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

Optimal. Leaf size=372 \[ \frac {2 (d+e x)^{5/2} \left (a+b \sec ^{-1}(c x)\right )}{5 e}+\frac {4 b d^3 \sqrt {1-c^2 x^2} \sqrt {\frac {c (d+e x)}{c d+e}} \Pi \left (2;\sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {2}}\right )|\frac {2 e}{c d+e}\right )}{5 c e x \sqrt {1-\frac {1}{c^2 x^2}} \sqrt {d+e x}}+\frac {28 b d \sqrt {1-c^2 x^2} \sqrt {d+e x} E\left (\sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {2}}\right )|\frac {2 e}{c d+e}\right )}{15 c^2 x \sqrt {1-\frac {1}{c^2 x^2}} \sqrt {\frac {c (d+e x)}{c d+e}}}+\frac {4 b \sqrt {1-c^2 x^2} \left (2 c^2 d^2+e^2\right ) \sqrt {\frac {c (d+e x)}{c d+e}} F\left (\sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {2}}\right )|\frac {2 e}{c d+e}\right )}{15 c^4 x \sqrt {1-\frac {1}{c^2 x^2}} \sqrt {d+e x}}+\frac {4 b e \left (1-c^2 x^2\right ) \sqrt {d+e x}}{15 c^3 x \sqrt {1-\frac {1}{c^2 x^2}}} \]

[Out]

2/5*(e*x+d)^(5/2)*(a+b*arcsec(c*x))/e+4/15*b*e*(-c^2*x^2+1)*(e*x+d)^(1/2)/c^3/x/(1-1/c^2/x^2)^(1/2)+28/15*b*d*
EllipticE(1/2*(-c*x+1)^(1/2)*2^(1/2),2^(1/2)*(e/(c*d+e))^(1/2))*(e*x+d)^(1/2)*(-c^2*x^2+1)^(1/2)/c^2/x/(1-1/c^
2/x^2)^(1/2)/(c*(e*x+d)/(c*d+e))^(1/2)+4/15*b*(2*c^2*d^2+e^2)*EllipticF(1/2*(-c*x+1)^(1/2)*2^(1/2),2^(1/2)*(e/
(c*d+e))^(1/2))*(c*(e*x+d)/(c*d+e))^(1/2)*(-c^2*x^2+1)^(1/2)/c^4/x/(1-1/c^2/x^2)^(1/2)/(e*x+d)^(1/2)+4/5*b*d^3
*EllipticPi(1/2*(-c*x+1)^(1/2)*2^(1/2),2,2^(1/2)*(e/(c*d+e))^(1/2))*(c*(e*x+d)/(c*d+e))^(1/2)*(-c^2*x^2+1)^(1/
2)/c/e/x/(1-1/c^2/x^2)^(1/2)/(e*x+d)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.76, antiderivative size = 372, normalized size of antiderivative = 1.00, number of steps used = 22, number of rules used = 13, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.722, Rules used = {5226, 1574, 958, 719, 419, 933, 168, 538, 537, 844, 424, 931, 1584} \[ \frac {2 (d+e x)^{5/2} \left (a+b \sec ^{-1}(c x)\right )}{5 e}+\frac {4 b \sqrt {1-c^2 x^2} \left (2 c^2 d^2+e^2\right ) \sqrt {\frac {c (d+e x)}{c d+e}} F\left (\sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {2}}\right )|\frac {2 e}{c d+e}\right )}{15 c^4 x \sqrt {1-\frac {1}{c^2 x^2}} \sqrt {d+e x}}+\frac {4 b d^3 \sqrt {1-c^2 x^2} \sqrt {\frac {c (d+e x)}{c d+e}} \Pi \left (2;\sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {2}}\right )|\frac {2 e}{c d+e}\right )}{5 c e x \sqrt {1-\frac {1}{c^2 x^2}} \sqrt {d+e x}}+\frac {4 b e \left (1-c^2 x^2\right ) \sqrt {d+e x}}{15 c^3 x \sqrt {1-\frac {1}{c^2 x^2}}}+\frac {28 b d \sqrt {1-c^2 x^2} \sqrt {d+e x} E\left (\sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {2}}\right )|\frac {2 e}{c d+e}\right )}{15 c^2 x \sqrt {1-\frac {1}{c^2 x^2}} \sqrt {\frac {c (d+e x)}{c d+e}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*b*e*Sqrt[d + e*x]*(1 - c^2*x^2))/(15*c^3*Sqrt[1 - 1/(c^2*x^2)]*x) + (2*(d + e*x)^(5/2)*(a + b*ArcSec[c*x]))
/(5*e) + (28*b*d*Sqrt[d + e*x]*Sqrt[1 - c^2*x^2]*EllipticE[ArcSin[Sqrt[1 - c*x]/Sqrt[2]], (2*e)/(c*d + e)])/(1
5*c^2*Sqrt[1 - 1/(c^2*x^2)]*x*Sqrt[(c*(d + e*x))/(c*d + e)]) + (4*b*(2*c^2*d^2 + e^2)*Sqrt[(c*(d + e*x))/(c*d
+ e)]*Sqrt[1 - c^2*x^2]*EllipticF[ArcSin[Sqrt[1 - c*x]/Sqrt[2]], (2*e)/(c*d + e)])/(15*c^4*Sqrt[1 - 1/(c^2*x^2
)]*x*Sqrt[d + e*x]) + (4*b*d^3*Sqrt[(c*(d + e*x))/(c*d + e)]*Sqrt[1 - c^2*x^2]*EllipticPi[2, ArcSin[Sqrt[1 - c
*x]/Sqrt[2]], (2*e)/(c*d + e)])/(5*c*e*Sqrt[1 - 1/(c^2*x^2)]*x*Sqrt[d + e*x])

Rule 168

Int[1/(((a_.) + (b_.)*(x_))*Sqrt[(c_.) + (d_.)*(x_)]*Sqrt[(e_.) + (f_.)*(x_)]*Sqrt[(g_.) + (h_.)*(x_)]), x_Sym
bol] :> Dist[-2, Subst[Int[1/(Simp[b*c - a*d - b*x^2, x]*Sqrt[Simp[(d*e - c*f)/d + (f*x^2)/d, x]]*Sqrt[Simp[(d
*g - c*h)/d + (h*x^2)/d, x]]), x], x, Sqrt[c + d*x]], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x] && GtQ[(d*e - c
*f)/d, 0]

Rule 419

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(1*EllipticF[ArcSin[Rt[-(d/c),
2]*x], (b*c)/(a*d)])/(Sqrt[a]*Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] &
& GtQ[a, 0] &&  !(NegQ[b/a] && SimplerSqrtQ[-(b/a), -(d/c)])

Rule 424

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]*EllipticE[ArcSin[Rt[-(d/c)
, 2]*x], (b*c)/(a*d)])/(Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[
a, 0]

Rule 537

Int[1/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Simp[(1*Ellipt
icPi[(b*c)/(a*d), ArcSin[Rt[-(d/c), 2]*x], (c*f)/(d*e)])/(a*Sqrt[c]*Sqrt[e]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b,
c, d, e, f}, x] &&  !GtQ[d/c, 0] && GtQ[c, 0] && GtQ[e, 0] &&  !( !GtQ[f/e, 0] && SimplerSqrtQ[-(f/e), -(d/c)
])

Rule 538

Int[1/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Dist[Sqrt[1 +
(d*x^2)/c]/Sqrt[c + d*x^2], Int[1/((a + b*x^2)*Sqrt[1 + (d*x^2)/c]*Sqrt[e + f*x^2]), x], x] /; FreeQ[{a, b, c,
d, e, f}, x] &&  !GtQ[c, 0]

Rule 719

Int[((d_) + (e_.)*(x_))^(m_)/Sqrt[(a_) + (c_.)*(x_)^2], x_Symbol] :> Dist[(2*a*Rt[-(c/a), 2]*(d + e*x)^m*Sqrt[
1 + (c*x^2)/a])/(c*Sqrt[a + c*x^2]*((c*(d + e*x))/(c*d - a*e*Rt[-(c/a), 2]))^m), Subst[Int[(1 + (2*a*e*Rt[-(c/
a), 2]*x^2)/(c*d - a*e*Rt[-(c/a), 2]))^m/Sqrt[1 - x^2], x], x, Sqrt[(1 - Rt[-(c/a), 2]*x)/2]], x] /; FreeQ[{a,
c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ[m^2, 1/4]

Rule 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rule 931

Int[((d_.) + (e_.)*(x_))^(m_)/(Sqrt[(f_.) + (g_.)*(x_)]*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> Simp[(2*e^2*(
d + e*x)^(m - 2)*Sqrt[f + g*x]*Sqrt[a + c*x^2])/(c*g*(2*m - 1)), x] - Dist[1/(c*g*(2*m - 1)), Int[((d + e*x)^(
m - 3)*Simp[a*e^2*(d*g + 2*e*f*(m - 2)) - c*d^3*g*(2*m - 1) + e*(e*(a*e*g*(2*m - 3)) + c*d*(2*e*f - 3*d*g*(2*m
- 1)))*x + 2*e^2*(c*e*f - 3*c*d*g)*(m - 1)*x^2, x])/(Sqrt[f + g*x]*Sqrt[a + c*x^2]), x], x] /; FreeQ[{a, c, d
, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[c*d^2 + a*e^2, 0] && IntegerQ[2*m] && GeQ[m, 2]

Rule 933

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(f_.) + (g_.)*(x_)]*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> With[{q = Rt[-(c
/a), 2]}, Dist[Sqrt[1 + (c*x^2)/a]/Sqrt[a + c*x^2], Int[1/((d + e*x)*Sqrt[f + g*x]*Sqrt[1 - q*x]*Sqrt[1 + q*x]
), x], x]] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[c*d^2 + a*e^2, 0] &&  !GtQ[a, 0]

Rule 958

Int[((f_.) + (g_.)*(x_))^(n_)/(((d_.) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> Int[ExpandIntegra
nd[1/(Sqrt[f + g*x]*Sqrt[a + c*x^2]), (f + g*x)^(n + 1/2)/(d + e*x), x], x] /; FreeQ[{a, c, d, e, f, g}, x] &&
NeQ[e*f - d*g, 0] && NeQ[c*d^2 + a*e^2, 0] && IntegerQ[n + 1/2]

Rule 1574

Int[(x_)^(m_.)*((a_.) + (c_.)*(x_)^(mn2_.))^(p_)*((d_) + (e_.)*(x_)^(n_.))^(q_.), x_Symbol] :> Dist[(x^(2*n*Fr
acPart[p])*(a + c/x^(2*n))^FracPart[p])/(c + a*x^(2*n))^FracPart[p], Int[x^(m - 2*n*p)*(d + e*x^n)^q*(c + a*x^
(2*n))^p, x], x] /; FreeQ[{a, c, d, e, m, n, p, q}, x] && EqQ[mn2, -2*n] &&  !IntegerQ[p] &&  !IntegerQ[q] &&
PosQ[n]

Rule 1584

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 5226

Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(a + b
*ArcSec[c*x]))/(e*(m + 1)), x] - Dist[b/(c*e*(m + 1)), Int[(d + e*x)^(m + 1)/(x^2*Sqrt[1 - 1/(c^2*x^2)]), x],
x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[m, -1]

Rubi steps

\begin {align*} \int (d+e x)^{3/2} \left (a+b \sec ^{-1}(c x)\right ) \, dx &=\frac {2 (d+e x)^{5/2} \left (a+b \sec ^{-1}(c x)\right )}{5 e}-\frac {(2 b) \int \frac {(d+e x)^{5/2}}{\sqrt {1-\frac {1}{c^2 x^2}} x^2} \, dx}{5 c e}\\ &=\frac {2 (d+e x)^{5/2} \left (a+b \sec ^{-1}(c x)\right )}{5 e}-\frac {\left (2 b \sqrt {-\frac {1}{c^2}+x^2}\right ) \int \frac {(d+e x)^{5/2}}{x \sqrt {-\frac {1}{c^2}+x^2}} \, dx}{5 c e \sqrt {1-\frac {1}{c^2 x^2}} x}\\ &=\frac {2 (d+e x)^{5/2} \left (a+b \sec ^{-1}(c x)\right )}{5 e}-\frac {\left (2 b \sqrt {-\frac {1}{c^2}+x^2}\right ) \int \left (\frac {3 d^2 e}{\sqrt {d+e x} \sqrt {-\frac {1}{c^2}+x^2}}+\frac {d^3}{x \sqrt {d+e x} \sqrt {-\frac {1}{c^2}+x^2}}+\frac {3 d e^2 x}{\sqrt {d+e x} \sqrt {-\frac {1}{c^2}+x^2}}+\frac {e^3 x^2}{\sqrt {d+e x} \sqrt {-\frac {1}{c^2}+x^2}}\right ) \, dx}{5 c e \sqrt {1-\frac {1}{c^2 x^2}} x}\\ &=\frac {2 (d+e x)^{5/2} \left (a+b \sec ^{-1}(c x)\right )}{5 e}-\frac {\left (6 b d^2 \sqrt {-\frac {1}{c^2}+x^2}\right ) \int \frac {1}{\sqrt {d+e x} \sqrt {-\frac {1}{c^2}+x^2}} \, dx}{5 c \sqrt {1-\frac {1}{c^2 x^2}} x}-\frac {\left (2 b d^3 \sqrt {-\frac {1}{c^2}+x^2}\right ) \int \frac {1}{x \sqrt {d+e x} \sqrt {-\frac {1}{c^2}+x^2}} \, dx}{5 c e \sqrt {1-\frac {1}{c^2 x^2}} x}-\frac {\left (6 b d e \sqrt {-\frac {1}{c^2}+x^2}\right ) \int \frac {x}{\sqrt {d+e x} \sqrt {-\frac {1}{c^2}+x^2}} \, dx}{5 c \sqrt {1-\frac {1}{c^2 x^2}} x}-\frac {\left (2 b e^2 \sqrt {-\frac {1}{c^2}+x^2}\right ) \int \frac {x^2}{\sqrt {d+e x} \sqrt {-\frac {1}{c^2}+x^2}} \, dx}{5 c \sqrt {1-\frac {1}{c^2 x^2}} x}\\ &=\frac {4 b e \sqrt {d+e x} \left (1-c^2 x^2\right )}{15 c^3 \sqrt {1-\frac {1}{c^2 x^2}} x}+\frac {2 (d+e x)^{5/2} \left (a+b \sec ^{-1}(c x)\right )}{5 e}-\frac {\left (6 b d \sqrt {-\frac {1}{c^2}+x^2}\right ) \int \frac {\sqrt {d+e x}}{\sqrt {-\frac {1}{c^2}+x^2}} \, dx}{5 c \sqrt {1-\frac {1}{c^2 x^2}} x}+\frac {\left (6 b d^2 \sqrt {-\frac {1}{c^2}+x^2}\right ) \int \frac {1}{\sqrt {d+e x} \sqrt {-\frac {1}{c^2}+x^2}} \, dx}{5 c \sqrt {1-\frac {1}{c^2 x^2}} x}+\frac {\left (2 b e \sqrt {-\frac {1}{c^2}+x^2}\right ) \int \frac {-\frac {e x}{c^2}+2 d x^2}{x \sqrt {d+e x} \sqrt {-\frac {1}{c^2}+x^2}} \, dx}{15 c \sqrt {1-\frac {1}{c^2 x^2}} x}-\frac {\left (2 b d^3 \sqrt {1-c^2 x^2}\right ) \int \frac {1}{x \sqrt {1-c x} \sqrt {1+c x} \sqrt {d+e x}} \, dx}{5 c e \sqrt {1-\frac {1}{c^2 x^2}} x}+\frac {\left (12 b d^2 \sqrt {\frac {d+e x}{d+\frac {e}{c}}} \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {1-\frac {2 e x^2}{c \left (d+\frac {e}{c}\right )}}} \, dx,x,\frac {\sqrt {1-c x}}{\sqrt {2}}\right )}{5 c^2 \sqrt {1-\frac {1}{c^2 x^2}} x \sqrt {d+e x}}\\ &=\frac {4 b e \sqrt {d+e x} \left (1-c^2 x^2\right )}{15 c^3 \sqrt {1-\frac {1}{c^2 x^2}} x}+\frac {2 (d+e x)^{5/2} \left (a+b \sec ^{-1}(c x)\right )}{5 e}+\frac {12 b d^2 \sqrt {\frac {c (d+e x)}{c d+e}} \sqrt {1-c^2 x^2} F\left (\sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {2}}\right )|\frac {2 e}{c d+e}\right )}{5 c^2 \sqrt {1-\frac {1}{c^2 x^2}} x \sqrt {d+e x}}+\frac {\left (2 b e \sqrt {-\frac {1}{c^2}+x^2}\right ) \int \frac {-\frac {e}{c^2}+2 d x}{\sqrt {d+e x} \sqrt {-\frac {1}{c^2}+x^2}} \, dx}{15 c \sqrt {1-\frac {1}{c^2 x^2}} x}+\frac {\left (4 b d^3 \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1-x^2\right ) \sqrt {2-x^2} \sqrt {d+\frac {e}{c}-\frac {e x^2}{c}}} \, dx,x,\sqrt {1-c x}\right )}{5 c e \sqrt {1-\frac {1}{c^2 x^2}} x}+\frac {\left (12 b d \sqrt {d+e x} \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {1-\frac {2 e x^2}{c \left (d+\frac {e}{c}\right )}}}{\sqrt {1-x^2}} \, dx,x,\frac {\sqrt {1-c x}}{\sqrt {2}}\right )}{5 c^2 \sqrt {1-\frac {1}{c^2 x^2}} x \sqrt {\frac {d+e x}{d+\frac {e}{c}}}}-\frac {\left (12 b d^2 \sqrt {\frac {d+e x}{d+\frac {e}{c}}} \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {1-\frac {2 e x^2}{c \left (d+\frac {e}{c}\right )}}} \, dx,x,\frac {\sqrt {1-c x}}{\sqrt {2}}\right )}{5 c^2 \sqrt {1-\frac {1}{c^2 x^2}} x \sqrt {d+e x}}\\ &=\frac {4 b e \sqrt {d+e x} \left (1-c^2 x^2\right )}{15 c^3 \sqrt {1-\frac {1}{c^2 x^2}} x}+\frac {2 (d+e x)^{5/2} \left (a+b \sec ^{-1}(c x)\right )}{5 e}+\frac {12 b d \sqrt {d+e x} \sqrt {1-c^2 x^2} E\left (\sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {2}}\right )|\frac {2 e}{c d+e}\right )}{5 c^2 \sqrt {1-\frac {1}{c^2 x^2}} x \sqrt {\frac {c (d+e x)}{c d+e}}}+\frac {\left (4 b d \sqrt {-\frac {1}{c^2}+x^2}\right ) \int \frac {\sqrt {d+e x}}{\sqrt {-\frac {1}{c^2}+x^2}} \, dx}{15 c \sqrt {1-\frac {1}{c^2 x^2}} x}+\frac {\left (2 b \left (-2 d^2-\frac {e^2}{c^2}\right ) \sqrt {-\frac {1}{c^2}+x^2}\right ) \int \frac {1}{\sqrt {d+e x} \sqrt {-\frac {1}{c^2}+x^2}} \, dx}{15 c \sqrt {1-\frac {1}{c^2 x^2}} x}+\frac {\left (4 b d^3 \sqrt {\frac {c (d+e x)}{c d+e}} \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1-x^2\right ) \sqrt {2-x^2} \sqrt {1-\frac {e x^2}{c \left (d+\frac {e}{c}\right )}}} \, dx,x,\sqrt {1-c x}\right )}{5 c e \sqrt {1-\frac {1}{c^2 x^2}} x \sqrt {d+e x}}\\ &=\frac {4 b e \sqrt {d+e x} \left (1-c^2 x^2\right )}{15 c^3 \sqrt {1-\frac {1}{c^2 x^2}} x}+\frac {2 (d+e x)^{5/2} \left (a+b \sec ^{-1}(c x)\right )}{5 e}+\frac {12 b d \sqrt {d+e x} \sqrt {1-c^2 x^2} E\left (\sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {2}}\right )|\frac {2 e}{c d+e}\right )}{5 c^2 \sqrt {1-\frac {1}{c^2 x^2}} x \sqrt {\frac {c (d+e x)}{c d+e}}}+\frac {4 b d^3 \sqrt {\frac {c (d+e x)}{c d+e}} \sqrt {1-c^2 x^2} \Pi \left (2;\sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {2}}\right )|\frac {2 e}{c d+e}\right )}{5 c e \sqrt {1-\frac {1}{c^2 x^2}} x \sqrt {d+e x}}-\frac {\left (8 b d \sqrt {d+e x} \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {1-\frac {2 e x^2}{c \left (d+\frac {e}{c}\right )}}}{\sqrt {1-x^2}} \, dx,x,\frac {\sqrt {1-c x}}{\sqrt {2}}\right )}{15 c^2 \sqrt {1-\frac {1}{c^2 x^2}} x \sqrt {\frac {d+e x}{d+\frac {e}{c}}}}-\frac {\left (4 b \left (-2 d^2-\frac {e^2}{c^2}\right ) \sqrt {\frac {d+e x}{d+\frac {e}{c}}} \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {1-\frac {2 e x^2}{c \left (d+\frac {e}{c}\right )}}} \, dx,x,\frac {\sqrt {1-c x}}{\sqrt {2}}\right )}{15 c^2 \sqrt {1-\frac {1}{c^2 x^2}} x \sqrt {d+e x}}\\ &=\frac {4 b e \sqrt {d+e x} \left (1-c^2 x^2\right )}{15 c^3 \sqrt {1-\frac {1}{c^2 x^2}} x}+\frac {2 (d+e x)^{5/2} \left (a+b \sec ^{-1}(c x)\right )}{5 e}+\frac {28 b d \sqrt {d+e x} \sqrt {1-c^2 x^2} E\left (\sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {2}}\right )|\frac {2 e}{c d+e}\right )}{15 c^2 \sqrt {1-\frac {1}{c^2 x^2}} x \sqrt {\frac {c (d+e x)}{c d+e}}}+\frac {4 b \left (2 c^2 d^2+e^2\right ) \sqrt {\frac {c (d+e x)}{c d+e}} \sqrt {1-c^2 x^2} F\left (\sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {2}}\right )|\frac {2 e}{c d+e}\right )}{15 c^4 \sqrt {1-\frac {1}{c^2 x^2}} x \sqrt {d+e x}}+\frac {4 b d^3 \sqrt {\frac {c (d+e x)}{c d+e}} \sqrt {1-c^2 x^2} \Pi \left (2;\sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {2}}\right )|\frac {2 e}{c d+e}\right )}{5 c e \sqrt {1-\frac {1}{c^2 x^2}} x \sqrt {d+e x}}\\ \end {align*}

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Mathematica [C]  time = 1.48, size = 333, normalized size = 0.90 \[ \frac {1}{15} \left (\frac {6 a (d+e x)^{5/2}}{e}-\frac {4 b e x \sqrt {1-\frac {1}{c^2 x^2}} \sqrt {d+e x}}{c}+\frac {4 i b \sqrt {\frac {e (c x+1)}{e-c d}} \sqrt {\frac {e-c e x}{c d+e}} \left (\left (9 c^2 d^2-7 c d e+e^2\right ) F\left (i \sinh ^{-1}\left (\sqrt {-\frac {c}{c d+e}} \sqrt {d+e x}\right )|\frac {c d+e}{c d-e}\right )-3 c^2 d^2 \Pi \left (\frac {e}{c d}+1;i \sinh ^{-1}\left (\sqrt {-\frac {c}{c d+e}} \sqrt {d+e x}\right )|\frac {c d+e}{c d-e}\right )-7 c d (c d-e) E\left (i \sinh ^{-1}\left (\sqrt {-\frac {c}{c d+e}} \sqrt {d+e x}\right )|\frac {c d+e}{c d-e}\right )\right )}{c^3 e x \sqrt {1-\frac {1}{c^2 x^2}} \sqrt {-\frac {c}{c d+e}}}+\frac {6 b \sec ^{-1}(c x) (d+e x)^{5/2}}{e}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-4*b*e*Sqrt[1 - 1/(c^2*x^2)]*x*Sqrt[d + e*x])/c + (6*a*(d + e*x)^(5/2))/e + (6*b*(d + e*x)^(5/2)*ArcSec[c*x]
)/e + ((4*I)*b*Sqrt[(e*(1 + c*x))/(-(c*d) + e)]*Sqrt[(e - c*e*x)/(c*d + e)]*(-7*c*d*(c*d - e)*EllipticE[I*ArcS
inh[Sqrt[-(c/(c*d + e))]*Sqrt[d + e*x]], (c*d + e)/(c*d - e)] + (9*c^2*d^2 - 7*c*d*e + e^2)*EllipticF[I*ArcSin
h[Sqrt[-(c/(c*d + e))]*Sqrt[d + e*x]], (c*d + e)/(c*d - e)] - 3*c^2*d^2*EllipticPi[1 + e/(c*d), I*ArcSinh[Sqrt
[-(c/(c*d + e))]*Sqrt[d + e*x]], (c*d + e)/(c*d - e)]))/(c^3*e*Sqrt[-(c/(c*d + e))]*Sqrt[1 - 1/(c^2*x^2)]*x))/
15

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.20, size = 810, normalized size = 2.18 \[ \frac {\frac {2 \left (e x +d \right )^{\frac {5}{2}} a}{5}+2 b \left (\frac {\left (e x +d \right )^{\frac {5}{2}} \mathrm {arcsec}\left (c x \right )}{5}-\frac {2 \left (\sqrt {\frac {c}{d c -e}}\, \left (e x +d \right )^{\frac {5}{2}} c^{2}+9 d^{2} \sqrt {-\frac {\left (e x +d \right ) c -d c +e}{d c -e}}\, \sqrt {-\frac {\left (e x +d \right ) c -d c -e}{d c +e}}\, \EllipticF \left (\sqrt {e x +d}\, \sqrt {\frac {c}{d c -e}}, \sqrt {\frac {d c -e}{d c +e}}\right ) c^{2}-7 \sqrt {-\frac {\left (e x +d \right ) c -d c +e}{d c -e}}\, \sqrt {-\frac {\left (e x +d \right ) c -d c -e}{d c +e}}\, \EllipticE \left (\sqrt {e x +d}\, \sqrt {\frac {c}{d c -e}}, \sqrt {\frac {d c -e}{d c +e}}\right ) c^{2} d^{2}-3 d^{2} \sqrt {-\frac {\left (e x +d \right ) c -d c +e}{d c -e}}\, \sqrt {-\frac {\left (e x +d \right ) c -d c -e}{d c +e}}\, \EllipticPi \left (\sqrt {e x +d}\, \sqrt {\frac {c}{d c -e}}, \frac {d c -e}{c d}, \frac {\sqrt {\frac {c}{d c +e}}}{\sqrt {\frac {c}{d c -e}}}\right ) c^{2}-2 \sqrt {\frac {c}{d c -e}}\, \left (e x +d \right )^{\frac {3}{2}} c^{2} d +7 \sqrt {-\frac {\left (e x +d \right ) c -d c +e}{d c -e}}\, \sqrt {-\frac {\left (e x +d \right ) c -d c -e}{d c +e}}\, \EllipticF \left (\sqrt {e x +d}\, \sqrt {\frac {c}{d c -e}}, \sqrt {\frac {d c -e}{d c +e}}\right ) c d e -7 \sqrt {-\frac {\left (e x +d \right ) c -d c +e}{d c -e}}\, \sqrt {-\frac {\left (e x +d \right ) c -d c -e}{d c +e}}\, \EllipticE \left (\sqrt {e x +d}\, \sqrt {\frac {c}{d c -e}}, \sqrt {\frac {d c -e}{d c +e}}\right ) c d e +\sqrt {\frac {c}{d c -e}}\, \sqrt {e x +d}\, c^{2} d^{2}+\sqrt {-\frac {\left (e x +d \right ) c -d c +e}{d c -e}}\, \sqrt {-\frac {\left (e x +d \right ) c -d c -e}{d c +e}}\, \EllipticF \left (\sqrt {e x +d}\, \sqrt {\frac {c}{d c -e}}, \sqrt {\frac {d c -e}{d c +e}}\right ) e^{2}-\sqrt {\frac {c}{d c -e}}\, \sqrt {e x +d}\, e^{2}\right )}{15 c^{3} \sqrt {\frac {c}{d c -e}}\, x \sqrt {\frac {c^{2} \left (e x +d \right )^{2}-2 c^{2} d \left (e x +d \right )+c^{2} d^{2}-e^{2}}{c^{2} e^{2} x^{2}}}}\right )}{e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/e*(1/5*(e*x+d)^(5/2)*a+b*(1/5*(e*x+d)^(5/2)*arcsec(c*x)-2/15/c^3*((c/(c*d-e))^(1/2)*(e*x+d)^(5/2)*c^2+9*d^2*
(-((e*x+d)*c-d*c+e)/(c*d-e))^(1/2)*(-((e*x+d)*c-d*c-e)/(c*d+e))^(1/2)*EllipticF((e*x+d)^(1/2)*(c/(c*d-e))^(1/2
),((c*d-e)/(c*d+e))^(1/2))*c^2-7*(-((e*x+d)*c-d*c+e)/(c*d-e))^(1/2)*(-((e*x+d)*c-d*c-e)/(c*d+e))^(1/2)*Ellipti
cE((e*x+d)^(1/2)*(c/(c*d-e))^(1/2),((c*d-e)/(c*d+e))^(1/2))*c^2*d^2-3*d^2*(-((e*x+d)*c-d*c+e)/(c*d-e))^(1/2)*(
-((e*x+d)*c-d*c-e)/(c*d+e))^(1/2)*EllipticPi((e*x+d)^(1/2)*(c/(c*d-e))^(1/2),1/c*(c*d-e)/d,(c/(c*d+e))^(1/2)/(
c/(c*d-e))^(1/2))*c^2-2*(c/(c*d-e))^(1/2)*(e*x+d)^(3/2)*c^2*d+7*(-((e*x+d)*c-d*c+e)/(c*d-e))^(1/2)*(-((e*x+d)*
c-d*c-e)/(c*d+e))^(1/2)*EllipticF((e*x+d)^(1/2)*(c/(c*d-e))^(1/2),((c*d-e)/(c*d+e))^(1/2))*c*d*e-7*(-((e*x+d)*
c-d*c+e)/(c*d-e))^(1/2)*(-((e*x+d)*c-d*c-e)/(c*d+e))^(1/2)*EllipticE((e*x+d)^(1/2)*(c/(c*d-e))^(1/2),((c*d-e)/
(c*d+e))^(1/2))*c*d*e+(c/(c*d-e))^(1/2)*(e*x+d)^(1/2)*c^2*d^2+(-((e*x+d)*c-d*c+e)/(c*d-e))^(1/2)*(-((e*x+d)*c-
d*c-e)/(c*d+e))^(1/2)*EllipticF((e*x+d)^(1/2)*(c/(c*d-e))^(1/2),((c*d-e)/(c*d+e))^(1/2))*e^2-(c/(c*d-e))^(1/2)
*(e*x+d)^(1/2)*e^2)/(c/(c*d-e))^(1/2)/x/((c^2*(e*x+d)^2-2*c^2*d*(e*x+d)+c^2*d^2-e^2)/c^2/e^2/x^2)^(1/2)))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(3/2)*(a+b*arcsec(c*x)),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(c*d-e>0)', see `assume?` for m
ore details)Is c*d-e positive or negative?

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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