Integrand size = 24, antiderivative size = 207 \[ \int \frac {(a+b x)^2}{(c+d x) (e+f x)^{9/2}} \, dx=\frac {2 (b e-a f)^2}{7 f^2 (d e-c f) (e+f x)^{7/2}}-\frac {2 (b e-a f) (b d e-2 b c f+a d f)}{5 f^2 (d e-c f)^2 (e+f x)^{5/2}}+\frac {2 (b c-a d)^2}{3 (d e-c f)^3 (e+f x)^{3/2}}+\frac {2 d (b c-a d)^2}{(d e-c f)^4 \sqrt {e+f x}}-\frac {2 d^{3/2} (b c-a d)^2 \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{(d e-c f)^{9/2}} \]
2/7*(-a*f+b*e)^2/f^2/(-c*f+d*e)/(f*x+e)^(7/2)-2/5*(-a*f+b*e)*(a*d*f-2*b*c* f+b*d*e)/f^2/(-c*f+d*e)^2/(f*x+e)^(5/2)+2/3*(-a*d+b*c)^2/(-c*f+d*e)^3/(f*x +e)^(3/2)-2*d^(3/2)*(-a*d+b*c)^2*arctanh(d^(1/2)*(f*x+e)^(1/2)/(-c*f+d*e)^ (1/2))/(-c*f+d*e)^(9/2)+2*d*(-a*d+b*c)^2/(-c*f+d*e)^4/(f*x+e)^(1/2)
Time = 0.61 (sec) , antiderivative size = 378, normalized size of antiderivative = 1.83 \[ \int \frac {(a+b x)^2}{(c+d x) (e+f x)^{9/2}} \, dx=\frac {-2 b^2 \left (3 d^3 e^4 (2 e+7 f x)-3 c d^2 e^3 f (13 e+28 f x)+c^3 f^3 \left (8 e^2+28 e f x+35 f^2 x^2\right )-5 c^2 d f^2 \left (16 e^3+56 e^2 f x+70 e f^2 x^2+21 f^3 x^3\right )\right )-4 a b f \left (15 d^3 e^4+3 c^3 f^3 (2 e+7 f x)-c^2 d f^2 \left (32 e^2+112 e f x+35 f^2 x^2\right )+c d^2 f \left (116 e^3+406 e^2 f x+350 e f^2 x^2+105 f^3 x^3\right )\right )+2 a^2 f^2 \left (-15 c^3 f^3+3 c^2 d f^2 (22 e+7 f x)-c d^2 f \left (122 e^2+112 e f x+35 f^2 x^2\right )+d^3 \left (176 e^3+406 e^2 f x+350 e f^2 x^2+105 f^3 x^3\right )\right )}{105 f^2 (d e-c f)^4 (e+f x)^{7/2}}+\frac {2 d^{3/2} (b c-a d)^2 \arctan \left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {-d e+c f}}\right )}{(-d e+c f)^{9/2}} \]
(-2*b^2*(3*d^3*e^4*(2*e + 7*f*x) - 3*c*d^2*e^3*f*(13*e + 28*f*x) + c^3*f^3 *(8*e^2 + 28*e*f*x + 35*f^2*x^2) - 5*c^2*d*f^2*(16*e^3 + 56*e^2*f*x + 70*e *f^2*x^2 + 21*f^3*x^3)) - 4*a*b*f*(15*d^3*e^4 + 3*c^3*f^3*(2*e + 7*f*x) - c^2*d*f^2*(32*e^2 + 112*e*f*x + 35*f^2*x^2) + c*d^2*f*(116*e^3 + 406*e^2*f *x + 350*e*f^2*x^2 + 105*f^3*x^3)) + 2*a^2*f^2*(-15*c^3*f^3 + 3*c^2*d*f^2* (22*e + 7*f*x) - c*d^2*f*(122*e^2 + 112*e*f*x + 35*f^2*x^2) + d^3*(176*e^3 + 406*e^2*f*x + 350*e*f^2*x^2 + 105*f^3*x^3)))/(105*f^2*(d*e - c*f)^4*(e + f*x)^(7/2)) + (2*d^(3/2)*(b*c - a*d)^2*ArcTan[(Sqrt[d]*Sqrt[e + f*x])/Sq rt[-(d*e) + c*f]])/(-(d*e) + c*f)^(9/2)
Time = 0.45 (sec) , antiderivative size = 207, 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 = {98, 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.
\(\displaystyle \int \frac {(a+b x)^2}{(c+d x) (e+f x)^{9/2}} \, dx\) |
\(\Big \downarrow \) 98 |
\(\displaystyle \int \left (\frac {d^2 (a d-b c)^2}{(c+d x) \sqrt {e+f x} (d e-c f)^4}+\frac {f (b c-a d)^2}{(e+f x)^{5/2} (c f-d e)^3}-\frac {d f (a d-b c)^2}{(e+f x)^{3/2} (c f-d e)^4}+\frac {(a f-b e) (-a d f+2 b c f-b d e)}{f (e+f x)^{7/2} (c f-d e)^2}+\frac {(a f-b e)^2}{f (e+f x)^{9/2} (c f-d e)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 d^{3/2} (b c-a d)^2 \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{(d e-c f)^{9/2}}-\frac {2 (b e-a f) (a d f-2 b c f+b d e)}{5 f^2 (e+f x)^{5/2} (d e-c f)^2}+\frac {2 (b e-a f)^2}{7 f^2 (e+f x)^{7/2} (d e-c f)}+\frac {2 d (b c-a d)^2}{\sqrt {e+f x} (d e-c f)^4}+\frac {2 (b c-a d)^2}{3 (e+f x)^{3/2} (d e-c f)^3}\) |
(2*(b*e - a*f)^2)/(7*f^2*(d*e - c*f)*(e + f*x)^(7/2)) - (2*(b*e - a*f)*(b* d*e - 2*b*c*f + a*d*f))/(5*f^2*(d*e - c*f)^2*(e + f*x)^(5/2)) + (2*(b*c - a*d)^2)/(3*(d*e - c*f)^3*(e + f*x)^(3/2)) + (2*d*(b*c - a*d)^2)/((d*e - c* f)^4*Sqrt[e + f*x]) - (2*d^(3/2)*(b*c - a*d)^2*ArcTanh[(Sqrt[d]*Sqrt[e + f *x])/Sqrt[d*e - c*f]])/(d*e - c*f)^(9/2)
3.18.81.3.1 Defintions of rubi rules used
Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_))/((a_.) + (b_.)*(x _)), x_] :> Int[ExpandIntegrand[(e + f*x)^FractionalPart[p], (c + d*x)^n*(( e + f*x)^IntegerPart[p]/(a + b*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[n, 0] && LtQ[p, -1] && FractionQ[p]
Time = 1.06 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.25
method | result | size |
derivativedivides | \(\frac {\frac {2 d^{2} f^{2} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{\left (c f -d e \right )^{4} \sqrt {\left (c f -d e \right ) d}}-\frac {2 \left (a^{2} f^{2}-2 a b f e +b^{2} e^{2}\right )}{7 \left (c f -d e \right ) \left (f x +e \right )^{\frac {7}{2}}}-\frac {2 \left (-a^{2} d \,f^{2}+2 a b c \,f^{2}-2 b^{2} c e f +b^{2} d \,e^{2}\right )}{5 \left (c f -d e \right )^{2} \left (f x +e \right )^{\frac {5}{2}}}-\frac {2 f^{2} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}{3 \left (c f -d e \right )^{3} \left (f x +e \right )^{\frac {3}{2}}}+\frac {2 f^{2} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) d}{\left (c f -d e \right )^{4} \sqrt {f x +e}}}{f^{2}}\) | \(258\) |
default | \(\frac {\frac {2 d^{2} f^{2} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{\left (c f -d e \right )^{4} \sqrt {\left (c f -d e \right ) d}}-\frac {2 \left (a^{2} f^{2}-2 a b f e +b^{2} e^{2}\right )}{7 \left (c f -d e \right ) \left (f x +e \right )^{\frac {7}{2}}}-\frac {2 \left (-a^{2} d \,f^{2}+2 a b c \,f^{2}-2 b^{2} c e f +b^{2} d \,e^{2}\right )}{5 \left (c f -d e \right )^{2} \left (f x +e \right )^{\frac {5}{2}}}-\frac {2 f^{2} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}{3 \left (c f -d e \right )^{3} \left (f x +e \right )^{\frac {3}{2}}}+\frac {2 f^{2} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) d}{\left (c f -d e \right )^{4} \sqrt {f x +e}}}{f^{2}}\) | \(258\) |
pseudoelliptic | \(-\frac {2 \left (-7 d^{2} f^{2} \left (f x +e \right )^{\frac {7}{2}} \left (a d -b c \right )^{2} \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )+\left (\left (-7 a^{2} d^{3} x^{3}+\frac {7 a c \,x^{2} \left (6 b x +a \right ) d^{2}}{3}-\frac {7 \left (5 b^{2} x^{2}+\frac {10}{3} a b x +a^{2}\right ) x \,c^{2} d}{5}+c^{3} \left (\frac {14}{5} a b x +a^{2}+\frac {7}{3} b^{2} x^{2}\right )\right ) f^{5}-\frac {22 \left (\frac {175 a^{2} d^{3} x^{2}}{33}-\frac {56 x \left (\frac {25 b x}{4}+a \right ) c a \,d^{2}}{33}+c^{2} \left (\frac {175}{33} b^{2} x^{2}+\frac {112}{33} a b x +a^{2}\right ) d -\frac {2 \left (\frac {7 b x}{3}+a \right ) b \,c^{3}}{11}\right ) e \,f^{4}}{5}+\frac {122 e^{2} \left (-\frac {203 a^{2} d^{3} x}{61}+a c \left (\frac {406 b x}{61}+a \right ) d^{2}-\frac {32 b \left (\frac {35 b x}{8}+a \right ) c^{2} d}{61}+\frac {4 b^{2} c^{3}}{61}\right ) f^{3}}{15}-\frac {176 e^{3} \left (a^{2} d^{2}-\frac {29 b c \left (-\frac {21 b x}{58}+a \right ) d}{22}+\frac {5 b^{2} c^{2}}{11}\right ) d \,f^{2}}{15}+2 e^{4} b \left (\left (\frac {7 b x}{10}+a \right ) d -\frac {13 b c}{10}\right ) d^{2} f +\frac {2 b^{2} d^{3} e^{5}}{5}\right ) \sqrt {\left (c f -d e \right ) d}\right )}{7 \sqrt {\left (c f -d e \right ) d}\, \left (f x +e \right )^{\frac {7}{2}} f^{2} \left (c f -d e \right )^{4}}\) | \(361\) |
2/f^2*(-1/7*(a^2*f^2-2*a*b*e*f+b^2*e^2)/(c*f-d*e)/(f*x+e)^(7/2)-1/5*(-a^2* d*f^2+2*a*b*c*f^2-2*b^2*c*e*f+b^2*d*e^2)/(c*f-d*e)^2/(f*x+e)^(5/2)-1/3*f^2 *(a^2*d^2-2*a*b*c*d+b^2*c^2)/(c*f-d*e)^3/(f*x+e)^(3/2)+f^2*(a^2*d^2-2*a*b* c*d+b^2*c^2)/(c*f-d*e)^4*d/(f*x+e)^(1/2)+d^2*f^2*(a^2*d^2-2*a*b*c*d+b^2*c^ 2)/(c*f-d*e)^4/((c*f-d*e)*d)^(1/2)*arctan(d*(f*x+e)^(1/2)/((c*f-d*e)*d)^(1 /2)))
Leaf count of result is larger than twice the leaf count of optimal. 894 vs. \(2 (183) = 366\).
Time = 0.32 (sec) , antiderivative size = 1799, normalized size of antiderivative = 8.69 \[ \int \frac {(a+b x)^2}{(c+d x) (e+f x)^{9/2}} \, dx=\text {Too large to display} \]
[1/105*(105*((b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*f^6*x^4 + 4*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*e*f^5*x^3 + 6*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*e ^2*f^4*x^2 + 4*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*e^3*f^3*x + (b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*e^4*f^2)*sqrt(d/(d*e - c*f))*log((d*f*x + 2*d*e - c*f - 2*(d*e - c*f)*sqrt(f*x + e)*sqrt(d/(d*e - c*f)))/(d*x + c)) - 2*(6* b^2*d^3*e^5 + 15*a^2*c^3*f^5 - 105*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*f^5 *x^3 - 3*(13*b^2*c*d^2 - 10*a*b*d^3)*e^4*f - 8*(10*b^2*c^2*d - 29*a*b*c*d^ 2 + 22*a^2*d^3)*e^3*f^2 + 2*(4*b^2*c^3 - 32*a*b*c^2*d + 61*a^2*c*d^2)*e^2* f^3 + 6*(2*a*b*c^3 - 11*a^2*c^2*d)*e*f^4 - 35*(10*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*e*f^4 - (b^2*c^3 - 2*a*b*c^2*d + a^2*c*d^2)*f^5)*x^2 + 7*(3*b^ 2*d^3*e^4*f - 12*b^2*c*d^2*e^3*f^2 - 2*(20*b^2*c^2*d - 58*a*b*c*d^2 + 29*a ^2*d^3)*e^2*f^3 + 4*(b^2*c^3 - 8*a*b*c^2*d + 4*a^2*c*d^2)*e*f^4 + 3*(2*a*b *c^3 - a^2*c^2*d)*f^5)*x)*sqrt(f*x + e))/(d^4*e^8*f^2 - 4*c*d^3*e^7*f^3 + 6*c^2*d^2*e^6*f^4 - 4*c^3*d*e^5*f^5 + c^4*e^4*f^6 + (d^4*e^4*f^6 - 4*c*d^3 *e^3*f^7 + 6*c^2*d^2*e^2*f^8 - 4*c^3*d*e*f^9 + c^4*f^10)*x^4 + 4*(d^4*e^5* f^5 - 4*c*d^3*e^4*f^6 + 6*c^2*d^2*e^3*f^7 - 4*c^3*d*e^2*f^8 + c^4*e*f^9)*x ^3 + 6*(d^4*e^6*f^4 - 4*c*d^3*e^5*f^5 + 6*c^2*d^2*e^4*f^6 - 4*c^3*d*e^3*f^ 7 + c^4*e^2*f^8)*x^2 + 4*(d^4*e^7*f^3 - 4*c*d^3*e^6*f^4 + 6*c^2*d^2*e^5*f^ 5 - 4*c^3*d*e^4*f^6 + c^4*e^3*f^7)*x), -2/105*(105*((b^2*c^2*d - 2*a*b*c*d ^2 + a^2*d^3)*f^6*x^4 + 4*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*e*f^5*x^3...
Time = 8.19 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.20 \[ \int \frac {(a+b x)^2}{(c+d x) (e+f x)^{9/2}} \, dx=\begin {cases} \frac {2 \left (\frac {d f \left (a d - b c\right )^{2}}{\sqrt {e + f x} \left (c f - d e\right )^{4}} + \frac {d f \left (a d - b c\right )^{2} \operatorname {atan}{\left (\frac {\sqrt {e + f x}}{\sqrt {\frac {c f - d e}{d}}} \right )}}{\sqrt {\frac {c f - d e}{d}} \left (c f - d e\right )^{4}} - \frac {f \left (a d - b c\right )^{2}}{3 \left (e + f x\right )^{\frac {3}{2}} \left (c f - d e\right )^{3}} + \frac {\left (a f - b e\right ) \left (a d f - 2 b c f + b d e\right )}{5 f \left (e + f x\right )^{\frac {5}{2}} \left (c f - d e\right )^{2}} - \frac {\left (a f - b e\right )^{2}}{7 f \left (e + f x\right )^{\frac {7}{2}} \left (c f - d e\right )}\right )}{f} & \text {for}\: f \neq 0 \\\frac {\frac {b^{2} x^{2}}{2 d} + \frac {x \left (2 a b d - b^{2} c\right )}{d^{2}} + \frac {\left (a d - b c\right )^{2} \left (\begin {cases} \frac {x}{c} & \text {for}\: d = 0 \\\frac {\log {\left (c + d x \right )}}{d} & \text {otherwise} \end {cases}\right )}{d^{2}}}{e^{\frac {9}{2}}} & \text {otherwise} \end {cases} \]
Piecewise((2*(d*f*(a*d - b*c)**2/(sqrt(e + f*x)*(c*f - d*e)**4) + d*f*(a*d - b*c)**2*atan(sqrt(e + f*x)/sqrt((c*f - d*e)/d))/(sqrt((c*f - d*e)/d)*(c *f - d*e)**4) - f*(a*d - b*c)**2/(3*(e + f*x)**(3/2)*(c*f - d*e)**3) + (a* f - b*e)*(a*d*f - 2*b*c*f + b*d*e)/(5*f*(e + f*x)**(5/2)*(c*f - d*e)**2) - (a*f - b*e)**2/(7*f*(e + f*x)**(7/2)*(c*f - d*e)))/f, Ne(f, 0)), ((b**2*x **2/(2*d) + x*(2*a*b*d - b**2*c)/d**2 + (a*d - b*c)**2*Piecewise((x/c, Eq( d, 0)), (log(c + d*x)/d, True))/d**2)/e**(9/2), True))
Exception generated. \[ \int \frac {(a+b x)^2}{(c+d x) (e+f x)^{9/2}} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(c*f-d*e>0)', see `assume?` for m ore detail
Leaf count of result is larger than twice the leaf count of optimal. 682 vs. \(2 (183) = 366\).
Time = 0.31 (sec) , antiderivative size = 682, normalized size of antiderivative = 3.29 \[ \int \frac {(a+b x)^2}{(c+d x) (e+f x)^{9/2}} \, dx=\frac {2 \, {\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )} \arctan \left (\frac {\sqrt {f x + e} d}{\sqrt {-d^{2} e + c d f}}\right )}{{\left (d^{4} e^{4} - 4 \, c d^{3} e^{3} f + 6 \, c^{2} d^{2} e^{2} f^{2} - 4 \, c^{3} d e f^{3} + c^{4} f^{4}\right )} \sqrt {-d^{2} e + c d f}} - \frac {2 \, {\left (21 \, {\left (f x + e\right )} b^{2} d^{3} e^{4} - 15 \, b^{2} d^{3} e^{5} - 84 \, {\left (f x + e\right )} b^{2} c d^{2} e^{3} f + 45 \, b^{2} c d^{2} e^{4} f + 30 \, a b d^{3} e^{4} f - 105 \, {\left (f x + e\right )}^{3} b^{2} c^{2} d f^{2} + 210 \, {\left (f x + e\right )}^{3} a b c d^{2} f^{2} - 105 \, {\left (f x + e\right )}^{3} a^{2} d^{3} f^{2} - 35 \, {\left (f x + e\right )}^{2} b^{2} c^{2} d e f^{2} + 70 \, {\left (f x + e\right )}^{2} a b c d^{2} e f^{2} - 35 \, {\left (f x + e\right )}^{2} a^{2} d^{3} e f^{2} + 105 \, {\left (f x + e\right )} b^{2} c^{2} d e^{2} f^{2} + 42 \, {\left (f x + e\right )} a b c d^{2} e^{2} f^{2} - 21 \, {\left (f x + e\right )} a^{2} d^{3} e^{2} f^{2} - 45 \, b^{2} c^{2} d e^{3} f^{2} - 90 \, a b c d^{2} e^{3} f^{2} - 15 \, a^{2} d^{3} e^{3} f^{2} + 35 \, {\left (f x + e\right )}^{2} b^{2} c^{3} f^{3} - 70 \, {\left (f x + e\right )}^{2} a b c^{2} d f^{3} + 35 \, {\left (f x + e\right )}^{2} a^{2} c d^{2} f^{3} - 42 \, {\left (f x + e\right )} b^{2} c^{3} e f^{3} - 84 \, {\left (f x + e\right )} a b c^{2} d e f^{3} + 42 \, {\left (f x + e\right )} a^{2} c d^{2} e f^{3} + 15 \, b^{2} c^{3} e^{2} f^{3} + 90 \, a b c^{2} d e^{2} f^{3} + 45 \, a^{2} c d^{2} e^{2} f^{3} + 42 \, {\left (f x + e\right )} a b c^{3} f^{4} - 21 \, {\left (f x + e\right )} a^{2} c^{2} d f^{4} - 30 \, a b c^{3} e f^{4} - 45 \, a^{2} c^{2} d e f^{4} + 15 \, a^{2} c^{3} f^{5}\right )}}{105 \, {\left (d^{4} e^{4} f^{2} - 4 \, c d^{3} e^{3} f^{3} + 6 \, c^{2} d^{2} e^{2} f^{4} - 4 \, c^{3} d e f^{5} + c^{4} f^{6}\right )} {\left (f x + e\right )}^{\frac {7}{2}}} \]
2*(b^2*c^2*d^2 - 2*a*b*c*d^3 + a^2*d^4)*arctan(sqrt(f*x + e)*d/sqrt(-d^2*e + c*d*f))/((d^4*e^4 - 4*c*d^3*e^3*f + 6*c^2*d^2*e^2*f^2 - 4*c^3*d*e*f^3 + c^4*f^4)*sqrt(-d^2*e + c*d*f)) - 2/105*(21*(f*x + e)*b^2*d^3*e^4 - 15*b^2 *d^3*e^5 - 84*(f*x + e)*b^2*c*d^2*e^3*f + 45*b^2*c*d^2*e^4*f + 30*a*b*d^3* e^4*f - 105*(f*x + e)^3*b^2*c^2*d*f^2 + 210*(f*x + e)^3*a*b*c*d^2*f^2 - 10 5*(f*x + e)^3*a^2*d^3*f^2 - 35*(f*x + e)^2*b^2*c^2*d*e*f^2 + 70*(f*x + e)^ 2*a*b*c*d^2*e*f^2 - 35*(f*x + e)^2*a^2*d^3*e*f^2 + 105*(f*x + e)*b^2*c^2*d *e^2*f^2 + 42*(f*x + e)*a*b*c*d^2*e^2*f^2 - 21*(f*x + e)*a^2*d^3*e^2*f^2 - 45*b^2*c^2*d*e^3*f^2 - 90*a*b*c*d^2*e^3*f^2 - 15*a^2*d^3*e^3*f^2 + 35*(f* x + e)^2*b^2*c^3*f^3 - 70*(f*x + e)^2*a*b*c^2*d*f^3 + 35*(f*x + e)^2*a^2*c *d^2*f^3 - 42*(f*x + e)*b^2*c^3*e*f^3 - 84*(f*x + e)*a*b*c^2*d*e*f^3 + 42* (f*x + e)*a^2*c*d^2*e*f^3 + 15*b^2*c^3*e^2*f^3 + 90*a*b*c^2*d*e^2*f^3 + 45 *a^2*c*d^2*e^2*f^3 + 42*(f*x + e)*a*b*c^3*f^4 - 21*(f*x + e)*a^2*c^2*d*f^4 - 30*a*b*c^3*e*f^4 - 45*a^2*c^2*d*e*f^4 + 15*a^2*c^3*f^5)/((d^4*e^4*f^2 - 4*c*d^3*e^3*f^3 + 6*c^2*d^2*e^2*f^4 - 4*c^3*d*e*f^5 + c^4*f^6)*(f*x + e)^ (7/2))
Time = 1.74 (sec) , antiderivative size = 327, normalized size of antiderivative = 1.58 \[ \int \frac {(a+b x)^2}{(c+d x) (e+f x)^{9/2}} \, dx=\frac {2\,d^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {d}\,\sqrt {e+f\,x}\,{\left (a\,d-b\,c\right )}^2\,\left (c^4\,f^4-4\,c^3\,d\,e\,f^3+6\,c^2\,d^2\,e^2\,f^2-4\,c\,d^3\,e^3\,f+d^4\,e^4\right )}{{\left (c\,f-d\,e\right )}^{9/2}\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}\right )\,{\left (a\,d-b\,c\right )}^2}{{\left (c\,f-d\,e\right )}^{9/2}}-\frac {\frac {2\,\left (a^2\,f^2-2\,a\,b\,e\,f+b^2\,e^2\right )}{7\,\left (c\,f-d\,e\right )}+\frac {2\,{\left (e+f\,x\right )}^2\,\left (a^2\,d^2\,f^2-2\,a\,b\,c\,d\,f^2+b^2\,c^2\,f^2\right )}{3\,{\left (c\,f-d\,e\right )}^3}-\frac {2\,\left (e+f\,x\right )\,\left (d\,a^2\,f^2-2\,c\,a\,b\,f^2-d\,b^2\,e^2+2\,c\,b^2\,e\,f\right )}{5\,{\left (c\,f-d\,e\right )}^2}-\frac {2\,d\,{\left (e+f\,x\right )}^3\,\left (a^2\,d^2\,f^2-2\,a\,b\,c\,d\,f^2+b^2\,c^2\,f^2\right )}{{\left (c\,f-d\,e\right )}^4}}{f^2\,{\left (e+f\,x\right )}^{7/2}} \]
(2*d^(3/2)*atan((d^(1/2)*(e + f*x)^(1/2)*(a*d - b*c)^2*(c^4*f^4 + d^4*e^4 + 6*c^2*d^2*e^2*f^2 - 4*c*d^3*e^3*f - 4*c^3*d*e*f^3))/((c*f - d*e)^(9/2)*( a^2*d^2 + b^2*c^2 - 2*a*b*c*d)))*(a*d - b*c)^2)/(c*f - d*e)^(9/2) - ((2*(a ^2*f^2 + b^2*e^2 - 2*a*b*e*f))/(7*(c*f - d*e)) + (2*(e + f*x)^2*(a^2*d^2*f ^2 + b^2*c^2*f^2 - 2*a*b*c*d*f^2))/(3*(c*f - d*e)^3) - (2*(e + f*x)*(a^2*d *f^2 - b^2*d*e^2 - 2*a*b*c*f^2 + 2*b^2*c*e*f))/(5*(c*f - d*e)^2) - (2*d*(e + f*x)^3*(a^2*d^2*f^2 + b^2*c^2*f^2 - 2*a*b*c*d*f^2))/(c*f - d*e)^4)/(f^2 *(e + f*x)^(7/2))