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

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

[Out]

2/7*x^(7/2)/b+2/5*x^(5/2)*(b*x-arctanh(tanh(b*x+a)))/b^2+2/3*x^(3/2)*(b*x-arctanh(tanh(b*x+a)))^2/b^3-2*arctan
h(b^(1/2)*x^(1/2)/(b*x-arctanh(tanh(b*x+a)))^(1/2))*(b*x-arctanh(tanh(b*x+a)))^(7/2)/b^(9/2)+2*(b*x-arctanh(ta
nh(b*x+a)))^3*x^(1/2)/b^4

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Rubi [A]  time = 0.13, antiderivative size = 143, normalized size of antiderivative = 1.00, 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*}

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Mathematica [A]  time = 0.07, size = 129, normalized size = 0.90 \[ \frac {2 \left (-406 b^{5/2} x^{5/2} \tanh ^{-1}(\tanh (a+b x))+350 b^{3/2} x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2-105 \sqrt {b} \sqrt {x} \tanh ^{-1}(\tanh (a+b x))^3+105 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{7/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))-b x}}\right )+176 b^{7/2} x^{7/2}\right )}{105 b^{9/2}} \]

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))

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fricas [A]  time = 0.50, size = 153, normalized size = 1.07 \[ \left [\frac {105 \, a^{3} \sqrt {-\frac {a}{b}} \log \left (\frac {b x + 2 \, b \sqrt {x} \sqrt {-\frac {a}{b}} - a}{b x + a}\right ) + 2 \, {\left (15 \, b^{3} x^{3} - 21 \, a b^{2} x^{2} + 35 \, a^{2} b x - 105 \, a^{3}\right )} \sqrt {x}}{105 \, b^{4}}, \frac {2 \, {\left (105 \, a^{3} \sqrt {\frac {a}{b}} \arctan \left (\frac {b \sqrt {x} \sqrt {\frac {a}{b}}}{a}\right ) + {\left (15 \, b^{3} x^{3} - 21 \, a b^{2} x^{2} + 35 \, a^{2} b x - 105 \, a^{3}\right )} \sqrt {x}\right )}}{105 \, b^{4}}\right ] \]

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]

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giac [A]  time = 0.14, size = 70, normalized size = 0.49 \[ \frac {2 \, a^{4} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{4}} + \frac {2 \, {\left (15 \, b^{6} x^{\frac {7}{2}} - 21 \, a b^{5} x^{\frac {5}{2}} + 35 \, a^{2} b^{4} x^{\frac {3}{2}} - 105 \, a^{3} b^{3} \sqrt {x}\right )}}{105 \, b^{7}} \]

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

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maple [B]  time = 0.27, size = 481, normalized size = 3.36 \[ \frac {2 x^{\frac {7}{2}}}{7 b}-\frac {2 x^{\frac {5}{2}} a}{5 b^{2}}-\frac {2 x^{\frac {5}{2}} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{5 b^{2}}+\frac {2 a^{2} x^{\frac {3}{2}}}{3 b^{3}}+\frac {4 x^{\frac {3}{2}} a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{3 b^{3}}+\frac {2 x^{\frac {3}{2}} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{3 b^{3}}-\frac {2 \sqrt {x}\, a^{3}}{b^{4}}-\frac {6 a^{2} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right ) \sqrt {x}}{b^{4}}-\frac {6 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2} \sqrt {x}}{b^{4}}-\frac {2 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{3} \sqrt {x}}{b^{4}}+\frac {2 \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right ) a^{4}}{b^{4} \sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}+\frac {8 \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right ) a^{3} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{b^{4} \sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}+\frac {12 \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right ) a^{2} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{b^{4} \sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}+\frac {8 \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right ) a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{3}}{b^{4} \sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}+\frac {2 \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right ) \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{4}}{b^{4} \sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}} \]

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*a^2*x^(3/2)+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

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maxima [A]  time = 0.43, size = 65, normalized size = 0.45 \[ \frac {2 \, a^{4} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{4}} + \frac {2 \, {\left (15 \, b^{3} x^{\frac {7}{2}} - 21 \, a b^{2} x^{\frac {5}{2}} + 35 \, a^{2} b x^{\frac {3}{2}} - 105 \, a^{3} \sqrt {x}\right )}}{105 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.72, size = 475, normalized size = 3.32 \[ \frac {2\,x^{7/2}}{7\,b}+\frac {x^{5/2}\,\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 )}{5\,b^2}+\frac {x^{3/2}\,{\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 )}^2}{6\,b^3}+\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 )}^3}{4\,b^4}+\frac {\sqrt {2}\,\ln \left (\frac {64\,b^{19/2}\,\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 )\,{\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 )}^{7/2}}{16\,b^{9/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(7/2)/atanh(tanh(a + b*x)),x)

[Out]

(2*x^(7/2))/(7*b) + (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))/(5*b^2) + (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)/(6*b^3) + (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)/(4*b^4) + (2^(1/2)*log((64*b^(19/2)*(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)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^(1/2) + 2*2^(1/2)*b*x
))/((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)))*(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)^(7/2))/(16*b^(9/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{\frac {7}{2}}}{\operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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