Integrand size = 24, antiderivative size = 173 \[ \int \frac {(a+b x)^2}{(c+d x) (e+f x)^{7/2}} \, dx=\frac {2 (b e-a f)^2}{5 f^2 (d e-c f) (e+f x)^{5/2}}-\frac {2 (b e-a f) (b d e-2 b c f+a d f)}{3 f^2 (d e-c f)^2 (e+f x)^{3/2}}+\frac {2 (b c-a d)^2}{(d e-c f)^3 \sqrt {e+f x}}-\frac {2 \sqrt {d} (b c-a d)^2 \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{(d e-c f)^{7/2}} \]
2/5*(-a*f+b*e)^2/f^2/(-c*f+d*e)/(f*x+e)^(5/2)-2/3*(-a*f+b*e)*(a*d*f-2*b*c* f+b*d*e)/f^2/(-c*f+d*e)^2/(f*x+e)^(3/2)-2*(-a*d+b*c)^2*arctanh(d^(1/2)*(f* x+e)^(1/2)/(-c*f+d*e)^(1/2))*d^(1/2)/(-c*f+d*e)^(7/2)+2*(-a*d+b*c)^2/(-c*f +d*e)^3/(f*x+e)^(1/2)
Time = 0.41 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.46 \[ \int \frac {(a+b x)^2}{(c+d x) (e+f x)^{7/2}} \, dx=-\frac {2 \left (b^2 \left (-d^2 e^3 (2 e+5 f x)+3 c d e^2 f (3 e+5 f x)+c^2 f^2 \left (8 e^2+20 e f x+15 f^2 x^2\right )\right )-2 a b f \left (3 d^2 e^3-c^2 f^2 (2 e+5 f x)+c d f \left (14 e^2+35 e f x+15 f^2 x^2\right )\right )+a^2 f^2 \left (3 c^2 f^2-c d f (11 e+5 f x)+d^2 \left (23 e^2+35 e f x+15 f^2 x^2\right )\right )\right )}{15 f^2 (-d e+c f)^3 (e+f x)^{5/2}}-\frac {2 \sqrt {d} (b c-a d)^2 \arctan \left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {-d e+c f}}\right )}{(-d e+c f)^{7/2}} \]
(-2*(b^2*(-(d^2*e^3*(2*e + 5*f*x)) + 3*c*d*e^2*f*(3*e + 5*f*x) + c^2*f^2*( 8*e^2 + 20*e*f*x + 15*f^2*x^2)) - 2*a*b*f*(3*d^2*e^3 - c^2*f^2*(2*e + 5*f* x) + c*d*f*(14*e^2 + 35*e*f*x + 15*f^2*x^2)) + a^2*f^2*(3*c^2*f^2 - c*d*f* (11*e + 5*f*x) + d^2*(23*e^2 + 35*e*f*x + 15*f^2*x^2))))/(15*f^2*(-(d*e) + c*f)^3*(e + f*x)^(5/2)) - (2*Sqrt[d]*(b*c - a*d)^2*ArcTan[(Sqrt[d]*Sqrt[e + f*x])/Sqrt[-(d*e) + c*f]])/(-(d*e) + c*f)^(7/2)
Time = 0.39 (sec) , antiderivative size = 173, 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.
| \(\displaystyle \int \frac {(a+b x)^2}{(c+d x) (e+f x)^{7/2}} \, dx\) |
| \(\Big \downarrow \) 98 |
| \(\displaystyle \int \left (\frac {f (b c-a d)^2}{(e+f x)^{3/2} (c f-d e)^3}+\frac {d (a d-b c)^2}{(c+d x) \sqrt {e+f x} (d e-c f)^3}+\frac {(a f-b e) (-a d f+2 b c f-b d e)}{f (e+f x)^{5/2} (c f-d e)^2}+\frac {(a f-b e)^2}{f (e+f x)^{7/2} (c f-d e)}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {2 \sqrt {d} (b c-a d)^2 \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{(d e-c f)^{7/2}}-\frac {2 (b e-a f) (a d f-2 b c f+b d e)}{3 f^2 (e+f x)^{3/2} (d e-c f)^2}+\frac {2 (b e-a f)^2}{5 f^2 (e+f x)^{5/2} (d e-c f)}+\frac {2 (b c-a d)^2}{\sqrt {e+f x} (d e-c f)^3}\) |
(2*(b*e - a*f)^2)/(5*f^2*(d*e - c*f)*(e + f*x)^(5/2)) - (2*(b*e - a*f)*(b* d*e - 2*b*c*f + a*d*f))/(3*f^2*(d*e - c*f)^2*(e + f*x)^(3/2)) + (2*(b*c - a*d)^2)/((d*e - c*f)^3*Sqrt[e + f*x]) - (2*Sqrt[d]*(b*c - a*d)^2*ArcTanh[( Sqrt[d]*Sqrt[e + f*x])/Sqrt[d*e - c*f]])/(d*e - c*f)^(7/2)
3.18.80.3.1 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 = 3.02 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.24
| method | result | size |
| derivativedivides | \(\frac {-\frac {2 d \,f^{2} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{\left (c f -d e \right )^{3} \sqrt {\left (c f -d e \right ) d}}-\frac {2 \left (a^{2} f^{2}-2 a b f e +b^{2} e^{2}\right )}{5 \left (c f -d e \right ) \left (f x +e \right )^{\frac {5}{2}}}-\frac {2 \left (-a^{2} d \,f^{2}+2 a b c \,f^{2}-2 b^{2} c e f +b^{2} d \,e^{2}\right )}{3 \left (c f -d e \right )^{2} \left (f x +e \right )^{\frac {3}{2}}}-\frac {2 f^{2} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}{\left (c f -d e \right )^{3} \sqrt {f x +e}}}{f^{2}}\) | \(214\) |
| default | \(\frac {-\frac {2 d \,f^{2} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{\left (c f -d e \right )^{3} \sqrt {\left (c f -d e \right ) d}}-\frac {2 \left (a^{2} f^{2}-2 a b f e +b^{2} e^{2}\right )}{5 \left (c f -d e \right ) \left (f x +e \right )^{\frac {5}{2}}}-\frac {2 \left (-a^{2} d \,f^{2}+2 a b c \,f^{2}-2 b^{2} c e f +b^{2} d \,e^{2}\right )}{3 \left (c f -d e \right )^{2} \left (f x +e \right )^{\frac {3}{2}}}-\frac {2 f^{2} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}{\left (c f -d e \right )^{3} \sqrt {f x +e}}}{f^{2}}\) | \(214\) |
| pseudoelliptic | \(-\frac {2 \left (5 d \,f^{2} \left (f x +e \right )^{\frac {5}{2}} \left (a d -b c \right )^{2} \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )+\sqrt {\left (c f -d e \right ) d}\, \left (\left (5 a^{2} d^{2} x^{2}-\frac {5 a c x \left (6 b x +a \right ) d}{3}+c^{2} \left (5 b^{2} x^{2}+\frac {10}{3} a b x +a^{2}\right )\right ) f^{4}-\frac {11 e \left (-\frac {35 a^{2} d^{2} x}{11}+a c \left (\frac {70 b x}{11}+a \right ) d -\frac {4 b \,c^{2} \left (5 b x +a \right )}{11}\right ) f^{3}}{3}+\frac {23 e^{2} \left (a^{2} d^{2}-\frac {28 b \left (-\frac {15 b x}{28}+a \right ) c d}{23}+\frac {8 b^{2} c^{2}}{23}\right ) f^{2}}{3}-2 e^{3} \left (\left (\frac {5 b x}{6}+a \right ) d -\frac {3 b c}{2}\right ) b d f -\frac {2 b^{2} d^{2} e^{4}}{3}\right )\right )}{5 \sqrt {\left (c f -d e \right ) d}\, \left (f x +e \right )^{\frac {5}{2}} f^{2} \left (c f -d e \right )^{3}}\) | \(249\) |
2/f^2*(-1/5*(a^2*f^2-2*a*b*e*f+b^2*e^2)/(c*f-d*e)/(f*x+e)^(5/2)-1/3*(-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)^(3/2)-f^2*(a^ 2*d^2-2*a*b*c*d+b^2*c^2)/(c*f-d*e)^3/(f*x+e)^(1/2)-d*f^2*(a^2*d^2-2*a*b*c* d+b^2*c^2)/(c*f-d*e)^3/((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. 581 vs. \(2 (153) = 306\).
Time = 0.28 (sec) , antiderivative size = 1173, normalized size of antiderivative = 6.78 \[ \int \frac {(a+b x)^2}{(c+d x) (e+f x)^{7/2}} \, dx=\left [-\frac {15 \, {\left ({\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} f^{5} x^{3} + 3 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} e f^{4} x^{2} + 3 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} e^{2} f^{3} x + {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} e^{3} f^{2}\right )} \sqrt {\frac {d}{d e - c f}} \log \left (\frac {d f x + 2 \, d e - c f + 2 \, {\left (d e - c f\right )} \sqrt {f x + e} \sqrt {\frac {d}{d e - c f}}}{d x + c}\right ) + 2 \, {\left (2 \, b^{2} d^{2} e^{4} - 3 \, a^{2} c^{2} f^{4} - 15 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} f^{4} x^{2} - 3 \, {\left (3 \, b^{2} c d - 2 \, a b d^{2}\right )} e^{3} f - {\left (8 \, b^{2} c^{2} - 28 \, a b c d + 23 \, a^{2} d^{2}\right )} e^{2} f^{2} - {\left (4 \, a b c^{2} - 11 \, a^{2} c d\right )} e f^{3} + 5 \, {\left (b^{2} d^{2} e^{3} f - 3 \, b^{2} c d e^{2} f^{2} - {\left (4 \, b^{2} c^{2} - 14 \, a b c d + 7 \, a^{2} d^{2}\right )} e f^{3} - {\left (2 \, a b c^{2} - a^{2} c d\right )} f^{4}\right )} x\right )} \sqrt {f x + e}}{15 \, {\left (d^{3} e^{6} f^{2} - 3 \, c d^{2} e^{5} f^{3} + 3 \, c^{2} d e^{4} f^{4} - c^{3} e^{3} f^{5} + {\left (d^{3} e^{3} f^{5} - 3 \, c d^{2} e^{2} f^{6} + 3 \, c^{2} d e f^{7} - c^{3} f^{8}\right )} x^{3} + 3 \, {\left (d^{3} e^{4} f^{4} - 3 \, c d^{2} e^{3} f^{5} + 3 \, c^{2} d e^{2} f^{6} - c^{3} e f^{7}\right )} x^{2} + 3 \, {\left (d^{3} e^{5} f^{3} - 3 \, c d^{2} e^{4} f^{4} + 3 \, c^{2} d e^{3} f^{5} - c^{3} e^{2} f^{6}\right )} x\right )}}, -\frac {2 \, {\left (15 \, {\left ({\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} f^{5} x^{3} + 3 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} e f^{4} x^{2} + 3 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} e^{2} f^{3} x + {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} e^{3} f^{2}\right )} \sqrt {-\frac {d}{d e - c f}} \arctan \left (-\frac {{\left (d e - c f\right )} \sqrt {f x + e} \sqrt {-\frac {d}{d e - c f}}}{d f x + d e}\right ) + {\left (2 \, b^{2} d^{2} e^{4} - 3 \, a^{2} c^{2} f^{4} - 15 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} f^{4} x^{2} - 3 \, {\left (3 \, b^{2} c d - 2 \, a b d^{2}\right )} e^{3} f - {\left (8 \, b^{2} c^{2} - 28 \, a b c d + 23 \, a^{2} d^{2}\right )} e^{2} f^{2} - {\left (4 \, a b c^{2} - 11 \, a^{2} c d\right )} e f^{3} + 5 \, {\left (b^{2} d^{2} e^{3} f - 3 \, b^{2} c d e^{2} f^{2} - {\left (4 \, b^{2} c^{2} - 14 \, a b c d + 7 \, a^{2} d^{2}\right )} e f^{3} - {\left (2 \, a b c^{2} - a^{2} c d\right )} f^{4}\right )} x\right )} \sqrt {f x + e}\right )}}{15 \, {\left (d^{3} e^{6} f^{2} - 3 \, c d^{2} e^{5} f^{3} + 3 \, c^{2} d e^{4} f^{4} - c^{3} e^{3} f^{5} + {\left (d^{3} e^{3} f^{5} - 3 \, c d^{2} e^{2} f^{6} + 3 \, c^{2} d e f^{7} - c^{3} f^{8}\right )} x^{3} + 3 \, {\left (d^{3} e^{4} f^{4} - 3 \, c d^{2} e^{3} f^{5} + 3 \, c^{2} d e^{2} f^{6} - c^{3} e f^{7}\right )} x^{2} + 3 \, {\left (d^{3} e^{5} f^{3} - 3 \, c d^{2} e^{4} f^{4} + 3 \, c^{2} d e^{3} f^{5} - c^{3} e^{2} f^{6}\right )} x\right )}}\right ] \]
[-1/15*(15*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*f^5*x^3 + 3*(b^2*c^2 - 2*a*b*c *d + a^2*d^2)*e*f^4*x^2 + 3*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*e^2*f^3*x + (b ^2*c^2 - 2*a*b*c*d + a^2*d^2)*e^3*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*(2*b^2*d^2*e^4 - 3*a^2*c^2*f^4 - 15*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*f^4* x^2 - 3*(3*b^2*c*d - 2*a*b*d^2)*e^3*f - (8*b^2*c^2 - 28*a*b*c*d + 23*a^2*d ^2)*e^2*f^2 - (4*a*b*c^2 - 11*a^2*c*d)*e*f^3 + 5*(b^2*d^2*e^3*f - 3*b^2*c* d*e^2*f^2 - (4*b^2*c^2 - 14*a*b*c*d + 7*a^2*d^2)*e*f^3 - (2*a*b*c^2 - a^2* c*d)*f^4)*x)*sqrt(f*x + e))/(d^3*e^6*f^2 - 3*c*d^2*e^5*f^3 + 3*c^2*d*e^4*f ^4 - c^3*e^3*f^5 + (d^3*e^3*f^5 - 3*c*d^2*e^2*f^6 + 3*c^2*d*e*f^7 - c^3*f^ 8)*x^3 + 3*(d^3*e^4*f^4 - 3*c*d^2*e^3*f^5 + 3*c^2*d*e^2*f^6 - c^3*e*f^7)*x ^2 + 3*(d^3*e^5*f^3 - 3*c*d^2*e^4*f^4 + 3*c^2*d*e^3*f^5 - c^3*e^2*f^6)*x), -2/15*(15*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*f^5*x^3 + 3*(b^2*c^2 - 2*a*b*c *d + a^2*d^2)*e*f^4*x^2 + 3*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*e^2*f^3*x + (b ^2*c^2 - 2*a*b*c*d + a^2*d^2)*e^3*f^2)*sqrt(-d/(d*e - c*f))*arctan(-(d*e - c*f)*sqrt(f*x + e)*sqrt(-d/(d*e - c*f))/(d*f*x + d*e)) + (2*b^2*d^2*e^4 - 3*a^2*c^2*f^4 - 15*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*f^4*x^2 - 3*(3*b^2*c*d - 2*a*b*d^2)*e^3*f - (8*b^2*c^2 - 28*a*b*c*d + 23*a^2*d^2)*e^2*f^2 - (4*a *b*c^2 - 11*a^2*c*d)*e*f^3 + 5*(b^2*d^2*e^3*f - 3*b^2*c*d*e^2*f^2 - (4*b^2 *c^2 - 14*a*b*c*d + 7*a^2*d^2)*e*f^3 - (2*a*b*c^2 - a^2*c*d)*f^4)*x)*sq...
Time = 7.32 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.24 \[ \int \frac {(a+b x)^2}{(c+d x) (e+f x)^{7/2}} \, dx=\begin {cases} \frac {2 \left (- \frac {f \left (a d - b c\right )^{2}}{\sqrt {e + f x} \left (c f - d e\right )^{3}} - \frac {f \left (a d - b c\right )^{2} \operatorname {atan}{\left (\frac {\sqrt {e + f x}}{\sqrt {\frac {c f - d e}{d}}} \right )}}{\sqrt {\frac {c f - d e}{d}} \left (c f - d e\right )^{3}} + \frac {\left (a f - b e\right ) \left (a d f - 2 b c f + b d e\right )}{3 f \left (e + f x\right )^{\frac {3}{2}} \left (c f - d e\right )^{2}} - \frac {\left (a f - b e\right )^{2}}{5 f \left (e + f x\right )^{\frac {5}{2}} \left (c f - d e\right )}\right )}{f} & \text {for}\: f \neq 0 \\\frac {\frac {b^{2} x^{2}}{2 d} + \frac {x \left (2 a b d - b^{2} c\right )}{d^{2}} + \frac {\left (a d - b c\right )^{2} \left (\begin {cases} \frac {x}{c} & \text {for}\: d = 0 \\\frac {\log {\left (c + d x \right )}}{d} & \text {otherwise} \end {cases}\right )}{d^{2}}}{e^{\frac {7}{2}}} & \text {otherwise} \end {cases} \]
Piecewise((2*(-f*(a*d - b*c)**2/(sqrt(e + f*x)*(c*f - d*e)**3) - 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)**3) + (a*f - b*e)*(a*d*f - 2*b*c*f + b*d*e)/(3*f*(e + f*x)**(3/2)*( c*f - d*e)**2) - (a*f - b*e)**2/(5*f*(e + f*x)**(5/2)*(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*Pie cewise((x/c, Eq(d, 0)), (log(c + d*x)/d, True))/d**2)/e**(7/2), True))
Exception generated. \[ \int \frac {(a+b x)^2}{(c+d x) (e+f x)^{7/2}} \, dx=\text {Exception raised: ValueError} \]
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. 423 vs. \(2 (153) = 306\).
Time = 0.28 (sec) , antiderivative size = 423, normalized size of antiderivative = 2.45 \[ \int \frac {(a+b x)^2}{(c+d x) (e+f x)^{7/2}} \, dx=\frac {2 \, {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} \arctan \left (\frac {\sqrt {f x + e} d}{\sqrt {-d^{2} e + c d f}}\right )}{{\left (d^{3} e^{3} - 3 \, c d^{2} e^{2} f + 3 \, c^{2} d e f^{2} - c^{3} f^{3}\right )} \sqrt {-d^{2} e + c d f}} - \frac {2 \, {\left (5 \, {\left (f x + e\right )} b^{2} d^{2} e^{3} - 3 \, b^{2} d^{2} e^{4} - 15 \, {\left (f x + e\right )} b^{2} c d e^{2} f + 6 \, b^{2} c d e^{3} f + 6 \, a b d^{2} e^{3} f - 15 \, {\left (f x + e\right )}^{2} b^{2} c^{2} f^{2} + 30 \, {\left (f x + e\right )}^{2} a b c d f^{2} - 15 \, {\left (f x + e\right )}^{2} a^{2} d^{2} f^{2} + 10 \, {\left (f x + e\right )} b^{2} c^{2} e f^{2} + 10 \, {\left (f x + e\right )} a b c d e f^{2} - 5 \, {\left (f x + e\right )} a^{2} d^{2} e f^{2} - 3 \, b^{2} c^{2} e^{2} f^{2} - 12 \, a b c d e^{2} f^{2} - 3 \, a^{2} d^{2} e^{2} f^{2} - 10 \, {\left (f x + e\right )} a b c^{2} f^{3} + 5 \, {\left (f x + e\right )} a^{2} c d f^{3} + 6 \, a b c^{2} e f^{3} + 6 \, a^{2} c d e f^{3} - 3 \, a^{2} c^{2} f^{4}\right )}}{15 \, {\left (d^{3} e^{3} f^{2} - 3 \, c d^{2} e^{2} f^{3} + 3 \, c^{2} d e f^{4} - c^{3} f^{5}\right )} {\left (f x + e\right )}^{\frac {5}{2}}} \]
2*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*arctan(sqrt(f*x + e)*d/sqrt(-d^2*e + c*d*f))/((d^3*e^3 - 3*c*d^2*e^2*f + 3*c^2*d*e*f^2 - c^3*f^3)*sqrt(-d^2*e + c*d*f)) - 2/15*(5*(f*x + e)*b^2*d^2*e^3 - 3*b^2*d^2*e^4 - 15*(f*x + e)*b ^2*c*d*e^2*f + 6*b^2*c*d*e^3*f + 6*a*b*d^2*e^3*f - 15*(f*x + e)^2*b^2*c^2* f^2 + 30*(f*x + e)^2*a*b*c*d*f^2 - 15*(f*x + e)^2*a^2*d^2*f^2 + 10*(f*x + e)*b^2*c^2*e*f^2 + 10*(f*x + e)*a*b*c*d*e*f^2 - 5*(f*x + e)*a^2*d^2*e*f^2 - 3*b^2*c^2*e^2*f^2 - 12*a*b*c*d*e^2*f^2 - 3*a^2*d^2*e^2*f^2 - 10*(f*x + e )*a*b*c^2*f^3 + 5*(f*x + e)*a^2*c*d*f^3 + 6*a*b*c^2*e*f^3 + 6*a^2*c*d*e*f^ 3 - 3*a^2*c^2*f^4)/((d^3*e^3*f^2 - 3*c*d^2*e^2*f^3 + 3*c^2*d*e*f^4 - c^3*f ^5)*(f*x + e)^(5/2))
Time = 1.66 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.53 \[ \int \frac {(a+b x)^2}{(c+d x) (e+f x)^{7/2}} \, dx=-\frac {\frac {2\,\left (a^2\,f^2-2\,a\,b\,e\,f+b^2\,e^2\right )}{5\,\left (c\,f-d\,e\right )}+\frac {2\,{\left (e+f\,x\right )}^2\,\left (a^2\,d^2\,f^2-2\,a\,b\,c\,d\,f^2+b^2\,c^2\,f^2\right )}{{\left (c\,f-d\,e\right )}^3}-\frac {2\,\left (e+f\,x\right )\,\left (d\,a^2\,f^2-2\,c\,a\,b\,f^2-d\,b^2\,e^2+2\,c\,b^2\,e\,f\right )}{3\,{\left (c\,f-d\,e\right )}^2}}{f^2\,{\left (e+f\,x\right )}^{5/2}}-\frac {2\,\sqrt {d}\,\mathrm {atan}\left (\frac {\sqrt {d}\,\sqrt {e+f\,x}\,{\left (a\,d-b\,c\right )}^2\,\left (c^3\,f^3-3\,c^2\,d\,e\,f^2+3\,c\,d^2\,e^2\,f-d^3\,e^3\right )}{{\left (c\,f-d\,e\right )}^{7/2}\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}\right )\,{\left (a\,d-b\,c\right )}^2}{{\left (c\,f-d\,e\right )}^{7/2}} \]
- ((2*(a^2*f^2 + b^2*e^2 - 2*a*b*e*f))/(5*(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))/(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))/(3*(c*f - d*e)^2))/(f^2 *(e + f*x)^(5/2)) - (2*d^(1/2)*atan((d^(1/2)*(e + f*x)^(1/2)*(a*d - b*c)^2 *(c^3*f^3 - d^3*e^3 + 3*c*d^2*e^2*f - 3*c^2*d*e*f^2))/((c*f - d*e)^(7/2)*( a^2*d^2 + b^2*c^2 - 2*a*b*c*d)))*(a*d - b*c)^2)/(c*f - d*e)^(7/2)