Integrand size = 27, antiderivative size = 150 \[ \int \frac {x^3 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=-\frac {b x}{6 c^3 d^2 \sqrt {1-c^2 x^2} \sqrt {d-c^2 d x^2}}+\frac {a+b \arcsin (c x)}{3 c^4 d \left (d-c^2 d x^2\right )^{3/2}}-\frac {a+b \arcsin (c x)}{c^4 d^2 \sqrt {d-c^2 d x^2}}+\frac {5 b \sqrt {1-c^2 x^2} \text {arctanh}(c x)}{6 c^4 d^2 \sqrt {d-c^2 d x^2}} \] Output:
-1/6*b*x/c^3/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^4/d/(-c^2*d*x^2+d)^(3/2)-(a+b*arcsin(c*x))/c^4/d^2/(-c^2*d*x^2+d)^( 1/2)+5/6*b*(-c^2*x^2+1)^(1/2)*arctanh(c*x)/c^4/d^2/(-c^2*d*x^2+d)^(1/2)
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 0.17 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.95 \[ \int \frac {x^3 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=\frac {\sqrt {d-c^2 d x^2} \left (\sqrt {-c^2} \left (-4 a+6 a c^2 x^2-b c x \sqrt {1-c^2 x^2}+2 b \left (-2+3 c^2 x^2\right ) \arcsin (c x)\right )-5 i b c \left (1-c^2 x^2\right )^{3/2} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {-c^2} x\right ),1\right )\right )}{6 c^4 \sqrt {-c^2} d^3 \left (-1+c^2 x^2\right )^2} \] Input:
Integrate[(x^3*(a + b*ArcSin[c*x]))/(d - c^2*d*x^2)^(5/2),x]
Output:
(Sqrt[d - c^2*d*x^2]*(Sqrt[-c^2]*(-4*a + 6*a*c^2*x^2 - b*c*x*Sqrt[1 - c^2* x^2] + 2*b*(-2 + 3*c^2*x^2)*ArcSin[c*x]) - (5*I)*b*c*(1 - c^2*x^2)^(3/2)*E llipticF[I*ArcSinh[Sqrt[-c^2]*x], 1]))/(6*c^4*Sqrt[-c^2]*d^3*(-1 + c^2*x^2 )^2)
Time = 0.46 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.89, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {5194, 27, 298, 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.
| \(\displaystyle \int \frac {x^3 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 5194 |
| \(\displaystyle -\frac {b c \sqrt {d-c^2 d x^2} \int -\frac {2-3 c^2 x^2}{3 c^4 d^3 \left (1-c^2 x^2\right )^2}dx}{\sqrt {1-c^2 x^2}}-\frac {a+b \arcsin (c x)}{c^4 d^2 \sqrt {d-c^2 d x^2}}+\frac {a+b \arcsin (c x)}{3 c^4 d \left (d-c^2 d x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b \sqrt {d-c^2 d x^2} \int \frac {2-3 c^2 x^2}{\left (1-c^2 x^2\right )^2}dx}{3 c^3 d^3 \sqrt {1-c^2 x^2}}-\frac {a+b \arcsin (c x)}{c^4 d^2 \sqrt {d-c^2 d x^2}}+\frac {a+b \arcsin (c x)}{3 c^4 d \left (d-c^2 d x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 298 |
| \(\displaystyle \frac {b \sqrt {d-c^2 d x^2} \left (\frac {5}{2} \int \frac {1}{1-c^2 x^2}dx-\frac {x}{2 \left (1-c^2 x^2\right )}\right )}{3 c^3 d^3 \sqrt {1-c^2 x^2}}-\frac {a+b \arcsin (c x)}{c^4 d^2 \sqrt {d-c^2 d x^2}}+\frac {a+b \arcsin (c x)}{3 c^4 d \left (d-c^2 d x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {a+b \arcsin (c x)}{c^4 d^2 \sqrt {d-c^2 d x^2}}+\frac {a+b \arcsin (c x)}{3 c^4 d \left (d-c^2 d x^2\right )^{3/2}}+\frac {b \left (\frac {5 \text {arctanh}(c x)}{2 c}-\frac {x}{2 \left (1-c^2 x^2\right )}\right ) \sqrt {d-c^2 d x^2}}{3 c^3 d^3 \sqrt {1-c^2 x^2}}\) |
Input:
Int[(x^3*(a + b*ArcSin[c*x]))/(d - c^2*d*x^2)^(5/2),x]
Output:
(a + b*ArcSin[c*x])/(3*c^4*d*(d - c^2*d*x^2)^(3/2)) - (a + b*ArcSin[c*x])/ (c^4*d^2*Sqrt[d - c^2*d*x^2]) + (b*Sqrt[d - c^2*d*x^2]*(-1/2*x/(1 - c^2*x^ 2) + (5*ArcTanh[c*x])/(2*c)))/(3*c^3*d^3*Sqrt[1 - c^2*x^2])
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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])
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-( b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 2*p + 3))/(2*a*b*(p + 1)) Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 0])
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(p_) , x_Symbol] :> With[{u = IntHide[x^m*(d + e*x^2)^p, x]}, Simp[(a + b*ArcSin [c*x]) u, x] - Simp[b*c*Simp[Sqrt[d + e*x^2]/Sqrt[1 - c^2*x^2]] Int[Sim plifyIntegrand[u/Sqrt[d + e*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && IntegerQ[p - 1/2] && NeQ[p, -2^(-1)] && (IGtQ[(m + 1)/2, 0] || ILtQ[(m + 2*p + 3)/2, 0])
Result contains complex when optimal does not.
Time = 0.50 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.68
| method | result | size |
| default | \(a \left (\frac {x^{2}}{c^{2} d \left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}-\frac {2}{3 d \,c^{4} \left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}\right )+b \left (\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (6 c^{2} x^{2} \arcsin \left (c x \right )-c x \sqrt {-c^{2} x^{2}+1}-4 \arcsin \left (c x \right )\right )}{6 \left (c^{2} x^{2}-1\right )^{2} d^{3} c^{4}}-\frac {5 \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}+i\right )}{6 d^{3} c^{4} \left (c^{2} x^{2}-1\right )}+\frac {5 \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}-i\right )}{6 d^{3} c^{4} \left (c^{2} x^{2}-1\right )}\right )\) | \(252\) |
| parts | \(a \left (\frac {x^{2}}{c^{2} d \left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}-\frac {2}{3 d \,c^{4} \left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}\right )+b \left (\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (6 c^{2} x^{2} \arcsin \left (c x \right )-c x \sqrt {-c^{2} x^{2}+1}-4 \arcsin \left (c x \right )\right )}{6 \left (c^{2} x^{2}-1\right )^{2} d^{3} c^{4}}-\frac {5 \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}+i\right )}{6 d^{3} c^{4} \left (c^{2} x^{2}-1\right )}+\frac {5 \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}-i\right )}{6 d^{3} c^{4} \left (c^{2} x^{2}-1\right )}\right )\) | \(252\) |
Input:
int(x^3*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(5/2),x,method=_RETURNVERBOSE)
Output:
a*(x^2/c^2/d/(-c^2*d*x^2+d)^(3/2)-2/3/d/c^4/(-c^2*d*x^2+d)^(3/2))+b*(1/6*( -d*(c^2*x^2-1))^(1/2)*(6*c^2*x^2*arcsin(c*x)-c*x*(-c^2*x^2+1)^(1/2)-4*arcs in(c*x))/(c^2*x^2-1)^2/d^3/c^4-5/6*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/ 2)/d^3/c^4/(c^2*x^2-1)*ln(I*c*x+(-c^2*x^2+1)^(1/2)+I)+5/6*(-d*(c^2*x^2-1)) ^(1/2)*(-c^2*x^2+1)^(1/2)/d^3/c^4/(c^2*x^2-1)*ln(I*c*x+(-c^2*x^2+1)^(1/2)- I))
Time = 0.16 (sec) , antiderivative size = 421, normalized size of antiderivative = 2.81 \[ \int \frac {x^3 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=\left [-\frac {4 \, \sqrt {-c^{2} d x^{2} + d} \sqrt {-c^{2} x^{2} + 1} b c x - 5 \, {\left (b c^{4} x^{4} - 2 \, b c^{2} x^{2} + b\right )} \sqrt {d} \log \left (-\frac {c^{6} d x^{6} + 5 \, c^{4} d x^{4} - 5 \, c^{2} d x^{2} - 4 \, {\left (c^{3} x^{3} + c x\right )} \sqrt {-c^{2} d x^{2} + d} \sqrt {-c^{2} x^{2} + 1} \sqrt {d} - d}{c^{6} x^{6} - 3 \, c^{4} x^{4} + 3 \, c^{2} x^{2} - 1}\right ) - 8 \, {\left (3 \, a c^{2} x^{2} + {\left (3 \, b c^{2} x^{2} - 2 \, b\right )} \arcsin \left (c x\right ) - 2 \, a\right )} \sqrt {-c^{2} d x^{2} + d}}{24 \, {\left (c^{8} d^{3} x^{4} - 2 \, c^{6} d^{3} x^{2} + c^{4} d^{3}\right )}}, -\frac {2 \, \sqrt {-c^{2} d x^{2} + d} \sqrt {-c^{2} x^{2} + 1} b c x - 5 \, {\left (b c^{4} x^{4} - 2 \, b c^{2} x^{2} + b\right )} \sqrt {-d} \arctan \left (\frac {2 \, \sqrt {-c^{2} d x^{2} + d} \sqrt {-c^{2} x^{2} + 1} c \sqrt {-d} x}{c^{4} d x^{4} - d}\right ) - 4 \, {\left (3 \, a c^{2} x^{2} + {\left (3 \, b c^{2} x^{2} - 2 \, b\right )} \arcsin \left (c x\right ) - 2 \, a\right )} \sqrt {-c^{2} d x^{2} + d}}{12 \, {\left (c^{8} d^{3} x^{4} - 2 \, c^{6} d^{3} x^{2} + c^{4} d^{3}\right )}}\right ] \] Input:
integrate(x^3*(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 - 5*(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*(3*a*c^2*x^2 + (3*b*c^2*x^2 - 2*b)*arc sin(c*x) - 2*a)*sqrt(-c^2*d*x^2 + d))/(c^8*d^3*x^4 - 2*c^6*d^3*x^2 + c^4*d ^3), -1/12*(2*sqrt(-c^2*d*x^2 + d)*sqrt(-c^2*x^2 + 1)*b*c*x - 5*(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*(3*a*c^2*x^2 + (3*b*c^2*x^2 - 2*b)*a rcsin(c*x) - 2*a)*sqrt(-c^2*d*x^2 + d))/(c^8*d^3*x^4 - 2*c^6*d^3*x^2 + c^4 *d^3)]
\[ \int \frac {x^3 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=\int \frac {x^{3} \left (a + b \operatorname {asin}{\left (c x \right )}\right )}{\left (- d \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {5}{2}}}\, dx \] Input:
integrate(x**3*(a+b*asin(c*x))/(-c**2*d*x**2+d)**(5/2),x)
Output:
Integral(x**3*(a + b*asin(c*x))/(-d*(c*x - 1)*(c*x + 1))**(5/2), x)
Time = 0.14 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.07 \[ \int \frac {x^3 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=\frac {1}{12} \, b c {\left (\frac {2 \, x}{c^{6} d^{\frac {5}{2}} x^{2} - c^{4} d^{\frac {5}{2}}} + \frac {5 \, \log \left (c x + 1\right )}{c^{5} d^{\frac {5}{2}}} - \frac {5 \, \log \left (c x - 1\right )}{c^{5} d^{\frac {5}{2}}}\right )} + \frac {1}{3} \, b {\left (\frac {3 \, x^{2}}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} c^{2} d} - \frac {2}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} c^{4} d}\right )} \arcsin \left (c x\right ) + \frac {1}{3} \, a {\left (\frac {3 \, x^{2}}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} c^{2} d} - \frac {2}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} c^{4} d}\right )} \] Input:
integrate(x^3*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(5/2),x, algorithm="maxima" )
Output:
1/12*b*c*(2*x/(c^6*d^(5/2)*x^2 - c^4*d^(5/2)) + 5*log(c*x + 1)/(c^5*d^(5/2 )) - 5*log(c*x - 1)/(c^5*d^(5/2))) + 1/3*b*(3*x^2/((-c^2*d*x^2 + d)^(3/2)* c^2*d) - 2/((-c^2*d*x^2 + d)^(3/2)*c^4*d))*arcsin(c*x) + 1/3*a*(3*x^2/((-c ^2*d*x^2 + d)^(3/2)*c^2*d) - 2/((-c^2*d*x^2 + d)^(3/2)*c^4*d))
Exception generated. \[ \int \frac {x^3 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(x^3*(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:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value
Timed out. \[ \int \frac {x^3 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=\int \frac {x^3\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}{{\left (d-c^2\,d\,x^2\right )}^{5/2}} \,d x \] Input:
int((x^3*(a + b*asin(c*x)))/(d - c^2*d*x^2)^(5/2),x)
Output:
int((x^3*(a + b*asin(c*x)))/(d - c^2*d*x^2)^(5/2), x)
\[ \int \frac {x^3 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=\frac {3 \sqrt {-c^{2} x^{2}+1}\, \left (\int \frac {\mathit {asin} \left (c x \right ) x^{3}}{\sqrt {-c^{2} x^{2}+1}\, c^{4} x^{4}-2 \sqrt {-c^{2} x^{2}+1}\, c^{2} x^{2}+\sqrt {-c^{2} x^{2}+1}}d x \right ) b \,c^{6} x^{2}-3 \sqrt {-c^{2} x^{2}+1}\, \left (\int \frac {\mathit {asin} \left (c x \right ) x^{3}}{\sqrt {-c^{2} x^{2}+1}\, c^{4} x^{4}-2 \sqrt {-c^{2} x^{2}+1}\, c^{2} x^{2}+\sqrt {-c^{2} x^{2}+1}}d x \right ) b \,c^{4}-3 a \,c^{2} x^{2}+2 a}{3 \sqrt {d}\, \sqrt {-c^{2} x^{2}+1}\, c^{4} d^{2} \left (c^{2} x^{2}-1\right )} \] Input:
int(x^3*(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**3)/(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**6*x**2 - 3*sqrt( - c**2*x**2 + 1)*int((asin(c*x)*x**3)/(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 - 3*a*c**2*x**2 + 2*a)/(3*sqrt(d)*sqrt( - c**2*x**2 + 1) *c**4*d**2*(c**2*x**2 - 1))