Integrand size = 24, antiderivative size = 140 \[ \int \frac {(a+b x)^2}{(c+d x) (e+f x)^{5/2}} \, dx=\frac {2 (b e-a f)^2}{3 f^2 (d e-c f) (e+f x)^{3/2}}-\frac {2 (b e-a f) (b d e-2 b c f+a d f)}{f^2 (d e-c f)^2 \sqrt {e+f x}}-\frac {2 (b c-a d)^2 \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{\sqrt {d} (d e-c f)^{5/2}} \]
2/3*(-a*f+b*e)^2/f^2/(-c*f+d*e)/(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))/(-c*f+d*e)^(5/2)/d^(1/2)-2*(-a*f+b*e)*( a*d*f-2*b*c*f+b*d*e)/f^2/(-c*f+d*e)^2/(f*x+e)^(1/2)
Time = 0.26 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.96 \[ \int \frac {(a+b x)^2}{(c+d x) (e+f x)^{5/2}} \, dx=-\frac {2 (b e-a f) (b d e (2 e+3 f x)-b c f (5 e+6 f x)+a f (4 d e-c f+3 d f x))}{3 f^2 (d e-c f)^2 (e+f x)^{3/2}}+\frac {2 (b c-a d)^2 \arctan \left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {-d e+c f}}\right )}{\sqrt {d} (-d e+c f)^{5/2}} \]
(-2*(b*e - a*f)*(b*d*e*(2*e + 3*f*x) - b*c*f*(5*e + 6*f*x) + a*f*(4*d*e - c*f + 3*d*f*x)))/(3*f^2*(d*e - c*f)^2*(e + f*x)^(3/2)) + (2*(b*c - a*d)^2* ArcTan[(Sqrt[d]*Sqrt[e + f*x])/Sqrt[-(d*e) + c*f]])/(Sqrt[d]*(-(d*e) + c*f )^(5/2))
Time = 0.33 (sec) , antiderivative size = 140, 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.
| \(\displaystyle \int \frac {(a+b x)^2}{(c+d x) (e+f x)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 98 |
| \(\displaystyle \int \left (\frac {(b c-a d)^2}{(c+d x) \sqrt {e+f x} (d e-c f)^2}+\frac {(a f-b e) (-a d f+2 b c f-b d e)}{f (e+f x)^{3/2} (c f-d e)^2}+\frac {(a f-b e)^2}{f (e+f x)^{5/2} (c f-d e)}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {2 (b c-a d)^2 \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{\sqrt {d} (d e-c f)^{5/2}}-\frac {2 (b e-a f) (a d f-2 b c f+b d e)}{f^2 \sqrt {e+f x} (d e-c f)^2}+\frac {2 (b e-a f)^2}{3 f^2 (e+f x)^{3/2} (d e-c f)}\) |
(2*(b*e - a*f)^2)/(3*f^2*(d*e - c*f)*(e + f*x)^(3/2)) - (2*(b*e - a*f)*(b* d*e - 2*b*c*f + a*d*f))/(f^2*(d*e - c*f)^2*Sqrt[e + f*x]) - (2*(b*c - a*d) ^2*ArcTanh[(Sqrt[d]*Sqrt[e + f*x])/Sqrt[d*e - c*f]])/(Sqrt[d]*(d*e - c*f)^ (5/2))
3.18.79.3.1 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 = 3.08 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.04
| method | result | size |
| pseudoelliptic | \(\frac {2 \left (a d -b c \right )^{2} f^{2} \left (f x +e \right )^{\frac {3}{2}} \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )-\frac {2 \left (\left (6 b c x +a \left (-3 d x +c \right )\right ) f^{2}-4 e \left (\frac {\left (3 d x -5 c \right ) b}{4}+a d \right ) f -2 b d \,e^{2}\right ) \left (a f -b e \right ) \sqrt {\left (c f -d e \right ) d}}{3}}{f^{2} \left (c f -d e \right )^{2} \sqrt {\left (c f -d e \right ) d}\, \left (f x +e \right )^{\frac {3}{2}}}\) | \(146\) |
| derivativedivides | \(\frac {\frac {2 f^{2} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{\left (c f -d e \right )^{2} \sqrt {\left (c f -d e \right ) d}}-\frac {2 \left (a^{2} f^{2}-2 a b f e +b^{2} e^{2}\right )}{3 \left (c f -d e \right ) \left (f x +e \right )^{\frac {3}{2}}}-\frac {2 \left (-a^{2} d \,f^{2}+2 a b c \,f^{2}-2 b^{2} c e f +b^{2} d \,e^{2}\right )}{\left (c f -d e \right )^{2} \sqrt {f x +e}}}{f^{2}}\) | \(169\) |
| default | \(\frac {\frac {2 f^{2} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{\left (c f -d e \right )^{2} \sqrt {\left (c f -d e \right ) d}}-\frac {2 \left (a^{2} f^{2}-2 a b f e +b^{2} e^{2}\right )}{3 \left (c f -d e \right ) \left (f x +e \right )^{\frac {3}{2}}}-\frac {2 \left (-a^{2} d \,f^{2}+2 a b c \,f^{2}-2 b^{2} c e f +b^{2} d \,e^{2}\right )}{\left (c f -d e \right )^{2} \sqrt {f x +e}}}{f^{2}}\) | \(169\) |
2*((a*d-b*c)^2*f^2*(f*x+e)^(3/2)*arctan(d*(f*x+e)^(1/2)/((c*f-d*e)*d)^(1/2 ))-1/3*((6*b*c*x+a*(-3*d*x+c))*f^2-4*e*(1/4*(3*d*x-5*c)*b+a*d)*f-2*b*d*e^2 )*(a*f-b*e)*((c*f-d*e)*d)^(1/2))/((c*f-d*e)*d)^(1/2)/(f*x+e)^(3/2)/f^2/(c* f-d*e)^2
Leaf count of result is larger than twice the leaf count of optimal. 481 vs. \(2 (124) = 248\).
Time = 0.25 (sec) , antiderivative size = 975, normalized size of antiderivative = 6.96 \[ \int \frac {(a+b x)^2}{(c+d x) (e+f x)^{5/2}} \, dx=\left [\frac {3 \, {\left ({\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} f^{4} x^{2} + 2 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} e f^{3} x + {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} e^{2} f^{2}\right )} \sqrt {d^{2} e - c d f} \log \left (\frac {d f x + 2 \, d e - c f - 2 \, \sqrt {d^{2} e - c d f} \sqrt {f x + e}}{d x + c}\right ) - 2 \, {\left (2 \, b^{2} d^{3} e^{4} - a^{2} c^{2} d f^{4} - {\left (7 \, b^{2} c d^{2} - 2 \, a b d^{3}\right )} e^{3} f + {\left (5 \, b^{2} c^{2} d + 2 \, a b c d^{2} - 4 \, a^{2} d^{3}\right )} e^{2} f^{2} - {\left (4 \, a b c^{2} d - 5 \, a^{2} c d^{2}\right )} e f^{3} + 3 \, {\left (b^{2} d^{3} e^{3} f - 3 \, b^{2} c d^{2} e^{2} f^{2} + {\left (2 \, b^{2} c^{2} d + 2 \, a b c d^{2} - a^{2} d^{3}\right )} e f^{3} - {\left (2 \, a b c^{2} d - a^{2} c d^{2}\right )} f^{4}\right )} x\right )} \sqrt {f x + e}}{3 \, {\left (d^{4} e^{5} f^{2} - 3 \, c d^{3} e^{4} f^{3} + 3 \, c^{2} d^{2} e^{3} f^{4} - c^{3} d e^{2} f^{5} + {\left (d^{4} e^{3} f^{4} - 3 \, c d^{3} e^{2} f^{5} + 3 \, c^{2} d^{2} e f^{6} - c^{3} d f^{7}\right )} x^{2} + 2 \, {\left (d^{4} e^{4} f^{3} - 3 \, c d^{3} e^{3} f^{4} + 3 \, c^{2} d^{2} e^{2} f^{5} - c^{3} d e f^{6}\right )} x\right )}}, \frac {2 \, {\left (3 \, {\left ({\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} f^{4} x^{2} + 2 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} e f^{3} x + {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} e^{2} f^{2}\right )} \sqrt {-d^{2} e + c d f} \arctan \left (\frac {\sqrt {-d^{2} e + c d f} \sqrt {f x + e}}{d f x + d e}\right ) - {\left (2 \, b^{2} d^{3} e^{4} - a^{2} c^{2} d f^{4} - {\left (7 \, b^{2} c d^{2} - 2 \, a b d^{3}\right )} e^{3} f + {\left (5 \, b^{2} c^{2} d + 2 \, a b c d^{2} - 4 \, a^{2} d^{3}\right )} e^{2} f^{2} - {\left (4 \, a b c^{2} d - 5 \, a^{2} c d^{2}\right )} e f^{3} + 3 \, {\left (b^{2} d^{3} e^{3} f - 3 \, b^{2} c d^{2} e^{2} f^{2} + {\left (2 \, b^{2} c^{2} d + 2 \, a b c d^{2} - a^{2} d^{3}\right )} e f^{3} - {\left (2 \, a b c^{2} d - a^{2} c d^{2}\right )} f^{4}\right )} x\right )} \sqrt {f x + e}\right )}}{3 \, {\left (d^{4} e^{5} f^{2} - 3 \, c d^{3} e^{4} f^{3} + 3 \, c^{2} d^{2} e^{3} f^{4} - c^{3} d e^{2} f^{5} + {\left (d^{4} e^{3} f^{4} - 3 \, c d^{3} e^{2} f^{5} + 3 \, c^{2} d^{2} e f^{6} - c^{3} d f^{7}\right )} x^{2} + 2 \, {\left (d^{4} e^{4} f^{3} - 3 \, c d^{3} e^{3} f^{4} + 3 \, c^{2} d^{2} e^{2} f^{5} - c^{3} d e f^{6}\right )} x\right )}}\right ] \]
[1/3*(3*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*f^4*x^2 + 2*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*e*f^3*x + (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*e^2*f^2)*sqrt(d^2*e - c*d*f)*log((d*f*x + 2*d*e - c*f - 2*sqrt(d^2*e - c*d*f)*sqrt(f*x + e))/(d *x + c)) - 2*(2*b^2*d^3*e^4 - a^2*c^2*d*f^4 - (7*b^2*c*d^2 - 2*a*b*d^3)*e^ 3*f + (5*b^2*c^2*d + 2*a*b*c*d^2 - 4*a^2*d^3)*e^2*f^2 - (4*a*b*c^2*d - 5*a ^2*c*d^2)*e*f^3 + 3*(b^2*d^3*e^3*f - 3*b^2*c*d^2*e^2*f^2 + (2*b^2*c^2*d + 2*a*b*c*d^2 - a^2*d^3)*e*f^3 - (2*a*b*c^2*d - a^2*c*d^2)*f^4)*x)*sqrt(f*x + e))/(d^4*e^5*f^2 - 3*c*d^3*e^4*f^3 + 3*c^2*d^2*e^3*f^4 - c^3*d*e^2*f^5 + (d^4*e^3*f^4 - 3*c*d^3*e^2*f^5 + 3*c^2*d^2*e*f^6 - c^3*d*f^7)*x^2 + 2*(d^ 4*e^4*f^3 - 3*c*d^3*e^3*f^4 + 3*c^2*d^2*e^2*f^5 - c^3*d*e*f^6)*x), 2/3*(3* ((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*f^4*x^2 + 2*(b^2*c^2 - 2*a*b*c*d + a^2*d^ 2)*e*f^3*x + (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*e^2*f^2)*sqrt(-d^2*e + c*d*f) *arctan(sqrt(-d^2*e + c*d*f)*sqrt(f*x + e)/(d*f*x + d*e)) - (2*b^2*d^3*e^4 - a^2*c^2*d*f^4 - (7*b^2*c*d^2 - 2*a*b*d^3)*e^3*f + (5*b^2*c^2*d + 2*a*b* c*d^2 - 4*a^2*d^3)*e^2*f^2 - (4*a*b*c^2*d - 5*a^2*c*d^2)*e*f^3 + 3*(b^2*d^ 3*e^3*f - 3*b^2*c*d^2*e^2*f^2 + (2*b^2*c^2*d + 2*a*b*c*d^2 - a^2*d^3)*e*f^ 3 - (2*a*b*c^2*d - a^2*c*d^2)*f^4)*x)*sqrt(f*x + e))/(d^4*e^5*f^2 - 3*c*d^ 3*e^4*f^3 + 3*c^2*d^2*e^3*f^4 - c^3*d*e^2*f^5 + (d^4*e^3*f^4 - 3*c*d^3*e^2 *f^5 + 3*c^2*d^2*e*f^6 - c^3*d*f^7)*x^2 + 2*(d^4*e^4*f^3 - 3*c*d^3*e^3*f^4 + 3*c^2*d^2*e^2*f^5 - c^3*d*e*f^6)*x)]
Time = 6.33 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.32 \[ \int \frac {(a+b x)^2}{(c+d x) (e+f x)^{5/2}} \, dx=\begin {cases} \frac {2 \left (\frac {\left (a f - b e\right ) \left (a d f - 2 b c f + b d e\right )}{f \sqrt {e + f x} \left (c f - d e\right )^{2}} - \frac {\left (a f - b e\right )^{2}}{3 f \left (e + f x\right )^{\frac {3}{2}} \left (c f - d e\right )} + \frac {f \left (a d - b c\right )^{2} \operatorname {atan}{\left (\frac {\sqrt {e + f x}}{\sqrt {\frac {c f - d e}{d}}} \right )}}{d \sqrt {\frac {c f - d e}{d}} \left (c f - d e\right )^{2}}\right )}{f} & \text {for}\: f \neq 0 \\\frac {\frac {b^{2} x^{2}}{2 d} + \frac {x \left (2 a b d - b^{2} c\right )}{d^{2}} + \frac {\left (a d - b c\right )^{2} \left (\begin {cases} \frac {x}{c} & \text {for}\: d = 0 \\\frac {\log {\left (c + d x \right )}}{d} & \text {otherwise} \end {cases}\right )}{d^{2}}}{e^{\frac {5}{2}}} & \text {otherwise} \end {cases} \]
Piecewise((2*((a*f - b*e)*(a*d*f - 2*b*c*f + b*d*e)/(f*sqrt(e + f*x)*(c*f - d*e)**2) - (a*f - b*e)**2/(3*f*(e + f*x)**(3/2)*(c*f - d*e)) + f*(a*d - b*c)**2*atan(sqrt(e + f*x)/sqrt((c*f - d*e)/d))/(d*sqrt((c*f - d*e)/d)*(c* f - d*e)**2))/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**(5/2), True))
Exception generated. \[ \int \frac {(a+b x)^2}{(c+d x) (e+f x)^{5/2}} \, dx=\text {Exception raised: ValueError} \]
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
Time = 0.28 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.64 \[ \int \frac {(a+b x)^2}{(c+d x) (e+f x)^{5/2}} \, dx=\frac {2 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {f x + e} d}{\sqrt {-d^{2} e + c d f}}\right )}{{\left (d^{2} e^{2} - 2 \, c d e f + c^{2} f^{2}\right )} \sqrt {-d^{2} e + c d f}} - \frac {2 \, {\left (3 \, {\left (f x + e\right )} b^{2} d e^{2} - b^{2} d e^{3} - 6 \, {\left (f x + e\right )} b^{2} c e f + b^{2} c e^{2} f + 2 \, a b d e^{2} f + 6 \, {\left (f x + e\right )} a b c f^{2} - 3 \, {\left (f x + e\right )} a^{2} d f^{2} - 2 \, a b c e f^{2} - a^{2} d e f^{2} + a^{2} c f^{3}\right )}}{3 \, {\left (d^{2} e^{2} f^{2} - 2 \, c d e f^{3} + c^{2} f^{4}\right )} {\left (f x + e\right )}^{\frac {3}{2}}} \]
2*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*arctan(sqrt(f*x + e)*d/sqrt(-d^2*e + c*d *f))/((d^2*e^2 - 2*c*d*e*f + c^2*f^2)*sqrt(-d^2*e + c*d*f)) - 2/3*(3*(f*x + e)*b^2*d*e^2 - b^2*d*e^3 - 6*(f*x + e)*b^2*c*e*f + b^2*c*e^2*f + 2*a*b*d *e^2*f + 6*(f*x + e)*a*b*c*f^2 - 3*(f*x + e)*a^2*d*f^2 - 2*a*b*c*e*f^2 - a ^2*d*e*f^2 + a^2*c*f^3)/((d^2*e^2*f^2 - 2*c*d*e*f^3 + c^2*f^4)*(f*x + e)^( 3/2))
Time = 1.76 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.45 \[ \int \frac {(a+b x)^2}{(c+d x) (e+f x)^{5/2}} \, dx=\frac {2\,\mathrm {atan}\left (\frac {2\,\sqrt {d}\,\sqrt {e+f\,x}\,{\left (a\,d-b\,c\right )}^2\,\left (c^2\,f^2-2\,c\,d\,e\,f+d^2\,e^2\right )}{{\left (c\,f-d\,e\right )}^{5/2}\,\left (2\,a^2\,d^2-4\,a\,b\,c\,d+2\,b^2\,c^2\right )}\right )\,{\left (a\,d-b\,c\right )}^2}{\sqrt {d}\,{\left (c\,f-d\,e\right )}^{5/2}}-\frac {\frac {2\,\left (a^2\,f^2-2\,a\,b\,e\,f+b^2\,e^2\right )}{3\,\left (c\,f-d\,e\right )}-\frac {2\,\left (e+f\,x\right )\,\left (d\,a^2\,f^2-2\,c\,a\,b\,f^2-d\,b^2\,e^2+2\,c\,b^2\,e\,f\right )}{{\left (c\,f-d\,e\right )}^2}}{f^2\,{\left (e+f\,x\right )}^{3/2}} \]
(2*atan((2*d^(1/2)*(e + f*x)^(1/2)*(a*d - b*c)^2*(c^2*f^2 + d^2*e^2 - 2*c* d*e*f))/((c*f - d*e)^(5/2)*(2*a^2*d^2 + 2*b^2*c^2 - 4*a*b*c*d)))*(a*d - b* c)^2)/(d^(1/2)*(c*f - d*e)^(5/2)) - ((2*(a^2*f^2 + b^2*e^2 - 2*a*b*e*f))/( 3*(c*f - d*e)) - (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))/(c*f - d*e)^2)/(f^2*(e + f*x)^(3/2))