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

Optimal. Leaf size=317 \[ -\frac{2 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )}{5 d (d x)^{5/2}}-\frac{b c^{5/4} \log \left (\sqrt{c} \sqrt{d} x-\sqrt{2} \sqrt [4]{c} \sqrt{d x}+\sqrt{d}\right )}{5 \sqrt{2} d^{7/2}}+\frac{b c^{5/4} \log \left (\sqrt{c} \sqrt{d} x+\sqrt{2} \sqrt [4]{c} \sqrt{d x}+\sqrt{d}\right )}{5 \sqrt{2} d^{7/2}}-\frac{2 b c^{5/4} \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d x}}{\sqrt{d}}\right )}{5 d^{7/2}}+\frac{\sqrt{2} b c^{5/4} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{d x}}{\sqrt{d}}\right )}{5 d^{7/2}}-\frac{\sqrt{2} b c^{5/4} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{d x}}{\sqrt{d}}+1\right )}{5 d^{7/2}}+\frac{2 b c^{5/4} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d x}}{\sqrt{d}}\right )}{5 d^{7/2}}-\frac{8 b c}{5 d^3 \sqrt{d x}} \]

[Out]

(-8*b*c)/(5*d^3*Sqrt[d*x]) - (2*b*c^(5/4)*ArcTan[(c^(1/4)*Sqrt[d*x])/Sqrt[d]])/(5*d^(7/2)) + (Sqrt[2]*b*c^(5/4
)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[d*x])/Sqrt[d]])/(5*d^(7/2)) - (Sqrt[2]*b*c^(5/4)*ArcTan[1 + (Sqrt[2]*c^(1/4
)*Sqrt[d*x])/Sqrt[d]])/(5*d^(7/2)) - (2*(a + b*ArcTanh[c*x^2]))/(5*d*(d*x)^(5/2)) + (2*b*c^(5/4)*ArcTanh[(c^(1
/4)*Sqrt[d*x])/Sqrt[d]])/(5*d^(7/2)) - (b*c^(5/4)*Log[Sqrt[d] + Sqrt[c]*Sqrt[d]*x - Sqrt[2]*c^(1/4)*Sqrt[d*x]]
)/(5*Sqrt[2]*d^(7/2)) + (b*c^(5/4)*Log[Sqrt[d] + Sqrt[c]*Sqrt[d]*x + Sqrt[2]*c^(1/4)*Sqrt[d*x]])/(5*Sqrt[2]*d^
(7/2))

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Rubi [A]  time = 0.277762, antiderivative size = 317, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 14, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.778, Rules used = {6097, 16, 325, 329, 301, 297, 1162, 617, 204, 1165, 628, 298, 205, 208} \[ -\frac{2 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )}{5 d (d x)^{5/2}}-\frac{b c^{5/4} \log \left (\sqrt{c} \sqrt{d} x-\sqrt{2} \sqrt [4]{c} \sqrt{d x}+\sqrt{d}\right )}{5 \sqrt{2} d^{7/2}}+\frac{b c^{5/4} \log \left (\sqrt{c} \sqrt{d} x+\sqrt{2} \sqrt [4]{c} \sqrt{d x}+\sqrt{d}\right )}{5 \sqrt{2} d^{7/2}}-\frac{2 b c^{5/4} \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d x}}{\sqrt{d}}\right )}{5 d^{7/2}}+\frac{\sqrt{2} b c^{5/4} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{d x}}{\sqrt{d}}\right )}{5 d^{7/2}}-\frac{\sqrt{2} b c^{5/4} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{d x}}{\sqrt{d}}+1\right )}{5 d^{7/2}}+\frac{2 b c^{5/4} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d x}}{\sqrt{d}}\right )}{5 d^{7/2}}-\frac{8 b c}{5 d^3 \sqrt{d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*b*c)/(5*d^3*Sqrt[d*x]) - (2*b*c^(5/4)*ArcTan[(c^(1/4)*Sqrt[d*x])/Sqrt[d]])/(5*d^(7/2)) + (Sqrt[2]*b*c^(5/4
)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[d*x])/Sqrt[d]])/(5*d^(7/2)) - (Sqrt[2]*b*c^(5/4)*ArcTan[1 + (Sqrt[2]*c^(1/4
)*Sqrt[d*x])/Sqrt[d]])/(5*d^(7/2)) - (2*(a + b*ArcTanh[c*x^2]))/(5*d*(d*x)^(5/2)) + (2*b*c^(5/4)*ArcTanh[(c^(1
/4)*Sqrt[d*x])/Sqrt[d]])/(5*d^(7/2)) - (b*c^(5/4)*Log[Sqrt[d] + Sqrt[c]*Sqrt[d]*x - Sqrt[2]*c^(1/4)*Sqrt[d*x]]
)/(5*Sqrt[2]*d^(7/2)) + (b*c^(5/4)*Log[Sqrt[d] + Sqrt[c]*Sqrt[d]*x + Sqrt[2]*c^(1/4)*Sqrt[d*x]])/(5*Sqrt[2]*d^
(7/2))

Rule 6097

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcTa
nh[c*x^n]))/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[(x^(n - 1)*(d*x)^(m + 1))/(1 - c^2*x^(2*n)), x], x
] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1]

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

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 301

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

Rule 297

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

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

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}\left (c x^2\right )}{(d x)^{7/2}} \, dx &=-\frac{2 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )}{5 d (d x)^{5/2}}+\frac{(4 b c) \int \frac{x}{(d x)^{5/2} \left (1-c^2 x^4\right )} \, dx}{5 d}\\ &=-\frac{2 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )}{5 d (d x)^{5/2}}+\frac{(4 b c) \int \frac{1}{(d x)^{3/2} \left (1-c^2 x^4\right )} \, dx}{5 d^2}\\ &=-\frac{8 b c}{5 d^3 \sqrt{d x}}-\frac{2 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )}{5 d (d x)^{5/2}}+\frac{\left (4 b c^3\right ) \int \frac{(d x)^{5/2}}{1-c^2 x^4} \, dx}{5 d^6}\\ &=-\frac{8 b c}{5 d^3 \sqrt{d x}}-\frac{2 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )}{5 d (d x)^{5/2}}+\frac{\left (8 b c^3\right ) \operatorname{Subst}\left (\int \frac{x^6}{1-\frac{c^2 x^8}{d^4}} \, dx,x,\sqrt{d x}\right )}{5 d^7}\\ &=-\frac{8 b c}{5 d^3 \sqrt{d x}}-\frac{2 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )}{5 d (d x)^{5/2}}+\frac{\left (4 b c^2\right ) \operatorname{Subst}\left (\int \frac{x^2}{d^2-c x^4} \, dx,x,\sqrt{d x}\right )}{5 d^3}-\frac{\left (4 b c^2\right ) \operatorname{Subst}\left (\int \frac{x^2}{d^2+c x^4} \, dx,x,\sqrt{d x}\right )}{5 d^3}\\ &=-\frac{8 b c}{5 d^3 \sqrt{d x}}-\frac{2 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )}{5 d (d x)^{5/2}}+\frac{\left (2 b c^{3/2}\right ) \operatorname{Subst}\left (\int \frac{1}{d-\sqrt{c} x^2} \, dx,x,\sqrt{d x}\right )}{5 d^3}-\frac{\left (2 b c^{3/2}\right ) \operatorname{Subst}\left (\int \frac{1}{d+\sqrt{c} x^2} \, dx,x,\sqrt{d x}\right )}{5 d^3}+\frac{\left (2 b c^{3/2}\right ) \operatorname{Subst}\left (\int \frac{d-\sqrt{c} x^2}{d^2+c x^4} \, dx,x,\sqrt{d x}\right )}{5 d^3}-\frac{\left (2 b c^{3/2}\right ) \operatorname{Subst}\left (\int \frac{d+\sqrt{c} x^2}{d^2+c x^4} \, dx,x,\sqrt{d x}\right )}{5 d^3}\\ &=-\frac{8 b c}{5 d^3 \sqrt{d x}}-\frac{2 b c^{5/4} \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d x}}{\sqrt{d}}\right )}{5 d^{7/2}}-\frac{2 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )}{5 d (d x)^{5/2}}+\frac{2 b c^{5/4} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d x}}{\sqrt{d}}\right )}{5 d^{7/2}}-\frac{\left (b c^{5/4}\right ) \operatorname{Subst}\left (\int \frac{\frac{\sqrt{2} \sqrt{d}}{\sqrt [4]{c}}+2 x}{-\frac{d}{\sqrt{c}}-\frac{\sqrt{2} \sqrt{d} x}{\sqrt [4]{c}}-x^2} \, dx,x,\sqrt{d x}\right )}{5 \sqrt{2} d^{7/2}}-\frac{\left (b c^{5/4}\right ) \operatorname{Subst}\left (\int \frac{\frac{\sqrt{2} \sqrt{d}}{\sqrt [4]{c}}-2 x}{-\frac{d}{\sqrt{c}}+\frac{\sqrt{2} \sqrt{d} x}{\sqrt [4]{c}}-x^2} \, dx,x,\sqrt{d x}\right )}{5 \sqrt{2} d^{7/2}}-\frac{(b c) \operatorname{Subst}\left (\int \frac{1}{\frac{d}{\sqrt{c}}-\frac{\sqrt{2} \sqrt{d} x}{\sqrt [4]{c}}+x^2} \, dx,x,\sqrt{d x}\right )}{5 d^3}-\frac{(b c) \operatorname{Subst}\left (\int \frac{1}{\frac{d}{\sqrt{c}}+\frac{\sqrt{2} \sqrt{d} x}{\sqrt [4]{c}}+x^2} \, dx,x,\sqrt{d x}\right )}{5 d^3}\\ &=-\frac{8 b c}{5 d^3 \sqrt{d x}}-\frac{2 b c^{5/4} \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d x}}{\sqrt{d}}\right )}{5 d^{7/2}}-\frac{2 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )}{5 d (d x)^{5/2}}+\frac{2 b c^{5/4} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d x}}{\sqrt{d}}\right )}{5 d^{7/2}}-\frac{b c^{5/4} \log \left (\sqrt{d}+\sqrt{c} \sqrt{d} x-\sqrt{2} \sqrt [4]{c} \sqrt{d x}\right )}{5 \sqrt{2} d^{7/2}}+\frac{b c^{5/4} \log \left (\sqrt{d}+\sqrt{c} \sqrt{d} x+\sqrt{2} \sqrt [4]{c} \sqrt{d x}\right )}{5 \sqrt{2} d^{7/2}}-\frac{\left (\sqrt{2} b c^{5/4}\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{d x}}{\sqrt{d}}\right )}{5 d^{7/2}}+\frac{\left (\sqrt{2} b c^{5/4}\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\frac{\sqrt{2} \sqrt [4]{c} \sqrt{d x}}{\sqrt{d}}\right )}{5 d^{7/2}}\\ &=-\frac{8 b c}{5 d^3 \sqrt{d x}}-\frac{2 b c^{5/4} \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d x}}{\sqrt{d}}\right )}{5 d^{7/2}}+\frac{\sqrt{2} b c^{5/4} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{d x}}{\sqrt{d}}\right )}{5 d^{7/2}}-\frac{\sqrt{2} b c^{5/4} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{c} \sqrt{d x}}{\sqrt{d}}\right )}{5 d^{7/2}}-\frac{2 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )}{5 d (d x)^{5/2}}+\frac{2 b c^{5/4} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d x}}{\sqrt{d}}\right )}{5 d^{7/2}}-\frac{b c^{5/4} \log \left (\sqrt{d}+\sqrt{c} \sqrt{d} x-\sqrt{2} \sqrt [4]{c} \sqrt{d x}\right )}{5 \sqrt{2} d^{7/2}}+\frac{b c^{5/4} \log \left (\sqrt{d}+\sqrt{c} \sqrt{d} x+\sqrt{2} \sqrt [4]{c} \sqrt{d x}\right )}{5 \sqrt{2} d^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.0857637, size = 275, normalized size = 0.87 \[ \frac{x \left (-4 a-2 b c^{5/4} x^{5/2} \log \left (1-\sqrt [4]{c} \sqrt{x}\right )+2 b c^{5/4} x^{5/2} \log \left (\sqrt [4]{c} \sqrt{x}+1\right )-\sqrt{2} b c^{5/4} x^{5/2} \log \left (\sqrt{c} x-\sqrt{2} \sqrt [4]{c} \sqrt{x}+1\right )+\sqrt{2} b c^{5/4} x^{5/2} \log \left (\sqrt{c} x+\sqrt{2} \sqrt [4]{c} \sqrt{x}+1\right )+2 \sqrt{2} b c^{5/4} x^{5/2} \tan ^{-1}\left (1-\sqrt{2} \sqrt [4]{c} \sqrt{x}\right )-2 \sqrt{2} b c^{5/4} x^{5/2} \tan ^{-1}\left (\sqrt{2} \sqrt [4]{c} \sqrt{x}+1\right )-4 b c^{5/4} x^{5/2} \tan ^{-1}\left (\sqrt [4]{c} \sqrt{x}\right )-16 b c x^2-4 b \tanh ^{-1}\left (c x^2\right )\right )}{10 (d x)^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(-4*a - 16*b*c*x^2 + 2*Sqrt[2]*b*c^(5/4)*x^(5/2)*ArcTan[1 - Sqrt[2]*c^(1/4)*Sqrt[x]] - 2*Sqrt[2]*b*c^(5/4)*
x^(5/2)*ArcTan[1 + Sqrt[2]*c^(1/4)*Sqrt[x]] - 4*b*c^(5/4)*x^(5/2)*ArcTan[c^(1/4)*Sqrt[x]] - 4*b*ArcTanh[c*x^2]
- 2*b*c^(5/4)*x^(5/2)*Log[1 - c^(1/4)*Sqrt[x]] + 2*b*c^(5/4)*x^(5/2)*Log[1 + c^(1/4)*Sqrt[x]] - Sqrt[2]*b*c^(
5/4)*x^(5/2)*Log[1 - Sqrt[2]*c^(1/4)*Sqrt[x] + Sqrt[c]*x] + Sqrt[2]*b*c^(5/4)*x^(5/2)*Log[1 + Sqrt[2]*c^(1/4)*
Sqrt[x] + Sqrt[c]*x]))/(10*(d*x)^(7/2))

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Maple [A]  time = 0.016, size = 292, normalized size = 0.9 \begin{align*} -{\frac{2\,a}{5\,d} \left ( dx \right ) ^{-{\frac{5}{2}}}}-{\frac{2\,b{\it Artanh} \left ( c{x}^{2} \right ) }{5\,d} \left ( dx \right ) ^{-{\frac{5}{2}}}}-{\frac{bc\sqrt{2}}{10\,{d}^{3}}\ln \left ({ \left ( dx-\sqrt [4]{{\frac{{d}^{2}}{c}}}\sqrt{dx}\sqrt{2}+\sqrt{{\frac{{d}^{2}}{c}}} \right ) \left ( dx+\sqrt [4]{{\frac{{d}^{2}}{c}}}\sqrt{dx}\sqrt{2}+\sqrt{{\frac{{d}^{2}}{c}}} \right ) ^{-1}} \right ){\frac{1}{\sqrt [4]{{\frac{{d}^{2}}{c}}}}}}-{\frac{bc\sqrt{2}}{5\,{d}^{3}}\arctan \left ({\sqrt{2}\sqrt{dx}{\frac{1}{\sqrt [4]{{\frac{{d}^{2}}{c}}}}}}+1 \right ){\frac{1}{\sqrt [4]{{\frac{{d}^{2}}{c}}}}}}-{\frac{bc\sqrt{2}}{5\,{d}^{3}}\arctan \left ({\sqrt{2}\sqrt{dx}{\frac{1}{\sqrt [4]{{\frac{{d}^{2}}{c}}}}}}-1 \right ){\frac{1}{\sqrt [4]{{\frac{{d}^{2}}{c}}}}}}-{\frac{8\,bc}{5\,{d}^{3}}{\frac{1}{\sqrt{dx}}}}-{\frac{2\,bc}{5\,{d}^{3}}\arctan \left ({\sqrt{dx}{\frac{1}{\sqrt [4]{{\frac{{d}^{2}}{c}}}}}} \right ){\frac{1}{\sqrt [4]{{\frac{{d}^{2}}{c}}}}}}+{\frac{bc}{5\,{d}^{3}}\ln \left ({ \left ( \sqrt{dx}+\sqrt [4]{{\frac{{d}^{2}}{c}}} \right ) \left ( \sqrt{dx}-\sqrt [4]{{\frac{{d}^{2}}{c}}} \right ) ^{-1}} \right ){\frac{1}{\sqrt [4]{{\frac{{d}^{2}}{c}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x^2))/(d*x)^(7/2),x)

[Out]

-2/5/d*a/(d*x)^(5/2)-2/5/d*b/(d*x)^(5/2)*arctanh(c*x^2)-1/10/d^3*b*c/(d^2/c)^(1/4)*2^(1/2)*ln((d*x-(d^2/c)^(1/
4)*(d*x)^(1/2)*2^(1/2)+(d^2/c)^(1/2))/(d*x+(d^2/c)^(1/4)*(d*x)^(1/2)*2^(1/2)+(d^2/c)^(1/2)))-1/5/d^3*b*c/(d^2/
c)^(1/4)*2^(1/2)*arctan(2^(1/2)/(d^2/c)^(1/4)*(d*x)^(1/2)+1)-1/5/d^3*b*c/(d^2/c)^(1/4)*2^(1/2)*arctan(2^(1/2)/
(d^2/c)^(1/4)*(d*x)^(1/2)-1)-8/5*b*c/d^3/(d*x)^(1/2)-2/5/d^3*b*c/(d^2/c)^(1/4)*arctan((d*x)^(1/2)/(d^2/c)^(1/4
))+1/5/d^3*b*c/(d^2/c)^(1/4)*ln(((d*x)^(1/2)+(d^2/c)^(1/4))/((d*x)^(1/2)-(d^2/c)^(1/4)))

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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^2))/(d*x)^(7/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.23702, size = 105, normalized size = 0.33 \begin{align*} -\frac{{\left (8 \, b c x^{2} + b \log \left (-\frac{c x^{2} + 1}{c x^{2} - 1}\right ) + 2 \, a\right )} \sqrt{d x}}{5 \, d^{4} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x^2))/(d*x)^(7/2),x, algorithm="fricas")

[Out]

-1/5*(8*b*c*x^2 + b*log(-(c*x^2 + 1)/(c*x^2 - 1)) + 2*a)*sqrt(d*x)/(d^4*x^3)

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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**2))/(d*x)**(7/2),x)

[Out]

Timed out

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Giac [B]  time = 9.77957, size = 717, normalized size = 2.26 \begin{align*} -\frac{1}{10} \, b c^{3}{\left (\frac{2 \, \sqrt{2} \left (c^{3} d^{2}\right )^{\frac{3}{4}} \arctan \left (\frac{\sqrt{2}{\left (\sqrt{2} \left (\frac{d^{2}}{c}\right )^{\frac{1}{4}} + 2 \, \sqrt{d x}\right )}}{2 \, \left (\frac{d^{2}}{c}\right )^{\frac{1}{4}}}\right )}{c^{4} d^{5}} + \frac{2 \, \sqrt{2} \left (c^{3} d^{2}\right )^{\frac{3}{4}} \arctan \left (-\frac{\sqrt{2}{\left (\sqrt{2} \left (\frac{d^{2}}{c}\right )^{\frac{1}{4}} - 2 \, \sqrt{d x}\right )}}{2 \, \left (\frac{d^{2}}{c}\right )^{\frac{1}{4}}}\right )}{c^{4} d^{5}} - \frac{2 \, \sqrt{2} \left (-c^{3} d^{2}\right )^{\frac{3}{4}} \arctan \left (\frac{\sqrt{2}{\left (\sqrt{2} \left (-\frac{d^{2}}{c}\right )^{\frac{1}{4}} + 2 \, \sqrt{d x}\right )}}{2 \, \left (-\frac{d^{2}}{c}\right )^{\frac{1}{4}}}\right )}{c^{4} d^{5}} - \frac{2 \, \sqrt{2} \left (-c^{3} d^{2}\right )^{\frac{3}{4}} \arctan \left (-\frac{\sqrt{2}{\left (\sqrt{2} \left (-\frac{d^{2}}{c}\right )^{\frac{1}{4}} - 2 \, \sqrt{d x}\right )}}{2 \, \left (-\frac{d^{2}}{c}\right )^{\frac{1}{4}}}\right )}{c^{4} d^{5}} - \frac{\sqrt{2} \left (c^{3} d^{2}\right )^{\frac{3}{4}} \log \left (d x + \sqrt{2} \sqrt{d x} \left (\frac{d^{2}}{c}\right )^{\frac{1}{4}} + \sqrt{\frac{d^{2}}{c}}\right )}{c^{4} d^{5}} + \frac{\sqrt{2} \left (c^{3} d^{2}\right )^{\frac{3}{4}} \log \left (d x - \sqrt{2} \sqrt{d x} \left (\frac{d^{2}}{c}\right )^{\frac{1}{4}} + \sqrt{\frac{d^{2}}{c}}\right )}{c^{4} d^{5}} + \frac{\sqrt{2} \left (-c^{3} d^{2}\right )^{\frac{3}{4}} \log \left (d x + \sqrt{2} \sqrt{d x} \left (-\frac{d^{2}}{c}\right )^{\frac{1}{4}} + \sqrt{-\frac{d^{2}}{c}}\right )}{c^{4} d^{5}} - \frac{\sqrt{2} \left (-c^{3} d^{2}\right )^{\frac{3}{4}} \log \left (d x - \sqrt{2} \sqrt{d x} \left (-\frac{d^{2}}{c}\right )^{\frac{1}{4}} + \sqrt{-\frac{d^{2}}{c}}\right )}{c^{4} d^{5}}\right )} - \frac{\frac{b \log \left (-\frac{c d^{2} x^{2} + d^{2}}{c d^{2} x^{2} - d^{2}}\right )}{\sqrt{d x} d^{2} x^{2}} + \frac{2 \,{\left (4 \, b c d^{2} x^{2} + a d^{2}\right )}}{\sqrt{d x} d^{4} x^{2}}}{5 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x^2))/(d*x)^(7/2),x, algorithm="giac")

[Out]

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