3.207 \(\int \frac{x^{7/2}}{\tanh ^{-1}(\tanh (a+b x))^3} \, dx\)

Optimal. Leaf size=135 \[ -\frac{7 x^{5/2}}{4 b^2 \tanh ^{-1}(\tanh (a+b x))}+\frac{35 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{4 b^4}-\frac{35 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{3/2}}{4 b^{9/2}}-\frac{x^{7/2}}{2 b \tanh ^{-1}(\tanh (a+b x))^2}+\frac{35 x^{3/2}}{12 b^3} \]

[Out]

(35*x^(3/2))/(12*b^3) + (35*Sqrt[x]*(b*x - ArcTanh[Tanh[a + b*x]]))/(4*b^4) - (35*ArcTanh[(Sqrt[b]*Sqrt[x])/Sq
rt[b*x - ArcTanh[Tanh[a + b*x]]]]*(b*x - ArcTanh[Tanh[a + b*x]])^(3/2))/(4*b^(9/2)) - x^(7/2)/(2*b*ArcTanh[Tan
h[a + b*x]]^2) - (7*x^(5/2))/(4*b^2*ArcTanh[Tanh[a + b*x]])

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Rubi [A]  time = 0.0970262, 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} \[ -\frac{7 x^{5/2}}{4 b^2 \tanh ^{-1}(\tanh (a+b x))}+\frac{35 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{4 b^4}-\frac{35 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{3/2}}{4 b^{9/2}}-\frac{x^{7/2}}{2 b \tanh ^{-1}(\tanh (a+b x))^2}+\frac{35 x^{3/2}}{12 b^3} \]

Antiderivative was successfully verified.

[In]

Int[x^(7/2)/ArcTanh[Tanh[a + b*x]]^3,x]

[Out]

(35*x^(3/2))/(12*b^3) + (35*Sqrt[x]*(b*x - ArcTanh[Tanh[a + b*x]]))/(4*b^4) - (35*ArcTanh[(Sqrt[b]*Sqrt[x])/Sq
rt[b*x - ArcTanh[Tanh[a + b*x]]]]*(b*x - ArcTanh[Tanh[a + b*x]])^(3/2))/(4*b^(9/2)) - x^(7/2)/(2*b*ArcTanh[Tan
h[a + b*x]]^2) - (7*x^(5/2))/(4*b^2*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))^3} \, dx &=-\frac{x^{7/2}}{2 b \tanh ^{-1}(\tanh (a+b x))^2}+\frac{7 \int \frac{x^{5/2}}{\tanh ^{-1}(\tanh (a+b x))^2} \, dx}{4 b}\\ &=-\frac{x^{7/2}}{2 b \tanh ^{-1}(\tanh (a+b x))^2}-\frac{7 x^{5/2}}{4 b^2 \tanh ^{-1}(\tanh (a+b x))}+\frac{35 \int \frac{x^{3/2}}{\tanh ^{-1}(\tanh (a+b x))} \, dx}{8 b^2}\\ &=\frac{35 x^{3/2}}{12 b^3}-\frac{x^{7/2}}{2 b \tanh ^{-1}(\tanh (a+b x))^2}-\frac{7 x^{5/2}}{4 b^2 \tanh ^{-1}(\tanh (a+b x))}-\frac{\left (35 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )\right ) \int \frac{\sqrt{x}}{\tanh ^{-1}(\tanh (a+b x))} \, dx}{8 b^3}\\ &=\frac{35 x^{3/2}}{12 b^3}+\frac{35 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{4 b^4}-\frac{x^{7/2}}{2 b \tanh ^{-1}(\tanh (a+b x))^2}-\frac{7 x^{5/2}}{4 b^2 \tanh ^{-1}(\tanh (a+b x))}+\frac{\left (35 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^2\right ) \int \frac{1}{\sqrt{x} \tanh ^{-1}(\tanh (a+b x))} \, dx}{8 b^4}\\ &=\frac{35 x^{3/2}}{12 b^3}+\frac{35 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{4 b^4}-\frac{35 \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 )^{3/2}}{4 b^{9/2}}-\frac{x^{7/2}}{2 b \tanh ^{-1}(\tanh (a+b x))^2}-\frac{7 x^{5/2}}{4 b^2 \tanh ^{-1}(\tanh (a+b x))}\\ \end{align*}

Mathematica [A]  time = 0.111725, size = 147, normalized size = 1.09 \[ -\frac{21 b^{5/2} x^{5/2} \tanh ^{-1}(\tanh (a+b x))-140 b^{3/2} x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2+105 \sqrt{b} \sqrt{x} \tanh ^{-1}(\tanh (a+b x))^3-105 \tanh ^{-1}(\tanh (a+b x))^2 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))-b x}}\right )+6 b^{7/2} x^{7/2}}{12 b^{9/2} \tanh ^{-1}(\tanh (a+b x))^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(7/2)/ArcTanh[Tanh[a + b*x]]^3,x]

[Out]

-(6*b^(7/2)*x^(7/2) + 21*b^(5/2)*x^(5/2)*ArcTanh[Tanh[a + b*x]] - 140*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]]]]*ArcTanh[Tanh[a + b*x]]^2*(-(b*x) + ArcTanh[Tanh[a + b*x]])^(3/2))/(12*b^(9/2)*ArcTanh[Tanh[a + b*x]]^2
)

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Maple [B]  time = 0.14, size = 418, normalized size = 3.1 \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))^3,x)

[Out]

2/3*x^(3/2)/b^3-6/b^4*a*x^(1/2)-6/b^4*(arctanh(tanh(b*x+a))-b*x-a)*x^(1/2)-13/4/b^3/arctanh(tanh(b*x+a))^2*x^(
3/2)*a^2-13/2/b^3/arctanh(tanh(b*x+a))^2*x^(3/2)*a*(arctanh(tanh(b*x+a))-b*x-a)-13/4/b^3/arctanh(tanh(b*x+a))^
2*x^(3/2)*(arctanh(tanh(b*x+a))-b*x-a)^2-11/4/b^4/arctanh(tanh(b*x+a))^2*x^(1/2)*a^3-33/4/b^4/arctanh(tanh(b*x
+a))^2*x^(1/2)*a^2*(arctanh(tanh(b*x+a))-b*x-a)-33/4/b^4/arctanh(tanh(b*x+a))^2*x^(1/2)*a*(arctanh(tanh(b*x+a)
)-b*x-a)^2-11/4/b^4/arctanh(tanh(b*x+a))^2*x^(1/2)*(arctanh(tanh(b*x+a))-b*x-a)^3+35/4/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+35/2/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*(arctanh(tanh(b*x+a))-b*x-a)+35/4/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)^2

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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))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.09214, size = 509, normalized size = 3.77 \begin{align*} \left [\frac{105 \,{\left (a b^{2} x^{2} + 2 \, 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 (8 \, b^{3} x^{3} - 56 \, a b^{2} x^{2} - 175 \, a^{2} b x - 105 \, a^{3}\right )} \sqrt{x}}{24 \,{\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}}, \frac{105 \,{\left (a b^{2} x^{2} + 2 \, a^{2} b x + a^{3}\right )} \sqrt{\frac{a}{b}} \arctan \left (\frac{b \sqrt{x} \sqrt{\frac{a}{b}}}{a}\right ) +{\left (8 \, b^{3} x^{3} - 56 \, a b^{2} x^{2} - 175 \, a^{2} b x - 105 \, a^{3}\right )} \sqrt{x}}{12 \,{\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} 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))^3,x, algorithm="fricas")

[Out]

[1/24*(105*(a*b^2*x^2 + 2*a^2*b*x + a^3)*sqrt(-a/b)*log((b*x + 2*b*sqrt(x)*sqrt(-a/b) - a)/(b*x + a)) + 2*(8*b
^3*x^3 - 56*a*b^2*x^2 - 175*a^2*b*x - 105*a^3)*sqrt(x))/(b^6*x^2 + 2*a*b^5*x + a^2*b^4), 1/12*(105*(a*b^2*x^2
+ 2*a^2*b*x + a^3)*sqrt(a/b)*arctan(b*sqrt(x)*sqrt(a/b)/a) + (8*b^3*x^3 - 56*a*b^2*x^2 - 175*a^2*b*x - 105*a^3
)*sqrt(x))/(b^6*x^2 + 2*a*b^5*x + a^2*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))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.1437, size = 104, normalized size = 0.77 \begin{align*} \frac{35 \, a^{2} \arctan \left (\frac{b \sqrt{x}}{\sqrt{a b}}\right )}{4 \, \sqrt{a b} b^{4}} - \frac{13 \, a^{2} b x^{\frac{3}{2}} + 11 \, a^{3} \sqrt{x}}{4 \,{\left (b x + a\right )}^{2} b^{4}} + \frac{2 \,{\left (b^{6} x^{\frac{3}{2}} - 9 \, a b^{5} \sqrt{x}\right )}}{3 \, b^{9}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)/arctanh(tanh(b*x+a))^3,x, algorithm="giac")

[Out]

35/4*a^2*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^4) - 1/4*(13*a^2*b*x^(3/2) + 11*a^3*sqrt(x))/((b*x + a)^2*b^
4) + 2/3*(b^6*x^(3/2) - 9*a*b^5*sqrt(x))/b^9