Integrand size = 15, antiderivative size = 201 \[ \int \frac {1}{x^{5/2} \text {arctanh}(\tanh (a+b x))^3} \, dx=-\frac {35 b^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\text {arctanh}(\tanh (a+b x))}}\right )}{4 (b x-\text {arctanh}(\tanh (a+b x)))^{9/2}}+\frac {35 b}{4 \sqrt {x} (b x-\text {arctanh}(\tanh (a+b x)))^4}+\frac {35}{12 x^{3/2} (b x-\text {arctanh}(\tanh (a+b x)))^3}+\frac {7}{4 b x^{5/2} (b x-\text {arctanh}(\tanh (a+b x)))^2}+\frac {5}{4 b^2 x^{7/2} (b x-\text {arctanh}(\tanh (a+b x)))}-\frac {1}{2 b x^{5/2} \text {arctanh}(\tanh (a+b x))^2}+\frac {5}{4 b^2 x^{7/2} \text {arctanh}(\tanh (a+b x))} \]
-35/4*b^(3/2)*arctanh(b^(1/2)*x^(1/2)/(b*x-arctanh(tanh(b*x+a)))^(1/2))/(b *x-arctanh(tanh(b*x+a)))^(9/2)+35/12/x^(3/2)/(b*x-arctanh(tanh(b*x+a)))^3+ 7/4/b/x^(5/2)/(b*x-arctanh(tanh(b*x+a)))^2+5/4/b^2/x^(7/2)/(b*x-arctanh(ta nh(b*x+a)))-1/2/b/x^(5/2)/arctanh(tanh(b*x+a))^2+5/4/b^2/x^(7/2)/arctanh(t anh(b*x+a))+35/4*b/(b*x-arctanh(tanh(b*x+a)))^4/x^(1/2)
Time = 0.20 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.78 \[ \int \frac {1}{x^{5/2} \text {arctanh}(\tanh (a+b x))^3} \, dx=\frac {1}{12} \left (\frac {105 b^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {-b x+\text {arctanh}(\tanh (a+b x))}}\right )}{(-b x+\text {arctanh}(\tanh (a+b x)))^{9/2}}+\frac {80 b x-8 \text {arctanh}(\tanh (a+b x))}{x^{3/2} (-b x+\text {arctanh}(\tanh (a+b x)))^4}+\frac {33 b^2 \sqrt {x}}{\text {arctanh}(\tanh (a+b x)) (-b x+\text {arctanh}(\tanh (a+b x)))^4}+\frac {6 b^2 \sqrt {x}}{\text {arctanh}(\tanh (a+b x))^2 (-b x+\text {arctanh}(\tanh (a+b x)))^3}\right ) \]
((105*b^(3/2)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[-(b*x) + ArcTanh[Tanh[a + b*x] ]]])/(-(b*x) + ArcTanh[Tanh[a + b*x]])^(9/2) + (80*b*x - 8*ArcTanh[Tanh[a + b*x]])/(x^(3/2)*(-(b*x) + ArcTanh[Tanh[a + b*x]])^4) + (33*b^2*Sqrt[x])/ (ArcTanh[Tanh[a + b*x]]*(-(b*x) + ArcTanh[Tanh[a + b*x]])^4) + (6*b^2*Sqrt [x])/(ArcTanh[Tanh[a + b*x]]^2*(-(b*x) + ArcTanh[Tanh[a + b*x]])^3))/12
Time = 0.57 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.28, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {2599, 2599, 2594, 2594, 2594, 2594, 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.
\(\displaystyle \int \frac {1}{x^{5/2} \text {arctanh}(\tanh (a+b x))^3} \, dx\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle -\frac {5 \int \frac {1}{x^{7/2} \text {arctanh}(\tanh (a+b x))^2}dx}{4 b}-\frac {1}{2 b x^{5/2} \text {arctanh}(\tanh (a+b x))^2}\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle -\frac {5 \left (-\frac {7 \int \frac {1}{x^{9/2} \text {arctanh}(\tanh (a+b x))}dx}{2 b}-\frac {1}{b x^{7/2} \text {arctanh}(\tanh (a+b x))}\right )}{4 b}-\frac {1}{2 b x^{5/2} \text {arctanh}(\tanh (a+b x))^2}\) |
\(\Big \downarrow \) 2594 |
\(\displaystyle -\frac {5 \left (-\frac {7 \left (\frac {b \int \frac {1}{x^{7/2} \text {arctanh}(\tanh (a+b x))}dx}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {2}{7 x^{7/2} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{2 b}-\frac {1}{b x^{7/2} \text {arctanh}(\tanh (a+b x))}\right )}{4 b}-\frac {1}{2 b x^{5/2} \text {arctanh}(\tanh (a+b x))^2}\) |
\(\Big \downarrow \) 2594 |
\(\displaystyle -\frac {5 \left (-\frac {7 \left (\frac {b \left (\frac {b \int \frac {1}{x^{5/2} \text {arctanh}(\tanh (a+b x))}dx}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {2}{5 x^{5/2} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {2}{7 x^{7/2} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{2 b}-\frac {1}{b x^{7/2} \text {arctanh}(\tanh (a+b x))}\right )}{4 b}-\frac {1}{2 b x^{5/2} \text {arctanh}(\tanh (a+b x))^2}\) |
\(\Big \downarrow \) 2594 |
\(\displaystyle -\frac {5 \left (-\frac {7 \left (\frac {b \left (\frac {b \left (\frac {b \int \frac {1}{x^{3/2} \text {arctanh}(\tanh (a+b x))}dx}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {2}{3 x^{3/2} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {2}{5 x^{5/2} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {2}{7 x^{7/2} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{2 b}-\frac {1}{b x^{7/2} \text {arctanh}(\tanh (a+b x))}\right )}{4 b}-\frac {1}{2 b x^{5/2} \text {arctanh}(\tanh (a+b x))^2}\) |
\(\Big \downarrow \) 2594 |
\(\displaystyle -\frac {5 \left (-\frac {7 \left (\frac {b \left (\frac {b \left (\frac {b \left (\frac {b \int \frac {1}{\sqrt {x} \text {arctanh}(\tanh (a+b x))}dx}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {2}{\sqrt {x} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {2}{3 x^{3/2} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {2}{5 x^{5/2} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {2}{7 x^{7/2} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{2 b}-\frac {1}{b x^{7/2} \text {arctanh}(\tanh (a+b x))}\right )}{4 b}-\frac {1}{2 b x^{5/2} \text {arctanh}(\tanh (a+b x))^2}\) |
\(\Big \downarrow \) 2593 |
\(\displaystyle -\frac {5 \left (-\frac {7 \left (\frac {b \left (\frac {b \left (\frac {2}{3 x^{3/2} (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {b \left (\frac {2}{\sqrt {x} (b x-\text {arctanh}(\tanh (a+b x)))}-\frac {2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\text {arctanh}(\tanh (a+b x))}}\right )}{(b x-\text {arctanh}(\tanh (a+b x)))^{3/2}}\right )}{b x-\text {arctanh}(\tanh (a+b x))}\right )}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {2}{5 x^{5/2} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {2}{7 x^{7/2} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{2 b}-\frac {1}{b x^{7/2} \text {arctanh}(\tanh (a+b x))}\right )}{4 b}-\frac {1}{2 b x^{5/2} \text {arctanh}(\tanh (a+b x))^2}\) |
(-5*((-7*(2/(7*x^(7/2)*(b*x - ArcTanh[Tanh[a + b*x]])) + (b*(2/(5*x^(5/2)* (b*x - ArcTanh[Tanh[a + b*x]])) + (b*(2/(3*x^(3/2)*(b*x - ArcTanh[Tanh[a + b*x]])) + (b*((-2*Sqrt[b]*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[b*x - ArcTanh[Ta nh[a + b*x]]]])/(b*x - ArcTanh[Tanh[a + b*x]])^(3/2) + 2/(Sqrt[x]*(b*x - A rcTanh[Tanh[a + b*x]]))))/(b*x - ArcTanh[Tanh[a + b*x]])))/(b*x - ArcTanh[ Tanh[a + b*x]])))/(b*x - ArcTanh[Tanh[a + b*x]])))/(2*b) - 1/(b*x^(7/2)*Ar cTanh[Tanh[a + b*x]])))/(4*b) - 1/(2*b*x^(5/2)*ArcTanh[Tanh[a + b*x]]^2)
3.3.13.3.1 Defintions of rubi rules used
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[(v_)^(n_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[ D[v, x]]}, Simp[v^(n + 1)/((n + 1)*(b*u - a*v)), x] - Simp[a*((n + 1)/((n + 1)*(b*u - a*v))) Int[v^(n + 1)/u, x], x] /; NeQ[b*u - a*v, 0]] /; Piecew iseLinearQ[u, v, x] && LtQ[n, -1]
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 = 0.73 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.68
method | result | size |
derivativedivides | \(-\frac {2}{3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{3} x^{\frac {3}{2}}}+\frac {6 b}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{4} \sqrt {x}}+\frac {2 b^{2} \left (\frac {\frac {11 b \,x^{\frac {3}{2}}}{8}+\left (\frac {13 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}{8}-\frac {13 b x}{8}\right ) \sqrt {x}}{\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{2}}+\frac {35 \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right )}{8 \sqrt {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right )}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{4}}\) | \(136\) |
default | \(-\frac {2}{3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{3} x^{\frac {3}{2}}}+\frac {6 b}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{4} \sqrt {x}}+\frac {2 b^{2} \left (\frac {\frac {11 b \,x^{\frac {3}{2}}}{8}+\left (\frac {13 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}{8}-\frac {13 b x}{8}\right ) \sqrt {x}}{\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{2}}+\frac {35 \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right )}{8 \sqrt {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right )}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{4}}\) | \(136\) |
risch | \(\text {Expression too large to display}\) | \(12319\) |
-2/3/(arctanh(tanh(b*x+a))-b*x)^3/x^(3/2)+6/(arctanh(tanh(b*x+a))-b*x)^4*b /x^(1/2)+2/(arctanh(tanh(b*x+a))-b*x)^4*b^2*((11/8*b*x^(3/2)+(13/8*arctanh (tanh(b*x+a))-13/8*b*x)*x^(1/2))/arctanh(tanh(b*x+a))^2+35/8/((arctanh(tan h(b*x+a))-b*x)*b)^(1/2)*arctan(b*x^(1/2)/((arctanh(tanh(b*x+a))-b*x)*b)^(1 /2)))
Time = 0.26 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.24 \[ \int \frac {1}{x^{5/2} \text {arctanh}(\tanh (a+b x))^3} \, dx=\left [\frac {105 \, {\left (b^{3} x^{4} + 2 \, a b^{2} x^{3} + a^{2} b x^{2}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x + 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - a}{b x + a}\right ) + 2 \, {\left (105 \, b^{3} x^{3} + 175 \, a b^{2} x^{2} + 56 \, a^{2} b x - 8 \, a^{3}\right )} \sqrt {x}}{24 \, {\left (a^{4} b^{2} x^{4} + 2 \, a^{5} b x^{3} + a^{6} x^{2}\right )}}, -\frac {105 \, {\left (b^{3} x^{4} + 2 \, a b^{2} x^{3} + a^{2} b x^{2}\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}}}{b \sqrt {x}}\right ) - {\left (105 \, b^{3} x^{3} + 175 \, a b^{2} x^{2} + 56 \, a^{2} b x - 8 \, a^{3}\right )} \sqrt {x}}{12 \, {\left (a^{4} b^{2} x^{4} + 2 \, a^{5} b x^{3} + a^{6} x^{2}\right )}}\right ] \]
[1/24*(105*(b^3*x^4 + 2*a*b^2*x^3 + a^2*b*x^2)*sqrt(-b/a)*log((b*x + 2*a*s qrt(x)*sqrt(-b/a) - a)/(b*x + a)) + 2*(105*b^3*x^3 + 175*a*b^2*x^2 + 56*a^ 2*b*x - 8*a^3)*sqrt(x))/(a^4*b^2*x^4 + 2*a^5*b*x^3 + a^6*x^2), -1/12*(105* (b^3*x^4 + 2*a*b^2*x^3 + a^2*b*x^2)*sqrt(b/a)*arctan(a*sqrt(b/a)/(b*sqrt(x ))) - (105*b^3*x^3 + 175*a*b^2*x^2 + 56*a^2*b*x - 8*a^3)*sqrt(x))/(a^4*b^2 *x^4 + 2*a^5*b*x^3 + a^6*x^2)]
\[ \int \frac {1}{x^{5/2} \text {arctanh}(\tanh (a+b x))^3} \, dx=\int \frac {1}{x^{\frac {5}{2}} \operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]
Time = 0.31 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.43 \[ \int \frac {1}{x^{5/2} \text {arctanh}(\tanh (a+b x))^3} \, dx=\frac {105 \, b^{3} x^{3} + 175 \, a b^{2} x^{2} + 56 \, a^{2} b x - 8 \, a^{3}}{12 \, {\left (a^{4} b^{2} x^{\frac {7}{2}} + 2 \, a^{5} b x^{\frac {5}{2}} + a^{6} x^{\frac {3}{2}}\right )}} + \frac {35 \, b^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a^{4}} \]
1/12*(105*b^3*x^3 + 175*a*b^2*x^2 + 56*a^2*b*x - 8*a^3)/(a^4*b^2*x^(7/2) + 2*a^5*b*x^(5/2) + a^6*x^(3/2)) + 35/4*b^2*arctan(b*sqrt(x)/sqrt(a*b))/(sq rt(a*b)*a^4)
Time = 0.27 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.35 \[ \int \frac {1}{x^{5/2} \text {arctanh}(\tanh (a+b x))^3} \, dx=\frac {35 \, b^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a^{4}} + \frac {2 \, {\left (9 \, b x - a\right )}}{3 \, a^{4} x^{\frac {3}{2}}} + \frac {11 \, b^{3} x^{\frac {3}{2}} + 13 \, a b^{2} \sqrt {x}}{4 \, {\left (b x + a\right )}^{2} a^{4}} \]
35/4*b^2*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^4) + 2/3*(9*b*x - a)/(a^ 4*x^(3/2)) + 1/4*(11*b^3*x^(3/2) + 13*a*b^2*sqrt(x))/((b*x + a)^2*a^4)
Time = 5.99 (sec) , antiderivative size = 1362, normalized size of antiderivative = 6.78 \[ \int \frac {1}{x^{5/2} \text {arctanh}(\tanh (a+b x))^3} \, dx=\text {Too large to display} \]
(x^(1/2)*((2*(2*b*(3*log(2/(exp(2*a)*exp(2*b*x) + 1)) - 3*log((2*exp(2*a)* exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 6*b*x) - 14*b*(log(2/(exp(2*a)*ex p(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*(2*a*b - b*(2*a - 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*b*x))*(log(2/(exp(2*a)*e xp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)) + (56*b^2*x)/(3*(2*a*b - b*(2*a - log((2*exp(2*a)*exp(2*b*x))/(ex p(2*a)*exp(2*b*x) + 1)) + log(2/(exp(2*a)*exp(2*b*x) + 1)) + 2*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))))/(2*b*x^2 - x*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - l og((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x))^2 - (x^(1/ 2)*((280*b)/(3*(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) - (280*b^2*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)^4))/(2*b*x^2 - 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)) + (70*2^(1/2)*b^(3/2 )*log((b^(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)*(2^(1/2)*(log(2/(exp(2*a)*e xp(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*...