3.4.93 \(\int x^4 (a+b \arcsin (c+d x^2)) \, dx\) [393]

3.4.93.1 Optimal result

Integrand size = 16, antiderivative size = 336 \[ \int x^4 \left (a+b \arcsin \left (c+d x^2\right )\right ) \, dx=-\frac {16 b c x \sqrt {1-c^2-2 c d x^2-d^2 x^4}}{75 d^2}+\frac {2 b x^3 \sqrt {1-c^2-2 c d x^2-d^2 x^4}}{25 d}+\frac {1}{5} x^5 \left (a+b \arcsin \left (c+d x^2\right )\right )-\frac {2 b \sqrt {1-c} (1+c) \left (9+23 c^2\right ) \sqrt {1-\frac {d x^2}{1-c}} \sqrt {1+\frac {d x^2}{1+c}} E\left (\arcsin \left (\frac {\sqrt {d} x}{\sqrt {1-c}}\right )|-\frac {1-c}{1+c}\right )}{75 d^{5/2} \sqrt {1-c^2-2 c d x^2-d^2 x^4}}+\frac {2 b \sqrt {1-c} (1+c) \left (9+8 c+15 c^2\right ) \sqrt {1-\frac {d x^2}{1-c}} \sqrt {1+\frac {d x^2}{1+c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {d} x}{\sqrt {1-c}}\right ),-\frac {1-c}{1+c}\right )}{75 d^{5/2} \sqrt {1-c^2-2 c d x^2-d^2 x^4}} \]

output

1/5*x^5*(a+b*arcsin(d*x^2+c))-2/75*b*(1+c)*(23*c^2+9)*EllipticE(x*d^(1/2)/ 
(1-c)^(1/2),((-1+c)/(1+c))^(1/2))*(1-c)^(1/2)*(1-d*x^2/(1-c))^(1/2)*(1+d*x 
^2/(1+c))^(1/2)/d^(5/2)/(-d^2*x^4-2*c*d*x^2-c^2+1)^(1/2)+2/75*b*(1+c)*(15* 
c^2+8*c+9)*EllipticF(x*d^(1/2)/(1-c)^(1/2),((-1+c)/(1+c))^(1/2))*(1-c)^(1/ 
2)*(1-d*x^2/(1-c))^(1/2)*(1+d*x^2/(1+c))^(1/2)/d^(5/2)/(-d^2*x^4-2*c*d*x^2 
-c^2+1)^(1/2)-16/75*b*c*x*(-d^2*x^4-2*c*d*x^2-c^2+1)^(1/2)/d^2+2/25*b*x^3* 
(-d^2*x^4-2*c*d*x^2-c^2+1)^(1/2)/d
 

3.4.93.2 Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.43 (sec) , antiderivative size = 349, normalized size of antiderivative = 1.04 \[ \int x^4 \left (a+b \arcsin \left (c+d x^2\right )\right ) \, dx=\frac {\sqrt {\frac {d}{1+c}} x \left (15 a d^2 x^4 \sqrt {1-c^2-2 c d x^2-d^2 x^4}+2 b \left (-8 c+8 c^3+3 d x^2+13 c^2 d x^2+2 c d^2 x^4-3 d^3 x^6\right )+15 b d^2 x^4 \sqrt {1-c^2-2 c d x^2-d^2 x^4} \arcsin \left (c+d x^2\right )\right )+2 i b \left (-9+9 c-23 c^2+23 c^3\right ) \sqrt {\frac {-1+c+d x^2}{-1+c}} \sqrt {\frac {1+c+d x^2}{1+c}} E\left (i \text {arcsinh}\left (\sqrt {\frac {d}{1+c}} x\right )|\frac {1+c}{-1+c}\right )-2 i b \left (-9+17 c-23 c^2+15 c^3\right ) \sqrt {\frac {-1+c+d x^2}{-1+c}} \sqrt {\frac {1+c+d x^2}{1+c}} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {\frac {d}{1+c}} x\right ),\frac {1+c}{-1+c}\right )}{75 d^2 \sqrt {\frac {d}{1+c}} \sqrt {1-c^2-2 c d x^2-d^2 x^4}} \]

input

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

output

(Sqrt[d/(1 + c)]*x*(15*a*d^2*x^4*Sqrt[1 - c^2 - 2*c*d*x^2 - d^2*x^4] + 2*b 
*(-8*c + 8*c^3 + 3*d*x^2 + 13*c^2*d*x^2 + 2*c*d^2*x^4 - 3*d^3*x^6) + 15*b* 
d^2*x^4*Sqrt[1 - c^2 - 2*c*d*x^2 - d^2*x^4]*ArcSin[c + d*x^2]) + (2*I)*b*( 
-9 + 9*c - 23*c^2 + 23*c^3)*Sqrt[(-1 + c + d*x^2)/(-1 + c)]*Sqrt[(1 + c + 
d*x^2)/(1 + c)]*EllipticE[I*ArcSinh[Sqrt[d/(1 + c)]*x], (1 + c)/(-1 + c)] 
- (2*I)*b*(-9 + 17*c - 23*c^2 + 15*c^3)*Sqrt[(-1 + c + d*x^2)/(-1 + c)]*Sq 
rt[(1 + c + d*x^2)/(1 + c)]*EllipticF[I*ArcSinh[Sqrt[d/(1 + c)]*x], (1 + c 
)/(-1 + c)])/(75*d^2*Sqrt[d/(1 + c)]*Sqrt[1 - c^2 - 2*c*d*x^2 - d^2*x^4])
 

3.4.93.3 Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 289, normalized size of antiderivative = 0.86, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {5341, 27, 1442, 1602, 25, 27, 1514, 399, 321, 327}

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.

\(\displaystyle \int x^4 \left (a+b \arcsin \left (c+d x^2\right )\right ) \, dx\)

\(\Big \downarrow \) 5341

\(\displaystyle \frac {1}{5} x^5 \left (a+b \arcsin \left (c+d x^2\right )\right )-\frac {1}{5} b \int \frac {2 d x^6}{\sqrt {-d^2 x^4-2 c d x^2-c^2+1}}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} x^5 \left (a+b \arcsin \left (c+d x^2\right )\right )-\frac {2}{5} b d \int \frac {x^6}{\sqrt {-d^2 x^4-2 c d x^2-c^2+1}}dx\)

\(\Big \downarrow \) 1442

\(\displaystyle \frac {1}{5} x^5 \left (a+b \arcsin \left (c+d x^2\right )\right )-\frac {2}{5} b d \left (\frac {\int \frac {x^2 \left (3 \left (1-c^2\right )-8 c d x^2\right )}{\sqrt {-d^2 x^4-2 c d x^2-c^2+1}}dx}{5 d^2}-\frac {x^3 \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}{5 d^2}\right )\)

\(\Big \downarrow \) 1602

\(\displaystyle \frac {1}{5} x^5 \left (a+b \arcsin \left (c+d x^2\right )\right )-\frac {2}{5} b d \left (\frac {\frac {\int -\frac {d \left (8 c \left (1-c^2\right )-\left (23 c^2+9\right ) d x^2\right )}{\sqrt {-d^2 x^4-2 c d x^2-c^2+1}}dx}{3 d^2}+\frac {8 c x \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}{3 d}}{5 d^2}-\frac {x^3 \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}{5 d^2}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{5} x^5 \left (a+b \arcsin \left (c+d x^2\right )\right )-\frac {2}{5} b d \left (\frac {\frac {8 c x \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}{3 d}-\frac {\int \frac {d \left (8 c \left (1-c^2\right )-\left (23 c^2+9\right ) d x^2\right )}{\sqrt {-d^2 x^4-2 c d x^2-c^2+1}}dx}{3 d^2}}{5 d^2}-\frac {x^3 \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}{5 d^2}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} x^5 \left (a+b \arcsin \left (c+d x^2\right )\right )-\frac {2}{5} b d \left (\frac {\frac {8 c x \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}{3 d}-\frac {\int \frac {8 c \left (1-c^2\right )-\left (23 c^2+9\right ) d x^2}{\sqrt {-d^2 x^4-2 c d x^2-c^2+1}}dx}{3 d}}{5 d^2}-\frac {x^3 \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}{5 d^2}\right )\)

\(\Big \downarrow \) 1514

\(\displaystyle \frac {1}{5} x^5 \left (a+b \arcsin \left (c+d x^2\right )\right )-\frac {2}{5} b d \left (\frac {\frac {8 c x \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}{3 d}-\frac {\sqrt {1-\frac {d x^2}{1-c}} \sqrt {\frac {d x^2}{c+1}+1} \int \frac {8 c \left (1-c^2\right )-\left (23 c^2+9\right ) d x^2}{\sqrt {1-\frac {d x^2}{1-c}} \sqrt {\frac {d x^2}{c+1}+1}}dx}{3 d \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}}{5 d^2}-\frac {x^3 \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}{5 d^2}\right )\)

\(\Big \downarrow \) 399

\(\displaystyle \frac {1}{5} x^5 \left (a+b \arcsin \left (c+d x^2\right )\right )-\frac {2}{5} b d \left (\frac {\frac {8 c x \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}{3 d}-\frac {\sqrt {1-\frac {d x^2}{1-c}} \sqrt {\frac {d x^2}{c+1}+1} \left ((c+1) \left (15 c^2+8 c+9\right ) \int \frac {1}{\sqrt {1-\frac {d x^2}{1-c}} \sqrt {\frac {d x^2}{c+1}+1}}dx-(c+1) \left (23 c^2+9\right ) \int \frac {\sqrt {\frac {d x^2}{c+1}+1}}{\sqrt {1-\frac {d x^2}{1-c}}}dx\right )}{3 d \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}}{5 d^2}-\frac {x^3 \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}{5 d^2}\right )\)

\(\Big \downarrow \) 321

\(\displaystyle \frac {1}{5} x^5 \left (a+b \arcsin \left (c+d x^2\right )\right )-\frac {2}{5} b d \left (\frac {\frac {8 c x \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}{3 d}-\frac {\sqrt {1-\frac {d x^2}{1-c}} \sqrt {\frac {d x^2}{c+1}+1} \left (\frac {\sqrt {1-c} (c+1) \left (15 c^2+8 c+9\right ) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {d} x}{\sqrt {1-c}}\right ),-\frac {1-c}{c+1}\right )}{\sqrt {d}}-(c+1) \left (23 c^2+9\right ) \int \frac {\sqrt {\frac {d x^2}{c+1}+1}}{\sqrt {1-\frac {d x^2}{1-c}}}dx\right )}{3 d \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}}{5 d^2}-\frac {x^3 \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}{5 d^2}\right )\)

\(\Big \downarrow \) 327

\(\displaystyle \frac {1}{5} x^5 \left (a+b \arcsin \left (c+d x^2\right )\right )-\frac {2}{5} b d \left (\frac {\frac {8 c x \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}{3 d}-\frac {\sqrt {1-\frac {d x^2}{1-c}} \sqrt {\frac {d x^2}{c+1}+1} \left (\frac {\sqrt {1-c} (c+1) \left (15 c^2+8 c+9\right ) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {d} x}{\sqrt {1-c}}\right ),-\frac {1-c}{c+1}\right )}{\sqrt {d}}-\frac {\sqrt {1-c} (c+1) \left (23 c^2+9\right ) E\left (\arcsin \left (\frac {\sqrt {d} x}{\sqrt {1-c}}\right )|-\frac {1-c}{c+1}\right )}{\sqrt {d}}\right )}{3 d \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}}{5 d^2}-\frac {x^3 \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}{5 d^2}\right )\)

input

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

output

(x^5*(a + b*ArcSin[c + d*x^2]))/5 - (2*b*d*(-1/5*(x^3*Sqrt[1 - c^2 - 2*c*d 
*x^2 - d^2*x^4])/d^2 + ((8*c*x*Sqrt[1 - c^2 - 2*c*d*x^2 - d^2*x^4])/(3*d) 
- (Sqrt[1 - (d*x^2)/(1 - c)]*Sqrt[1 + (d*x^2)/(1 + c)]*(-((Sqrt[1 - c]*(1 
+ c)*(9 + 23*c^2)*EllipticE[ArcSin[(Sqrt[d]*x)/Sqrt[1 - c]], -((1 - c)/(1 
+ c))])/Sqrt[d]) + (Sqrt[1 - c]*(1 + c)*(9 + 8*c + 15*c^2)*EllipticF[ArcSi 
n[(Sqrt[d]*x)/Sqrt[1 - c]], -((1 - c)/(1 + c))])/Sqrt[d]))/(3*d*Sqrt[1 - c 
^2 - 2*c*d*x^2 - d^2*x^4]))/(5*d^2)))/5
 

3.4.93.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 321

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> S 
imp[(1/(Sqrt[a]*Sqrt[c]*Rt[-d/c, 2]))*EllipticF[ArcSin[Rt[-d/c, 2]*x], b*(c 
/(a*d))], 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 327

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

rule 399

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

rule 1442

Int[((d_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] 
:> Simp[d^3*(d*x)^(m - 3)*((a + b*x^2 + c*x^4)^(p + 1)/(c*(m + 4*p + 1))), 
x] - Simp[d^4/(c*(m + 4*p + 1))   Int[(d*x)^(m - 4)*Simp[a*(m - 3) + b*(m + 
 2*p - 1)*x^2, x]*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, p}, x 
] && NeQ[b^2 - 4*a*c, 0] && GtQ[m, 3] && NeQ[m + 4*p + 1, 0] && IntegerQ[2* 
p] && (IntegerQ[p] || IntegerQ[m])
 

rule 1514

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbo 
l] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[Sqrt[1 + 2*c*(x^2/(b - q))]*(Sqrt 
[1 + 2*c*(x^2/(b + q))]/Sqrt[a + b*x^2 + c*x^4])   Int[(d + e*x^2)/(Sqrt[1 
+ 2*c*(x^2/(b - q))]*Sqrt[1 + 2*c*(x^2/(b + q))]), x], x]] /; FreeQ[{a, b, 
c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NegQ[c/a]
 

rule 1602

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*( 
x_)^4)^(p_), x_Symbol] :> Simp[e*f*(f*x)^(m - 1)*((a + b*x^2 + c*x^4)^(p + 
1)/(c*(m + 4*p + 3))), x] - Simp[f^2/(c*(m + 4*p + 3))   Int[(f*x)^(m - 2)* 
(a + b*x^2 + c*x^4)^p*Simp[a*e*(m - 1) + (b*e*(m + 2*p + 1) - c*d*(m + 4*p 
+ 3))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && NeQ[b^2 - 4*a*c 
, 0] && GtQ[m, 1] && NeQ[m + 4*p + 3, 0] && IntegerQ[2*p] && (IntegerQ[p] | 
| IntegerQ[m])
 

rule 5341

Int[((a_.) + ArcSin[u_]*(b_.))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Sim 
p[(c + d*x)^(m + 1)*((a + b*ArcSin[u])/(d*(m + 1))), x] - Simp[b/(d*(m + 1) 
)   Int[SimplifyIntegrand[(c + d*x)^(m + 1)*(D[u, x]/Sqrt[1 - u^2]), x], x] 
, x] /; FreeQ[{a, b, c, d, m}, x] && NeQ[m, -1] && InverseFunctionFreeQ[u, 
x] &&  !FunctionOfQ[(c + d*x)^(m + 1), u, x] &&  !FunctionOfExponentialQ[u, 
 x]
 

3.4.93.4 Maple [A] (verified)

Time = 1.19 (sec) , antiderivative size = 346, normalized size of antiderivative = 1.03

method	result	size
default	\(\frac {a \,x^{5}}{5}+b \left (\frac {x^{5} \arcsin \left (d \,x^{2}+c \right )}{5}-\frac {2 d \left (-\frac {x^{3} \sqrt {-d^{2} x^{4}-2 c d \,x^{2}-c^{2}+1}}{5 d^{2}}+\frac {8 c x \sqrt {-d^{2} x^{4}-2 c d \,x^{2}-c^{2}+1}}{15 d^{3}}-\frac {8 c \left (-c^{2}+1\right ) \sqrt {1+\frac {d \,x^{2}}{-1+c}}\, \sqrt {1+\frac {d \,x^{2}}{1+c}}\, \operatorname {EllipticF}\left (x \sqrt {-\frac {d}{-1+c}}, \sqrt {-1+\frac {2 c}{1+c}}\right )}{15 d^{3} \sqrt {-\frac {d}{-1+c}}\, \sqrt {-d^{2} x^{4}-2 c d \,x^{2}-c^{2}+1}}-\frac {2 \left (\frac {-3 c^{2}+3}{5 d^{2}}+\frac {32 c^{2}}{15 d^{2}}\right ) \left (-c^{2}+1\right ) \sqrt {1+\frac {d \,x^{2}}{-1+c}}\, \sqrt {1+\frac {d \,x^{2}}{1+c}}\, \left (\operatorname {EllipticF}\left (x \sqrt {-\frac {d}{-1+c}}, \sqrt {-1+\frac {2 c}{1+c}}\right )-\operatorname {EllipticE}\left (x \sqrt {-\frac {d}{-1+c}}, \sqrt {-1+\frac {2 c}{1+c}}\right )\right )}{\sqrt {-\frac {d}{-1+c}}\, \sqrt {-d^{2} x^{4}-2 c d \,x^{2}-c^{2}+1}\, \left (-2 c d +2 d \right )}\right )}{5}\right )\)	\(346\)
parts	\(\frac {a \,x^{5}}{5}+b \left (\frac {x^{5} \arcsin \left (d \,x^{2}+c \right )}{5}-\frac {2 d \left (-\frac {x^{3} \sqrt {-d^{2} x^{4}-2 c d \,x^{2}-c^{2}+1}}{5 d^{2}}+\frac {8 c x \sqrt {-d^{2} x^{4}-2 c d \,x^{2}-c^{2}+1}}{15 d^{3}}-\frac {8 c \left (-c^{2}+1\right ) \sqrt {1+\frac {d \,x^{2}}{-1+c}}\, \sqrt {1+\frac {d \,x^{2}}{1+c}}\, \operatorname {EllipticF}\left (x \sqrt {-\frac {d}{-1+c}}, \sqrt {-1+\frac {2 c}{1+c}}\right )}{15 d^{3} \sqrt {-\frac {d}{-1+c}}\, \sqrt {-d^{2} x^{4}-2 c d \,x^{2}-c^{2}+1}}-\frac {2 \left (\frac {-3 c^{2}+3}{5 d^{2}}+\frac {32 c^{2}}{15 d^{2}}\right ) \left (-c^{2}+1\right ) \sqrt {1+\frac {d \,x^{2}}{-1+c}}\, \sqrt {1+\frac {d \,x^{2}}{1+c}}\, \left (\operatorname {EllipticF}\left (x \sqrt {-\frac {d}{-1+c}}, \sqrt {-1+\frac {2 c}{1+c}}\right )-\operatorname {EllipticE}\left (x \sqrt {-\frac {d}{-1+c}}, \sqrt {-1+\frac {2 c}{1+c}}\right )\right )}{\sqrt {-\frac {d}{-1+c}}\, \sqrt {-d^{2} x^{4}-2 c d \,x^{2}-c^{2}+1}\, \left (-2 c d +2 d \right )}\right )}{5}\right )\)	\(346\)

input

int(x^4*(a+b*arcsin(d*x^2+c)),x,method=_RETURNVERBOSE)
 

output

1/5*a*x^5+b*(1/5*x^5*arcsin(d*x^2+c)-2/5*d*(-1/5*x^3/d^2*(-d^2*x^4-2*c*d*x 
^2-c^2+1)^(1/2)+8/15*c/d^3*x*(-d^2*x^4-2*c*d*x^2-c^2+1)^(1/2)-8/15*c/d^3*( 
-c^2+1)/(-d/(-1+c))^(1/2)*(1+d/(-1+c)*x^2)^(1/2)*(1+d*x^2/(1+c))^(1/2)/(-d 
^2*x^4-2*c*d*x^2-c^2+1)^(1/2)*EllipticF(x*(-d/(-1+c))^(1/2),(-1+2*c/(1+c)) 
^(1/2))-2*(1/5/d^2*(-3*c^2+3)+32/15*c^2/d^2)*(-c^2+1)/(-d/(-1+c))^(1/2)*(1 
+d/(-1+c)*x^2)^(1/2)*(1+d*x^2/(1+c))^(1/2)/(-d^2*x^4-2*c*d*x^2-c^2+1)^(1/2 
)/(-2*c*d+2*d)*(EllipticF(x*(-d/(-1+c))^(1/2),(-1+2*c/(1+c))^(1/2))-Ellipt 
icE(x*(-d/(-1+c))^(1/2),(-1+2*c/(1+c))^(1/2)))))
 

3.4.93.5 Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.26 \[ \int x^4 \left (a+b \arcsin \left (c+d x^2\right )\right ) \, dx=\frac {15 \, b d^{3} x^{6} \arcsin \left (d x^{2} + c\right ) + 15 \, a d^{3} x^{6} + 2 \, {\left (3 \, b d^{2} x^{4} - 8 \, b c d x^{2} + 23 \, b c^{2} + 9 \, b\right )} \sqrt {-d^{2} x^{4} - 2 \, c d x^{2} - c^{2} + 1}}{75 \, d^{3} x} \]

input

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

output

1/75*(15*b*d^3*x^6*arcsin(d*x^2 + c) + 15*a*d^3*x^6 + 2*(3*b*d^2*x^4 - 8*b 
*c*d*x^2 + 23*b*c^2 + 9*b)*sqrt(-d^2*x^4 - 2*c*d*x^2 - c^2 + 1))/(d^3*x)
 

3.4.93.6 Sympy [F]

\[ \int x^4 \left (a+b \arcsin \left (c+d x^2\right )\right ) \, dx=\int x^{4} \left (a + b \operatorname {asin}{\left (c + d x^{2} \right )}\right )\, dx \]

input

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

output

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

3.4.93.7 Maxima [F(-2)]

Exception generated. \[ \int x^4 \left (a+b \arcsin \left (c+d x^2\right )\right ) \, dx=\text {Exception raised: ValueError} \]

input

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

output

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(c-1>0)', see `assume?` for more 
details)Is
 

3.4.93.8 Giac [F]

\[ \int x^4 \left (a+b \arcsin \left (c+d x^2\right )\right ) \, dx=\int { {\left (b \arcsin \left (d x^{2} + c\right ) + a\right )} x^{4} \,d x } \]

input

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

output

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

3.4.93.9 Mupad [F(-1)]

Timed out. \[ \int x^4 \left (a+b \arcsin \left (c+d x^2\right )\right ) \, dx=\int x^4\,\left (a+b\,\mathrm {asin}\left (d\,x^2+c\right )\right ) \,d x \]

input

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

output

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