Integrand size = 21, antiderivative size = 141 \[ \int x^3 \sqrt {a+\frac {b}{c+d x^2}} \, dx=\frac {(b-4 a c) \left (c+d x^2\right ) \sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}{8 a d^2}+\frac {\left (c+d x^2\right )^2 \sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}{4 d^2}-\frac {b (b+4 a c) \text {arctanh}\left (\frac {\sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}{\sqrt {a}}\right )}{8 a^{3/2} d^2} \]
-1/8*b*(4*a*c+b)*arctanh(((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)/a^(1/2))/a^(3/2 )/d^2+1/8*(-4*a*c+b)*(d*x^2+c)*((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)/a/d^2+1/4 *(d*x^2+c)^2*((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)/d^2
Time = 0.14 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.76 \[ \int x^3 \sqrt {a+\frac {b}{c+d x^2}} \, dx=\frac {\left (c+d x^2\right ) \sqrt {\frac {b+a c+a d x^2}{c+d x^2}} \left (b-2 a c+2 a d x^2\right )}{8 a d^2}-\frac {b (b+4 a c) \text {arctanh}\left (\frac {\sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}{\sqrt {a}}\right )}{8 a^{3/2} d^2} \]
((c + d*x^2)*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)]*(b - 2*a*c + 2*a*d*x^2) )/(8*a*d^2) - (b*(b + 4*a*c)*ArcTanh[Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)] /Sqrt[a]])/(8*a^(3/2)*d^2)
Time = 0.35 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.02, 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, 360, 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.
| \(\displaystyle \int x^3 \sqrt {a+\frac {b}{c+d x^2}} \, dx\) |
| \(\Big \downarrow \) 2057 |
| \(\displaystyle \int x^3 \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}dx\) |
| \(\Big \downarrow \) 2053 |
| \(\displaystyle \frac {1}{2} \int x^2 \sqrt {\frac {a d x^2+b+a c}{d x^2+c}}dx^2\) |
| \(\Big \downarrow \) 2052 |
| \(\displaystyle -b d \int -\frac {x^4 \left (-c x^4+b+a c\right )}{d^3 \left (a-x^4\right )^3}d\sqrt {\frac {a d x^2+b+a c}{d x^2+c}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle b d \int \frac {x^4 \left (-c x^4+b+a c\right )}{d^3 \left (a-x^4\right )^3}d\sqrt {\frac {a d x^2+b+a c}{d x^2+c}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b \int \frac {x^4 \left (-c x^4+b+a c\right )}{\left (a-x^4\right )^3}d\sqrt {\frac {a d x^2+b+a c}{d x^2+c}}}{d^2}\) |
| \(\Big \downarrow \) 360 |
| \(\displaystyle \frac {b \left (\frac {b \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{4 \left (a-x^4\right )^2}-\frac {1}{4} \int \frac {b-4 c x^4}{\left (a-x^4\right )^2}d\sqrt {\frac {a d x^2+b+a c}{d x^2+c}}\right )}{d^2}\) |
| \(\Big \downarrow \) 298 |
| \(\displaystyle \frac {b \left (\frac {1}{4} \left (-\frac {(4 a c+b) \int \frac {1}{a-x^4}d\sqrt {\frac {a d x^2+b+a c}{d x^2+c}}}{2 a}-\frac {(b-4 a c) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{2 a \left (a-x^4\right )}\right )+\frac {b \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{4 \left (a-x^4\right )^2}\right )}{d^2}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {b \left (\frac {1}{4} \left (-\frac {(4 a c+b) \text {arctanh}\left (\frac {\sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{\sqrt {a}}\right )}{2 a^{3/2}}-\frac {(b-4 a c) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{2 a \left (a-x^4\right )}\right )+\frac {b \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{4 \left (a-x^4\right )^2}\right )}{d^2}\) |
(b*((b*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)])/(4*(a - x^4)^2) + (-1/2*((b - 4*a*c)*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)])/(a*(a - x^4)) - ((b + 4*a* c)*ArcTanh[Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)]/Sqrt[a]])/(2*a^(3/2)))/4) )/d^2
3.4.19.3.1 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)^(-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_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] : > Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p + 1))), x] + Simp[1/(2*b^(m/2 + 1)*(p + 1)) Int[(a + b*x^2)^(p + 1)*Expan dToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)] - (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[m/2, 0] & & (IntegerQ[p] || EqQ[m + 2*p + 1, 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 = 193, normalized size of antiderivative = 1.37
| method | result | size |
| risch | \(-\frac {\left (-2 a d \,x^{2}+2 a c -b \right ) \left (d \,x^{2}+c \right ) \sqrt {\frac {a d \,x^{2}+a c +b}{d \,x^{2}+c}}}{8 d^{2} a}-\frac {b \left (4 a c +b \right ) \ln \left (\frac {a c d +\frac {1}{2} b d +a \,d^{2} x^{2}}{\sqrt {a \,d^{2}}}+\sqrt {a \,c^{2}+b c +\left (2 a c d +b d \right ) x^{2}+a \,d^{2} x^{4}}\right ) \sqrt {\frac {a d \,x^{2}+a c +b}{d \,x^{2}+c}}\, \sqrt {\left (a d \,x^{2}+a c +b \right ) \left (d \,x^{2}+c \right )}}{16 d a \sqrt {a \,d^{2}}\, \left (a d \,x^{2}+a c +b \right )}\) | \(193\) |
| default | \(-\frac {\sqrt {\frac {a d \,x^{2}+a c +b}{d \,x^{2}+c}}\, \left (d \,x^{2}+c \right ) \left (-4 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}\, a d \,x^{2}+4 \ln \left (\frac {2 a \,d^{2} x^{2}+2 a c d +2 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}+b d}{2 \sqrt {a \,d^{2}}}\right ) a b c d +4 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}\, a c +\ln \left (\frac {2 a \,d^{2} x^{2}+2 a c d +2 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}+b d}{2 \sqrt {a \,d^{2}}}\right ) b^{2} d -2 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}\, b \right )}{16 d^{2} \sqrt {\left (a d \,x^{2}+a c +b \right ) \left (d \,x^{2}+c \right )}\, a \sqrt {a \,d^{2}}}\) | \(353\) |
-1/8/d^2*(-2*a*d*x^2+2*a*c-b)*(d*x^2+c)/a*((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2 )-1/16*b/d*(4*a*c+b)/a*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)/(a*d*x^2+a*c+b)
Time = 0.31 (sec) , antiderivative size = 325, normalized size of antiderivative = 2.30 \[ \int x^3 \sqrt {a+\frac {b}{c+d x^2}} \, dx=\left [\frac {{\left (4 \, a b c + b^{2}\right )} \sqrt {a} \log \left (8 \, a^{2} d^{2} x^{4} + 8 \, a^{2} c^{2} + 8 \, {\left (2 \, a^{2} c + a b\right )} d x^{2} + 8 \, a b c + b^{2} - 4 \, {\left (2 \, a d^{2} x^{4} + {\left (4 \, a c + b\right )} d x^{2} + 2 \, a c^{2} + b c\right )} \sqrt {a} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}\right ) + 4 \, {\left (2 \, a^{2} d^{2} x^{4} + a b d x^{2} - 2 \, a^{2} c^{2} + a b c\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{32 \, a^{2} d^{2}}, \frac {{\left (4 \, a b c + b^{2}\right )} \sqrt {-a} \arctan \left (\frac {{\left (2 \, a d x^{2} + 2 \, a c + b\right )} \sqrt {-a} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{2 \, {\left (a^{2} d x^{2} + a^{2} c + a b\right )}}\right ) + 2 \, {\left (2 \, a^{2} d^{2} x^{4} + a b d x^{2} - 2 \, a^{2} c^{2} + a b c\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{16 \, a^{2} d^{2}}\right ] \]
[1/32*((4*a*b*c + 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 + a*b*d*x^2 - 2*a^2*c^2 + a*b*c)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)))/( a^2*d^2), 1/16*((4*a*b*c + 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 + a*b*d*x^2 - 2*a^2*c^2 + a*b*c)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)))/(a^2*d^2)]
\[ \int x^3 \sqrt {a+\frac {b}{c+d x^2}} \, dx=\int x^{3} \sqrt {\frac {a c + a d x^{2} + b}{c + d x^{2}}}\, dx \]
Time = 0.29 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.55 \[ \int x^3 \sqrt {a+\frac {b}{c+d x^2}} \, dx=-\frac {{\left (4 \, a b c - b^{2}\right )} \left (\frac {a d x^{2} + a c + b}{d x^{2} + c}\right )^{\frac {3}{2}} - {\left (4 \, a^{2} b c + a b^{2}\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{8 \, {\left (a^{3} d^{2} - \frac {2 \, {\left (a d x^{2} + a c + b\right )} a^{2} d^{2}}{d x^{2} + c} + \frac {{\left (a d x^{2} + a c + b\right )}^{2} a d^{2}}{{\left (d x^{2} + c\right )}^{2}}\right )}} + \frac {{\left (4 \, a c + b\right )} b \log \left (-\frac {\sqrt {a} - \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{\sqrt {a} + \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}\right )}{16 \, a^{\frac {3}{2}} d^{2}} \]
-1/8*((4*a*b*c - b^2)*((a*d*x^2 + a*c + b)/(d*x^2 + c))^(3/2) - (4*a^2*b*c + a*b^2)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)))/(a^3*d^2 - 2*(a*d*x^2 + a *c + b)*a^2*d^2/(d*x^2 + c) + (a*d*x^2 + a*c + b)^2*a*d^2/(d*x^2 + c)^2) + 1/16*(4*a*c + 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^(3/2)*d^2)
Time = 0.42 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.12 \[ \int x^3 \sqrt {a+\frac {b}{c+d x^2}} \, dx=\frac {1}{16} \, {\left (2 \, \sqrt {a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c} {\left (\frac {2 \, x^{2}}{d} - \frac {2 \, a c d - b d}{a d^{3}}\right )} + \frac {{\left (4 \, a b c + b^{2}\right )} \log \left ({\left | 2 \, a c d + 2 \, {\left (\sqrt {a d^{2}} x^{2} - \sqrt {a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c}\right )} \sqrt {a} {\left | d \right |} + b d \right |}\right )}{a^{\frac {3}{2}} d {\left | d \right |}}\right )} \mathrm {sgn}\left (d x^{2} + c\right ) \]
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/d - ( 2*a*c*d - b*d)/(a*d^3)) + (4*a*b*c + b^2)*log(abs(2*a*c*d + 2*(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))*sqrt(a)*abs( d) + b*d))/(a^(3/2)*d*abs(d)))*sgn(d*x^2 + c)
Timed out. \[ \int x^3 \sqrt {a+\frac {b}{c+d x^2}} \, dx=\int x^3\,\sqrt {a+\frac {b}{d\,x^2+c}} \,d x \]