Integrand size = 24, antiderivative size = 88 \[ \int \frac {x^3 \sqrt {c+d x^2}}{a+b x^2} \, dx=-\frac {a \sqrt {c+d x^2}}{b^2}+\frac {\left (c+d x^2\right )^{3/2}}{3 b d}+\frac {a \sqrt {b c-a d} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{b^{5/2}} \]
1/3*(d*x^2+c)^(3/2)/b/d+a*arctanh(b^(1/2)*(d*x^2+c)^(1/2)/(-a*d+b*c)^(1/2) )*(-a*d+b*c)^(1/2)/b^(5/2)-a*(d*x^2+c)^(1/2)/b^2
Time = 0.12 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.97 \[ \int \frac {x^3 \sqrt {c+d x^2}}{a+b x^2} \, dx=\frac {\sqrt {c+d x^2} \left (-3 a d+b \left (c+d x^2\right )\right )}{3 b^2 d}+\frac {a \sqrt {-b c+a d} \arctan \left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {-b c+a d}}\right )}{b^{5/2}} \]
(Sqrt[c + d*x^2]*(-3*a*d + b*(c + d*x^2)))/(3*b^2*d) + (a*Sqrt[-(b*c) + a* d]*ArcTan[(Sqrt[b]*Sqrt[c + d*x^2])/Sqrt[-(b*c) + a*d]])/b^(5/2)
Time = 0.23 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.11, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {354, 90, 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.
\(\displaystyle \int \frac {x^3 \sqrt {c+d x^2}}{a+b x^2} \, dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int \frac {x^2 \sqrt {d x^2+c}}{b x^2+a}dx^2\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {1}{2} \left (\frac {2 \left (c+d x^2\right )^{3/2}}{3 b d}-\frac {a \int \frac {\sqrt {d x^2+c}}{b x^2+a}dx^2}{b}\right )\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{2} \left (\frac {2 \left (c+d x^2\right )^{3/2}}{3 b d}-\frac {a \left (\frac {(b c-a d) \int \frac {1}{\left (b x^2+a\right ) \sqrt {d x^2+c}}dx^2}{b}+\frac {2 \sqrt {c+d x^2}}{b}\right )}{b}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} \left (\frac {2 \left (c+d x^2\right )^{3/2}}{3 b d}-\frac {a \left (\frac {2 (b c-a d) \int \frac {1}{\frac {b x^4}{d}+a-\frac {b c}{d}}d\sqrt {d x^2+c}}{b d}+\frac {2 \sqrt {c+d x^2}}{b}\right )}{b}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{2} \left (\frac {2 \left (c+d x^2\right )^{3/2}}{3 b d}-\frac {a \left (\frac {2 \sqrt {c+d x^2}}{b}-\frac {2 \sqrt {b c-a d} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{b^{3/2}}\right )}{b}\right )\) |
((2*(c + d*x^2)^(3/2))/(3*b*d) - (a*((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
3.7.77.3.1 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*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p + 2)) Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0]
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 = 3.17 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.94
method | result | size |
pseudoelliptic | \(\frac {-\frac {\sqrt {d \,x^{2}+c}\, \left (-b d \,x^{2}+3 a d -b c \right )}{3}+\frac {a d \left (a d -b c \right ) \arctan \left (\frac {b \sqrt {d \,x^{2}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}}}{d \,b^{2}}\) | \(83\) |
risch | \(-\frac {\left (-b d \,x^{2}+3 a d -b c \right ) \sqrt {d \,x^{2}+c}}{3 d \,b^{2}}+\frac {a \left (a d -b c \right ) \left (-\frac {\ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {d \left (x -\frac {\sqrt {-a b}}{b}\right )^{2}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x -\frac {\sqrt {-a b}}{b}}\right )}{2 b \sqrt {-\frac {a d -b c}{b}}}-\frac {\ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {d \left (x +\frac {\sqrt {-a b}}{b}\right )^{2}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x +\frac {\sqrt {-a b}}{b}}\right )}{2 b \sqrt {-\frac {a d -b c}{b}}}\right )}{b^{2}}\) | \(347\) |
default | \(\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}}}{3 b d}-\frac {a \left (\sqrt {d \left (x -\frac {\sqrt {-a b}}{b}\right )^{2}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}+\frac {\sqrt {d}\, \sqrt {-a b}\, \ln \left (\frac {\frac {d \sqrt {-a b}}{b}+d \left (x -\frac {\sqrt {-a b}}{b}\right )}{\sqrt {d}}+\sqrt {d \left (x -\frac {\sqrt {-a b}}{b}\right )^{2}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}\right )}{b}+\frac {\left (a d -b c \right ) \ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {d \left (x -\frac {\sqrt {-a b}}{b}\right )^{2}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x -\frac {\sqrt {-a b}}{b}}\right )}{b \sqrt {-\frac {a d -b c}{b}}}\right )}{2 b^{2}}-\frac {a \left (\sqrt {d \left (x +\frac {\sqrt {-a b}}{b}\right )^{2}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}-\frac {\sqrt {d}\, \sqrt {-a b}\, \ln \left (\frac {-\frac {d \sqrt {-a b}}{b}+d \left (x +\frac {\sqrt {-a b}}{b}\right )}{\sqrt {d}}+\sqrt {d \left (x +\frac {\sqrt {-a b}}{b}\right )^{2}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}\right )}{b}+\frac {\left (a d -b c \right ) \ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {d \left (x +\frac {\sqrt {-a b}}{b}\right )^{2}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x +\frac {\sqrt {-a b}}{b}}\right )}{b \sqrt {-\frac {a d -b c}{b}}}\right )}{2 b^{2}}\) | \(666\) |
1/d/b^2*(-1/3*(d*x^2+c)^(1/2)*(-b*d*x^2+3*a*d-b*c)+a*d*(a*d-b*c)/((a*d-b*c )*b)^(1/2)*arctan(b*(d*x^2+c)^(1/2)/((a*d-b*c)*b)^(1/2)))
Time = 0.28 (sec) , antiderivative size = 295, normalized size of antiderivative = 3.35 \[ \int \frac {x^3 \sqrt {c+d x^2}}{a+b x^2} \, dx=\left [\frac {3 \, a d \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b^{2} d^{2} x^{4} + 8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2} + 2 \, {\left (4 \, b^{2} c d - 3 \, a b d^{2}\right )} x^{2} + 4 \, {\left (b^{2} d x^{2} + 2 \, b^{2} c - a b d\right )} \sqrt {d x^{2} + c} \sqrt {\frac {b c - a d}{b}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) + 4 \, {\left (b d x^{2} + b c - 3 \, a d\right )} \sqrt {d x^{2} + c}}{12 \, b^{2} d}, \frac {3 \, a d \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {{\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {b c - a d}{b}}}{2 \, {\left (b c^{2} - a c d + {\left (b c d - a d^{2}\right )} x^{2}\right )}}\right ) + 2 \, {\left (b d x^{2} + b c - 3 \, a d\right )} \sqrt {d x^{2} + c}}{6 \, b^{2} d}\right ] \]
[1/12*(3*a*d*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*(b *d*x^2 + b*c - 3*a*d)*sqrt(d*x^2 + c))/(b^2*d), 1/6*(3*a*d*sqrt(-(b*c - a* d)/b)*arctan(-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*(b*d*x^2 + b*c - 3*a*d)*sq rt(d*x^2 + c))/(b^2*d)]
Time = 2.45 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.35 \[ \int \frac {x^3 \sqrt {c+d x^2}}{a+b x^2} \, dx=\begin {cases} \frac {2 \left (- \frac {a d^{2} \sqrt {c + d x^{2}}}{2 b^{2}} + \frac {a d^{2} \left (a d - b c\right ) \operatorname {atan}{\left (\frac {\sqrt {c + d x^{2}}}{\sqrt {\frac {a d - b c}{b}}} \right )}}{2 b^{3} \sqrt {\frac {a d - b c}{b}}} + \frac {d \left (c + d x^{2}\right )^{\frac {3}{2}}}{6 b}\right )}{d^{2}} & \text {for}\: d \neq 0 \\\sqrt {c} \left (- \frac {a \left (\begin {cases} \frac {x^{2}}{a} & \text {for}\: b = 0 \\\frac {\log {\left (a + b x^{2} \right )}}{b} & \text {otherwise} \end {cases}\right )}{2 b} + \frac {x^{2}}{2 b}\right ) & \text {otherwise} \end {cases} \]
Piecewise((2*(-a*d**2*sqrt(c + d*x**2)/(2*b**2) + a*d**2*(a*d - b*c)*atan( sqrt(c + d*x**2)/sqrt((a*d - b*c)/b))/(2*b**3*sqrt((a*d - b*c)/b)) + d*(c + d*x**2)**(3/2)/(6*b))/d**2, Ne(d, 0)), (sqrt(c)*(-a*Piecewise((x**2/a, E q(b, 0)), (log(a + b*x**2)/b, True))/(2*b) + x**2/(2*b)), True))
Exception generated. \[ \int \frac {x^3 \sqrt {c+d x^2}}{a+b x^2} \, dx=\text {Exception raised: ValueError} \]
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 = 96, normalized size of antiderivative = 1.09 \[ \int \frac {x^3 \sqrt {c+d x^2}}{a+b x^2} \, dx=-\frac {{\left (a b c - a^{2} d\right )} \arctan \left (\frac {\sqrt {d x^{2} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d} b^{2}} + \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} d^{2} - 3 \, \sqrt {d x^{2} + c} a b d^{3}}{3 \, b^{3} d^{3}} \]
-(a*b*c - a^2*d)*arctan(sqrt(d*x^2 + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(-b^2 *c + a*b*d)*b^2) + 1/3*((d*x^2 + c)^(3/2)*b^2*d^2 - 3*sqrt(d*x^2 + c)*a*b* d^3)/(b^3*d^3)
Time = 5.26 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.98 \[ \int \frac {x^3 \sqrt {c+d x^2}}{a+b x^2} \, dx=\frac {{\left (d\,x^2+c\right )}^{3/2}}{3\,b\,d}-\frac {a\,\sqrt {d\,x^2+c}}{b^2}+\frac {a\,\mathrm {atan}\left (\frac {a\,\sqrt {b}\,\sqrt {d\,x^2+c}\,\sqrt {a\,d-b\,c}}{a^2\,d-a\,b\,c}\right )\,\sqrt {a\,d-b\,c}}{b^{5/2}} \]