Integrand size = 15, antiderivative size = 110 \[ \int \frac {x^{5/2}}{\text {arctanh}(\tanh (a+b x))^3} \, dx=\frac {15 \sqrt {x}}{4 b^3}-\frac {15 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\text {arctanh}(\tanh (a+b x))}}\right ) \sqrt {b x-\text {arctanh}(\tanh (a+b x))}}{4 b^{7/2}}-\frac {x^{5/2}}{2 b \text {arctanh}(\tanh (a+b x))^2}-\frac {5 x^{3/2}}{4 b^2 \text {arctanh}(\tanh (a+b x))} \]
-1/2*x^(5/2)/b/arctanh(tanh(b*x+a))^2-5/4*x^(3/2)/b^2/arctanh(tanh(b*x+a)) +15/4*x^(1/2)/b^3-15/4*arctanh(b^(1/2)*x^(1/2)/(b*x-arctanh(tanh(b*x+a)))^ (1/2))*(b*x-arctanh(tanh(b*x+a)))^(1/2)/b^(7/2)
Time = 0.08 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.95 \[ \int \frac {x^{5/2}}{\text {arctanh}(\tanh (a+b x))^3} \, dx=\frac {1}{4} \left (\frac {15 \sqrt {x}}{b^3}-\frac {2 x^{5/2}}{b \text {arctanh}(\tanh (a+b x))^2}-\frac {5 x^{3/2}}{b^2 \text {arctanh}(\tanh (a+b x))}-\frac {15 \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {-b x+\text {arctanh}(\tanh (a+b x))}}\right ) \sqrt {-b x+\text {arctanh}(\tanh (a+b x))}}{b^{7/2}}\right ) \]
((15*Sqrt[x])/b^3 - (2*x^(5/2))/(b*ArcTanh[Tanh[a + b*x]]^2) - (5*x^(3/2)) /(b^2*ArcTanh[Tanh[a + b*x]]) - (15*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[-(b*x) + ArcTanh[Tanh[a + b*x]]]]*Sqrt[-(b*x) + ArcTanh[Tanh[a + b*x]]])/b^(7/2))/ 4
Time = 0.32 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.09, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {2599, 2599, 2590, 2593}
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.
| \(\displaystyle \int \frac {x^{5/2}}{\text {arctanh}(\tanh (a+b x))^3} \, dx\) |
| \(\Big \downarrow \) 2599 |
| \(\displaystyle \frac {5 \int \frac {x^{3/2}}{\text {arctanh}(\tanh (a+b x))^2}dx}{4 b}-\frac {x^{5/2}}{2 b \text {arctanh}(\tanh (a+b x))^2}\) |
| \(\Big \downarrow \) 2599 |
| \(\displaystyle \frac {5 \left (\frac {3 \int \frac {\sqrt {x}}{\text {arctanh}(\tanh (a+b x))}dx}{2 b}-\frac {x^{3/2}}{b \text {arctanh}(\tanh (a+b x))}\right )}{4 b}-\frac {x^{5/2}}{2 b \text {arctanh}(\tanh (a+b x))^2}\) |
| \(\Big \downarrow \) 2590 |
| \(\displaystyle \frac {5 \left (\frac {3 \left (\frac {(b x-\text {arctanh}(\tanh (a+b x))) \int \frac {1}{\sqrt {x} \text {arctanh}(\tanh (a+b x))}dx}{b}+\frac {2 \sqrt {x}}{b}\right )}{2 b}-\frac {x^{3/2}}{b \text {arctanh}(\tanh (a+b x))}\right )}{4 b}-\frac {x^{5/2}}{2 b \text {arctanh}(\tanh (a+b x))^2}\) |
| \(\Big \downarrow \) 2593 |
| \(\displaystyle \frac {5 \left (\frac {3 \left (\frac {2 \sqrt {x}}{b}-\frac {2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\text {arctanh}(\tanh (a+b x))}}\right ) \sqrt {b x-\text {arctanh}(\tanh (a+b x))}}{b^{3/2}}\right )}{2 b}-\frac {x^{3/2}}{b \text {arctanh}(\tanh (a+b x))}\right )}{4 b}-\frac {x^{5/2}}{2 b \text {arctanh}(\tanh (a+b x))^2}\) |
(5*((3*((2*Sqrt[x])/b - (2*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[b*x - ArcTanh[Ta nh[a + b*x]]]]*Sqrt[b*x - ArcTanh[Tanh[a + b*x]]])/b^(3/2)))/(2*b) - x^(3/ 2)/(b*ArcTanh[Tanh[a + b*x]])))/(4*b) - x^(5/2)/(2*b*ArcTanh[Tanh[a + b*x] ]^2)
3.3.8.3.1 Defintions of rubi rules used
Int[(v_)^(n_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[ D[v, x]]}, Simp[v^n/(a*n), x] - Simp[(b*u - a*v)/a Int[v^(n - 1)/u, x], x ] /; NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x] && GtQ[n, 0] && NeQ[n, 1]
Int[1/((u_)*Sqrt[v_]), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simpli fy[D[v, x]]}, Simp[-2*(ArcTanh[Sqrt[v]/Rt[-(b*u - a*v)/a, 2]]/(a*Rt[-(b*u - a*v)/a, 2])), x] /; NeQ[b*u - a*v, 0] && NegQ[(b*u - a*v)/a]] /; Piecewise LinearQ[u, v, x]
Int[(u_)^(m_)*(v_)^(n_.), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Sim plify[D[v, x]]}, Simp[u^(m + 1)*(v^n/(a*(m + 1))), x] - Simp[b*(n/(a*(m + 1 ))) Int[u^(m + 1)*v^(n - 1), x], x] /; NeQ[b*u - a*v, 0]] /; FreeQ[{m, n} , x] && PiecewiseLinearQ[u, v, x] && NeQ[m, -1] && ((LtQ[m, -1] && GtQ[n, 0 ] && !(ILtQ[m + n, -2] && (FractionQ[m] || GeQ[2*n + m + 1, 0]))) || (IGtQ [n, 0] && IGtQ[m, 0] && LeQ[n, m]) || (IGtQ[n, 0] && !IntegerQ[m]) || (ILt Q[m, 0] && !IntegerQ[n]))
Time = 0.58 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.39
| method | result | size |
| derivativedivides | \(\frac {2 \sqrt {x}}{b^{3}}-\frac {2 \left (\frac {\left (-\frac {9 a b}{8}-\frac {9 b \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )}{8}\right ) x^{\frac {3}{2}}+\left (-\frac {7 a^{2}}{8}-\frac {7 a \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )}{4}-\frac {7 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{8}\right ) \sqrt {x}}{\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{2}}+\frac {15 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right )}{8 \sqrt {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right )}{b^{3}}\) | \(153\) |
| default | \(\frac {2 \sqrt {x}}{b^{3}}-\frac {2 \left (\frac {\left (-\frac {9 a b}{8}-\frac {9 b \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )}{8}\right ) x^{\frac {3}{2}}+\left (-\frac {7 a^{2}}{8}-\frac {7 a \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )}{4}-\frac {7 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{8}\right ) \sqrt {x}}{\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{2}}+\frac {15 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right )}{8 \sqrt {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right )}{b^{3}}\) | \(153\) |
| risch | \(\text {Expression too large to display}\) | \(1354\) |
2*x^(1/2)/b^3-2/b^3*(((-9/8*a*b-9/8*b*(arctanh(tanh(b*x+a))-b*x-a))*x^(3/2 )+(-7/8*a^2-7/4*a*(arctanh(tanh(b*x+a))-b*x-a)-7/8*(arctanh(tanh(b*x+a))-b *x-a)^2)*x^(1/2))/arctanh(tanh(b*x+a))^2+15/8*(arctanh(tanh(b*x+a))-b*x)/( (arctanh(tanh(b*x+a))-b*x)*b)^(1/2)*arctan(b*x^(1/2)/((arctanh(tanh(b*x+a) )-b*x)*b)^(1/2)))
Time = 0.27 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.82 \[ \int \frac {x^{5/2}}{\text {arctanh}(\tanh (a+b x))^3} \, dx=\left [\frac {15 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x - 2 \, b \sqrt {x} \sqrt {-\frac {a}{b}} - a}{b x + a}\right ) + 2 \, {\left (8 \, b^{2} x^{2} + 25 \, a b x + 15 \, a^{2}\right )} \sqrt {x}}{8 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}}, -\frac {15 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b \sqrt {x} \sqrt {\frac {a}{b}}}{a}\right ) - {\left (8 \, b^{2} x^{2} + 25 \, a b x + 15 \, a^{2}\right )} \sqrt {x}}{4 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}}\right ] \]
[1/8*(15*(b^2*x^2 + 2*a*b*x + a^2)*sqrt(-a/b)*log((b*x - 2*b*sqrt(x)*sqrt( -a/b) - a)/(b*x + a)) + 2*(8*b^2*x^2 + 25*a*b*x + 15*a^2)*sqrt(x))/(b^5*x^ 2 + 2*a*b^4*x + a^2*b^3), -1/4*(15*(b^2*x^2 + 2*a*b*x + a^2)*sqrt(a/b)*arc tan(b*sqrt(x)*sqrt(a/b)/a) - (8*b^2*x^2 + 25*a*b*x + 15*a^2)*sqrt(x))/(b^5 *x^2 + 2*a*b^4*x + a^2*b^3)]
\[ \int \frac {x^{5/2}}{\text {arctanh}(\tanh (a+b x))^3} \, dx=\int \frac {x^{\frac {5}{2}}}{\operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]
Time = 0.30 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.66 \[ \int \frac {x^{5/2}}{\text {arctanh}(\tanh (a+b x))^3} \, dx=\frac {8 \, b^{2} x^{\frac {5}{2}} + 25 \, a b x^{\frac {3}{2}} + 15 \, a^{2} \sqrt {x}}{4 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}} - \frac {15 \, a \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} b^{3}} \]
1/4*(8*b^2*x^(5/2) + 25*a*b*x^(3/2) + 15*a^2*sqrt(x))/(b^5*x^2 + 2*a*b^4*x + a^2*b^3) - 15/4*a*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^3)
Time = 0.27 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.54 \[ \int \frac {x^{5/2}}{\text {arctanh}(\tanh (a+b x))^3} \, dx=-\frac {15 \, a \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} b^{3}} + \frac {2 \, \sqrt {x}}{b^{3}} + \frac {9 \, a b x^{\frac {3}{2}} + 7 \, a^{2} \sqrt {x}}{4 \, {\left (b x + a\right )}^{2} b^{3}} \]
-15/4*a*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^3) + 2*sqrt(x)/b^3 + 1/4* (9*a*b*x^(3/2) + 7*a^2*sqrt(x))/((b*x + a)^2*b^3)
Time = 4.68 (sec) , antiderivative size = 511, normalized size of antiderivative = 4.65 \[ \int \frac {x^{5/2}}{\text {arctanh}(\tanh (a+b x))^3} \, dx=\frac {2\,\sqrt {x}}{b^3}-\frac {9\,\sqrt {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 )}{4\,b^3\,\left (\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\right )}+\frac {15\,\sqrt {2}\,\ln \left (\frac {64\,b^{15/2}\,\left (\sqrt {2}\,\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 )-4\,\sqrt {b}\,\sqrt {x}\,\sqrt {\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}+2\,\sqrt {2}\,b\,x\right )}{\left (\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\right )\,\sqrt {\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 )\,\sqrt {\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}}{16\,b^{7/2}}-\frac {\sqrt {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 )}^2}{2\,b^3\,{\left (\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\right )}^2} \]
(2*x^(1/2))/b^3 - (9*x^(1/2)*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*ex p(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x))/(4*b^3*(log((2*exp (2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) - log(2/(exp(2*a)*exp(2*b*x) + 1)))) + (15*2^(1/2)*log((64*b^(15/2)*(2^(1/2)*(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) - 4*b^(1/2)*x^(1/2)*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*ex p(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^(1/2) + 2*2^(1/2)*b*x))/((lo g((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) - log(2/(exp(2*a)*exp (2*b*x) + 1)))*(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)^(1/2)))*(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) ^(1/2))/(16*b^(7/2)) - (x^(1/2)*(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)/(2*b^3*(log(( 2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) - log(2/(exp(2*a)*exp(2* b*x) + 1)))^2)