )\right )^{\frac {1}{4}} x \,a^{\frac {3}{4}}+\ln \left (\frac {-a^{\frac {1}{4}} x^{2}-\left (x^{5} \left (a \,x^{3}+b \right )\right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x^{2}-\left (x^{5} \left (a \,x^{3}+b \right )\right )^{\frac {1}{4}}}\right ) b +2 \arctan \left (\frac {\left (x^{5} \left (a \,x^{3}+b \right )\right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x^{2}}\right ) b}{12 a^{\frac {3}{4}}}\) | \(103\) |
1/12*(4*(x^5*(a*x^3+b))^(1/4)*x*a^(3/4)+ln((-a^(1/4)*x^2-(x^5*(a*x^3+b))^( 1/4))/(a^(1/4)*x^2-(x^5*(a*x^3+b))^(1/4)))*b+2*arctan((x^5*(a*x^3+b))^(1/4 )/a^(1/4)/x^2)*b)/a^(3/4)
Timed out. \[ \int \sqrt [4]{b x^5+a x^8} \, dx=\text {Timed out} \]
\[ \int \sqrt [4]{b x^5+a x^8} \, dx=\int \sqrt [4]{a x^{8} + b x^{5}}\, dx \]
\[ \int \sqrt [4]{b x^5+a x^8} \, dx=\int { {\left (a x^{8} + b x^{5}\right )}^{\frac {1}{4}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 213 vs. \(2 (71) = 142\).
Time = 0.28 (sec) , antiderivative size = 213, normalized size of antiderivative = 2.34 \[ \int \sqrt [4]{b x^5+a x^8} \, dx=\frac {8 \, {\left (a + \frac {b}{x^{3}}\right )}^{\frac {1}{4}} b x^{3} + \frac {2 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (a + \frac {b}{x^{3}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{a} + \frac {2 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (a + \frac {b}{x^{3}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{a} + \frac {\sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{2} \log \left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a + \frac {b}{x^{3}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a + \frac {b}{x^{3}}}\right )}{a} + \frac {\sqrt {2} b^{2} \log \left (-\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a + \frac {b}{x^{3}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a + \frac {b}{x^{3}}}\right )}{\left (-a\right )^{\frac {3}{4}}}}{24 \, b} \]
1/24*(8*(a + b/x^3)^(1/4)*b*x^3 + 2*sqrt(2)*(-a)^(1/4)*b^2*arctan(1/2*sqrt (2)*(sqrt(2)*(-a)^(1/4) + 2*(a + b/x^3)^(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^3)^(1/4 ))/(-a)^(1/4))/a + sqrt(2)*(-a)^(1/4)*b^2*log(sqrt(2)*(-a)^(1/4)*(a + b/x^ 3)^(1/4) + sqrt(-a) + sqrt(a + b/x^3))/a + sqrt(2)*b^2*log(-sqrt(2)*(-a)^( 1/4)*(a + b/x^3)^(1/4) + sqrt(-a) + sqrt(a + b/x^3))/(-a)^(3/4))/b
Time = 5.94 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.46 \[ \int \sqrt [4]{b x^5+a x^8} \, dx=\frac {4\,x\,{\left (a\,x^8+b\,x^5\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {3}{4};\ \frac {7}{4};\ -\frac {a\,x^3}{b}\right )}{9\,{\left (\frac {a\,x^3}{b}+1\right )}^{1/4}} \]