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

Optimal. Leaf size=106 \[ \frac{2 (d x)^{5/2} \left (a+b \tanh ^{-1}(c x)\right )}{5 d}+\frac{2 b d^{3/2} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{d x}}{\sqrt{d}}\right )}{5 c^{5/2}}-\frac{2 b d^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d x}}{\sqrt{d}}\right )}{5 c^{5/2}}+\frac{4 b (d x)^{3/2}}{15 c} \]

[Out]

(4*b*(d*x)^(3/2))/(15*c) + (2*b*d^(3/2)*ArcTan[(Sqrt[c]*Sqrt[d*x])/Sqrt[d]])/(5*c^(5/2)) + (2*(d*x)^(5/2)*(a +
b*ArcTanh[c*x]))/(5*d) - (2*b*d^(3/2)*ArcTanh[(Sqrt[c]*Sqrt[d*x])/Sqrt[d]])/(5*c^(5/2))

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Rubi [A]  time = 0.0672078, antiderivative size = 106, 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, 321, 329, 298, 205, 208} \[ \frac{2 (d x)^{5/2} \left (a+b \tanh ^{-1}(c x)\right )}{5 d}+\frac{2 b d^{3/2} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{d x}}{\sqrt{d}}\right )}{5 c^{5/2}}-\frac{2 b d^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d x}}{\sqrt{d}}\right )}{5 c^{5/2}}+\frac{4 b (d x)^{3/2}}{15 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*b*(d*x)^(3/2))/(15*c) + (2*b*d^(3/2)*ArcTan[(Sqrt[c]*Sqrt[d*x])/Sqrt[d]])/(5*c^(5/2)) + (2*(d*x)^(5/2)*(a +
b*ArcTanh[c*x]))/(5*d) - (2*b*d^(3/2)*ArcTanh[(Sqrt[c]*Sqrt[d*x])/Sqrt[d]])/(5*c^(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 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && 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 (d x)^{3/2} \left (a+b \tanh ^{-1}(c x)\right ) \, dx &=\frac{2 (d x)^{5/2} \left (a+b \tanh ^{-1}(c x)\right )}{5 d}-\frac{(2 b c) \int \frac{(d x)^{5/2}}{1-c^2 x^2} \, dx}{5 d}\\ &=\frac{4 b (d x)^{3/2}}{15 c}+\frac{2 (d x)^{5/2} \left (a+b \tanh ^{-1}(c x)\right )}{5 d}-\frac{(2 b d) \int \frac{\sqrt{d x}}{1-c^2 x^2} \, dx}{5 c}\\ &=\frac{4 b (d x)^{3/2}}{15 c}+\frac{2 (d x)^{5/2} \left (a+b \tanh ^{-1}(c x)\right )}{5 d}-\frac{(4 b) \operatorname{Subst}\left (\int \frac{x^2}{1-\frac{c^2 x^4}{d^2}} \, dx,x,\sqrt{d x}\right )}{5 c}\\ &=\frac{4 b (d x)^{3/2}}{15 c}+\frac{2 (d x)^{5/2} \left (a+b \tanh ^{-1}(c x)\right )}{5 d}-\frac{\left (2 b d^2\right ) \operatorname{Subst}\left (\int \frac{1}{d-c x^2} \, dx,x,\sqrt{d x}\right )}{5 c^2}+\frac{\left (2 b d^2\right ) \operatorname{Subst}\left (\int \frac{1}{d+c x^2} \, dx,x,\sqrt{d x}\right )}{5 c^2}\\ &=\frac{4 b (d x)^{3/2}}{15 c}+\frac{2 b d^{3/2} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{d x}}{\sqrt{d}}\right )}{5 c^{5/2}}+\frac{2 (d x)^{5/2} \left (a+b \tanh ^{-1}(c x)\right )}{5 d}-\frac{2 b d^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d x}}{\sqrt{d}}\right )}{5 c^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.055221, size = 115, normalized size = 1.08 \[ \frac{(d x)^{3/2} \left (6 a c^{5/2} x^{5/2}+4 b c^{3/2} x^{3/2}+6 b c^{5/2} x^{5/2} \tanh ^{-1}(c x)+3 b \log \left (1-\sqrt{c} \sqrt{x}\right )-3 b \log \left (\sqrt{c} \sqrt{x}+1\right )+6 b \tan ^{-1}\left (\sqrt{c} \sqrt{x}\right )\right )}{15 c^{5/2} x^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((d*x)^(3/2)*(4*b*c^(3/2)*x^(3/2) + 6*a*c^(5/2)*x^(5/2) + 6*b*ArcTan[Sqrt[c]*Sqrt[x]] + 6*b*c^(5/2)*x^(5/2)*Ar
cTanh[c*x] + 3*b*Log[1 - Sqrt[c]*Sqrt[x]] - 3*b*Log[1 + Sqrt[c]*Sqrt[x]]))/(15*c^(5/2)*x^(3/2))

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Maple [A]  time = 0.011, size = 93, normalized size = 0.9 \begin{align*}{\frac{2\,a}{5\,d} \left ( dx \right ) ^{{\frac{5}{2}}}}+{\frac{2\,b{\it Artanh} \left ( cx \right ) }{5\,d} \left ( dx \right ) ^{{\frac{5}{2}}}}+{\frac{4\,b}{15\,c} \left ( dx \right ) ^{{\frac{3}{2}}}}+{\frac{2\,{d}^{2}b}{5\,{c}^{2}}\arctan \left ({c\sqrt{dx}{\frac{1}{\sqrt{cd}}}} \right ){\frac{1}{\sqrt{cd}}}}-{\frac{2\,{d}^{2}b}{5\,{c}^{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((d*x)^(3/2)*(a+b*arctanh(c*x)),x)

[Out]

2/5/d*(d*x)^(5/2)*a+2/5/d*b*(d*x)^(5/2)*arctanh(c*x)+4/15*b*(d*x)^(3/2)/c+2/5*d^2*b/c^2/(c*d)^(1/2)*arctan(c*(
d*x)^(1/2)/(c*d)^(1/2))-2/5*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((d*x)^(3/2)*(a+b*arctanh(c*x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.2526, size = 586, normalized size = 5.53 \begin{align*} \left [\frac{6 \, b d \sqrt{\frac{d}{c}} \arctan \left (\frac{\sqrt{d x} c \sqrt{\frac{d}{c}}}{d}\right ) + 3 \, b d \sqrt{\frac{d}{c}} \log \left (\frac{c d x - 2 \, \sqrt{d x} c \sqrt{\frac{d}{c}} + d}{c x - 1}\right ) +{\left (3 \, b c^{2} d x^{2} \log \left (-\frac{c x + 1}{c x - 1}\right ) + 6 \, a c^{2} d x^{2} + 4 \, b c d x\right )} \sqrt{d x}}{15 \, c^{2}}, \frac{6 \, b d \sqrt{-\frac{d}{c}} \arctan \left (\frac{\sqrt{d x} c \sqrt{-\frac{d}{c}}}{d}\right ) + 3 \, b d \sqrt{-\frac{d}{c}} \log \left (\frac{c d x + 2 \, \sqrt{d x} c \sqrt{-\frac{d}{c}} - d}{c x + 1}\right ) +{\left (3 \, b c^{2} d x^{2} \log \left (-\frac{c x + 1}{c x - 1}\right ) + 6 \, a c^{2} d x^{2} + 4 \, b c d x\right )} \sqrt{d x}}{15 \, c^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/15*(6*b*d*sqrt(d/c)*arctan(sqrt(d*x)*c*sqrt(d/c)/d) + 3*b*d*sqrt(d/c)*log((c*d*x - 2*sqrt(d*x)*c*sqrt(d/c)
+ d)/(c*x - 1)) + (3*b*c^2*d*x^2*log(-(c*x + 1)/(c*x - 1)) + 6*a*c^2*d*x^2 + 4*b*c*d*x)*sqrt(d*x))/c^2, 1/15*(
6*b*d*sqrt(-d/c)*arctan(sqrt(d*x)*c*sqrt(-d/c)/d) + 3*b*d*sqrt(-d/c)*log((c*d*x + 2*sqrt(d*x)*c*sqrt(-d/c) - d
)/(c*x + 1)) + (3*b*c^2*d*x^2*log(-(c*x + 1)/(c*x - 1)) + 6*a*c^2*d*x^2 + 4*b*c*d*x)*sqrt(d*x))/c^2]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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