Integrand size = 15, antiderivative size = 73 \[ \int \frac {\sqrt [4]{a+b x}}{x^{5/4}} \, dx=-\frac {4 \sqrt [4]{a+b x}}{\sqrt [4]{x}}-2 \sqrt [4]{b} \arctan \left (\frac {\sqrt [4]{b} \sqrt [4]{x}}{\sqrt [4]{a+b x}}\right )+2 \sqrt [4]{b} \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt [4]{x}}{\sqrt [4]{a+b x}}\right ) \] Output:
-4*(b*x+a)^(1/4)/x^(1/4)-2*b^(1/4)*arctan(b^(1/4)*x^(1/4)/(b*x+a)^(1/4))+2 *b^(1/4)*arctanh(b^(1/4)*x^(1/4)/(b*x+a)^(1/4))
Time = 0.19 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt [4]{a+b x}}{x^{5/4}} \, dx=-\frac {4 \sqrt [4]{a+b x}}{\sqrt [4]{x}}-2 \sqrt [4]{b} \arctan \left (\frac {\sqrt [4]{b} \sqrt [4]{x}}{\sqrt [4]{a+b x}}\right )+2 \sqrt [4]{b} \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt [4]{x}}{\sqrt [4]{a+b x}}\right ) \] Input:
Integrate[(a + b*x)^(1/4)/x^(5/4),x]
Output:
(-4*(a + b*x)^(1/4))/x^(1/4) - 2*b^(1/4)*ArcTan[(b^(1/4)*x^(1/4))/(a + b*x )^(1/4)] + 2*b^(1/4)*ArcTanh[(b^(1/4)*x^(1/4))/(a + b*x)^(1/4)]
Time = 0.18 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.11, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {57, 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.
\(\displaystyle \int \frac {\sqrt [4]{a+b x}}{x^{5/4}} \, dx\) |
\(\Big \downarrow \) 57 |
\(\displaystyle b \int \frac {1}{\sqrt [4]{x} (a+b x)^{3/4}}dx-\frac {4 \sqrt [4]{a+b x}}{\sqrt [4]{x}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle 4 b \int \frac {\sqrt {x}}{(a+b x)^{3/4}}d\sqrt [4]{x}-\frac {4 \sqrt [4]{a+b x}}{\sqrt [4]{x}}\) |
\(\Big \downarrow \) 854 |
\(\displaystyle 4 b \int \frac {\sqrt {x}}{1-b x}d\frac {\sqrt [4]{x}}{\sqrt [4]{a+b x}}-\frac {4 \sqrt [4]{a+b x}}{\sqrt [4]{x}}\) |
\(\Big \downarrow \) 827 |
\(\displaystyle 4 b \left (\frac {\int \frac {1}{1-\sqrt {b} \sqrt {x}}d\frac {\sqrt [4]{x}}{\sqrt [4]{a+b x}}}{2 \sqrt {b}}-\frac {\int \frac {1}{\sqrt {b} \sqrt {x}+1}d\frac {\sqrt [4]{x}}{\sqrt [4]{a+b x}}}{2 \sqrt {b}}\right )-\frac {4 \sqrt [4]{a+b x}}{\sqrt [4]{x}}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle 4 b \left (\frac {\int \frac {1}{1-\sqrt {b} \sqrt {x}}d\frac {\sqrt [4]{x}}{\sqrt [4]{a+b x}}}{2 \sqrt {b}}-\frac {\arctan \left (\frac {\sqrt [4]{b} \sqrt [4]{x}}{\sqrt [4]{a+b x}}\right )}{2 b^{3/4}}\right )-\frac {4 \sqrt [4]{a+b x}}{\sqrt [4]{x}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle 4 b \left (\frac {\text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt [4]{x}}{\sqrt [4]{a+b x}}\right )}{2 b^{3/4}}-\frac {\arctan \left (\frac {\sqrt [4]{b} \sqrt [4]{x}}{\sqrt [4]{a+b x}}\right )}{2 b^{3/4}}\right )-\frac {4 \sqrt [4]{a+b x}}{\sqrt [4]{x}}\) |
Input:
Int[(a + b*x)^(1/4)/x^(5/4),x]
Output:
(-4*(a + b*x)^(1/4))/x^(1/4) + 4*b*(-1/2*ArcTan[(b^(1/4)*x^(1/4))/(a + b*x )^(1/4)]/b^(3/4) + ArcTanh[(b^(1/4)*x^(1/4))/(a + b*x)^(1/4)]/(2*b^(3/4)))
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] & & GtQ[n, 0] && LtQ[m, -1] && !(IntegerQ[n] && !IntegerQ[m]) && !(ILeQ[m + n + 2, 0] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c , d, m, n, x]
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]
\[\int \frac {\left (b x +a \right )^{\frac {1}{4}}}{x^{\frac {5}{4}}}d x\]
Input:
int((b*x+a)^(1/4)/x^(5/4),x)
Output:
int((b*x+a)^(1/4)/x^(5/4),x)
Result contains complex when optimal does not.
Time = 0.08 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.81 \[ \int \frac {\sqrt [4]{a+b x}}{x^{5/4}} \, dx=\frac {b^{\frac {1}{4}} x \log \left (\frac {b^{\frac {1}{4}} x + {\left (b x + a\right )}^{\frac {1}{4}} x^{\frac {3}{4}}}{x}\right ) - b^{\frac {1}{4}} x \log \left (-\frac {b^{\frac {1}{4}} x - {\left (b x + a\right )}^{\frac {1}{4}} x^{\frac {3}{4}}}{x}\right ) + i \, b^{\frac {1}{4}} x \log \left (\frac {i \, b^{\frac {1}{4}} x + {\left (b x + a\right )}^{\frac {1}{4}} x^{\frac {3}{4}}}{x}\right ) - i \, b^{\frac {1}{4}} x \log \left (\frac {-i \, b^{\frac {1}{4}} x + {\left (b x + a\right )}^{\frac {1}{4}} x^{\frac {3}{4}}}{x}\right ) - 4 \, {\left (b x + a\right )}^{\frac {1}{4}} x^{\frac {3}{4}}}{x} \] Input:
integrate((b*x+a)^(1/4)/x^(5/4),x, algorithm="fricas")
Output:
(b^(1/4)*x*log((b^(1/4)*x + (b*x + a)^(1/4)*x^(3/4))/x) - b^(1/4)*x*log(-( b^(1/4)*x - (b*x + a)^(1/4)*x^(3/4))/x) + I*b^(1/4)*x*log((I*b^(1/4)*x + ( b*x + a)^(1/4)*x^(3/4))/x) - I*b^(1/4)*x*log((-I*b^(1/4)*x + (b*x + a)^(1/ 4)*x^(3/4))/x) - 4*(b*x + a)^(1/4)*x^(3/4))/x
Result contains complex when optimal does not.
Time = 1.22 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.56 \[ \int \frac {\sqrt [4]{a+b x}}{x^{5/4}} \, dx=\frac {\sqrt [4]{a} \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b x e^{i \pi }}{a}} \right )}}{\sqrt [4]{x} \Gamma \left (\frac {3}{4}\right )} \] Input:
integrate((b*x+a)**(1/4)/x**(5/4),x)
Output:
a**(1/4)*gamma(-1/4)*hyper((-1/4, -1/4), (3/4,), b*x*exp_polar(I*pi)/a)/(x **(1/4)*gamma(3/4))
Time = 0.11 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.01 \[ \int \frac {\sqrt [4]{a+b x}}{x^{5/4}} \, dx=2 \, b^{\frac {1}{4}} \arctan \left (\frac {{\left (b x + a\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}} x^{\frac {1}{4}}}\right ) - b^{\frac {1}{4}} \log \left (-\frac {b^{\frac {1}{4}} - \frac {{\left (b x + a\right )}^{\frac {1}{4}}}{x^{\frac {1}{4}}}}{b^{\frac {1}{4}} + \frac {{\left (b x + a\right )}^{\frac {1}{4}}}{x^{\frac {1}{4}}}}\right ) - \frac {4 \, {\left (b x + a\right )}^{\frac {1}{4}}}{x^{\frac {1}{4}}} \] Input:
integrate((b*x+a)^(1/4)/x^(5/4),x, algorithm="maxima")
Output:
2*b^(1/4)*arctan((b*x + a)^(1/4)/(b^(1/4)*x^(1/4))) - b^(1/4)*log(-(b^(1/4 ) - (b*x + a)^(1/4)/x^(1/4))/(b^(1/4) + (b*x + a)^(1/4)/x^(1/4))) - 4*(b*x + a)^(1/4)/x^(1/4)
\[ \int \frac {\sqrt [4]{a+b x}}{x^{5/4}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {1}{4}}}{x^{\frac {5}{4}}} \,d x } \] Input:
integrate((b*x+a)^(1/4)/x^(5/4),x, algorithm="giac")
Output:
integrate((b*x + a)^(1/4)/x^(5/4), x)
Timed out. \[ \int \frac {\sqrt [4]{a+b x}}{x^{5/4}} \, dx=\int \frac {{\left (a+b\,x\right )}^{1/4}}{x^{5/4}} \,d x \] Input:
int((a + b*x)^(1/4)/x^(5/4),x)
Output:
int((a + b*x)^(1/4)/x^(5/4), x)
\[ \int \frac {\sqrt [4]{a+b x}}{x^{5/4}} \, dx=\int \frac {\left (b x +a \right )^{\frac {1}{4}}}{x^{\frac {5}{4}}}d x \] Input:
int((b*x+a)^(1/4)/x^(5/4),x)
Output:
int((a + b*x)**(1/4)/(x**(1/4)*x),x)