3.2.50 \(\int \frac {x^5 (a+b \text {sech}^{-1}(c x))}{\sqrt {d+e x^2}} \, dx\) [150]

Optimal. Leaf size=356 \[ \frac {b \left (19 c^2 d-9 e\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \sqrt {d+e x^2}}{120 c^4 e^2}-\frac {b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \left (d+e x^2\right )^{3/2}}{20 c^2 e^2}+\frac {d^2 \sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{e^3}-\frac {2 d \left (d+e x^2\right )^{3/2} \left (a+b \text {sech}^{-1}(c x)\right )}{3 e^3}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \text {sech}^{-1}(c x)\right )}{5 e^3}-\frac {b \left (45 c^4 d^2-10 c^2 d e+9 e^2\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \text {ArcTan}\left (\frac {\sqrt {e} \sqrt {1-c^2 x^2}}{c \sqrt {d+e x^2}}\right )}{120 c^5 e^{5/2}}-\frac {8 b d^{5/2} \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {1-c^2 x^2}}\right )}{15 e^3} \]

[Out]

-2/3*d*(e*x^2+d)^(3/2)*(a+b*arcsech(c*x))/e^3+1/5*(e*x^2+d)^(5/2)*(a+b*arcsech(c*x))/e^3-1/120*b*(45*c^4*d^2-1
0*c^2*d*e+9*e^2)*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^5/e^(5
/2)-8/15*b*d^(5/2)*arctanh((e*x^2+d)^(1/2)/d^(1/2)/(-c^2*x^2+1)^(1/2))*(1/(c*x+1))^(1/2)*(c*x+1)^(1/2)/e^3-1/2
0*b*(e*x^2+d)^(3/2)*(1/(c*x+1))^(1/2)*(c*x+1)^(1/2)*(-c^2*x^2+1)^(1/2)/c^2/e^2+d^2*(a+b*arcsech(c*x))*(e*x^2+d
)^(1/2)/e^3+1/120*b*(19*c^2*d-9*e)*(1/(c*x+1))^(1/2)*(c*x+1)^(1/2)*(-c^2*x^2+1)^(1/2)*(e*x^2+d)^(1/2)/c^4/e^2

________________________________________________________________________________________

Rubi [A]
time = 0.77, antiderivative size = 356, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 12, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.522, Rules used = {272, 45, 6436, 12, 1629, 159, 163, 65, 223, 209, 95, 213} \begin {gather*} \frac {d^2 \sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{e^3}-\frac {2 d \left (d+e x^2\right )^{3/2} \left (a+b \text {sech}^{-1}(c x)\right )}{3 e^3}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \text {sech}^{-1}(c x)\right )}{5 e^3}-\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \left (45 c^4 d^2-10 c^2 d e+9 e^2\right ) \text {ArcTan}\left (\frac {\sqrt {e} \sqrt {1-c^2 x^2}}{c \sqrt {d+e x^2}}\right )}{120 c^5 e^{5/2}}-\frac {8 b d^{5/2} \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {1-c^2 x^2}}\right )}{15 e^3}-\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2} \left (d+e x^2\right )^{3/2}}{20 c^2 e^2}+\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2} \left (19 c^2 d-9 e\right ) \sqrt {d+e x^2}}{120 c^4 e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^5*(a + b*ArcSech[c*x]))/Sqrt[d + e*x^2],x]

[Out]

(b*(19*c^2*d - 9*e)*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*Sqrt[1 - c^2*x^2]*Sqrt[d + e*x^2])/(120*c^4*e^2) - (b*S
qrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*Sqrt[1 - c^2*x^2]*(d + e*x^2)^(3/2))/(20*c^2*e^2) + (d^2*Sqrt[d + e*x^2]*(a
+ b*ArcSech[c*x]))/e^3 - (2*d*(d + e*x^2)^(3/2)*(a + b*ArcSech[c*x]))/(3*e^3) + ((d + e*x^2)^(5/2)*(a + b*ArcS
ech[c*x]))/(5*e^3) - (b*(45*c^4*d^2 - 10*c^2*d*e + 9*e^2)*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*ArcTan[(Sqrt[e]*S
qrt[1 - c^2*x^2])/(c*Sqrt[d + e*x^2])])/(120*c^5*e^(5/2)) - (8*b*d^(5/2)*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*Ar
cTanh[Sqrt[d + e*x^2]/(Sqrt[d]*Sqrt[1 - c^2*x^2])])/(15*e^3)

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 159

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[h*(a + b*x)^m*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(m + n + p + 2))), x] + Dist[1/(d*f*(m + n
+ p + 2)), Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2) - h*(b*c*e*m + a*(d*e*(
n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x]
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] && IntegersQ[2*m, 2
*n, 2*p]

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 1629

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> With[
{q = Expon[Px, x], k = Coeff[Px, x, Expon[Px, x]]}, Simp[k*(a + b*x)^(m + q - 1)*(c + d*x)^(n + 1)*((e + f*x)^
(p + 1)/(d*f*b^(q - 1)*(m + n + p + q + 1))), x] + Dist[1/(d*f*b^q*(m + n + p + q + 1)), Int[(a + b*x)^m*(c +
d*x)^n*(e + f*x)^p*ExpandToSum[d*f*b^q*(m + n + p + q + 1)*Px - d*f*k*(m + n + p + q + 1)*(a + b*x)^q + k*(a +
b*x)^(q - 2)*(a^2*d*f*(m + n + p + q + 1) - b*(b*c*e*(m + q - 1) + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*
(2*(m + q) + n + p) - b*(d*e*(m + q + n) + c*f*(m + q + p)))*x), x], x], x] /; NeQ[m + n + p + q + 1, 0]] /; F
reeQ[{a, b, c, d, e, f, m, n, p}, x] && PolyQ[Px, x]

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*} \int \frac {x^5 \left (a+b \text {sech}^{-1}(c x)\right )}{\sqrt {d+e x^2}} \, dx &=\frac {d^2 \sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{e^3}-\frac {2 d \left (d+e x^2\right )^{3/2} \left (a+b \text {sech}^{-1}(c x)\right )}{3 e^3}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \text {sech}^{-1}(c x)\right )}{5 e^3}+\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {\sqrt {d+e x^2} \left (8 d^2-4 d e x^2+3 e^2 x^4\right )}{15 e^3 x \sqrt {1-c^2 x^2}} \, dx\\ &=\frac {d^2 \sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{e^3}-\frac {2 d \left (d+e x^2\right )^{3/2} \left (a+b \text {sech}^{-1}(c x)\right )}{3 e^3}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \text {sech}^{-1}(c x)\right )}{5 e^3}+\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {\sqrt {d+e x^2} \left (8 d^2-4 d e x^2+3 e^2 x^4\right )}{x \sqrt {1-c^2 x^2}} \, dx}{15 e^3}\\ &=\frac {d^2 \sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{e^3}-\frac {2 d \left (d+e x^2\right )^{3/2} \left (a+b \text {sech}^{-1}(c x)\right )}{3 e^3}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \text {sech}^{-1}(c x)\right )}{5 e^3}+\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \text {Subst}\left (\int \frac {\sqrt {d+e x} \left (8 d^2-4 d e x+3 e^2 x^2\right )}{x \sqrt {1-c^2 x}} \, dx,x,x^2\right )}{30 e^3}\\ &=-\frac {b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \left (d+e x^2\right )^{3/2}}{20 c^2 e^2}+\frac {d^2 \sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{e^3}-\frac {2 d \left (d+e x^2\right )^{3/2} \left (a+b \text {sech}^{-1}(c x)\right )}{3 e^3}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \text {sech}^{-1}(c x)\right )}{5 e^3}-\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \text {Subst}\left (\int \frac {\sqrt {d+e x} \left (-16 c^2 d^2 e+\frac {1}{2} \left (19 c^2 d-9 e\right ) e^2 x\right )}{x \sqrt {1-c^2 x}} \, dx,x,x^2\right )}{60 c^2 e^4}\\ &=\frac {b \left (19 c^2 d-9 e\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \sqrt {d+e x^2}}{120 c^4 e^2}-\frac {b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \left (d+e x^2\right )^{3/2}}{20 c^2 e^2}+\frac {d^2 \sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{e^3}-\frac {2 d \left (d+e x^2\right )^{3/2} \left (a+b \text {sech}^{-1}(c x)\right )}{3 e^3}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \text {sech}^{-1}(c x)\right )}{5 e^3}+\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \text {Subst}\left (\int \frac {16 c^4 d^3 e+\frac {1}{4} e^2 \left (45 c^4 d^2-10 c^2 d e+9 e^2\right ) x}{x \sqrt {1-c^2 x} \sqrt {d+e x}} \, dx,x,x^2\right )}{60 c^4 e^4}\\ &=\frac {b \left (19 c^2 d-9 e\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \sqrt {d+e x^2}}{120 c^4 e^2}-\frac {b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \left (d+e x^2\right )^{3/2}}{20 c^2 e^2}+\frac {d^2 \sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{e^3}-\frac {2 d \left (d+e x^2\right )^{3/2} \left (a+b \text {sech}^{-1}(c x)\right )}{3 e^3}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \text {sech}^{-1}(c x)\right )}{5 e^3}+\frac {\left (4 b d^3 \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 )}{15 e^3}+\frac {\left (b \left (45 c^4 d^2-10 c^2 d e+9 e^2\right ) \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 )}{240 c^4 e^2}\\ &=\frac {b \left (19 c^2 d-9 e\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \sqrt {d+e x^2}}{120 c^4 e^2}-\frac {b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \left (d+e x^2\right )^{3/2}}{20 c^2 e^2}+\frac {d^2 \sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{e^3}-\frac {2 d \left (d+e x^2\right )^{3/2} \left (a+b \text {sech}^{-1}(c x)\right )}{3 e^3}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \text {sech}^{-1}(c x)\right )}{5 e^3}+\frac {\left (8 b d^3 \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 )}{15 e^3}-\frac {\left (b \left (45 c^4 d^2-10 c^2 d e+9 e^2\right ) \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 )}{120 c^6 e^2}\\ &=\frac {b \left (19 c^2 d-9 e\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \sqrt {d+e x^2}}{120 c^4 e^2}-\frac {b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \left (d+e x^2\right )^{3/2}}{20 c^2 e^2}+\frac {d^2 \sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{e^3}-\frac {2 d \left (d+e x^2\right )^{3/2} \left (a+b \text {sech}^{-1}(c x)\right )}{3 e^3}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \text {sech}^{-1}(c x)\right )}{5 e^3}-\frac {8 b d^{5/2} \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {1-c^2 x^2}}\right )}{15 e^3}-\frac {\left (b \left (45 c^4 d^2-10 c^2 d e+9 e^2\right ) \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 )}{120 c^6 e^2}\\ &=\frac {b \left (19 c^2 d-9 e\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \sqrt {d+e x^2}}{120 c^4 e^2}-\frac {b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \left (d+e x^2\right )^{3/2}}{20 c^2 e^2}+\frac {d^2 \sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{e^3}-\frac {2 d \left (d+e x^2\right )^{3/2} \left (a+b \text {sech}^{-1}(c x)\right )}{3 e^3}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \text {sech}^{-1}(c x)\right )}{5 e^3}-\frac {b \left (45 c^4 d^2-10 c^2 d e+9 e^2\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \tan ^{-1}\left (\frac {\sqrt {e} \sqrt {1-c^2 x^2}}{c \sqrt {d+e x^2}}\right )}{120 c^5 e^{5/2}}-\frac {8 b d^{5/2} \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {1-c^2 x^2}}\right )}{15 e^3}\\ \end {align*}

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Mathematica [A]
time = 21.05, size = 366, normalized size = 1.03 \begin {gather*} \frac {\sqrt {d+e x^2} \left (8 a c^4 \left (8 d^2-4 d e x^2+3 e^2 x^4\right )-b e \sqrt {\frac {1-c x}{1+c x}} (1+c x) \left (9 e+c^2 \left (-13 d+6 e x^2\right )\right )+8 b c^4 \left (8 d^2-4 d e x^2+3 e^2 x^4\right ) \text {sech}^{-1}(c x)\right )}{120 c^4 e^3}+\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} \left (45 c^4 d^2-10 c^2 d e+9 e^2\right ) \sqrt {\frac {c^2 \left (d+e x^2\right )}{c^2 d+e}} \text {ArcSin}\left (\frac {c \sqrt {e} \sqrt {1-c^2 x^2}}{\sqrt {-c^2} \sqrt {-c^2 d-e}}\right )+64 c^7 d^{5/2} \sqrt {-d-e x^2} \text {ArcTan}\left (\frac {\sqrt {d} \sqrt {1-c^2 x^2}}{\sqrt {-d-e x^2}}\right )\right )}{120 c^7 e^3 (-1+c x) \sqrt {d+e x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^5*(a + b*ArcSech[c*x]))/Sqrt[d + e*x^2],x]

[Out]

(Sqrt[d + e*x^2]*(8*a*c^4*(8*d^2 - 4*d*e*x^2 + 3*e^2*x^4) - b*e*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x)*(9*e + c^2
*(-13*d + 6*e*x^2)) + 8*b*c^4*(8*d^2 - 4*d*e*x^2 + 3*e^2*x^4)*ArcSech[c*x]))/(120*c^4*e^3) + (b*Sqrt[(1 - c*x)
/(1 + c*x)]*Sqrt[1 - c^2*x^2]*(Sqrt[-c^2]*Sqrt[-(c^2*d) - e]*Sqrt[e]*(45*c^4*d^2 - 10*c^2*d*e + 9*e^2)*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])] + 64*c^7*d^
(5/2)*Sqrt[-d - e*x^2]*ArcTan[(Sqrt[d]*Sqrt[1 - c^2*x^2])/Sqrt[-d - e*x^2]]))/(120*c^7*e^3*(-1 + c*x)*Sqrt[d +
e*x^2])

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Maple [F]
time = 1.36, size = 0, normalized size = 0.00 \[\int \frac {x^{5} \left (a +b \,\mathrm {arcsech}\left (c x \right )\right )}{\sqrt {e \,x^{2}+d}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(a+b*arcsech(c*x))/(e*x^2+d)^(1/2),x)

[Out]

int(x^5*(a+b*arcsech(c*x))/(e*x^2+d)^(1/2),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*arcsech(c*x))/(e*x^2+d)^(1/2),x, algorithm="maxima")

[Out]

1/15*(3*sqrt(x^2*e + d)*x^4*e^(-1) - 4*sqrt(x^2*e + d)*d*x^2*e^(-2) + 8*sqrt(x^2*e + d)*d^2*e^(-3))*a + 1/15*(
(3*x^6*e^3 - d*x^4*e^2 + 4*d^2*x^2*e + 8*d^3)*e^(-3)*log(sqrt(c*x + 1)*sqrt(-c*x + 1) + 1)/sqrt(x^2*e + d) - 1
5*integrate(1/15*(30*(c^2*x^6*e^3 - x^4*e^3)*x^5*log(sqrt(x)) + 15*(c^2*x^6*e^3*log(c) - x^4*e^3*log(c))*x^5 +
(30*(c^2*x^6*e^3 - x^4*e^3)*x^5*log(sqrt(x)) + (3*c^2*x^6*(5*log(c) + 1)*e^3 + 4*c^2*d^2*x^2*e + 8*c^2*d^3 -
(c^2*d*e^2 + 15*e^3*log(c))*x^4)*x^5)*e^(1/2*log(c*x + 1) + 1/2*log(-c*x + 1)))/((c^2*x^6*e^3 - x^4*e^3 + (c^2
*x^6*e^3 - x^4*e^3)*e^(1/2*log(c*x + 1) + 1/2*log(-c*x + 1)))*sqrt(x^2*e + d)), x))*b

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 737 vs. \(2 (234) = 468\).
time = 1.19, size = 1509, normalized size = 4.24 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*arcsech(c*x))/(e*x^2+d)^(1/2),x, algorithm="fricas")

[Out]

[1/240*(32*b*c^5*d^(5/2)*log((c^4*d^2*x^4 - 8*c^2*d^2*x^2 + x^4*cosh(1)^2 + x^4*sinh(1)^2 + 4*(c^3*d*x^3 - c*x
^3*cosh(1) - c*x^3*sinh(1) - 2*c*d*x)*sqrt(x^2*cosh(1) + x^2*sinh(1) + d)*sqrt(d)*sqrt(-(c^2*x^2 - 1)/(c^2*x^2
)) + 8*d^2 - 2*(3*c^2*d*x^4 - 4*d*x^2)*cosh(1) - 2*(3*c^2*d*x^4 - x^4*cosh(1) - 4*d*x^2)*sinh(1))/x^4) - (45*b
*c^4*d^2 - 10*b*c^2*d*cosh(1) + 9*b*cosh(1)^2 + 9*b*sinh(1)^2 - 2*(5*b*c^2*d - 9*b*cosh(1))*sinh(1))*sqrt(cosh
(1) + sinh(1))*arctan(1/2*(c^2*d*x + (2*c^2*x^3 - x)*cosh(1) + (2*c^2*x^3 - x)*sinh(1))*sqrt(x^2*cosh(1) + x^2
*sinh(1) + d)*sqrt(-(c^2*x^2 - 1)/(c^2*x^2))*sqrt(cosh(1) + sinh(1))/((c^2*x^4 - x^2)*cosh(1)^2 + (c^2*x^4 - x
^2)*sinh(1)^2 + (c^2*d*x^2 - d)*cosh(1) + (c^2*d*x^2 + 2*(c^2*x^4 - x^2)*cosh(1) - d)*sinh(1))) + 16*(3*b*c^5*
x^4*cosh(1)^2 + 3*b*c^5*x^4*sinh(1)^2 - 4*b*c^5*d*x^2*cosh(1) + 8*b*c^5*d^2 + 2*(3*b*c^5*x^4*cosh(1) - 2*b*c^5
*d*x^2)*sinh(1))*sqrt(x^2*cosh(1) + x^2*sinh(1) + d)*log((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + 1)/(c*x)) + 2*(
24*a*c^5*x^4*cosh(1)^2 + 24*a*c^5*x^4*sinh(1)^2 - 32*a*c^5*d*x^2*cosh(1) + 64*a*c^5*d^2 + 16*(3*a*c^5*x^4*cosh
(1) - 2*a*c^5*d*x^2)*sinh(1) + (13*b*c^4*d*x*cosh(1) - 3*(2*b*c^4*x^3 + 3*b*c^2*x)*cosh(1)^2 - 3*(2*b*c^4*x^3
+ 3*b*c^2*x)*sinh(1)^2 + (13*b*c^4*d*x - 6*(2*b*c^4*x^3 + 3*b*c^2*x)*cosh(1))*sinh(1))*sqrt(-(c^2*x^2 - 1)/(c^
2*x^2)))*sqrt(x^2*cosh(1) + x^2*sinh(1) + d))/(c^5*cosh(1)^3 + 3*c^5*cosh(1)^2*sinh(1) + 3*c^5*cosh(1)*sinh(1)
^2 + c^5*sinh(1)^3), -1/240*(64*b*c^5*sqrt(-d)*d^2*arctan(-1/2*(c^3*d*x^3 - c*x^3*cosh(1) - c*x^3*sinh(1) - 2*
c*d*x)*sqrt(x^2*cosh(1) + x^2*sinh(1) + d)*sqrt(-d)*sqrt(-(c^2*x^2 - 1)/(c^2*x^2))/(c^2*d^2*x^2 - d^2 + (c^2*d
*x^4 - d*x^2)*cosh(1) + (c^2*d*x^4 - d*x^2)*sinh(1))) + (45*b*c^4*d^2 - 10*b*c^2*d*cosh(1) + 9*b*cosh(1)^2 + 9
*b*sinh(1)^2 - 2*(5*b*c^2*d - 9*b*cosh(1))*sinh(1))*sqrt(cosh(1) + sinh(1))*arctan(1/2*(c^2*d*x + (2*c^2*x^3 -
x)*cosh(1) + (2*c^2*x^3 - x)*sinh(1))*sqrt(x^2*cosh(1) + x^2*sinh(1) + d)*sqrt(-(c^2*x^2 - 1)/(c^2*x^2))*sqrt
(cosh(1) + sinh(1))/((c^2*x^4 - x^2)*cosh(1)^2 + (c^2*x^4 - x^2)*sinh(1)^2 + (c^2*d*x^2 - d)*cosh(1) + (c^2*d*
x^2 + 2*(c^2*x^4 - x^2)*cosh(1) - d)*sinh(1))) - 16*(3*b*c^5*x^4*cosh(1)^2 + 3*b*c^5*x^4*sinh(1)^2 - 4*b*c^5*d
*x^2*cosh(1) + 8*b*c^5*d^2 + 2*(3*b*c^5*x^4*cosh(1) - 2*b*c^5*d*x^2)*sinh(1))*sqrt(x^2*cosh(1) + x^2*sinh(1) +
d)*log((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + 1)/(c*x)) - 2*(24*a*c^5*x^4*cosh(1)^2 + 24*a*c^5*x^4*sinh(1)^2 -
32*a*c^5*d*x^2*cosh(1) + 64*a*c^5*d^2 + 16*(3*a*c^5*x^4*cosh(1) - 2*a*c^5*d*x^2)*sinh(1) + (13*b*c^4*d*x*cosh
(1) - 3*(2*b*c^4*x^3 + 3*b*c^2*x)*cosh(1)^2 - 3*(2*b*c^4*x^3 + 3*b*c^2*x)*sinh(1)^2 + (13*b*c^4*d*x - 6*(2*b*c
^4*x^3 + 3*b*c^2*x)*cosh(1))*sinh(1))*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)))*sqrt(x^2*cosh(1) + x^2*sinh(1) + d))/(c^
5*cosh(1)^3 + 3*c^5*cosh(1)^2*sinh(1) + 3*c^5*cosh(1)*sinh(1)^2 + c^5*sinh(1)^3)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{5} \left (a + b \operatorname {asech}{\left (c x \right )}\right )}{\sqrt {d + e x^{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(a+b*asech(c*x))/(e*x**2+d)**(1/2),x)

[Out]

Integral(x**5*(a + b*asech(c*x))/sqrt(d + e*x**2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*arcsech(c*x))/(e*x^2+d)^(1/2),x, algorithm="giac")

[Out]

integrate((b*arcsech(c*x) + a)*x^5/sqrt(e*x^2 + d), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^5\,\left (a+b\,\mathrm {acosh}\left (\frac {1}{c\,x}\right )\right )}{\sqrt {e\,x^2+d}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^5*(a + b*acosh(1/(c*x))))/(d + e*x^2)^(1/2),x)

[Out]

int((x^5*(a + b*acosh(1/(c*x))))/(d + e*x^2)^(1/2), x)

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