Integrand size = 17, antiderivative size = 110 \[ \int \frac {1}{x^{7/2} \sqrt {\text {arctanh}(\tanh (a+b x))}} \, dx=\frac {16 b^2 \sqrt {\text {arctanh}(\tanh (a+b x))}}{15 \sqrt {x} (b x-\text {arctanh}(\tanh (a+b x)))^3}+\frac {8 b \sqrt {\text {arctanh}(\tanh (a+b x))}}{15 x^{3/2} (b x-\text {arctanh}(\tanh (a+b x)))^2}+\frac {2 \sqrt {\text {arctanh}(\tanh (a+b x))}}{5 x^{5/2} (b x-\text {arctanh}(\tanh (a+b x)))} \] Output:
16/15*b^2*arctanh(tanh(b*x+a))^(1/2)/x^(1/2)/(b*x-arctanh(tanh(b*x+a)))^3+ 8/15*b*arctanh(tanh(b*x+a))^(1/2)/x^(3/2)/(b*x-arctanh(tanh(b*x+a)))^2+2/5 *arctanh(tanh(b*x+a))^(1/2)/x^(5/2)/(b*x-arctanh(tanh(b*x+a)))
Time = 0.04 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.60 \[ \int \frac {1}{x^{7/2} \sqrt {\text {arctanh}(\tanh (a+b x))}} \, dx=\frac {2 \sqrt {\text {arctanh}(\tanh (a+b x))} \left (15 b^2 x^2-10 b x \text {arctanh}(\tanh (a+b x))+3 \text {arctanh}(\tanh (a+b x))^2\right )}{15 x^{5/2} (b x-\text {arctanh}(\tanh (a+b x)))^3} \] Input:
Integrate[1/(x^(7/2)*Sqrt[ArcTanh[Tanh[a + b*x]]]),x]
Output:
(2*Sqrt[ArcTanh[Tanh[a + b*x]]]*(15*b^2*x^2 - 10*b*x*ArcTanh[Tanh[a + b*x] ] + 3*ArcTanh[Tanh[a + b*x]]^2))/(15*x^(5/2)*(b*x - ArcTanh[Tanh[a + b*x]] )^3)
Time = 0.30 (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.
\(\displaystyle \int \frac {1}{x^{7/2} \sqrt {\text {arctanh}(\tanh (a+b x))}} \, dx\) |
\(\Big \downarrow \) 2602 |
\(\displaystyle \frac {4 b \int \frac {1}{x^{5/2} \sqrt {\text {arctanh}(\tanh (a+b x))}}dx}{5 (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {2 \sqrt {\text {arctanh}(\tanh (a+b x))}}{5 x^{5/2} (b x-\text {arctanh}(\tanh (a+b x)))}\) |
\(\Big \downarrow \) 2602 |
\(\displaystyle \frac {4 b \left (\frac {2 b \int \frac {1}{x^{3/2} \sqrt {\text {arctanh}(\tanh (a+b x))}}dx}{3 (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {2 \sqrt {\text {arctanh}(\tanh (a+b x))}}{3 x^{3/2} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{5 (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {2 \sqrt {\text {arctanh}(\tanh (a+b x))}}{5 x^{5/2} (b x-\text {arctanh}(\tanh (a+b x)))}\) |
\(\Big \downarrow \) 2598 |
\(\displaystyle \frac {4 b \left (\frac {2 \sqrt {\text {arctanh}(\tanh (a+b x))}}{3 x^{3/2} (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {4 b \sqrt {\text {arctanh}(\tanh (a+b x))}}{3 \sqrt {x} (b x-\text {arctanh}(\tanh (a+b x)))^2}\right )}{5 (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {2 \sqrt {\text {arctanh}(\tanh (a+b x))}}{5 x^{5/2} (b x-\text {arctanh}(\tanh (a+b x)))}\) |
Input:
Int[1/(x^(7/2)*Sqrt[ArcTanh[Tanh[a + b*x]]]),x]
Output:
(4*b*((4*b*Sqrt[ArcTanh[Tanh[a + b*x]]])/(3*Sqrt[x]*(b*x - ArcTanh[Tanh[a + b*x]])^2) + (2*Sqrt[ArcTanh[Tanh[a + b*x]]])/(3*x^(3/2)*(b*x - ArcTanh[T anh[a + b*x]]))))/(5*(b*x - ArcTanh[Tanh[a + b*x]])) + (2*Sqrt[ArcTanh[Tan h[a + b*x]]])/(5*x^(5/2)*(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 | \(-\frac {2 \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{5 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {5}{2}}}-\frac {8 b \left (-\frac {\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {3}{2}}}+\frac {2 b \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{2} \sqrt {x}}\right )}{5 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )}\) | \(105\) |
default | \(-\frac {2 \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{5 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {5}{2}}}-\frac {8 b \left (-\frac {\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {3}{2}}}+\frac {2 b \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{2} \sqrt {x}}\right )}{5 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )}\) | \(105\) |
Input:
int(1/x^(7/2)/arctanh(tanh(b*x+a))^(1/2),x,method=_RETURNVERBOSE)
Output:
-2/5/(arctanh(tanh(b*x+a))-b*x)/x^(5/2)*arctanh(tanh(b*x+a))^(1/2)-8/5*b/( arctanh(tanh(b*x+a))-b*x)*(-1/3/(arctanh(tanh(b*x+a))-b*x)/x^(3/2)*arctanh (tanh(b*x+a))^(1/2)+2/3*b/(arctanh(tanh(b*x+a))-b*x)^2/x^(1/2)*arctanh(tan h(b*x+a))^(1/2))
Time = 0.08 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.31 \[ \int \frac {1}{x^{7/2} \sqrt {\text {arctanh}(\tanh (a+b x))}} \, dx=-\frac {2 \, {\left (8 \, b^{2} x^{2} - 4 \, a b x + 3 \, a^{2}\right )} \sqrt {b x + a}}{15 \, a^{3} x^{\frac {5}{2}}} \] Input:
integrate(1/x^(7/2)/arctanh(tanh(b*x+a))^(1/2),x, algorithm="fricas")
Output:
-2/15*(8*b^2*x^2 - 4*a*b*x + 3*a^2)*sqrt(b*x + a)/(a^3*x^(5/2))
Timed out. \[ \int \frac {1}{x^{7/2} \sqrt {\text {arctanh}(\tanh (a+b x))}} \, dx=\text {Timed out} \] Input:
integrate(1/x**(7/2)/atanh(tanh(b*x+a))**(1/2),x)
Output:
Timed out
Time = 0.12 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.41 \[ \int \frac {1}{x^{7/2} \sqrt {\text {arctanh}(\tanh (a+b x))}} \, dx=-\frac {2 \, {\left (8 \, b^{3} x^{3} + 4 \, a b^{2} x^{2} - a^{2} b x + 3 \, a^{3}\right )}}{15 \, \sqrt {b x + a} a^{3} x^{\frac {5}{2}}} \] Input:
integrate(1/x^(7/2)/arctanh(tanh(b*x+a))^(1/2),x, algorithm="maxima")
Output:
-2/15*(8*b^3*x^3 + 4*a*b^2*x^2 - a^2*b*x + 3*a^3)/(sqrt(b*x + a)*a^3*x^(5/ 2))
Time = 0.13 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.70 \[ \int \frac {1}{x^{7/2} \sqrt {\text {arctanh}(\tanh (a+b x))}} \, dx=\frac {32 \, {\left (10 \, {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{4} - 5 \, a {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{2} + a^{2}\right )} b^{\frac {5}{2}}}{15 \, {\left ({\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{2} - a\right )}^{5}} \] Input:
integrate(1/x^(7/2)/arctanh(tanh(b*x+a))^(1/2),x, algorithm="giac")
Output:
32/15*(10*(sqrt(b)*sqrt(x) - sqrt(b*x + a))^4 - 5*a*(sqrt(b)*sqrt(x) - sqr t(b*x + a))^2 + a^2)*b^(5/2)/((sqrt(b)*sqrt(x) - sqrt(b*x + a))^2 - a)^5
Time = 3.59 (sec) , antiderivative size = 227, normalized size of antiderivative = 2.06 \[ \int \frac {1}{x^{7/2} \sqrt {\text {arctanh}(\tanh (a+b x))}} \, dx=\frac {\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}\,\left (\frac {4}{5\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}+\frac {128\,b^2\,x^2}{15\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3}+\frac {32\,b\,x}{15\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}\right )}{x^{5/2}} \] Input:
int(1/(x^(7/2)*atanh(tanh(a + b*x))^(1/2)),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)*(4/(5*(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)) + (128*b^2*x ^2)/(15*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(e xp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^3) + (32*b*x)/(15*(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)))/x^(5/2)
\[ \int \frac {1}{x^{7/2} \sqrt {\text {arctanh}(\tanh (a+b x))}} \, dx=\int \frac {\sqrt {x}\, \sqrt {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )}}{\mathit {atanh} \left (\tanh \left (b x +a \right )\right ) x^{4}}d x \] Input:
int(1/x^(7/2)/atanh(tanh(b*x+a))^(1/2),x)
Output:
int((sqrt(x)*sqrt(atanh(tanh(a + b*x))))/(atanh(tanh(a + b*x))*x**4),x)