Integrand size = 22, antiderivative size = 49 \[ \int \frac {\left (-b+a x^3\right ) \sqrt {x+x^4}}{x^3} \, dx=\frac {\left (2 b+a x^3\right ) \sqrt {x+x^4}}{3 x^2}+\frac {1}{3} (a-2 b) \text {arctanh}\left (\frac {x^2}{\sqrt {x+x^4}}\right ) \]
Time = 0.21 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.51 \[ \int \frac {\left (-b+a x^3\right ) \sqrt {x+x^4}}{x^3} \, dx=\frac {\left (2 b+a x^3\right ) \sqrt {x+x^4}}{3 x^2}+\frac {(a-2 b) \sqrt {x+x^4} \log \left (x^{3/2}+\sqrt {1+x^3}\right )}{3 \sqrt {x} \sqrt {1+x^3}} \]
((2*b + a*x^3)*Sqrt[x + x^4])/(3*x^2) + ((a - 2*b)*Sqrt[x + x^4]*Log[x^(3/ 2) + Sqrt[1 + x^3]])/(3*Sqrt[x]*Sqrt[1 + x^3])
Time = 0.22 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.16, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {1944, 1910, 1935, 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.
\(\displaystyle \int \frac {\sqrt {x^4+x} \left (a x^3-b\right )}{x^3} \, dx\) |
\(\Big \downarrow \) 1944 |
\(\displaystyle (a-2 b) \int \sqrt {x^4+x}dx+\frac {2 b \left (x^4+x\right )^{3/2}}{3 x^3}\) |
\(\Big \downarrow \) 1910 |
\(\displaystyle (a-2 b) \left (\frac {1}{2} \int \frac {x}{\sqrt {x^4+x}}dx+\frac {1}{3} \sqrt {x^4+x} x\right )+\frac {2 b \left (x^4+x\right )^{3/2}}{3 x^3}\) |
\(\Big \downarrow \) 1935 |
\(\displaystyle (a-2 b) \left (\frac {1}{3} \int \frac {1}{1-\frac {x^4}{x^4+x}}d\frac {x^2}{\sqrt {x^4+x}}+\frac {1}{3} \sqrt {x^4+x} x\right )+\frac {2 b \left (x^4+x\right )^{3/2}}{3 x^3}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle (a-2 b) \left (\frac {1}{3} \text {arctanh}\left (\frac {x^2}{\sqrt {x^4+x}}\right )+\frac {1}{3} \sqrt {x^4+x} x\right )+\frac {2 b \left (x^4+x\right )^{3/2}}{3 x^3}\) |
(2*b*(x + x^4)^(3/2))/(3*x^3) + (a - 2*b)*((x*Sqrt[x + x^4])/3 + ArcTanh[x ^2/Sqrt[x + x^4]]/3)
3.7.23.3.1 Defintions of rubi rules used
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[((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[(x_)^(m_.)/Sqrt[(a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.)], x_Symbol] :> Simp [-2/(n - j) Subst[Int[1/(1 - a*x^2), x], x, x^(j/2)/Sqrt[a*x^j + b*x^n]], x] /; FreeQ[{a, b, j, n}, x] && EqQ[m, j/2 - 1] && NeQ[n, j]
Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> Simp[c*e^(j - 1)*(e*x)^(m - j + 1)*((a*x^j + b*x^(j + n))^(p + 1)/(a*(m + j*p + 1))), x] + Simp[(a*d*(m + j*p + 1) - b *c*(m + n + p*(j + n) + 1))/(a*e^n*(m + j*p + 1)) Int[(e*x)^(m + n)*(a*x^ j + b*x^(j + n))^p, x], x] /; FreeQ[{a, b, c, d, e, j, p}, x] && EqQ[jn, j + n] && !IntegerQ[p] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && (LtQ[m + j*p, -1 ] || (IntegersQ[m - 1/2, p - 1/2] && LtQ[p, 0] && LtQ[m, (-n)*p - 1])) && ( GtQ[e, 0] || IntegersQ[j, n]) && NeQ[m + j*p + 1, 0] && NeQ[m - n + j*p + 1 , 0]
Time = 3.17 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.98
method | result | size |
trager | \(\frac {\left (a \,x^{3}+2 b \right ) \sqrt {x^{4}+x}}{3 x^{2}}-\frac {\left (a -2 b \right ) \ln \left (2 x^{3}-2 x \sqrt {x^{4}+x}+1\right )}{6}\) | \(48\) |
risch | \(\frac {\left (x^{3}+1\right ) \left (a \,x^{3}+2 b \right )}{3 x \sqrt {x \left (x^{3}+1\right )}}+\frac {\left (\frac {a}{2}-b \right ) \ln \left (-2 x^{3}-2 x \sqrt {x^{4}+x}-1\right )}{3}\) | \(57\) |
meijerg | \(-\frac {a \left (-2 \sqrt {\pi }\, x^{\frac {3}{2}} \sqrt {x^{3}+1}-2 \sqrt {\pi }\, \operatorname {arcsinh}\left (x^{\frac {3}{2}}\right )\right )}{6 \sqrt {\pi }}+\frac {b \left (\frac {4 \sqrt {\pi }\, \sqrt {x^{3}+1}}{x^{\frac {3}{2}}}-4 \sqrt {\pi }\, \operatorname {arcsinh}\left (x^{\frac {3}{2}}\right )\right )}{6 \sqrt {\pi }}\) | \(64\) |
pseudoelliptic | \(\frac {\left (2 a \,x^{3}+4 b \right ) \sqrt {x^{4}+x}-\left (a -2 b \right ) \left (\ln \left (\frac {-x^{2}+\sqrt {x^{4}+x}}{x^{2}}\right )-\ln \left (\frac {x^{2}+\sqrt {x^{4}+x}}{x^{2}}\right )\right ) x^{2}}{6 x^{2}}\) | \(72\) |
default | \(a \left (\frac {x \sqrt {x^{4}+x}}{3}+\frac {\ln \left (\frac {x^{2}+\sqrt {x^{4}+x}}{x^{2}}\right )}{6}-\frac {\ln \left (\frac {-x^{2}+\sqrt {x^{4}+x}}{x^{2}}\right )}{6}\right )-b \left (-\frac {2 \sqrt {x^{4}+x}}{3 x^{2}}+\frac {\ln \left (-2 x^{3}-2 x \sqrt {x^{4}+x}-1\right )}{3}\right )\) | \(89\) |
elliptic | \(\frac {2 b \sqrt {x^{4}+x}}{3 x^{2}}+\frac {a x \sqrt {x^{4}+x}}{3}-\frac {2 \left (\frac {a}{2}-b \right ) \left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {\left (\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) x}{\left (\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (1+x \right )}}\, \left (1+x \right )^{2} \sqrt {-\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{\left (\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \left (1+x \right )}}\, \sqrt {-\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{\left (\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (1+x \right )}}\, \left (-\operatorname {EllipticF}\left (\sqrt {\frac {\left (\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) x}{\left (\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (1+x \right )}}, \sqrt {\frac {\left (-\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) \left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )}{\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (-\frac {3}{2}-\frac {i \sqrt {3}}{2}\right )}}\right )+\operatorname {EllipticPi}\left (\sqrt {\frac {\left (\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) x}{\left (\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (1+x \right )}}, \frac {\frac {1}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}+\frac {i \sqrt {3}}{2}}, \sqrt {\frac {\left (-\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) \left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )}{\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (-\frac {3}{2}-\frac {i \sqrt {3}}{2}\right )}}\right )\right )}{\left (\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {x \left (1+x \right ) \left (x -\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (x -\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )}}\) | \(322\) |
Time = 0.29 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.04 \[ \int \frac {\left (-b+a x^3\right ) \sqrt {x+x^4}}{x^3} \, dx=-\frac {{\left (a - 2 \, b\right )} x^{2} \log \left (2 \, x^{3} - 2 \, \sqrt {x^{4} + x} x + 1\right ) - 2 \, {\left (a x^{3} + 2 \, b\right )} \sqrt {x^{4} + x}}{6 \, x^{2}} \]
\[ \int \frac {\left (-b+a x^3\right ) \sqrt {x+x^4}}{x^3} \, dx=\int \frac {\sqrt {x \left (x + 1\right ) \left (x^{2} - x + 1\right )} \left (a x^{3} - b\right )}{x^{3}}\, dx \]
\[ \int \frac {\left (-b+a x^3\right ) \sqrt {x+x^4}}{x^3} \, dx=\int { \frac {{\left (a x^{3} - b\right )} \sqrt {x^{4} + x}}{x^{3}} \,d x } \]
Time = 0.29 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.16 \[ \int \frac {\left (-b+a x^3\right ) \sqrt {x+x^4}}{x^3} \, dx=\frac {1}{3} \, \sqrt {x^{4} + x} a x + \frac {1}{6} \, {\left (a - 2 \, b\right )} \log \left (\sqrt {\frac {1}{x^{3}} + 1} + 1\right ) - \frac {1}{6} \, {\left (a - 2 \, b\right )} \log \left ({\left | \sqrt {\frac {1}{x^{3}} + 1} - 1 \right |}\right ) + \frac {2}{3} \, b \sqrt {\frac {1}{x^{3}} + 1} \]
1/3*sqrt(x^4 + x)*a*x + 1/6*(a - 2*b)*log(sqrt(1/x^3 + 1) + 1) - 1/6*(a - 2*b)*log(abs(sqrt(1/x^3 + 1) - 1)) + 2/3*b*sqrt(1/x^3 + 1)
Timed out. \[ \int \frac {\left (-b+a x^3\right ) \sqrt {x+x^4}}{x^3} \, dx=-\int \frac {\left (b-a\,x^3\right )\,\sqrt {x^4+x}}{x^3} \,d x \]