\(\int x^3 (a+b \arcsin (c x))^2 \, dx\) [12]

Optimal result

Integrand size = 14, antiderivative size = 122 \[ \int x^3 (a+b \arcsin (c x))^2 \, dx=-\frac {3 b^2 x^2}{32 c^2}-\frac {b^2 x^4}{32}+\frac {3 b x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{16 c^3}+\frac {b x^3 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{8 c}-\frac {3 (a+b \arcsin (c x))^2}{32 c^4}+\frac {1}{4} x^4 (a+b \arcsin (c x))^2 \] Output:

-3/32*b^2*x^2/c^2-1/32*b^2*x^4+3/16*b*x*(-c^2*x^2+1)^(1/2)*(a+b*arcsin(c*x 
))/c^3+1/8*b*x^3*(-c^2*x^2+1)^(1/2)*(a+b*arcsin(c*x))/c-3/32*(a+b*arcsin(c 
*x))^2/c^4+1/4*x^4*(a+b*arcsin(c*x))^2
 

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.12 \[ \int x^3 (a+b \arcsin (c x))^2 \, dx=\frac {1}{4} x^4 (a+b \arcsin (c x))^2-\frac {1}{2} b c \left (\frac {b x^4}{16 c}-\frac {x^3 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{4 c^2}+\frac {3 \left (\frac {b x^2}{4 c}-\frac {x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{2 c^2}+\frac {(a+b \arcsin (c x))^2}{4 b c^3}\right )}{4 c^2}\right ) \] Input:

Integrate[x^3*(a + b*ArcSin[c*x])^2,x]
 

Output:

(x^4*(a + b*ArcSin[c*x])^2)/4 - (b*c*((b*x^4)/(16*c) - (x^3*Sqrt[1 - c^2*x 
^2]*(a + b*ArcSin[c*x]))/(4*c^2) + (3*((b*x^2)/(4*c) - (x*Sqrt[1 - c^2*x^2 
]*(a + b*ArcSin[c*x]))/(2*c^2) + (a + b*ArcSin[c*x])^2/(4*b*c^3)))/(4*c^2) 
))/2
 

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.12, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {5138, 5210, 15, 5210, 15, 5152}

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.

\(\displaystyle \int x^3 (a+b \arcsin (c x))^2 \, dx\)

\(\Big \downarrow \) 5138

\(\displaystyle \frac {1}{4} x^4 (a+b \arcsin (c x))^2-\frac {1}{2} b c \int \frac {x^4 (a+b \arcsin (c x))}{\sqrt {1-c^2 x^2}}dx\)

\(\Big \downarrow \) 5210

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

\(\Big \downarrow \) 15

\(\displaystyle \frac {1}{4} x^4 (a+b \arcsin (c x))^2-\frac {1}{2} b c \left (\frac {3 \int \frac {x^2 (a+b \arcsin (c x))}{\sqrt {1-c^2 x^2}}dx}{4 c^2}-\frac {x^3 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{4 c^2}+\frac {b x^4}{16 c}\right )\)

\(\Big \downarrow \) 5210

\(\displaystyle \frac {1}{4} x^4 (a+b \arcsin (c x))^2-\frac {1}{2} b c \left (\frac {3 \left (\frac {\int \frac {a+b \arcsin (c x)}{\sqrt {1-c^2 x^2}}dx}{2 c^2}+\frac {b \int xdx}{2 c}-\frac {x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{2 c^2}\right )}{4 c^2}-\frac {x^3 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{4 c^2}+\frac {b x^4}{16 c}\right )\)

\(\Big \downarrow \) 15

\(\displaystyle \frac {1}{4} x^4 (a+b \arcsin (c x))^2-\frac {1}{2} b c \left (\frac {3 \left (\frac {\int \frac {a+b \arcsin (c x)}{\sqrt {1-c^2 x^2}}dx}{2 c^2}-\frac {x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{2 c^2}+\frac {b x^2}{4 c}\right )}{4 c^2}-\frac {x^3 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{4 c^2}+\frac {b x^4}{16 c}\right )\)

\(\Big \downarrow \) 5152

\(\displaystyle \frac {1}{4} x^4 (a+b \arcsin (c x))^2-\frac {1}{2} b c \left (-\frac {x^3 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{4 c^2}+\frac {3 \left (\frac {(a+b \arcsin (c x))^2}{4 b c^3}-\frac {x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{2 c^2}+\frac {b x^2}{4 c}\right )}{4 c^2}+\frac {b x^4}{16 c}\right )\)

Input:

Int[x^3*(a + b*ArcSin[c*x])^2,x]
 

Output:

(x^4*(a + b*ArcSin[c*x])^2)/4 - (b*c*((b*x^4)/(16*c) - (x^3*Sqrt[1 - c^2*x 
^2]*(a + b*ArcSin[c*x]))/(4*c^2) + (3*((b*x^2)/(4*c) - (x*Sqrt[1 - c^2*x^2 
]*(a + b*ArcSin[c*x]))/(2*c^2) + (a + b*ArcSin[c*x])^2/(4*b*c^3)))/(4*c^2) 
))/2
 

Defintions of rubi rules used

rule 15

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

rule 5138

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] 
:> Simp[(d*x)^(m + 1)*((a + b*ArcSin[c*x])^n/(d*(m + 1))), x] - Simp[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 5152

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

rule 5210

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

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.35

method	result	size
parts	\(\frac {a^{2} x^{4}}{4}+\frac {b^{2} \left (\frac {c^{4} x^{4} \arcsin \left (c x \right )^{2}}{4}-\frac {\arcsin \left (c x \right ) \left (-2 c^{3} x^{3} \sqrt {-c^{2} x^{2}+1}-3 c x \sqrt {-c^{2} x^{2}+1}+3 \arcsin \left (c x \right )\right )}{16}+\frac {3 \arcsin \left (c x \right )^{2}}{32}-\frac {\left (2 c^{2} x^{2}+3\right )^{2}}{128}\right )}{c^{4}}+\frac {2 a b \left (\frac {c^{4} x^{4} \arcsin \left (c x \right )}{4}+\frac {c^{3} x^{3} \sqrt {-c^{2} x^{2}+1}}{16}+\frac {3 c x \sqrt {-c^{2} x^{2}+1}}{32}-\frac {3 \arcsin \left (c x \right )}{32}\right )}{c^{4}}\)	\(165\)
derivativedivides	\(\frac {\frac {a^{2} c^{4} x^{4}}{4}+b^{2} \left (\frac {c^{4} x^{4} \arcsin \left (c x \right )^{2}}{4}-\frac {\arcsin \left (c x \right ) \left (-2 c^{3} x^{3} \sqrt {-c^{2} x^{2}+1}-3 c x \sqrt {-c^{2} x^{2}+1}+3 \arcsin \left (c x \right )\right )}{16}+\frac {3 \arcsin \left (c x \right )^{2}}{32}-\frac {\left (2 c^{2} x^{2}+3\right )^{2}}{128}\right )+2 a b \left (\frac {c^{4} x^{4} \arcsin \left (c x \right )}{4}+\frac {c^{3} x^{3} \sqrt {-c^{2} x^{2}+1}}{16}+\frac {3 c x \sqrt {-c^{2} x^{2}+1}}{32}-\frac {3 \arcsin \left (c x \right )}{32}\right )}{c^{4}}\)	\(166\)
default	\(\frac {\frac {a^{2} c^{4} x^{4}}{4}+b^{2} \left (\frac {c^{4} x^{4} \arcsin \left (c x \right )^{2}}{4}-\frac {\arcsin \left (c x \right ) \left (-2 c^{3} x^{3} \sqrt {-c^{2} x^{2}+1}-3 c x \sqrt {-c^{2} x^{2}+1}+3 \arcsin \left (c x \right )\right )}{16}+\frac {3 \arcsin \left (c x \right )^{2}}{32}-\frac {\left (2 c^{2} x^{2}+3\right )^{2}}{128}\right )+2 a b \left (\frac {c^{4} x^{4} \arcsin \left (c x \right )}{4}+\frac {c^{3} x^{3} \sqrt {-c^{2} x^{2}+1}}{16}+\frac {3 c x \sqrt {-c^{2} x^{2}+1}}{32}-\frac {3 \arcsin \left (c x \right )}{32}\right )}{c^{4}}\)	\(166\)
orering	\(\frac {\left (37 c^{4} x^{4}+21 c^{2} x^{2}-60\right ) \left (a +b \arcsin \left (c x \right )\right )^{2}}{64 c^{4}}-\frac {\left (9 c^{4} x^{4}+11 c^{2} x^{2}-24\right ) \left (3 x^{2} \left (a +b \arcsin \left (c x \right )\right )^{2}+\frac {2 x^{3} \left (a +b \arcsin \left (c x \right )\right ) b c}{\sqrt {-c^{2} x^{2}+1}}\right )}{64 c^{4} x^{2}}+\frac {\left (c^{2} x^{2}+3\right ) \left (c x -1\right ) \left (c x +1\right ) \left (6 x \left (a +b \arcsin \left (c x \right )\right )^{2}+\frac {12 x^{2} \left (a +b \arcsin \left (c x \right )\right ) b c}{\sqrt {-c^{2} x^{2}+1}}+\frac {2 x^{3} b^{2} c^{2}}{-c^{2} x^{2}+1}+\frac {2 x^{4} \left (a +b \arcsin \left (c x \right )\right ) b \,c^{3}}{\left (-c^{2} x^{2}+1\right )^{\frac {3}{2}}}\right )}{64 x \,c^{4}}\)	\(224\)

Input:

int(x^3*(a+b*arcsin(c*x))^2,x,method=_RETURNVERBOSE)
 

Output:

1/4*a^2*x^4+b^2/c^4*(1/4*c^4*x^4*arcsin(c*x)^2-1/16*arcsin(c*x)*(-2*c^3*x^ 
3*(-c^2*x^2+1)^(1/2)-3*c*x*(-c^2*x^2+1)^(1/2)+3*arcsin(c*x))+3/32*arcsin(c 
*x)^2-1/128*(2*c^2*x^2+3)^2)+2*a*b/c^4*(1/4*c^4*x^4*arcsin(c*x)+1/16*c^3*x 
^3*(-c^2*x^2+1)^(1/2)+3/32*c*x*(-c^2*x^2+1)^(1/2)-3/32*arcsin(c*x))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.11 \[ \int x^3 (a+b \arcsin (c x))^2 \, dx=\frac {{\left (8 \, a^{2} - b^{2}\right )} c^{4} x^{4} - 3 \, b^{2} c^{2} x^{2} + {\left (8 \, b^{2} c^{4} x^{4} - 3 \, b^{2}\right )} \arcsin \left (c x\right )^{2} + 2 \, {\left (8 \, a b c^{4} x^{4} - 3 \, a b\right )} \arcsin \left (c x\right ) + 2 \, {\left (2 \, a b c^{3} x^{3} + 3 \, a b c x + {\left (2 \, b^{2} c^{3} x^{3} + 3 \, b^{2} c x\right )} \arcsin \left (c x\right )\right )} \sqrt {-c^{2} x^{2} + 1}}{32 \, c^{4}} \] Input:

integrate(x^3*(a+b*arcsin(c*x))^2,x, algorithm="fricas")
 

Output:

1/32*((8*a^2 - b^2)*c^4*x^4 - 3*b^2*c^2*x^2 + (8*b^2*c^4*x^4 - 3*b^2)*arcs 
in(c*x)^2 + 2*(8*a*b*c^4*x^4 - 3*a*b)*arcsin(c*x) + 2*(2*a*b*c^3*x^3 + 3*a 
*b*c*x + (2*b^2*c^3*x^3 + 3*b^2*c*x)*arcsin(c*x))*sqrt(-c^2*x^2 + 1))/c^4
 

Sympy [A] (verification not implemented)

Time = 0.40 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.65 \[ \int x^3 (a+b \arcsin (c x))^2 \, dx=\begin {cases} \frac {a^{2} x^{4}}{4} + \frac {a b x^{4} \operatorname {asin}{\left (c x \right )}}{2} + \frac {a b x^{3} \sqrt {- c^{2} x^{2} + 1}}{8 c} + \frac {3 a b x \sqrt {- c^{2} x^{2} + 1}}{16 c^{3}} - \frac {3 a b \operatorname {asin}{\left (c x \right )}}{16 c^{4}} + \frac {b^{2} x^{4} \operatorname {asin}^{2}{\left (c x \right )}}{4} - \frac {b^{2} x^{4}}{32} + \frac {b^{2} x^{3} \sqrt {- c^{2} x^{2} + 1} \operatorname {asin}{\left (c x \right )}}{8 c} - \frac {3 b^{2} x^{2}}{32 c^{2}} + \frac {3 b^{2} x \sqrt {- c^{2} x^{2} + 1} \operatorname {asin}{\left (c x \right )}}{16 c^{3}} - \frac {3 b^{2} \operatorname {asin}^{2}{\left (c x \right )}}{32 c^{4}} & \text {for}\: c \neq 0 \\\frac {a^{2} x^{4}}{4} & \text {otherwise} \end {cases} \] Input:

integrate(x**3*(a+b*asin(c*x))**2,x)
 

Output:

Piecewise((a**2*x**4/4 + a*b*x**4*asin(c*x)/2 + a*b*x**3*sqrt(-c**2*x**2 + 
 1)/(8*c) + 3*a*b*x*sqrt(-c**2*x**2 + 1)/(16*c**3) - 3*a*b*asin(c*x)/(16*c 
**4) + b**2*x**4*asin(c*x)**2/4 - b**2*x**4/32 + b**2*x**3*sqrt(-c**2*x**2 
 + 1)*asin(c*x)/(8*c) - 3*b**2*x**2/(32*c**2) + 3*b**2*x*sqrt(-c**2*x**2 + 
 1)*asin(c*x)/(16*c**3) - 3*b**2*asin(c*x)**2/(32*c**4), Ne(c, 0)), (a**2* 
x**4/4, True))
 

Maxima [F]

\[ \int x^3 (a+b \arcsin (c x))^2 \, dx=\int { {\left (b \arcsin \left (c x\right ) + a\right )}^{2} x^{3} \,d x } \] Input:

integrate(x^3*(a+b*arcsin(c*x))^2,x, algorithm="maxima")
 

Output:

1/4*a^2*x^4 + 1/16*(8*x^4*arcsin(c*x) + (2*sqrt(-c^2*x^2 + 1)*x^3/c^2 + 3* 
sqrt(-c^2*x^2 + 1)*x/c^4 - 3*arcsin(c*x)/c^5)*c)*a*b + 1/4*(x^4*arctan2(c* 
x, sqrt(c*x + 1)*sqrt(-c*x + 1))^2 + 4*c*integrate(1/2*sqrt(c*x + 1)*sqrt( 
-c*x + 1)*x^4*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))/(c^2*x^2 - 1), x) 
)*b^2
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 257 vs. \(2 (106) = 212\).

Time = 0.13 (sec) , antiderivative size = 257, normalized size of antiderivative = 2.11 \[ \int x^3 (a+b \arcsin (c x))^2 \, dx=\frac {1}{4} \, a^{2} x^{4} - \frac {{\left (-c^{2} x^{2} + 1\right )}^{\frac {3}{2}} b^{2} x \arcsin \left (c x\right )}{8 \, c^{3}} + \frac {{\left (c^{2} x^{2} - 1\right )}^{2} b^{2} \arcsin \left (c x\right )^{2}}{4 \, c^{4}} - \frac {{\left (-c^{2} x^{2} + 1\right )}^{\frac {3}{2}} a b x}{8 \, c^{3}} + \frac {5 \, \sqrt {-c^{2} x^{2} + 1} b^{2} x \arcsin \left (c x\right )}{16 \, c^{3}} + \frac {{\left (c^{2} x^{2} - 1\right )}^{2} a b \arcsin \left (c x\right )}{2 \, c^{4}} + \frac {{\left (c^{2} x^{2} - 1\right )} b^{2} \arcsin \left (c x\right )^{2}}{2 \, c^{4}} + \frac {5 \, \sqrt {-c^{2} x^{2} + 1} a b x}{16 \, c^{3}} - \frac {{\left (c^{2} x^{2} - 1\right )}^{2} b^{2}}{32 \, c^{4}} + \frac {{\left (c^{2} x^{2} - 1\right )} a b \arcsin \left (c x\right )}{c^{4}} + \frac {5 \, b^{2} \arcsin \left (c x\right )^{2}}{32 \, c^{4}} - \frac {5 \, {\left (c^{2} x^{2} - 1\right )} b^{2}}{32 \, c^{4}} + \frac {5 \, a b \arcsin \left (c x\right )}{16 \, c^{4}} - \frac {17 \, b^{2}}{256 \, c^{4}} \] Input:

integrate(x^3*(a+b*arcsin(c*x))^2,x, algorithm="giac")
 

Output:

1/4*a^2*x^4 - 1/8*(-c^2*x^2 + 1)^(3/2)*b^2*x*arcsin(c*x)/c^3 + 1/4*(c^2*x^ 
2 - 1)^2*b^2*arcsin(c*x)^2/c^4 - 1/8*(-c^2*x^2 + 1)^(3/2)*a*b*x/c^3 + 5/16 
*sqrt(-c^2*x^2 + 1)*b^2*x*arcsin(c*x)/c^3 + 1/2*(c^2*x^2 - 1)^2*a*b*arcsin 
(c*x)/c^4 + 1/2*(c^2*x^2 - 1)*b^2*arcsin(c*x)^2/c^4 + 5/16*sqrt(-c^2*x^2 + 
 1)*a*b*x/c^3 - 1/32*(c^2*x^2 - 1)^2*b^2/c^4 + (c^2*x^2 - 1)*a*b*arcsin(c* 
x)/c^4 + 5/32*b^2*arcsin(c*x)^2/c^4 - 5/32*(c^2*x^2 - 1)*b^2/c^4 + 5/16*a* 
b*arcsin(c*x)/c^4 - 17/256*b^2/c^4
 

Mupad [F(-1)]

Timed out. \[ \int x^3 (a+b \arcsin (c x))^2 \, dx=\int x^3\,{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2 \,d x \] Input:

int(x^3*(a + b*asin(c*x))^2,x)
 

Output:

int(x^3*(a + b*asin(c*x))^2, x)
 

Reduce [F]

\[ \int x^3 (a+b \arcsin (c x))^2 \, dx=\frac {8 \mathit {asin} \left (c x \right ) a b \,c^{4} x^{4}-3 \mathit {asin} \left (c x \right ) a b +2 \sqrt {-c^{2} x^{2}+1}\, a b \,c^{3} x^{3}+3 \sqrt {-c^{2} x^{2}+1}\, a b c x +16 \left (\int \mathit {asin} \left (c x \right )^{2} x^{3}d x \right ) b^{2} c^{4}+4 a^{2} c^{4} x^{4}}{16 c^{4}} \] Input:

int(x^3*(a+b*asin(c*x))^2,x)
 

Output:

(8*asin(c*x)*a*b*c**4*x**4 - 3*asin(c*x)*a*b + 2*sqrt( - c**2*x**2 + 1)*a* 
b*c**3*x**3 + 3*sqrt( - c**2*x**2 + 1)*a*b*c*x + 16*int(asin(c*x)**2*x**3, 
x)*b**2*c**4 + 4*a**2*c**4*x**4)/(16*c**4)