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

Optimal. Leaf size=107 \[ -\frac{2 \left (a+b \tanh ^{-1}(c x)\right )}{3 d (d x)^{3/2}}-\frac{2 b c^{3/2} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{d x}}{\sqrt{d}}\right )}{3 d^{5/2}}+\frac{2 b c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d x}}{\sqrt{d}}\right )}{3 d^{5/2}}-\frac{4 b c}{3 d^2 \sqrt{d x}} \]

[Out]

(-4*b*c)/(3*d^2*Sqrt[d*x]) - (2*b*c^(3/2)*ArcTan[(Sqrt[c]*Sqrt[d*x])/Sqrt[d]])/(3*d^(5/2)) - (2*(a + b*ArcTanh
[c*x]))/(3*d*(d*x)^(3/2)) + (2*b*c^(3/2)*ArcTanh[(Sqrt[c]*Sqrt[d*x])/Sqrt[d]])/(3*d^(5/2))

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Rubi [A]  time = 0.0633332, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {5916, 325, 329, 298, 205, 208} \[ -\frac{2 \left (a+b \tanh ^{-1}(c x)\right )}{3 d (d x)^{3/2}}-\frac{2 b c^{3/2} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{d x}}{\sqrt{d}}\right )}{3 d^{5/2}}+\frac{2 b c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d x}}{\sqrt{d}}\right )}{3 d^{5/2}}-\frac{4 b c}{3 d^2 \sqrt{d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*b*c)/(3*d^2*Sqrt[d*x]) - (2*b*c^(3/2)*ArcTan[(Sqrt[c]*Sqrt[d*x])/Sqrt[d]])/(3*d^(5/2)) - (2*(a + b*ArcTanh
[c*x]))/(3*d*(d*x)^(3/2)) + (2*b*c^(3/2)*ArcTanh[(Sqrt[c]*Sqrt[d*x])/Sqrt[d]])/(3*d^(5/2))

Rule 5916

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcT
anh[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcTanh[c*x])^(p - 1))/(1 -
c^2*x^2), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[p, 0] && (EqQ[p, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
&  !GtQ[a/b, 0]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{a+b \tanh ^{-1}(c x)}{(d x)^{5/2}} \, dx &=-\frac{2 \left (a+b \tanh ^{-1}(c x)\right )}{3 d (d x)^{3/2}}+\frac{(2 b c) \int \frac{1}{(d x)^{3/2} \left (1-c^2 x^2\right )} \, dx}{3 d}\\ &=-\frac{4 b c}{3 d^2 \sqrt{d x}}-\frac{2 \left (a+b \tanh ^{-1}(c x)\right )}{3 d (d x)^{3/2}}+\frac{\left (2 b c^3\right ) \int \frac{\sqrt{d x}}{1-c^2 x^2} \, dx}{3 d^3}\\ &=-\frac{4 b c}{3 d^2 \sqrt{d x}}-\frac{2 \left (a+b \tanh ^{-1}(c x)\right )}{3 d (d x)^{3/2}}+\frac{\left (4 b c^3\right ) \operatorname{Subst}\left (\int \frac{x^2}{1-\frac{c^2 x^4}{d^2}} \, dx,x,\sqrt{d x}\right )}{3 d^4}\\ &=-\frac{4 b c}{3 d^2 \sqrt{d x}}-\frac{2 \left (a+b \tanh ^{-1}(c x)\right )}{3 d (d x)^{3/2}}+\frac{\left (2 b c^2\right ) \operatorname{Subst}\left (\int \frac{1}{d-c x^2} \, dx,x,\sqrt{d x}\right )}{3 d^2}-\frac{\left (2 b c^2\right ) \operatorname{Subst}\left (\int \frac{1}{d+c x^2} \, dx,x,\sqrt{d x}\right )}{3 d^2}\\ &=-\frac{4 b c}{3 d^2 \sqrt{d x}}-\frac{2 b c^{3/2} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{d x}}{\sqrt{d}}\right )}{3 d^{5/2}}-\frac{2 \left (a+b \tanh ^{-1}(c x)\right )}{3 d (d x)^{3/2}}+\frac{2 b c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d x}}{\sqrt{d}}\right )}{3 d^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.0604537, size = 107, normalized size = 1. \[ -\frac{x \left (2 a+b c^{3/2} x^{3/2} \log \left (1-\sqrt{c} \sqrt{x}\right )-b c^{3/2} x^{3/2} \log \left (\sqrt{c} \sqrt{x}+1\right )+2 b c^{3/2} x^{3/2} \tan ^{-1}\left (\sqrt{c} \sqrt{x}\right )+4 b c x+2 b \tanh ^{-1}(c x)\right )}{3 (d x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x*(2*a + 4*b*c*x + 2*b*c^(3/2)*x^(3/2)*ArcTan[Sqrt[c]*Sqrt[x]] + 2*b*ArcTanh[c*x] + b*c^(3/2)*x^(3/2)*Log[1
- Sqrt[c]*Sqrt[x]] - b*c^(3/2)*x^(3/2)*Log[1 + Sqrt[c]*Sqrt[x]]))/(3*(d*x)^(5/2))

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Maple [A]  time = 0.014, size = 94, normalized size = 0.9 \begin{align*} -{\frac{2\,a}{3\,d} \left ( dx \right ) ^{-{\frac{3}{2}}}}-{\frac{2\,b{\it Artanh} \left ( cx \right ) }{3\,d} \left ( dx \right ) ^{-{\frac{3}{2}}}}-{\frac{4\,bc}{3\,{d}^{2}}{\frac{1}{\sqrt{dx}}}}-{\frac{2\,b{c}^{2}}{3\,{d}^{2}}\arctan \left ({c\sqrt{dx}{\frac{1}{\sqrt{cd}}}} \right ){\frac{1}{\sqrt{cd}}}}+{\frac{2\,b{c}^{2}}{3\,{d}^{2}}{\it Artanh} \left ({c\sqrt{dx}{\frac{1}{\sqrt{cd}}}} \right ){\frac{1}{\sqrt{cd}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x))/(d*x)^(5/2),x)

[Out]

-2/3/d*a/(d*x)^(3/2)-2/3/d*b/(d*x)^(3/2)*arctanh(c*x)-4/3*b*c/d^2/(d*x)^(1/2)-2/3/d^2*b*c^2/(c*d)^(1/2)*arctan
(c*(d*x)^(1/2)/(c*d)^(1/2))+2/3/d^2*b*c^2/(c*d)^(1/2)*arctanh(c*(d*x)^(1/2)/(c*d)^(1/2))

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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((a+b*arctanh(c*x))/(d*x)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.2123, size = 558, normalized size = 5.21 \begin{align*} \left [\frac{2 \, b c d x^{2} \sqrt{\frac{c}{d}} \arctan \left (\frac{\sqrt{d x} \sqrt{\frac{c}{d}}}{c x}\right ) + b c d x^{2} \sqrt{\frac{c}{d}} \log \left (\frac{c x + 2 \, \sqrt{d x} \sqrt{\frac{c}{d}} + 1}{c x - 1}\right ) -{\left (4 \, b c x + b \log \left (-\frac{c x + 1}{c x - 1}\right ) + 2 \, a\right )} \sqrt{d x}}{3 \, d^{3} x^{2}}, -\frac{2 \, b c d x^{2} \sqrt{-\frac{c}{d}} \arctan \left (\frac{\sqrt{d x} \sqrt{-\frac{c}{d}}}{c x}\right ) - b c d x^{2} \sqrt{-\frac{c}{d}} \log \left (\frac{c x - 2 \, \sqrt{d x} \sqrt{-\frac{c}{d}} - 1}{c x + 1}\right ) +{\left (4 \, b c x + b \log \left (-\frac{c x + 1}{c x - 1}\right ) + 2 \, a\right )} \sqrt{d x}}{3 \, d^{3} x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))/(d*x)^(5/2),x, algorithm="fricas")

[Out]

[1/3*(2*b*c*d*x^2*sqrt(c/d)*arctan(sqrt(d*x)*sqrt(c/d)/(c*x)) + b*c*d*x^2*sqrt(c/d)*log((c*x + 2*sqrt(d*x)*sqr
t(c/d) + 1)/(c*x - 1)) - (4*b*c*x + b*log(-(c*x + 1)/(c*x - 1)) + 2*a)*sqrt(d*x))/(d^3*x^2), -1/3*(2*b*c*d*x^2
*sqrt(-c/d)*arctan(sqrt(d*x)*sqrt(-c/d)/(c*x)) - b*c*d*x^2*sqrt(-c/d)*log((c*x - 2*sqrt(d*x)*sqrt(-c/d) - 1)/(
c*x + 1)) + (4*b*c*x + b*log(-(c*x + 1)/(c*x - 1)) + 2*a)*sqrt(d*x))/(d^3*x^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a + b \operatorname{atanh}{\left (c x \right )}}{\left (d x\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(c*x))/(d*x)**(5/2),x)

[Out]

Integral((a + b*atanh(c*x))/(d*x)**(5/2), x)

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Giac [A]  time = 1.34339, size = 163, normalized size = 1.52 \begin{align*} -\frac{2}{3} \, b c^{3}{\left (\frac{\arctan \left (\frac{\sqrt{d x} c}{\sqrt{c d}}\right )}{\sqrt{c d} c d^{2}} + \frac{\arctan \left (\frac{\sqrt{d x} c}{\sqrt{-c d}}\right )}{\sqrt{-c d} c d^{2}}\right )} - \frac{\frac{b \log \left (-\frac{c d x + d}{c d x - d}\right )}{\sqrt{d x} d x} + \frac{2 \,{\left (2 \, b c d x + a d\right )}}{\sqrt{d x} d^{2} x}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))/(d*x)^(5/2),x, algorithm="giac")

[Out]

-2/3*b*c^3*(arctan(sqrt(d*x)*c/sqrt(c*d))/(sqrt(c*d)*c*d^2) + arctan(sqrt(d*x)*c/sqrt(-c*d))/(sqrt(-c*d)*c*d^2
)) - 1/3*(b*log(-(c*d*x + d)/(c*d*x - d))/(sqrt(d*x)*d*x) + 2*(2*b*c*d*x + a*d)/(sqrt(d*x)*d^2*x))/d