3.191 \(\int \frac{x^{7/2}}{\tanh ^{-1}(\tanh (a+b x))} \, dx\)

Optimal. Leaf size=143 \[ \frac{2 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{5 b^2}+\frac{2 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{3 b^3}+\frac{2 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}{b^4}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{7/2}}{b^{9/2}}+\frac{2 x^{7/2}}{7 b} \]

[Out]

(2*x^(7/2))/(7*b) + (2*x^(5/2)*(b*x - ArcTanh[Tanh[a + b*x]]))/(5*b^2) + (2*x^(3/2)*(b*x - ArcTanh[Tanh[a + b*
x]])^2)/(3*b^3) + (2*Sqrt[x]*(b*x - ArcTanh[Tanh[a + b*x]])^3)/b^4 - (2*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[b*x - A
rcTanh[Tanh[a + b*x]]]]*(b*x - ArcTanh[Tanh[a + b*x]])^(7/2))/b^(9/2)

________________________________________________________________________________________

Rubi [A]  time = 0.127289, antiderivative size = 143, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {2159, 2162} \[ \frac{2 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{5 b^2}+\frac{2 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{3 b^3}+\frac{2 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}{b^4}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{7/2}}{b^{9/2}}+\frac{2 x^{7/2}}{7 b} \]

Antiderivative was successfully verified.

[In]

Int[x^(7/2)/ArcTanh[Tanh[a + b*x]],x]

[Out]

(2*x^(7/2))/(7*b) + (2*x^(5/2)*(b*x - ArcTanh[Tanh[a + b*x]]))/(5*b^2) + (2*x^(3/2)*(b*x - ArcTanh[Tanh[a + b*
x]])^2)/(3*b^3) + (2*Sqrt[x]*(b*x - ArcTanh[Tanh[a + b*x]])^3)/b^4 - (2*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[b*x - A
rcTanh[Tanh[a + b*x]]]]*(b*x - ArcTanh[Tanh[a + b*x]])^(7/2))/b^(9/2)

Rule 2159

Int[(v_)^(n_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[v^n/(a*n), x] - Dis
t[(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] && Ne
Q[n, 1]

Rule 2162

Int[1/((u_)*Sqrt[v_]), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[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]] /;
PiecewiseLinearQ[u, v, x]

Rubi steps

\begin{align*} \int \frac{x^{7/2}}{\tanh ^{-1}(\tanh (a+b x))} \, dx &=\frac{2 x^{7/2}}{7 b}-\frac{\left (-b x+\tanh ^{-1}(\tanh (a+b x))\right ) \int \frac{x^{5/2}}{\tanh ^{-1}(\tanh (a+b x))} \, dx}{b}\\ &=\frac{2 x^{7/2}}{7 b}+\frac{2 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{5 b^2}+\frac{\left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^2 \int \frac{x^{3/2}}{\tanh ^{-1}(\tanh (a+b x))} \, dx}{b^2}\\ &=\frac{2 x^{7/2}}{7 b}+\frac{2 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{5 b^2}+\frac{2 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{3 b^3}-\frac{\left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^3 \int \frac{\sqrt{x}}{\tanh ^{-1}(\tanh (a+b x))} \, dx}{b^3}\\ &=\frac{2 x^{7/2}}{7 b}+\frac{2 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{5 b^2}+\frac{2 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{3 b^3}+\frac{2 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}{b^4}+\frac{\left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^4 \int \frac{1}{\sqrt{x} \tanh ^{-1}(\tanh (a+b x))} \, dx}{b^4}\\ &=\frac{2 x^{7/2}}{7 b}+\frac{2 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{5 b^2}+\frac{2 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{3 b^3}+\frac{2 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}{b^4}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{7/2}}{b^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.0584617, size = 129, normalized size = 0.9 \[ \frac{2 \left (-406 b^{5/2} x^{5/2} \tanh ^{-1}(\tanh (a+b x))+350 b^{3/2} x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2-105 \sqrt{b} \sqrt{x} \tanh ^{-1}(\tanh (a+b x))^3+105 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{7/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))-b x}}\right )+176 b^{7/2} x^{7/2}\right )}{105 b^{9/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(7/2)/ArcTanh[Tanh[a + b*x]],x]

[Out]

(2*(176*b^(7/2)*x^(7/2) - 406*b^(5/2)*x^(5/2)*ArcTanh[Tanh[a + b*x]] + 350*b^(3/2)*x^(3/2)*ArcTanh[Tanh[a + b*
x]]^2 - 105*Sqrt[b]*Sqrt[x]*ArcTanh[Tanh[a + b*x]]^3 + 105*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[-(b*x) + ArcTanh[Tanh
[a + b*x]]]]*(-(b*x) + ArcTanh[Tanh[a + b*x]])^(7/2)))/(105*b^(9/2))

________________________________________________________________________________________

Maple [B]  time = 0.132, size = 481, normalized size = 3.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(7/2)/arctanh(tanh(b*x+a)),x)

[Out]

2/7*x^(7/2)/b-2/5/b^2*x^(5/2)*a-2/5/b^2*x^(5/2)*(arctanh(tanh(b*x+a))-b*x-a)+2/3/b^3*x^(3/2)*a^2+4/3/b^3*x^(3/
2)*a*(arctanh(tanh(b*x+a))-b*x-a)+2/3/b^3*x^(3/2)*(arctanh(tanh(b*x+a))-b*x-a)^2-2/b^4*x^(1/2)*a^3-6/b^4*a^2*(
arctanh(tanh(b*x+a))-b*x-a)*x^(1/2)-6/b^4*a*(arctanh(tanh(b*x+a))-b*x-a)^2*x^(1/2)-2/b^4*(arctanh(tanh(b*x+a))
-b*x-a)^3*x^(1/2)+2/b^4/((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))*a^4+8/b^4/((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))*a^
3*(arctanh(tanh(b*x+a))-b*x-a)+12/b^4/((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))*a^2*(arctanh(tanh(b*x+a))-b*x-a)^2+8/b^4/((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))*a*(arctanh(tanh(b*x+a))-b*x-a)^3+2/b^4/((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))*(arctanh(tanh(b*x+a))-b*x-a)^4

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)/arctanh(tanh(b*x+a)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 2.09325, size = 366, normalized size = 2.56 \begin{align*} \left [\frac{105 \, a^{3} \sqrt{-\frac{a}{b}} \log \left (\frac{b x + 2 \, b \sqrt{x} \sqrt{-\frac{a}{b}} - a}{b x + a}\right ) + 2 \,{\left (15 \, b^{3} x^{3} - 21 \, a b^{2} x^{2} + 35 \, a^{2} b x - 105 \, a^{3}\right )} \sqrt{x}}{105 \, b^{4}}, \frac{2 \,{\left (105 \, a^{3} \sqrt{\frac{a}{b}} \arctan \left (\frac{b \sqrt{x} \sqrt{\frac{a}{b}}}{a}\right ) +{\left (15 \, b^{3} x^{3} - 21 \, a b^{2} x^{2} + 35 \, a^{2} b x - 105 \, a^{3}\right )} \sqrt{x}\right )}}{105 \, b^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)/arctanh(tanh(b*x+a)),x, algorithm="fricas")

[Out]

[1/105*(105*a^3*sqrt(-a/b)*log((b*x + 2*b*sqrt(x)*sqrt(-a/b) - a)/(b*x + a)) + 2*(15*b^3*x^3 - 21*a*b^2*x^2 +
35*a^2*b*x - 105*a^3)*sqrt(x))/b^4, 2/105*(105*a^3*sqrt(a/b)*arctan(b*sqrt(x)*sqrt(a/b)/a) + (15*b^3*x^3 - 21*
a*b^2*x^2 + 35*a^2*b*x - 105*a^3)*sqrt(x))/b^4]

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(7/2)/atanh(tanh(b*x+a)),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.12817, size = 95, normalized size = 0.66 \begin{align*} \frac{2 \, a^{4} \arctan \left (\frac{b \sqrt{x}}{\sqrt{a b}}\right )}{\sqrt{a b} b^{4}} + \frac{2 \,{\left (15 \, b^{6} x^{\frac{7}{2}} - 21 \, a b^{5} x^{\frac{5}{2}} + 35 \, a^{2} b^{4} x^{\frac{3}{2}} - 105 \, a^{3} b^{3} \sqrt{x}\right )}}{105 \, b^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)/arctanh(tanh(b*x+a)),x, algorithm="giac")

[Out]

2*a^4*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^4) + 2/105*(15*b^6*x^(7/2) - 21*a*b^5*x^(5/2) + 35*a^2*b^4*x^(3
/2) - 105*a^3*b^3*sqrt(x))/b^7