Integrand size = 24, antiderivative size = 163 \[ \int \frac {x^3 \left (c+d x^2\right )^{3/2}}{\left (a+b x^2\right )^2} \, dx=\frac {(2 b c-5 a d) \sqrt {c+d x^2}}{2 b^3}+\frac {(2 b c-5 a d) \left (c+d x^2\right )^{3/2}}{6 b^2 (b c-a d)}+\frac {a \left (c+d x^2\right )^{5/2}}{2 b (b c-a d) \left (a+b x^2\right )}-\frac {(2 b c-5 a d) \sqrt {b c-a d} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{2 b^{7/2}} \]
1/6*(-5*a*d+2*b*c)*(d*x^2+c)^(3/2)/b^2/(-a*d+b*c)+1/2*a*(d*x^2+c)^(5/2)/b/ (-a*d+b*c)/(b*x^2+a)-1/2*(-5*a*d+2*b*c)*arctanh(b^(1/2)*(d*x^2+c)^(1/2)/(- a*d+b*c)^(1/2))*(-a*d+b*c)^(1/2)/b^(7/2)+1/2*(-5*a*d+2*b*c)*(d*x^2+c)^(1/2 )/b^3
Time = 0.28 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.77 \[ \int \frac {x^3 \left (c+d x^2\right )^{3/2}}{\left (a+b x^2\right )^2} \, dx=\frac {\sqrt {c+d x^2} \left (-15 a^2 d+a b \left (11 c-10 d x^2\right )+2 b^2 x^2 \left (4 c+d x^2\right )\right )}{6 b^3 \left (a+b x^2\right )}-\frac {(2 b c-5 a d) \sqrt {-b c+a d} \arctan \left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {-b c+a d}}\right )}{2 b^{7/2}} \]
(Sqrt[c + d*x^2]*(-15*a^2*d + a*b*(11*c - 10*d*x^2) + 2*b^2*x^2*(4*c + d*x ^2)))/(6*b^3*(a + b*x^2)) - ((2*b*c - 5*a*d)*Sqrt[-(b*c) + a*d]*ArcTan[(Sq rt[b]*Sqrt[c + d*x^2])/Sqrt[-(b*c) + a*d]])/(2*b^(7/2))
Time = 0.26 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {354, 87, 60, 60, 73, 221}
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.
| \(\displaystyle \int \frac {x^3 \left (c+d x^2\right )^{3/2}}{\left (a+b x^2\right )^2} \, dx\) |
| \(\Big \downarrow \) 354 |
| \(\displaystyle \frac {1}{2} \int \frac {x^2 \left (d x^2+c\right )^{3/2}}{\left (b x^2+a\right )^2}dx^2\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle \frac {1}{2} \left (\frac {(2 b c-5 a d) \int \frac {\left (d x^2+c\right )^{3/2}}{b x^2+a}dx^2}{2 b (b c-a d)}+\frac {a \left (c+d x^2\right )^{5/2}}{b \left (a+b x^2\right ) (b c-a d)}\right )\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {1}{2} \left (\frac {(2 b c-5 a d) \left (\frac {(b c-a d) \int \frac {\sqrt {d x^2+c}}{b x^2+a}dx^2}{b}+\frac {2 \left (c+d x^2\right )^{3/2}}{3 b}\right )}{2 b (b c-a d)}+\frac {a \left (c+d x^2\right )^{5/2}}{b \left (a+b x^2\right ) (b c-a d)}\right )\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {1}{2} \left (\frac {(2 b c-5 a d) \left (\frac {(b c-a d) \left (\frac {(b c-a d) \int \frac {1}{\left (b x^2+a\right ) \sqrt {d x^2+c}}dx^2}{b}+\frac {2 \sqrt {c+d x^2}}{b}\right )}{b}+\frac {2 \left (c+d x^2\right )^{3/2}}{3 b}\right )}{2 b (b c-a d)}+\frac {a \left (c+d x^2\right )^{5/2}}{b \left (a+b x^2\right ) (b c-a d)}\right )\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {1}{2} \left (\frac {(2 b c-5 a d) \left (\frac {(b c-a d) \left (\frac {2 (b c-a d) \int \frac {1}{\frac {b x^4}{d}+a-\frac {b c}{d}}d\sqrt {d x^2+c}}{b d}+\frac {2 \sqrt {c+d x^2}}{b}\right )}{b}+\frac {2 \left (c+d x^2\right )^{3/2}}{3 b}\right )}{2 b (b c-a d)}+\frac {a \left (c+d x^2\right )^{5/2}}{b \left (a+b x^2\right ) (b c-a d)}\right )\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {1}{2} \left (\frac {(2 b c-5 a d) \left (\frac {(b c-a d) \left (\frac {2 \sqrt {c+d x^2}}{b}-\frac {2 \sqrt {b c-a d} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{b^{3/2}}\right )}{b}+\frac {2 \left (c+d x^2\right )^{3/2}}{3 b}\right )}{2 b (b c-a d)}+\frac {a \left (c+d x^2\right )^{5/2}}{b \left (a+b x^2\right ) (b c-a d)}\right )\) |
((a*(c + d*x^2)^(5/2))/(b*(b*c - a*d)*(a + b*x^2)) + ((2*b*c - 5*a*d)*((2* (c + d*x^2)^(3/2))/(3*b) + ((b*c - a*d)*((2*Sqrt[c + d*x^2])/b - (2*Sqrt[b *c - a*d]*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^2])/Sqrt[b*c - a*d]])/b^(3/2)))/b) )/(2*b*(b*c - a*d)))/2
3.8.41.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/2]
Time = 3.10 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.82
| method | result | size |
| pseudoelliptic | \(-\frac {5 \left (-\left (b \,x^{2}+a \right ) \left (a d -\frac {2 b c}{5}\right ) \left (a d -b c \right ) \arctan \left (\frac {b \sqrt {d \,x^{2}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )+\sqrt {\left (a d -b c \right ) b}\, \left (-\frac {8 x^{2} \left (\frac {d \,x^{2}}{4}+c \right ) b^{2}}{15}-\frac {11 \left (-\frac {10 d \,x^{2}}{11}+c \right ) a b}{15}+a^{2} d \right ) \sqrt {d \,x^{2}+c}\right )}{2 \sqrt {\left (a d -b c \right ) b}\, b^{3} \left (b \,x^{2}+a \right )}\) | \(133\) |
| risch | \(-\frac {\left (-b d \,x^{2}+6 a d -4 b c \right ) \sqrt {d \,x^{2}+c}}{3 b^{3}}+\frac {-\frac {\left (\frac {3}{2} a^{2} d^{2}-2 a b c d +\frac {1}{2} b^{2} c^{2}\right ) \ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {d \left (x +\frac {\sqrt {-a b}}{b}\right )^{2}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x +\frac {\sqrt {-a b}}{b}}\right )}{b \sqrt {-\frac {a d -b c}{b}}}-\frac {\left (\frac {3}{2} a^{2} d^{2}-2 a b c d +\frac {1}{2} b^{2} c^{2}\right ) \ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {d \left (x -\frac {\sqrt {-a b}}{b}\right )^{2}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x -\frac {\sqrt {-a b}}{b}}\right )}{b \sqrt {-\frac {a d -b c}{b}}}-\frac {\sqrt {-a b}\, \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \left (\frac {b \sqrt {d \left (x +\frac {\sqrt {-a b}}{b}\right )^{2}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{\left (a d -b c \right ) \left (x +\frac {\sqrt {-a b}}{b}\right )}+\frac {d \sqrt {-a b}\, \ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {d \left (x +\frac {\sqrt {-a b}}{b}\right )^{2}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x +\frac {\sqrt {-a b}}{b}}\right )}{\left (a d -b c \right ) \sqrt {-\frac {a d -b c}{b}}}\right )}{4 b^{2}}+\frac {\sqrt {-a b}\, \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \left (\frac {b \sqrt {d \left (x -\frac {\sqrt {-a b}}{b}\right )^{2}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{\left (a d -b c \right ) \left (x -\frac {\sqrt {-a b}}{b}\right )}-\frac {d \sqrt {-a b}\, \ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {d \left (x -\frac {\sqrt {-a b}}{b}\right )^{2}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x -\frac {\sqrt {-a b}}{b}}\right )}{\left (a d -b c \right ) \sqrt {-\frac {a d -b c}{b}}}\right )}{4 b^{2}}}{b^{3}}\) | \(939\) |
| default | \(\text {Expression too large to display}\) | \(3381\) |
-5/2*(-(b*x^2+a)*(a*d-2/5*b*c)*(a*d-b*c)*arctan(b*(d*x^2+c)^(1/2)/((a*d-b* c)*b)^(1/2))+((a*d-b*c)*b)^(1/2)*(-8/15*x^2*(1/4*d*x^2+c)*b^2-11/15*(-10/1 1*d*x^2+c)*a*b+a^2*d)*(d*x^2+c)^(1/2))/((a*d-b*c)*b)^(1/2)/b^3/(b*x^2+a)
Time = 0.30 (sec) , antiderivative size = 413, normalized size of antiderivative = 2.53 \[ \int \frac {x^3 \left (c+d x^2\right )^{3/2}}{\left (a+b x^2\right )^2} \, dx=\left [-\frac {3 \, {\left (2 \, a b c - 5 \, a^{2} d + {\left (2 \, b^{2} c - 5 \, a b d\right )} x^{2}\right )} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b^{2} d^{2} x^{4} + 8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2} + 2 \, {\left (4 \, b^{2} c d - 3 \, a b d^{2}\right )} x^{2} + 4 \, {\left (b^{2} d x^{2} + 2 \, b^{2} c - a b d\right )} \sqrt {d x^{2} + c} \sqrt {\frac {b c - a d}{b}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) - 4 \, {\left (2 \, b^{2} d x^{4} + 11 \, a b c - 15 \, a^{2} d + 2 \, {\left (4 \, b^{2} c - 5 \, a b d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{24 \, {\left (b^{4} x^{2} + a b^{3}\right )}}, -\frac {3 \, {\left (2 \, a b c - 5 \, a^{2} d + {\left (2 \, b^{2} c - 5 \, a b d\right )} x^{2}\right )} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {{\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {b c - a d}{b}}}{2 \, {\left (b c^{2} - a c d + {\left (b c d - a d^{2}\right )} x^{2}\right )}}\right ) - 2 \, {\left (2 \, b^{2} d x^{4} + 11 \, a b c - 15 \, a^{2} d + 2 \, {\left (4 \, b^{2} c - 5 \, a b d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{12 \, {\left (b^{4} x^{2} + a b^{3}\right )}}\right ] \]
[-1/24*(3*(2*a*b*c - 5*a^2*d + (2*b^2*c - 5*a*b*d)*x^2)*sqrt((b*c - a*d)/b )*log((b^2*d^2*x^4 + 8*b^2*c^2 - 8*a*b*c*d + a^2*d^2 + 2*(4*b^2*c*d - 3*a* b*d^2)*x^2 + 4*(b^2*d*x^2 + 2*b^2*c - a*b*d)*sqrt(d*x^2 + c)*sqrt((b*c - a *d)/b))/(b^2*x^4 + 2*a*b*x^2 + a^2)) - 4*(2*b^2*d*x^4 + 11*a*b*c - 15*a^2* d + 2*(4*b^2*c - 5*a*b*d)*x^2)*sqrt(d*x^2 + c))/(b^4*x^2 + a*b^3), -1/12*( 3*(2*a*b*c - 5*a^2*d + (2*b^2*c - 5*a*b*d)*x^2)*sqrt(-(b*c - a*d)/b)*arcta n(-1/2*(b*d*x^2 + 2*b*c - a*d)*sqrt(d*x^2 + c)*sqrt(-(b*c - a*d)/b)/(b*c^2 - a*c*d + (b*c*d - a*d^2)*x^2)) - 2*(2*b^2*d*x^4 + 11*a*b*c - 15*a^2*d + 2*(4*b^2*c - 5*a*b*d)*x^2)*sqrt(d*x^2 + c))/(b^4*x^2 + a*b^3)]
\[ \int \frac {x^3 \left (c+d x^2\right )^{3/2}}{\left (a+b x^2\right )^2} \, dx=\int \frac {x^{3} \left (c + d x^{2}\right )^{\frac {3}{2}}}{\left (a + b x^{2}\right )^{2}}\, dx \]
Exception generated. \[ \int \frac {x^3 \left (c+d x^2\right )^{3/2}}{\left (a+b x^2\right )^2} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for m ore detail
Time = 0.28 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.06 \[ \int \frac {x^3 \left (c+d x^2\right )^{3/2}}{\left (a+b x^2\right )^2} \, dx=\frac {{\left (2 \, b^{2} c^{2} - 7 \, a b c d + 5 \, a^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {d x^{2} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{2 \, \sqrt {-b^{2} c + a b d} b^{3}} + \frac {\sqrt {d x^{2} + c} a b c d - \sqrt {d x^{2} + c} a^{2} d^{2}}{2 \, {\left ({\left (d x^{2} + c\right )} b - b c + a d\right )} b^{3}} + \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{4} + 3 \, \sqrt {d x^{2} + c} b^{4} c - 6 \, \sqrt {d x^{2} + c} a b^{3} d}{3 \, b^{6}} \]
1/2*(2*b^2*c^2 - 7*a*b*c*d + 5*a^2*d^2)*arctan(sqrt(d*x^2 + c)*b/sqrt(-b^2 *c + a*b*d))/(sqrt(-b^2*c + a*b*d)*b^3) + 1/2*(sqrt(d*x^2 + c)*a*b*c*d - s qrt(d*x^2 + c)*a^2*d^2)/(((d*x^2 + c)*b - b*c + a*d)*b^3) + 1/3*((d*x^2 + c)^(3/2)*b^4 + 3*sqrt(d*x^2 + c)*b^4*c - 6*sqrt(d*x^2 + c)*a*b^3*d)/b^6
Time = 5.59 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.12 \[ \int \frac {x^3 \left (c+d x^2\right )^{3/2}}{\left (a+b x^2\right )^2} \, dx=\frac {{\left (d\,x^2+c\right )}^{3/2}}{3\,b^2}-\sqrt {d\,x^2+c}\,\left (\frac {c}{b^2}-\frac {2\,b^2\,c-2\,a\,b\,d}{b^4}\right )-\frac {\left (\frac {a^2\,d^2}{2}-\frac {a\,b\,c\,d}{2}\right )\,\sqrt {d\,x^2+c}}{b^4\,\left (d\,x^2+c\right )-b^4\,c+a\,b^3\,d}+\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d\,x^2+c}\,\sqrt {a\,d-b\,c}\,\left (5\,a\,d-2\,b\,c\right )}{5\,a^2\,d^2-7\,a\,b\,c\,d+2\,b^2\,c^2}\right )\,\sqrt {a\,d-b\,c}\,\left (5\,a\,d-2\,b\,c\right )}{2\,b^{7/2}} \]
(c + d*x^2)^(3/2)/(3*b^2) - (c + d*x^2)^(1/2)*(c/b^2 - (2*b^2*c - 2*a*b*d) /b^4) - (((a^2*d^2)/2 - (a*b*c*d)/2)*(c + d*x^2)^(1/2))/(b^4*(c + d*x^2) - b^4*c + a*b^3*d) + (atan((b^(1/2)*(c + d*x^2)^(1/2)*(a*d - b*c)^(1/2)*(5* a*d - 2*b*c))/(5*a^2*d^2 + 2*b^2*c^2 - 7*a*b*c*d))*(a*d - b*c)^(1/2)*(5*a* d - 2*b*c))/(2*b^(7/2))