\(\int (c e+d e x)^4 (a+b \arcsin (c+d x)) \, dx\) [177]

Optimal result

Integrand size = 21, antiderivative size = 106 \[ \int (c e+d e x)^4 (a+b \arcsin (c+d x)) \, dx=\frac {b e^4 \sqrt {1-(c+d x)^2}}{5 d}-\frac {2 b e^4 \left (1-(c+d x)^2\right )^{3/2}}{15 d}+\frac {b e^4 \left (1-(c+d x)^2\right )^{5/2}}{25 d}+\frac {e^4 (c+d x)^5 (a+b \arcsin (c+d x))}{5 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {4889, 12, 4723, 272, 45} \[ \int (c e+d e x)^4 (a+b \arcsin (c+d x)) \, dx=\frac {e^4 (c+d x)^5 (a+b \arcsin (c+d x))}{5 d}+\frac {b e^4 \left (1-(c+d x)^2\right )^{5/2}}{25 d}-\frac {2 b e^4 \left (1-(c+d x)^2\right )^{3/2}}{15 d}+\frac {b e^4 \sqrt {1-(c+d x)^2}}{5 d} \]

[In]

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

[Out]

(b*e^4*Sqrt[1 - (c + d*x)^2])/(5*d) - (2*b*e^4*(1 - (c + d*x)^2)^(3/2))/(15*d) + (b*e^4*(1 - (c + d*x)^2)^(5/2
))/(25*d) + (e^4*(c + d*x)^5*(a + b*ArcSin[c + d*x]))/(5*d)

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 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 4723

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 4889

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*} \text {integral}& = \frac {\text {Subst}\left (\int e^4 x^4 (a+b \arcsin (x)) \, dx,x,c+d x\right )}{d} \\ & = \frac {e^4 \text {Subst}\left (\int x^4 (a+b \arcsin (x)) \, dx,x,c+d x\right )}{d} \\ & = \frac {e^4 (c+d x)^5 (a+b \arcsin (c+d x))}{5 d}-\frac {\left (b e^4\right ) \text {Subst}\left (\int \frac {x^5}{\sqrt {1-x^2}} \, dx,x,c+d x\right )}{5 d} \\ & = \frac {e^4 (c+d x)^5 (a+b \arcsin (c+d x))}{5 d}-\frac {\left (b e^4\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt {1-x}} \, dx,x,(c+d x)^2\right )}{10 d} \\ & = \frac {e^4 (c+d x)^5 (a+b \arcsin (c+d x))}{5 d}-\frac {\left (b e^4\right ) \text {Subst}\left (\int \left (\frac {1}{\sqrt {1-x}}-2 \sqrt {1-x}+(1-x)^{3/2}\right ) \, dx,x,(c+d x)^2\right )}{10 d} \\ & = \frac {b e^4 \sqrt {1-(c+d x)^2}}{5 d}-\frac {2 b e^4 \left (1-(c+d x)^2\right )^{3/2}}{15 d}+\frac {b e^4 \left (1-(c+d x)^2\right )^{5/2}}{25 d}+\frac {e^4 (c+d x)^5 (a+b \arcsin (c+d x))}{5 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.73 \[ \int (c e+d e x)^4 (a+b \arcsin (c+d x)) \, dx=\frac {e^4 \left (-\frac {1}{75} b \sqrt {1-(c+d x)^2} \left (-15+10 \left (1-(c+d x)^2\right )-3 \left (-1+(c+d x)^2\right )^2\right )+\frac {1}{5} (c+d x)^5 (a+b \arcsin (c+d x))\right )}{d} \]

[In]

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

[Out]

(e^4*(-1/75*(b*Sqrt[1 - (c + d*x)^2]*(-15 + 10*(1 - (c + d*x)^2) - 3*(-1 + (c + d*x)^2)^2)) + ((c + d*x)^5*(a
+ b*ArcSin[c + d*x]))/5))/d

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.93

method	result	size
derivativedivides	\(\frac {\frac {e^{4} a \left (d x +c \right )^{5}}{5}+e^{4} b \left (\frac {\left (d x +c \right )^{5} \arcsin \left (d x +c \right )}{5}+\frac {\left (d x +c \right )^{4} \sqrt {1-\left (d x +c \right )^{2}}}{25}+\frac {4 \left (d x +c \right )^{2} \sqrt {1-\left (d x +c \right )^{2}}}{75}+\frac {8 \sqrt {1-\left (d x +c \right )^{2}}}{75}\right )}{d}\)	\(99\)
default	\(\frac {\frac {e^{4} a \left (d x +c \right )^{5}}{5}+e^{4} b \left (\frac {\left (d x +c \right )^{5} \arcsin \left (d x +c \right )}{5}+\frac {\left (d x +c \right )^{4} \sqrt {1-\left (d x +c \right )^{2}}}{25}+\frac {4 \left (d x +c \right )^{2} \sqrt {1-\left (d x +c \right )^{2}}}{75}+\frac {8 \sqrt {1-\left (d x +c \right )^{2}}}{75}\right )}{d}\)	\(99\)
parts	\(\frac {e^{4} a \left (d x +c \right )^{5}}{5 d}+\frac {e^{4} b \left (\frac {\left (d x +c \right )^{5} \arcsin \left (d x +c \right )}{5}+\frac {\left (d x +c \right )^{4} \sqrt {1-\left (d x +c \right )^{2}}}{25}+\frac {4 \left (d x +c \right )^{2} \sqrt {1-\left (d x +c \right )^{2}}}{75}+\frac {8 \sqrt {1-\left (d x +c \right )^{2}}}{75}\right )}{d}\)	\(101\)

[In]

int((d*e*x+c*e)^4*(a+b*arcsin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d*(1/5*e^4*a*(d*x+c)^5+e^4*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)))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 262 vs. \(2 (92) = 184\).

Time = 0.29 (sec) , antiderivative size = 262, normalized size of antiderivative = 2.47 \[ \int (c e+d e x)^4 (a+b \arcsin (c+d x)) \, dx=\frac {15 \, a d^{5} e^{4} x^{5} + 75 \, a c d^{4} e^{4} x^{4} + 150 \, a c^{2} d^{3} e^{4} x^{3} + 150 \, a c^{3} d^{2} e^{4} x^{2} + 75 \, a c^{4} d e^{4} x + 15 \, {\left (b d^{5} e^{4} x^{5} + 5 \, b c d^{4} e^{4} x^{4} + 10 \, b c^{2} d^{3} e^{4} x^{3} + 10 \, b c^{3} d^{2} e^{4} x^{2} + 5 \, b c^{4} d e^{4} x + b c^{5} e^{4}\right )} \arcsin \left (d x + c\right ) + {\left (3 \, b d^{4} e^{4} x^{4} + 12 \, b c d^{3} e^{4} x^{3} + 2 \, {\left (9 \, b c^{2} + 2 \, b\right )} d^{2} e^{4} x^{2} + 4 \, {\left (3 \, b c^{3} + 2 \, b c\right )} d e^{4} x + {\left (3 \, b c^{4} + 4 \, b c^{2} + 8 \, b\right )} e^{4}\right )} \sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1}}{75 \, d} \]

[In]

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

[Out]

1/75*(15*a*d^5*e^4*x^5 + 75*a*c*d^4*e^4*x^4 + 150*a*c^2*d^3*e^4*x^3 + 150*a*c^3*d^2*e^4*x^2 + 75*a*c^4*d*e^4*x
+ 15*(b*d^5*e^4*x^5 + 5*b*c*d^4*e^4*x^4 + 10*b*c^2*d^3*e^4*x^3 + 10*b*c^3*d^2*e^4*x^2 + 5*b*c^4*d*e^4*x + b*c
^5*e^4)*arcsin(d*x + c) + (3*b*d^4*e^4*x^4 + 12*b*c*d^3*e^4*x^3 + 2*(9*b*c^2 + 2*b)*d^2*e^4*x^2 + 4*(3*b*c^3 +
2*b*c)*d*e^4*x + (3*b*c^4 + 4*b*c^2 + 8*b)*e^4)*sqrt(-d^2*x^2 - 2*c*d*x - c^2 + 1))/d

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 527 vs. \(2 (85) = 170\).

Time = 0.38 (sec) , antiderivative size = 527, normalized size of antiderivative = 4.97 \[ \int (c e+d e x)^4 (a+b \arcsin (c+d x)) \, dx=\begin {cases} a c^{4} e^{4} x + 2 a c^{3} d e^{4} x^{2} + 2 a c^{2} d^{2} e^{4} x^{3} + a c d^{3} e^{4} x^{4} + \frac {a d^{4} e^{4} x^{5}}{5} + \frac {b c^{5} e^{4} \operatorname {asin}{\left (c + d x \right )}}{5 d} + b c^{4} e^{4} x \operatorname {asin}{\left (c + d x \right )} + \frac {b c^{4} e^{4} \sqrt {- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{25 d} + 2 b c^{3} d e^{4} x^{2} \operatorname {asin}{\left (c + d x \right )} + \frac {4 b c^{3} e^{4} x \sqrt {- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{25} + 2 b c^{2} d^{2} e^{4} x^{3} \operatorname {asin}{\left (c + d x \right )} + \frac {6 b c^{2} d e^{4} x^{2} \sqrt {- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{25} + \frac {4 b c^{2} e^{4} \sqrt {- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{75 d} + b c d^{3} e^{4} x^{4} \operatorname {asin}{\left (c + d x \right )} + \frac {4 b c d^{2} e^{4} x^{3} \sqrt {- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{25} + \frac {8 b c e^{4} x \sqrt {- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{75} + \frac {b d^{4} e^{4} x^{5} \operatorname {asin}{\left (c + d x \right )}}{5} + \frac {b d^{3} e^{4} x^{4} \sqrt {- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{25} + \frac {4 b d e^{4} x^{2} \sqrt {- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{75} + \frac {8 b e^{4} \sqrt {- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{75 d} & \text {for}\: d \neq 0 \\c^{4} e^{4} x \left (a + b \operatorname {asin}{\left (c \right )}\right ) & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((a*c**4*e**4*x + 2*a*c**3*d*e**4*x**2 + 2*a*c**2*d**2*e**4*x**3 + a*c*d**3*e**4*x**4 + a*d**4*e**4*x
**5/5 + b*c**5*e**4*asin(c + d*x)/(5*d) + b*c**4*e**4*x*asin(c + d*x) + b*c**4*e**4*sqrt(-c**2 - 2*c*d*x - d**
2*x**2 + 1)/(25*d) + 2*b*c**3*d*e**4*x**2*asin(c + d*x) + 4*b*c**3*e**4*x*sqrt(-c**2 - 2*c*d*x - d**2*x**2 + 1
)/25 + 2*b*c**2*d**2*e**4*x**3*asin(c + d*x) + 6*b*c**2*d*e**4*x**2*sqrt(-c**2 - 2*c*d*x - d**2*x**2 + 1)/25 +
4*b*c**2*e**4*sqrt(-c**2 - 2*c*d*x - d**2*x**2 + 1)/(75*d) + b*c*d**3*e**4*x**4*asin(c + d*x) + 4*b*c*d**2*e*
*4*x**3*sqrt(-c**2 - 2*c*d*x - d**2*x**2 + 1)/25 + 8*b*c*e**4*x*sqrt(-c**2 - 2*c*d*x - d**2*x**2 + 1)/75 + b*d
**4*e**4*x**5*asin(c + d*x)/5 + b*d**3*e**4*x**4*sqrt(-c**2 - 2*c*d*x - d**2*x**2 + 1)/25 + 4*b*d*e**4*x**2*sq
rt(-c**2 - 2*c*d*x - d**2*x**2 + 1)/75 + 8*b*e**4*sqrt(-c**2 - 2*c*d*x - d**2*x**2 + 1)/(75*d), Ne(d, 0)), (c*
*4*e**4*x*(a + b*asin(c)), True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1280 vs. \(2 (92) = 184\).

Time = 0.28 (sec) , antiderivative size = 1280, normalized size of antiderivative = 12.08 \[ \int (c e+d e x)^4 (a+b \arcsin (c+d x)) \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/5*a*d^4*e^4*x^5 + a*c*d^3*e^4*x^4 + 2*a*c^2*d^2*e^4*x^3 + 2*a*c^3*d*e^4*x^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))*
b*c^3*d*e^4 + 1/3*(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*s
qrt(-d^2*x^2 - 2*c*d*x - c^2 + 1)*(c^2 - 1)/d^4))*b*c^2*d^2*e^4 + 1/24*(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*arcsin(-(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*a
rcsin(-(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*s
qrt(-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)*b*c*d^3*e^4 + 1/600*(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 + 1
26*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)*b*d^4*e^4 + a*c^4*e^4*x + ((d*x + c)*arcsin(d*x + c)
+ sqrt(-(d*x + c)^2 + 1))*b*c^4*e^4/d

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.64 \[ \int (c e+d e x)^4 (a+b \arcsin (c+d x)) \, dx=\frac {{\left (d x + c\right )}^{5} a e^{4}}{5 \, d} + \frac {{\left ({\left (d x + c\right )}^{2} - 1\right )}^{2} {\left (d x + c\right )} b e^{4} \arcsin \left (d x + c\right )}{5 \, d} + \frac {2 \, {\left ({\left (d x + c\right )}^{2} - 1\right )} {\left (d x + c\right )} b e^{4} \arcsin \left (d x + c\right )}{5 \, d} + \frac {{\left ({\left (d x + c\right )}^{2} - 1\right )}^{2} \sqrt {-{\left (d x + c\right )}^{2} + 1} b e^{4}}{25 \, d} + \frac {{\left (d x + c\right )} b e^{4} \arcsin \left (d x + c\right )}{5 \, d} - \frac {2 \, {\left (-{\left (d x + c\right )}^{2} + 1\right )}^{\frac {3}{2}} b e^{4}}{15 \, d} + \frac {\sqrt {-{\left (d x + c\right )}^{2} + 1} b e^{4}}{5 \, d} \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int (c e+d e x)^4 (a+b \arcsin (c+d x)) \, dx=\int {\left (c\,e+d\,e\,x\right )}^4\,\left (a+b\,\mathrm {asin}\left (c+d\,x\right )\right ) \,d x \]

[In]

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

[Out]

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