Integrand size = 17, antiderivative size = 110 \[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^{9/2}} \, dx=\frac {16 b^2 \text {arctanh}(\tanh (a+b x))^{3/2}}{105 x^{3/2} (b x-\text {arctanh}(\tanh (a+b x)))^3}+\frac {8 b \text {arctanh}(\tanh (a+b x))^{3/2}}{35 x^{5/2} (b x-\text {arctanh}(\tanh (a+b x)))^2}+\frac {2 \text {arctanh}(\tanh (a+b x))^{3/2}}{7 x^{7/2} (b x-\text {arctanh}(\tanh (a+b x)))} \] Output:
16/105*b^2*arctanh(tanh(b*x+a))^(3/2)/x^(3/2)/(b*x-arctanh(tanh(b*x+a)))^3 +8/35*b*arctanh(tanh(b*x+a))^(3/2)/x^(5/2)/(b*x-arctanh(tanh(b*x+a)))^2+2/ 7*arctanh(tanh(b*x+a))^(3/2)/x^(7/2)/(b*x-arctanh(tanh(b*x+a)))
Time = 0.05 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.60 \[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^{9/2}} \, dx=\frac {2 \text {arctanh}(\tanh (a+b x))^{3/2} \left (35 b^2 x^2-42 b x \text {arctanh}(\tanh (a+b x))+15 \text {arctanh}(\tanh (a+b x))^2\right )}{105 x^{7/2} (b x-\text {arctanh}(\tanh (a+b x)))^3} \] Input:
Integrate[Sqrt[ArcTanh[Tanh[a + b*x]]]/x^(9/2),x]
Output:
(2*ArcTanh[Tanh[a + b*x]]^(3/2)*(35*b^2*x^2 - 42*b*x*ArcTanh[Tanh[a + b*x] ] + 15*ArcTanh[Tanh[a + b*x]]^2))/(105*x^(7/2)*(b*x - ArcTanh[Tanh[a + b*x ]])^3)
Time = 0.50 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.16, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {2602, 2602, 2598}
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.
\(\displaystyle \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^{9/2}} \, dx\) |
\(\Big \downarrow \) 2602 |
\(\displaystyle \frac {4 b \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^{7/2}}dx}{7 (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {2 \text {arctanh}(\tanh (a+b x))^{3/2}}{7 x^{7/2} (b x-\text {arctanh}(\tanh (a+b x)))}\) |
\(\Big \downarrow \) 2602 |
\(\displaystyle \frac {4 b \left (\frac {2 b \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^{5/2}}dx}{5 (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {2 \text {arctanh}(\tanh (a+b x))^{3/2}}{5 x^{5/2} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{7 (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {2 \text {arctanh}(\tanh (a+b x))^{3/2}}{7 x^{7/2} (b x-\text {arctanh}(\tanh (a+b x)))}\) |
\(\Big \downarrow \) 2598 |
\(\displaystyle \frac {2 \text {arctanh}(\tanh (a+b x))^{3/2}}{7 x^{7/2} (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {4 b \left (\frac {2 \text {arctanh}(\tanh (a+b x))^{3/2}}{5 x^{5/2} (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {4 b \text {arctanh}(\tanh (a+b x))^{3/2}}{15 x^{3/2} (b x-\text {arctanh}(\tanh (a+b x)))^2}\right )}{7 (b x-\text {arctanh}(\tanh (a+b x)))}\) |
Input:
Int[Sqrt[ArcTanh[Tanh[a + b*x]]]/x^(9/2),x]
Output:
(2*ArcTanh[Tanh[a + b*x]]^(3/2))/(7*x^(7/2)*(b*x - ArcTanh[Tanh[a + b*x]]) ) + (4*b*((4*b*ArcTanh[Tanh[a + b*x]]^(3/2))/(15*x^(3/2)*(b*x - ArcTanh[Ta nh[a + b*x]])^2) + (2*ArcTanh[Tanh[a + b*x]]^(3/2))/(5*x^(5/2)*(b*x - ArcT anh[Tanh[a + b*x]]))))/(7*(b*x - ArcTanh[Tanh[a + b*x]]))
Int[(u_)^(m_)*(v_)^(n_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simp lify[D[v, x]]}, Simp[(-u^(m + 1))*(v^(n + 1)/((m + 1)*(b*u - a*v))), x] /; NeQ[b*u - a*v, 0]] /; FreeQ[{m, n}, x] && PiecewiseLinearQ[u, v, x] && EqQ[ m + n + 2, 0] && NeQ[m, -1]
Int[(u_)^(m_)*(v_)^(n_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simp lify[D[v, x]]}, Simp[(-u^(m + 1))*(v^(n + 1)/((m + 1)*(b*u - a*v))), x] + S imp[b*((m + n + 2)/((m + 1)*(b*u - a*v))) Int[u^(m + 1)*v^n, x], x] /; Ne Q[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x] && NeQ[m + n + 2, 0] && LtQ[m , -1]
Time = 0.42 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.95
method | result | size |
derivativedivides | \(-\frac {2 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{7 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {7}{2}}}-\frac {8 b \left (-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{5 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {5}{2}}}+\frac {2 b \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{15 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{2} x^{\frac {3}{2}}}\right )}{7 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )}\) | \(105\) |
default | \(-\frac {2 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{7 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {7}{2}}}-\frac {8 b \left (-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{5 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {5}{2}}}+\frac {2 b \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{15 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{2} x^{\frac {3}{2}}}\right )}{7 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )}\) | \(105\) |
Input:
int(arctanh(tanh(b*x+a))^(1/2)/x^(9/2),x,method=_RETURNVERBOSE)
Output:
-2/7/(arctanh(tanh(b*x+a))-b*x)/x^(7/2)*arctanh(tanh(b*x+a))^(3/2)-8/7*b/( arctanh(tanh(b*x+a))-b*x)*(-1/5/(arctanh(tanh(b*x+a))-b*x)/x^(5/2)*arctanh (tanh(b*x+a))^(3/2)+2/15*b/(arctanh(tanh(b*x+a))-b*x)^2/x^(3/2)*arctanh(ta nh(b*x+a))^(3/2))
Time = 0.07 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.41 \[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^{9/2}} \, dx=-\frac {2 \, {\left (8 \, b^{3} x^{3} - 4 \, a b^{2} x^{2} + 3 \, a^{2} b x + 15 \, a^{3}\right )} \sqrt {b x + a}}{105 \, a^{3} x^{\frac {7}{2}}} \] Input:
integrate(arctanh(tanh(b*x+a))^(1/2)/x^(9/2),x, algorithm="fricas")
Output:
-2/105*(8*b^3*x^3 - 4*a*b^2*x^2 + 3*a^2*b*x + 15*a^3)*sqrt(b*x + a)/(a^3*x ^(7/2))
Timed out. \[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^{9/2}} \, dx=\text {Timed out} \] Input:
integrate(atanh(tanh(b*x+a))**(1/2)/x**(9/2),x)
Output:
Timed out
Time = 0.12 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.41 \[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^{9/2}} \, dx=-\frac {2 \, {\left (8 \, b^{3} x^{3} - 4 \, a b^{2} x^{2} + 3 \, a^{2} b x + 15 \, a^{3}\right )} \sqrt {b x + a}}{105 \, a^{3} x^{\frac {7}{2}}} \] Input:
integrate(arctanh(tanh(b*x+a))^(1/2)/x^(9/2),x, algorithm="maxima")
Output:
-2/105*(8*b^3*x^3 - 4*a*b^2*x^2 + 3*a^2*b*x + 15*a^3)*sqrt(b*x + a)/(a^3*x ^(7/2))
Time = 0.12 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.25 \[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^{9/2}} \, dx=\frac {32 \, {\left (70 \, b^{\frac {7}{2}} {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{8} + 35 \, a b^{\frac {7}{2}} {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{6} + 21 \, a^{2} b^{\frac {7}{2}} {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{4} - 7 \, a^{3} b^{\frac {7}{2}} {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{2} + a^{4} b^{\frac {7}{2}}\right )}}{105 \, {\left ({\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{2} - a\right )}^{7}} \] Input:
integrate(arctanh(tanh(b*x+a))^(1/2)/x^(9/2),x, algorithm="giac")
Output:
32/105*(70*b^(7/2)*(sqrt(b)*sqrt(x) - sqrt(b*x + a))^8 + 35*a*b^(7/2)*(sqr t(b)*sqrt(x) - sqrt(b*x + a))^6 + 21*a^2*b^(7/2)*(sqrt(b)*sqrt(x) - sqrt(b *x + a))^4 - 7*a^3*b^(7/2)*(sqrt(b)*sqrt(x) - sqrt(b*x + a))^2 + a^4*b^(7/ 2))/((sqrt(b)*sqrt(x) - sqrt(b*x + a))^2 - a)^7
Time = 3.84 (sec) , antiderivative size = 234, normalized size of antiderivative = 2.13 \[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^{9/2}} \, dx=\frac {\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}\,\left (\frac {32\,b^2\,x^2}{105\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}+\frac {128\,b^3\,x^3}{105\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3}+\frac {4\,b\,x}{35\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}-\frac {2}{7}\right )}{x^{7/2}} \] Input:
int(atanh(tanh(a + b*x))^(1/2)/x^(9/2),x)
Output:
((log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(2/(exp(2* a)*exp(2*b*x) + 1))/2)^(1/2)*((32*b^2*x^2)/(105*(log(2/(exp(2*a)*exp(2*b*x ) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^ 2) + (128*b^3*x^3)/(105*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a )*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^3) + (4*b*x)/(35*(log(2/ (exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b *x) + 1)) + 2*b*x)) - 2/7))/x^(7/2)
\[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^{9/2}} \, dx=\frac {-2 \sqrt {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )}+\sqrt {x}\, \left (\int \frac {\sqrt {x}\, \sqrt {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )}}{\mathit {atanh} \left (\tanh \left (b x +a \right )\right ) x^{4}}d x \right ) b \,x^{3}}{7 \sqrt {x}\, x^{3}} \] Input:
int(atanh(tanh(b*x+a))^(1/2)/x^(9/2),x)
Output:
( - 2*sqrt(atanh(tanh(a + b*x))) + sqrt(x)*int((sqrt(x)*sqrt(atanh(tanh(a + b*x))))/(atanh(tanh(a + b*x))*x**4),x)*b*x**3)/(7*sqrt(x)*x**3)