3.2.21 \(\int x^3 (d+e x^2)^{3/2} (a+b \sec ^{-1}(c x)) \, dx\) [121]

Optimal. Leaf size=374 \[ \frac {b \left (3 c^4 d^2-38 c^2 d e-25 e^2\right ) x \sqrt {-1+c^2 x^2} \sqrt {d+e x^2}}{560 c^5 e \sqrt {c^2 x^2}}-\frac {b \left (13 c^2 d+25 e\right ) x \sqrt {-1+c^2 x^2} \left (d+e x^2\right )^{3/2}}{840 c^3 e \sqrt {c^2 x^2}}-\frac {b x \sqrt {-1+c^2 x^2} \left (d+e x^2\right )^{5/2}}{42 c e \sqrt {c^2 x^2}}-\frac {d \left (d+e x^2\right )^{5/2} \left (a+b \sec ^{-1}(c x)\right )}{5 e^2}+\frac {\left (d+e x^2\right )^{7/2} \left (a+b \sec ^{-1}(c x)\right )}{7 e^2}-\frac {2 b c d^{7/2} x \text {ArcTan}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {-1+c^2 x^2}}\right )}{35 e^2 \sqrt {c^2 x^2}}+\frac {b \left (35 c^6 d^3-35 c^4 d^2 e-63 c^2 d e^2-25 e^3\right ) x \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {-1+c^2 x^2}}{c \sqrt {d+e x^2}}\right )}{560 c^6 e^{3/2} \sqrt {c^2 x^2}} \]

[Out]

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

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Rubi [A]
time = 0.37, antiderivative size = 374, 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 = {272, 45, 5346, 12, 587, 159, 163, 65, 223, 212, 95, 210} \begin {gather*} -\frac {d \left (d+e x^2\right )^{5/2} \left (a+b \sec ^{-1}(c x)\right )}{5 e^2}+\frac {\left (d+e x^2\right )^{7/2} \left (a+b \sec ^{-1}(c x)\right )}{7 e^2}-\frac {2 b c d^{7/2} x \text {ArcTan}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {c^2 x^2-1}}\right )}{35 e^2 \sqrt {c^2 x^2}}-\frac {b x \sqrt {c^2 x^2-1} \left (d+e x^2\right )^{5/2}}{42 c e \sqrt {c^2 x^2}}-\frac {b x \sqrt {c^2 x^2-1} \left (13 c^2 d+25 e\right ) \left (d+e x^2\right )^{3/2}}{840 c^3 e \sqrt {c^2 x^2}}+\frac {b x \left (35 c^6 d^3-35 c^4 d^2 e-63 c^2 d e^2-25 e^3\right ) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {c^2 x^2-1}}{c \sqrt {d+e x^2}}\right )}{560 c^6 e^{3/2} \sqrt {c^2 x^2}}+\frac {b x \sqrt {c^2 x^2-1} \left (3 c^4 d^2-38 c^2 d e-25 e^2\right ) \sqrt {d+e x^2}}{560 c^5 e \sqrt {c^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(3*c^4*d^2 - 38*c^2*d*e - 25*e^2)*x*Sqrt[-1 + c^2*x^2]*Sqrt[d + e*x^2])/(560*c^5*e*Sqrt[c^2*x^2]) - (b*(13*
c^2*d + 25*e)*x*Sqrt[-1 + c^2*x^2]*(d + e*x^2)^(3/2))/(840*c^3*e*Sqrt[c^2*x^2]) - (b*x*Sqrt[-1 + c^2*x^2]*(d +
e*x^2)^(5/2))/(42*c*e*Sqrt[c^2*x^2]) - (d*(d + e*x^2)^(5/2)*(a + b*ArcSec[c*x]))/(5*e^2) + ((d + e*x^2)^(7/2)
*(a + b*ArcSec[c*x]))/(7*e^2) - (2*b*c*d^(7/2)*x*ArcTan[Sqrt[d + e*x^2]/(Sqrt[d]*Sqrt[-1 + c^2*x^2])])/(35*e^2
*Sqrt[c^2*x^2]) + (b*(35*c^6*d^3 - 35*c^4*d^2*e - 63*c^2*d*e^2 - 25*e^3)*x*ArcTanh[(Sqrt[e]*Sqrt[-1 + c^2*x^2]
)/(c*Sqrt[d + e*x^2])])/(560*c^6*e^(3/2)*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 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 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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[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 5346

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

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Mathematica [A]
time = 3.18, size = 295, normalized size = 0.79 \begin {gather*} -\frac {\sqrt {d+e x^2} \left (48 a c^5 \left (2 d-5 e x^2\right ) \left (d+e x^2\right )^2+b e \sqrt {1-\frac {1}{c^2 x^2}} x \left (75 e^2+2 c^2 e \left (82 d+25 e x^2\right )+c^4 \left (57 d^2+106 d e x^2+40 e^2 x^4\right )\right )+48 b c^5 \left (2 d-5 e x^2\right ) \left (d+e x^2\right )^2 \sec ^{-1}(c x)\right )}{1680 c^5 e^2}-\frac {b \sqrt {1-\frac {1}{c^2 x^2}} x \left (-32 c^7 d^{7/2} \text {ArcTan}\left (\frac {\sqrt {d} \sqrt {-1+c^2 x^2}}{\sqrt {d+e x^2}}\right )+\sqrt {e} \left (-35 c^6 d^3+35 c^4 d^2 e+63 c^2 d e^2+25 e^3\right ) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {-1+c^2 x^2}}{c \sqrt {d+e x^2}}\right )\right )}{560 c^6 e^2 \sqrt {-1+c^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/1680*(Sqrt[d + e*x^2]*(48*a*c^5*(2*d - 5*e*x^2)*(d + e*x^2)^2 + b*e*Sqrt[1 - 1/(c^2*x^2)]*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^5*(2*d - 5*e*x^2)*(d + e*x^2)^2*ArcSe
c[c*x]))/(c^5*e^2) - (b*Sqrt[1 - 1/(c^2*x^2)]*x*(-32*c^7*d^(7/2)*ArcTan[(Sqrt[d]*Sqrt[-1 + c^2*x^2])/Sqrt[d +
e*x^2]] + Sqrt[e]*(-35*c^6*d^3 + 35*c^4*d^2*e + 63*c^2*d*e^2 + 25*e^3)*ArcTanh[(Sqrt[e]*Sqrt[-1 + c^2*x^2])/(c
*Sqrt[d + e*x^2])]))/(560*c^6*e^2*Sqrt[-1 + c^2*x^2])

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Maple [F]
time = 2.09, size = 0, normalized size = 0.00 \[\int x^{3} \left (e \,x^{2}+d \right )^{\frac {3}{2}} \left (a +b \,\mathrm {arcsec}\left (c x \right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Fricas [A]
time = 8.71, size = 867, normalized size = 2.32 \begin {gather*} \left [\frac {{\left (96 \, b c^{7} \sqrt {-d} d^{3} \log \left (\frac {c^{4} d^{2} x^{4} - 8 \, c^{2} d^{2} x^{2} + x^{4} e^{2} + 4 \, {\left (c^{2} d x^{2} - x^{2} e - 2 \, d\right )} \sqrt {c^{2} x^{2} - 1} \sqrt {x^{2} e + d} \sqrt {-d} + 8 \, d^{2} - 2 \, {\left (3 \, c^{2} d x^{4} - 4 \, d x^{2}\right )} e}{x^{4}}\right ) - 3 \, {\left (35 \, b c^{6} d^{3} - 35 \, b c^{4} d^{2} e - 63 \, b c^{2} d e^{2} - 25 \, b e^{3}\right )} e^{\frac {1}{2}} \log \left (c^{4} d^{2} - 4 \, {\left (c^{3} d + {\left (2 \, c^{3} x^{2} - c\right )} e\right )} \sqrt {c^{2} x^{2} - 1} \sqrt {x^{2} e + d} e^{\frac {1}{2}} + {\left (8 \, c^{4} x^{4} - 8 \, c^{2} x^{2} + 1\right )} e^{2} + 2 \, {\left (4 \, c^{4} d x^{2} - 3 \, c^{2} d\right )} e\right ) + 4 \, {\left (240 \, a c^{7} x^{6} e^{3} + 384 \, a c^{7} d x^{4} e^{2} + 48 \, a c^{7} d^{2} x^{2} e - 96 \, a c^{7} d^{3} + 48 \, {\left (5 \, b c^{7} x^{6} e^{3} + 8 \, b c^{7} d x^{4} e^{2} + b c^{7} d^{2} x^{2} e - 2 \, b c^{7} d^{3}\right )} \operatorname {arcsec}\left (c x\right ) - {\left (57 \, b c^{5} d^{2} e + 5 \, {\left (8 \, b c^{5} x^{4} + 10 \, b c^{3} x^{2} + 15 \, b c\right )} e^{3} + 2 \, {\left (53 \, b c^{5} d x^{2} + 82 \, b c^{3} d\right )} e^{2}\right )} \sqrt {c^{2} x^{2} - 1}\right )} \sqrt {x^{2} e + d}\right )} e^{\left (-2\right )}}{6720 \, c^{7}}, -\frac {{\left (192 \, b c^{7} d^{\frac {7}{2}} \arctan \left (-\frac {{\left (c^{2} d x^{2} - x^{2} e - 2 \, d\right )} \sqrt {c^{2} x^{2} - 1} \sqrt {x^{2} e + d} \sqrt {d}}{2 \, {\left (c^{2} d^{2} x^{2} - d^{2} + {\left (c^{2} d x^{4} - d x^{2}\right )} e\right )}}\right ) + 3 \, {\left (35 \, b c^{6} d^{3} - 35 \, b c^{4} d^{2} e - 63 \, b c^{2} d e^{2} - 25 \, b e^{3}\right )} e^{\frac {1}{2}} \log \left (c^{4} d^{2} - 4 \, {\left (c^{3} d + {\left (2 \, c^{3} x^{2} - c\right )} e\right )} \sqrt {c^{2} x^{2} - 1} \sqrt {x^{2} e + d} e^{\frac {1}{2}} + {\left (8 \, c^{4} x^{4} - 8 \, c^{2} x^{2} + 1\right )} e^{2} + 2 \, {\left (4 \, c^{4} d x^{2} - 3 \, c^{2} d\right )} e\right ) - 4 \, {\left (240 \, a c^{7} x^{6} e^{3} + 384 \, a c^{7} d x^{4} e^{2} + 48 \, a c^{7} d^{2} x^{2} e - 96 \, a c^{7} d^{3} + 48 \, {\left (5 \, b c^{7} x^{6} e^{3} + 8 \, b c^{7} d x^{4} e^{2} + b c^{7} d^{2} x^{2} e - 2 \, b c^{7} d^{3}\right )} \operatorname {arcsec}\left (c x\right ) - {\left (57 \, b c^{5} d^{2} e + 5 \, {\left (8 \, b c^{5} x^{4} + 10 \, b c^{3} x^{2} + 15 \, b c\right )} e^{3} + 2 \, {\left (53 \, b c^{5} d x^{2} + 82 \, b c^{3} d\right )} e^{2}\right )} \sqrt {c^{2} x^{2} - 1}\right )} \sqrt {x^{2} e + d}\right )} e^{\left (-2\right )}}{6720 \, c^{7}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6720*(96*b*c^7*sqrt(-d)*d^3*log((c^4*d^2*x^4 - 8*c^2*d^2*x^2 + x^4*e^2 + 4*(c^2*d*x^2 - x^2*e - 2*d)*sqrt(c
^2*x^2 - 1)*sqrt(x^2*e + d)*sqrt(-d) + 8*d^2 - 2*(3*c^2*d*x^4 - 4*d*x^2)*e)/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)*e^(1/2)*log(c^4*d^2 - 4*(c^3*d + (2*c^3*x^2 - c)*e)*sqrt(c^2*x^2 - 1)*sqrt(
x^2*e + d)*e^(1/2) + (8*c^4*x^4 - 8*c^2*x^2 + 1)*e^2 + 2*(4*c^4*d*x^2 - 3*c^2*d)*e) + 4*(240*a*c^7*x^6*e^3 + 3
84*a*c^7*d*x^4*e^2 + 48*a*c^7*d^2*x^2*e - 96*a*c^7*d^3 + 48*(5*b*c^7*x^6*e^3 + 8*b*c^7*d*x^4*e^2 + b*c^7*d^2*x
^2*e - 2*b*c^7*d^3)*arcsec(c*x) - (57*b*c^5*d^2*e + 5*(8*b*c^5*x^4 + 10*b*c^3*x^2 + 15*b*c)*e^3 + 2*(53*b*c^5*
d*x^2 + 82*b*c^3*d)*e^2)*sqrt(c^2*x^2 - 1))*sqrt(x^2*e + d))*e^(-2)/c^7, -1/6720*(192*b*c^7*d^(7/2)*arctan(-1/
2*(c^2*d*x^2 - x^2*e - 2*d)*sqrt(c^2*x^2 - 1)*sqrt(x^2*e + d)*sqrt(d)/(c^2*d^2*x^2 - d^2 + (c^2*d*x^4 - d*x^2)
*e)) + 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)*e^(1/2)*log(c^4*d^2 - 4*(c^3*d + (2*c^3*x
^2 - c)*e)*sqrt(c^2*x^2 - 1)*sqrt(x^2*e + d)*e^(1/2) + (8*c^4*x^4 - 8*c^2*x^2 + 1)*e^2 + 2*(4*c^4*d*x^2 - 3*c^
2*d)*e) - 4*(240*a*c^7*x^6*e^3 + 384*a*c^7*d*x^4*e^2 + 48*a*c^7*d^2*x^2*e - 96*a*c^7*d^3 + 48*(5*b*c^7*x^6*e^3
+ 8*b*c^7*d*x^4*e^2 + b*c^7*d^2*x^2*e - 2*b*c^7*d^3)*arcsec(c*x) - (57*b*c^5*d^2*e + 5*(8*b*c^5*x^4 + 10*b*c^
3*x^2 + 15*b*c)*e^3 + 2*(53*b*c^5*d*x^2 + 82*b*c^3*d)*e^2)*sqrt(c^2*x^2 - 1))*sqrt(x^2*e + d))*e^(-2)/c^7]

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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