Integrand size = 17, antiderivative size = 166 \[ \int \frac {x^{7/2}}{\text {arctanh}(\tanh (a+b x))^{3/2}} \, dx=\frac {35 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\text {arctanh}(\tanh (a+b x))}}\right ) (b x-\text {arctanh}(\tanh (a+b x)))^3}{8 b^{9/2}}-\frac {2 x^{7/2}}{b \sqrt {\text {arctanh}(\tanh (a+b x))}}+\frac {7 x^{5/2} \sqrt {\text {arctanh}(\tanh (a+b x))}}{3 b^2}+\frac {35 x^{3/2} (b x-\text {arctanh}(\tanh (a+b x))) \sqrt {\text {arctanh}(\tanh (a+b x))}}{12 b^3}+\frac {35 \sqrt {x} (b x-\text {arctanh}(\tanh (a+b x)))^2 \sqrt {\text {arctanh}(\tanh (a+b x))}}{8 b^4} \]
35/8*arctanh(b^(1/2)*x^(1/2)/arctanh(tanh(b*x+a))^(1/2))*(b*x-arctanh(tanh (b*x+a)))^3/b^(9/2)-2*x^(7/2)/b/arctanh(tanh(b*x+a))^(1/2)+7/3*x^(5/2)*arc tanh(tanh(b*x+a))^(1/2)/b^2+35/12*x^(3/2)*(b*x-arctanh(tanh(b*x+a)))*arcta nh(tanh(b*x+a))^(1/2)/b^3+35/8*(b*x-arctanh(tanh(b*x+a)))^2*x^(1/2)*arctan h(tanh(b*x+a))^(1/2)/b^4
Time = 0.10 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.73 \[ \int \frac {x^{7/2}}{\text {arctanh}(\tanh (a+b x))^{3/2}} \, dx=\frac {\sqrt {x} \left (-48 b^3 x^3+231 b^2 x^2 \text {arctanh}(\tanh (a+b x))-280 b x \text {arctanh}(\tanh (a+b x))^2+105 \text {arctanh}(\tanh (a+b x))^3\right )}{24 b^4 \sqrt {\text {arctanh}(\tanh (a+b x))}}+\frac {35 (b x-\text {arctanh}(\tanh (a+b x)))^3 \log \left (b \sqrt {x}+\sqrt {b} \sqrt {\text {arctanh}(\tanh (a+b x))}\right )}{8 b^{9/2}} \]
(Sqrt[x]*(-48*b^3*x^3 + 231*b^2*x^2*ArcTanh[Tanh[a + b*x]] - 280*b*x*ArcTa nh[Tanh[a + b*x]]^2 + 105*ArcTanh[Tanh[a + b*x]]^3))/(24*b^4*Sqrt[ArcTanh[ Tanh[a + b*x]]]) + (35*(b*x - ArcTanh[Tanh[a + b*x]])^3*Log[b*Sqrt[x] + Sq rt[b]*Sqrt[ArcTanh[Tanh[a + b*x]]]])/(8*b^(9/2))
Time = 0.41 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.07, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {2599, 2600, 2600, 2600, 2596}
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.
\(\displaystyle \int \frac {x^{7/2}}{\text {arctanh}(\tanh (a+b x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {7 \int \frac {x^{5/2}}{\sqrt {\text {arctanh}(\tanh (a+b x))}}dx}{b}-\frac {2 x^{7/2}}{b \sqrt {\text {arctanh}(\tanh (a+b x))}}\) |
\(\Big \downarrow \) 2600 |
\(\displaystyle \frac {7 \left (\frac {5 (b x-\text {arctanh}(\tanh (a+b x))) \int \frac {x^{3/2}}{\sqrt {\text {arctanh}(\tanh (a+b x))}}dx}{6 b}+\frac {x^{5/2} \sqrt {\text {arctanh}(\tanh (a+b x))}}{3 b}\right )}{b}-\frac {2 x^{7/2}}{b \sqrt {\text {arctanh}(\tanh (a+b x))}}\) |
\(\Big \downarrow \) 2600 |
\(\displaystyle \frac {7 \left (\frac {5 (b x-\text {arctanh}(\tanh (a+b x))) \left (\frac {3 (b x-\text {arctanh}(\tanh (a+b x))) \int \frac {\sqrt {x}}{\sqrt {\text {arctanh}(\tanh (a+b x))}}dx}{4 b}+\frac {x^{3/2} \sqrt {\text {arctanh}(\tanh (a+b x))}}{2 b}\right )}{6 b}+\frac {x^{5/2} \sqrt {\text {arctanh}(\tanh (a+b x))}}{3 b}\right )}{b}-\frac {2 x^{7/2}}{b \sqrt {\text {arctanh}(\tanh (a+b x))}}\) |
\(\Big \downarrow \) 2600 |
\(\displaystyle \frac {7 \left (\frac {5 (b x-\text {arctanh}(\tanh (a+b x))) \left (\frac {3 (b x-\text {arctanh}(\tanh (a+b x))) \left (\frac {(b x-\text {arctanh}(\tanh (a+b x))) \int \frac {1}{\sqrt {x} \sqrt {\text {arctanh}(\tanh (a+b x))}}dx}{2 b}+\frac {\sqrt {x} \sqrt {\text {arctanh}(\tanh (a+b x))}}{b}\right )}{4 b}+\frac {x^{3/2} \sqrt {\text {arctanh}(\tanh (a+b x))}}{2 b}\right )}{6 b}+\frac {x^{5/2} \sqrt {\text {arctanh}(\tanh (a+b x))}}{3 b}\right )}{b}-\frac {2 x^{7/2}}{b \sqrt {\text {arctanh}(\tanh (a+b x))}}\) |
\(\Big \downarrow \) 2596 |
\(\displaystyle \frac {7 \left (\frac {5 \left (\frac {3 \left (\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\text {arctanh}(\tanh (a+b x))}}\right ) (b x-\text {arctanh}(\tanh (a+b x)))}{b^{3/2}}+\frac {\sqrt {x} \sqrt {\text {arctanh}(\tanh (a+b x))}}{b}\right ) (b x-\text {arctanh}(\tanh (a+b x)))}{4 b}+\frac {x^{3/2} \sqrt {\text {arctanh}(\tanh (a+b x))}}{2 b}\right ) (b x-\text {arctanh}(\tanh (a+b x)))}{6 b}+\frac {x^{5/2} \sqrt {\text {arctanh}(\tanh (a+b x))}}{3 b}\right )}{b}-\frac {2 x^{7/2}}{b \sqrt {\text {arctanh}(\tanh (a+b x))}}\) |
(7*((5*((3*((ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[ArcTanh[Tanh[a + b*x]]]]*(b*x - ArcTanh[Tanh[a + b*x]]))/b^(3/2) + (Sqrt[x]*Sqrt[ArcTanh[Tanh[a + b*x]]] )/b)*(b*x - ArcTanh[Tanh[a + b*x]]))/(4*b) + (x^(3/2)*Sqrt[ArcTanh[Tanh[a + b*x]]])/(2*b))*(b*x - ArcTanh[Tanh[a + b*x]]))/(6*b) + (x^(5/2)*Sqrt[Arc Tanh[Tanh[a + b*x]]])/(3*b)))/b - (2*x^(7/2))/(b*Sqrt[ArcTanh[Tanh[a + b*x ]]])
3.3.49.3.1 Defintions of rubi rules used
Int[1/(Sqrt[u_]*Sqrt[v_]), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Si mplify[D[v, x]]}, Simp[(2/Rt[a*b, 2])*ArcTanh[Rt[a*b, 2]*(Sqrt[u]/(a*Sqrt[v ]))], x] /; NeQ[b*u - a*v, 0] && PosQ[a*b]] /; PiecewiseLinearQ[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]))
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 + n + 1))), x] - Simp[n*((b*u - a*v)/(a*(m + n + 1))) Int[u^m*v^(n - 1), x], x] /; NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x] && NeQ[m + n + 2, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !IntegerQ[n] || LtQ[0, m, n])) && !ILtQ[m + n, -2]
Time = 0.18 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.88
method | result | size |
derivativedivides | \(\frac {x^{\frac {7}{2}}}{3 b \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}-\frac {7 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \left (\frac {x^{\frac {5}{2}}}{4 b \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}-\frac {5 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \left (\frac {x^{\frac {3}{2}}}{2 b \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}-\frac {3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \left (-\frac {\sqrt {x}}{b \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}+\frac {\ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}\right )}{b^{\frac {3}{2}}}\right )}{2 b}\right )}{4 b}\right )}{3 b}\) | \(146\) |
default | \(\frac {x^{\frac {7}{2}}}{3 b \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}-\frac {7 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \left (\frac {x^{\frac {5}{2}}}{4 b \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}-\frac {5 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \left (\frac {x^{\frac {3}{2}}}{2 b \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}-\frac {3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \left (-\frac {\sqrt {x}}{b \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}+\frac {\ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}\right )}{b^{\frac {3}{2}}}\right )}{2 b}\right )}{4 b}\right )}{3 b}\) | \(146\) |
1/3*x^(7/2)/b/arctanh(tanh(b*x+a))^(1/2)-7/3*(arctanh(tanh(b*x+a))-b*x)/b* (1/4*x^(5/2)/b/arctanh(tanh(b*x+a))^(1/2)-5/4*(arctanh(tanh(b*x+a))-b*x)/b *(1/2*x^(3/2)/b/arctanh(tanh(b*x+a))^(1/2)-3/2*(arctanh(tanh(b*x+a))-b*x)/ b*(-x^(1/2)/b/arctanh(tanh(b*x+a))^(1/2)+1/b^(3/2)*ln(b^(1/2)*x^(1/2)+arct anh(tanh(b*x+a))^(1/2)))))
Time = 0.26 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.18 \[ \int \frac {x^{7/2}}{\text {arctanh}(\tanh (a+b x))^{3/2}} \, dx=\left [\frac {105 \, {\left (a^{3} b x + a^{4}\right )} \sqrt {b} \log \left (2 \, b x - 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) + 2 \, {\left (8 \, b^{4} x^{3} - 14 \, a b^{3} x^{2} + 35 \, a^{2} b^{2} x + 105 \, a^{3} b\right )} \sqrt {b x + a} \sqrt {x}}{48 \, {\left (b^{6} x + a b^{5}\right )}}, \frac {105 \, {\left (a^{3} b x + a^{4}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) + {\left (8 \, b^{4} x^{3} - 14 \, a b^{3} x^{2} + 35 \, a^{2} b^{2} x + 105 \, a^{3} b\right )} \sqrt {b x + a} \sqrt {x}}{24 \, {\left (b^{6} x + a b^{5}\right )}}\right ] \]
[1/48*(105*(a^3*b*x + a^4)*sqrt(b)*log(2*b*x - 2*sqrt(b*x + a)*sqrt(b)*sqr t(x) + a) + 2*(8*b^4*x^3 - 14*a*b^3*x^2 + 35*a^2*b^2*x + 105*a^3*b)*sqrt(b *x + a)*sqrt(x))/(b^6*x + a*b^5), 1/24*(105*(a^3*b*x + a^4)*sqrt(-b)*arcta n(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) + (8*b^4*x^3 - 14*a*b^3*x^2 + 35*a^2 *b^2*x + 105*a^3*b)*sqrt(b*x + a)*sqrt(x))/(b^6*x + a*b^5)]
Timed out. \[ \int \frac {x^{7/2}}{\text {arctanh}(\tanh (a+b x))^{3/2}} \, dx=\text {Timed out} \]
\[ \int \frac {x^{7/2}}{\text {arctanh}(\tanh (a+b x))^{3/2}} \, dx=\int { \frac {x^{\frac {7}{2}}}{\operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{\frac {3}{2}}} \,d x } \]
Time = 0.30 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.45 \[ \int \frac {x^{7/2}}{\text {arctanh}(\tanh (a+b x))^{3/2}} \, dx=\frac {{\left ({\left (2 \, x {\left (\frac {4 \, x}{b} - \frac {7 \, a}{b^{2}}\right )} + \frac {35 \, a^{2}}{b^{3}}\right )} x + \frac {105 \, a^{3}}{b^{4}}\right )} \sqrt {x}}{24 \, \sqrt {b x + a}} + \frac {35 \, a^{3} \log \left ({\left | -\sqrt {b} \sqrt {x} + \sqrt {b x + a} \right |}\right )}{8 \, b^{\frac {9}{2}}} \]
1/24*((2*x*(4*x/b - 7*a/b^2) + 35*a^2/b^3)*x + 105*a^3/b^4)*sqrt(x)/sqrt(b *x + a) + 35/8*a^3*log(abs(-sqrt(b)*sqrt(x) + sqrt(b*x + a)))/b^(9/2)
Timed out. \[ \int \frac {x^{7/2}}{\text {arctanh}(\tanh (a+b x))^{3/2}} \, dx=\int \frac {x^{7/2}}{{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^{3/2}} \,d x \]