Integrand size = 21, antiderivative size = 148 \[ \int \frac {x^3}{\sqrt {a+\frac {b}{c+d x^2}}} \, dx=-\frac {(3 b+4 a c) \left (c+d x^2\right ) \sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}{8 a^2 d^2}+\frac {\left (c+d x^2\right )^2 \sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}{4 a d^2}+\frac {b (3 b+4 a c) \text {arctanh}\left (\frac {\sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}{\sqrt {a}}\right )}{8 a^{5/2} d^2} \]
1/8*b*(4*a*c+3*b)*arctanh(((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)/a^(1/2))/a^(5/ 2)/d^2-1/8*(4*a*c+3*b)*(d*x^2+c)*((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)/a^2/d^2 +1/4*(d*x^2+c)^2*((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)/a/d^2
Time = 0.14 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.73 \[ \int \frac {x^3}{\sqrt {a+\frac {b}{c+d x^2}}} \, dx=\frac {\sqrt {a} \left (c+d x^2\right ) \sqrt {\frac {b+a c+a d x^2}{c+d x^2}} \left (-3 b-2 a c+2 a d x^2\right )+b (3 b+4 a c) \text {arctanh}\left (\frac {\sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}{\sqrt {a}}\right )}{8 a^{5/2} d^2} \]
(Sqrt[a]*(c + d*x^2)*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)]*(-3*b - 2*a*c + 2*a*d*x^2) + b*(3*b + 4*a*c)*ArcTanh[Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2) ]/Sqrt[a]])/(8*a^(5/2)*d^2)
Time = 0.33 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.99, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {2057, 2053, 2052, 25, 27, 298, 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.
| \(\displaystyle \int \frac {x^3}{\sqrt {a+\frac {b}{c+d x^2}}} \, dx\) |
| \(\Big \downarrow \) 2057 |
| \(\displaystyle \int \frac {x^3}{\sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}dx\) |
| \(\Big \downarrow \) 2053 |
| \(\displaystyle \frac {1}{2} \int \frac {x^2}{\sqrt {\frac {a d x^2+b+a c}{d x^2+c}}}dx^2\) |
| \(\Big \downarrow \) 2052 |
| \(\displaystyle -b d \int -\frac {-c x^4+b+a c}{d^3 \left (a-x^4\right )^3}d\sqrt {\frac {a d x^2+b+a c}{d x^2+c}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle b d \int \frac {-c x^4+b+a c}{d^3 \left (a-x^4\right )^3}d\sqrt {\frac {a d x^2+b+a c}{d x^2+c}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b \int \frac {-c x^4+b+a c}{\left (a-x^4\right )^3}d\sqrt {\frac {a d x^2+b+a c}{d x^2+c}}}{d^2}\) |
| \(\Big \downarrow \) 298 |
| \(\displaystyle \frac {b \left (\frac {(4 a c+3 b) \int \frac {1}{\left (a-x^4\right )^2}d\sqrt {\frac {a d x^2+b+a c}{d x^2+c}}}{4 a}+\frac {b \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{4 a \left (a-x^4\right )^2}\right )}{d^2}\) |
| \(\Big \downarrow \) 215 |
| \(\displaystyle \frac {b \left (\frac {(4 a c+3 b) \left (\frac {\int \frac {1}{a-x^4}d\sqrt {\frac {a d x^2+b+a c}{d x^2+c}}}{2 a}+\frac {\sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{2 a \left (a-x^4\right )}\right )}{4 a}+\frac {b \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{4 a \left (a-x^4\right )^2}\right )}{d^2}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {b \left (\frac {(4 a c+3 b) \left (\frac {\text {arctanh}\left (\frac {\sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{\sqrt {a}}\right )}{2 a^{3/2}}+\frac {\sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{2 a \left (a-x^4\right )}\right )}{4 a}+\frac {b \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{4 a \left (a-x^4\right )^2}\right )}{d^2}\) |
(b*((b*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)])/(4*a*(a - x^4)^2) + ((3*b + 4*a*c)*(Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)]/(2*a*(a - x^4)) + ArcTanh[Sq rt[(b + a*c + a*d*x^2)/(c + d*x^2)]/Sqrt[a]]/(2*a^(3/2))))/(4*a)))/d^2
3.4.45.3.1 Defintions of rubi rules used
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)^(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])
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[(x_)^(m_.)*(((e_.)*((a_.) + (b_.)*(x_)))/((c_) + (d_.)*(x_)))^(p_), x_S ymbol] :> With[{q = Denominator[p]}, Simp[q*e*(b*c - a*d) Subst[Int[x^(q* (p + 1) - 1)*(((-a)*e + c*x^q)^m/(b*e - d*x^q)^(m + 2)), x], x, (e*((a + b* x)/(c + d*x)))^(1/q)], x]] /; FreeQ[{a, b, c, d, e, m}, x] && FractionQ[p] && IntegerQ[m]
Int[(x_)^(m_.)*(((e_.)*((a_.) + (b_.)*(x_)^(n_.)))/((c_) + (d_.)*(x_)^(n_.) ))^(p_), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(e*( (a + b*x)/(c + d*x)))^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
Int[(u_.)*((a_) + (b_.)/((c_) + (d_.)*(x_)^(n_)))^(p_), x_Symbol] :> Int[u* ((b + a*c + a*d*x^n)/(c + d*x^n))^p, x] /; FreeQ[{a, b, c, d, n, p}, x]
Time = 0.12 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.32
| method | result | size |
| risch | \(-\frac {\left (-2 a d \,x^{2}+2 a c +3 b \right ) \left (a d \,x^{2}+a c +b \right )}{8 d^{2} a^{2} \sqrt {\frac {a d \,x^{2}+a c +b}{d \,x^{2}+c}}}+\frac {b \left (4 a c +3 b \right ) \ln \left (\frac {a c d +\frac {1}{2} b d +a \,d^{2} x^{2}}{\sqrt {a \,d^{2}}}+\sqrt {a \,c^{2}+b c +\left (2 a c d +b d \right ) x^{2}+a \,d^{2} x^{4}}\right ) \sqrt {\left (a d \,x^{2}+a c +b \right ) \left (d \,x^{2}+c \right )}}{16 d \,a^{2} \sqrt {a \,d^{2}}\, \sqrt {\frac {a d \,x^{2}+a c +b}{d \,x^{2}+c}}\, \left (d \,x^{2}+c \right )}\) | \(195\) |
| default | \(-\frac {\sqrt {\frac {a d \,x^{2}+a c +b}{d \,x^{2}+c}}\, \left (d \,x^{2}+c \right ) \left (-4 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}\, a d \,x^{2}-4 \ln \left (\frac {2 a \,d^{2} x^{2}+2 a c d +2 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}+b d}{2 \sqrt {a \,d^{2}}}\right ) a b c d +4 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}\, a c -3 \ln \left (\frac {2 a \,d^{2} x^{2}+2 a c d +2 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}+b d}{2 \sqrt {a \,d^{2}}}\right ) b^{2} d +6 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}\, b \right )}{16 d^{2} \sqrt {\left (a d \,x^{2}+a c +b \right ) \left (d \,x^{2}+c \right )}\, a^{2} \sqrt {a \,d^{2}}}\) | \(354\) |
-1/8/d^2*(-2*a*d*x^2+2*a*c+3*b)*(a*d*x^2+a*c+b)/a^2/((a*d*x^2+a*c+b)/(d*x^ 2+c))^(1/2)+1/16*b/d*(4*a*c+3*b)/a^2*ln((a*c*d+1/2*b*d+a*d^2*x^2)/(a*d^2)^ (1/2)+(a*c^2+b*c+(2*a*c*d+b*d)*x^2+a*d^2*x^4)^(1/2))/(a*d^2)^(1/2)/((a*d*x ^2+a*c+b)/(d*x^2+c))^(1/2)*((a*d*x^2+a*c+b)*(d*x^2+c))^(1/2)/(d*x^2+c)
Time = 0.31 (sec) , antiderivative size = 333, normalized size of antiderivative = 2.25 \[ \int \frac {x^3}{\sqrt {a+\frac {b}{c+d x^2}}} \, dx=\left [\frac {{\left (4 \, a b c + 3 \, b^{2}\right )} \sqrt {a} \log \left (8 \, a^{2} d^{2} x^{4} + 8 \, a^{2} c^{2} + 8 \, {\left (2 \, a^{2} c + a b\right )} d x^{2} + 8 \, a b c + b^{2} + 4 \, {\left (2 \, a d^{2} x^{4} + {\left (4 \, a c + b\right )} d x^{2} + 2 \, a c^{2} + b c\right )} \sqrt {a} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}\right ) + 4 \, {\left (2 \, a^{2} d^{2} x^{4} - 3 \, a b d x^{2} - 2 \, a^{2} c^{2} - 3 \, a b c\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{32 \, a^{3} d^{2}}, -\frac {{\left (4 \, a b c + 3 \, b^{2}\right )} \sqrt {-a} \arctan \left (\frac {{\left (2 \, a d x^{2} + 2 \, a c + b\right )} \sqrt {-a} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{2 \, {\left (a^{2} d x^{2} + a^{2} c + a b\right )}}\right ) - 2 \, {\left (2 \, a^{2} d^{2} x^{4} - 3 \, a b d x^{2} - 2 \, a^{2} c^{2} - 3 \, a b c\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{16 \, a^{3} d^{2}}\right ] \]
[1/32*((4*a*b*c + 3*b^2)*sqrt(a)*log(8*a^2*d^2*x^4 + 8*a^2*c^2 + 8*(2*a^2* c + a*b)*d*x^2 + 8*a*b*c + b^2 + 4*(2*a*d^2*x^4 + (4*a*c + b)*d*x^2 + 2*a* c^2 + b*c)*sqrt(a)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c))) + 4*(2*a^2*d^2*x ^4 - 3*a*b*d*x^2 - 2*a^2*c^2 - 3*a*b*c)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)))/(a^3*d^2), -1/16*((4*a*b*c + 3*b^2)*sqrt(-a)*arctan(1/2*(2*a*d*x^2 + 2*a*c + b)*sqrt(-a)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c))/(a^2*d*x^2 + a^2 *c + a*b)) - 2*(2*a^2*d^2*x^4 - 3*a*b*d*x^2 - 2*a^2*c^2 - 3*a*b*c)*sqrt((a *d*x^2 + a*c + b)/(d*x^2 + c)))/(a^3*d^2)]
\[ \int \frac {x^3}{\sqrt {a+\frac {b}{c+d x^2}}} \, dx=\int \frac {x^{3}}{\sqrt {\frac {a c + a d x^{2} + b}{c + d x^{2}}}}\, dx \]
Time = 0.28 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.51 \[ \int \frac {x^3}{\sqrt {a+\frac {b}{c+d x^2}}} \, dx=-\frac {{\left (4 \, a b c + 3 \, b^{2}\right )} \left (\frac {a d x^{2} + a c + b}{d x^{2} + c}\right )^{\frac {3}{2}} - {\left (4 \, a^{2} b c + 5 \, a b^{2}\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{8 \, {\left (a^{4} d^{2} - \frac {2 \, {\left (a d x^{2} + a c + b\right )} a^{3} d^{2}}{d x^{2} + c} + \frac {{\left (a d x^{2} + a c + b\right )}^{2} a^{2} d^{2}}{{\left (d x^{2} + c\right )}^{2}}\right )}} - \frac {{\left (4 \, a c + 3 \, b\right )} b \log \left (-\frac {\sqrt {a} - \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{\sqrt {a} + \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}\right )}{16 \, a^{\frac {5}{2}} d^{2}} \]
-1/8*((4*a*b*c + 3*b^2)*((a*d*x^2 + a*c + b)/(d*x^2 + c))^(3/2) - (4*a^2*b *c + 5*a*b^2)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)))/(a^4*d^2 - 2*(a*d*x^2 + a*c + b)*a^3*d^2/(d*x^2 + c) + (a*d*x^2 + a*c + b)^2*a^2*d^2/(d*x^2 + c )^2) - 1/16*(4*a*c + 3*b)*b*log(-(sqrt(a) - sqrt((a*d*x^2 + a*c + b)/(d*x^ 2 + c)))/(sqrt(a) + sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c))))/(a^(5/2)*d^2)
Time = 0.39 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.12 \[ \int \frac {x^3}{\sqrt {a+\frac {b}{c+d x^2}}} \, dx=\frac {2 \, \sqrt {a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c} {\left (\frac {2 \, x^{2}}{a d} - \frac {2 \, a c d + 3 \, b d}{a^{2} d^{3}}\right )} - \frac {{\left (4 \, a b c + 3 \, b^{2}\right )} \log \left ({\left | 2 \, a c d + 2 \, {\left (\sqrt {a d^{2}} x^{2} - \sqrt {a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c}\right )} \sqrt {a} {\left | d \right |} + b d \right |}\right )}{a^{\frac {5}{2}} d {\left | d \right |}}}{16 \, \mathrm {sgn}\left (d x^{2} + c\right )} \]
1/16*(2*sqrt(a*d^2*x^4 + 2*a*c*d*x^2 + b*d*x^2 + a*c^2 + b*c)*(2*x^2/(a*d) - (2*a*c*d + 3*b*d)/(a^2*d^3)) - (4*a*b*c + 3*b^2)*log(abs(2*a*c*d + 2*(s qrt(a*d^2)*x^2 - sqrt(a*d^2*x^4 + 2*a*c*d*x^2 + b*d*x^2 + a*c^2 + b*c))*sq rt(a)*abs(d) + b*d))/(a^(5/2)*d*abs(d)))/sgn(d*x^2 + c)
Timed out. \[ \int \frac {x^3}{\sqrt {a+\frac {b}{c+d x^2}}} \, dx=\int \frac {x^3}{\sqrt {a+\frac {b}{d\,x^2+c}}} \,d x \]