\(\int (d+e x) (a+b \arcsin (c x))^2 \, dx\) [11]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [A] (verification not implemented)
Maxima [F]
Giac [A] (verification not implemented)
Mupad [F(-1)]
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 16, antiderivative size = 142 \[ \int (d+e x) (a+b \arcsin (c x))^2 \, dx=-2 b^2 d x-\frac {1}{4} b^2 e x^2+\frac {2 b d \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{c}+\frac {b e x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{2 c}-\frac {d^2 (a+b \arcsin (c x))^2}{2 e}-\frac {e (a+b \arcsin (c x))^2}{4 c^2}+\frac {(d+e x)^2 (a+b \arcsin (c x))^2}{2 e} \] Output:

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

Mathematica [A] (verified)

Time = 0.55 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.08 \[ \int (d+e x) (a+b \arcsin (c x))^2 \, dx=\frac {(d+e x)^2 (a+b \arcsin (c x))^2}{2 e}-\frac {b \left (2 b d e x+\frac {1}{4} b e^2 x^2-\frac {2 d e \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{c}-\frac {e^2 x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{2 c}+\frac {d^2 (a+b \arcsin (c x))^2}{2 b}+\frac {e^2 (a+b \arcsin (c x))^2}{4 b c^2}\right )}{e} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.15, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {5242, 5262, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int (d+e x) (a+b \arcsin (c x))^2 \, dx\)

\(\Big \downarrow \) 5242

\(\displaystyle \frac {(d+e x)^2 (a+b \arcsin (c x))^2}{2 e}-\frac {b c \int \frac {(d+e x)^2 (a+b \arcsin (c x))}{\sqrt {1-c^2 x^2}}dx}{e}\)

\(\Big \downarrow \) 5262

\(\displaystyle \frac {(d+e x)^2 (a+b \arcsin (c x))^2}{2 e}-\frac {b c \int \left (\frac {(a+b \arcsin (c x)) d^2}{\sqrt {1-c^2 x^2}}+\frac {2 e x (a+b \arcsin (c x)) d}{\sqrt {1-c^2 x^2}}+\frac {e^2 x^2 (a+b \arcsin (c x))}{\sqrt {1-c^2 x^2}}\right )dx}{e}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {(d+e x)^2 (a+b \arcsin (c x))^2}{2 e}-\frac {b c \left (\frac {e^2 (a+b \arcsin (c x))^2}{4 b c^3}-\frac {2 d e \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{c^2}-\frac {e^2 x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{2 c^2}+\frac {d^2 (a+b \arcsin (c x))^2}{2 b c}+\frac {2 b d e x}{c}+\frac {b e^2 x^2}{4 c}\right )}{e}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 5242
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_))^(m_.), x_S 
ymbol] :> Simp[(d + e*x)^(m + 1)*((a + b*ArcSin[c*x])^n/(e*(m + 1))), x] - 
Simp[b*c*(n/(e*(m + 1)))   Int[(d + e*x)^(m + 1)*((a + b*ArcSin[c*x])^(n - 
1)/Sqrt[1 - c^2*x^2]), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[n, 0] 
 && NeQ[m, -1]
 

rule 5262
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.)*((d_ 
) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(d + e*x^2)^p*(a + 
 b*ArcSin[c*x])^n, (f + g*x)^m, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] & 
& EqQ[c^2*d + e, 0] && IGtQ[m, 0] && IntegerQ[p + 1/2] && GtQ[d, 0] && IGtQ 
[n, 0] && (m == 1 || p > 0 || (n == 1 && p > -1) || (m == 2 && p < -2))
 
Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.33

method result size
parts \(a^{2} \left (\frac {1}{2} e \,x^{2}+d x \right )+\frac {b^{2} \left (\frac {e \left (2 \arcsin \left (c x \right )^{2} x^{2} c^{2}+2 \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}\, c x -\arcsin \left (c x \right )^{2}-c^{2} x^{2}\right )}{4 c}+d \left (\arcsin \left (c x \right )^{2} c x -2 c x +2 \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}\right )\right )}{c}+\frac {2 a b \left (\frac {c \arcsin \left (c x \right ) e \,x^{2}}{2}+\arcsin \left (c x \right ) c x d -\frac {e \left (-\frac {c x \sqrt {-c^{2} x^{2}+1}}{2}+\frac {\arcsin \left (c x \right )}{2}\right )-2 d c \sqrt {-c^{2} x^{2}+1}}{2 c}\right )}{c}\) \(189\)
derivativedivides \(\frac {\frac {a^{2} \left (c^{2} d x +\frac {1}{2} c^{2} e \,x^{2}\right )}{c}+\frac {b^{2} \left (d c \left (\arcsin \left (c x \right )^{2} c x -2 c x +2 \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}\right )+\frac {e \left (2 \arcsin \left (c x \right )^{2} x^{2} c^{2}+2 \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}\, c x -\arcsin \left (c x \right )^{2}-c^{2} x^{2}\right )}{4}\right )}{c}+\frac {2 a b \left (\arcsin \left (c x \right ) d \,c^{2} x +\frac {\arcsin \left (c x \right ) c^{2} e \,x^{2}}{2}-\frac {e \left (-\frac {c x \sqrt {-c^{2} x^{2}+1}}{2}+\frac {\arcsin \left (c x \right )}{2}\right )}{2}+d c \sqrt {-c^{2} x^{2}+1}\right )}{c}}{c}\) \(198\)
default \(\frac {\frac {a^{2} \left (c^{2} d x +\frac {1}{2} c^{2} e \,x^{2}\right )}{c}+\frac {b^{2} \left (d c \left (\arcsin \left (c x \right )^{2} c x -2 c x +2 \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}\right )+\frac {e \left (2 \arcsin \left (c x \right )^{2} x^{2} c^{2}+2 \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}\, c x -\arcsin \left (c x \right )^{2}-c^{2} x^{2}\right )}{4}\right )}{c}+\frac {2 a b \left (\arcsin \left (c x \right ) d \,c^{2} x +\frac {\arcsin \left (c x \right ) c^{2} e \,x^{2}}{2}-\frac {e \left (-\frac {c x \sqrt {-c^{2} x^{2}+1}}{2}+\frac {\arcsin \left (c x \right )}{2}\right )}{2}+d c \sqrt {-c^{2} x^{2}+1}\right )}{c}}{c}\) \(198\)
orering \(\frac {\left (7 x^{4} e^{3} c^{2}+33 c^{2} d \,e^{2} x^{3}+20 x^{2} e \,d^{2} c^{2}+8 c^{2} d^{3} x -6 x^{2} e^{3}-30 d \,e^{2} x -10 d^{2} e \right ) \left (a +b \arcsin \left (c x \right )\right )^{2}}{8 \left (e x +d \right )^{2} c^{2}}-\frac {\left (3 c^{2} e^{2} x^{4}+17 d \,c^{2} e \,x^{3}-4 e^{2} x^{2}-26 d e x -8 d^{2}\right ) \left (e \left (a +b \arcsin \left (c x \right )\right )^{2}+\frac {2 \left (e x +d \right ) \left (a +b \arcsin \left (c x \right )\right ) b c}{\sqrt {-c^{2} x^{2}+1}}\right )}{8 c^{2} \left (e x +d \right )^{2}}+\frac {x \left (e x +8 d \right ) \left (c x -1\right ) \left (c x +1\right ) \left (\frac {4 e \left (a +b \arcsin \left (c x \right )\right ) b c}{\sqrt {-c^{2} x^{2}+1}}+\frac {2 \left (e x +d \right ) b^{2} c^{2}}{-c^{2} x^{2}+1}+\frac {2 \left (e x +d \right ) \left (a +b \arcsin \left (c x \right )\right ) b \,c^{3} x}{\left (-c^{2} x^{2}+1\right )^{\frac {3}{2}}}\right )}{8 c^{2} \left (e x +d \right )}\) \(297\)

Input:

int((e*x+d)*(a+b*arcsin(c*x))^2,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.10 \[ \int (d+e x) (a+b \arcsin (c x))^2 \, dx=\frac {{\left (2 \, a^{2} - b^{2}\right )} c^{2} e x^{2} + 4 \, {\left (a^{2} - 2 \, b^{2}\right )} c^{2} d x + {\left (2 \, b^{2} c^{2} e x^{2} + 4 \, b^{2} c^{2} d x - b^{2} e\right )} \arcsin \left (c x\right )^{2} + 2 \, {\left (2 \, a b c^{2} e x^{2} + 4 \, a b c^{2} d x - a b e\right )} \arcsin \left (c x\right ) + 2 \, {\left (a b c e x + 4 \, a b c d + {\left (b^{2} c e x + 4 \, b^{2} c d\right )} \arcsin \left (c x\right )\right )} \sqrt {-c^{2} x^{2} + 1}}{4 \, c^{2}} \] Input:

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

Output:

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

Sympy [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.64 \[ \int (d+e x) (a+b \arcsin (c x))^2 \, dx=\begin {cases} a^{2} d x + \frac {a^{2} e x^{2}}{2} + 2 a b d x \operatorname {asin}{\left (c x \right )} + a b e x^{2} \operatorname {asin}{\left (c x \right )} + \frac {2 a b d \sqrt {- c^{2} x^{2} + 1}}{c} + \frac {a b e x \sqrt {- c^{2} x^{2} + 1}}{2 c} - \frac {a b e \operatorname {asin}{\left (c x \right )}}{2 c^{2}} + b^{2} d x \operatorname {asin}^{2}{\left (c x \right )} - 2 b^{2} d x + \frac {b^{2} e x^{2} \operatorname {asin}^{2}{\left (c x \right )}}{2} - \frac {b^{2} e x^{2}}{4} + \frac {2 b^{2} d \sqrt {- c^{2} x^{2} + 1} \operatorname {asin}{\left (c x \right )}}{c} + \frac {b^{2} e x \sqrt {- c^{2} x^{2} + 1} \operatorname {asin}{\left (c x \right )}}{2 c} - \frac {b^{2} e \operatorname {asin}^{2}{\left (c x \right )}}{4 c^{2}} & \text {for}\: c \neq 0 \\a^{2} \left (d x + \frac {e x^{2}}{2}\right ) & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((a**2*d*x + a**2*e*x**2/2 + 2*a*b*d*x*asin(c*x) + a*b*e*x**2*asi 
n(c*x) + 2*a*b*d*sqrt(-c**2*x**2 + 1)/c + a*b*e*x*sqrt(-c**2*x**2 + 1)/(2* 
c) - a*b*e*asin(c*x)/(2*c**2) + b**2*d*x*asin(c*x)**2 - 2*b**2*d*x + b**2* 
e*x**2*asin(c*x)**2/2 - b**2*e*x**2/4 + 2*b**2*d*sqrt(-c**2*x**2 + 1)*asin 
(c*x)/c + b**2*e*x*sqrt(-c**2*x**2 + 1)*asin(c*x)/(2*c) - b**2*e*asin(c*x) 
**2/(4*c**2), Ne(c, 0)), (a**2*(d*x + e*x**2/2), True))
 

Maxima [F]

\[ \int (d+e x) (a+b \arcsin (c x))^2 \, dx=\int { {\left (e x + d\right )} {\left (b \arcsin \left (c x\right ) + a\right )}^{2} \,d x } \] Input:

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

Output:

b^2*d*x*arcsin(c*x)^2 + 1/2*a^2*e*x^2 + 1/2*(2*x^2*arcsin(c*x) + c*(sqrt(- 
c^2*x^2 + 1)*x/c^2 - arcsin(c*x)/c^3))*a*b*e + 1/2*(x^2*arctan2(c*x, sqrt( 
c*x + 1)*sqrt(-c*x + 1))^2 + 2*c*integrate(sqrt(c*x + 1)*sqrt(-c*x + 1)*x^ 
2*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))/(c^2*x^2 - 1), x))*b^2*e - 2* 
b^2*d*(x - sqrt(-c^2*x^2 + 1)*arcsin(c*x)/c) + a^2*d*x + 2*(c*x*arcsin(c*x 
) + sqrt(-c^2*x^2 + 1))*a*b*d/c
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.72 \[ \int (d+e x) (a+b \arcsin (c x))^2 \, dx=b^{2} d x \arcsin \left (c x\right )^{2} + 2 \, a b d x \arcsin \left (c x\right ) + \frac {\sqrt {-c^{2} x^{2} + 1} b^{2} e x \arcsin \left (c x\right )}{2 \, c} + a^{2} d x - 2 \, b^{2} d x + \frac {{\left (c^{2} x^{2} - 1\right )} b^{2} e \arcsin \left (c x\right )^{2}}{2 \, c^{2}} + \frac {\sqrt {-c^{2} x^{2} + 1} a b e x}{2 \, c} + \frac {2 \, \sqrt {-c^{2} x^{2} + 1} b^{2} d \arcsin \left (c x\right )}{c} + \frac {{\left (c^{2} x^{2} - 1\right )} a b e \arcsin \left (c x\right )}{c^{2}} + \frac {b^{2} e \arcsin \left (c x\right )^{2}}{4 \, c^{2}} + \frac {2 \, \sqrt {-c^{2} x^{2} + 1} a b d}{c} + \frac {{\left (c^{2} x^{2} - 1\right )} a^{2} e}{2 \, c^{2}} - \frac {{\left (c^{2} x^{2} - 1\right )} b^{2} e}{4 \, c^{2}} + \frac {a b e \arcsin \left (c x\right )}{2 \, c^{2}} - \frac {b^{2} e}{8 \, c^{2}} \] Input:

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

Output:

b^2*d*x*arcsin(c*x)^2 + 2*a*b*d*x*arcsin(c*x) + 1/2*sqrt(-c^2*x^2 + 1)*b^2 
*e*x*arcsin(c*x)/c + a^2*d*x - 2*b^2*d*x + 1/2*(c^2*x^2 - 1)*b^2*e*arcsin( 
c*x)^2/c^2 + 1/2*sqrt(-c^2*x^2 + 1)*a*b*e*x/c + 2*sqrt(-c^2*x^2 + 1)*b^2*d 
*arcsin(c*x)/c + (c^2*x^2 - 1)*a*b*e*arcsin(c*x)/c^2 + 1/4*b^2*e*arcsin(c* 
x)^2/c^2 + 2*sqrt(-c^2*x^2 + 1)*a*b*d/c + 1/2*(c^2*x^2 - 1)*a^2*e/c^2 - 1/ 
4*(c^2*x^2 - 1)*b^2*e/c^2 + 1/2*a*b*e*arcsin(c*x)/c^2 - 1/8*b^2*e/c^2
 

Mupad [F(-1)]

Timed out. \[ \int (d+e x) (a+b \arcsin (c x))^2 \, dx=\int {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2\,\left (d+e\,x\right ) \,d x \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.54 \[ \int (d+e x) (a+b \arcsin (c x))^2 \, dx=\frac {4 \mathit {asin} \left (c x \right )^{2} b^{2} c^{2} d x +2 \mathit {asin} \left (c x \right )^{2} b^{2} c^{2} e \,x^{2}-\mathit {asin} \left (c x \right )^{2} b^{2} e +8 \sqrt {-c^{2} x^{2}+1}\, \mathit {asin} \left (c x \right ) b^{2} c d +2 \sqrt {-c^{2} x^{2}+1}\, \mathit {asin} \left (c x \right ) b^{2} c e x +8 \mathit {asin} \left (c x \right ) a b \,c^{2} d x +4 \mathit {asin} \left (c x \right ) a b \,c^{2} e \,x^{2}-2 \mathit {asin} \left (c x \right ) a b e +8 \sqrt {-c^{2} x^{2}+1}\, a b c d +2 \sqrt {-c^{2} x^{2}+1}\, a b c e x +4 a^{2} c^{2} d x +2 a^{2} c^{2} e \,x^{2}-8 b^{2} c^{2} d x -b^{2} c^{2} e \,x^{2}+b^{2} e}{4 c^{2}} \] Input:

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

Output:

(4*asin(c*x)**2*b**2*c**2*d*x + 2*asin(c*x)**2*b**2*c**2*e*x**2 - asin(c*x 
)**2*b**2*e + 8*sqrt( - c**2*x**2 + 1)*asin(c*x)*b**2*c*d + 2*sqrt( - c**2 
*x**2 + 1)*asin(c*x)*b**2*c*e*x + 8*asin(c*x)*a*b*c**2*d*x + 4*asin(c*x)*a 
*b*c**2*e*x**2 - 2*asin(c*x)*a*b*e + 8*sqrt( - c**2*x**2 + 1)*a*b*c*d + 2* 
sqrt( - c**2*x**2 + 1)*a*b*c*e*x + 4*a**2*c**2*d*x + 2*a**2*c**2*e*x**2 - 
8*b**2*c**2*d*x - b**2*c**2*e*x**2 + b**2*e)/(4*c**2)