Integrand size = 15, antiderivative size = 128 \[ \int \frac {1}{x^{7/2} \text {arctanh}(\tanh (a+b x))} \, dx=-\frac {2 b^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\text {arctanh}(\tanh (a+b x))}}\right )}{(b x-\text {arctanh}(\tanh (a+b x)))^{7/2}}+\frac {2 b^2}{\sqrt {x} (b x-\text {arctanh}(\tanh (a+b x)))^3}+\frac {2 b}{3 x^{3/2} (b x-\text {arctanh}(\tanh (a+b x)))^2}+\frac {2}{5 x^{5/2} (b x-\text {arctanh}(\tanh (a+b x)))} \]
-2*b^(5/2)*arctanh(b^(1/2)*x^(1/2)/(b*x-arctanh(tanh(b*x+a)))^(1/2))/(b*x- arctanh(tanh(b*x+a)))^(7/2)+2/3*b/x^(3/2)/(b*x-arctanh(tanh(b*x+a)))^2+2/5 /x^(5/2)/(b*x-arctanh(tanh(b*x+a)))+2*b^2/(b*x-arctanh(tanh(b*x+a)))^3/x^( 1/2)
Time = 0.12 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.84 \[ \int \frac {1}{x^{7/2} \text {arctanh}(\tanh (a+b x))} \, dx=-\frac {2 b^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {-b x+\text {arctanh}(\tanh (a+b x))}}\right )}{(-b x+\text {arctanh}(\tanh (a+b x)))^{7/2}}+\frac {2 \left (23 b^2 x^2-11 b x \text {arctanh}(\tanh (a+b x))+3 \text {arctanh}(\tanh (a+b x))^2\right )}{15 x^{5/2} (b x-\text {arctanh}(\tanh (a+b x)))^3} \]
(-2*b^(5/2)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[-(b*x) + ArcTanh[Tanh[a + b*x]]] ])/(-(b*x) + ArcTanh[Tanh[a + b*x]])^(7/2) + (2*(23*b^2*x^2 - 11*b*x*ArcTa nh[Tanh[a + b*x]] + 3*ArcTanh[Tanh[a + b*x]]^2))/(15*x^(5/2)*(b*x - ArcTan h[Tanh[a + b*x]])^3)
Time = 0.36 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.25, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {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.
\(\displaystyle \int \frac {1}{x^{7/2} \text {arctanh}(\tanh (a+b x))} \, dx\) |
\(\Big \downarrow \) 2594 |
\(\displaystyle \frac {b \int \frac {1}{x^{5/2} \text {arctanh}(\tanh (a+b x))}dx}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {2}{5 x^{5/2} (b x-\text {arctanh}(\tanh (a+b x)))}\) |
\(\Big \downarrow \) 2594 |
\(\displaystyle \frac {b \left (\frac {b \int \frac {1}{x^{3/2} \text {arctanh}(\tanh (a+b x))}dx}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {2}{3 x^{3/2} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {2}{5 x^{5/2} (b x-\text {arctanh}(\tanh (a+b x)))}\) |
\(\Big \downarrow \) 2594 |
\(\displaystyle \frac {b \left (\frac {b \left (\frac {b \int \frac {1}{\sqrt {x} \text {arctanh}(\tanh (a+b x))}dx}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {2}{\sqrt {x} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {2}{3 x^{3/2} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {2}{5 x^{5/2} (b x-\text {arctanh}(\tanh (a+b x)))}\) |
\(\Big \downarrow \) 2593 |
\(\displaystyle \frac {b \left (\frac {2}{3 x^{3/2} (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {b \left (\frac {2}{\sqrt {x} (b x-\text {arctanh}(\tanh (a+b x)))}-\frac {2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\text {arctanh}(\tanh (a+b x))}}\right )}{(b x-\text {arctanh}(\tanh (a+b x)))^{3/2}}\right )}{b x-\text {arctanh}(\tanh (a+b x))}\right )}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {2}{5 x^{5/2} (b x-\text {arctanh}(\tanh (a+b x)))}\) |
2/(5*x^(5/2)*(b*x - ArcTanh[Tanh[a + b*x]])) + (b*(2/(3*x^(3/2)*(b*x - Arc Tanh[Tanh[a + b*x]])) + (b*((-2*Sqrt[b]*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[b*x - ArcTanh[Tanh[a + b*x]]]])/(b*x - ArcTanh[Tanh[a + b*x]])^(3/2) + 2/(Sqr t[x]*(b*x - ArcTanh[Tanh[a + b*x]]))))/(b*x - ArcTanh[Tanh[a + b*x]])))/(b *x - ArcTanh[Tanh[a + b*x]])
3.2.98.3.1 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]
Time = 0.11 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.94
method | result | size |
derivativedivides | \(-\frac {2 b^{3} \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right )}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{3} \sqrt {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) b}}-\frac {2}{5 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {5}{2}}}-\frac {2 b^{2}}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{3} \sqrt {x}}+\frac {2 b}{3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{2} x^{\frac {3}{2}}}\) | \(120\) |
default | \(-\frac {2 b^{3} \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right )}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{3} \sqrt {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) b}}-\frac {2}{5 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {5}{2}}}-\frac {2 b^{2}}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{3} \sqrt {x}}+\frac {2 b}{3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{2} x^{\frac {3}{2}}}\) | \(120\) |
-2*b^3/(arctanh(tanh(b*x+a))-b*x)^3/((arctanh(tanh(b*x+a))-b*x)*b)^(1/2)*a rctan(b*x^(1/2)/((arctanh(tanh(b*x+a))-b*x)*b)^(1/2))-2/5/(arctanh(tanh(b* x+a))-b*x)/x^(5/2)-2/(arctanh(tanh(b*x+a))-b*x)^3*b^2/x^(1/2)+2/3/(arctanh (tanh(b*x+a))-b*x)^2*b/x^(3/2)
Time = 0.25 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.12 \[ \int \frac {1}{x^{7/2} \text {arctanh}(\tanh (a+b x))} \, dx=\left [\frac {15 \, b^{2} x^{3} \sqrt {-\frac {b}{a}} \log \left (\frac {b x - 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - a}{b x + a}\right ) - 2 \, {\left (15 \, b^{2} x^{2} - 5 \, a b x + 3 \, a^{2}\right )} \sqrt {x}}{15 \, a^{3} x^{3}}, \frac {2 \, {\left (15 \, b^{2} x^{3} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}}}{b \sqrt {x}}\right ) - {\left (15 \, b^{2} x^{2} - 5 \, a b x + 3 \, a^{2}\right )} \sqrt {x}\right )}}{15 \, a^{3} x^{3}}\right ] \]
[1/15*(15*b^2*x^3*sqrt(-b/a)*log((b*x - 2*a*sqrt(x)*sqrt(-b/a) - a)/(b*x + a)) - 2*(15*b^2*x^2 - 5*a*b*x + 3*a^2)*sqrt(x))/(a^3*x^3), 2/15*(15*b^2*x ^3*sqrt(b/a)*arctan(a*sqrt(b/a)/(b*sqrt(x))) - (15*b^2*x^2 - 5*a*b*x + 3*a ^2)*sqrt(x))/(a^3*x^3)]
\[ \int \frac {1}{x^{7/2} \text {arctanh}(\tanh (a+b x))} \, dx=\int \frac {1}{x^{\frac {7}{2}} \operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]
Time = 0.29 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.41 \[ \int \frac {1}{x^{7/2} \text {arctanh}(\tanh (a+b x))} \, dx=-\frac {2 \, b^{3} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3}} - \frac {2 \, {\left (15 \, b^{2} x^{2} - 5 \, a b x + 3 \, a^{2}\right )}}{15 \, a^{3} x^{\frac {5}{2}}} \]
-2*b^3*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^3) - 2/15*(15*b^2*x^2 - 5* a*b*x + 3*a^2)/(a^3*x^(5/2))
Time = 0.27 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.41 \[ \int \frac {1}{x^{7/2} \text {arctanh}(\tanh (a+b x))} \, dx=-\frac {2 \, b^{3} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3}} - \frac {2 \, {\left (15 \, b^{2} x^{2} - 5 \, a b x + 3 \, a^{2}\right )}}{15 \, a^{3} x^{\frac {5}{2}}} \]
-2*b^3*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^3) - 2/15*(15*b^2*x^2 - 5* a*b*x + 3*a^2)/(a^3*x^(5/2))
Time = 4.38 (sec) , antiderivative size = 822, normalized size of antiderivative = 6.42 \[ \int \frac {1}{x^{7/2} \text {arctanh}(\tanh (a+b x))} \, dx=\text {Too large to display} \]
4/(5*x^(5/2)*(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)) + (8*b)/(3*x^(3/2)*(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) + (16*b^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)^3) + (8*2^(1/2 )*b^(5/2)*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/(ex p(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)) - lo g((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^(1/2) + 2*2^ (1/2)*b*x)*((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)^6 + 60*a^2*(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)^4 - 160*a^3*(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)^3 + 240*a^4*(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)^2 + 64*a^6 - 12*a*(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)^5 - 192*a^5*(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)))/(2*(log((2*exp(2*a)*ex...