Integrand size = 17, antiderivative size = 110 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^{13/2}} \, dx=\frac {16 b^2 \text {arctanh}(\tanh (a+b x))^{7/2}}{693 x^{7/2} (b x-\text {arctanh}(\tanh (a+b x)))^3}+\frac {8 b \text {arctanh}(\tanh (a+b x))^{7/2}}{99 x^{9/2} (b x-\text {arctanh}(\tanh (a+b x)))^2}+\frac {2 \text {arctanh}(\tanh (a+b x))^{7/2}}{11 x^{11/2} (b x-\text {arctanh}(\tanh (a+b x)))} \]
16/693*b^2*arctanh(tanh(b*x+a))^(7/2)/x^(7/2)/(b*x-arctanh(tanh(b*x+a)))^3 +8/99*b*arctanh(tanh(b*x+a))^(7/2)/x^(9/2)/(b*x-arctanh(tanh(b*x+a)))^2+2/ 11*arctanh(tanh(b*x+a))^(7/2)/x^(11/2)/(b*x-arctanh(tanh(b*x+a)))
Time = 0.05 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.60 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^{13/2}} \, dx=\frac {2 \text {arctanh}(\tanh (a+b x))^{7/2} \left (99 b^2 x^2-154 b x \text {arctanh}(\tanh (a+b x))+63 \text {arctanh}(\tanh (a+b x))^2\right )}{693 x^{11/2} (b x-\text {arctanh}(\tanh (a+b x)))^3} \]
(2*ArcTanh[Tanh[a + b*x]]^(7/2)*(99*b^2*x^2 - 154*b*x*ArcTanh[Tanh[a + b*x ]] + 63*ArcTanh[Tanh[a + b*x]]^2))/(693*x^(11/2)*(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.
\(\displaystyle \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^{13/2}} \, dx\) |
\(\Big \downarrow \) 2602 |
\(\displaystyle \frac {4 b \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^{11/2}}dx}{11 (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {2 \text {arctanh}(\tanh (a+b x))^{7/2}}{11 x^{11/2} (b x-\text {arctanh}(\tanh (a+b x)))}\) |
\(\Big \downarrow \) 2602 |
\(\displaystyle \frac {4 b \left (\frac {2 b \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^{9/2}}dx}{9 (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {2 \text {arctanh}(\tanh (a+b x))^{7/2}}{9 x^{9/2} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{11 (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {2 \text {arctanh}(\tanh (a+b x))^{7/2}}{11 x^{11/2} (b x-\text {arctanh}(\tanh (a+b x)))}\) |
\(\Big \downarrow \) 2598 |
\(\displaystyle \frac {2 \text {arctanh}(\tanh (a+b x))^{7/2}}{11 x^{11/2} (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {4 b \left (\frac {2 \text {arctanh}(\tanh (a+b x))^{7/2}}{9 x^{9/2} (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {4 b \text {arctanh}(\tanh (a+b x))^{7/2}}{63 x^{7/2} (b x-\text {arctanh}(\tanh (a+b x)))^2}\right )}{11 (b x-\text {arctanh}(\tanh (a+b x)))}\) |
(2*ArcTanh[Tanh[a + b*x]]^(7/2))/(11*x^(11/2)*(b*x - ArcTanh[Tanh[a + b*x] ])) + (4*b*((4*b*ArcTanh[Tanh[a + b*x]]^(7/2))/(63*x^(7/2)*(b*x - ArcTanh[ Tanh[a + b*x]])^2) + (2*ArcTanh[Tanh[a + b*x]]^(7/2))/(9*x^(9/2)*(b*x - Ar cTanh[Tanh[a + b*x]]))))/(11*(b*x - ArcTanh[Tanh[a + b*x]]))
3.3.39.3.1 Defintions of rubi rules used
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 = 0.21 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.95
method | result | size |
derivativedivides | \(-\frac {2 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{11 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {11}{2}}}-\frac {8 b \left (-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{9 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {9}{2}}}+\frac {2 b \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{63 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{2} x^{\frac {7}{2}}}\right )}{11 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )}\) | \(105\) |
default | \(-\frac {2 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{11 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {11}{2}}}-\frac {8 b \left (-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{9 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {9}{2}}}+\frac {2 b \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{63 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{2} x^{\frac {7}{2}}}\right )}{11 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )}\) | \(105\) |
-2/11/(arctanh(tanh(b*x+a))-b*x)/x^(11/2)*arctanh(tanh(b*x+a))^(7/2)-8/11* b/(arctanh(tanh(b*x+a))-b*x)*(-1/9/(arctanh(tanh(b*x+a))-b*x)/x^(9/2)*arct anh(tanh(b*x+a))^(7/2)+2/63*b/(arctanh(tanh(b*x+a))-b*x)^2/x^(7/2)*arctanh (tanh(b*x+a))^(7/2))
Time = 0.24 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.61 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^{13/2}} \, dx=-\frac {2 \, {\left (8 \, b^{5} x^{5} - 4 \, a b^{4} x^{4} + 3 \, a^{2} b^{3} x^{3} + 113 \, a^{3} b^{2} x^{2} + 161 \, a^{4} b x + 63 \, a^{5}\right )} \sqrt {b x + a}}{693 \, a^{3} x^{\frac {11}{2}}} \]
-2/693*(8*b^5*x^5 - 4*a*b^4*x^4 + 3*a^2*b^3*x^3 + 113*a^3*b^2*x^2 + 161*a^ 4*b*x + 63*a^5)*sqrt(b*x + a)/(a^3*x^(11/2))
Timed out. \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^{13/2}} \, dx=\text {Timed out} \]
Time = 0.29 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.41 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^{13/2}} \, dx=-\frac {2 \, {\left (8 \, b^{3} x^{3} - 20 \, a b^{2} x^{2} + 35 \, a^{2} b x + 63 \, a^{3}\right )} {\left (b x + a\right )}^{\frac {5}{2}}}{693 \, a^{3} x^{\frac {11}{2}}} \]
Time = 0.31 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.71 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^{13/2}} \, dx=-\frac {\sqrt {2} {\left (\frac {99 \, \sqrt {2} b^{11}}{a} + 4 \, {\left (\frac {2 \, \sqrt {2} {\left (b x + a\right )} b^{11}}{a^{3}} - \frac {11 \, \sqrt {2} b^{11}}{a^{2}}\right )} {\left (b x + a\right )}\right )} {\left (b x + a\right )}^{\frac {7}{2}} b}{693 \, {\left ({\left (b x + a\right )} b - a b\right )}^{\frac {11}{2}} {\left | b \right |}} \]
-1/693*sqrt(2)*(99*sqrt(2)*b^11/a + 4*(2*sqrt(2)*(b*x + a)*b^11/a^3 - 11*s qrt(2)*b^11/a^2)*(b*x + a))*(b*x + a)^(7/2)*b/(((b*x + a)*b - a*b)^(11/2)* abs(b))
Time = 4.79 (sec) , antiderivative size = 353, normalized size of antiderivative = 3.21 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^{13/2}} \, dx=\frac {\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}\,\left (\frac {23\,b\,x\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}{99}-\frac {226\,b^2\,x^2}{693}-\frac {{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}{22}+\frac {4\,b^3\,x^3}{231\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}+\frac {32\,b^4\,x^4}{693\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}+\frac {128\,b^5\,x^5}{693\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3}\right )}{x^{11/2}} \]
((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)*((23*b*x*(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))/99 - (226 *b^2*x^2)/693 - (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)^2/22 + (4*b^3*x^3)/(231*(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)) + (32*b^4*x^4)/(693*(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^5*x^5)/(693*(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)))/x^(11/2)