3.188 \(\int (c e+d e x)^4 (a+b \sin ^{-1}(c+d x))^2 \, dx\)

Optimal. Leaf size=203 \[ \frac {e^4 (c+d x)^5 \left (a+b \sin ^{-1}(c+d x)\right )^2}{5 d}+\frac {2 b e^4 \sqrt {1-(c+d x)^2} (c+d x)^4 \left (a+b \sin ^{-1}(c+d x)\right )}{25 d}+\frac {8 b e^4 \sqrt {1-(c+d x)^2} (c+d x)^2 \left (a+b \sin ^{-1}(c+d x)\right )}{75 d}+\frac {16 b e^4 \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )}{75 d}-\frac {2 b^2 e^4 (c+d x)^5}{125 d}-\frac {8 b^2 e^4 (c+d x)^3}{225 d}-\frac {16}{75} b^2 e^4 x \]

[Out]

-16/75*b^2*e^4*x-8/225*b^2*e^4*(d*x+c)^3/d-2/125*b^2*e^4*(d*x+c)^5/d+1/5*e^4*(d*x+c)^5*(a+b*arcsin(d*x+c))^2/d
+16/75*b*e^4*(a+b*arcsin(d*x+c))*(1-(d*x+c)^2)^(1/2)/d+8/75*b*e^4*(d*x+c)^2*(a+b*arcsin(d*x+c))*(1-(d*x+c)^2)^
(1/2)/d+2/25*b*e^4*(d*x+c)^4*(a+b*arcsin(d*x+c))*(1-(d*x+c)^2)^(1/2)/d

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Rubi [A]  time = 0.30, antiderivative size = 203, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {4805, 12, 4627, 4707, 4677, 8, 30} \[ \frac {e^4 (c+d x)^5 \left (a+b \sin ^{-1}(c+d x)\right )^2}{5 d}+\frac {2 b e^4 \sqrt {1-(c+d x)^2} (c+d x)^4 \left (a+b \sin ^{-1}(c+d x)\right )}{25 d}+\frac {8 b e^4 \sqrt {1-(c+d x)^2} (c+d x)^2 \left (a+b \sin ^{-1}(c+d x)\right )}{75 d}+\frac {16 b e^4 \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )}{75 d}-\frac {2 b^2 e^4 (c+d x)^5}{125 d}-\frac {8 b^2 e^4 (c+d x)^3}{225 d}-\frac {16}{75} b^2 e^4 x \]

Antiderivative was successfully verified.

[In]

Int[(c*e + d*e*x)^4*(a + b*ArcSin[c + d*x])^2,x]

[Out]

(-16*b^2*e^4*x)/75 - (8*b^2*e^4*(c + d*x)^3)/(225*d) - (2*b^2*e^4*(c + d*x)^5)/(125*d) + (16*b*e^4*Sqrt[1 - (c
+ d*x)^2]*(a + b*ArcSin[c + d*x]))/(75*d) + (8*b*e^4*(c + d*x)^2*Sqrt[1 - (c + d*x)^2]*(a + b*ArcSin[c + d*x]
))/(75*d) + (2*b*e^4*(c + d*x)^4*Sqrt[1 - (c + d*x)^2]*(a + b*ArcSin[c + d*x]))/(25*d) + (e^4*(c + d*x)^5*(a +
b*ArcSin[c + d*x])^2)/(5*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 4627

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcSi
n[c*x])^n)/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcSin[c*x])^(n - 1))/Sqrt[1
- c^2*x^2], x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rule 4677

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*(x_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x^2)^
(p + 1)*(a + b*ArcSin[c*x])^n)/(2*e*(p + 1)), x] + Dist[(b*n*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*c*(p + 1
)*(1 - c^2*x^2)^FracPart[p]), Int[(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcSin[c*x])^(n - 1), x], x] /; FreeQ[{a, b,
c, d, e, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && NeQ[p, -1]

Rule 4707

Int[(((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[
(f*(f*x)^(m - 1)*Sqrt[d + e*x^2]*(a + b*ArcSin[c*x])^n)/(e*m), x] + (Dist[(f^2*(m - 1))/(c^2*m), Int[((f*x)^(m
- 2)*(a + b*ArcSin[c*x])^n)/Sqrt[d + e*x^2], x], x] + Dist[(b*f*n*Sqrt[1 - c^2*x^2])/(c*m*Sqrt[d + e*x^2]), I
nt[(f*x)^(m - 1)*(a + b*ArcSin[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[c^2*d + e, 0] &&
GtQ[n, 0] && GtQ[m, 1] && IntegerQ[m]

Rule 4805

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[I
nt[((d*e - c*f)/d + (f*x)/d)^m*(a + b*ArcSin[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]

Rubi steps

\begin {align*} \int (c e+d e x)^4 \left (a+b \sin ^{-1}(c+d x)\right )^2 \, dx &=\frac {\operatorname {Subst}\left (\int e^4 x^4 \left (a+b \sin ^{-1}(x)\right )^2 \, dx,x,c+d x\right )}{d}\\ &=\frac {e^4 \operatorname {Subst}\left (\int x^4 \left (a+b \sin ^{-1}(x)\right )^2 \, dx,x,c+d x\right )}{d}\\ &=\frac {e^4 (c+d x)^5 \left (a+b \sin ^{-1}(c+d x)\right )^2}{5 d}-\frac {\left (2 b e^4\right ) \operatorname {Subst}\left (\int \frac {x^5 \left (a+b \sin ^{-1}(x)\right )}{\sqrt {1-x^2}} \, dx,x,c+d x\right )}{5 d}\\ &=\frac {2 b e^4 (c+d x)^4 \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )}{25 d}+\frac {e^4 (c+d x)^5 \left (a+b \sin ^{-1}(c+d x)\right )^2}{5 d}-\frac {\left (8 b e^4\right ) \operatorname {Subst}\left (\int \frac {x^3 \left (a+b \sin ^{-1}(x)\right )}{\sqrt {1-x^2}} \, dx,x,c+d x\right )}{25 d}-\frac {\left (2 b^2 e^4\right ) \operatorname {Subst}\left (\int x^4 \, dx,x,c+d x\right )}{25 d}\\ &=-\frac {2 b^2 e^4 (c+d x)^5}{125 d}+\frac {8 b e^4 (c+d x)^2 \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )}{75 d}+\frac {2 b e^4 (c+d x)^4 \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )}{25 d}+\frac {e^4 (c+d x)^5 \left (a+b \sin ^{-1}(c+d x)\right )^2}{5 d}-\frac {\left (16 b e^4\right ) \operatorname {Subst}\left (\int \frac {x \left (a+b \sin ^{-1}(x)\right )}{\sqrt {1-x^2}} \, dx,x,c+d x\right )}{75 d}-\frac {\left (8 b^2 e^4\right ) \operatorname {Subst}\left (\int x^2 \, dx,x,c+d x\right )}{75 d}\\ &=-\frac {8 b^2 e^4 (c+d x)^3}{225 d}-\frac {2 b^2 e^4 (c+d x)^5}{125 d}+\frac {16 b e^4 \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )}{75 d}+\frac {8 b e^4 (c+d x)^2 \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )}{75 d}+\frac {2 b e^4 (c+d x)^4 \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )}{25 d}+\frac {e^4 (c+d x)^5 \left (a+b \sin ^{-1}(c+d x)\right )^2}{5 d}-\frac {\left (16 b^2 e^4\right ) \operatorname {Subst}(\int 1 \, dx,x,c+d x)}{75 d}\\ &=-\frac {16}{75} b^2 e^4 x-\frac {8 b^2 e^4 (c+d x)^3}{225 d}-\frac {2 b^2 e^4 (c+d x)^5}{125 d}+\frac {16 b e^4 \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )}{75 d}+\frac {8 b e^4 (c+d x)^2 \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )}{75 d}+\frac {2 b e^4 (c+d x)^4 \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )}{25 d}+\frac {e^4 (c+d x)^5 \left (a+b \sin ^{-1}(c+d x)\right )^2}{5 d}\\ \end {align*}

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Mathematica [A]  time = 0.50, size = 164, normalized size = 0.81 \[ \frac {e^4 \left ((c+d x)^5 \left (a+b \sin ^{-1}(c+d x)\right )^2-\frac {2}{25} b \left (-5 \sqrt {1-(c+d x)^2} (c+d x)^4 \left (a+b \sin ^{-1}(c+d x)\right )-\frac {20}{3} \sqrt {1-(c+d x)^2} (c+d x)^2 \left (a+b \sin ^{-1}(c+d x)\right )+\frac {40}{3} \left (b d x-\sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )\right )+b (c+d x)^5+\frac {20}{9} b (c+d x)^3\right )\right )}{5 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*e + d*e*x)^4*(a + b*ArcSin[c + d*x])^2,x]

[Out]

(e^4*((c + d*x)^5*(a + b*ArcSin[c + d*x])^2 - (2*b*((20*b*(c + d*x)^3)/9 + b*(c + d*x)^5 - (20*(c + d*x)^2*Sqr
t[1 - (c + d*x)^2]*(a + b*ArcSin[c + d*x]))/3 - 5*(c + d*x)^4*Sqrt[1 - (c + d*x)^2]*(a + b*ArcSin[c + d*x]) +
(40*(b*d*x - Sqrt[1 - (c + d*x)^2]*(a + b*ArcSin[c + d*x])))/3))/25))/(5*d)

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fricas [B]  time = 0.46, size = 567, normalized size = 2.79 \[ \frac {9 \, {\left (25 \, a^{2} - 2 \, b^{2}\right )} d^{5} e^{4} x^{5} + 45 \, {\left (25 \, a^{2} - 2 \, b^{2}\right )} c d^{4} e^{4} x^{4} + 10 \, {\left (9 \, {\left (25 \, a^{2} - 2 \, b^{2}\right )} c^{2} - 4 \, b^{2}\right )} d^{3} e^{4} x^{3} + 30 \, {\left (3 \, {\left (25 \, a^{2} - 2 \, b^{2}\right )} c^{3} - 4 \, b^{2} c\right )} d^{2} e^{4} x^{2} + 15 \, {\left (3 \, {\left (25 \, a^{2} - 2 \, b^{2}\right )} c^{4} - 8 \, b^{2} c^{2} - 16 \, b^{2}\right )} d e^{4} x + 225 \, {\left (b^{2} d^{5} e^{4} x^{5} + 5 \, b^{2} c d^{4} e^{4} x^{4} + 10 \, b^{2} c^{2} d^{3} e^{4} x^{3} + 10 \, b^{2} c^{3} d^{2} e^{4} x^{2} + 5 \, b^{2} c^{4} d e^{4} x + b^{2} c^{5} e^{4}\right )} \arcsin \left (d x + c\right )^{2} + 450 \, {\left (a b d^{5} e^{4} x^{5} + 5 \, a b c d^{4} e^{4} x^{4} + 10 \, a b c^{2} d^{3} e^{4} x^{3} + 10 \, a b c^{3} d^{2} e^{4} x^{2} + 5 \, a b c^{4} d e^{4} x + a b c^{5} e^{4}\right )} \arcsin \left (d x + c\right ) + 30 \, {\left (3 \, a b d^{4} e^{4} x^{4} + 12 \, a b c d^{3} e^{4} x^{3} + 2 \, {\left (9 \, a b c^{2} + 2 \, a b\right )} d^{2} e^{4} x^{2} + 4 \, {\left (3 \, a b c^{3} + 2 \, a b c\right )} d e^{4} x + {\left (3 \, a b c^{4} + 4 \, a b c^{2} + 8 \, a b\right )} e^{4} + {\left (3 \, b^{2} d^{4} e^{4} x^{4} + 12 \, b^{2} c d^{3} e^{4} x^{3} + 2 \, {\left (9 \, b^{2} c^{2} + 2 \, b^{2}\right )} d^{2} e^{4} x^{2} + 4 \, {\left (3 \, b^{2} c^{3} + 2 \, b^{2} c\right )} d e^{4} x + {\left (3 \, b^{2} c^{4} + 4 \, b^{2} c^{2} + 8 \, b^{2}\right )} e^{4}\right )} \arcsin \left (d x + c\right )\right )} \sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1}}{1125 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^4*(a+b*arcsin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/1125*(9*(25*a^2 - 2*b^2)*d^5*e^4*x^5 + 45*(25*a^2 - 2*b^2)*c*d^4*e^4*x^4 + 10*(9*(25*a^2 - 2*b^2)*c^2 - 4*b^
2)*d^3*e^4*x^3 + 30*(3*(25*a^2 - 2*b^2)*c^3 - 4*b^2*c)*d^2*e^4*x^2 + 15*(3*(25*a^2 - 2*b^2)*c^4 - 8*b^2*c^2 -
16*b^2)*d*e^4*x + 225*(b^2*d^5*e^4*x^5 + 5*b^2*c*d^4*e^4*x^4 + 10*b^2*c^2*d^3*e^4*x^3 + 10*b^2*c^3*d^2*e^4*x^2
+ 5*b^2*c^4*d*e^4*x + b^2*c^5*e^4)*arcsin(d*x + c)^2 + 450*(a*b*d^5*e^4*x^5 + 5*a*b*c*d^4*e^4*x^4 + 10*a*b*c^
2*d^3*e^4*x^3 + 10*a*b*c^3*d^2*e^4*x^2 + 5*a*b*c^4*d*e^4*x + a*b*c^5*e^4)*arcsin(d*x + c) + 30*(3*a*b*d^4*e^4*
x^4 + 12*a*b*c*d^3*e^4*x^3 + 2*(9*a*b*c^2 + 2*a*b)*d^2*e^4*x^2 + 4*(3*a*b*c^3 + 2*a*b*c)*d*e^4*x + (3*a*b*c^4
+ 4*a*b*c^2 + 8*a*b)*e^4 + (3*b^2*d^4*e^4*x^4 + 12*b^2*c*d^3*e^4*x^3 + 2*(9*b^2*c^2 + 2*b^2)*d^2*e^4*x^2 + 4*(
3*b^2*c^3 + 2*b^2*c)*d*e^4*x + (3*b^2*c^4 + 4*b^2*c^2 + 8*b^2)*e^4)*arcsin(d*x + c))*sqrt(-d^2*x^2 - 2*c*d*x -
c^2 + 1))/d

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giac [B]  time = 0.48, size = 427, normalized size = 2.10 \[ \frac {{\left (d x + c\right )}^{5} a^{2} e^{4}}{5 \, d} + \frac {{\left ({\left (d x + c\right )}^{2} - 1\right )}^{2} {\left (d x + c\right )} b^{2} \arcsin \left (d x + c\right )^{2} e^{4}}{5 \, d} + \frac {2 \, {\left ({\left (d x + c\right )}^{2} - 1\right )}^{2} {\left (d x + c\right )} a b \arcsin \left (d x + c\right ) e^{4}}{5 \, d} + \frac {2 \, {\left ({\left (d x + c\right )}^{2} - 1\right )} {\left (d x + c\right )} b^{2} \arcsin \left (d x + c\right )^{2} e^{4}}{5 \, d} + \frac {2 \, {\left ({\left (d x + c\right )}^{2} - 1\right )}^{2} \sqrt {-{\left (d x + c\right )}^{2} + 1} b^{2} \arcsin \left (d x + c\right ) e^{4}}{25 \, d} - \frac {2 \, {\left ({\left (d x + c\right )}^{2} - 1\right )}^{2} {\left (d x + c\right )} b^{2} e^{4}}{125 \, d} + \frac {4 \, {\left ({\left (d x + c\right )}^{2} - 1\right )} {\left (d x + c\right )} a b \arcsin \left (d x + c\right ) e^{4}}{5 \, d} + \frac {{\left (d x + c\right )} b^{2} \arcsin \left (d x + c\right )^{2} e^{4}}{5 \, d} + \frac {2 \, {\left ({\left (d x + c\right )}^{2} - 1\right )}^{2} \sqrt {-{\left (d x + c\right )}^{2} + 1} a b e^{4}}{25 \, d} - \frac {4 \, {\left (-{\left (d x + c\right )}^{2} + 1\right )}^{\frac {3}{2}} b^{2} \arcsin \left (d x + c\right ) e^{4}}{15 \, d} - \frac {76 \, {\left ({\left (d x + c\right )}^{2} - 1\right )} {\left (d x + c\right )} b^{2} e^{4}}{1125 \, d} + \frac {2 \, {\left (d x + c\right )} a b \arcsin \left (d x + c\right ) e^{4}}{5 \, d} - \frac {4 \, {\left (-{\left (d x + c\right )}^{2} + 1\right )}^{\frac {3}{2}} a b e^{4}}{15 \, d} + \frac {2 \, \sqrt {-{\left (d x + c\right )}^{2} + 1} b^{2} \arcsin \left (d x + c\right ) e^{4}}{5 \, d} - \frac {298 \, {\left (d x + c\right )} b^{2} e^{4}}{1125 \, d} + \frac {2 \, \sqrt {-{\left (d x + c\right )}^{2} + 1} a b e^{4}}{5 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^4*(a+b*arcsin(d*x+c))^2,x, algorithm="giac")

[Out]

1/5*(d*x + c)^5*a^2*e^4/d + 1/5*((d*x + c)^2 - 1)^2*(d*x + c)*b^2*arcsin(d*x + c)^2*e^4/d + 2/5*((d*x + c)^2 -
1)^2*(d*x + c)*a*b*arcsin(d*x + c)*e^4/d + 2/5*((d*x + c)^2 - 1)*(d*x + c)*b^2*arcsin(d*x + c)^2*e^4/d + 2/25
*((d*x + c)^2 - 1)^2*sqrt(-(d*x + c)^2 + 1)*b^2*arcsin(d*x + c)*e^4/d - 2/125*((d*x + c)^2 - 1)^2*(d*x + c)*b^
2*e^4/d + 4/5*((d*x + c)^2 - 1)*(d*x + c)*a*b*arcsin(d*x + c)*e^4/d + 1/5*(d*x + c)*b^2*arcsin(d*x + c)^2*e^4/
d + 2/25*((d*x + c)^2 - 1)^2*sqrt(-(d*x + c)^2 + 1)*a*b*e^4/d - 4/15*(-(d*x + c)^2 + 1)^(3/2)*b^2*arcsin(d*x +
c)*e^4/d - 76/1125*((d*x + c)^2 - 1)*(d*x + c)*b^2*e^4/d + 2/5*(d*x + c)*a*b*arcsin(d*x + c)*e^4/d - 4/15*(-(
d*x + c)^2 + 1)^(3/2)*a*b*e^4/d + 2/5*sqrt(-(d*x + c)^2 + 1)*b^2*arcsin(d*x + c)*e^4/d - 298/1125*(d*x + c)*b^
2*e^4/d + 2/5*sqrt(-(d*x + c)^2 + 1)*a*b*e^4/d

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maple [A]  time = 0.04, size = 194, normalized size = 0.96 \[ \frac {\frac {e^{4} \left (d x +c \right )^{5} a^{2}}{5}+e^{4} b^{2} \left (\frac {\arcsin \left (d x +c \right )^{2} \left (d x +c \right )^{5}}{5}+\frac {2 \arcsin \left (d x +c \right ) \left (3 \left (d x +c \right )^{4}+4 \left (d x +c \right )^{2}+8\right ) \sqrt {1-\left (d x +c \right )^{2}}}{75}-\frac {2 \left (d x +c \right )^{5}}{125}-\frac {8 \left (d x +c \right )^{3}}{225}-\frac {16 d x}{75}-\frac {16 c}{75}\right )+2 e^{4} a b \left (\frac {\left (d x +c \right )^{5} \arcsin \left (d x +c \right )}{5}+\frac {\left (d x +c \right )^{4} \sqrt {1-\left (d x +c \right )^{2}}}{25}+\frac {4 \left (d x +c \right )^{2} \sqrt {1-\left (d x +c \right )^{2}}}{75}+\frac {8 \sqrt {1-\left (d x +c \right )^{2}}}{75}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*e*x+c*e)^4*(a+b*arcsin(d*x+c))^2,x)

[Out]

1/d*(1/5*e^4*(d*x+c)^5*a^2+e^4*b^2*(1/5*arcsin(d*x+c)^2*(d*x+c)^5+2/75*arcsin(d*x+c)*(3*(d*x+c)^4+4*(d*x+c)^2+
8)*(1-(d*x+c)^2)^(1/2)-2/125*(d*x+c)^5-8/225*(d*x+c)^3-16/75*d*x-16/75*c)+2*e^4*a*b*(1/5*(d*x+c)^5*arcsin(d*x+
c)+1/25*(d*x+c)^4*(1-(d*x+c)^2)^(1/2)+4/75*(d*x+c)^2*(1-(d*x+c)^2)^(1/2)+8/75*(1-(d*x+c)^2)^(1/2)))

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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((d*e*x+c*e)^4*(a+b*arcsin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/5*a^2*d^4*e^4*x^5 + a^2*c*d^3*e^4*x^4 + 2*a^2*c^2*d^2*e^4*x^3 + 2*a^2*c^3*d*e^4*x^2 + 2*(2*x^2*arcsin(d*x +
c) + d*(3*c^2*arcsin(-(d^2*x + c*d)/sqrt(c^2*d^2 - (c^2 - 1)*d^2))/d^3 + sqrt(-d^2*x^2 - 2*c*d*x - c^2 + 1)*x/
d^2 - (c^2 - 1)*arcsin(-(d^2*x + c*d)/sqrt(c^2*d^2 - (c^2 - 1)*d^2))/d^3 - 3*sqrt(-d^2*x^2 - 2*c*d*x - c^2 + 1
)*c/d^3))*a*b*c^3*d*e^4 + 2/3*(6*x^3*arcsin(d*x + c) + d*(2*sqrt(-d^2*x^2 - 2*c*d*x - c^2 + 1)*x^2/d^2 - 15*c^
3*arcsin(-(d^2*x + c*d)/sqrt(c^2*d^2 - (c^2 - 1)*d^2))/d^4 - 5*sqrt(-d^2*x^2 - 2*c*d*x - c^2 + 1)*c*x/d^3 + 9*
(c^2 - 1)*c*arcsin(-(d^2*x + c*d)/sqrt(c^2*d^2 - (c^2 - 1)*d^2))/d^4 + 15*sqrt(-d^2*x^2 - 2*c*d*x - c^2 + 1)*c
^2/d^4 - 4*sqrt(-d^2*x^2 - 2*c*d*x - c^2 + 1)*(c^2 - 1)/d^4))*a*b*c^2*d^2*e^4 + 1/12*(24*x^4*arcsin(d*x + c) +
(6*sqrt(-d^2*x^2 - 2*c*d*x - c^2 + 1)*x^3/d^2 - 14*sqrt(-d^2*x^2 - 2*c*d*x - c^2 + 1)*c*x^2/d^3 + 105*c^4*arc
sin(-(d^2*x + c*d)/sqrt(c^2*d^2 - (c^2 - 1)*d^2))/d^5 + 35*sqrt(-d^2*x^2 - 2*c*d*x - c^2 + 1)*c^2*x/d^4 - 90*(
c^2 - 1)*c^2*arcsin(-(d^2*x + c*d)/sqrt(c^2*d^2 - (c^2 - 1)*d^2))/d^5 - 105*sqrt(-d^2*x^2 - 2*c*d*x - c^2 + 1)
*c^3/d^5 - 9*sqrt(-d^2*x^2 - 2*c*d*x - c^2 + 1)*(c^2 - 1)*x/d^4 + 9*(c^2 - 1)^2*arcsin(-(d^2*x + c*d)/sqrt(c^2
*d^2 - (c^2 - 1)*d^2))/d^5 + 55*sqrt(-d^2*x^2 - 2*c*d*x - c^2 + 1)*(c^2 - 1)*c/d^5)*d)*a*b*c*d^3*e^4 + 1/300*(
120*x^5*arcsin(d*x + c) + (24*sqrt(-d^2*x^2 - 2*c*d*x - c^2 + 1)*x^4/d^2 - 54*sqrt(-d^2*x^2 - 2*c*d*x - c^2 +
1)*c*x^3/d^3 + 126*sqrt(-d^2*x^2 - 2*c*d*x - c^2 + 1)*c^2*x^2/d^4 - 945*c^5*arcsin(-(d^2*x + c*d)/sqrt(c^2*d^2
- (c^2 - 1)*d^2))/d^6 - 315*sqrt(-d^2*x^2 - 2*c*d*x - c^2 + 1)*c^3*x/d^5 - 32*sqrt(-d^2*x^2 - 2*c*d*x - c^2 +
1)*(c^2 - 1)*x^2/d^4 + 1050*(c^2 - 1)*c^3*arcsin(-(d^2*x + c*d)/sqrt(c^2*d^2 - (c^2 - 1)*d^2))/d^6 + 945*sqrt
(-d^2*x^2 - 2*c*d*x - c^2 + 1)*c^4/d^6 + 161*sqrt(-d^2*x^2 - 2*c*d*x - c^2 + 1)*(c^2 - 1)*c*x/d^5 - 225*(c^2 -
1)^2*c*arcsin(-(d^2*x + c*d)/sqrt(c^2*d^2 - (c^2 - 1)*d^2))/d^6 - 735*sqrt(-d^2*x^2 - 2*c*d*x - c^2 + 1)*(c^2
- 1)*c^2/d^6 + 64*sqrt(-d^2*x^2 - 2*c*d*x - c^2 + 1)*(c^2 - 1)^2/d^6)*d)*a*b*d^4*e^4 + a^2*c^4*e^4*x + 2*((d*
x + c)*arcsin(d*x + c) + sqrt(-(d*x + c)^2 + 1))*a*b*c^4*e^4/d + 1/5*(b^2*d^4*e^4*x^5 + 5*b^2*c*d^3*e^4*x^4 +
10*b^2*c^2*d^2*e^4*x^3 + 10*b^2*c^3*d*e^4*x^2 + 5*b^2*c^4*e^4*x)*arctan2(d*x + c, sqrt(d*x + c + 1)*sqrt(-d*x
- c + 1))^2 + integrate(2/5*(b^2*d^5*e^4*x^5 + 5*b^2*c*d^4*e^4*x^4 + 10*b^2*c^2*d^3*e^4*x^3 + 10*b^2*c^3*d^2*e
^4*x^2 + 5*b^2*c^4*d*e^4*x)*sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)*arctan2(d*x + c, sqrt(d*x + c + 1)*sqrt(-d*x
- c + 1))/(d^2*x^2 + 2*c*d*x + c^2 - 1), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (c\,e+d\,e\,x\right )}^4\,{\left (a+b\,\mathrm {asin}\left (c+d\,x\right )\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*e + d*e*x)^4*(a + b*asin(c + d*x))^2,x)

[Out]

int((c*e + d*e*x)^4*(a + b*asin(c + d*x))^2, x)

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sympy [A]  time = 6.66, size = 1268, normalized size = 6.25 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)**4*(a+b*asin(d*x+c))**2,x)

[Out]

Piecewise((a**2*c**4*e**4*x + 2*a**2*c**3*d*e**4*x**2 + 2*a**2*c**2*d**2*e**4*x**3 + a**2*c*d**3*e**4*x**4 + a
**2*d**4*e**4*x**5/5 + 2*a*b*c**5*e**4*asin(c + d*x)/(5*d) + 2*a*b*c**4*e**4*x*asin(c + d*x) + 2*a*b*c**4*e**4
*sqrt(-c**2 - 2*c*d*x - d**2*x**2 + 1)/(25*d) + 4*a*b*c**3*d*e**4*x**2*asin(c + d*x) + 8*a*b*c**3*e**4*x*sqrt(
-c**2 - 2*c*d*x - d**2*x**2 + 1)/25 + 4*a*b*c**2*d**2*e**4*x**3*asin(c + d*x) + 12*a*b*c**2*d*e**4*x**2*sqrt(-
c**2 - 2*c*d*x - d**2*x**2 + 1)/25 + 8*a*b*c**2*e**4*sqrt(-c**2 - 2*c*d*x - d**2*x**2 + 1)/(75*d) + 2*a*b*c*d*
*3*e**4*x**4*asin(c + d*x) + 8*a*b*c*d**2*e**4*x**3*sqrt(-c**2 - 2*c*d*x - d**2*x**2 + 1)/25 + 16*a*b*c*e**4*x
*sqrt(-c**2 - 2*c*d*x - d**2*x**2 + 1)/75 + 2*a*b*d**4*e**4*x**5*asin(c + d*x)/5 + 2*a*b*d**3*e**4*x**4*sqrt(-
c**2 - 2*c*d*x - d**2*x**2 + 1)/25 + 8*a*b*d*e**4*x**2*sqrt(-c**2 - 2*c*d*x - d**2*x**2 + 1)/75 + 16*a*b*e**4*
sqrt(-c**2 - 2*c*d*x - d**2*x**2 + 1)/(75*d) + b**2*c**5*e**4*asin(c + d*x)**2/(5*d) + b**2*c**4*e**4*x*asin(c
+ d*x)**2 - 2*b**2*c**4*e**4*x/25 + 2*b**2*c**4*e**4*sqrt(-c**2 - 2*c*d*x - d**2*x**2 + 1)*asin(c + d*x)/(25*
d) + 2*b**2*c**3*d*e**4*x**2*asin(c + d*x)**2 - 4*b**2*c**3*d*e**4*x**2/25 + 8*b**2*c**3*e**4*x*sqrt(-c**2 - 2
*c*d*x - d**2*x**2 + 1)*asin(c + d*x)/25 + 2*b**2*c**2*d**2*e**4*x**3*asin(c + d*x)**2 - 4*b**2*c**2*d**2*e**4
*x**3/25 + 12*b**2*c**2*d*e**4*x**2*sqrt(-c**2 - 2*c*d*x - d**2*x**2 + 1)*asin(c + d*x)/25 - 8*b**2*c**2*e**4*
x/75 + 8*b**2*c**2*e**4*sqrt(-c**2 - 2*c*d*x - d**2*x**2 + 1)*asin(c + d*x)/(75*d) + b**2*c*d**3*e**4*x**4*asi
n(c + d*x)**2 - 2*b**2*c*d**3*e**4*x**4/25 + 8*b**2*c*d**2*e**4*x**3*sqrt(-c**2 - 2*c*d*x - d**2*x**2 + 1)*asi
n(c + d*x)/25 - 8*b**2*c*d*e**4*x**2/75 + 16*b**2*c*e**4*x*sqrt(-c**2 - 2*c*d*x - d**2*x**2 + 1)*asin(c + d*x)
/75 + b**2*d**4*e**4*x**5*asin(c + d*x)**2/5 - 2*b**2*d**4*e**4*x**5/125 + 2*b**2*d**3*e**4*x**4*sqrt(-c**2 -
2*c*d*x - d**2*x**2 + 1)*asin(c + d*x)/25 - 8*b**2*d**2*e**4*x**3/225 + 8*b**2*d*e**4*x**2*sqrt(-c**2 - 2*c*d*
x - d**2*x**2 + 1)*asin(c + d*x)/75 - 16*b**2*e**4*x/75 + 16*b**2*e**4*sqrt(-c**2 - 2*c*d*x - d**2*x**2 + 1)*a
sin(c + d*x)/(75*d), Ne(d, 0)), (c**4*e**4*x*(a + b*asin(c))**2, True))

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