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

Optimal. Leaf size=85 \[ -\frac{2 \left (a+b \tanh ^{-1}(c x)\right )}{d \sqrt{d x}}+\frac{2 b \sqrt{c} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{d x}}{\sqrt{d}}\right )}{d^{3/2}}+\frac{2 b \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d x}}{\sqrt{d}}\right )}{d^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0546279, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {5916, 329, 212, 208, 205} \[ -\frac{2 \left (a+b \tanh ^{-1}(c x)\right )}{d \sqrt{d x}}+\frac{2 b \sqrt{c} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{d x}}{\sqrt{d}}\right )}{d^{3/2}}+\frac{2 b \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d x}}{\sqrt{d}}\right )}{d^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 212

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

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]

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]

Rubi steps

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

Mathematica [A]  time = 0.0429033, size = 99, normalized size = 1.16 \[ \frac{x \left (-2 a-b \sqrt{c} \sqrt{x} \log \left (1-\sqrt{c} \sqrt{x}\right )+b \sqrt{c} \sqrt{x} \log \left (\sqrt{c} \sqrt{x}+1\right )+2 b \sqrt{c} \sqrt{x} \tan ^{-1}\left (\sqrt{c} \sqrt{x}\right )-2 b \tanh ^{-1}(c x)\right )}{(d x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.013, size = 78, normalized size = 0.9 \begin{align*} -2\,{\frac{a}{d\sqrt{dx}}}-2\,{\frac{b{\it Artanh} \left ( cx \right ) }{d\sqrt{dx}}}+2\,{\frac{bc}{d\sqrt{cd}}\arctan \left ({\frac{c\sqrt{dx}}{\sqrt{cd}}} \right ) }+2\,{\frac{bc}{d\sqrt{cd}}{\it Artanh} \left ({\frac{c\sqrt{dx}}{\sqrt{cd}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x))/(d*x)^(3/2),x)

[Out]

-2/d*a/(d*x)^(1/2)-2/d*b/(d*x)^(1/2)*arctanh(c*x)+2/d*b*c/(c*d)^(1/2)*arctan(c*(d*x)^(1/2)/(c*d)^(1/2))+2/d*b*
c/(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)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.16046, size = 494, normalized size = 5.81 \begin{align*} \left [-\frac{2 \, b d x \sqrt{\frac{c}{d}} \arctan \left (\frac{\sqrt{d x} \sqrt{\frac{c}{d}}}{c x}\right ) - b d x \sqrt{\frac{c}{d}} \log \left (\frac{c x + 2 \, \sqrt{d x} \sqrt{\frac{c}{d}} + 1}{c x - 1}\right ) + \sqrt{d x}{\left (b \log \left (-\frac{c x + 1}{c x - 1}\right ) + 2 \, a\right )}}{d^{2} x}, -\frac{2 \, b d x \sqrt{-\frac{c}{d}} \arctan \left (\frac{\sqrt{d x} \sqrt{-\frac{c}{d}}}{c x}\right ) - b d x \sqrt{-\frac{c}{d}} \log \left (\frac{c x + 2 \, \sqrt{d x} \sqrt{-\frac{c}{d}} - 1}{c x + 1}\right ) + \sqrt{d x}{\left (b \log \left (-\frac{c x + 1}{c x - 1}\right ) + 2 \, a\right )}}{d^{2} x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-(2*b*d*x*sqrt(c/d)*arctan(sqrt(d*x)*sqrt(c/d)/(c*x)) - b*d*x*sqrt(c/d)*log((c*x + 2*sqrt(d*x)*sqrt(c/d) + 1)
/(c*x - 1)) + sqrt(d*x)*(b*log(-(c*x + 1)/(c*x - 1)) + 2*a))/(d^2*x), -(2*b*d*x*sqrt(-c/d)*arctan(sqrt(d*x)*sq
rt(-c/d)/(c*x)) - b*d*x*sqrt(-c/d)*log((c*x + 2*sqrt(d*x)*sqrt(-c/d) - 1)/(c*x + 1)) + sqrt(d*x)*(b*log(-(c*x
+ 1)/(c*x - 1)) + 2*a))/(d^2*x)]

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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{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(c*x))/(d*x)**(3/2),x)

[Out]

Integral((a + b*atanh(c*x))/(d*x)**(3/2), x)

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Giac [A]  time = 1.27011, size = 127, normalized size = 1.49 \begin{align*} 2 \, b c d{\left (\frac{\arctan \left (\frac{\sqrt{d x} c}{\sqrt{c d}}\right )}{\sqrt{c d} d^{2}} - \frac{\arctan \left (\frac{\sqrt{d x} c}{\sqrt{-c d}}\right )}{\sqrt{-c d} d^{2}}\right )} - \frac{\frac{b \log \left (-\frac{c d x + d}{c d x - d}\right )}{\sqrt{d x}} + \frac{2 \, a}{\sqrt{d x}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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