3.140 \(\int x^3 (d+e x^2)^{3/2} (a+b \text {sech}^{-1}(c x)) \, dx\)

Optimal. Leaf size=418 \[ -\frac {d \left (d+e x^2\right )^{5/2} \left (a+b \text {sech}^{-1}(c x)\right )}{5 e^2}+\frac {\left (d+e x^2\right )^{7/2} \left (a+b \text {sech}^{-1}(c x)\right )}{7 e^2}+\frac {2 b d^{7/2} \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {1-c^2 x^2}}\right )}{35 e^2}-\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2} \left (d+e x^2\right )^{5/2}}{42 c^2 e}-\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2} \left (13 c^2 d+25 e\right ) \left (d+e x^2\right )^{3/2}}{840 c^4 e}+\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2} \left (3 c^4 d^2-38 c^2 d e-25 e^2\right ) \sqrt {d+e x^2}}{560 c^6 e}+\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \left (35 c^6 d^3-35 c^4 d^2 e-63 c^2 d e^2-25 e^3\right ) \tan ^{-1}\left (\frac {\sqrt {e} \sqrt {1-c^2 x^2}}{c \sqrt {d+e x^2}}\right )}{560 c^7 e^{3/2}} \]

[Out]

-1/5*d*(e*x^2+d)^(5/2)*(a+b*arcsech(c*x))/e^2+1/7*(e*x^2+d)^(7/2)*(a+b*arcsech(c*x))/e^2+1/560*b*(35*c^6*d^3-3
5*c^4*d^2*e-63*c^2*d*e^2-25*e^3)*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^7/e^(3/2)+2/35*b*d^(7/2)*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^2-1/840*b*(13*c^2*d+25*e)*(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^4/e-
1/42*b*(e*x^2+d)^(5/2)*(1/(c*x+1))^(1/2)*(c*x+1)^(1/2)*(-c^2*x^2+1)^(1/2)/c^2/e+1/560*b*(3*c^4*d^2-38*c^2*d*e-
25*e^2)*(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^6/e

________________________________________________________________________________________

Rubi [A]  time = 0.53, antiderivative size = 418, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 12, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.522, Rules used = {266, 43, 6301, 12, 573, 154, 157, 63, 217, 203, 93, 207} \[ -\frac {d \left (d+e x^2\right )^{5/2} \left (a+b \text {sech}^{-1}(c x)\right )}{5 e^2}+\frac {\left (d+e x^2\right )^{7/2} \left (a+b \text {sech}^{-1}(c x)\right )}{7 e^2}+\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2} \left (3 c^4 d^2-38 c^2 d e-25 e^2\right ) \sqrt {d+e x^2}}{560 c^6 e}+\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \left (-35 c^4 d^2 e+35 c^6 d^3-63 c^2 d e^2-25 e^3\right ) \tan ^{-1}\left (\frac {\sqrt {e} \sqrt {1-c^2 x^2}}{c \sqrt {d+e x^2}}\right )}{560 c^7 e^{3/2}}+\frac {2 b d^{7/2} \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {1-c^2 x^2}}\right )}{35 e^2}-\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2} \left (d+e x^2\right )^{5/2}}{42 c^2 e}-\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2} \left (13 c^2 d+25 e\right ) \left (d+e x^2\right )^{3/2}}{840 c^4 e} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(d + e*x^2)^(3/2)*(a + b*ArcSech[c*x]),x]

[Out]

(b*(3*c^4*d^2 - 38*c^2*d*e - 25*e^2)*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*Sqrt[1 - c^2*x^2]*Sqrt[d + e*x^2])/(56
0*c^6*e) - (b*(13*c^2*d + 25*e)*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*Sqrt[1 - c^2*x^2]*(d + e*x^2)^(3/2))/(840*c
^4*e) - (b*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*Sqrt[1 - c^2*x^2]*(d + e*x^2)^(5/2))/(42*c^2*e) - (d*(d + e*x^2)
^(5/2)*(a + b*ArcSech[c*x]))/(5*e^2) + ((d + e*x^2)^(7/2)*(a + b*ArcSech[c*x]))/(7*e^2) + (b*(35*c^6*d^3 - 35*
c^4*d^2*e - 63*c^2*d*e^2 - 25*e^3)*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*ArcTan[(Sqrt[e]*Sqrt[1 - c^2*x^2])/(c*Sq
rt[d + e*x^2])])/(560*c^7*e^(3/2)) + (2*b*d^(7/2)*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*ArcTanh[Sqrt[d + e*x^2]/(
Sqrt[d]*Sqrt[1 - c^2*x^2])])/(35*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 43

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 63

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 93

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 154

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 157

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 203

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

Rule 207

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

Rule 217

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 266

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 573

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 6301

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 x^3 \left (d+e x^2\right )^{3/2} \left (a+b \text {sech}^{-1}(c x)\right ) \, dx &=-\frac {d \left (d+e x^2\right )^{5/2} \left (a+b \text {sech}^{-1}(c x)\right )}{5 e^2}+\frac {\left (d+e x^2\right )^{7/2} \left (a+b \text {sech}^{-1}(c x)\right )}{7 e^2}+\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {\left (d+e x^2\right )^{5/2} \left (-2 d+5 e x^2\right )}{35 e^2 x \sqrt {1-c^2 x^2}} \, dx\\ &=-\frac {d \left (d+e x^2\right )^{5/2} \left (a+b \text {sech}^{-1}(c x)\right )}{5 e^2}+\frac {\left (d+e x^2\right )^{7/2} \left (a+b \text {sech}^{-1}(c x)\right )}{7 e^2}+\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {\left (d+e x^2\right )^{5/2} \left (-2 d+5 e x^2\right )}{x \sqrt {1-c^2 x^2}} \, dx}{35 e^2}\\ &=-\frac {d \left (d+e x^2\right )^{5/2} \left (a+b \text {sech}^{-1}(c x)\right )}{5 e^2}+\frac {\left (d+e x^2\right )^{7/2} \left (a+b \text {sech}^{-1}(c x)\right )}{7 e^2}+\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \operatorname {Subst}\left (\int \frac {(d+e x)^{5/2} (-2 d+5 e x)}{x \sqrt {1-c^2 x}} \, dx,x,x^2\right )}{70 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 )^{5/2}}{42 c^2 e}-\frac {d \left (d+e x^2\right )^{5/2} \left (a+b \text {sech}^{-1}(c x)\right )}{5 e^2}+\frac {\left (d+e x^2\right )^{7/2} \left (a+b \text {sech}^{-1}(c x)\right )}{7 e^2}-\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \operatorname {Subst}\left (\int \frac {(d+e x)^{3/2} \left (6 c^2 d^2-\frac {1}{2} e \left (13 c^2 d+25 e\right ) x\right )}{x \sqrt {1-c^2 x}} \, dx,x,x^2\right )}{210 c^2 e^2}\\ &=-\frac {b \left (13 c^2 d+25 e\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \left (d+e x^2\right )^{3/2}}{840 c^4 e}-\frac {b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \left (d+e x^2\right )^{5/2}}{42 c^2 e}-\frac {d \left (d+e x^2\right )^{5/2} \left (a+b \text {sech}^{-1}(c x)\right )}{5 e^2}+\frac {\left (d+e x^2\right )^{7/2} \left (a+b \text {sech}^{-1}(c x)\right )}{7 e^2}+\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {d+e x} \left (-12 c^4 d^3-\frac {3}{4} e \left (3 c^4 d^2-38 c^2 d e-25 e^2\right ) x\right )}{x \sqrt {1-c^2 x}} \, dx,x,x^2\right )}{420 c^4 e^2}\\ &=\frac {b \left (3 c^4 d^2-38 c^2 d e-25 e^2\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \sqrt {d+e x^2}}{560 c^6 e}-\frac {b \left (13 c^2 d+25 e\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \left (d+e x^2\right )^{3/2}}{840 c^4 e}-\frac {b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \left (d+e x^2\right )^{5/2}}{42 c^2 e}-\frac {d \left (d+e x^2\right )^{5/2} \left (a+b \text {sech}^{-1}(c x)\right )}{5 e^2}+\frac {\left (d+e x^2\right )^{7/2} \left (a+b \text {sech}^{-1}(c x)\right )}{7 e^2}-\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \operatorname {Subst}\left (\int \frac {12 c^6 d^4+\frac {3}{8} e \left (35 c^6 d^3-35 c^4 d^2 e-63 c^2 d e^2-25 e^3\right ) x}{x \sqrt {1-c^2 x} \sqrt {d+e x}} \, dx,x,x^2\right )}{420 c^6 e^2}\\ &=\frac {b \left (3 c^4 d^2-38 c^2 d e-25 e^2\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \sqrt {d+e x^2}}{560 c^6 e}-\frac {b \left (13 c^2 d+25 e\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \left (d+e x^2\right )^{3/2}}{840 c^4 e}-\frac {b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \left (d+e x^2\right )^{5/2}}{42 c^2 e}-\frac {d \left (d+e x^2\right )^{5/2} \left (a+b \text {sech}^{-1}(c x)\right )}{5 e^2}+\frac {\left (d+e x^2\right )^{7/2} \left (a+b \text {sech}^{-1}(c x)\right )}{7 e^2}-\frac {\left (b d^4 \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1-c^2 x} \sqrt {d+e x}} \, dx,x,x^2\right )}{35 e^2}-\frac {\left (b \left (35 c^6 d^3-35 c^4 d^2 e-63 c^2 d e^2-25 e^3\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-c^2 x} \sqrt {d+e x}} \, dx,x,x^2\right )}{1120 c^6 e}\\ &=\frac {b \left (3 c^4 d^2-38 c^2 d e-25 e^2\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \sqrt {d+e x^2}}{560 c^6 e}-\frac {b \left (13 c^2 d+25 e\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \left (d+e x^2\right )^{3/2}}{840 c^4 e}-\frac {b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \left (d+e x^2\right )^{5/2}}{42 c^2 e}-\frac {d \left (d+e x^2\right )^{5/2} \left (a+b \text {sech}^{-1}(c x)\right )}{5 e^2}+\frac {\left (d+e x^2\right )^{7/2} \left (a+b \text {sech}^{-1}(c x)\right )}{7 e^2}-\frac {\left (2 b d^4 \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \operatorname {Subst}\left (\int \frac {1}{-d+x^2} \, dx,x,\frac {\sqrt {d+e x^2}}{\sqrt {1-c^2 x^2}}\right )}{35 e^2}+\frac {\left (b \left (35 c^6 d^3-35 c^4 d^2 e-63 c^2 d e^2-25 e^3\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \operatorname {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 )}{560 c^8 e}\\ &=\frac {b \left (3 c^4 d^2-38 c^2 d e-25 e^2\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \sqrt {d+e x^2}}{560 c^6 e}-\frac {b \left (13 c^2 d+25 e\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \left (d+e x^2\right )^{3/2}}{840 c^4 e}-\frac {b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \left (d+e x^2\right )^{5/2}}{42 c^2 e}-\frac {d \left (d+e x^2\right )^{5/2} \left (a+b \text {sech}^{-1}(c x)\right )}{5 e^2}+\frac {\left (d+e x^2\right )^{7/2} \left (a+b \text {sech}^{-1}(c x)\right )}{7 e^2}+\frac {2 b d^{7/2} \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 )}{35 e^2}+\frac {\left (b \left (35 c^6 d^3-35 c^4 d^2 e-63 c^2 d e^2-25 e^3\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \operatorname {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 )}{560 c^8 e}\\ &=\frac {b \left (3 c^4 d^2-38 c^2 d e-25 e^2\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \sqrt {d+e x^2}}{560 c^6 e}-\frac {b \left (13 c^2 d+25 e\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \left (d+e x^2\right )^{3/2}}{840 c^4 e}-\frac {b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \left (d+e x^2\right )^{5/2}}{42 c^2 e}-\frac {d \left (d+e x^2\right )^{5/2} \left (a+b \text {sech}^{-1}(c x)\right )}{5 e^2}+\frac {\left (d+e x^2\right )^{7/2} \left (a+b \text {sech}^{-1}(c x)\right )}{7 e^2}+\frac {b \left (35 c^6 d^3-35 c^4 d^2 e-63 c^2 d e^2-25 e^3\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 )}{560 c^7 e^{3/2}}+\frac {2 b d^{7/2} \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 )}{35 e^2}\\ \end {align*}

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Mathematica [A]  time = 2.99, size = 382, normalized size = 0.91 \[ \frac {b \sqrt {\frac {1-c x}{c x+1}} \sqrt {c^2 x^2-1} \left (32 c^9 d^{7/2} \sqrt {d+e x^2} \tan ^{-1}\left (\frac {\sqrt {d} \sqrt {c^2 x^2-1}}{\sqrt {d+e x^2}}\right )+\sqrt {c^2} \sqrt {e} \sqrt {c^2 d+e} \left (35 c^6 d^3-35 c^4 d^2 e-63 c^2 d e^2-25 e^3\right ) \sqrt {\frac {c^2 \left (d+e x^2\right )}{c^2 d+e}} \sinh ^{-1}\left (\frac {c \sqrt {e} \sqrt {c^2 x^2-1}}{\sqrt {c^2} \sqrt {c^2 d+e}}\right )\right )}{560 c^9 e^2 (c x-1) \sqrt {d+e x^2}}-\frac {\sqrt {d+e x^2} \left (48 a c^6 \left (2 d-5 e x^2\right ) \left (d+e x^2\right )^2+48 b c^6 \text {sech}^{-1}(c x) \left (2 d-5 e x^2\right ) \left (d+e x^2\right )^2+b e \sqrt {\frac {1-c x}{c x+1}} (c x+1) \left (c^4 \left (57 d^2+106 d e x^2+40 e^2 x^4\right )+2 c^2 e \left (82 d+25 e x^2\right )+75 e^2\right )\right )}{1680 c^6 e^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*(d + e*x^2)^(3/2)*(a + b*ArcSech[c*x]),x]

[Out]

-1/1680*(Sqrt[d + e*x^2]*(48*a*c^6*(2*d - 5*e*x^2)*(d + e*x^2)^2 + b*e*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x)*(75
*e^2 + 2*c^2*e*(82*d + 25*e*x^2) + c^4*(57*d^2 + 106*d*e*x^2 + 40*e^2*x^4)) + 48*b*c^6*(2*d - 5*e*x^2)*(d + e*
x^2)^2*ArcSech[c*x]))/(c^6*e^2) + (b*Sqrt[(1 - c*x)/(1 + c*x)]*Sqrt[-1 + c^2*x^2]*(Sqrt[c^2]*Sqrt[e]*Sqrt[c^2*
d + e]*(35*c^6*d^3 - 35*c^4*d^2*e - 63*c^2*d*e^2 - 25*e^3)*Sqrt[(c^2*(d + e*x^2))/(c^2*d + e)]*ArcSinh[(c*Sqrt
[e]*Sqrt[-1 + c^2*x^2])/(Sqrt[c^2]*Sqrt[c^2*d + e])] + 32*c^9*d^(7/2)*Sqrt[d + e*x^2]*ArcTan[(Sqrt[d]*Sqrt[-1
+ c^2*x^2])/Sqrt[d + e*x^2]]))/(560*c^9*e^2*(-1 + c*x)*Sqrt[d + e*x^2])

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fricas [A]  time = 3.98, size = 1989, normalized size = 4.76 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(e*x^2+d)^(3/2)*(a+b*arcsech(c*x)),x, algorithm="fricas")

[Out]

[1/6720*(96*b*c^7*d^(7/2)*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) + 3*(35*b*c^6*d^3 - 35*b*c^4*d
^2*e - 63*b*c^2*d*e^2 - 25*b*e^3)*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) + 192*(5
*b*c^7*e^3*x^6 + 8*b*c^7*d*e^2*x^4 + b*c^7*d^2*e*x^2 - 2*b*c^7*d^3)*sqrt(e*x^2 + d)*log((c*x*sqrt(-(c^2*x^2 -
1)/(c^2*x^2)) + 1)/(c*x)) + 4*(240*a*c^7*e^3*x^6 + 384*a*c^7*d*e^2*x^4 + 48*a*c^7*d^2*e*x^2 - 96*a*c^7*d^3 - (
40*b*c^6*e^3*x^5 + 2*(53*b*c^6*d*e^2 + 25*b*c^4*e^3)*x^3 + (57*b*c^6*d^2*e + 164*b*c^4*d*e^2 + 75*b*c^2*e^3)*x
)*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)))*sqrt(e*x^2 + d))/(c^7*e^2), 1/3360*(48*b*c^7*d^(7/2)*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) + 3*(35*b*c^6*d^3 - 35*b*c^4*d^2*e - 63*b*c^2*d*e^2 - 25*b*e^3)*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)) + 96*(5*b*c^7*e^3*x^6 + 8*b*c^7*d*e^2*x^4 + b*c^7*d^2*e*x^2 - 2*b*c^7*d^3)*sqrt(e*x^2 +
d)*log((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + 1)/(c*x)) + 2*(240*a*c^7*e^3*x^6 + 384*a*c^7*d*e^2*x^4 + 48*a*c^7
*d^2*e*x^2 - 96*a*c^7*d^3 - (40*b*c^6*e^3*x^5 + 2*(53*b*c^6*d*e^2 + 25*b*c^4*e^3)*x^3 + (57*b*c^6*d^2*e + 164*
b*c^4*d*e^2 + 75*b*c^2*e^3)*x)*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)))*sqrt(e*x^2 + d))/(c^7*e^2), 1/6720*(192*b*c^7*s
qrt(-d)*d^3*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)) + 3*(35*b*c^6*d^3 - 35*b*c^4*d^2*e - 63*b*c^2*d*e^2 - 25*b*e^3)*sqr
t(-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) + 192*(5*b*c^7*e^3*x^6 + 8*b*c^7*d*e^2*x^4 +
b*c^7*d^2*e*x^2 - 2*b*c^7*d^3)*sqrt(e*x^2 + d)*log((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + 1)/(c*x)) + 4*(240*a*
c^7*e^3*x^6 + 384*a*c^7*d*e^2*x^4 + 48*a*c^7*d^2*e*x^2 - 96*a*c^7*d^3 - (40*b*c^6*e^3*x^5 + 2*(53*b*c^6*d*e^2
+ 25*b*c^4*e^3)*x^3 + (57*b*c^6*d^2*e + 164*b*c^4*d*e^2 + 75*b*c^2*e^3)*x)*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)))*sqr
t(e*x^2 + d))/(c^7*e^2), 1/3360*(96*b*c^7*sqrt(-d)*d^3*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)) + 3*(35*b*c^6*d^3 - 35*b
*c^4*d^2*e - 63*b*c^2*d*e^2 - 25*b*e^3)*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)) + 96*(5*b*c^7*e^3*x^6 + 8*b*c^7*d
*e^2*x^4 + b*c^7*d^2*e*x^2 - 2*b*c^7*d^3)*sqrt(e*x^2 + d)*log((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + 1)/(c*x))
+ 2*(240*a*c^7*e^3*x^6 + 384*a*c^7*d*e^2*x^4 + 48*a*c^7*d^2*e*x^2 - 96*a*c^7*d^3 - (40*b*c^6*e^3*x^5 + 2*(53*b
*c^6*d*e^2 + 25*b*c^4*e^3)*x^3 + (57*b*c^6*d^2*e + 164*b*c^4*d*e^2 + 75*b*c^2*e^3)*x)*sqrt(-(c^2*x^2 - 1)/(c^2
*x^2)))*sqrt(e*x^2 + d))/(c^7*e^2)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (e x^{2} + d\right )}^{\frac {3}{2}} {\left (b \operatorname {arsech}\left (c x\right ) + a\right )} x^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(e*x^2+d)^(3/2)*(a+b*arcsech(c*x)),x, algorithm="giac")

[Out]

integrate((e*x^2 + d)^(3/2)*(b*arcsech(c*x) + a)*x^3, x)

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maple [F]  time = 3.37, size = 0, normalized size = 0.00 \[ \int x^{3} \left (e \,x^{2}+d \right )^{\frac {3}{2}} \left (a +b \,\mathrm {arcsech}\left (c x \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(e*x^2+d)^(3/2)*(a+b*arcsech(c*x)),x)

[Out]

int(x^3*(e*x^2+d)^(3/2)*(a+b*arcsech(c*x)),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{35} \, {\left (\frac {5 \, {\left (e x^{2} + d\right )}^{\frac {5}{2}} x^{2}}{e} - \frac {2 \, {\left (e x^{2} + d\right )}^{\frac {5}{2}} d}{e^{2}}\right )} a + \frac {1}{105} \, b {\left (\frac {{\left ({\left (15 \, e^{3} x^{6} + 3 \, d e^{2} x^{4} - 4 \, d^{2} e x^{2} + 8 \, d^{3}\right )} x^{5} + 7 \, {\left (3 \, d e^{2} x^{6} + d^{2} e x^{4} - 2 \, d^{3} x^{2}\right )} x^{3}\right )} \sqrt {e x^{2} + d} \log \left (\sqrt {c x + 1} \sqrt {-c x + 1} + 1\right )}{e^{2} x^{5}} - 105 \, \int \frac {{\left (105 \, {\left (c^{2} e^{3} x^{6} \log \relax (c) - e^{3} x^{4} \log \relax (c)\right )} x^{5} + 105 \, {\left (c^{2} d e^{2} x^{6} \log \relax (c) - d e^{2} x^{4} \log \relax (c)\right )} x^{3} + {\left ({\left (15 \, {\left (7 \, e^{3} \log \relax (c) + e^{3}\right )} c^{2} x^{6} - 4 \, c^{2} d^{2} e x^{2} + 8 \, c^{2} d^{3} + 3 \, {\left (c^{2} d e^{2} - 35 \, e^{3} \log \relax (c)\right )} x^{4}\right )} x^{5} + 7 \, {\left (3 \, {\left (5 \, d e^{2} \log \relax (c) + d e^{2}\right )} c^{2} x^{6} - 2 \, c^{2} d^{3} x^{2} + {\left (c^{2} d^{2} e - 15 \, d e^{2} \log \relax (c)\right )} x^{4}\right )} x^{3} + 210 \, {\left ({\left (c^{2} e^{3} x^{6} - e^{3} x^{4}\right )} x^{5} + {\left (c^{2} d e^{2} x^{6} - d e^{2} x^{4}\right )} x^{3}\right )} \log \left (\sqrt {x}\right )\right )} e^{\left (\frac {1}{2} \, \log \left (c x + 1\right ) + \frac {1}{2} \, \log \left (-c x + 1\right )\right )} + 210 \, {\left ({\left (c^{2} e^{3} x^{6} - e^{3} x^{4}\right )} x^{5} + {\left (c^{2} d e^{2} x^{6} - d e^{2} x^{4}\right )} x^{3}\right )} \log \left (\sqrt {x}\right )\right )} \sqrt {e x^{2} + d}}{105 \, {\left (c^{2} e^{2} x^{6} - e^{2} x^{4} + {\left (c^{2} e^{2} x^{6} - e^{2} x^{4}\right )} e^{\left (\frac {1}{2} \, \log \left (c x + 1\right ) + \frac {1}{2} \, \log \left (-c x + 1\right )\right )}\right )}}\,{d x}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(e*x^2+d)^(3/2)*(a+b*arcsech(c*x)),x, algorithm="maxima")

[Out]

1/35*(5*(e*x^2 + d)^(5/2)*x^2/e - 2*(e*x^2 + d)^(5/2)*d/e^2)*a + 1/105*b*(((15*e^3*x^6 + 3*d*e^2*x^4 - 4*d^2*e
*x^2 + 8*d^3)*x^5 + 7*(3*d*e^2*x^6 + d^2*e*x^4 - 2*d^3*x^2)*x^3)*sqrt(e*x^2 + d)*log(sqrt(c*x + 1)*sqrt(-c*x +
1) + 1)/(e^2*x^5) - 105*integrate(1/105*(105*(c^2*e^3*x^6*log(c) - e^3*x^4*log(c))*x^5 + 105*(c^2*d*e^2*x^6*l
og(c) - d*e^2*x^4*log(c))*x^3 + ((15*(7*e^3*log(c) + e^3)*c^2*x^6 - 4*c^2*d^2*e*x^2 + 8*c^2*d^3 + 3*(c^2*d*e^2
- 35*e^3*log(c))*x^4)*x^5 + 7*(3*(5*d*e^2*log(c) + d*e^2)*c^2*x^6 - 2*c^2*d^3*x^2 + (c^2*d^2*e - 15*d*e^2*log
(c))*x^4)*x^3 + 210*((c^2*e^3*x^6 - e^3*x^4)*x^5 + (c^2*d*e^2*x^6 - d*e^2*x^4)*x^3)*log(sqrt(x)))*e^(1/2*log(c
*x + 1) + 1/2*log(-c*x + 1)) + 210*((c^2*e^3*x^6 - e^3*x^4)*x^5 + (c^2*d*e^2*x^6 - d*e^2*x^4)*x^3)*log(sqrt(x)
))*sqrt(e*x^2 + d)/(c^2*e^2*x^6 - e^2*x^4 + (c^2*e^2*x^6 - e^2*x^4)*e^(1/2*log(c*x + 1) + 1/2*log(-c*x + 1))),
x))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x^3\,{\left (e\,x^2+d\right )}^{3/2}\,\left (a+b\,\mathrm {acosh}\left (\frac {1}{c\,x}\right )\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(d + e*x^2)^(3/2)*(a + b*acosh(1/(c*x))),x)

[Out]

int(x^3*(d + e*x^2)^(3/2)*(a + b*acosh(1/(c*x))), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(e*x**2+d)**(3/2)*(a+b*asech(c*x)),x)

[Out]

Timed out

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