Integrand size = 22, antiderivative size = 140 \[ \int \frac {x}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}} \, dx=-\frac {5 d}{6 (b c-a d)^2 \left (c+d x^2\right )^{3/2}}-\frac {1}{2 (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}}-\frac {5 b d}{2 (b c-a d)^3 \sqrt {c+d x^2}}+\frac {5 b^{3/2} d \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{2 (b c-a d)^{7/2}} \]
-5/6*d/(-a*d+b*c)^2/(d*x^2+c)^(3/2)-1/2/(-a*d+b*c)/(b*x^2+a)/(d*x^2+c)^(3/ 2)+5/2*b^(3/2)*d*arctanh(b^(1/2)*(d*x^2+c)^(1/2)/(-a*d+b*c)^(1/2))/(-a*d+b *c)^(7/2)-5/2*b*d/(-a*d+b*c)^3/(d*x^2+c)^(1/2)
Time = 0.32 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.98 \[ \int \frac {x}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}} \, dx=\frac {2 a^2 d^2-2 a b d \left (7 c+5 d x^2\right )-b^2 \left (3 c^2+20 c d x^2+15 d^2 x^4\right )}{6 (b c-a d)^3 \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}}+\frac {5 b^{3/2} d \arctan \left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {-b c+a d}}\right )}{2 (-b c+a d)^{7/2}} \]
(2*a^2*d^2 - 2*a*b*d*(7*c + 5*d*x^2) - b^2*(3*c^2 + 20*c*d*x^2 + 15*d^2*x^ 4))/(6*(b*c - a*d)^3*(a + b*x^2)*(c + d*x^2)^(3/2)) + (5*b^(3/2)*d*ArcTan[ (Sqrt[b]*Sqrt[c + d*x^2])/Sqrt[-(b*c) + a*d]])/(2*(-(b*c) + a*d)^(7/2))
Time = 0.25 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.16, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {353, 52, 61, 61, 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.
| \(\displaystyle \int \frac {x}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 353 |
| \(\displaystyle \frac {1}{2} \int \frac {1}{\left (b x^2+a\right )^2 \left (d x^2+c\right )^{5/2}}dx^2\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {1}{2} \left (-\frac {5 d \int \frac {1}{\left (b x^2+a\right ) \left (d x^2+c\right )^{5/2}}dx^2}{2 (b c-a d)}-\frac {1}{\left (a+b x^2\right ) \left (c+d x^2\right )^{3/2} (b c-a d)}\right )\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {1}{2} \left (-\frac {5 d \left (\frac {b \int \frac {1}{\left (b x^2+a\right ) \left (d x^2+c\right )^{3/2}}dx^2}{b c-a d}+\frac {2}{3 \left (c+d x^2\right )^{3/2} (b c-a d)}\right )}{2 (b c-a d)}-\frac {1}{\left (a+b x^2\right ) \left (c+d x^2\right )^{3/2} (b c-a d)}\right )\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {1}{2} \left (-\frac {5 d \left (\frac {b \left (\frac {b \int \frac {1}{\left (b x^2+a\right ) \sqrt {d x^2+c}}dx^2}{b c-a d}+\frac {2}{\sqrt {c+d x^2} (b c-a d)}\right )}{b c-a d}+\frac {2}{3 \left (c+d x^2\right )^{3/2} (b c-a d)}\right )}{2 (b c-a d)}-\frac {1}{\left (a+b x^2\right ) \left (c+d x^2\right )^{3/2} (b c-a d)}\right )\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {1}{2} \left (-\frac {5 d \left (\frac {b \left (\frac {2 b \int \frac {1}{\frac {b x^4}{d}+a-\frac {b c}{d}}d\sqrt {d x^2+c}}{d (b c-a d)}+\frac {2}{\sqrt {c+d x^2} (b c-a d)}\right )}{b c-a d}+\frac {2}{3 \left (c+d x^2\right )^{3/2} (b c-a d)}\right )}{2 (b c-a d)}-\frac {1}{\left (a+b x^2\right ) \left (c+d x^2\right )^{3/2} (b c-a d)}\right )\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {1}{2} \left (-\frac {5 d \left (\frac {b \left (\frac {2}{\sqrt {c+d x^2} (b c-a d)}-\frac {2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{(b c-a d)^{3/2}}\right )}{b c-a d}+\frac {2}{3 \left (c+d x^2\right )^{3/2} (b c-a d)}\right )}{2 (b c-a d)}-\frac {1}{\left (a+b x^2\right ) \left (c+d x^2\right )^{3/2} (b c-a d)}\right )\) |
(-(1/((b*c - a*d)*(a + b*x^2)*(c + d*x^2)^(3/2))) - (5*d*(2/(3*(b*c - a*d) *(c + d*x^2)^(3/2)) + (b*(2/((b*c - a*d)*Sqrt[c + d*x^2]) - (2*Sqrt[b]*Arc Tanh[(Sqrt[b]*Sqrt[c + d*x^2])/Sqrt[b*c - a*d]])/(b*c - a*d)^(3/2)))/(b*c - a*d)))/(2*(b*c - a*d)))/2
3.8.79.3.1 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 + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0 ] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[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_)^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_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[1/2 Subst[Int[(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]
Time = 3.10 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.96
| method | result | size |
| pseudoelliptic | \(d \left (\frac {\sqrt {d \,x^{2}+c}\, b^{2}}{2 \left (b \,x^{2}+a \right ) d \left (a d -b c \right )^{3}}+\frac {5 \arctan \left (\frac {b \sqrt {d \,x^{2}+c}}{\sqrt {\left (a d -b c \right ) b}}\right ) b^{2}}{2 \sqrt {\left (a d -b c \right ) b}\, \left (a d -b c \right )^{3}}-\frac {1}{3 \left (a d -b c \right )^{2} \left (d \,x^{2}+c \right )^{\frac {3}{2}}}+\frac {2 b}{\left (a d -b c \right )^{3} \sqrt {d \,x^{2}+c}}\right )\) | \(134\) |
| default | \(\text {Expression too large to display}\) | \(2101\) |
d*(1/2*(d*x^2+c)^(1/2)*b^2/(b*x^2+a)/d/(a*d-b*c)^3+5/2/((a*d-b*c)*b)^(1/2) *arctan(b*(d*x^2+c)^(1/2)/((a*d-b*c)*b)^(1/2))/(a*d-b*c)^3*b^2-1/3/(a*d-b* c)^2/(d*x^2+c)^(3/2)+2/(a*d-b*c)^3*b/(d*x^2+c)^(1/2))
Leaf count of result is larger than twice the leaf count of optimal. 405 vs. \(2 (116) = 232\).
Time = 0.45 (sec) , antiderivative size = 895, normalized size of antiderivative = 6.39 \[ \int \frac {x}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}} \, dx=\left [-\frac {15 \, {\left (b^{2} d^{3} x^{6} + a b c^{2} d + {\left (2 \, b^{2} c d^{2} + a b d^{3}\right )} x^{4} + {\left (b^{2} c^{2} d + 2 \, a b c d^{2}\right )} x^{2}\right )} \sqrt {\frac {b}{b c - a d}} \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 (2 \, b^{2} c^{2} - 3 \, a b c d + a^{2} d^{2} + {\left (b^{2} c d - a b d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c} \sqrt {\frac {b}{b c - a d}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) + 4 \, {\left (15 \, b^{2} d^{2} x^{4} + 3 \, b^{2} c^{2} + 14 \, a b c d - 2 \, a^{2} d^{2} + 10 \, {\left (2 \, b^{2} c d + a b d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{24 \, {\left (a b^{3} c^{5} - 3 \, a^{2} b^{2} c^{4} d + 3 \, a^{3} b c^{3} d^{2} - a^{4} c^{2} d^{3} + {\left (b^{4} c^{3} d^{2} - 3 \, a b^{3} c^{2} d^{3} + 3 \, a^{2} b^{2} c d^{4} - a^{3} b d^{5}\right )} x^{6} + {\left (2 \, b^{4} c^{4} d - 5 \, a b^{3} c^{3} d^{2} + 3 \, a^{2} b^{2} c^{2} d^{3} + a^{3} b c d^{4} - a^{4} d^{5}\right )} x^{4} + {\left (b^{4} c^{5} - a b^{3} c^{4} d - 3 \, a^{2} b^{2} c^{3} d^{2} + 5 \, a^{3} b c^{2} d^{3} - 2 \, a^{4} c d^{4}\right )} x^{2}\right )}}, -\frac {15 \, {\left (b^{2} d^{3} x^{6} + a b c^{2} d + {\left (2 \, b^{2} c d^{2} + a b d^{3}\right )} x^{4} + {\left (b^{2} c^{2} d + 2 \, a b c d^{2}\right )} x^{2}\right )} \sqrt {-\frac {b}{b c - a d}} \arctan \left (\frac {{\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {b}{b c - a d}}}{2 \, {\left (b d x^{2} + b c\right )}}\right ) + 2 \, {\left (15 \, b^{2} d^{2} x^{4} + 3 \, b^{2} c^{2} + 14 \, a b c d - 2 \, a^{2} d^{2} + 10 \, {\left (2 \, b^{2} c d + a b d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{12 \, {\left (a b^{3} c^{5} - 3 \, a^{2} b^{2} c^{4} d + 3 \, a^{3} b c^{3} d^{2} - a^{4} c^{2} d^{3} + {\left (b^{4} c^{3} d^{2} - 3 \, a b^{3} c^{2} d^{3} + 3 \, a^{2} b^{2} c d^{4} - a^{3} b d^{5}\right )} x^{6} + {\left (2 \, b^{4} c^{4} d - 5 \, a b^{3} c^{3} d^{2} + 3 \, a^{2} b^{2} c^{2} d^{3} + a^{3} b c d^{4} - a^{4} d^{5}\right )} x^{4} + {\left (b^{4} c^{5} - a b^{3} c^{4} d - 3 \, a^{2} b^{2} c^{3} d^{2} + 5 \, a^{3} b c^{2} d^{3} - 2 \, a^{4} c d^{4}\right )} x^{2}\right )}}\right ] \]
[-1/24*(15*(b^2*d^3*x^6 + a*b*c^2*d + (2*b^2*c*d^2 + a*b*d^3)*x^4 + (b^2*c ^2*d + 2*a*b*c*d^2)*x^2)*sqrt(b/(b*c - a*d))*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*(2*b^2*c^2 - 3*a *b*c*d + a^2*d^2 + (b^2*c*d - a*b*d^2)*x^2)*sqrt(d*x^2 + c)*sqrt(b/(b*c - a*d)))/(b^2*x^4 + 2*a*b*x^2 + a^2)) + 4*(15*b^2*d^2*x^4 + 3*b^2*c^2 + 14*a *b*c*d - 2*a^2*d^2 + 10*(2*b^2*c*d + a*b*d^2)*x^2)*sqrt(d*x^2 + c))/(a*b^3 *c^5 - 3*a^2*b^2*c^4*d + 3*a^3*b*c^3*d^2 - a^4*c^2*d^3 + (b^4*c^3*d^2 - 3* a*b^3*c^2*d^3 + 3*a^2*b^2*c*d^4 - a^3*b*d^5)*x^6 + (2*b^4*c^4*d - 5*a*b^3* c^3*d^2 + 3*a^2*b^2*c^2*d^3 + a^3*b*c*d^4 - a^4*d^5)*x^4 + (b^4*c^5 - a*b^ 3*c^4*d - 3*a^2*b^2*c^3*d^2 + 5*a^3*b*c^2*d^3 - 2*a^4*c*d^4)*x^2), -1/12*( 15*(b^2*d^3*x^6 + a*b*c^2*d + (2*b^2*c*d^2 + a*b*d^3)*x^4 + (b^2*c^2*d + 2 *a*b*c*d^2)*x^2)*sqrt(-b/(b*c - a*d))*arctan(1/2*(b*d*x^2 + 2*b*c - a*d)*s qrt(d*x^2 + c)*sqrt(-b/(b*c - a*d))/(b*d*x^2 + b*c)) + 2*(15*b^2*d^2*x^4 + 3*b^2*c^2 + 14*a*b*c*d - 2*a^2*d^2 + 10*(2*b^2*c*d + a*b*d^2)*x^2)*sqrt(d *x^2 + c))/(a*b^3*c^5 - 3*a^2*b^2*c^4*d + 3*a^3*b*c^3*d^2 - a^4*c^2*d^3 + (b^4*c^3*d^2 - 3*a*b^3*c^2*d^3 + 3*a^2*b^2*c*d^4 - a^3*b*d^5)*x^6 + (2*b^4 *c^4*d - 5*a*b^3*c^3*d^2 + 3*a^2*b^2*c^2*d^3 + a^3*b*c*d^4 - a^4*d^5)*x^4 + (b^4*c^5 - a*b^3*c^4*d - 3*a^2*b^2*c^3*d^2 + 5*a^3*b*c^2*d^3 - 2*a^4*c*d ^4)*x^2)]
\[ \int \frac {x}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}} \, dx=\int \frac {x}{\left (a + b x^{2}\right )^{2} \left (c + d x^{2}\right )^{\frac {5}{2}}}\, dx \]
Exception generated. \[ \int \frac {x}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}} \, dx=\text {Exception raised: ValueError} \]
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 = 226, normalized size of antiderivative = 1.61 \[ \int \frac {x}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}} \, dx=-\frac {5 \, b^{2} d \arctan \left (\frac {\sqrt {d x^{2} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{2 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {-b^{2} c + a b d}} - \frac {\sqrt {d x^{2} + c} b^{2} d}{2 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} {\left ({\left (d x^{2} + c\right )} b - b c + a d\right )}} - \frac {6 \, {\left (d x^{2} + c\right )} b d + b c d - a d^{2}}{3 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} {\left (d x^{2} + c\right )}^{\frac {3}{2}}} \]
-5/2*b^2*d*arctan(sqrt(d*x^2 + c)*b/sqrt(-b^2*c + a*b*d))/((b^3*c^3 - 3*a* b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*sqrt(-b^2*c + a*b*d)) - 1/2*sqrt(d*x^ 2 + c)*b^2*d/((b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*((d*x^2 + c)*b - b*c + a*d)) - 1/3*(6*(d*x^2 + c)*b*d + b*c*d - a*d^2)/((b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*(d*x^2 + c)^(3/2))
Time = 5.73 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.22 \[ \int \frac {x}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}} \, dx=\frac {\frac {5\,b^2\,d\,{\left (d\,x^2+c\right )}^2}{2\,{\left (a\,d-b\,c\right )}^3}-\frac {d}{3\,\left (a\,d-b\,c\right )}+\frac {5\,b\,d\,\left (d\,x^2+c\right )}{3\,{\left (a\,d-b\,c\right )}^2}}{b\,{\left (d\,x^2+c\right )}^{5/2}+{\left (d\,x^2+c\right )}^{3/2}\,\left (a\,d-b\,c\right )}+\frac {5\,b^{3/2}\,d\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d\,x^2+c}\,\left (a^3\,d^3-3\,a^2\,b\,c\,d^2+3\,a\,b^2\,c^2\,d-b^3\,c^3\right )}{{\left (a\,d-b\,c\right )}^{7/2}}\right )}{2\,{\left (a\,d-b\,c\right )}^{7/2}} \]
((5*b^2*d*(c + d*x^2)^2)/(2*(a*d - b*c)^3) - d/(3*(a*d - b*c)) + (5*b*d*(c + d*x^2))/(3*(a*d - b*c)^2))/(b*(c + d*x^2)^(5/2) + (c + d*x^2)^(3/2)*(a* d - b*c)) + (5*b^(3/2)*d*atan((b^(1/2)*(c + d*x^2)^(1/2)*(a^3*d^3 - b^3*c^ 3 + 3*a*b^2*c^2*d - 3*a^2*b*c*d^2))/(a*d - b*c)^(7/2)))/(2*(a*d - b*c)^(7/ 2))