Integrand size = 24, antiderivative size = 609 \[ \int \frac {x^{5/2}}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^2} \, dx=-\frac {d x^{3/2}}{(b c-a d)^2 \left (c+d x^2\right )}-\frac {x^{3/2}}{2 (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )}-\frac {\sqrt [4]{b} (3 b c+5 a d) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} \sqrt [4]{a} (b c-a d)^3}+\frac {\sqrt [4]{b} (3 b c+5 a d) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} \sqrt [4]{a} (b c-a d)^3}+\frac {\sqrt [4]{d} (5 b c+3 a d) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{4 \sqrt {2} \sqrt [4]{c} (b c-a d)^3}-\frac {\sqrt [4]{d} (5 b c+3 a d) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{4 \sqrt {2} \sqrt [4]{c} (b c-a d)^3}+\frac {\sqrt [4]{b} (3 b c+5 a d) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{8 \sqrt {2} \sqrt [4]{a} (b c-a d)^3}-\frac {\sqrt [4]{b} (3 b c+5 a d) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{8 \sqrt {2} \sqrt [4]{a} (b c-a d)^3}-\frac {\sqrt [4]{d} (5 b c+3 a d) \log \left (\sqrt {c}-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {d} x\right )}{8 \sqrt {2} \sqrt [4]{c} (b c-a d)^3}+\frac {\sqrt [4]{d} (5 b c+3 a d) \log \left (\sqrt {c}+\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {d} x\right )}{8 \sqrt {2} \sqrt [4]{c} (b c-a d)^3} \]
-d*x^(3/2)/(-a*d+b*c)^2/(d*x^2+c)-1/2*x^(3/2)/(-a*d+b*c)/(b*x^2+a)/(d*x^2+ c)-1/8*b^(1/4)*(5*a*d+3*b*c)*arctan(1-b^(1/4)*2^(1/2)*x^(1/2)/a^(1/4))/a^( 1/4)/(-a*d+b*c)^3*2^(1/2)+1/8*b^(1/4)*(5*a*d+3*b*c)*arctan(1+b^(1/4)*2^(1/ 2)*x^(1/2)/a^(1/4))/a^(1/4)/(-a*d+b*c)^3*2^(1/2)+1/8*d^(1/4)*(3*a*d+5*b*c) *arctan(1-d^(1/4)*2^(1/2)*x^(1/2)/c^(1/4))/c^(1/4)/(-a*d+b*c)^3*2^(1/2)-1/ 8*d^(1/4)*(3*a*d+5*b*c)*arctan(1+d^(1/4)*2^(1/2)*x^(1/2)/c^(1/4))/c^(1/4)/ (-a*d+b*c)^3*2^(1/2)+1/16*b^(1/4)*(5*a*d+3*b*c)*ln(a^(1/2)+x*b^(1/2)-a^(1/ 4)*b^(1/4)*2^(1/2)*x^(1/2))/a^(1/4)/(-a*d+b*c)^3*2^(1/2)-1/16*b^(1/4)*(5*a *d+3*b*c)*ln(a^(1/2)+x*b^(1/2)+a^(1/4)*b^(1/4)*2^(1/2)*x^(1/2))/a^(1/4)/(- a*d+b*c)^3*2^(1/2)-1/16*d^(1/4)*(3*a*d+5*b*c)*ln(c^(1/2)+x*d^(1/2)-c^(1/4) *d^(1/4)*2^(1/2)*x^(1/2))/c^(1/4)/(-a*d+b*c)^3*2^(1/2)+1/16*d^(1/4)*(3*a*d +5*b*c)*ln(c^(1/2)+x*d^(1/2)+c^(1/4)*d^(1/4)*2^(1/2)*x^(1/2))/c^(1/4)/(-a* d+b*c)^3*2^(1/2)
Time = 1.52 (sec) , antiderivative size = 340, normalized size of antiderivative = 0.56 \[ \int \frac {x^{5/2}}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^2} \, dx=\frac {1}{8} \left (-\frac {4 x^{3/2} \left (a d+b \left (c+2 d x^2\right )\right )}{(b c-a d)^2 \left (a+b x^2\right ) \left (c+d x^2\right )}+\frac {\sqrt {2} \sqrt [4]{b} (3 b c+5 a d) \arctan \left (\frac {\sqrt {a}-\sqrt {b} x}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}\right )}{\sqrt [4]{a} (-b c+a d)^3}+\frac {\sqrt {2} \sqrt [4]{d} (5 b c+3 a d) \arctan \left (\frac {\sqrt {c}-\sqrt {d} x}{\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}}\right )}{\sqrt [4]{c} (b c-a d)^3}+\frac {\sqrt {2} \sqrt [4]{b} (3 b c+5 a d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}{\sqrt {a}+\sqrt {b} x}\right )}{\sqrt [4]{a} (-b c+a d)^3}+\frac {\sqrt {2} \sqrt [4]{d} (5 b c+3 a d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}}{\sqrt {c}+\sqrt {d} x}\right )}{\sqrt [4]{c} (b c-a d)^3}\right ) \]
((-4*x^(3/2)*(a*d + b*(c + 2*d*x^2)))/((b*c - a*d)^2*(a + b*x^2)*(c + d*x^ 2)) + (Sqrt[2]*b^(1/4)*(3*b*c + 5*a*d)*ArcTan[(Sqrt[a] - Sqrt[b]*x)/(Sqrt[ 2]*a^(1/4)*b^(1/4)*Sqrt[x])])/(a^(1/4)*(-(b*c) + a*d)^3) + (Sqrt[2]*d^(1/4 )*(5*b*c + 3*a*d)*ArcTan[(Sqrt[c] - Sqrt[d]*x)/(Sqrt[2]*c^(1/4)*d^(1/4)*Sq rt[x])])/(c^(1/4)*(b*c - a*d)^3) + (Sqrt[2]*b^(1/4)*(3*b*c + 5*a*d)*ArcTan h[(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x])/(Sqrt[a] + Sqrt[b]*x)])/(a^(1/4)*(-(b* c) + a*d)^3) + (Sqrt[2]*d^(1/4)*(5*b*c + 3*a*d)*ArcTanh[(Sqrt[2]*c^(1/4)*d ^(1/4)*Sqrt[x])/(Sqrt[c] + Sqrt[d]*x)])/(c^(1/4)*(b*c - a*d)^3))/8
Time = 0.78 (sec) , antiderivative size = 638, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {368, 971, 1049, 27, 1054, 2009}
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.
\(\displaystyle \int \frac {x^{5/2}}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^2} \, dx\) |
\(\Big \downarrow \) 368 |
\(\displaystyle 2 \int \frac {x^3}{\left (b x^2+a\right )^2 \left (d x^2+c\right )^2}d\sqrt {x}\) |
\(\Big \downarrow \) 971 |
\(\displaystyle 2 \left (\frac {\int \frac {x \left (3 c-5 d x^2\right )}{\left (b x^2+a\right ) \left (d x^2+c\right )^2}d\sqrt {x}}{4 (b c-a d)}-\frac {x^{3/2}}{4 \left (a+b x^2\right ) \left (c+d x^2\right ) (b c-a d)}\right )\) |
\(\Big \downarrow \) 1049 |
\(\displaystyle 2 \left (\frac {\frac {\int \frac {4 c x \left (3 (b c+a d)-2 b d x^2\right )}{\left (b x^2+a\right ) \left (d x^2+c\right )}d\sqrt {x}}{4 c (b c-a d)}-\frac {2 d x^{3/2}}{\left (c+d x^2\right ) (b c-a d)}}{4 (b c-a d)}-\frac {x^{3/2}}{4 \left (a+b x^2\right ) \left (c+d x^2\right ) (b c-a d)}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 \left (\frac {\frac {\int \frac {x \left (3 (b c+a d)-2 b d x^2\right )}{\left (b x^2+a\right ) \left (d x^2+c\right )}d\sqrt {x}}{b c-a d}-\frac {2 d x^{3/2}}{\left (c+d x^2\right ) (b c-a d)}}{4 (b c-a d)}-\frac {x^{3/2}}{4 \left (a+b x^2\right ) \left (c+d x^2\right ) (b c-a d)}\right )\) |
\(\Big \downarrow \) 1054 |
\(\displaystyle 2 \left (\frac {\frac {\int \left (\frac {b (3 b c+5 a d) x}{(b c-a d) \left (b x^2+a\right )}-\frac {d (5 b c+3 a d) x}{(b c-a d) \left (d x^2+c\right )}\right )d\sqrt {x}}{b c-a d}-\frac {2 d x^{3/2}}{\left (c+d x^2\right ) (b c-a d)}}{4 (b c-a d)}-\frac {x^{3/2}}{4 \left (a+b x^2\right ) \left (c+d x^2\right ) (b c-a d)}\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \left (\frac {\frac {-\frac {\sqrt [4]{b} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right ) (5 a d+3 b c)}{2 \sqrt {2} \sqrt [4]{a} (b c-a d)}+\frac {\sqrt [4]{b} \arctan \left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right ) (5 a d+3 b c)}{2 \sqrt {2} \sqrt [4]{a} (b c-a d)}+\frac {\sqrt [4]{d} (3 a d+5 b c) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{2 \sqrt {2} \sqrt [4]{c} (b c-a d)}-\frac {\sqrt [4]{d} (3 a d+5 b c) \arctan \left (\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}+1\right )}{2 \sqrt {2} \sqrt [4]{c} (b c-a d)}+\frac {\sqrt [4]{b} (5 a d+3 b c) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{4 \sqrt {2} \sqrt [4]{a} (b c-a d)}-\frac {\sqrt [4]{b} (5 a d+3 b c) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{4 \sqrt {2} \sqrt [4]{a} (b c-a d)}-\frac {\sqrt [4]{d} (3 a d+5 b c) \log \left (-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{4 \sqrt {2} \sqrt [4]{c} (b c-a d)}+\frac {\sqrt [4]{d} (3 a d+5 b c) \log \left (\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{4 \sqrt {2} \sqrt [4]{c} (b c-a d)}}{b c-a d}-\frac {2 d x^{3/2}}{\left (c+d x^2\right ) (b c-a d)}}{4 (b c-a d)}-\frac {x^{3/2}}{4 \left (a+b x^2\right ) \left (c+d x^2\right ) (b c-a d)}\right )\) |
2*(-1/4*x^(3/2)/((b*c - a*d)*(a + b*x^2)*(c + d*x^2)) + ((-2*d*x^(3/2))/(( b*c - a*d)*(c + d*x^2)) + (-1/2*(b^(1/4)*(3*b*c + 5*a*d)*ArcTan[1 - (Sqrt[ 2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(Sqrt[2]*a^(1/4)*(b*c - a*d)) + (b^(1/4)*(3* b*c + 5*a*d)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(2*Sqrt[2]*a^( 1/4)*(b*c - a*d)) + (d^(1/4)*(5*b*c + 3*a*d)*ArcTan[1 - (Sqrt[2]*d^(1/4)*S qrt[x])/c^(1/4)])/(2*Sqrt[2]*c^(1/4)*(b*c - a*d)) - (d^(1/4)*(5*b*c + 3*a* d)*ArcTan[1 + (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/(2*Sqrt[2]*c^(1/4)*(b*c - a*d)) + (b^(1/4)*(3*b*c + 5*a*d)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*S qrt[x] + Sqrt[b]*x])/(4*Sqrt[2]*a^(1/4)*(b*c - a*d)) - (b^(1/4)*(3*b*c + 5 *a*d)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(4*Sqrt[ 2]*a^(1/4)*(b*c - a*d)) - (d^(1/4)*(5*b*c + 3*a*d)*Log[Sqrt[c] - Sqrt[2]*c ^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x])/(4*Sqrt[2]*c^(1/4)*(b*c - a*d)) + (d^ (1/4)*(5*b*c + 3*a*d)*Log[Sqrt[c] + Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt [d]*x])/(4*Sqrt[2]*c^(1/4)*(b*c - a*d)))/(b*c - a*d))/(4*(b*c - a*d)))
3.5.89.3.1 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[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_) , x_Symbol] :> With[{k = Denominator[m]}, Simp[k/e Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(k*2)/e^2))^p*(c + d*(x^(k*2)/e^2))^q, x], x, (e*x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && FractionQ[m ] && IntegerQ[p]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ ))^(q_), x_Symbol] :> Simp[e^(n - 1)*(e*x)^(m - n + 1)*(a + b*x^n)^(p + 1)* ((c + d*x^n)^(q + 1)/(n*(b*c - a*d)*(p + 1))), x] - Simp[e^n/(n*(b*c - a*d) *(p + 1)) Int[(e*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(m - n + 1) + d*(m + n*(p + q + 1) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e , q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && GeQ[n, m - n + 1] && GtQ[m - n + 1, 0] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]
Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ ))^(q_)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Simp[(-(b*e - a*f))*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*g*n*(b*c - a*d)*(p + 1))) , x] + Simp[1/(a*n*(b*c - a*d)*(p + 1)) Int[(g*x)^m*(a + b*x^n)^(p + 1)*( c + d*x^n)^q*Simp[c*(b*e - a*f)*(m + 1) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m, q}, x] && IGtQ[n, 0] && LtQ[p, -1]
Int[(((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((e_) + (f_.)*(x_)^(n _)))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegrand[(g*x)^m*(a + b*x^n)^p*((e + f*x^n)/(c + d*x^n)), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && IGtQ[n, 0]
Time = 4.13 (sec) , antiderivative size = 302, normalized size of antiderivative = 0.50
method | result | size |
derivativedivides | \(-\frac {2 b \left (\frac {\left (\frac {a d}{4}-\frac {b c}{4}\right ) x^{\frac {3}{2}}}{b \,x^{2}+a}+\frac {\left (\frac {5 a d}{4}+\frac {3 b c}{4}\right ) \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}{x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{8 b \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{\left (a d -b c \right )^{3}}+\frac {2 d \left (\frac {\left (-\frac {a d}{4}+\frac {b c}{4}\right ) x^{\frac {3}{2}}}{d \,x^{2}+c}+\frac {\left (\frac {5 b c}{4}+\frac {3 a d}{4}\right ) \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {c}{d}}}{x +\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {c}{d}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )\right )}{8 d \left (\frac {c}{d}\right )^{\frac {1}{4}}}\right )}{\left (a d -b c \right )^{3}}\) | \(302\) |
default | \(-\frac {2 b \left (\frac {\left (\frac {a d}{4}-\frac {b c}{4}\right ) x^{\frac {3}{2}}}{b \,x^{2}+a}+\frac {\left (\frac {5 a d}{4}+\frac {3 b c}{4}\right ) \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}{x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{8 b \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{\left (a d -b c \right )^{3}}+\frac {2 d \left (\frac {\left (-\frac {a d}{4}+\frac {b c}{4}\right ) x^{\frac {3}{2}}}{d \,x^{2}+c}+\frac {\left (\frac {5 b c}{4}+\frac {3 a d}{4}\right ) \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {c}{d}}}{x +\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {c}{d}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )\right )}{8 d \left (\frac {c}{d}\right )^{\frac {1}{4}}}\right )}{\left (a d -b c \right )^{3}}\) | \(302\) |
-2*b/(a*d-b*c)^3*((1/4*a*d-1/4*b*c)*x^(3/2)/(b*x^2+a)+1/8*(5/4*a*d+3/4*b*c )/b/(a/b)^(1/4)*2^(1/2)*(ln((x-(a/b)^(1/4)*x^(1/2)*2^(1/2)+(a/b)^(1/2))/(x +(a/b)^(1/4)*x^(1/2)*2^(1/2)+(a/b)^(1/2)))+2*arctan(2^(1/2)/(a/b)^(1/4)*x^ (1/2)+1)+2*arctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)-1)))+2*d/(a*d-b*c)^3*((-1/4* a*d+1/4*b*c)*x^(3/2)/(d*x^2+c)+1/8*(5/4*b*c+3/4*a*d)/d/(c/d)^(1/4)*2^(1/2) *(ln((x-(c/d)^(1/4)*x^(1/2)*2^(1/2)+(c/d)^(1/2))/(x+(c/d)^(1/4)*x^(1/2)*2^ (1/2)+(c/d)^(1/2)))+2*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)+1)+2*arctan(2^(1/ 2)/(c/d)^(1/4)*x^(1/2)-1)))
Result contains complex when optimal does not.
Time = 14.89 (sec) , antiderivative size = 5884, normalized size of antiderivative = 9.66 \[ \int \frac {x^{5/2}}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^2} \, dx=\text {Too large to display} \]
Timed out. \[ \int \frac {x^{5/2}}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^2} \, dx=\text {Timed out} \]
Time = 0.30 (sec) , antiderivative size = 567, normalized size of antiderivative = 0.93 \[ \int \frac {x^{5/2}}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^2} \, dx=\frac {{\left (3 \, b^{2} c + 5 \, a b d\right )} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} + 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} - 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} - \frac {\sqrt {2} \log \left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}}\right )}}{16 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )}} - \frac {{\left (5 \, b c d + 3 \, a d^{2}\right )} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} + 2 \, \sqrt {d} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {c} \sqrt {d}}}\right )}{\sqrt {\sqrt {c} \sqrt {d}} \sqrt {d}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} - 2 \, \sqrt {d} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {c} \sqrt {d}}}\right )}{\sqrt {\sqrt {c} \sqrt {d}} \sqrt {d}} - \frac {\sqrt {2} \log \left (\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} \sqrt {x} + \sqrt {d} x + \sqrt {c}\right )}{c^{\frac {1}{4}} d^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} \sqrt {x} + \sqrt {d} x + \sqrt {c}\right )}{c^{\frac {1}{4}} d^{\frac {3}{4}}}\right )}}{16 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )}} - \frac {2 \, b d x^{\frac {7}{2}} + {\left (b c + a d\right )} x^{\frac {3}{2}}}{2 \, {\left (a b^{2} c^{3} - 2 \, a^{2} b c^{2} d + a^{3} c d^{2} + {\left (b^{3} c^{2} d - 2 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} x^{4} + {\left (b^{3} c^{3} - a b^{2} c^{2} d - a^{2} b c d^{2} + a^{3} d^{3}\right )} x^{2}\right )}} \]
1/16*(3*b^2*c + 5*a*b*d)*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*a^(1/4)*b^ (1/4) + 2*sqrt(b)*sqrt(x))/sqrt(sqrt(a)*sqrt(b)))/(sqrt(sqrt(a)*sqrt(b))*s qrt(b)) + 2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*a^(1/4)*b^(1/4) - 2*sqrt( b)*sqrt(x))/sqrt(sqrt(a)*sqrt(b)))/(sqrt(sqrt(a)*sqrt(b))*sqrt(b)) - sqrt( 2)*log(sqrt(2)*a^(1/4)*b^(1/4)*sqrt(x) + sqrt(b)*x + sqrt(a))/(a^(1/4)*b^( 3/4)) + sqrt(2)*log(-sqrt(2)*a^(1/4)*b^(1/4)*sqrt(x) + sqrt(b)*x + sqrt(a) )/(a^(1/4)*b^(3/4)))/(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3) - 1/16*(5*b*c*d + 3*a*d^2)*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*c^(1/4)*d ^(1/4) + 2*sqrt(d)*sqrt(x))/sqrt(sqrt(c)*sqrt(d)))/(sqrt(sqrt(c)*sqrt(d))* sqrt(d)) + 2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*c^(1/4)*d^(1/4) - 2*sqrt (d)*sqrt(x))/sqrt(sqrt(c)*sqrt(d)))/(sqrt(sqrt(c)*sqrt(d))*sqrt(d)) - sqrt (2)*log(sqrt(2)*c^(1/4)*d^(1/4)*sqrt(x) + sqrt(d)*x + sqrt(c))/(c^(1/4)*d^ (3/4)) + sqrt(2)*log(-sqrt(2)*c^(1/4)*d^(1/4)*sqrt(x) + sqrt(d)*x + sqrt(c ))/(c^(1/4)*d^(3/4)))/(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3) - 1/2*(2*b*d*x^(7/2) + (b*c + a*d)*x^(3/2))/(a*b^2*c^3 - 2*a^2*b*c^2*d + a ^3*c*d^2 + (b^3*c^2*d - 2*a*b^2*c*d^2 + a^2*b*d^3)*x^4 + (b^3*c^3 - a*b^2* c^2*d - a^2*b*c*d^2 + a^3*d^3)*x^2)
Leaf count of result is larger than twice the leaf count of optimal. 952 vs. \(2 (459) = 918\).
Time = 0.46 (sec) , antiderivative size = 952, normalized size of antiderivative = 1.56 \[ \int \frac {x^{5/2}}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^2} \, dx=\text {Too large to display} \]
1/4*(3*(a*b^3)^(3/4)*b*c + 5*(a*b^3)^(3/4)*a*d)*arctan(1/2*sqrt(2)*(sqrt(2 )*(a/b)^(1/4) + 2*sqrt(x))/(a/b)^(1/4))/(sqrt(2)*a*b^5*c^3 - 3*sqrt(2)*a^2 *b^4*c^2*d + 3*sqrt(2)*a^3*b^3*c*d^2 - sqrt(2)*a^4*b^2*d^3) + 1/4*(3*(a*b^ 3)^(3/4)*b*c + 5*(a*b^3)^(3/4)*a*d)*arctan(-1/2*sqrt(2)*(sqrt(2)*(a/b)^(1/ 4) - 2*sqrt(x))/(a/b)^(1/4))/(sqrt(2)*a*b^5*c^3 - 3*sqrt(2)*a^2*b^4*c^2*d + 3*sqrt(2)*a^3*b^3*c*d^2 - sqrt(2)*a^4*b^2*d^3) - 1/4*(5*(c*d^3)^(3/4)*b* c + 3*(c*d^3)^(3/4)*a*d)*arctan(1/2*sqrt(2)*(sqrt(2)*(c/d)^(1/4) + 2*sqrt( x))/(c/d)^(1/4))/(sqrt(2)*b^3*c^4*d^2 - 3*sqrt(2)*a*b^2*c^3*d^3 + 3*sqrt(2 )*a^2*b*c^2*d^4 - sqrt(2)*a^3*c*d^5) - 1/4*(5*(c*d^3)^(3/4)*b*c + 3*(c*d^3 )^(3/4)*a*d)*arctan(-1/2*sqrt(2)*(sqrt(2)*(c/d)^(1/4) - 2*sqrt(x))/(c/d)^( 1/4))/(sqrt(2)*b^3*c^4*d^2 - 3*sqrt(2)*a*b^2*c^3*d^3 + 3*sqrt(2)*a^2*b*c^2 *d^4 - sqrt(2)*a^3*c*d^5) - 1/8*(3*(a*b^3)^(3/4)*b*c + 5*(a*b^3)^(3/4)*a*d )*log(sqrt(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))/(sqrt(2)*a*b^5*c^3 - 3* sqrt(2)*a^2*b^4*c^2*d + 3*sqrt(2)*a^3*b^3*c*d^2 - sqrt(2)*a^4*b^2*d^3) + 1 /8*(3*(a*b^3)^(3/4)*b*c + 5*(a*b^3)^(3/4)*a*d)*log(-sqrt(2)*sqrt(x)*(a/b)^ (1/4) + x + sqrt(a/b))/(sqrt(2)*a*b^5*c^3 - 3*sqrt(2)*a^2*b^4*c^2*d + 3*sq rt(2)*a^3*b^3*c*d^2 - sqrt(2)*a^4*b^2*d^3) + 1/8*(5*(c*d^3)^(3/4)*b*c + 3* (c*d^3)^(3/4)*a*d)*log(sqrt(2)*sqrt(x)*(c/d)^(1/4) + x + sqrt(c/d))/(sqrt( 2)*b^3*c^4*d^2 - 3*sqrt(2)*a*b^2*c^3*d^3 + 3*sqrt(2)*a^2*b*c^2*d^4 - sqrt( 2)*a^3*c*d^5) - 1/8*(5*(c*d^3)^(3/4)*b*c + 3*(c*d^3)^(3/4)*a*d)*log(-sq...
Time = 9.85 (sec) , antiderivative size = 30956, normalized size of antiderivative = 50.83 \[ \int \frac {x^{5/2}}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^2} \, dx=\text {Too large to display} \]
2*atan((((-(81*b^5*c^4 + 625*a^4*b*d^4 + 1500*a^3*b^2*c*d^3 + 1350*a^2*b^3 *c^2*d^2 + 540*a*b^4*c^3*d)/(4096*a^13*d^12 + 4096*a*b^12*c^12 - 49152*a^2 *b^11*c^11*d + 270336*a^3*b^10*c^10*d^2 - 901120*a^4*b^9*c^9*d^3 + 2027520 *a^5*b^8*c^8*d^4 - 3244032*a^6*b^7*c^7*d^5 + 3784704*a^7*b^6*c^6*d^6 - 324 4032*a^8*b^5*c^5*d^7 + 2027520*a^9*b^4*c^4*d^8 - 901120*a^10*b^3*c^3*d^9 + 270336*a^11*b^2*c^2*d^10 - 49152*a^12*b*c*d^11))^(3/4)*(((864*a*b^20*c^17 *d^4 + 864*a^17*b^4*c*d^20 - 5184*a^2*b^19*c^16*d^5 + 3200*a^3*b^18*c^15*d ^6 + 56640*a^4*b^17*c^14*d^7 - 220800*a^5*b^16*c^13*d^8 + 369088*a^6*b^15* c^12*d^9 - 240768*a^7*b^14*c^11*d^10 - 158400*a^8*b^13*c^10*d^11 + 390720* a^9*b^12*c^9*d^12 - 158400*a^10*b^11*c^8*d^13 - 240768*a^11*b^10*c^7*d^14 + 369088*a^12*b^9*c^6*d^15 - 220800*a^13*b^8*c^5*d^16 + 56640*a^14*b^7*c^4 *d^17 + 3200*a^15*b^6*c^3*d^18 - 5184*a^16*b^5*c^2*d^19)*1i)/(a^14*d^14 + b^14*c^14 + 91*a^2*b^12*c^12*d^2 - 364*a^3*b^11*c^11*d^3 + 1001*a^4*b^10*c ^10*d^4 - 2002*a^5*b^9*c^9*d^5 + 3003*a^6*b^8*c^8*d^6 - 3432*a^7*b^7*c^7*d ^7 + 3003*a^8*b^6*c^6*d^8 - 2002*a^9*b^5*c^5*d^9 + 1001*a^10*b^4*c^4*d^10 - 364*a^11*b^3*c^3*d^11 + 91*a^12*b^2*c^2*d^12 - 14*a*b^13*c^13*d - 14*a^1 3*b*c*d^13) - (x^(1/2)*(-(81*b^5*c^4 + 625*a^4*b*d^4 + 1500*a^3*b^2*c*d^3 + 1350*a^2*b^3*c^2*d^2 + 540*a*b^4*c^3*d)/(4096*a^13*d^12 + 4096*a*b^12*c^ 12 - 49152*a^2*b^11*c^11*d + 270336*a^3*b^10*c^10*d^2 - 901120*a^4*b^9*c^9 *d^3 + 2027520*a^5*b^8*c^8*d^4 - 3244032*a^6*b^7*c^7*d^5 + 3784704*a^7*...