Integrand size = 17, antiderivative size = 51 \[ \int \frac {1}{(4-3 x) \sqrt [4]{-2+3 x}} \, dx=-\frac {1}{3} 2^{3/4} \arctan \left (\frac {\sqrt [4]{-2+3 x}}{\sqrt [4]{2}}\right )+\frac {1}{3} 2^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{-2+3 x}}{\sqrt [4]{2}}\right ) \] Output:
-1/3*2^(3/4)*arctan(1/2*(-2+3*x)^(1/4)*2^(3/4))+1/3*2^(3/4)*arctanh(1/2*(- 2+3*x)^(1/4)*2^(3/4))
Time = 0.06 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.71 \[ \int \frac {1}{(4-3 x) \sqrt [4]{-2+3 x}} \, dx=\frac {1}{3} 2^{3/4} \left (-\arctan \left (\sqrt [4]{-1+\frac {3 x}{2}}\right )+\text {arctanh}\left (\sqrt [4]{-1+\frac {3 x}{2}}\right )\right ) \] Input:
Integrate[1/((4 - 3*x)*(-2 + 3*x)^(1/4)),x]
Output:
(2^(3/4)*(-ArcTan[(-1 + (3*x)/2)^(1/4)] + ArcTanh[(-1 + (3*x)/2)^(1/4)]))/ 3
Time = 0.16 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.08, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {73, 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.
\(\displaystyle \int \frac {1}{(4-3 x) \sqrt [4]{3 x-2}} \, dx\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {4}{3} \int \frac {\sqrt {3 x-2}}{4-3 x}d\sqrt [4]{3 x-2}\) |
\(\Big \downarrow \) 827 |
\(\displaystyle \frac {4}{3} \left (\frac {1}{2} \int \frac {1}{\sqrt {2}-\sqrt {3 x-2}}d\sqrt [4]{3 x-2}-\frac {1}{2} \int \frac {1}{\sqrt {3 x-2}+\sqrt {2}}d\sqrt [4]{3 x-2}\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {4}{3} \left (\frac {1}{2} \int \frac {1}{\sqrt {2}-\sqrt {3 x-2}}d\sqrt [4]{3 x-2}-\frac {\arctan \left (\frac {\sqrt [4]{3 x-2}}{\sqrt [4]{2}}\right )}{2 \sqrt [4]{2}}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {4}{3} \left (\frac {\text {arctanh}\left (\frac {\sqrt [4]{3 x-2}}{\sqrt [4]{2}}\right )}{2 \sqrt [4]{2}}-\frac {\arctan \left (\frac {\sqrt [4]{3 x-2}}{\sqrt [4]{2}}\right )}{2 \sqrt [4]{2}}\right )\) |
Input:
Int[1/((4 - 3*x)*(-2 + 3*x)^(1/4)),x]
Output:
(4*(-1/2*ArcTan[(-2 + 3*x)^(1/4)/2^(1/4)]/2^(1/4) + ArcTanh[(-2 + 3*x)^(1/ 4)/2^(1/4)]/(2*2^(1/4))))/3
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]
Time = 0.76 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.02
method | result | size |
derivativedivides | \(-\frac {2^{\frac {3}{4}} \left (2 \arctan \left (\frac {\left (-2+3 x \right )^{\frac {1}{4}} 2^{\frac {3}{4}}}{2}\right )-\ln \left (\frac {\left (-2+3 x \right )^{\frac {1}{4}}+2^{\frac {1}{4}}}{\left (-2+3 x \right )^{\frac {1}{4}}-2^{\frac {1}{4}}}\right )\right )}{6}\) | \(52\) |
default | \(-\frac {2^{\frac {3}{4}} \left (2 \arctan \left (\frac {\left (-2+3 x \right )^{\frac {1}{4}} 2^{\frac {3}{4}}}{2}\right )-\ln \left (\frac {\left (-2+3 x \right )^{\frac {1}{4}}+2^{\frac {1}{4}}}{\left (-2+3 x \right )^{\frac {1}{4}}-2^{\frac {1}{4}}}\right )\right )}{6}\) | \(52\) |
pseudoelliptic | \(\frac {2^{\frac {3}{4}} \left (-2 \arctan \left (\frac {\left (-2+3 x \right )^{\frac {1}{4}} 2^{\frac {3}{4}}}{2}\right )+\ln \left (\frac {-\left (-2+3 x \right )^{\frac {1}{4}}-2^{\frac {1}{4}}}{-\left (-2+3 x \right )^{\frac {1}{4}}+2^{\frac {1}{4}}}\right )\right )}{6}\) | \(54\) |
trager | \(\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right ) \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right ) \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2} \left (-2+3 x \right )^{\frac {3}{4}}+2 \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2} \sqrt {-2+3 x}-4 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right ) \left (-2+3 x \right )^{\frac {1}{4}}-6 x}{3 x -4}\right )}{6}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right ) \ln \left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{3} \left (-2+3 x \right )^{\frac {3}{4}}+2 \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2} \sqrt {-2+3 x}+4 \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right ) \left (-2+3 x \right )^{\frac {1}{4}}+6 x}{3 x -4}\right )}{6}\) | \(168\) |
Input:
int(1/(4-3*x)/(-2+3*x)^(1/4),x,method=_RETURNVERBOSE)
Output:
-1/6*2^(3/4)*(2*arctan(1/2*(-2+3*x)^(1/4)*2^(3/4))-ln(((-2+3*x)^(1/4)+2^(1 /4))/((-2+3*x)^(1/4)-2^(1/4))))
Time = 0.10 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.08 \[ \int \frac {1}{(4-3 x) \sqrt [4]{-2+3 x}} \, dx=-\frac {1}{3} \cdot 2^{\frac {3}{4}} \arctan \left (\frac {1}{2} \cdot 2^{\frac {3}{4}} {\left (3 \, x - 2\right )}^{\frac {1}{4}}\right ) + \frac {1}{6} \cdot 2^{\frac {3}{4}} \log \left (2^{\frac {1}{4}} + {\left (3 \, x - 2\right )}^{\frac {1}{4}}\right ) - \frac {1}{6} \cdot 2^{\frac {3}{4}} \log \left (-2^{\frac {1}{4}} + {\left (3 \, x - 2\right )}^{\frac {1}{4}}\right ) \] Input:
integrate(1/(4-3*x)/(-2+3*x)^(1/4),x, algorithm="fricas")
Output:
-1/3*2^(3/4)*arctan(1/2*2^(3/4)*(3*x - 2)^(1/4)) + 1/6*2^(3/4)*log(2^(1/4) + (3*x - 2)^(1/4)) - 1/6*2^(3/4)*log(-2^(1/4) + (3*x - 2)^(1/4))
Result contains complex when optimal does not.
Time = 0.96 (sec) , antiderivative size = 192, normalized size of antiderivative = 3.76 \[ \int \frac {1}{(4-3 x) \sqrt [4]{-2+3 x}} \, dx=- \frac {2^{\frac {3}{4}} i \log {\left (- \frac {2^{\frac {3}{4}} \cdot \sqrt [4]{3} \sqrt [4]{x - \frac {2}{3}} e^{\frac {i \pi }{2}}}{2} + 1 \right )} \Gamma \left (\frac {3}{4}\right )}{8 \Gamma \left (\frac {7}{4}\right )} + \frac {2^{\frac {3}{4}} \log {\left (- \frac {2^{\frac {3}{4}} \cdot \sqrt [4]{3} \sqrt [4]{x - \frac {2}{3}} e^{i \pi }}{2} + 1 \right )} \Gamma \left (\frac {3}{4}\right )}{8 \Gamma \left (\frac {7}{4}\right )} + \frac {2^{\frac {3}{4}} i \log {\left (- \frac {2^{\frac {3}{4}} \cdot \sqrt [4]{3} \sqrt [4]{x - \frac {2}{3}} e^{\frac {3 i \pi }{2}}}{2} + 1 \right )} \Gamma \left (\frac {3}{4}\right )}{8 \Gamma \left (\frac {7}{4}\right )} - \frac {2^{\frac {3}{4}} \log {\left (- \frac {2^{\frac {3}{4}} \cdot \sqrt [4]{3} \sqrt [4]{x - \frac {2}{3}} e^{2 i \pi }}{2} + 1 \right )} \Gamma \left (\frac {3}{4}\right )}{8 \Gamma \left (\frac {7}{4}\right )} \] Input:
integrate(1/(4-3*x)/(-2+3*x)**(1/4),x)
Output:
-2**(3/4)*I*log(-2**(3/4)*3**(1/4)*(x - 2/3)**(1/4)*exp_polar(I*pi/2)/2 + 1)*gamma(3/4)/(8*gamma(7/4)) + 2**(3/4)*log(-2**(3/4)*3**(1/4)*(x - 2/3)** (1/4)*exp_polar(I*pi)/2 + 1)*gamma(3/4)/(8*gamma(7/4)) + 2**(3/4)*I*log(-2 **(3/4)*3**(1/4)*(x - 2/3)**(1/4)*exp_polar(3*I*pi/2)/2 + 1)*gamma(3/4)/(8 *gamma(7/4)) - 2**(3/4)*log(-2**(3/4)*3**(1/4)*(x - 2/3)**(1/4)*exp_polar( 2*I*pi)/2 + 1)*gamma(3/4)/(8*gamma(7/4))
Time = 0.15 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.04 \[ \int \frac {1}{(4-3 x) \sqrt [4]{-2+3 x}} \, dx=-\frac {1}{3} \cdot 2^{\frac {3}{4}} \arctan \left (\frac {1}{2} \cdot 2^{\frac {3}{4}} {\left (3 \, x - 2\right )}^{\frac {1}{4}}\right ) - \frac {1}{6} \cdot 2^{\frac {3}{4}} \log \left (-\frac {2^{\frac {1}{4}} - {\left (3 \, x - 2\right )}^{\frac {1}{4}}}{2^{\frac {1}{4}} + {\left (3 \, x - 2\right )}^{\frac {1}{4}}}\right ) \] Input:
integrate(1/(4-3*x)/(-2+3*x)^(1/4),x, algorithm="maxima")
Output:
-1/3*2^(3/4)*arctan(1/2*2^(3/4)*(3*x - 2)^(1/4)) - 1/6*2^(3/4)*log(-(2^(1/ 4) - (3*x - 2)^(1/4))/(2^(1/4) + (3*x - 2)^(1/4)))
Time = 0.21 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.10 \[ \int \frac {1}{(4-3 x) \sqrt [4]{-2+3 x}} \, dx=-\frac {1}{3} \cdot 2^{\frac {3}{4}} \arctan \left (\frac {1}{2} \cdot 2^{\frac {3}{4}} {\left (3 \, x - 2\right )}^{\frac {1}{4}}\right ) + \frac {1}{6} \cdot 2^{\frac {3}{4}} \log \left (2^{\frac {1}{4}} + {\left (3 \, x - 2\right )}^{\frac {1}{4}}\right ) - \frac {1}{6} \cdot 2^{\frac {3}{4}} \log \left ({\left | -2^{\frac {1}{4}} + {\left (3 \, x - 2\right )}^{\frac {1}{4}} \right |}\right ) \] Input:
integrate(1/(4-3*x)/(-2+3*x)^(1/4),x, algorithm="giac")
Output:
-1/3*2^(3/4)*arctan(1/2*2^(3/4)*(3*x - 2)^(1/4)) + 1/6*2^(3/4)*log(2^(1/4) + (3*x - 2)^(1/4)) - 1/6*2^(3/4)*log(abs(-2^(1/4) + (3*x - 2)^(1/4)))
Time = 0.06 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.61 \[ \int \frac {1}{(4-3 x) \sqrt [4]{-2+3 x}} \, dx=\frac {8^{1/4}\,\mathrm {atanh}\left (\frac {{\left (24\,x-16\right )}^{1/4}}{2}\right )}{3}-\frac {8^{1/4}\,\mathrm {atan}\left (\frac {{\left (24\,x-16\right )}^{1/4}}{2}\right )}{3} \] Input:
int(-1/((3*x - 2)^(1/4)*(3*x - 4)),x)
Output:
(8^(1/4)*atanh((24*x - 16)^(1/4)/2))/3 - (8^(1/4)*atan((24*x - 16)^(1/4)/2 ))/3
Time = 0.16 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00 \[ \int \frac {1}{(4-3 x) \sqrt [4]{-2+3 x}} \, dx=\frac {\sqrt {2}\, 2^{\frac {1}{4}} \left (-2 \mathit {atan} \left (\frac {\left (3 x -2\right )^{\frac {1}{4}} 2^{\frac {3}{4}}}{2}\right )+\mathrm {log}\left (\left (3 x -2\right )^{\frac {1}{4}}+2^{\frac {1}{4}}\right )-\mathrm {log}\left (\left (3 x -2\right )^{\frac {1}{4}}-2^{\frac {1}{4}}\right )\right )}{6} \] Input:
int(1/(4-3*x)/(-2+3*x)^(1/4),x)
Output:
(sqrt(2)*2**(1/4)*( - 2*atan((3*x - 2)**(1/4)/2**(1/4)) + log((3*x - 2)**( 1/4) + 2**(1/4)) - log((3*x - 2)**(1/4) - 2**(1/4))))/6