\(\int \frac {x (a+b \arcsin (c x))}{(d-c^2 d x^2)^{5/2}} \, dx\) [132]

Optimal result

Integrand size = 25, antiderivative size = 119 \[ \int \frac {x (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=-\frac {b x}{6 c d^2 \sqrt {1-c^2 x^2} \sqrt {d-c^2 d x^2}}+\frac {a+b \arcsin (c x)}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}-\frac {b \sqrt {1-c^2 x^2} \text {arctanh}(c x)}{6 c^2 d^2 \sqrt {d-c^2 d x^2}} \] Output:

-1/6*b*x/c/d^2/(-c^2*x^2+1)^(1/2)/(-c^2*d*x^2+d)^(1/2)+1/3*(a+b*arcsin(c*x 
))/c^2/d/(-c^2*d*x^2+d)^(3/2)-1/6*b*(-c^2*x^2+1)^(1/2)*arctanh(c*x)/c^2/d^ 
2/(-c^2*d*x^2+d)^(1/2)
 

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.71 \[ \int \frac {x (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=\frac {-2 a+b c x \sqrt {1-c^2 x^2}-2 b \arcsin (c x)+b \left (1-c^2 x^2\right )^{3/2} \text {arctanh}(c x)}{6 c^2 d^2 \left (-1+c^2 x^2\right ) \sqrt {d-c^2 d x^2}} \] Input:

Integrate[(x*(a + b*ArcSin[c*x]))/(d - c^2*d*x^2)^(5/2),x]
 

Output:

(-2*a + b*c*x*Sqrt[1 - c^2*x^2] - 2*b*ArcSin[c*x] + b*(1 - c^2*x^2)^(3/2)* 
ArcTanh[c*x])/(6*c^2*d^2*(-1 + c^2*x^2)*Sqrt[d - c^2*d*x^2])
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.87, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {5182, 215, 219}

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.

\(\displaystyle \int \frac {x (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 5182

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

\(\Big \downarrow \) 215

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

\(\Big \downarrow \) 219

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

Input:

Int[(x*(a + b*ArcSin[c*x]))/(d - c^2*d*x^2)^(5/2),x]
 

Output:

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

Defintions of rubi rules used

rule 215

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) 
/(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1))   Int[(a + b*x^2)^(p + 1 
), x], x] /; FreeQ[{a, b}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[6 
*p])
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 5182

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] + Simp[b*(n/(2*c*(p + 1)))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^p]   I 
nt[(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]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.46 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.87

method	result	size
default	\(\frac {a}{3 c^{2} d \left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}+b \left (\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (-c x \sqrt {-c^{2} x^{2}+1}+2 \arcsin \left (c x \right )\right )}{6 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right ) c^{2}}+\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}+i\right )}{6 d^{3} c^{2} \left (c^{2} x^{2}-1\right )}-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}-i\right )}{6 d^{3} c^{2} \left (c^{2} x^{2}-1\right )}\right )\)	\(223\)
parts	\(\frac {a}{3 c^{2} d \left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}+b \left (\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (-c x \sqrt {-c^{2} x^{2}+1}+2 \arcsin \left (c x \right )\right )}{6 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right ) c^{2}}+\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}+i\right )}{6 d^{3} c^{2} \left (c^{2} x^{2}-1\right )}-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}-i\right )}{6 d^{3} c^{2} \left (c^{2} x^{2}-1\right )}\right )\)	\(223\)

Input:

int(x*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

1/3*a/c^2/d/(-c^2*d*x^2+d)^(3/2)+b*(1/6*(-d*(c^2*x^2-1))^(1/2)*(-c*x*(-c^2 
*x^2+1)^(1/2)+2*arcsin(c*x))/d^3/(c^4*x^4-2*c^2*x^2+1)/c^2+1/6*(-d*(c^2*x^ 
2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/d^3/c^2/(c^2*x^2-1)*ln(I*c*x+(-c^2*x^2+1)^( 
1/2)+I)-1/6*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/d^3/c^2/(c^2*x^2-1)* 
ln(I*c*x+(-c^2*x^2+1)^(1/2)-I))
 

Fricas [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 374, normalized size of antiderivative = 3.14 \[ \int \frac {x (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=\left [-\frac {4 \, \sqrt {-c^{2} d x^{2} + d} \sqrt {-c^{2} x^{2} + 1} b c x - {\left (b c^{4} x^{4} - 2 \, b c^{2} x^{2} + b\right )} \sqrt {d} \log \left (-\frac {c^{6} d x^{6} + 5 \, c^{4} d x^{4} - 5 \, c^{2} d x^{2} + 4 \, {\left (c^{3} x^{3} + c x\right )} \sqrt {-c^{2} d x^{2} + d} \sqrt {-c^{2} x^{2} + 1} \sqrt {d} - d}{c^{6} x^{6} - 3 \, c^{4} x^{4} + 3 \, c^{2} x^{2} - 1}\right ) - 8 \, \sqrt {-c^{2} d x^{2} + d} {\left (b \arcsin \left (c x\right ) + a\right )}}{24 \, {\left (c^{6} d^{3} x^{4} - 2 \, c^{4} d^{3} x^{2} + c^{2} d^{3}\right )}}, -\frac {2 \, \sqrt {-c^{2} d x^{2} + d} \sqrt {-c^{2} x^{2} + 1} b c x + {\left (b c^{4} x^{4} - 2 \, b c^{2} x^{2} + b\right )} \sqrt {-d} \arctan \left (\frac {2 \, \sqrt {-c^{2} d x^{2} + d} \sqrt {-c^{2} x^{2} + 1} c \sqrt {-d} x}{c^{4} d x^{4} - d}\right ) - 4 \, \sqrt {-c^{2} d x^{2} + d} {\left (b \arcsin \left (c x\right ) + a\right )}}{12 \, {\left (c^{6} d^{3} x^{4} - 2 \, c^{4} d^{3} x^{2} + c^{2} d^{3}\right )}}\right ] \] Input:

integrate(x*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(5/2),x, algorithm="fricas")
 

Output:

[-1/24*(4*sqrt(-c^2*d*x^2 + d)*sqrt(-c^2*x^2 + 1)*b*c*x - (b*c^4*x^4 - 2*b 
*c^2*x^2 + b)*sqrt(d)*log(-(c^6*d*x^6 + 5*c^4*d*x^4 - 5*c^2*d*x^2 + 4*(c^3 
*x^3 + c*x)*sqrt(-c^2*d*x^2 + d)*sqrt(-c^2*x^2 + 1)*sqrt(d) - d)/(c^6*x^6 
- 3*c^4*x^4 + 3*c^2*x^2 - 1)) - 8*sqrt(-c^2*d*x^2 + d)*(b*arcsin(c*x) + a) 
)/(c^6*d^3*x^4 - 2*c^4*d^3*x^2 + c^2*d^3), -1/12*(2*sqrt(-c^2*d*x^2 + d)*s 
qrt(-c^2*x^2 + 1)*b*c*x + (b*c^4*x^4 - 2*b*c^2*x^2 + b)*sqrt(-d)*arctan(2* 
sqrt(-c^2*d*x^2 + d)*sqrt(-c^2*x^2 + 1)*c*sqrt(-d)*x/(c^4*d*x^4 - d)) - 4* 
sqrt(-c^2*d*x^2 + d)*(b*arcsin(c*x) + a))/(c^6*d^3*x^4 - 2*c^4*d^3*x^2 + c 
^2*d^3)]
 

Sympy [F]

\[ \int \frac {x (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=\int \frac {x \left (a + b \operatorname {asin}{\left (c x \right )}\right )}{\left (- d \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(x*(a+b*asin(c*x))/(-c**2*d*x**2+d)**(5/2),x)
 

Output:

Integral(x*(a + b*asin(c*x))/(-d*(c*x - 1)*(c*x + 1))**(5/2), x)
 

Maxima [F]

\[ \int \frac {x (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=\int { \frac {{\left (b \arcsin \left (c x\right ) + a\right )} x}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(x*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(5/2),x, algorithm="maxima")
 

Output:

b*integrate(x*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))/((c^4*d^2*x^4 - 2 
*c^2*d^2*x^2 + d^2)*sqrt(c*x + 1)*sqrt(-c*x + 1)), x)/sqrt(d) + 1/3*a/((-c 
^2*d*x^2 + d)^(3/2)*c^2*d)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {x (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(x*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(5/2),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:index.cc index_m i_lex_is_greater E 
rror: Bad Argument Value
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=\int \frac {x\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}{{\left (d-c^2\,d\,x^2\right )}^{5/2}} \,d x \] Input:

int((x*(a + b*asin(c*x)))/(d - c^2*d*x^2)^(5/2),x)
 

Output:

int((x*(a + b*asin(c*x)))/(d - c^2*d*x^2)^(5/2), x)
 

Reduce [F]

\[ \int \frac {x (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=\frac {3 \sqrt {-c^{2} x^{2}+1}\, \left (\int \frac {\mathit {asin} \left (c x \right ) x}{\sqrt {-c^{2} x^{2}+1}\, c^{4} x^{4}-2 \sqrt {-c^{2} x^{2}+1}\, c^{2} x^{2}+\sqrt {-c^{2} x^{2}+1}}d x \right ) b \,c^{4} x^{2}-3 \sqrt {-c^{2} x^{2}+1}\, \left (\int \frac {\mathit {asin} \left (c x \right ) x}{\sqrt {-c^{2} x^{2}+1}\, c^{4} x^{4}-2 \sqrt {-c^{2} x^{2}+1}\, c^{2} x^{2}+\sqrt {-c^{2} x^{2}+1}}d x \right ) b \,c^{2}-a}{3 \sqrt {d}\, \sqrt {-c^{2} x^{2}+1}\, c^{2} d^{2} \left (c^{2} x^{2}-1\right )} \] Input:

int(x*(a+b*asin(c*x))/(-c^2*d*x^2+d)^(5/2),x)
                                                                                    
                                                                                    
 

Output:

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