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

Optimal. Leaf size=135 \[ \frac{7 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{3 b^3}+\frac{7 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{b^4}-\frac{7 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{5/2}}{b^{9/2}}-\frac{x^{7/2}}{b \tanh ^{-1}(\tanh (a+b x))}+\frac{7 x^{5/2}}{5 b^2} \]

[Out]

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

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Rubi [A]  time = 0.103377, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2168, 2159, 2162} \[ \frac{7 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{3 b^3}+\frac{7 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{b^4}-\frac{7 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{5/2}}{b^{9/2}}-\frac{x^{7/2}}{b \tanh ^{-1}(\tanh (a+b x))}+\frac{7 x^{5/2}}{5 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2168

Int[(u_)^(m_)*(v_)^(n_.), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[(u^(m + 1)*v^
n)/(a*(m + 1)), x] - Dist[(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] && (Fra
ctionQ[m] || GeQ[2*n + m + 1, 0]))) || (IGtQ[n, 0] && IGtQ[m, 0] && LeQ[n, m]) || (IGtQ[n, 0] &&  !IntegerQ[m]
) || (ILtQ[m, 0] &&  !IntegerQ[n]))

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

Mathematica [A]  time = 0.186507, size = 144, normalized size = 1.07 \[ -\frac{4 x^{3/2} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )}{3 b^3}+\frac{\sqrt{x} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^3}{b^4 \tanh ^{-1}(\tanh (a+b x))}+\frac{6 \sqrt{x} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^2}{b^4}-\frac{7 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{b^{9/2}}+\frac{2 x^{5/2}}{5 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^(5/2))/(5*b^2) - (4*x^(3/2)*(-(b*x) + ArcTanh[Tanh[a + b*x]]))/(3*b^3) + (6*Sqrt[x]*(-(b*x) + ArcTanh[Tan
h[a + b*x]])^2)/b^4 - (7*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[-(b*x) + ArcTanh[Tanh[a + b*x]]]]*(-(b*x) + ArcTanh[Tan
h[a + b*x]])^(5/2))/b^(9/2) + (Sqrt[x]*(-(b*x) + ArcTanh[Tanh[a + b*x]])^3)/(b^4*ArcTanh[Tanh[a + b*x]])

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Maple [B]  time = 0.136, size = 452, 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))^2,x)

[Out]

2/5*x^(5/2)/b^2-4/3/b^3*x^(3/2)*a-4/3/b^3*x^(3/2)*(arctanh(tanh(b*x+a))-b*x-a)+6/b^4*x^(1/2)*a^2+12/b^4*a*(arc
tanh(tanh(b*x+a))-b*x-a)*x^(1/2)+6/b^4*(arctanh(tanh(b*x+a))-b*x-a)^2*x^(1/2)+1/b^4*x^(1/2)/arctanh(tanh(b*x+a
))*a^3+3/b^4*x^(1/2)/arctanh(tanh(b*x+a))*a^2*(arctanh(tanh(b*x+a))-b*x-a)+3/b^4*x^(1/2)/arctanh(tanh(b*x+a))*
a*(arctanh(tanh(b*x+a))-b*x-a)^2+1/b^4*x^(1/2)/arctanh(tanh(b*x+a))*(arctanh(tanh(b*x+a))-b*x-a)^3-7/b^4/((arc
tanh(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^3-21/b^4/((arctanh(ta
nh(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)-21/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*(arcta
nh(tanh(b*x+a))-b*x-a)^2-7/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)^3

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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))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.10656, size = 427, normalized size = 3.16 \begin{align*} \left [\frac{105 \,{\left (a^{2} b x + a^{3}\right )} \sqrt{-\frac{a}{b}} \log \left (\frac{b x - 2 \, b \sqrt{x} \sqrt{-\frac{a}{b}} - a}{b x + a}\right ) + 2 \,{\left (6 \, b^{3} x^{3} - 14 \, a b^{2} x^{2} + 70 \, a^{2} b x + 105 \, a^{3}\right )} \sqrt{x}}{30 \,{\left (b^{5} x + a b^{4}\right )}}, -\frac{105 \,{\left (a^{2} b x + a^{3}\right )} \sqrt{\frac{a}{b}} \arctan \left (\frac{b \sqrt{x} \sqrt{\frac{a}{b}}}{a}\right ) -{\left (6 \, b^{3} x^{3} - 14 \, a b^{2} x^{2} + 70 \, a^{2} b x + 105 \, a^{3}\right )} \sqrt{x}}{15 \,{\left (b^{5} x + a b^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/30*(105*(a^2*b*x + a^3)*sqrt(-a/b)*log((b*x - 2*b*sqrt(x)*sqrt(-a/b) - a)/(b*x + a)) + 2*(6*b^3*x^3 - 14*a*
b^2*x^2 + 70*a^2*b*x + 105*a^3)*sqrt(x))/(b^5*x + a*b^4), -1/15*(105*(a^2*b*x + a^3)*sqrt(a/b)*arctan(b*sqrt(x
)*sqrt(a/b)/a) - (6*b^3*x^3 - 14*a*b^2*x^2 + 70*a^2*b*x + 105*a^3)*sqrt(x))/(b^5*x + a*b^4)]

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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))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.1208, size = 103, normalized size = 0.76 \begin{align*} -\frac{7 \, a^{3} \arctan \left (\frac{b \sqrt{x}}{\sqrt{a b}}\right )}{\sqrt{a b} b^{4}} + \frac{a^{3} \sqrt{x}}{{\left (b x + a\right )} b^{4}} + \frac{2 \,{\left (3 \, b^{8} x^{\frac{5}{2}} - 10 \, a b^{7} x^{\frac{3}{2}} + 45 \, a^{2} b^{6} \sqrt{x}\right )}}{15 \, b^{10}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-7*a^3*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^4) + a^3*sqrt(x)/((b*x + a)*b^4) + 2/15*(3*b^8*x^(5/2) - 10*a*
b^7*x^(3/2) + 45*a^2*b^6*sqrt(x))/b^10