3.131 \(\int \frac {x^5 (a+b \sec ^{-1}(c x))}{\sqrt {d+e x^2}} \, dx\)

Optimal. Leaf size=321 \[ \frac {d^2 \sqrt {d+e x^2} \left (a+b \sec ^{-1}(c x)\right )}{e^3}-\frac {2 d \left (d+e x^2\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{3 e^3}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \sec ^{-1}(c x)\right )}{5 e^3}+\frac {8 b c d^{5/2} x \tan ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {c^2 x^2-1}}\right )}{15 e^3 \sqrt {c^2 x^2}}-\frac {b x \sqrt {c^2 x^2-1} \left (d+e x^2\right )^{3/2}}{20 c e^2 \sqrt {c^2 x^2}}-\frac {b x \left (45 c^4 d^2-10 c^2 d e+9 e^2\right ) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {c^2 x^2-1}}{c \sqrt {d+e x^2}}\right )}{120 c^4 e^{5/2} \sqrt {c^2 x^2}}+\frac {b x \sqrt {c^2 x^2-1} \left (19 c^2 d-9 e\right ) \sqrt {d+e x^2}}{120 c^3 e^2 \sqrt {c^2 x^2}} \]

[Out]

-2/3*d*(e*x^2+d)^(3/2)*(a+b*arcsec(c*x))/e^3+1/5*(e*x^2+d)^(5/2)*(a+b*arcsec(c*x))/e^3+8/15*b*c*d^(5/2)*x*arct
an((e*x^2+d)^(1/2)/d^(1/2)/(c^2*x^2-1)^(1/2))/e^3/(c^2*x^2)^(1/2)-1/120*b*(45*c^4*d^2-10*c^2*d*e+9*e^2)*x*arct
anh(e^(1/2)*(c^2*x^2-1)^(1/2)/c/(e*x^2+d)^(1/2))/c^4/e^(5/2)/(c^2*x^2)^(1/2)-1/20*b*x*(e*x^2+d)^(3/2)*(c^2*x^2
-1)^(1/2)/c/e^2/(c^2*x^2)^(1/2)+d^2*(a+b*arcsec(c*x))*(e*x^2+d)^(1/2)/e^3+1/120*b*(19*c^2*d-9*e)*x*(c^2*x^2-1)
^(1/2)*(e*x^2+d)^(1/2)/c^3/e^2/(c^2*x^2)^(1/2)

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Rubi [A]  time = 1.01, antiderivative size = 321, 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 = {266, 43, 5238, 12, 1615, 154, 157, 63, 217, 206, 93, 204} \[ \frac {d^2 \sqrt {d+e x^2} \left (a+b \sec ^{-1}(c x)\right )}{e^3}-\frac {2 d \left (d+e x^2\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{3 e^3}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \sec ^{-1}(c x)\right )}{5 e^3}+\frac {8 b c d^{5/2} x \tan ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {c^2 x^2-1}}\right )}{15 e^3 \sqrt {c^2 x^2}}-\frac {b x \left (45 c^4 d^2-10 c^2 d e+9 e^2\right ) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {c^2 x^2-1}}{c \sqrt {d+e x^2}}\right )}{120 c^4 e^{5/2} \sqrt {c^2 x^2}}-\frac {b x \sqrt {c^2 x^2-1} \left (d+e x^2\right )^{3/2}}{20 c e^2 \sqrt {c^2 x^2}}+\frac {b x \sqrt {c^2 x^2-1} \left (19 c^2 d-9 e\right ) \sqrt {d+e x^2}}{120 c^3 e^2 \sqrt {c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(19*c^2*d - 9*e)*x*Sqrt[-1 + c^2*x^2]*Sqrt[d + e*x^2])/(120*c^3*e^2*Sqrt[c^2*x^2]) - (b*x*Sqrt[-1 + c^2*x^2
]*(d + e*x^2)^(3/2))/(20*c*e^2*Sqrt[c^2*x^2]) + (d^2*Sqrt[d + e*x^2]*(a + b*ArcSec[c*x]))/e^3 - (2*d*(d + e*x^
2)^(3/2)*(a + b*ArcSec[c*x]))/(3*e^3) + ((d + e*x^2)^(5/2)*(a + b*ArcSec[c*x]))/(5*e^3) + (8*b*c*d^(5/2)*x*Arc
Tan[Sqrt[d + e*x^2]/(Sqrt[d]*Sqrt[-1 + c^2*x^2])])/(15*e^3*Sqrt[c^2*x^2]) - (b*(45*c^4*d^2 - 10*c^2*d*e + 9*e^
2)*x*ArcTanh[(Sqrt[e]*Sqrt[-1 + c^2*x^2])/(c*Sqrt[d + e*x^2])])/(120*c^4*e^(5/2)*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 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 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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[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 1615

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] && IntegersQ[2*m, 2*n, 2*p]

Rule 5238

Int[((a_.) + ArcSec[(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*ArcSec[c*x], u, x] - Dist[(b*c*x)/Sqrt[c^2*x^2], Int[SimplifyI
ntegrand[u/(x*Sqrt[c^2*x^2 - 1]), 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])) || (I
LtQ[(m + 2*p + 1)/2, 0] &&  !ILtQ[(m - 1)/2, 0]))

Rubi steps

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

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Mathematica [A]  time = 0.75, size = 328, normalized size = 1.02 \[ \frac {\sqrt {d+e x^2} \left (8 a c^3 \left (8 d^2-4 d e x^2+3 e^2 x^4\right )+8 b c^3 \sec ^{-1}(c x) \left (8 d^2-4 d e x^2+3 e^2 x^4\right )+b e x \sqrt {1-\frac {1}{c^2 x^2}} \left (c^2 \left (13 d-6 e x^2\right )-9 e\right )\right )}{120 c^3 e^3}-\frac {b x \sqrt {1-\frac {1}{c^2 x^2}} \left (64 c^7 d^{5/2} \sqrt {d+e x^2} \tan ^{-1}\left (\frac {\sqrt {d} \sqrt {c^2 x^2-1}}{\sqrt {d+e x^2}}\right )+\sqrt {c^2} \sqrt {e} \sqrt {c^2 d+e} \left (45 c^4 d^2-10 c^2 d e+9 e^2\right ) \sqrt {\frac {c^2 \left (d+e x^2\right )}{c^2 d+e}} \sinh ^{-1}\left (\frac {c \sqrt {e} \sqrt {c^2 x^2-1}}{\sqrt {c^2} \sqrt {c^2 d+e}}\right )\right )}{120 c^6 e^3 \sqrt {c^2 x^2-1} \sqrt {d+e x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d + e*x^2]*(8*a*c^3*(8*d^2 - 4*d*e*x^2 + 3*e^2*x^4) + b*e*Sqrt[1 - 1/(c^2*x^2)]*x*(-9*e + c^2*(13*d - 6*
e*x^2)) + 8*b*c^3*(8*d^2 - 4*d*e*x^2 + 3*e^2*x^4)*ArcSec[c*x]))/(120*c^3*e^3) - (b*Sqrt[1 - 1/(c^2*x^2)]*x*(Sq
rt[c^2]*Sqrt[e]*Sqrt[c^2*d + 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)]*ArcSinh[
(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]*S
qrt[-1 + c^2*x^2])/Sqrt[d + e*x^2]]))/(120*c^6*e^3*Sqrt[-1 + c^2*x^2]*Sqrt[d + e*x^2])

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fricas [A]  time = 5.85, size = 1385, normalized size = 4.31 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/480*(64*b*c^5*sqrt(-d)*d^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*sqrt(c^2*x^2 -
1)*((c^2*d - e)*x^2 - 2*d)*sqrt(e*x^2 + d)*sqrt(-d) + 8*d^2)/x^4) + (45*b*c^4*d^2 - 10*b*c^2*d*e + 9*b*e^2)*sq
rt(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^3*e*x^2 + c^3*d - c*e)*sqrt
(c^2*x^2 - 1)*sqrt(e*x^2 + d)*sqrt(e) + e^2) + 4*(24*a*c^5*e^2*x^4 - 32*a*c^5*d*e*x^2 + 64*a*c^5*d^2 + 8*(3*b*
c^5*e^2*x^4 - 4*b*c^5*d*e*x^2 + 8*b*c^5*d^2)*arcsec(c*x) - (6*b*c^3*e^2*x^2 - 13*b*c^3*d*e + 9*b*c*e^2)*sqrt(c
^2*x^2 - 1))*sqrt(e*x^2 + d))/(c^5*e^3), 1/480*(128*b*c^5*d^(5/2)*arctan(-1/2*sqrt(c^2*x^2 - 1)*((c^2*d - e)*x
^2 - 2*d)*sqrt(e*x^2 + d)*sqrt(d)/(c^2*d*e*x^4 + (c^2*d^2 - d*e)*x^2 - d^2)) + (45*b*c^4*d^2 - 10*b*c^2*d*e +
9*b*e^2)*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^3*e*x^2 + c^3*d
- c*e)*sqrt(c^2*x^2 - 1)*sqrt(e*x^2 + d)*sqrt(e) + e^2) + 4*(24*a*c^5*e^2*x^4 - 32*a*c^5*d*e*x^2 + 64*a*c^5*d^
2 + 8*(3*b*c^5*e^2*x^4 - 4*b*c^5*d*e*x^2 + 8*b*c^5*d^2)*arcsec(c*x) - (6*b*c^3*e^2*x^2 - 13*b*c^3*d*e + 9*b*c*
e^2)*sqrt(c^2*x^2 - 1))*sqrt(e*x^2 + d))/(c^5*e^3), 1/240*(32*b*c^5*sqrt(-d)*d^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*sqrt(c^2*x^2 - 1)*((c^2*d - e)*x^2 - 2*d)*sqrt(e*x^2 + d)*sqrt(-d) + 8*d^2
)/x^4) + (45*b*c^4*d^2 - 10*b*c^2*d*e + 9*b*e^2)*sqrt(-e)*arctan(1/2*(2*c^2*e*x^2 + c^2*d - e)*sqrt(c^2*x^2 -
1)*sqrt(e*x^2 + d)*sqrt(-e)/(c^3*e^2*x^4 - c*d*e + (c^3*d*e - c*e^2)*x^2)) + 2*(24*a*c^5*e^2*x^4 - 32*a*c^5*d*
e*x^2 + 64*a*c^5*d^2 + 8*(3*b*c^5*e^2*x^4 - 4*b*c^5*d*e*x^2 + 8*b*c^5*d^2)*arcsec(c*x) - (6*b*c^3*e^2*x^2 - 13
*b*c^3*d*e + 9*b*c*e^2)*sqrt(c^2*x^2 - 1))*sqrt(e*x^2 + d))/(c^5*e^3), 1/240*(64*b*c^5*d^(5/2)*arctan(-1/2*sqr
t(c^2*x^2 - 1)*((c^2*d - e)*x^2 - 2*d)*sqrt(e*x^2 + d)*sqrt(d)/(c^2*d*e*x^4 + (c^2*d^2 - d*e)*x^2 - d^2)) + (4
5*b*c^4*d^2 - 10*b*c^2*d*e + 9*b*e^2)*sqrt(-e)*arctan(1/2*(2*c^2*e*x^2 + c^2*d - e)*sqrt(c^2*x^2 - 1)*sqrt(e*x
^2 + d)*sqrt(-e)/(c^3*e^2*x^4 - c*d*e + (c^3*d*e - c*e^2)*x^2)) + 2*(24*a*c^5*e^2*x^4 - 32*a*c^5*d*e*x^2 + 64*
a*c^5*d^2 + 8*(3*b*c^5*e^2*x^4 - 4*b*c^5*d*e*x^2 + 8*b*c^5*d^2)*arcsec(c*x) - (6*b*c^3*e^2*x^2 - 13*b*c^3*d*e
+ 9*b*c*e^2)*sqrt(c^2*x^2 - 1))*sqrt(e*x^2 + d))/(c^5*e^3)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \operatorname {arcsec}\left (c x\right ) + a\right )} x^{5}}{\sqrt {e x^{2} + d}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 5.36, size = 0, normalized size = 0.00 \[ \int \frac {x^{5} \left (a +b \,\mathrm {arcsec}\left (c x \right )\right )}{\sqrt {e \,x^{2}+d}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(3*sqrt(e*x^2 + d)*x^4/e - 4*sqrt(e*x^2 + d)*d*x^2/e^2 + 8*sqrt(e*x^2 + d)*d^2/e^3)*a - 1/15*(15*e^3*inte
grate((15*c^2*e^3*x^7*log(c) - 15*e^3*x^5*log(c) + (3*(5*c^2*log(c) + c^2)*e^3*x^7 + 4*c^2*d^2*e*x^3 + 8*c^2*d
^3*x - (c^2*d*e^2 + 15*e^3*log(c))*x^5)*e^(log(c*x + 1) + log(c*x - 1)) + 15*(c^2*e^3*x^7 - e^3*x^5 + (c^2*e^3
*x^7 - e^3*x^5)*e^(log(c*x + 1) + log(c*x - 1)))*log(x))/((c^2*e^3*x^2 - e^3 + (c^2*e^3*x^2 - e^3)*e^(log(c*x
+ 1) + log(c*x - 1)))*sqrt(e*x^2 + d)), x) - (3*e^2*x^4 - 4*d*e*x^2 + 8*d^2)*sqrt(e*x^2 + d)*arctan(sqrt(c*x +
1)*sqrt(c*x - 1)))*b/e^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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