3.257 \(\int \frac{x^{7/2}}{\tanh ^{-1}(\tanh (a+b x))^{5/2}} \, dx\)

Optimal. Leaf size=153 \[ -\frac{14 x^{5/2}}{3 b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{35 x^{3/2} \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{6 b^3}+\frac{35 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{4 b^4}+\frac{35 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{4 b^{9/2}}-\frac{2 x^{7/2}}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]

[Out]

(35*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[ArcTanh[Tanh[a + b*x]]]]*(b*x - ArcTanh[Tanh[a + b*x]])^2)/(4*b^(9/2)) - (2
*x^(7/2))/(3*b*ArcTanh[Tanh[a + b*x]]^(3/2)) - (14*x^(5/2))/(3*b^2*Sqrt[ArcTanh[Tanh[a + b*x]]]) + (35*x^(3/2)
*Sqrt[ArcTanh[Tanh[a + b*x]]])/(6*b^3) + (35*Sqrt[x]*(b*x - ArcTanh[Tanh[a + b*x]])*Sqrt[ArcTanh[Tanh[a + b*x]
]])/(4*b^4)

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Rubi [A]  time = 0.0959509, antiderivative size = 153, 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} \[ -\frac{14 x^{5/2}}{3 b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{35 x^{3/2} \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{6 b^3}+\frac{35 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{4 b^4}+\frac{35 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{4 b^{9/2}}-\frac{2 x^{7/2}}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^(7/2)/ArcTanh[Tanh[a + b*x]]^(5/2),x]

[Out]

(35*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[ArcTanh[Tanh[a + b*x]]]]*(b*x - ArcTanh[Tanh[a + b*x]])^2)/(4*b^(9/2)) - (2
*x^(7/2))/(3*b*ArcTanh[Tanh[a + b*x]]^(3/2)) - (14*x^(5/2))/(3*b^2*Sqrt[ArcTanh[Tanh[a + b*x]]]) + (35*x^(3/2)
*Sqrt[ArcTanh[Tanh[a + b*x]]])/(6*b^3) + (35*Sqrt[x]*(b*x - ArcTanh[Tanh[a + b*x]])*Sqrt[ArcTanh[Tanh[a + b*x]
]])/(4*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))^{5/2}} \, dx &=-\frac{2 x^{7/2}}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{7 \int \frac{x^{5/2}}{\tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx}{3 b}\\ &=-\frac{2 x^{7/2}}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{14 x^{5/2}}{3 b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{35 \int \frac{x^{3/2}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))}} \, dx}{3 b^2}\\ &=-\frac{2 x^{7/2}}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{14 x^{5/2}}{3 b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{35 x^{3/2} \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{6 b^3}-\frac{\left (35 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )\right ) \int \frac{\sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))}} \, dx}{4 b^3}\\ &=-\frac{2 x^{7/2}}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{14 x^{5/2}}{3 b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{35 x^{3/2} \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{6 b^3}+\frac{35 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{4 b^4}+\frac{\left (35 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^2\right ) \int \frac{1}{\sqrt{x} \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \, dx}{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 )^2}{4 b^{9/2}}-\frac{2 x^{7/2}}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{14 x^{5/2}}{3 b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{35 x^{3/2} \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{6 b^3}+\frac{35 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{4 b^4}\\ \end{align*}

Mathematica [A]  time = 0.105465, size = 121, normalized size = 0.79 \[ \frac{35 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^2 \log \left (\sqrt{b} \sqrt{\tanh ^{-1}(\tanh (a+b x))}+b \sqrt{x}\right )}{4 b^{9/2}}-\frac{\sqrt{x} \left (56 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))-175 b x \tanh ^{-1}(\tanh (a+b x))^2+105 \tanh ^{-1}(\tanh (a+b x))^3+8 b^3 x^3\right )}{12 b^4 \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(7/2)/ArcTanh[Tanh[a + b*x]]^(5/2),x]

[Out]

-(Sqrt[x]*(8*b^3*x^3 + 56*b^2*x^2*ArcTanh[Tanh[a + b*x]] - 175*b*x*ArcTanh[Tanh[a + b*x]]^2 + 105*ArcTanh[Tanh
[a + b*x]]^3))/(12*b^4*ArcTanh[Tanh[a + b*x]]^(3/2)) + (35*(-(b*x) + ArcTanh[Tanh[a + b*x]])^2*Log[b*Sqrt[x] +
Sqrt[b]*Sqrt[ArcTanh[Tanh[a + b*x]]]])/(4*b^(9/2))

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Maple [B]  time = 0.125, size = 348, normalized size = 2.3 \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))^(5/2),x)

[Out]

1/2*x^(7/2)/b/arctanh(tanh(b*x+a))^(3/2)-7/4/b^2*a*x^(5/2)/arctanh(tanh(b*x+a))^(3/2)-35/12/b^3*a^2*x^(3/2)/ar
ctanh(tanh(b*x+a))^(3/2)-35/4/b^4*a^2*x^(1/2)/arctanh(tanh(b*x+a))^(1/2)+35/4/b^(9/2)*a^2*ln(b^(1/2)*x^(1/2)+a
rctanh(tanh(b*x+a))^(1/2))-35/6/b^3*a*(arctanh(tanh(b*x+a))-b*x-a)*x^(3/2)/arctanh(tanh(b*x+a))^(3/2)-35/2/b^4
*a*(arctanh(tanh(b*x+a))-b*x-a)*x^(1/2)/arctanh(tanh(b*x+a))^(1/2)+35/2/b^(9/2)*a*(arctanh(tanh(b*x+a))-b*x-a)
*ln(b^(1/2)*x^(1/2)+arctanh(tanh(b*x+a))^(1/2))-7/4/b^2*(arctanh(tanh(b*x+a))-b*x-a)*x^(5/2)/arctanh(tanh(b*x+
a))^(3/2)-35/12/b^3*(arctanh(tanh(b*x+a))-b*x-a)^2*x^(3/2)/arctanh(tanh(b*x+a))^(3/2)-35/4/b^4*(arctanh(tanh(b
*x+a))-b*x-a)^2*x^(1/2)/arctanh(tanh(b*x+a))^(1/2)+35/4/b^(9/2)*(arctanh(tanh(b*x+a))-b*x-a)^2*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{5}{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))^(5/2),x, algorithm="maxima")

[Out]

integrate(x^(7/2)/arctanh(tanh(b*x + a))^(5/2), x)

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Fricas [A]  time = 2.15042, size = 575, normalized size = 3.76 \begin{align*} \left [\frac{105 \,{\left (a^{2} b^{2} x^{2} + 2 \, 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 (6 \, b^{4} x^{3} - 21 \, a b^{3} x^{2} - 140 \, a^{2} b^{2} x - 105 \, a^{3} b\right )} \sqrt{b x + a} \sqrt{x}}{24 \,{\left (b^{7} x^{2} + 2 \, a b^{6} x + a^{2} b^{5}\right )}}, -\frac{105 \,{\left (a^{2} b^{2} x^{2} + 2 \, a^{3} b x + a^{4}\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-b}}{b \sqrt{x}}\right ) -{\left (6 \, b^{4} x^{3} - 21 \, a b^{3} x^{2} - 140 \, a^{2} b^{2} x - 105 \, a^{3} b\right )} \sqrt{b x + a} \sqrt{x}}{12 \,{\left (b^{7} x^{2} + 2 \, a b^{6} x + a^{2} 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))^(5/2),x, algorithm="fricas")

[Out]

[1/24*(105*(a^2*b^2*x^2 + 2*a^3*b*x + a^4)*sqrt(b)*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) + 2*(6*b^4
*x^3 - 21*a*b^3*x^2 - 140*a^2*b^2*x - 105*a^3*b)*sqrt(b*x + a)*sqrt(x))/(b^7*x^2 + 2*a*b^6*x + a^2*b^5), -1/12
*(105*(a^2*b^2*x^2 + 2*a^3*b*x + a^4)*sqrt(-b)*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) - (6*b^4*x^3 - 21*a*
b^3*x^2 - 140*a^2*b^2*x - 105*a^3*b)*sqrt(b*x + a)*sqrt(x))/(b^7*x^2 + 2*a*b^6*x + a^2*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))**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 1.20858, size = 101, normalized size = 0.66 \begin{align*} \frac{{\left ({\left (3 \, x{\left (\frac{2 \, x}{b} - \frac{7 \, a}{b^{2}}\right )} - \frac{140 \, a^{2}}{b^{3}}\right )} x - \frac{105 \, a^{3}}{b^{4}}\right )} \sqrt{x}}{12 \,{\left (b x + a\right )}^{\frac{3}{2}}} - \frac{35 \, a^{2} \log \left ({\left | -\sqrt{b} \sqrt{x} + \sqrt{b x + a} \right |}\right )}{4 \, 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))^(5/2),x, algorithm="giac")

[Out]

1/12*((3*x*(2*x/b - 7*a/b^2) - 140*a^2/b^3)*x - 105*a^3/b^4)*sqrt(x)/(b*x + a)^(3/2) - 35/4*a^2*log(abs(-sqrt(
b)*sqrt(x) + sqrt(b*x + a)))/b^(9/2)