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

Optimal. Leaf size=166 \[ \frac{7 x^{5/2} \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{3 b^2}+\frac{35 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{12 b^3}+\frac{35 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{8 b^4}+\frac{35 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}{8 b^{9/2}}-\frac{2 x^{7/2}}{b \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \]

[Out]

(35*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[ArcTanh[Tanh[a + b*x]]]]*(b*x - ArcTanh[Tanh[a + b*x]])^3)/(8*b^(9/2)) - (2
*x^(7/2))/(b*Sqrt[ArcTanh[Tanh[a + b*x]]]) + (7*x^(5/2)*Sqrt[ArcTanh[Tanh[a + b*x]]])/(3*b^2) + (35*x^(3/2)*(b
*x - ArcTanh[Tanh[a + b*x]])*Sqrt[ArcTanh[Tanh[a + b*x]]])/(12*b^3) + (35*Sqrt[x]*(b*x - ArcTanh[Tanh[a + b*x]
])^2*Sqrt[ArcTanh[Tanh[a + b*x]]])/(8*b^4)

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Rubi [A]  time = 0.107418, antiderivative size = 166, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {2168, 2169, 2165} \[ \frac{7 x^{5/2} \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{3 b^2}+\frac{35 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{12 b^3}+\frac{35 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{8 b^4}+\frac{35 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}{8 b^{9/2}}-\frac{2 x^{7/2}}{b \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(35*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[ArcTanh[Tanh[a + b*x]]]]*(b*x - ArcTanh[Tanh[a + b*x]])^3)/(8*b^(9/2)) - (2
*x^(7/2))/(b*Sqrt[ArcTanh[Tanh[a + b*x]]]) + (7*x^(5/2)*Sqrt[ArcTanh[Tanh[a + b*x]]])/(3*b^2) + (35*x^(3/2)*(b
*x - ArcTanh[Tanh[a + b*x]])*Sqrt[ArcTanh[Tanh[a + b*x]]])/(12*b^3) + (35*Sqrt[x]*(b*x - ArcTanh[Tanh[a + b*x]
])^2*Sqrt[ArcTanh[Tanh[a + b*x]]])/(8*b^4)

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 2169

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 + n + 1)), x] - Dist[(n*(b*u - a*v))/(a*(m + n + 1)), Int[u^m*v^(n - 1), x], x] /; NeQ[b*u - a*v, 0]]
/; PiecewiseLinearQ[u, v, x] && NeQ[m + n + 2, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !In
tegerQ[n] || LtQ[0, m, n])) &&  !ILtQ[m + n, -2]

Rule 2165

Int[1/(Sqrt[u_]*Sqrt[v_]), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[(2*ArcTanh[(
Rt[a*b, 2]*Sqrt[u])/(a*Sqrt[v])])/Rt[a*b, 2], x] /; NeQ[b*u - a*v, 0] && PosQ[a*b]] /; PiecewiseLinearQ[u, v,
x]

Rubi steps

\begin{align*} \int \frac{x^{7/2}}{\tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx &=-\frac{2 x^{7/2}}{b \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{7 \int \frac{x^{5/2}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))}} \, dx}{b}\\ &=-\frac{2 x^{7/2}}{b \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{7 x^{5/2} \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{3 b^2}-\frac{\left (35 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )\right ) \int \frac{x^{3/2}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))}} \, dx}{6 b^2}\\ &=-\frac{2 x^{7/2}}{b \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{7 x^{5/2} \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{3 b^2}+\frac{35 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{12 b^3}+\frac{\left (35 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^2\right ) \int \frac{\sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))}} \, dx}{8 b^3}\\ &=-\frac{2 x^{7/2}}{b \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{7 x^{5/2} \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{3 b^2}+\frac{35 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{12 b^3}+\frac{35 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{8 b^4}-\frac{\left (35 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^3\right ) \int \frac{1}{\sqrt{x} \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \, dx}{16 b^4}\\ &=\frac{35 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}{8 b^{9/2}}-\frac{2 x^{7/2}}{b \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{7 x^{5/2} \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{3 b^2}+\frac{35 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{12 b^3}+\frac{35 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{8 b^4}\\ \end{align*}

Mathematica [A]  time = 0.102065, size = 122, normalized size = 0.73 \[ \frac{\sqrt{x} \left (231 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))-280 b x \tanh ^{-1}(\tanh (a+b x))^2+105 \tanh ^{-1}(\tanh (a+b x))^3-48 b^3 x^3\right )}{24 b^4 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{35 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \log \left (\sqrt{b} \sqrt{\tanh ^{-1}(\tanh (a+b x))}+b \sqrt{x}\right )}{8 b^{9/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x]*(-48*b^3*x^3 + 231*b^2*x^2*ArcTanh[Tanh[a + b*x]] - 280*b*x*ArcTanh[Tanh[a + b*x]]^2 + 105*ArcTanh[Ta
nh[a + b*x]]^3))/(24*b^4*Sqrt[ArcTanh[Tanh[a + b*x]]]) + (35*(b*x - ArcTanh[Tanh[a + b*x]])^3*Log[b*Sqrt[x] +
Sqrt[b]*Sqrt[ArcTanh[Tanh[a + b*x]]]])/(8*b^(9/2))

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Maple [B]  time = 0.124, size = 428, normalized size = 2.6 \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))^(3/2),x)

[Out]

1/3*x^(7/2)/b/arctanh(tanh(b*x+a))^(1/2)-7/12/b^2*a*x^(5/2)/arctanh(tanh(b*x+a))^(1/2)+35/24/b^3*a^2*x^(3/2)/a
rctanh(tanh(b*x+a))^(1/2)+35/8/b^4*a^3*x^(1/2)/arctanh(tanh(b*x+a))^(1/2)-35/8/b^(9/2)*a^3*ln(b^(1/2)*x^(1/2)+
arctanh(tanh(b*x+a))^(1/2))+105/8/b^4*a^2*(arctanh(tanh(b*x+a))-b*x-a)*x^(1/2)/arctanh(tanh(b*x+a))^(1/2)-105/
8/b^(9/2)*a^2*(arctanh(tanh(b*x+a))-b*x-a)*ln(b^(1/2)*x^(1/2)+arctanh(tanh(b*x+a))^(1/2))+35/12/b^3*a*(arctanh
(tanh(b*x+a))-b*x-a)*x^(3/2)/arctanh(tanh(b*x+a))^(1/2)+105/8/b^4*a*(arctanh(tanh(b*x+a))-b*x-a)^2*x^(1/2)/arc
tanh(tanh(b*x+a))^(1/2)-105/8/b^(9/2)*a*(arctanh(tanh(b*x+a))-b*x-a)^2*ln(b^(1/2)*x^(1/2)+arctanh(tanh(b*x+a))
^(1/2))-7/12/b^2*(arctanh(tanh(b*x+a))-b*x-a)*x^(5/2)/arctanh(tanh(b*x+a))^(1/2)+35/24/b^3*(arctanh(tanh(b*x+a
))-b*x-a)^2*x^(3/2)/arctanh(tanh(b*x+a))^(1/2)+35/8/b^4*(arctanh(tanh(b*x+a))-b*x-a)^3*x^(1/2)/arctanh(tanh(b*
x+a))^(1/2)-35/8/b^(9/2)*(arctanh(tanh(b*x+a))-b*x-a)^3*ln(b^(1/2)*x^(1/2)+arctanh(tanh(b*x+a))^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{\frac{7}{2}}}{\operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.15873, size = 485, normalized size = 2.92 \begin{align*} \left [\frac{105 \,{\left (a^{3} b x + a^{4}\right )} \sqrt{b} \log \left (2 \, b x - 2 \, \sqrt{b x + a} \sqrt{b} \sqrt{x} + a\right ) + 2 \,{\left (8 \, b^{4} x^{3} - 14 \, a b^{3} x^{2} + 35 \, a^{2} b^{2} x + 105 \, a^{3} b\right )} \sqrt{b x + a} \sqrt{x}}{48 \,{\left (b^{6} x + a b^{5}\right )}}, \frac{105 \,{\left (a^{3} b x + a^{4}\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-b}}{b \sqrt{x}}\right ) +{\left (8 \, b^{4} x^{3} - 14 \, a b^{3} x^{2} + 35 \, a^{2} b^{2} x + 105 \, a^{3} b\right )} \sqrt{b x + a} \sqrt{x}}{24 \,{\left (b^{6} x + a b^{5}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/48*(105*(a^3*b*x + a^4)*sqrt(b)*log(2*b*x - 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) + 2*(8*b^4*x^3 - 14*a*b^3*
x^2 + 35*a^2*b^2*x + 105*a^3*b)*sqrt(b*x + a)*sqrt(x))/(b^6*x + a*b^5), 1/24*(105*(a^3*b*x + a^4)*sqrt(-b)*arc
tan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) + (8*b^4*x^3 - 14*a*b^3*x^2 + 35*a^2*b^2*x + 105*a^3*b)*sqrt(b*x + a)*
sqrt(x))/(b^6*x + a*b^5)]

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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))**(3/2),x)

[Out]

Timed out

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Giac [A]  time = 1.1954, size = 101, normalized size = 0.61 \begin{align*} \frac{{\left ({\left (2 \, x{\left (\frac{4 \, x}{b} - \frac{7 \, a}{b^{2}}\right )} + \frac{35 \, a^{2}}{b^{3}}\right )} x + \frac{105 \, a^{3}}{b^{4}}\right )} \sqrt{x}}{24 \, \sqrt{b x + a}} + \frac{35 \, a^{3} \log \left ({\left | -\sqrt{b} \sqrt{x} + \sqrt{b x + a} \right |}\right )}{8 \, b^{\frac{9}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*((2*x*(4*x/b - 7*a/b^2) + 35*a^2/b^3)*x + 105*a^3/b^4)*sqrt(x)/sqrt(b*x + a) + 35/8*a^3*log(abs(-sqrt(b)*
sqrt(x) + sqrt(b*x + a)))/b^(9/2)