Integrand size = 24, antiderivative size = 376 \[ \int \frac {\left (c+d x^2\right )^3}{x^{9/2} \left (a+b x^2\right )^2} \, dx=-\frac {c^2 (11 b c-7 a d)}{14 a^2 b x^{7/2}}+\frac {c \left (11 b^2 c^2-21 a b c d+6 a^2 d^2\right )}{6 a^3 b x^{3/2}}+\frac {(b c-a d) \left (c+d x^2\right )^2}{2 a b x^{7/2} \left (a+b x^2\right )}-\frac {(b c-a d)^2 (11 b c+a d) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} a^{15/4} b^{5/4}}+\frac {(b c-a d)^2 (11 b c+a d) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} a^{15/4} b^{5/4}}-\frac {(b c-a d)^2 (11 b c+a d) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{8 \sqrt {2} a^{15/4} b^{5/4}}+\frac {(b c-a d)^2 (11 b c+a d) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{8 \sqrt {2} a^{15/4} b^{5/4}} \]
-1/14*c^2*(-7*a*d+11*b*c)/a^2/b/x^(7/2)+1/6*c*(6*a^2*d^2-21*a*b*c*d+11*b^2 *c^2)/a^3/b/x^(3/2)+1/2*(-a*d+b*c)*(d*x^2+c)^2/a/b/x^(7/2)/(b*x^2+a)-1/8*( -a*d+b*c)^2*(a*d+11*b*c)*arctan(1-b^(1/4)*2^(1/2)*x^(1/2)/a^(1/4))/a^(15/4 )/b^(5/4)*2^(1/2)+1/8*(-a*d+b*c)^2*(a*d+11*b*c)*arctan(1+b^(1/4)*2^(1/2)*x ^(1/2)/a^(1/4))/a^(15/4)/b^(5/4)*2^(1/2)-1/16*(-a*d+b*c)^2*(a*d+11*b*c)*ln (a^(1/2)+x*b^(1/2)-a^(1/4)*b^(1/4)*2^(1/2)*x^(1/2))/a^(15/4)/b^(5/4)*2^(1/ 2)+1/16*(-a*d+b*c)^2*(a*d+11*b*c)*ln(a^(1/2)+x*b^(1/2)+a^(1/4)*b^(1/4)*2^( 1/2)*x^(1/2))/a^(15/4)/b^(5/4)*2^(1/2)
Time = 0.54 (sec) , antiderivative size = 236, normalized size of antiderivative = 0.63 \[ \int \frac {\left (c+d x^2\right )^3}{x^{9/2} \left (a+b x^2\right )^2} \, dx=\frac {-\frac {4 a^{3/4} \sqrt [4]{b} \left (-77 b^3 c^3 x^4+21 a^3 d^3 x^4+a b^2 c^2 x^2 \left (-44 c+147 d x^2\right )+3 a^2 b c \left (4 c^2+28 c d x^2-21 d^2 x^4\right )\right )}{x^{7/2} \left (a+b x^2\right )}-21 \sqrt {2} (b c-a d)^2 (11 b c+a d) \arctan \left (\frac {\sqrt {a}-\sqrt {b} x}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}\right )+21 \sqrt {2} (b c-a d)^2 (11 b c+a d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}{\sqrt {a}+\sqrt {b} x}\right )}{168 a^{15/4} b^{5/4}} \]
((-4*a^(3/4)*b^(1/4)*(-77*b^3*c^3*x^4 + 21*a^3*d^3*x^4 + a*b^2*c^2*x^2*(-4 4*c + 147*d*x^2) + 3*a^2*b*c*(4*c^2 + 28*c*d*x^2 - 21*d^2*x^4)))/(x^(7/2)* (a + b*x^2)) - 21*Sqrt[2]*(b*c - a*d)^2*(11*b*c + a*d)*ArcTan[(Sqrt[a] - S qrt[b]*x)/(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x])] + 21*Sqrt[2]*(b*c - a*d)^2*(1 1*b*c + a*d)*ArcTanh[(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x])/(Sqrt[a] + Sqrt[b]* x)])/(168*a^(15/4)*b^(5/4))
Time = 0.57 (sec) , antiderivative size = 383, normalized size of antiderivative = 1.02, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {368, 968, 25, 1040, 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.
| \(\displaystyle \int \frac {\left (c+d x^2\right )^3}{x^{9/2} \left (a+b x^2\right )^2} \, dx\) |
| \(\Big \downarrow \) 368 |
| \(\displaystyle 2 \int \frac {\left (d x^2+c\right )^3}{x^4 \left (b x^2+a\right )^2}d\sqrt {x}\) |
| \(\Big \downarrow \) 968 |
| \(\displaystyle 2 \left (\frac {\left (c+d x^2\right )^2 (b c-a d)}{4 a b x^{7/2} \left (a+b x^2\right )}-\frac {\int -\frac {\left (d x^2+c\right ) \left (d (3 b c+a d) x^2+c (11 b c-7 a d)\right )}{x^4 \left (b x^2+a\right )}d\sqrt {x}}{4 a b}\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle 2 \left (\frac {\int \frac {\left (d x^2+c\right ) \left (d (3 b c+a d) x^2+c (11 b c-7 a d)\right )}{x^4 \left (b x^2+a\right )}d\sqrt {x}}{4 a b}+\frac {\left (c+d x^2\right )^2 (b c-a d)}{4 a b x^{7/2} \left (a+b x^2\right )}\right )\) |
| \(\Big \downarrow \) 1040 |
| \(\displaystyle 2 \left (\frac {\int \left (-\frac {(7 a d-11 b c) c^2}{a x^4}-\frac {\left (11 b^2 c^2-21 a b d c+6 a^2 d^2\right ) c}{a^2 x^2}+\frac {(a d-b c)^2 (11 b c+a d)}{a^2 \left (b x^2+a\right )}\right )d\sqrt {x}}{4 a b}+\frac {\left (c+d x^2\right )^2 (b c-a d)}{4 a b x^{7/2} \left (a+b x^2\right )}\right )\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 2 \left (\frac {-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right ) (b c-a d)^2 (a d+11 b c)}{2 \sqrt {2} a^{11/4} \sqrt [4]{b}}+\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right ) (b c-a d)^2 (a d+11 b c)}{2 \sqrt {2} a^{11/4} \sqrt [4]{b}}-\frac {(b c-a d)^2 (a d+11 b c) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{4 \sqrt {2} a^{11/4} \sqrt [4]{b}}+\frac {(b c-a d)^2 (a d+11 b c) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{4 \sqrt {2} a^{11/4} \sqrt [4]{b}}+\frac {c \left (6 a^2 d^2-21 a b c d+11 b^2 c^2\right )}{3 a^2 x^{3/2}}-\frac {c^2 (11 b c-7 a d)}{7 a x^{7/2}}}{4 a b}+\frac {\left (c+d x^2\right )^2 (b c-a d)}{4 a b x^{7/2} \left (a+b x^2\right )}\right )\) |
2*(((b*c - a*d)*(c + d*x^2)^2)/(4*a*b*x^(7/2)*(a + b*x^2)) + (-1/7*(c^2*(1 1*b*c - 7*a*d))/(a*x^(7/2)) + (c*(11*b^2*c^2 - 21*a*b*c*d + 6*a^2*d^2))/(3 *a^2*x^(3/2)) - ((b*c - a*d)^2*(11*b*c + a*d)*ArcTan[1 - (Sqrt[2]*b^(1/4)* Sqrt[x])/a^(1/4)])/(2*Sqrt[2]*a^(11/4)*b^(1/4)) + ((b*c - a*d)^2*(11*b*c + a*d)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(2*Sqrt[2]*a^(11/4)*b ^(1/4)) - ((b*c - a*d)^2*(11*b*c + a*d)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1 /4)*Sqrt[x] + Sqrt[b]*x])/(4*Sqrt[2]*a^(11/4)*b^(1/4)) + ((b*c - a*d)^2*(1 1*b*c + a*d)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/( 4*Sqrt[2]*a^(11/4)*b^(1/4)))/(4*a*b))
3.5.60.3.1 Defintions of rubi rules used
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[(-(c*b - a*d))*(e*x)^(m + 1)*(a + b*x^n)^(p + 1) *((c + d*x^n)^(q - 1)/(a*b*e*n*(p + 1))), x] + Simp[1/(a*b*n*(p + 1)) Int [(e*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 2)*Simp[c*(c*b*n*(p + 1) + (c *b - a*d)*(m + 1)) + d*(c*b*n*(p + 1) + (c*b - a*d)*(m + n*(q - 1) + 1))*x^ n, x], x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[ n, 0] && LtQ[p, -1] && GtQ[q, 1] && 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_))^(r_.), x_Symbol] :> Int[ExpandIntegrand[ (g*x)^m*(a + b*x^n)^p*(c + d*x^n)^q*(e + f*x^n)^r, x], x] /; FreeQ[{a, b, c , d, e, f, g, m, n}, x] && IGtQ[p, -2] && IGtQ[q, 0] && IGtQ[r, 0]
Time = 2.78 (sec) , antiderivative size = 201, normalized size of antiderivative = 0.53
| method | result | size |
| risch | \(-\frac {2 c^{2} \left (21 a d \,x^{2}-14 c b \,x^{2}+3 a c \right )}{21 a^{3} x^{\frac {7}{2}}}+\frac {\left (2 a^{2} d^{2}-4 a b c d +2 b^{2} c^{2}\right ) \left (-\frac {\left (a d -b c \right ) \sqrt {x}}{4 b \left (b \,x^{2}+a \right )}+\frac {\left (a d +11 b c \right ) \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}{x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{32 b a}\right )}{a^{3}}\) | \(201\) |
| derivativedivides | \(-\frac {2 c^{3}}{7 a^{2} x^{\frac {7}{2}}}-\frac {2 c^{2} \left (3 a d -2 b c \right )}{3 a^{3} x^{\frac {3}{2}}}+\frac {-\frac {\left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \sqrt {x}}{2 b \left (b \,x^{2}+a \right )}+\frac {\left (a^{3} d^{3}+9 a^{2} b c \,d^{2}-21 a \,b^{2} c^{2} d +11 b^{3} c^{3}\right ) \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}{x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{16 b a}}{a^{3}}\) | \(236\) |
| default | \(-\frac {2 c^{3}}{7 a^{2} x^{\frac {7}{2}}}-\frac {2 c^{2} \left (3 a d -2 b c \right )}{3 a^{3} x^{\frac {3}{2}}}+\frac {-\frac {\left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \sqrt {x}}{2 b \left (b \,x^{2}+a \right )}+\frac {\left (a^{3} d^{3}+9 a^{2} b c \,d^{2}-21 a \,b^{2} c^{2} d +11 b^{3} c^{3}\right ) \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}{x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{16 b a}}{a^{3}}\) | \(236\) |
-2/21*c^2*(21*a*d*x^2-14*b*c*x^2+3*a*c)/a^3/x^(7/2)+1/a^3*(2*a^2*d^2-4*a*b *c*d+2*b^2*c^2)*(-1/4*(a*d-b*c)/b*x^(1/2)/(b*x^2+a)+1/32*(a*d+11*b*c)/b*(a /b)^(1/4)/a*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)))
Result contains complex when optimal does not.
Time = 0.27 (sec) , antiderivative size = 1761, normalized size of antiderivative = 4.68 \[ \int \frac {\left (c+d x^2\right )^3}{x^{9/2} \left (a+b x^2\right )^2} \, dx=\text {Too large to display} \]
1/168*(21*(a^3*b^2*x^6 + a^4*b*x^4)*(-(14641*b^12*c^12 - 111804*a*b^11*c^1 1*d + 368082*a^2*b^10*c^10*d^2 - 676588*a^3*b^9*c^9*d^3 + 746703*a^4*b^8*c ^8*d^4 - 486648*a^5*b^7*c^7*d^5 + 160188*a^6*b^6*c^6*d^6 - 5688*a^7*b^5*c^ 5*d^7 - 10017*a^8*b^4*c^4*d^8 + 692*a^9*b^3*c^3*d^9 + 402*a^10*b^2*c^2*d^1 0 + 36*a^11*b*c*d^11 + a^12*d^12)/(a^15*b^5))^(1/4)*log(a^4*b*(-(14641*b^1 2*c^12 - 111804*a*b^11*c^11*d + 368082*a^2*b^10*c^10*d^2 - 676588*a^3*b^9* c^9*d^3 + 746703*a^4*b^8*c^8*d^4 - 486648*a^5*b^7*c^7*d^5 + 160188*a^6*b^6 *c^6*d^6 - 5688*a^7*b^5*c^5*d^7 - 10017*a^8*b^4*c^4*d^8 + 692*a^9*b^3*c^3* d^9 + 402*a^10*b^2*c^2*d^10 + 36*a^11*b*c*d^11 + a^12*d^12)/(a^15*b^5))^(1 /4) + (11*b^3*c^3 - 21*a*b^2*c^2*d + 9*a^2*b*c*d^2 + a^3*d^3)*sqrt(x)) - 2 1*(-I*a^3*b^2*x^6 - I*a^4*b*x^4)*(-(14641*b^12*c^12 - 111804*a*b^11*c^11*d + 368082*a^2*b^10*c^10*d^2 - 676588*a^3*b^9*c^9*d^3 + 746703*a^4*b^8*c^8* d^4 - 486648*a^5*b^7*c^7*d^5 + 160188*a^6*b^6*c^6*d^6 - 5688*a^7*b^5*c^5*d ^7 - 10017*a^8*b^4*c^4*d^8 + 692*a^9*b^3*c^3*d^9 + 402*a^10*b^2*c^2*d^10 + 36*a^11*b*c*d^11 + a^12*d^12)/(a^15*b^5))^(1/4)*log(I*a^4*b*(-(14641*b^12 *c^12 - 111804*a*b^11*c^11*d + 368082*a^2*b^10*c^10*d^2 - 676588*a^3*b^9*c ^9*d^3 + 746703*a^4*b^8*c^8*d^4 - 486648*a^5*b^7*c^7*d^5 + 160188*a^6*b^6* c^6*d^6 - 5688*a^7*b^5*c^5*d^7 - 10017*a^8*b^4*c^4*d^8 + 692*a^9*b^3*c^3*d ^9 + 402*a^10*b^2*c^2*d^10 + 36*a^11*b*c*d^11 + a^12*d^12)/(a^15*b^5))^(1/ 4) + (11*b^3*c^3 - 21*a*b^2*c^2*d + 9*a^2*b*c*d^2 + a^3*d^3)*sqrt(x)) -...
Timed out. \[ \int \frac {\left (c+d x^2\right )^3}{x^{9/2} \left (a+b x^2\right )^2} \, dx=\text {Timed out} \]
Time = 0.34 (sec) , antiderivative size = 424, normalized size of antiderivative = 1.13 \[ \int \frac {\left (c+d x^2\right )^3}{x^{9/2} \left (a+b x^2\right )^2} \, dx=-\frac {12 \, a^{2} b c^{3} - 7 \, {\left (11 \, b^{3} c^{3} - 21 \, a b^{2} c^{2} d + 9 \, a^{2} b c d^{2} - 3 \, a^{3} d^{3}\right )} x^{4} - 4 \, {\left (11 \, a b^{2} c^{3} - 21 \, a^{2} b c^{2} d\right )} x^{2}}{42 \, {\left (a^{3} b^{2} x^{\frac {11}{2}} + a^{4} b x^{\frac {7}{2}}\right )}} + \frac {\frac {2 \, \sqrt {2} {\left (11 \, b^{3} c^{3} - 21 \, a b^{2} c^{2} d + 9 \, a^{2} b c d^{2} + a^{3} d^{3}\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} + 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {b}}} + \frac {2 \, \sqrt {2} {\left (11 \, b^{3} c^{3} - 21 \, a b^{2} c^{2} d + 9 \, a^{2} b c d^{2} + a^{3} d^{3}\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} - 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {b}}} + \frac {\sqrt {2} {\left (11 \, b^{3} c^{3} - 21 \, a b^{2} c^{2} d + 9 \, a^{2} b c d^{2} + a^{3} d^{3}\right )} \log \left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {3}{4}} b^{\frac {1}{4}}} - \frac {\sqrt {2} {\left (11 \, b^{3} c^{3} - 21 \, a b^{2} c^{2} d + 9 \, a^{2} b c d^{2} + a^{3} d^{3}\right )} \log \left (-\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {3}{4}} b^{\frac {1}{4}}}}{16 \, a^{3} b} \]
-1/42*(12*a^2*b*c^3 - 7*(11*b^3*c^3 - 21*a*b^2*c^2*d + 9*a^2*b*c*d^2 - 3*a ^3*d^3)*x^4 - 4*(11*a*b^2*c^3 - 21*a^2*b*c^2*d)*x^2)/(a^3*b^2*x^(11/2) + a ^4*b*x^(7/2)) + 1/16*(2*sqrt(2)*(11*b^3*c^3 - 21*a*b^2*c^2*d + 9*a^2*b*c*d ^2 + a^3*d^3)*arctan(1/2*sqrt(2)*(sqrt(2)*a^(1/4)*b^(1/4) + 2*sqrt(b)*sqrt (x))/sqrt(sqrt(a)*sqrt(b)))/(sqrt(a)*sqrt(sqrt(a)*sqrt(b))) + 2*sqrt(2)*(1 1*b^3*c^3 - 21*a*b^2*c^2*d + 9*a^2*b*c*d^2 + a^3*d^3)*arctan(-1/2*sqrt(2)* (sqrt(2)*a^(1/4)*b^(1/4) - 2*sqrt(b)*sqrt(x))/sqrt(sqrt(a)*sqrt(b)))/(sqrt (a)*sqrt(sqrt(a)*sqrt(b))) + sqrt(2)*(11*b^3*c^3 - 21*a*b^2*c^2*d + 9*a^2* b*c*d^2 + a^3*d^3)*log(sqrt(2)*a^(1/4)*b^(1/4)*sqrt(x) + sqrt(b)*x + sqrt( a))/(a^(3/4)*b^(1/4)) - sqrt(2)*(11*b^3*c^3 - 21*a*b^2*c^2*d + 9*a^2*b*c*d ^2 + a^3*d^3)*log(-sqrt(2)*a^(1/4)*b^(1/4)*sqrt(x) + sqrt(b)*x + sqrt(a))/ (a^(3/4)*b^(1/4)))/(a^3*b)
Time = 0.32 (sec) , antiderivative size = 509, normalized size of antiderivative = 1.35 \[ \int \frac {\left (c+d x^2\right )^3}{x^{9/2} \left (a+b x^2\right )^2} \, dx=\frac {\sqrt {2} {\left (11 \, \left (a b^{3}\right )^{\frac {1}{4}} b^{3} c^{3} - 21 \, \left (a b^{3}\right )^{\frac {1}{4}} a b^{2} c^{2} d + 9 \, \left (a b^{3}\right )^{\frac {1}{4}} a^{2} b c d^{2} + \left (a b^{3}\right )^{\frac {1}{4}} a^{3} d^{3}\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{8 \, a^{4} b^{2}} + \frac {\sqrt {2} {\left (11 \, \left (a b^{3}\right )^{\frac {1}{4}} b^{3} c^{3} - 21 \, \left (a b^{3}\right )^{\frac {1}{4}} a b^{2} c^{2} d + 9 \, \left (a b^{3}\right )^{\frac {1}{4}} a^{2} b c d^{2} + \left (a b^{3}\right )^{\frac {1}{4}} a^{3} d^{3}\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{8 \, a^{4} b^{2}} + \frac {\sqrt {2} {\left (11 \, \left (a b^{3}\right )^{\frac {1}{4}} b^{3} c^{3} - 21 \, \left (a b^{3}\right )^{\frac {1}{4}} a b^{2} c^{2} d + 9 \, \left (a b^{3}\right )^{\frac {1}{4}} a^{2} b c d^{2} + \left (a b^{3}\right )^{\frac {1}{4}} a^{3} d^{3}\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {a}{b}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{b}}\right )}{16 \, a^{4} b^{2}} - \frac {\sqrt {2} {\left (11 \, \left (a b^{3}\right )^{\frac {1}{4}} b^{3} c^{3} - 21 \, \left (a b^{3}\right )^{\frac {1}{4}} a b^{2} c^{2} d + 9 \, \left (a b^{3}\right )^{\frac {1}{4}} a^{2} b c d^{2} + \left (a b^{3}\right )^{\frac {1}{4}} a^{3} d^{3}\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {a}{b}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{b}}\right )}{16 \, a^{4} b^{2}} + \frac {b^{3} c^{3} \sqrt {x} - 3 \, a b^{2} c^{2} d \sqrt {x} + 3 \, a^{2} b c d^{2} \sqrt {x} - a^{3} d^{3} \sqrt {x}}{2 \, {\left (b x^{2} + a\right )} a^{3} b} + \frac {2 \, {\left (14 \, b c^{3} x^{2} - 21 \, a c^{2} d x^{2} - 3 \, a c^{3}\right )}}{21 \, a^{3} x^{\frac {7}{2}}} \]
1/8*sqrt(2)*(11*(a*b^3)^(1/4)*b^3*c^3 - 21*(a*b^3)^(1/4)*a*b^2*c^2*d + 9*( a*b^3)^(1/4)*a^2*b*c*d^2 + (a*b^3)^(1/4)*a^3*d^3)*arctan(1/2*sqrt(2)*(sqrt (2)*(a/b)^(1/4) + 2*sqrt(x))/(a/b)^(1/4))/(a^4*b^2) + 1/8*sqrt(2)*(11*(a*b ^3)^(1/4)*b^3*c^3 - 21*(a*b^3)^(1/4)*a*b^2*c^2*d + 9*(a*b^3)^(1/4)*a^2*b*c *d^2 + (a*b^3)^(1/4)*a^3*d^3)*arctan(-1/2*sqrt(2)*(sqrt(2)*(a/b)^(1/4) - 2 *sqrt(x))/(a/b)^(1/4))/(a^4*b^2) + 1/16*sqrt(2)*(11*(a*b^3)^(1/4)*b^3*c^3 - 21*(a*b^3)^(1/4)*a*b^2*c^2*d + 9*(a*b^3)^(1/4)*a^2*b*c*d^2 + (a*b^3)^(1/ 4)*a^3*d^3)*log(sqrt(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))/(a^4*b^2) - 1 /16*sqrt(2)*(11*(a*b^3)^(1/4)*b^3*c^3 - 21*(a*b^3)^(1/4)*a*b^2*c^2*d + 9*( a*b^3)^(1/4)*a^2*b*c*d^2 + (a*b^3)^(1/4)*a^3*d^3)*log(-sqrt(2)*sqrt(x)*(a/ b)^(1/4) + x + sqrt(a/b))/(a^4*b^2) + 1/2*(b^3*c^3*sqrt(x) - 3*a*b^2*c^2*d *sqrt(x) + 3*a^2*b*c*d^2*sqrt(x) - a^3*d^3*sqrt(x))/((b*x^2 + a)*a^3*b) + 2/21*(14*b*c^3*x^2 - 21*a*c^2*d*x^2 - 3*a*c^3)/(a^3*x^(7/2))
Time = 5.94 (sec) , antiderivative size = 1746, normalized size of antiderivative = 4.64 \[ \int \frac {\left (c+d x^2\right )^3}{x^{9/2} \left (a+b x^2\right )^2} \, dx=\text {Too large to display} \]
(atan((((x^(1/2)*(3872*a^9*b^12*c^6 + 32*a^15*b^6*d^6 - 14784*a^10*b^11*c^ 5*d + 576*a^14*b^7*c*d^5 + 20448*a^11*b^10*c^4*d^2 - 11392*a^12*b^9*c^3*d^ 3 + 1248*a^13*b^8*c^2*d^4) - ((a*d - b*c)^2*(a*d + 11*b*c)*(2816*a^13*b^10 *c^3 + 256*a^16*b^7*d^3 - 5376*a^14*b^9*c^2*d + 2304*a^15*b^8*c*d^2))/(8*( -a)^(15/4)*b^(5/4)))*(a*d - b*c)^2*(a*d + 11*b*c)*1i)/(8*(-a)^(15/4)*b^(5/ 4)) + ((x^(1/2)*(3872*a^9*b^12*c^6 + 32*a^15*b^6*d^6 - 14784*a^10*b^11*c^5 *d + 576*a^14*b^7*c*d^5 + 20448*a^11*b^10*c^4*d^2 - 11392*a^12*b^9*c^3*d^3 + 1248*a^13*b^8*c^2*d^4) + ((a*d - b*c)^2*(a*d + 11*b*c)*(2816*a^13*b^10* c^3 + 256*a^16*b^7*d^3 - 5376*a^14*b^9*c^2*d + 2304*a^15*b^8*c*d^2))/(8*(- a)^(15/4)*b^(5/4)))*(a*d - b*c)^2*(a*d + 11*b*c)*1i)/(8*(-a)^(15/4)*b^(5/4 )))/(((x^(1/2)*(3872*a^9*b^12*c^6 + 32*a^15*b^6*d^6 - 14784*a^10*b^11*c^5* d + 576*a^14*b^7*c*d^5 + 20448*a^11*b^10*c^4*d^2 - 11392*a^12*b^9*c^3*d^3 + 1248*a^13*b^8*c^2*d^4) - ((a*d - b*c)^2*(a*d + 11*b*c)*(2816*a^13*b^10*c ^3 + 256*a^16*b^7*d^3 - 5376*a^14*b^9*c^2*d + 2304*a^15*b^8*c*d^2))/(8*(-a )^(15/4)*b^(5/4)))*(a*d - b*c)^2*(a*d + 11*b*c))/(8*(-a)^(15/4)*b^(5/4)) - ((x^(1/2)*(3872*a^9*b^12*c^6 + 32*a^15*b^6*d^6 - 14784*a^10*b^11*c^5*d + 576*a^14*b^7*c*d^5 + 20448*a^11*b^10*c^4*d^2 - 11392*a^12*b^9*c^3*d^3 + 12 48*a^13*b^8*c^2*d^4) + ((a*d - b*c)^2*(a*d + 11*b*c)*(2816*a^13*b^10*c^3 + 256*a^16*b^7*d^3 - 5376*a^14*b^9*c^2*d + 2304*a^15*b^8*c*d^2))/(8*(-a)^(1 5/4)*b^(5/4)))*(a*d - b*c)^2*(a*d + 11*b*c))/(8*(-a)^(15/4)*b^(5/4))))*...