Integrand size = 17, antiderivative size = 51 \[ \int \frac {1}{(4-3 x) (-2+3 x)^{3/4}} \, dx=\frac {1}{3} \sqrt [4]{2} \arctan \left (\frac {\sqrt [4]{-2+3 x}}{\sqrt [4]{2}}\right )+\frac {1}{3} \sqrt [4]{2} \text {arctanh}\left (\frac {\sqrt [4]{-2+3 x}}{\sqrt [4]{2}}\right ) \] Output:
1/3*2^(1/4)*arctan(1/2*(-2+3*x)^(1/4)*2^(3/4))+1/3*2^(1/4)*arctanh(1/2*(-2 +3*x)^(1/4)*2^(3/4))
Time = 0.06 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.67 \[ \int \frac {1}{(4-3 x) (-2+3 x)^{3/4}} \, dx=\frac {1}{3} \sqrt [4]{2} \left (\arctan \left (\sqrt [4]{-1+\frac {3 x}{2}}\right )+\text {arctanh}\left (\sqrt [4]{-1+\frac {3 x}{2}}\right )\right ) \] Input:
Integrate[1/((4 - 3*x)*(-2 + 3*x)^(3/4)),x]
Output:
(2^(1/4)*(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, 756, 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.
\(\displaystyle \int \frac {1}{(4-3 x) (3 x-2)^{3/4}} \, dx\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {4}{3} \int \frac {1}{4-3 x}d\sqrt [4]{3 x-2}\) |
\(\Big \downarrow \) 756 |
\(\displaystyle \frac {4}{3} \left (\frac {\int \frac {1}{\sqrt {2}-\sqrt {3 x-2}}d\sqrt [4]{3 x-2}}{2 \sqrt {2}}+\frac {\int \frac {1}{\sqrt {3 x-2}+\sqrt {2}}d\sqrt [4]{3 x-2}}{2 \sqrt {2}}\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {4}{3} \left (\frac {\int \frac {1}{\sqrt {2}-\sqrt {3 x-2}}d\sqrt [4]{3 x-2}}{2 \sqrt {2}}+\frac {\arctan \left (\frac {\sqrt [4]{3 x-2}}{\sqrt [4]{2}}\right )}{2\ 2^{3/4}}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {4}{3} \left (\frac {\arctan \left (\frac {\sqrt [4]{3 x-2}}{\sqrt [4]{2}}\right )}{2\ 2^{3/4}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{3 x-2}}{\sqrt [4]{2}}\right )}{2\ 2^{3/4}}\right )\) |
Input:
Int[1/((4 - 3*x)*(-2 + 3*x)^(3/4)),x]
Output:
(4*(ArcTan[(-2 + 3*x)^(1/4)/2^(1/4)]/(2*2^(3/4)) + ArcTanh[(-2 + 3*x)^(1/4 )/2^(1/4)]/(2*2^(3/4))))/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[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 ]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a) Int[1/(r - s*x^2), x], x] + Simp[r/(2*a) Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ[a /b, 0]
Time = 0.61 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.98
method | result | size |
derivativedivides | \(\frac {2^{\frac {1}{4}} \left (\ln \left (\frac {\left (-2+3 x \right )^{\frac {1}{4}}+2^{\frac {1}{4}}}{\left (-2+3 x \right )^{\frac {1}{4}}-2^{\frac {1}{4}}}\right )+2 \arctan \left (\frac {\left (-2+3 x \right )^{\frac {1}{4}} 2^{\frac {3}{4}}}{2}\right )\right )}{6}\) | \(50\) |
default | \(\frac {2^{\frac {1}{4}} \left (\ln \left (\frac {\left (-2+3 x \right )^{\frac {1}{4}}+2^{\frac {1}{4}}}{\left (-2+3 x \right )^{\frac {1}{4}}-2^{\frac {1}{4}}}\right )+2 \arctan \left (\frac {\left (-2+3 x \right )^{\frac {1}{4}} 2^{\frac {3}{4}}}{2}\right )\right )}{6}\) | \(50\) |
pseudoelliptic | \(\frac {2^{\frac {1}{4}} \left (\ln \left (\frac {-\left (-2+3 x \right )^{\frac {1}{4}}-2^{\frac {1}{4}}}{-\left (-2+3 x \right )^{\frac {1}{4}}+2^{\frac {1}{4}}}\right )+2 \arctan \left (\frac {\left (-2+3 x \right )^{\frac {1}{4}} 2^{\frac {3}{4}}}{2}\right )\right )}{6}\) | \(54\) |
trager | \(\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2}\right ) \ln \left (-\frac {2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2}\right ) \left (-2+3 x \right )^{\frac {3}{4}}-2 \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2} \sqrt {-2+3 x}-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2}\right ) \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2} \left (-2+3 x \right )^{\frac {1}{4}}+3 x}{3 x -4}\right )}{6}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right ) \ln \left (-\frac {2 \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right ) \left (-2+3 x \right )^{\frac {3}{4}}+2 \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2} \sqrt {-2+3 x}+2 \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{3} \left (-2+3 x \right )^{\frac {1}{4}}+3 x}{3 x -4}\right )}{6}\) | \(171\) |
Input:
int(1/(4-3*x)/(-2+3*x)^(3/4),x,method=_RETURNVERBOSE)
Output:
1/6*2^(1/4)*(ln(((-2+3*x)^(1/4)+2^(1/4))/((-2+3*x)^(1/4)-2^(1/4)))+2*arcta n(1/2*(-2+3*x)^(1/4)*2^(3/4)))
Time = 0.07 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.16 \[ \int \frac {1}{(4-3 x) (-2+3 x)^{3/4}} \, dx=\frac {1}{12} \cdot 8^{\frac {3}{4}} \arctan \left (\frac {1}{2} \cdot 8^{\frac {1}{4}} {\left (3 \, x - 2\right )}^{\frac {1}{4}}\right ) + \frac {1}{24} \cdot 8^{\frac {3}{4}} \log \left (8^{\frac {3}{4}} + 4 \, {\left (3 \, x - 2\right )}^{\frac {1}{4}}\right ) - \frac {1}{24} \cdot 8^{\frac {3}{4}} \log \left (-8^{\frac {3}{4}} + 4 \, {\left (3 \, x - 2\right )}^{\frac {1}{4}}\right ) \] Input:
integrate(1/(4-3*x)/(-2+3*x)^(3/4),x, algorithm="fricas")
Output:
1/12*8^(3/4)*arctan(1/2*8^(1/4)*(3*x - 2)^(1/4)) + 1/24*8^(3/4)*log(8^(3/4 ) + 4*(3*x - 2)^(1/4)) - 1/24*8^(3/4)*log(-8^(3/4) + 4*(3*x - 2)^(1/4))
Result contains complex when optimal does not.
Time = 0.97 (sec) , antiderivative size = 192, normalized size of antiderivative = 3.76 \[ \int \frac {1}{(4-3 x) (-2+3 x)^{3/4}} \, dx=\frac {\sqrt [4]{2} i \log {\left (- \frac {2^{\frac {3}{4}} \cdot \sqrt [4]{3} \sqrt [4]{x - \frac {2}{3}} e^{\frac {i \pi }{2}}}{2} + 1 \right )} \Gamma \left (\frac {1}{4}\right )}{24 \Gamma \left (\frac {5}{4}\right )} + \frac {\sqrt [4]{2} \log {\left (- \frac {2^{\frac {3}{4}} \cdot \sqrt [4]{3} \sqrt [4]{x - \frac {2}{3}} e^{i \pi }}{2} + 1 \right )} \Gamma \left (\frac {1}{4}\right )}{24 \Gamma \left (\frac {5}{4}\right )} - \frac {\sqrt [4]{2} i \log {\left (- \frac {2^{\frac {3}{4}} \cdot \sqrt [4]{3} \sqrt [4]{x - \frac {2}{3}} e^{\frac {3 i \pi }{2}}}{2} + 1 \right )} \Gamma \left (\frac {1}{4}\right )}{24 \Gamma \left (\frac {5}{4}\right )} - \frac {\sqrt [4]{2} \log {\left (- \frac {2^{\frac {3}{4}} \cdot \sqrt [4]{3} \sqrt [4]{x - \frac {2}{3}} e^{2 i \pi }}{2} + 1 \right )} \Gamma \left (\frac {1}{4}\right )}{24 \Gamma \left (\frac {5}{4}\right )} \] Input:
integrate(1/(4-3*x)/(-2+3*x)**(3/4),x)
Output:
2**(1/4)*I*log(-2**(3/4)*3**(1/4)*(x - 2/3)**(1/4)*exp_polar(I*pi/2)/2 + 1 )*gamma(1/4)/(24*gamma(5/4)) + 2**(1/4)*log(-2**(3/4)*3**(1/4)*(x - 2/3)** (1/4)*exp_polar(I*pi)/2 + 1)*gamma(1/4)/(24*gamma(5/4)) - 2**(1/4)*I*log(- 2**(3/4)*3**(1/4)*(x - 2/3)**(1/4)*exp_polar(3*I*pi/2)/2 + 1)*gamma(1/4)/( 24*gamma(5/4)) - 2**(1/4)*log(-2**(3/4)*3**(1/4)*(x - 2/3)**(1/4)*exp_pola r(2*I*pi)/2 + 1)*gamma(1/4)/(24*gamma(5/4))
Time = 0.11 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.04 \[ \int \frac {1}{(4-3 x) (-2+3 x)^{3/4}} \, dx=\frac {1}{3} \cdot 2^{\frac {1}{4}} \arctan \left (\frac {1}{2} \cdot 2^{\frac {3}{4}} {\left (3 \, x - 2\right )}^{\frac {1}{4}}\right ) - \frac {1}{6} \cdot 2^{\frac {1}{4}} \log \left (-\frac {2^{\frac {1}{4}} - {\left (3 \, x - 2\right )}^{\frac {1}{4}}}{2^{\frac {1}{4}} + {\left (3 \, x - 2\right )}^{\frac {1}{4}}}\right ) \] Input:
integrate(1/(4-3*x)/(-2+3*x)^(3/4),x, algorithm="maxima")
Output:
1/3*2^(1/4)*arctan(1/2*2^(3/4)*(3*x - 2)^(1/4)) - 1/6*2^(1/4)*log(-(2^(1/4 ) - (3*x - 2)^(1/4))/(2^(1/4) + (3*x - 2)^(1/4)))
Time = 0.20 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.10 \[ \int \frac {1}{(4-3 x) (-2+3 x)^{3/4}} \, dx=\frac {1}{3} \cdot 2^{\frac {1}{4}} \arctan \left (\frac {1}{2} \cdot 2^{\frac {3}{4}} {\left (3 \, x - 2\right )}^{\frac {1}{4}}\right ) + \frac {1}{6} \cdot 2^{\frac {1}{4}} \log \left (2^{\frac {1}{4}} + {\left (3 \, x - 2\right )}^{\frac {1}{4}}\right ) - \frac {1}{6} \cdot 2^{\frac {1}{4}} \log \left ({\left | -2^{\frac {1}{4}} + {\left (3 \, x - 2\right )}^{\frac {1}{4}} \right |}\right ) \] Input:
integrate(1/(4-3*x)/(-2+3*x)^(3/4),x, algorithm="giac")
Output:
1/3*2^(1/4)*arctan(1/2*2^(3/4)*(3*x - 2)^(1/4)) + 1/6*2^(1/4)*log(2^(1/4) + (3*x - 2)^(1/4)) - 1/6*2^(1/4)*log(abs(-2^(1/4) + (3*x - 2)^(1/4)))
Time = 0.16 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.61 \[ \int \frac {1}{(4-3 x) (-2+3 x)^{3/4}} \, dx=\frac {2^{1/4}\,\mathrm {atan}\left (\frac {{\left (24\,x-16\right )}^{1/4}}{2}\right )}{3}+\frac {2^{1/4}\,\mathrm {atanh}\left (\frac {{\left (24\,x-16\right )}^{1/4}}{2}\right )}{3} \] Input:
int(-1/((3*x - 2)^(3/4)*(3*x - 4)),x)
Output:
(2^(1/4)*atan((24*x - 16)^(1/4)/2))/3 + (2^(1/4)*atanh((24*x - 16)^(1/4)/2 ))/3
Time = 0.17 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.96 \[ \int \frac {1}{(4-3 x) (-2+3 x)^{3/4}} \, dx=\frac {2^{\frac {1}{4}} \left (2 \mathit {atan} \left (\frac {\left (3 x -2\right )^{\frac {1}{4}} 2^{\frac {3}{4}}}{2}\right )+\mathrm {log}\left (\left (3 x -2\right )^{\frac {1}{4}}+2^{\frac {1}{4}}\right )-\mathrm {log}\left (\left (3 x -2\right )^{\frac {1}{4}}-2^{\frac {1}{4}}\right )\right )}{6} \] Input:
int(1/(4-3*x)/(-2+3*x)^(3/4),x)
Output:
(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