\(\int \frac {x^3 (a+b \text {sech}^{-1}(c x))}{(d+e x^2)^{3/2}} \, dx\) [160]
Optimal result
Integrand size = 23, antiderivative size = 177 \[
\int \frac {x^3 \left (a+b \text {sech}^{-1}(c x)\right )}{\left (d+e x^2\right )^{3/2}} \, dx=\frac {d \left (a+b \text {sech}^{-1}(c x)\right )}{e^2 \sqrt {d+e x^2}}+\frac {\sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{e^2}-\frac {b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \arctan \left (\frac {\sqrt {e} \sqrt {1-c^2 x^2}}{c \sqrt {d+e x^2}}\right )}{c e^{3/2}}-\frac {2 b \sqrt {d} \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {1-c^2 x^2}}\right )}{e^2}
\]
[Out]
-b*arctan(e^(1/2)*(-c^2*x^2+1)^(1/2)/c/(e*x^2+d)^(1/2))*(1/(c*x+1))^(1/2)*(c*x+1)^(1/2)/c/e^(3/2)-2*b*arctanh(
(e*x^2+d)^(1/2)/d^(1/2)/(-c^2*x^2+1)^(1/2))*d^(1/2)*(1/(c*x+1))^(1/2)*(c*x+1)^(1/2)/e^2+d*(a+b*arcsech(c*x))/e
^2/(e*x^2+d)^(1/2)+(a+b*arcsech(c*x))*(e*x^2+d)^(1/2)/e^2
Rubi [A] (verified)
Time = 0.18 (sec) , antiderivative size = 177, 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, 6436, 12, 587, 163,
65, 223, 209, 95, 213} \[
\int \frac {x^3 \left (a+b \text {sech}^{-1}(c x)\right )}{\left (d+e x^2\right )^{3/2}} \, dx=\frac {\sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{e^2}+\frac {d \left (a+b \text {sech}^{-1}(c x)\right )}{e^2 \sqrt {d+e x^2}}-\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \arctan \left (\frac {\sqrt {e} \sqrt {1-c^2 x^2}}{c \sqrt {d+e x^2}}\right )}{c e^{3/2}}-\frac {2 b \sqrt {d} \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {1-c^2 x^2}}\right )}{e^2}
\]
[In]
Int[(x^3*(a + b*ArcSech[c*x]))/(d + e*x^2)^(3/2),x]
[Out]
(d*(a + b*ArcSech[c*x]))/(e^2*Sqrt[d + e*x^2]) + (Sqrt[d + e*x^2]*(a + b*ArcSech[c*x]))/e^2 - (b*Sqrt[(1 + c*x
)^(-1)]*Sqrt[1 + c*x]*ArcTan[(Sqrt[e]*Sqrt[1 - c^2*x^2])/(c*Sqrt[d + e*x^2])])/(c*e^(3/2)) - (2*b*Sqrt[d]*Sqrt
[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*ArcTanh[Sqrt[d + e*x^2]/(Sqrt[d]*Sqrt[1 - c^2*x^2])])/e^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 213
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[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 6436
Int[((a_.) + ArcSech[(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*ArcSech[c*x], u, x] + Dist[b*Sqrt[1 + c*x]*Sqrt[1/(1 + c*x)],
Int[SimplifyIntegrand[u/(x*Sqrt[1 - c*x]*Sqrt[1 + c*x]), 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 {sech}^{-1}(c x)\right )}{e^2 \sqrt {d+e x^2}}+\frac {\sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{e^2}+\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {2 d+e x^2}{e^2 x \sqrt {1-c^2 x^2} \sqrt {d+e x^2}} \, dx \\ & = \frac {d \left (a+b \text {sech}^{-1}(c x)\right )}{e^2 \sqrt {d+e x^2}}+\frac {\sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{e^2}+\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {2 d+e x^2}{x \sqrt {1-c^2 x^2} \sqrt {d+e x^2}} \, dx}{e^2} \\ & = \frac {d \left (a+b \text {sech}^{-1}(c x)\right )}{e^2 \sqrt {d+e x^2}}+\frac {\sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{e^2}+\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \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} \\ & = \frac {d \left (a+b \text {sech}^{-1}(c x)\right )}{e^2 \sqrt {d+e x^2}}+\frac {\sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{e^2}+\frac {\left (b d \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {1-c^2 x} \sqrt {d+e x}} \, dx,x,x^2\right )}{e^2}+\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-c^2 x} \sqrt {d+e x}} \, dx,x,x^2\right )}{2 e} \\ & = \frac {d \left (a+b \text {sech}^{-1}(c x)\right )}{e^2 \sqrt {d+e x^2}}+\frac {\sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{e^2}+\frac {\left (2 b d \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \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}-\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \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^2 e} \\ & = \frac {d \left (a+b \text {sech}^{-1}(c x)\right )}{e^2 \sqrt {d+e x^2}}+\frac {\sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{e^2}-\frac {2 b \sqrt {d} \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {1-c^2 x^2}}\right )}{e^2}-\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \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^2 e} \\ & = \frac {d \left (a+b \text {sech}^{-1}(c x)\right )}{e^2 \sqrt {d+e x^2}}+\frac {\sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{e^2}-\frac {b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \arctan \left (\frac {\sqrt {e} \sqrt {1-c^2 x^2}}{c \sqrt {d+e x^2}}\right )}{c e^{3/2}}-\frac {2 b \sqrt {d} \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {1-c^2 x^2}}\right )}{e^2} \\
\end{align*}
Mathematica [A] (verified)
Time = 23.17 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.41
\[
\int \frac {x^3 \left (a+b \text {sech}^{-1}(c x)\right )}{\left (d+e x^2\right )^{3/2}} \, dx=\frac {\left (2 d+e x^2\right ) \left (a+b \text {sech}^{-1}(c x)\right )}{e^2 \sqrt {d+e x^2}}+\frac {b \sqrt {\frac {1-c x}{1+c x}} \sqrt {1-c^2 x^2} \left (\sqrt {-c^2} \sqrt {-c^2 d-e} \sqrt {e} \sqrt {\frac {c^2 \left (d+e x^2\right )}{c^2 d+e}} \arcsin \left (\frac {c \sqrt {e} \sqrt {1-c^2 x^2}}{\sqrt {-c^2} \sqrt {-c^2 d-e}}\right )+2 c^3 \sqrt {d} \sqrt {-d-e x^2} \arctan \left (\frac {\sqrt {d} \sqrt {1-c^2 x^2}}{\sqrt {-d-e x^2}}\right )\right )}{c^3 e^2 (-1+c x) \sqrt {d+e x^2}}
\]
[In]
Integrate[(x^3*(a + b*ArcSech[c*x]))/(d + e*x^2)^(3/2),x]
[Out]
((2*d + e*x^2)*(a + b*ArcSech[c*x]))/(e^2*Sqrt[d + e*x^2]) + (b*Sqrt[(1 - c*x)/(1 + c*x)]*Sqrt[1 - c^2*x^2]*(S
qrt[-c^2]*Sqrt[-(c^2*d) - e]*Sqrt[e]*Sqrt[(c^2*(d + e*x^2))/(c^2*d + e)]*ArcSin[(c*Sqrt[e]*Sqrt[1 - c^2*x^2])/
(Sqrt[-c^2]*Sqrt[-(c^2*d) - e])] + 2*c^3*Sqrt[d]*Sqrt[-d - e*x^2]*ArcTan[(Sqrt[d]*Sqrt[1 - c^2*x^2])/Sqrt[-d -
e*x^2]]))/(c^3*e^2*(-1 + c*x)*Sqrt[d + e*x^2])
Maple [F]
\[\int \frac {x^{3} \left (a +b \,\operatorname {arcsech}\left (c x \right )\right )}{\left (e \,x^{2}+d \right )^{\frac {3}{2}}}d x\]
[In]
int(x^3*(a+b*arcsech(c*x))/(e*x^2+d)^(3/2),x)
[Out]
int(x^3*(a+b*arcsech(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. 310 vs. \(2 (117) = 234\).
Time = 0.36 (sec) , antiderivative size = 1311, normalized size of antiderivative = 7.41
\[
\int \frac {x^3 \left (a+b \text {sech}^{-1}(c x)\right )}{\left (d+e x^2\right )^{3/2}} \, dx=\text {Too large to display}
\]
[In]
integrate(x^3*(a+b*arcsech(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)*sq
rt(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)*lo
g((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)*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)) - 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]
\[
\int \frac {x^3 \left (a+b \text {sech}^{-1}(c x)\right )}{\left (d+e x^2\right )^{3/2}} \, dx=\int \frac {x^{3} \left (a + b \operatorname {asech}{\left (c x \right )}\right )}{\left (d + e x^{2}\right )^{\frac {3}{2}}}\, dx
\]
[In]
integrate(x**3*(a+b*asech(c*x))/(e*x**2+d)**(3/2),x)
[Out]
Integral(x**3*(a + b*asech(c*x))/(d + e*x**2)**(3/2), x)
Maxima [F(-2)]
Exception generated. \[
\int \frac {x^3 \left (a+b \text {sech}^{-1}(c x)\right )}{\left (d+e x^2\right )^{3/2}} \, dx=\text {Exception raised: ValueError}
\]
[In]
integrate(x^3*(a+b*arcsech(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]
\[
\int \frac {x^3 \left (a+b \text {sech}^{-1}(c x)\right )}{\left (d+e x^2\right )^{3/2}} \, dx=\int { \frac {{\left (b \operatorname {arsech}\left (c x\right ) + a\right )} x^{3}}{{\left (e x^{2} + d\right )}^{\frac {3}{2}}} \,d x }
\]
[In]
integrate(x^3*(a+b*arcsech(c*x))/(e*x^2+d)^(3/2),x, algorithm="giac")
[Out]
integrate((b*arcsech(c*x) + a)*x^3/(e*x^2 + d)^(3/2), x)
Mupad [F(-1)]
Timed out. \[
\int \frac {x^3 \left (a+b \text {sech}^{-1}(c x)\right )}{\left (d+e x^2\right )^{3/2}} \, dx=\int \frac {x^3\,\left (a+b\,\mathrm {acosh}\left (\frac {1}{c\,x}\right )\right )}{{\left (e\,x^2+d\right )}^{3/2}} \,d x
\]
[In]
int((x^3*(a + b*acosh(1/(c*x))))/(d + e*x^2)^(3/2),x)
[Out]
int((x^3*(a + b*acosh(1/(c*x))))/(d + e*x^2)^(3/2), x)