\(\int \frac {x^3 (a+b \text {csch}^{-1}(c x))}{(d+e x^2)^{3/2}} \, dx\) [148]
Optimal result
Integrand size = 23, antiderivative size = 160 \[
\int \frac {x^3 \left (a+b \text {csch}^{-1}(c x)\right )}{\left (d+e x^2\right )^{3/2}} \, dx=\frac {d \left (a+b \text {csch}^{-1}(c x)\right )}{e^2 \sqrt {d+e x^2}}+\frac {\sqrt {d+e x^2} \left (a+b \text {csch}^{-1}(c x)\right )}{e^2}+\frac {b x \arctan \left (\frac {\sqrt {e} \sqrt {-1-c^2 x^2}}{c \sqrt {d+e x^2}}\right )}{e^{3/2} \sqrt {-c^2 x^2}}+\frac {2 b c \sqrt {d} x \arctan \left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {-1-c^2 x^2}}\right )}{e^2 \sqrt {-c^2 x^2}}
\]
[Out]
b*x*arctan(e^(1/2)*(-c^2*x^2-1)^(1/2)/c/(e*x^2+d)^(1/2))/e^(3/2)/(-c^2*x^2)^(1/2)+2*b*c*x*arctan((e*x^2+d)^(1/
2)/d^(1/2)/(-c^2*x^2-1)^(1/2))*d^(1/2)/e^2/(-c^2*x^2)^(1/2)+d*(a+b*arccsch(c*x))/e^2/(e*x^2+d)^(1/2)+(a+b*arcc
sch(c*x))*(e*x^2+d)^(1/2)/e^2
Rubi [A] (verified)
Time = 0.20 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.00, number of
steps used = 9, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {272, 45, 6437, 12, 587, 163,
65, 223, 209, 95, 210} \[
\int \frac {x^3 \left (a+b \text {csch}^{-1}(c x)\right )}{\left (d+e x^2\right )^{3/2}} \, dx=\frac {\sqrt {d+e x^2} \left (a+b \text {csch}^{-1}(c x)\right )}{e^2}+\frac {d \left (a+b \text {csch}^{-1}(c x)\right )}{e^2 \sqrt {d+e x^2}}+\frac {b x \arctan \left (\frac {\sqrt {e} \sqrt {-c^2 x^2-1}}{c \sqrt {d+e x^2}}\right )}{e^{3/2} \sqrt {-c^2 x^2}}+\frac {2 b c \sqrt {d} x \arctan \left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {-c^2 x^2-1}}\right )}{e^2 \sqrt {-c^2 x^2}}
\]
[In]
Int[(x^3*(a + b*ArcCsch[c*x]))/(d + e*x^2)^(3/2),x]
[Out]
(d*(a + b*ArcCsch[c*x]))/(e^2*Sqrt[d + e*x^2]) + (Sqrt[d + e*x^2]*(a + b*ArcCsch[c*x]))/e^2 + (b*x*ArcTan[(Sqr
t[e]*Sqrt[-1 - c^2*x^2])/(c*Sqrt[d + e*x^2])])/(e^(3/2)*Sqrt[-(c^2*x^2)]) + (2*b*c*Sqrt[d]*x*ArcTan[Sqrt[d + e
*x^2]/(Sqrt[d]*Sqrt[-1 - c^2*x^2])])/(e^2*Sqrt[-(c^2*x^2)])
Rule 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]
Rule 45
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Rule 65
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]
Rule 95
Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]
Rule 163
Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/((a_.) + (b_.)*(x_)), x_Symbol]
:> Dist[h/b, Int[(c + d*x)^n*(e + f*x)^p, x], x] + Dist[(b*g - a*h)/b, Int[(c + d*x)^n*((e + f*x)^p/(a + b*x)
), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x]
Rule 209
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])
Rule 210
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])
Rule 223
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] && !GtQ[a, 0]
Rule 272
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
Rule 587
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_))^(r_.), x
_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q*(e + f*x)^r, x], x, x^n],
x] /; FreeQ[{a, b, c, d, e, f, m, n, p, q, r}, x] && IntegerQ[Simplify[(m + 1)/n]]
Rule 6437
Int[((a_.) + ArcCsch[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u
= IntHide[(f*x)^m*(d + e*x^2)^p, x]}, Dist[a + b*ArcCsch[c*x], u, x] - Dist[b*c*(x/Sqrt[(-c^2)*x^2]), Int[Simp
lifyIntegrand[u/(x*Sqrt[-1 - c^2*x^2]), x], x], x]] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && ((IGtQ[p, 0] &&
!(ILtQ[(m - 1)/2, 0] && GtQ[m + 2*p + 3, 0])) || (IGtQ[(m + 1)/2, 0] && !(ILtQ[p, 0] && GtQ[m + 2*p + 3, 0]))
|| (ILtQ[(m + 2*p + 1)/2, 0] && !ILtQ[(m - 1)/2, 0]))
Rubi steps \begin{align*}
\text {integral}& = \frac {d \left (a+b \text {csch}^{-1}(c x)\right )}{e^2 \sqrt {d+e x^2}}+\frac {\sqrt {d+e x^2} \left (a+b \text {csch}^{-1}(c x)\right )}{e^2}-\frac {(b c x) \int \frac {2 d+e x^2}{e^2 x \sqrt {-1-c^2 x^2} \sqrt {d+e x^2}} \, dx}{\sqrt {-c^2 x^2}} \\ & = \frac {d \left (a+b \text {csch}^{-1}(c x)\right )}{e^2 \sqrt {d+e x^2}}+\frac {\sqrt {d+e x^2} \left (a+b \text {csch}^{-1}(c x)\right )}{e^2}-\frac {(b c x) \int \frac {2 d+e x^2}{x \sqrt {-1-c^2 x^2} \sqrt {d+e x^2}} \, dx}{e^2 \sqrt {-c^2 x^2}} \\ & = \frac {d \left (a+b \text {csch}^{-1}(c x)\right )}{e^2 \sqrt {d+e x^2}}+\frac {\sqrt {d+e x^2} \left (a+b \text {csch}^{-1}(c x)\right )}{e^2}-\frac {(b c x) \text {Subst}\left (\int \frac {2 d+e x}{x \sqrt {-1-c^2 x} \sqrt {d+e x}} \, dx,x,x^2\right )}{2 e^2 \sqrt {-c^2 x^2}} \\ & = \frac {d \left (a+b \text {csch}^{-1}(c x)\right )}{e^2 \sqrt {d+e x^2}}+\frac {\sqrt {d+e x^2} \left (a+b \text {csch}^{-1}(c x)\right )}{e^2}-\frac {(b c d x) \text {Subst}\left (\int \frac {1}{x \sqrt {-1-c^2 x} \sqrt {d+e x}} \, dx,x,x^2\right )}{e^2 \sqrt {-c^2 x^2}}-\frac {(b c x) \text {Subst}\left (\int \frac {1}{\sqrt {-1-c^2 x} \sqrt {d+e x}} \, dx,x,x^2\right )}{2 e \sqrt {-c^2 x^2}} \\ & = \frac {d \left (a+b \text {csch}^{-1}(c x)\right )}{e^2 \sqrt {d+e x^2}}+\frac {\sqrt {d+e x^2} \left (a+b \text {csch}^{-1}(c x)\right )}{e^2}-\frac {(2 b c d x) \text {Subst}\left (\int \frac {1}{-d-x^2} \, dx,x,\frac {\sqrt {d+e x^2}}{\sqrt {-1-c^2 x^2}}\right )}{e^2 \sqrt {-c^2 x^2}}+\frac {(b x) \text {Subst}\left (\int \frac {1}{\sqrt {d-\frac {e}{c^2}-\frac {e x^2}{c^2}}} \, dx,x,\sqrt {-1-c^2 x^2}\right )}{c e \sqrt {-c^2 x^2}} \\ & = \frac {d \left (a+b \text {csch}^{-1}(c x)\right )}{e^2 \sqrt {d+e x^2}}+\frac {\sqrt {d+e x^2} \left (a+b \text {csch}^{-1}(c x)\right )}{e^2}+\frac {2 b c \sqrt {d} x \arctan \left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {-1-c^2 x^2}}\right )}{e^2 \sqrt {-c^2 x^2}}+\frac {(b x) \text {Subst}\left (\int \frac {1}{1+\frac {e x^2}{c^2}} \, dx,x,\frac {\sqrt {-1-c^2 x^2}}{\sqrt {d+e x^2}}\right )}{c e \sqrt {-c^2 x^2}} \\ & = \frac {d \left (a+b \text {csch}^{-1}(c x)\right )}{e^2 \sqrt {d+e x^2}}+\frac {\sqrt {d+e x^2} \left (a+b \text {csch}^{-1}(c x)\right )}{e^2}+\frac {b x \arctan \left (\frac {\sqrt {e} \sqrt {-1-c^2 x^2}}{c \sqrt {d+e x^2}}\right )}{e^{3/2} \sqrt {-c^2 x^2}}+\frac {2 b c \sqrt {d} x \arctan \left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {-1-c^2 x^2}}\right )}{e^2 \sqrt {-c^2 x^2}} \\
\end{align*}
Mathematica [C] (verified)
Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
Time = 1.47 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.19
\[
\int \frac {x^3 \left (a+b \text {csch}^{-1}(c x)\right )}{\left (d+e x^2\right )^{3/2}} \, dx=\frac {-2 b d \sqrt {1+\frac {d}{e x^2}} \sqrt {1+c^2 x^2} \operatorname {AppellF1}\left (1,\frac {1}{2},\frac {1}{2},2,-\frac {1}{c^2 x^2},-\frac {d}{e x^2}\right )+c x \left (b c e \sqrt {1+\frac {1}{c^2 x^2}} x^3 \sqrt {1+\frac {e x^2}{d}} \operatorname {AppellF1}\left (1,\frac {1}{2},\frac {1}{2},2,-c^2 x^2,-\frac {e x^2}{d}\right )+2 \sqrt {1+c^2 x^2} \left (2 d+e x^2\right ) \left (a+b \text {csch}^{-1}(c x)\right )\right )}{2 c e^2 x \sqrt {1+c^2 x^2} \sqrt {d+e x^2}}
\]
[In]
Integrate[(x^3*(a + b*ArcCsch[c*x]))/(d + e*x^2)^(3/2),x]
[Out]
(-2*b*d*Sqrt[1 + d/(e*x^2)]*Sqrt[1 + c^2*x^2]*AppellF1[1, 1/2, 1/2, 2, -(1/(c^2*x^2)), -(d/(e*x^2))] + c*x*(b*
c*e*Sqrt[1 + 1/(c^2*x^2)]*x^3*Sqrt[1 + (e*x^2)/d]*AppellF1[1, 1/2, 1/2, 2, -(c^2*x^2), -((e*x^2)/d)] + 2*Sqrt[
1 + c^2*x^2]*(2*d + e*x^2)*(a + b*ArcCsch[c*x])))/(2*c*e^2*x*Sqrt[1 + c^2*x^2]*Sqrt[d + e*x^2])
Maple [F]
\[\int \frac {x^{3} \left (a +b \,\operatorname {arccsch}\left (c x \right )\right )}{\left (e \,x^{2}+d \right )^{\frac {3}{2}}}d x\]
[In]
int(x^3*(a+b*arccsch(c*x))/(e*x^2+d)^(3/2),x)
[Out]
int(x^3*(a+b*arccsch(c*x))/(e*x^2+d)^(3/2),x)
Fricas [B] (verification not implemented)
Leaf count of result is larger than twice the leaf count of optimal. 303 vs. \(2 (136) = 272\).
Time = 0.35 (sec) , antiderivative size = 1274, normalized size of antiderivative = 7.96
\[
\int \frac {x^3 \left (a+b \text {csch}^{-1}(c x)\right )}{\left (d+e x^2\right )^{3/2}} \, dx=\text {Too large to display}
\]
[In]
integrate(x^3*(a+b*arccsch(c*x))/(e*x^2+d)^(3/2),x, algorithm="fricas")
[Out]
[1/4*((b*e*x^2 + b*d)*sqrt(e)*log(8*c^4*e^2*x^4 + c^4*d^2 + 6*c^2*d*e + 8*(c^4*d*e + c^2*e^2)*x^2 + 4*(2*c^4*e
*x^3 + (c^4*d + c^2*e)*x)*sqrt(e*x^2 + d)*sqrt(e)*sqrt((c^2*x^2 + 1)/(c^2*x^2)) + e^2) + 4*(b*c*e*x^2 + 2*b*c*
d)*sqrt(e*x^2 + d)*log((c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) + 1)/(c*x)) + 2*(b*c*e*x^2 + b*c*d)*sqrt(d)*log(((c^
4*d^2 + 6*c^2*d*e + e^2)*x^4 + 8*(c^2*d^2 + d*e)*x^2 - 4*((c^3*d + c*e)*x^3 + 2*c*d*x)*sqrt(e*x^2 + d)*sqrt(d)
*sqrt((c^2*x^2 + 1)/(c^2*x^2)) + 8*d^2)/x^4) + 4*(a*c*e*x^2 + 2*a*c*d)*sqrt(e*x^2 + d))/(c*e^3*x^2 + c*d*e^2),
-1/2*((b*e*x^2 + b*d)*sqrt(-e)*arctan(1/2*(2*c^2*e*x^3 + (c^2*d + e)*x)*sqrt(e*x^2 + d)*sqrt(-e)*sqrt((c^2*x^
2 + 1)/(c^2*x^2))/(c^2*e^2*x^4 + (c^2*d*e + e^2)*x^2 + d*e)) - 2*(b*c*e*x^2 + 2*b*c*d)*sqrt(e*x^2 + d)*log((c*
x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) + 1)/(c*x)) - (b*c*e*x^2 + b*c*d)*sqrt(d)*log(((c^4*d^2 + 6*c^2*d*e + e^2)*x^4
+ 8*(c^2*d^2 + d*e)*x^2 - 4*((c^3*d + c*e)*x^3 + 2*c*d*x)*sqrt(e*x^2 + d)*sqrt(d)*sqrt((c^2*x^2 + 1)/(c^2*x^2
)) + 8*d^2)/x^4) - 2*(a*c*e*x^2 + 2*a*c*d)*sqrt(e*x^2 + d))/(c*e^3*x^2 + c*d*e^2), 1/4*(4*(b*c*e*x^2 + b*c*d)*
sqrt(-d)*arctan(1/2*((c^3*d + c*e)*x^3 + 2*c*d*x)*sqrt(e*x^2 + d)*sqrt(-d)*sqrt((c^2*x^2 + 1)/(c^2*x^2))/(c^2*
d*e*x^4 + (c^2*d^2 + d*e)*x^2 + d^2)) + (b*e*x^2 + b*d)*sqrt(e)*log(8*c^4*e^2*x^4 + c^4*d^2 + 6*c^2*d*e + 8*(c
^4*d*e + c^2*e^2)*x^2 + 4*(2*c^4*e*x^3 + (c^4*d + c^2*e)*x)*sqrt(e*x^2 + d)*sqrt(e)*sqrt((c^2*x^2 + 1)/(c^2*x^
2)) + e^2) + 4*(b*c*e*x^2 + 2*b*c*d)*sqrt(e*x^2 + d)*log((c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) + 1)/(c*x)) + 4*(a
*c*e*x^2 + 2*a*c*d)*sqrt(e*x^2 + d))/(c*e^3*x^2 + c*d*e^2), 1/2*(2*(b*c*e*x^2 + b*c*d)*sqrt(-d)*arctan(1/2*((c
^3*d + c*e)*x^3 + 2*c*d*x)*sqrt(e*x^2 + d)*sqrt(-d)*sqrt((c^2*x^2 + 1)/(c^2*x^2))/(c^2*d*e*x^4 + (c^2*d^2 + d*
e)*x^2 + d^2)) - (b*e*x^2 + b*d)*sqrt(-e)*arctan(1/2*(2*c^2*e*x^3 + (c^2*d + e)*x)*sqrt(e*x^2 + d)*sqrt(-e)*sq
rt((c^2*x^2 + 1)/(c^2*x^2))/(c^2*e^2*x^4 + (c^2*d*e + e^2)*x^2 + d*e)) + 2*(b*c*e*x^2 + 2*b*c*d)*sqrt(e*x^2 +
d)*log((c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) + 1)/(c*x)) + 2*(a*c*e*x^2 + 2*a*c*d)*sqrt(e*x^2 + d))/(c*e^3*x^2 +
c*d*e^2)]
Sympy [F]
\[
\int \frac {x^3 \left (a+b \text {csch}^{-1}(c x)\right )}{\left (d+e x^2\right )^{3/2}} \, dx=\int \frac {x^{3} \left (a + b \operatorname {acsch}{\left (c x \right )}\right )}{\left (d + e x^{2}\right )^{\frac {3}{2}}}\, dx
\]
[In]
integrate(x**3*(a+b*acsch(c*x))/(e*x**2+d)**(3/2),x)
[Out]
Integral(x**3*(a + b*acsch(c*x))/(d + e*x**2)**(3/2), x)
Maxima [F(-2)]
Exception generated. \[
\int \frac {x^3 \left (a+b \text {csch}^{-1}(c x)\right )}{\left (d+e x^2\right )^{3/2}} \, dx=\text {Exception raised: ValueError}
\]
[In]
integrate(x^3*(a+b*arccsch(c*x))/(e*x^2+d)^(3/2),x, algorithm="maxima")
[Out]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more
details)Is e
Giac [F]
\[
\int \frac {x^3 \left (a+b \text {csch}^{-1}(c x)\right )}{\left (d+e x^2\right )^{3/2}} \, dx=\int { \frac {{\left (b \operatorname {arcsch}\left (c x\right ) + a\right )} x^{3}}{{\left (e x^{2} + d\right )}^{\frac {3}{2}}} \,d x }
\]
[In]
integrate(x^3*(a+b*arccsch(c*x))/(e*x^2+d)^(3/2),x, algorithm="giac")
[Out]
integrate((b*arccsch(c*x) + a)*x^3/(e*x^2 + d)^(3/2), x)
Mupad [F(-1)]
Timed out. \[
\int \frac {x^3 \left (a+b \text {csch}^{-1}(c x)\right )}{\left (d+e x^2\right )^{3/2}} \, dx=\int \frac {x^3\,\left (a+b\,\mathrm {asinh}\left (\frac {1}{c\,x}\right )\right )}{{\left (e\,x^2+d\right )}^{3/2}} \,d x
\]
[In]
int((x^3*(a + b*asinh(1/(c*x))))/(d + e*x^2)^(3/2),x)
[Out]
int((x^3*(a + b*asinh(1/(c*x))))/(d + e*x^2)^(3/2), x)