3.231 \(\int \frac {\tanh ^{-1}(\tanh (a+b x))^{3/2}}{x^{13/2}} \, dx\)

Optimal. Leaf size=148 \[ \frac {32 b^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{1155 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}+\frac {16 b^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{231 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {4 b \tanh ^{-1}(\tanh (a+b x))^{5/2}}{33 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{11 x^{11/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )} \]

[Out]

32/1155*b^3*arctanh(tanh(b*x+a))^(5/2)/x^(5/2)/(b*x-arctanh(tanh(b*x+a)))^4+16/231*b^2*arctanh(tanh(b*x+a))^(5
/2)/x^(7/2)/(b*x-arctanh(tanh(b*x+a)))^3+4/33*b*arctanh(tanh(b*x+a))^(5/2)/x^(9/2)/(b*x-arctanh(tanh(b*x+a)))^
2+2/11*arctanh(tanh(b*x+a))^(5/2)/x^(11/2)/(b*x-arctanh(tanh(b*x+a)))

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Rubi [A]  time = 0.08, antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2171, 2167} \[ \frac {32 b^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{1155 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}+\frac {16 b^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{231 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {4 b \tanh ^{-1}(\tanh (a+b x))^{5/2}}{33 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{11 x^{11/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]^(3/2)/x^(13/2),x]

[Out]

(32*b^3*ArcTanh[Tanh[a + b*x]]^(5/2))/(1155*x^(5/2)*(b*x - ArcTanh[Tanh[a + b*x]])^4) + (16*b^2*ArcTanh[Tanh[a
+ b*x]]^(5/2))/(231*x^(7/2)*(b*x - ArcTanh[Tanh[a + b*x]])^3) + (4*b*ArcTanh[Tanh[a + b*x]]^(5/2))/(33*x^(9/2
)*(b*x - ArcTanh[Tanh[a + b*x]])^2) + (2*ArcTanh[Tanh[a + b*x]]^(5/2))/(11*x^(11/2)*(b*x - ArcTanh[Tanh[a + b*
x]]))

Rule 2167

Int[(u_)^(m_)*(v_)^(n_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, -Simp[(u^(m + 1)*v^
(n + 1))/((m + 1)*(b*u - a*v)), x] /; NeQ[b*u - a*v, 0]] /; FreeQ[{m, n}, x] && PiecewiseLinearQ[u, v, x] && E
qQ[m + n + 2, 0] && NeQ[m, -1]

Rule 2171

Int[(u_)^(m_)*(v_)^(n_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, -Simp[(u^(m + 1)*v^
(n + 1))/((m + 1)*(b*u - a*v)), x] + Dist[(b*(m + n + 2))/((m + 1)*(b*u - a*v)), Int[u^(m + 1)*v^n, x], x] /;
NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x] && NeQ[m + n + 2, 0] && LtQ[m, -1]

Rubi steps

\begin {align*} \int \frac {\tanh ^{-1}(\tanh (a+b x))^{3/2}}{x^{13/2}} \, dx &=\frac {2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{11 x^{11/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {(6 b) \int \frac {\tanh ^{-1}(\tanh (a+b x))^{3/2}}{x^{11/2}} \, dx}{11 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac {4 b \tanh ^{-1}(\tanh (a+b x))^{5/2}}{33 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{11 x^{11/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {\left (8 b^2\right ) \int \frac {\tanh ^{-1}(\tanh (a+b x))^{3/2}}{x^{9/2}} \, dx}{33 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=\frac {16 b^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{231 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {4 b \tanh ^{-1}(\tanh (a+b x))^{5/2}}{33 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{11 x^{11/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {\left (16 b^3\right ) \int \frac {\tanh ^{-1}(\tanh (a+b x))^{3/2}}{x^{7/2}} \, dx}{231 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}\\ &=\frac {32 b^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{1155 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}+\frac {16 b^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{231 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {4 b \tanh ^{-1}(\tanh (a+b x))^{5/2}}{33 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{11 x^{11/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 82, normalized size = 0.55 \[ \frac {2 \tanh ^{-1}(\tanh (a+b x))^{5/2} \left (-495 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))+385 b x \tanh ^{-1}(\tanh (a+b x))^2-105 \tanh ^{-1}(\tanh (a+b x))^3+231 b^3 x^3\right )}{1155 x^{11/2} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^4} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[Tanh[a + b*x]]^(3/2)/x^(13/2),x]

[Out]

(2*ArcTanh[Tanh[a + b*x]]^(5/2)*(231*b^3*x^3 - 495*b^2*x^2*ArcTanh[Tanh[a + b*x]] + 385*b*x*ArcTanh[Tanh[a + b
*x]]^2 - 105*ArcTanh[Tanh[a + b*x]]^3))/(1155*x^(11/2)*(-(b*x) + ArcTanh[Tanh[a + b*x]])^4)

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fricas [A]  time = 0.51, size = 67, normalized size = 0.45 \[ \frac {2 \, {\left (16 \, b^{5} x^{5} - 8 \, a b^{4} x^{4} + 6 \, a^{2} b^{3} x^{3} - 5 \, a^{3} b^{2} x^{2} - 140 \, a^{4} b x - 105 \, a^{5}\right )} \sqrt {b x + a}}{1155 \, a^{4} x^{\frac {11}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(3/2)/x^(13/2),x, algorithm="fricas")

[Out]

2/1155*(16*b^5*x^5 - 8*a*b^4*x^4 + 6*a^2*b^3*x^3 - 5*a^3*b^2*x^2 - 140*a^4*b*x - 105*a^5)*sqrt(b*x + a)/(a^4*x
^(11/2))

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giac [A]  time = 0.17, size = 97, normalized size = 0.66 \[ -\frac {\sqrt {2} {\left (\frac {231 \, \sqrt {2} b^{11}}{a} - 2 \, {\left (\frac {99 \, \sqrt {2} b^{11}}{a^{2}} + 4 \, {\left (\frac {2 \, \sqrt {2} {\left (b x + a\right )} b^{11}}{a^{4}} - \frac {11 \, \sqrt {2} b^{11}}{a^{3}}\right )} {\left (b x + a\right )}\right )} {\left (b x + a\right )}\right )} {\left (b x + a\right )}^{\frac {5}{2}} b}{1155 \, {\left ({\left (b x + a\right )} b - a b\right )}^{\frac {11}{2}} {\left | b \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(3/2)/x^(13/2),x, algorithm="giac")

[Out]

-1/1155*sqrt(2)*(231*sqrt(2)*b^11/a - 2*(99*sqrt(2)*b^11/a^2 + 4*(2*sqrt(2)*(b*x + a)*b^11/a^4 - 11*sqrt(2)*b^
11/a^3)*(b*x + a))*(b*x + a))*(b*x + a)^(5/2)*b/(((b*x + a)*b - a*b)^(11/2)*abs(b))

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maple [A]  time = 0.39, size = 151, normalized size = 1.02 \[ -\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{11 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {11}{2}}}-\frac {12 b \left (-\frac {\arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{9 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {9}{2}}}-\frac {4 b \left (-\frac {\arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{7 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {7}{2}}}+\frac {2 b \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{35 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{2} x^{\frac {5}{2}}}\right )}{9 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )}\right )}{11 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(tanh(b*x+a))^(3/2)/x^(13/2),x)

[Out]

-2/11/(arctanh(tanh(b*x+a))-b*x)/x^(11/2)*arctanh(tanh(b*x+a))^(5/2)-12/11*b/(arctanh(tanh(b*x+a))-b*x)*(-1/9/
(arctanh(tanh(b*x+a))-b*x)/x^(9/2)*arctanh(tanh(b*x+a))^(5/2)-4/9*b/(arctanh(tanh(b*x+a))-b*x)*(-1/7/(arctanh(
tanh(b*x+a))-b*x)/x^(7/2)*arctanh(tanh(b*x+a))^(5/2)+2/35*b/(arctanh(tanh(b*x+a))-b*x)^2/x^(5/2)*arctanh(tanh(
b*x+a))^(5/2)))

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maxima [A]  time = 0.44, size = 56, normalized size = 0.38 \[ \frac {2 \, {\left (16 \, b^{4} x^{4} - 24 \, a b^{3} x^{3} + 30 \, a^{2} b^{2} x^{2} - 35 \, a^{3} b x - 105 \, a^{4}\right )} {\left (b x + a\right )}^{\frac {3}{2}}}{1155 \, a^{4} x^{\frac {11}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(3/2)/x^(13/2),x, algorithm="maxima")

[Out]

2/1155*(16*b^4*x^4 - 24*a*b^3*x^3 + 30*a^2*b^2*x^2 - 35*a^3*b*x - 105*a^4)*(b*x + a)^(3/2)/(a^4*x^(11/2))

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mupad [B]  time = 1.74, size = 348, normalized size = 2.35 \[ \frac {\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}\,\left (\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{11}-\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{11}-\frac {2\,b\,x}{33}+\frac {4\,b^2\,x^2}{231\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}+\frac {16\,b^3\,x^3}{385\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}+\frac {128\,b^4\,x^4}{1155\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3}+\frac {512\,b^5\,x^5}{1155\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^4}\right )}{x^{11/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(tanh(a + b*x))^(3/2)/x^(13/2),x)

[Out]

((log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(2/(exp(2*a)*exp(2*b*x) + 1))/2)^(1/2)*(log(2/
(exp(2*a)*exp(2*b*x) + 1))/11 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/11 - (2*b*x)/33 + (4*b^
2*x^2)/(231*(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
)) + (16*b^3*x^3)/(385*(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) + (128*b^4*x^4)/(1155*(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) + (512*b^5*x^5)/(1155*(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)))/x^(11/2)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(tanh(b*x+a))**(3/2)/x**(13/2),x)

[Out]

Timed out

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