Integrand size = 24, antiderivative size = 290 \[ \int \frac {x^{5/2} \left (a+b x^2\right )^2}{c+d x^2} \, dx=\frac {2 (b c-a d)^2 x^{3/2}}{3 d^3}-\frac {2 b (b c-2 a d) x^{7/2}}{7 d^2}+\frac {2 b^2 x^{11/2}}{11 d}+\frac {c^{3/4} (b c-a d)^2 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{\sqrt {2} d^{15/4}}-\frac {c^{3/4} (b c-a d)^2 \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{\sqrt {2} d^{15/4}}-\frac {c^{3/4} (b c-a d)^2 \log \left (\sqrt {c}-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {d} x\right )}{2 \sqrt {2} d^{15/4}}+\frac {c^{3/4} (b c-a d)^2 \log \left (\sqrt {c}+\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {d} x\right )}{2 \sqrt {2} d^{15/4}} \]
2/3*(-a*d+b*c)^2*x^(3/2)/d^3-2/7*b*(-2*a*d+b*c)*x^(7/2)/d^2+2/11*b^2*x^(11 /2)/d+1/2*c^(3/4)*(-a*d+b*c)^2*arctan(1-d^(1/4)*2^(1/2)*x^(1/2)/c^(1/4))/d ^(15/4)*2^(1/2)-1/2*c^(3/4)*(-a*d+b*c)^2*arctan(1+d^(1/4)*2^(1/2)*x^(1/2)/ c^(1/4))/d^(15/4)*2^(1/2)-1/4*c^(3/4)*(-a*d+b*c)^2*ln(c^(1/2)+x*d^(1/2)-c^ (1/4)*d^(1/4)*2^(1/2)*x^(1/2))/d^(15/4)*2^(1/2)+1/4*c^(3/4)*(-a*d+b*c)^2*l n(c^(1/2)+x*d^(1/2)+c^(1/4)*d^(1/4)*2^(1/2)*x^(1/2))/d^(15/4)*2^(1/2)
Time = 0.24 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.64 \[ \int \frac {x^{5/2} \left (a+b x^2\right )^2}{c+d x^2} \, dx=\frac {2 x^{3/2} \left (77 a^2 d^2+22 a b d \left (-7 c+3 d x^2\right )+b^2 \left (77 c^2-33 c d x^2+21 d^2 x^4\right )\right )}{231 d^3}+\frac {c^{3/4} (b c-a d)^2 \arctan \left (\frac {\sqrt {c}-\sqrt {d} x}{\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}}\right )}{\sqrt {2} d^{15/4}}+\frac {c^{3/4} (b c-a d)^2 \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}}{\sqrt {c}+\sqrt {d} x}\right )}{\sqrt {2} d^{15/4}} \]
(2*x^(3/2)*(77*a^2*d^2 + 22*a*b*d*(-7*c + 3*d*x^2) + b^2*(77*c^2 - 33*c*d* x^2 + 21*d^2*x^4)))/(231*d^3) + (c^(3/4)*(b*c - a*d)^2*ArcTan[(Sqrt[c] - S qrt[d]*x)/(Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x])])/(Sqrt[2]*d^(15/4)) + (c^(3/4 )*(b*c - a*d)^2*ArcTanh[(Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x])/(Sqrt[c] + Sqrt[ d]*x)])/(Sqrt[2]*d^(15/4))
Time = 0.47 (sec) , antiderivative size = 290, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {364, 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.
| \(\displaystyle \int \frac {x^{5/2} \left (a+b x^2\right )^2}{c+d x^2} \, dx\) |
| \(\Big \downarrow \) 364 |
| \(\displaystyle \int \left (\frac {x^{5/2} \left (a^2 d^2-2 a b c d+b^2 c^2\right )}{d^2 \left (c+d x^2\right )}-\frac {b x^{5/2} (b c-2 a d)}{d^2}+\frac {b^2 x^{9/2}}{d}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {c^{3/4} (b c-a d)^2 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{\sqrt {2} d^{15/4}}-\frac {c^{3/4} (b c-a d)^2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}+1\right )}{\sqrt {2} d^{15/4}}-\frac {c^{3/4} (b c-a d)^2 \log \left (-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{2 \sqrt {2} d^{15/4}}+\frac {c^{3/4} (b c-a d)^2 \log \left (\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{2 \sqrt {2} d^{15/4}}+\frac {2 x^{3/2} (b c-a d)^2}{3 d^3}-\frac {2 b x^{7/2} (b c-2 a d)}{7 d^2}+\frac {2 b^2 x^{11/2}}{11 d}\) |
(2*(b*c - a*d)^2*x^(3/2))/(3*d^3) - (2*b*(b*c - 2*a*d)*x^(7/2))/(7*d^2) + (2*b^2*x^(11/2))/(11*d) + (c^(3/4)*(b*c - a*d)^2*ArcTan[1 - (Sqrt[2]*d^(1/ 4)*Sqrt[x])/c^(1/4)])/(Sqrt[2]*d^(15/4)) - (c^(3/4)*(b*c - a*d)^2*ArcTan[1 + (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/(Sqrt[2]*d^(15/4)) - (c^(3/4)*(b*c - a*d)^2*Log[Sqrt[c] - Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x])/(2*Sq rt[2]*d^(15/4)) + (c^(3/4)*(b*c - a*d)^2*Log[Sqrt[c] + Sqrt[2]*c^(1/4)*d^( 1/4)*Sqrt[x] + Sqrt[d]*x])/(2*Sqrt[2]*d^(15/4))
3.5.16.3.1 Defintions of rubi rules used
Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_))/((c_) + (d_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*((a + b*x^2)^p/(c + d*x^2)), x], x ] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && (In tegerQ[m] || IGtQ[2*(m + 1), 0] || !RationalQ[m])
Time = 2.79 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.66
| method | result | size |
| risch | \(\frac {2 x^{\frac {3}{2}} \left (21 b^{2} d^{2} x^{4}+66 x^{2} a b \,d^{2}-33 x^{2} b^{2} c d +77 a^{2} d^{2}-154 a b c d +77 b^{2} c^{2}\right )}{231 d^{3}}-\frac {c \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {c}{d}}}{x +\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {c}{d}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )\right )}{4 d^{4} \left (\frac {c}{d}\right )^{\frac {1}{4}}}\) | \(191\) |
| derivativedivides | \(\frac {\frac {2 b^{2} d^{2} x^{\frac {11}{2}}}{11}+\frac {2 \left (2 a b \,d^{2}-b^{2} c d \right ) x^{\frac {7}{2}}}{7}+\frac {2 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) x^{\frac {3}{2}}}{3}}{d^{3}}-\frac {c \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {c}{d}}}{x +\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {c}{d}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )\right )}{4 d^{4} \left (\frac {c}{d}\right )^{\frac {1}{4}}}\) | \(192\) |
| default | \(\frac {\frac {2 b^{2} d^{2} x^{\frac {11}{2}}}{11}+\frac {2 \left (2 a b \,d^{2}-b^{2} c d \right ) x^{\frac {7}{2}}}{7}+\frac {2 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) x^{\frac {3}{2}}}{3}}{d^{3}}-\frac {c \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {c}{d}}}{x +\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {c}{d}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )\right )}{4 d^{4} \left (\frac {c}{d}\right )^{\frac {1}{4}}}\) | \(192\) |
2/231*x^(3/2)*(21*b^2*d^2*x^4+66*a*b*d^2*x^2-33*b^2*c*d*x^2+77*a^2*d^2-154 *a*b*c*d+77*b^2*c^2)/d^3-1/4*c*(a^2*d^2-2*a*b*c*d+b^2*c^2)/d^4/(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*arct an(2^(1/2)/(c/d)^(1/4)*x^(1/2)-1))
Result contains complex when optimal does not.
Time = 0.27 (sec) , antiderivative size = 1393, normalized size of antiderivative = 4.80 \[ \int \frac {x^{5/2} \left (a+b x^2\right )^2}{c+d x^2} \, dx=\text {Too large to display} \]
-1/462*(231*d^3*(-(b^8*c^11 - 8*a*b^7*c^10*d + 28*a^2*b^6*c^9*d^2 - 56*a^3 *b^5*c^8*d^3 + 70*a^4*b^4*c^7*d^4 - 56*a^5*b^3*c^6*d^5 + 28*a^6*b^2*c^5*d^ 6 - 8*a^7*b*c^4*d^7 + a^8*c^3*d^8)/d^15)^(1/4)*log(d^11*(-(b^8*c^11 - 8*a* b^7*c^10*d + 28*a^2*b^6*c^9*d^2 - 56*a^3*b^5*c^8*d^3 + 70*a^4*b^4*c^7*d^4 - 56*a^5*b^3*c^6*d^5 + 28*a^6*b^2*c^5*d^6 - 8*a^7*b*c^4*d^7 + a^8*c^3*d^8) /d^15)^(3/4) + (b^6*c^8 - 6*a*b^5*c^7*d + 15*a^2*b^4*c^6*d^2 - 20*a^3*b^3* c^5*d^3 + 15*a^4*b^2*c^4*d^4 - 6*a^5*b*c^3*d^5 + a^6*c^2*d^6)*sqrt(x)) - 2 31*I*d^3*(-(b^8*c^11 - 8*a*b^7*c^10*d + 28*a^2*b^6*c^9*d^2 - 56*a^3*b^5*c^ 8*d^3 + 70*a^4*b^4*c^7*d^4 - 56*a^5*b^3*c^6*d^5 + 28*a^6*b^2*c^5*d^6 - 8*a ^7*b*c^4*d^7 + a^8*c^3*d^8)/d^15)^(1/4)*log(I*d^11*(-(b^8*c^11 - 8*a*b^7*c ^10*d + 28*a^2*b^6*c^9*d^2 - 56*a^3*b^5*c^8*d^3 + 70*a^4*b^4*c^7*d^4 - 56* a^5*b^3*c^6*d^5 + 28*a^6*b^2*c^5*d^6 - 8*a^7*b*c^4*d^7 + a^8*c^3*d^8)/d^15 )^(3/4) + (b^6*c^8 - 6*a*b^5*c^7*d + 15*a^2*b^4*c^6*d^2 - 20*a^3*b^3*c^5*d ^3 + 15*a^4*b^2*c^4*d^4 - 6*a^5*b*c^3*d^5 + a^6*c^2*d^6)*sqrt(x)) + 231*I* d^3*(-(b^8*c^11 - 8*a*b^7*c^10*d + 28*a^2*b^6*c^9*d^2 - 56*a^3*b^5*c^8*d^3 + 70*a^4*b^4*c^7*d^4 - 56*a^5*b^3*c^6*d^5 + 28*a^6*b^2*c^5*d^6 - 8*a^7*b* c^4*d^7 + a^8*c^3*d^8)/d^15)^(1/4)*log(-I*d^11*(-(b^8*c^11 - 8*a*b^7*c^10* d + 28*a^2*b^6*c^9*d^2 - 56*a^3*b^5*c^8*d^3 + 70*a^4*b^4*c^7*d^4 - 56*a^5* b^3*c^6*d^5 + 28*a^6*b^2*c^5*d^6 - 8*a^7*b*c^4*d^7 + a^8*c^3*d^8)/d^15)^(3 /4) + (b^6*c^8 - 6*a*b^5*c^7*d + 15*a^2*b^4*c^6*d^2 - 20*a^3*b^3*c^5*d^...
Timed out. \[ \int \frac {x^{5/2} \left (a+b x^2\right )^2}{c+d x^2} \, dx=\text {Timed out} \]
Time = 0.31 (sec) , antiderivative size = 263, normalized size of antiderivative = 0.91 \[ \int \frac {x^{5/2} \left (a+b x^2\right )^2}{c+d x^2} \, dx=-\frac {{\left (b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2}\right )} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} + 2 \, \sqrt {d} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {c} \sqrt {d}}}\right )}{\sqrt {\sqrt {c} \sqrt {d}} \sqrt {d}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} - 2 \, \sqrt {d} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {c} \sqrt {d}}}\right )}{\sqrt {\sqrt {c} \sqrt {d}} \sqrt {d}} - \frac {\sqrt {2} \log \left (\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} \sqrt {x} + \sqrt {d} x + \sqrt {c}\right )}{c^{\frac {1}{4}} d^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} \sqrt {x} + \sqrt {d} x + \sqrt {c}\right )}{c^{\frac {1}{4}} d^{\frac {3}{4}}}\right )}}{4 \, d^{3}} + \frac {2 \, {\left (21 \, b^{2} d^{2} x^{\frac {11}{2}} - 33 \, {\left (b^{2} c d - 2 \, a b d^{2}\right )} x^{\frac {7}{2}} + 77 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x^{\frac {3}{2}}\right )}}{231 \, d^{3}} \]
-1/4*(b^2*c^3 - 2*a*b*c^2*d + a^2*c*d^2)*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sq rt(2)*c^(1/4)*d^(1/4) + 2*sqrt(d)*sqrt(x))/sqrt(sqrt(c)*sqrt(d)))/(sqrt(sq rt(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) + sqr t(d)*x + sqrt(c))/(c^(1/4)*d^(3/4)))/d^3 + 2/231*(21*b^2*d^2*x^(11/2) - 33 *(b^2*c*d - 2*a*b*d^2)*x^(7/2) + 77*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*x^(3/2 ))/d^3
Time = 0.31 (sec) , antiderivative size = 385, normalized size of antiderivative = 1.33 \[ \int \frac {x^{5/2} \left (a+b x^2\right )^2}{c+d x^2} \, dx=-\frac {\sqrt {2} {\left (\left (c d^{3}\right )^{\frac {3}{4}} b^{2} c^{2} - 2 \, \left (c d^{3}\right )^{\frac {3}{4}} a b c d + \left (c d^{3}\right )^{\frac {3}{4}} a^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {c}{d}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {c}{d}\right )^{\frac {1}{4}}}\right )}{2 \, d^{6}} - \frac {\sqrt {2} {\left (\left (c d^{3}\right )^{\frac {3}{4}} b^{2} c^{2} - 2 \, \left (c d^{3}\right )^{\frac {3}{4}} a b c d + \left (c d^{3}\right )^{\frac {3}{4}} a^{2} d^{2}\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {c}{d}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {c}{d}\right )^{\frac {1}{4}}}\right )}{2 \, d^{6}} + \frac {\sqrt {2} {\left (\left (c d^{3}\right )^{\frac {3}{4}} b^{2} c^{2} - 2 \, \left (c d^{3}\right )^{\frac {3}{4}} a b c d + \left (c d^{3}\right )^{\frac {3}{4}} a^{2} d^{2}\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {c}{d}\right )^{\frac {1}{4}} + x + \sqrt {\frac {c}{d}}\right )}{4 \, d^{6}} - \frac {\sqrt {2} {\left (\left (c d^{3}\right )^{\frac {3}{4}} b^{2} c^{2} - 2 \, \left (c d^{3}\right )^{\frac {3}{4}} a b c d + \left (c d^{3}\right )^{\frac {3}{4}} a^{2} d^{2}\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {c}{d}\right )^{\frac {1}{4}} + x + \sqrt {\frac {c}{d}}\right )}{4 \, d^{6}} + \frac {2 \, {\left (21 \, b^{2} d^{10} x^{\frac {11}{2}} - 33 \, b^{2} c d^{9} x^{\frac {7}{2}} + 66 \, a b d^{10} x^{\frac {7}{2}} + 77 \, b^{2} c^{2} d^{8} x^{\frac {3}{2}} - 154 \, a b c d^{9} x^{\frac {3}{2}} + 77 \, a^{2} d^{10} x^{\frac {3}{2}}\right )}}{231 \, d^{11}} \]
-1/2*sqrt(2)*((c*d^3)^(3/4)*b^2*c^2 - 2*(c*d^3)^(3/4)*a*b*c*d + (c*d^3)^(3 /4)*a^2*d^2)*arctan(1/2*sqrt(2)*(sqrt(2)*(c/d)^(1/4) + 2*sqrt(x))/(c/d)^(1 /4))/d^6 - 1/2*sqrt(2)*((c*d^3)^(3/4)*b^2*c^2 - 2*(c*d^3)^(3/4)*a*b*c*d + (c*d^3)^(3/4)*a^2*d^2)*arctan(-1/2*sqrt(2)*(sqrt(2)*(c/d)^(1/4) - 2*sqrt(x ))/(c/d)^(1/4))/d^6 + 1/4*sqrt(2)*((c*d^3)^(3/4)*b^2*c^2 - 2*(c*d^3)^(3/4) *a*b*c*d + (c*d^3)^(3/4)*a^2*d^2)*log(sqrt(2)*sqrt(x)*(c/d)^(1/4) + x + sq rt(c/d))/d^6 - 1/4*sqrt(2)*((c*d^3)^(3/4)*b^2*c^2 - 2*(c*d^3)^(3/4)*a*b*c* d + (c*d^3)^(3/4)*a^2*d^2)*log(-sqrt(2)*sqrt(x)*(c/d)^(1/4) + x + sqrt(c/d ))/d^6 + 2/231*(21*b^2*d^10*x^(11/2) - 33*b^2*c*d^9*x^(7/2) + 66*a*b*d^10* x^(7/2) + 77*b^2*c^2*d^8*x^(3/2) - 154*a*b*c*d^9*x^(3/2) + 77*a^2*d^10*x^( 3/2))/d^11
Time = 5.46 (sec) , antiderivative size = 435, normalized size of antiderivative = 1.50 \[ \int \frac {x^{5/2} \left (a+b x^2\right )^2}{c+d x^2} \, dx=x^{3/2}\,\left (\frac {2\,a^2}{3\,d}+\frac {c\,\left (\frac {2\,b^2\,c}{d^2}-\frac {4\,a\,b}{d}\right )}{3\,d}\right )-x^{7/2}\,\left (\frac {2\,b^2\,c}{7\,d^2}-\frac {4\,a\,b}{7\,d}\right )+\frac {2\,b^2\,x^{11/2}}{11\,d}-\frac {{\left (-c\right )}^{3/4}\,\mathrm {atan}\left (\frac {{\left (-c\right )}^{3/4}\,d^{1/4}\,\sqrt {x}\,{\left (a\,d-b\,c\right )}^2\,\left (a^4\,c^3\,d^4-4\,a^3\,b\,c^4\,d^3+6\,a^2\,b^2\,c^5\,d^2-4\,a\,b^3\,c^6\,d+b^4\,c^7\right )}{a^6\,c^4\,d^6-6\,a^5\,b\,c^5\,d^5+15\,a^4\,b^2\,c^6\,d^4-20\,a^3\,b^3\,c^7\,d^3+15\,a^2\,b^4\,c^8\,d^2-6\,a\,b^5\,c^9\,d+b^6\,c^{10}}\right )\,{\left (a\,d-b\,c\right )}^2}{d^{15/4}}-\frac {{\left (-c\right )}^{3/4}\,\mathrm {atan}\left (\frac {{\left (-c\right )}^{3/4}\,d^{1/4}\,\sqrt {x}\,{\left (a\,d-b\,c\right )}^2\,\left (a^4\,c^3\,d^4-4\,a^3\,b\,c^4\,d^3+6\,a^2\,b^2\,c^5\,d^2-4\,a\,b^3\,c^6\,d+b^4\,c^7\right )\,1{}\mathrm {i}}{a^6\,c^4\,d^6-6\,a^5\,b\,c^5\,d^5+15\,a^4\,b^2\,c^6\,d^4-20\,a^3\,b^3\,c^7\,d^3+15\,a^2\,b^4\,c^8\,d^2-6\,a\,b^5\,c^9\,d+b^6\,c^{10}}\right )\,{\left (a\,d-b\,c\right )}^2\,1{}\mathrm {i}}{d^{15/4}} \]
x^(3/2)*((2*a^2)/(3*d) + (c*((2*b^2*c)/d^2 - (4*a*b)/d))/(3*d)) - x^(7/2)* ((2*b^2*c)/(7*d^2) - (4*a*b)/(7*d)) + (2*b^2*x^(11/2))/(11*d) - ((-c)^(3/4 )*atan(((-c)^(3/4)*d^(1/4)*x^(1/2)*(a*d - b*c)^2*(b^4*c^7 + a^4*c^3*d^4 - 4*a^3*b*c^4*d^3 + 6*a^2*b^2*c^5*d^2 - 4*a*b^3*c^6*d))/(b^6*c^10 + a^6*c^4* d^6 - 6*a^5*b*c^5*d^5 + 15*a^2*b^4*c^8*d^2 - 20*a^3*b^3*c^7*d^3 + 15*a^4*b ^2*c^6*d^4 - 6*a*b^5*c^9*d))*(a*d - b*c)^2)/d^(15/4) - ((-c)^(3/4)*atan((( -c)^(3/4)*d^(1/4)*x^(1/2)*(a*d - b*c)^2*(b^4*c^7 + a^4*c^3*d^4 - 4*a^3*b*c ^4*d^3 + 6*a^2*b^2*c^5*d^2 - 4*a*b^3*c^6*d)*1i)/(b^6*c^10 + a^6*c^4*d^6 - 6*a^5*b*c^5*d^5 + 15*a^2*b^4*c^8*d^2 - 20*a^3*b^3*c^7*d^3 + 15*a^4*b^2*c^6 *d^4 - 6*a*b^5*c^9*d))*(a*d - b*c)^2*1i)/d^(15/4)