3.42 \(\int \frac{a+b \tanh ^{-1}(c x)}{(d x)^{9/2}} \, dx\)

Optimal. Leaf size=125 \[ -\frac{2 \left (a+b \tanh ^{-1}(c x)\right )}{7 d (d x)^{7/2}}-\frac{4 b c^3}{7 d^4 \sqrt{d x}}-\frac{2 b c^{7/2} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{d x}}{\sqrt{d}}\right )}{7 d^{9/2}}+\frac{2 b c^{7/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d x}}{\sqrt{d}}\right )}{7 d^{9/2}}-\frac{4 b c}{35 d^2 (d x)^{5/2}} \]

[Out]

(-4*b*c)/(35*d^2*(d*x)^(5/2)) - (4*b*c^3)/(7*d^4*Sqrt[d*x]) - (2*b*c^(7/2)*ArcTan[(Sqrt[c]*Sqrt[d*x])/Sqrt[d]]
)/(7*d^(9/2)) - (2*(a + b*ArcTanh[c*x]))/(7*d*(d*x)^(7/2)) + (2*b*c^(7/2)*ArcTanh[(Sqrt[c]*Sqrt[d*x])/Sqrt[d]]
)/(7*d^(9/2))

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Rubi [A]  time = 0.078228, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ -\frac{2 \left (a+b \tanh ^{-1}(c x)\right )}{7 d (d x)^{7/2}}-\frac{4 b c^3}{7 d^4 \sqrt{d x}}-\frac{2 b c^{7/2} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{d x}}{\sqrt{d}}\right )}{7 d^{9/2}}+\frac{2 b c^{7/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d x}}{\sqrt{d}}\right )}{7 d^{9/2}}-\frac{4 b c}{35 d^2 (d x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x])/(d*x)^(9/2),x]

[Out]

(-4*b*c)/(35*d^2*(d*x)^(5/2)) - (4*b*c^3)/(7*d^4*Sqrt[d*x]) - (2*b*c^(7/2)*ArcTan[(Sqrt[c]*Sqrt[d*x])/Sqrt[d]]
)/(7*d^(9/2)) - (2*(a + b*ArcTanh[c*x]))/(7*d*(d*x)^(7/2)) + (2*b*c^(7/2)*ArcTanh[(Sqrt[c]*Sqrt[d*x])/Sqrt[d]]
)/(7*d^(9/2))

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)^{9/2}} \, dx &=-\frac{2 \left (a+b \tanh ^{-1}(c x)\right )}{7 d (d x)^{7/2}}+\frac{(2 b c) \int \frac{1}{(d x)^{7/2} \left (1-c^2 x^2\right )} \, dx}{7 d}\\ &=-\frac{4 b c}{35 d^2 (d x)^{5/2}}-\frac{2 \left (a+b \tanh ^{-1}(c x)\right )}{7 d (d x)^{7/2}}+\frac{\left (2 b c^3\right ) \int \frac{1}{(d x)^{3/2} \left (1-c^2 x^2\right )} \, dx}{7 d^3}\\ &=-\frac{4 b c}{35 d^2 (d x)^{5/2}}-\frac{4 b c^3}{7 d^4 \sqrt{d x}}-\frac{2 \left (a+b \tanh ^{-1}(c x)\right )}{7 d (d x)^{7/2}}+\frac{\left (2 b c^5\right ) \int \frac{\sqrt{d x}}{1-c^2 x^2} \, dx}{7 d^5}\\ &=-\frac{4 b c}{35 d^2 (d x)^{5/2}}-\frac{4 b c^3}{7 d^4 \sqrt{d x}}-\frac{2 \left (a+b \tanh ^{-1}(c x)\right )}{7 d (d x)^{7/2}}+\frac{\left (4 b c^5\right ) \operatorname{Subst}\left (\int \frac{x^2}{1-\frac{c^2 x^4}{d^2}} \, dx,x,\sqrt{d x}\right )}{7 d^6}\\ &=-\frac{4 b c}{35 d^2 (d x)^{5/2}}-\frac{4 b c^3}{7 d^4 \sqrt{d x}}-\frac{2 \left (a+b \tanh ^{-1}(c x)\right )}{7 d (d x)^{7/2}}+\frac{\left (2 b c^4\right ) \operatorname{Subst}\left (\int \frac{1}{d-c x^2} \, dx,x,\sqrt{d x}\right )}{7 d^4}-\frac{\left (2 b c^4\right ) \operatorname{Subst}\left (\int \frac{1}{d+c x^2} \, dx,x,\sqrt{d x}\right )}{7 d^4}\\ &=-\frac{4 b c}{35 d^2 (d x)^{5/2}}-\frac{4 b c^3}{7 d^4 \sqrt{d x}}-\frac{2 b c^{7/2} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{d x}}{\sqrt{d}}\right )}{7 d^{9/2}}-\frac{2 \left (a+b \tanh ^{-1}(c x)\right )}{7 d (d x)^{7/2}}+\frac{2 b c^{7/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d x}}{\sqrt{d}}\right )}{7 d^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.0661833, size = 122, normalized size = 0.98 \[ -\frac{\sqrt{d x} \left (10 a+20 b c^3 x^3+5 b c^{7/2} x^{7/2} \log \left (1-\sqrt{c} \sqrt{x}\right )-5 b c^{7/2} x^{7/2} \log \left (\sqrt{c} \sqrt{x}+1\right )+10 b c^{7/2} x^{7/2} \tan ^{-1}\left (\sqrt{c} \sqrt{x}\right )+4 b c x+10 b \tanh ^{-1}(c x)\right )}{35 d^5 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTanh[c*x])/(d*x)^(9/2),x]

[Out]

-(Sqrt[d*x]*(10*a + 4*b*c*x + 20*b*c^3*x^3 + 10*b*c^(7/2)*x^(7/2)*ArcTan[Sqrt[c]*Sqrt[x]] + 10*b*ArcTanh[c*x]
+ 5*b*c^(7/2)*x^(7/2)*Log[1 - Sqrt[c]*Sqrt[x]] - 5*b*c^(7/2)*x^(7/2)*Log[1 + Sqrt[c]*Sqrt[x]]))/(35*d^5*x^4)

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Maple [A]  time = 0.015, size = 108, normalized size = 0.9 \begin{align*} -{\frac{2\,a}{7\,d} \left ( dx \right ) ^{-{\frac{7}{2}}}}-{\frac{2\,b{\it Artanh} \left ( cx \right ) }{7\,d} \left ( dx \right ) ^{-{\frac{7}{2}}}}-{\frac{2\,b{c}^{4}}{7\,{d}^{4}}\arctan \left ({c\sqrt{dx}{\frac{1}{\sqrt{cd}}}} \right ){\frac{1}{\sqrt{cd}}}}-{\frac{4\,bc}{35\,{d}^{2}} \left ( dx \right ) ^{-{\frac{5}{2}}}}-{\frac{4\,b{c}^{3}}{7\,{d}^{4}}{\frac{1}{\sqrt{dx}}}}+{\frac{2\,b{c}^{4}}{7\,{d}^{4}}{\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)^(9/2),x)

[Out]

-2/7/d*a/(d*x)^(7/2)-2/7/d*b/(d*x)^(7/2)*arctanh(c*x)-2/7/d^4*b*c^4/(c*d)^(1/2)*arctan(c*(d*x)^(1/2)/(c*d)^(1/
2))-4/35*b*c/d^2/(d*x)^(5/2)-4/7*b*c^3/d^4/(d*x)^(1/2)+2/7/d^4*b*c^4/(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)^(9/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.45178, size = 628, normalized size = 5.02 \begin{align*} \left [\frac{10 \, b c^{3} d x^{4} \sqrt{\frac{c}{d}} \arctan \left (\frac{\sqrt{d x} \sqrt{\frac{c}{d}}}{c x}\right ) + 5 \, b c^{3} d x^{4} \sqrt{\frac{c}{d}} \log \left (\frac{c x + 2 \, \sqrt{d x} \sqrt{\frac{c}{d}} + 1}{c x - 1}\right ) -{\left (20 \, b c^{3} x^{3} + 4 \, b c x + 5 \, b \log \left (-\frac{c x + 1}{c x - 1}\right ) + 10 \, a\right )} \sqrt{d x}}{35 \, d^{5} x^{4}}, -\frac{10 \, b c^{3} d x^{4} \sqrt{-\frac{c}{d}} \arctan \left (\frac{\sqrt{d x} \sqrt{-\frac{c}{d}}}{c x}\right ) - 5 \, b c^{3} d x^{4} \sqrt{-\frac{c}{d}} \log \left (\frac{c x - 2 \, \sqrt{d x} \sqrt{-\frac{c}{d}} - 1}{c x + 1}\right ) +{\left (20 \, b c^{3} x^{3} + 4 \, b c x + 5 \, b \log \left (-\frac{c x + 1}{c x - 1}\right ) + 10 \, a\right )} \sqrt{d x}}{35 \, d^{5} x^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))/(d*x)^(9/2),x, algorithm="fricas")

[Out]

[1/35*(10*b*c^3*d*x^4*sqrt(c/d)*arctan(sqrt(d*x)*sqrt(c/d)/(c*x)) + 5*b*c^3*d*x^4*sqrt(c/d)*log((c*x + 2*sqrt(
d*x)*sqrt(c/d) + 1)/(c*x - 1)) - (20*b*c^3*x^3 + 4*b*c*x + 5*b*log(-(c*x + 1)/(c*x - 1)) + 10*a)*sqrt(d*x))/(d
^5*x^4), -1/35*(10*b*c^3*d*x^4*sqrt(-c/d)*arctan(sqrt(d*x)*sqrt(-c/d)/(c*x)) - 5*b*c^3*d*x^4*sqrt(-c/d)*log((c
*x - 2*sqrt(d*x)*sqrt(-c/d) - 1)/(c*x + 1)) + (20*b*c^3*x^3 + 4*b*c*x + 5*b*log(-(c*x + 1)/(c*x - 1)) + 10*a)*
sqrt(d*x))/(d^5*x^4)]

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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((a+b*atanh(c*x))/(d*x)**(9/2),x)

[Out]

Timed out

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Giac [A]  time = 1.4231, size = 188, normalized size = 1.5 \begin{align*} -\frac{2}{7} \, b c^{5}{\left (\frac{\arctan \left (\frac{\sqrt{d x} c}{\sqrt{c d}}\right )}{\sqrt{c d} c d^{4}} + \frac{\arctan \left (\frac{\sqrt{d x} c}{\sqrt{-c d}}\right )}{\sqrt{-c d} c d^{4}}\right )} - \frac{\frac{5 \, b \log \left (-\frac{c d x + d}{c d x - d}\right )}{\sqrt{d x} d^{3} x^{3}} + \frac{2 \,{\left (10 \, b c^{3} d^{3} x^{3} + 2 \, b c d^{3} x + 5 \, a d^{3}\right )}}{\sqrt{d x} d^{6} x^{3}}}{35 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))/(d*x)^(9/2),x, algorithm="giac")

[Out]

-2/7*b*c^5*(arctan(sqrt(d*x)*c/sqrt(c*d))/(sqrt(c*d)*c*d^4) + arctan(sqrt(d*x)*c/sqrt(-c*d))/(sqrt(-c*d)*c*d^4
)) - 1/35*(5*b*log(-(c*d*x + d)/(c*d*x - d))/(sqrt(d*x)*d^3*x^3) + 2*(10*b*c^3*d^3*x^3 + 2*b*c*d^3*x + 5*a*d^3
)/(sqrt(d*x)*d^6*x^3))/d