Integrand size = 24, antiderivative size = 184 \[ \int \frac {(a+b x)^3}{(c+d x) \sqrt {e+f x}} \, dx=\frac {2 b \left (3 a^2 d^2 f^2-3 a b d f (d e+c f)+b^2 \left (d^2 e^2+c d e f+c^2 f^2\right )\right ) \sqrt {e+f x}}{d^3 f^3}-\frac {2 b^2 (2 b d e+b c f-3 a d f) (e+f x)^{3/2}}{3 d^2 f^3}+\frac {2 b^3 (e+f x)^{5/2}}{5 d f^3}+\frac {2 (b c-a d)^3 \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{d^{7/2} \sqrt {d e-c f}} \]
-2/3*b^2*(-3*a*d*f+b*c*f+2*b*d*e)*(f*x+e)^(3/2)/d^2/f^3+2/5*b^3*(f*x+e)^(5 /2)/d/f^3+2*(-a*d+b*c)^3*arctanh(d^(1/2)*(f*x+e)^(1/2)/(-c*f+d*e)^(1/2))/d ^(7/2)/(-c*f+d*e)^(1/2)+2*b*(3*a^2*d^2*f^2-3*a*b*d*f*(c*f+d*e)+b^2*(c^2*f^ 2+c*d*e*f+d^2*e^2))*(f*x+e)^(1/2)/d^3/f^3
Time = 0.24 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.85 \[ \int \frac {(a+b x)^3}{(c+d x) \sqrt {e+f x}} \, dx=\frac {2 b \sqrt {e+f x} \left (45 a^2 d^2 f^2+15 a b d f (-2 d e-3 c f+d f x)+b^2 \left (15 c^2 f^2-5 c d f (-2 e+f x)+d^2 \left (8 e^2-4 e f x+3 f^2 x^2\right )\right )\right )}{15 d^3 f^3}+\frac {2 (-b c+a d)^3 \arctan \left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {-d e+c f}}\right )}{d^{7/2} \sqrt {-d e+c f}} \]
(2*b*Sqrt[e + f*x]*(45*a^2*d^2*f^2 + 15*a*b*d*f*(-2*d*e - 3*c*f + d*f*x) + b^2*(15*c^2*f^2 - 5*c*d*f*(-2*e + f*x) + d^2*(8*e^2 - 4*e*f*x + 3*f^2*x^2 ))))/(15*d^3*f^3) + (2*(-(b*c) + a*d)^3*ArcTan[(Sqrt[d]*Sqrt[e + f*x])/Sqr t[-(d*e) + c*f]])/(d^(7/2)*Sqrt[-(d*e) + c*f])
Time = 0.33 (sec) , antiderivative size = 184, 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 = {99, 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.
| \(\displaystyle \int \frac {(a+b x)^3}{(c+d x) \sqrt {e+f x}} \, dx\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle \int \left (\frac {b \left (3 a^2 d^2 f^2-3 a b d f (c f+d e)+b^2 \left (c^2 f^2+c d e f+d^2 e^2\right )\right )}{d^3 f^2 \sqrt {e+f x}}-\frac {b^2 \sqrt {e+f x} (-3 a d f+b c f+2 b d e)}{d^2 f^2}+\frac {(a d-b c)^3}{d^3 (c+d x) \sqrt {e+f x}}+\frac {b^3 (e+f x)^{3/2}}{d f^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 b \sqrt {e+f x} \left (3 a^2 d^2 f^2-3 a b d f (c f+d e)+b^2 \left (c^2 f^2+c d e f+d^2 e^2\right )\right )}{d^3 f^3}+\frac {2 (b c-a d)^3 \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{d^{7/2} \sqrt {d e-c f}}-\frac {2 b^2 (e+f x)^{3/2} (-3 a d f+b c f+2 b d e)}{3 d^2 f^3}+\frac {2 b^3 (e+f x)^{5/2}}{5 d f^3}\) |
(2*b*(3*a^2*d^2*f^2 - 3*a*b*d*f*(d*e + c*f) + b^2*(d^2*e^2 + c*d*e*f + c^2 *f^2))*Sqrt[e + f*x])/(d^3*f^3) - (2*b^2*(2*b*d*e + b*c*f - 3*a*d*f)*(e + f*x)^(3/2))/(3*d^2*f^3) + (2*b^3*(e + f*x)^(5/2))/(5*d*f^3) + (2*(b*c - a* d)^3*ArcTanh[(Sqrt[d]*Sqrt[e + f*x])/Sqrt[d*e - c*f]])/(d^(7/2)*Sqrt[d*e - c*f])
3.18.85.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
Time = 1.01 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.86
| method | result | size |
| pseudoelliptic | \(\frac {6 b \sqrt {\left (c f -d e \right ) d}\, \left (\left (\frac {\left (f^{2} x^{2}-\frac {4}{3} e f x +\frac {8}{3} e^{2}\right ) b^{2}}{15}-\frac {2 f a \left (-\frac {f x}{2}+e \right ) b}{3}+a^{2} f^{2}\right ) d^{2}-\left (\frac {\left (f x -2 e \right ) b}{9}+a f \right ) b f c d +\frac {b^{2} c^{2} f^{2}}{3}\right ) \sqrt {f x +e}+2 f^{3} \left (a d -b c \right )^{3} \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{f^{3} d^{3} \sqrt {\left (c f -d e \right ) d}}\) | \(158\) |
| risch | \(\frac {2 b \left (3 b^{2} d^{2} f^{2} x^{2}+15 a b \,d^{2} f^{2} x -5 b^{2} c d \,f^{2} x -4 b^{2} d^{2} e f x +45 a^{2} d^{2} f^{2}-45 a b c d \,f^{2}-30 a b \,d^{2} e f +15 b^{2} c^{2} f^{2}+10 b^{2} c d e f +8 b^{2} d^{2} e^{2}\right ) \sqrt {f x +e}}{15 f^{3} d^{3}}+\frac {2 \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{d^{3} \sqrt {\left (c f -d e \right ) d}}\) | \(201\) |
| derivativedivides | \(\frac {\frac {2 b \left (\frac {b^{2} \left (f x +e \right )^{\frac {5}{2}} d^{2}}{5}+a b \,d^{2} f \left (f x +e \right )^{\frac {3}{2}}-\frac {b^{2} c d f \left (f x +e \right )^{\frac {3}{2}}}{3}-\frac {2 b^{2} d^{2} e \left (f x +e \right )^{\frac {3}{2}}}{3}+3 d^{2} f^{2} a^{2} \sqrt {f x +e}-3 a b c d \,f^{2} \sqrt {f x +e}-3 \sqrt {f x +e}\, a b \,d^{2} e f +b^{2} c^{2} f^{2} \sqrt {f x +e}+\sqrt {f x +e}\, b^{2} c d e f +\sqrt {f x +e}\, b^{2} d^{2} e^{2}\right )}{d^{3}}+\frac {2 f^{3} \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{d^{3} \sqrt {\left (c f -d e \right ) d}}}{f^{3}}\) | \(249\) |
| default | \(\frac {\frac {2 b \left (\frac {b^{2} \left (f x +e \right )^{\frac {5}{2}} d^{2}}{5}+a b \,d^{2} f \left (f x +e \right )^{\frac {3}{2}}-\frac {b^{2} c d f \left (f x +e \right )^{\frac {3}{2}}}{3}-\frac {2 b^{2} d^{2} e \left (f x +e \right )^{\frac {3}{2}}}{3}+3 d^{2} f^{2} a^{2} \sqrt {f x +e}-3 a b c d \,f^{2} \sqrt {f x +e}-3 \sqrt {f x +e}\, a b \,d^{2} e f +b^{2} c^{2} f^{2} \sqrt {f x +e}+\sqrt {f x +e}\, b^{2} c d e f +\sqrt {f x +e}\, b^{2} d^{2} e^{2}\right )}{d^{3}}+\frac {2 f^{3} \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{d^{3} \sqrt {\left (c f -d e \right ) d}}}{f^{3}}\) | \(249\) |
2/((c*f-d*e)*d)^(1/2)*(3*b*((c*f-d*e)*d)^(1/2)*((1/15*(f^2*x^2-4/3*e*f*x+8 /3*e^2)*b^2-2/3*f*a*(-1/2*f*x+e)*b+a^2*f^2)*d^2-(1/9*(f*x-2*e)*b+a*f)*b*f* c*d+1/3*b^2*c^2*f^2)*(f*x+e)^(1/2)+f^3*(a*d-b*c)^3*arctan(d*(f*x+e)^(1/2)/ ((c*f-d*e)*d)^(1/2)))/f^3/d^3
Time = 0.24 (sec) , antiderivative size = 653, normalized size of antiderivative = 3.55 \[ \int \frac {(a+b x)^3}{(c+d x) \sqrt {e+f x}} \, dx=\left [-\frac {15 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {d^{2} e - c d f} f^{3} \log \left (\frac {d f x + 2 \, d e - c f - 2 \, \sqrt {d^{2} e - c d f} \sqrt {f x + e}}{d x + c}\right ) - 2 \, {\left (8 \, b^{3} d^{4} e^{3} + 2 \, {\left (b^{3} c d^{3} - 15 \, a b^{2} d^{4}\right )} e^{2} f + 5 \, {\left (b^{3} c^{2} d^{2} - 3 \, a b^{2} c d^{3} + 9 \, a^{2} b d^{4}\right )} e f^{2} - 15 \, {\left (b^{3} c^{3} d - 3 \, a b^{2} c^{2} d^{2} + 3 \, a^{2} b c d^{3}\right )} f^{3} + 3 \, {\left (b^{3} d^{4} e f^{2} - b^{3} c d^{3} f^{3}\right )} x^{2} - {\left (4 \, b^{3} d^{4} e^{2} f + {\left (b^{3} c d^{3} - 15 \, a b^{2} d^{4}\right )} e f^{2} - 5 \, {\left (b^{3} c^{2} d^{2} - 3 \, a b^{2} c d^{3}\right )} f^{3}\right )} x\right )} \sqrt {f x + e}}{15 \, {\left (d^{5} e f^{3} - c d^{4} f^{4}\right )}}, -\frac {2 \, {\left (15 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {-d^{2} e + c d f} f^{3} \arctan \left (\frac {\sqrt {-d^{2} e + c d f} \sqrt {f x + e}}{d f x + d e}\right ) - {\left (8 \, b^{3} d^{4} e^{3} + 2 \, {\left (b^{3} c d^{3} - 15 \, a b^{2} d^{4}\right )} e^{2} f + 5 \, {\left (b^{3} c^{2} d^{2} - 3 \, a b^{2} c d^{3} + 9 \, a^{2} b d^{4}\right )} e f^{2} - 15 \, {\left (b^{3} c^{3} d - 3 \, a b^{2} c^{2} d^{2} + 3 \, a^{2} b c d^{3}\right )} f^{3} + 3 \, {\left (b^{3} d^{4} e f^{2} - b^{3} c d^{3} f^{3}\right )} x^{2} - {\left (4 \, b^{3} d^{4} e^{2} f + {\left (b^{3} c d^{3} - 15 \, a b^{2} d^{4}\right )} e f^{2} - 5 \, {\left (b^{3} c^{2} d^{2} - 3 \, a b^{2} c d^{3}\right )} f^{3}\right )} x\right )} \sqrt {f x + e}\right )}}{15 \, {\left (d^{5} e f^{3} - c d^{4} f^{4}\right )}}\right ] \]
[-1/15*(15*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*sqrt(d^2*e - c*d*f)*f^3*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*(8*b^3*d^4*e^3 + 2*(b^3*c*d^3 - 15*a*b^2*d^4)*e^2*f + 5* (b^3*c^2*d^2 - 3*a*b^2*c*d^3 + 9*a^2*b*d^4)*e*f^2 - 15*(b^3*c^3*d - 3*a*b^ 2*c^2*d^2 + 3*a^2*b*c*d^3)*f^3 + 3*(b^3*d^4*e*f^2 - b^3*c*d^3*f^3)*x^2 - ( 4*b^3*d^4*e^2*f + (b^3*c*d^3 - 15*a*b^2*d^4)*e*f^2 - 5*(b^3*c^2*d^2 - 3*a* b^2*c*d^3)*f^3)*x)*sqrt(f*x + e))/(d^5*e*f^3 - c*d^4*f^4), -2/15*(15*(b^3* c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*sqrt(-d^2*e + c*d*f)*f^3*ar ctan(sqrt(-d^2*e + c*d*f)*sqrt(f*x + e)/(d*f*x + d*e)) - (8*b^3*d^4*e^3 + 2*(b^3*c*d^3 - 15*a*b^2*d^4)*e^2*f + 5*(b^3*c^2*d^2 - 3*a*b^2*c*d^3 + 9*a^ 2*b*d^4)*e*f^2 - 15*(b^3*c^3*d - 3*a*b^2*c^2*d^2 + 3*a^2*b*c*d^3)*f^3 + 3* (b^3*d^4*e*f^2 - b^3*c*d^3*f^3)*x^2 - (4*b^3*d^4*e^2*f + (b^3*c*d^3 - 15*a *b^2*d^4)*e*f^2 - 5*(b^3*c^2*d^2 - 3*a*b^2*c*d^3)*f^3)*x)*sqrt(f*x + e))/( d^5*e*f^3 - c*d^4*f^4)]
Time = 4.11 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.61 \[ \int \frac {(a+b x)^3}{(c+d x) \sqrt {e+f x}} \, dx=\begin {cases} \frac {2 \left (\frac {b^{3} \left (e + f x\right )^{\frac {5}{2}}}{5 d f^{2}} + \frac {\left (e + f x\right )^{\frac {3}{2}} \cdot \left (3 a b^{2} d f - b^{3} c f - 2 b^{3} d e\right )}{3 d^{2} f^{2}} + \frac {\sqrt {e + f x} \left (3 a^{2} b d^{2} f^{2} - 3 a b^{2} c d f^{2} - 3 a b^{2} d^{2} e f + b^{3} c^{2} f^{2} + b^{3} c d e f + b^{3} d^{2} e^{2}\right )}{d^{3} f^{2}} + \frac {f \left (a d - b c\right )^{3} \operatorname {atan}{\left (\frac {\sqrt {e + f x}}{\sqrt {\frac {c f - d e}{d}}} \right )}}{d^{4} \sqrt {\frac {c f - d e}{d}}}\right )}{f} & \text {for}\: f \neq 0 \\\frac {\frac {b^{3} x^{3}}{3 d} + \frac {x^{2} \cdot \left (3 a b^{2} d - b^{3} c\right )}{2 d^{2}} + \frac {x \left (3 a^{2} b d^{2} - 3 a b^{2} c d + b^{3} c^{2}\right )}{d^{3}} + \frac {\left (a d - b c\right )^{3} \left (\begin {cases} \frac {x}{c} & \text {for}\: d = 0 \\\frac {\log {\left (c + d x \right )}}{d} & \text {otherwise} \end {cases}\right )}{d^{3}}}{\sqrt {e}} & \text {otherwise} \end {cases} \]
Piecewise((2*(b**3*(e + f*x)**(5/2)/(5*d*f**2) + (e + f*x)**(3/2)*(3*a*b** 2*d*f - b**3*c*f - 2*b**3*d*e)/(3*d**2*f**2) + sqrt(e + f*x)*(3*a**2*b*d** 2*f**2 - 3*a*b**2*c*d*f**2 - 3*a*b**2*d**2*e*f + b**3*c**2*f**2 + b**3*c*d *e*f + b**3*d**2*e**2)/(d**3*f**2) + f*(a*d - b*c)**3*atan(sqrt(e + f*x)/s qrt((c*f - d*e)/d))/(d**4*sqrt((c*f - d*e)/d)))/f, Ne(f, 0)), ((b**3*x**3/ (3*d) + x**2*(3*a*b**2*d - b**3*c)/(2*d**2) + x*(3*a**2*b*d**2 - 3*a*b**2* c*d + b**3*c**2)/d**3 + (a*d - b*c)**3*Piecewise((x/c, Eq(d, 0)), (log(c + d*x)/d, True))/d**3)/sqrt(e), True))
Exception generated. \[ \int \frac {(a+b x)^3}{(c+d x) \sqrt {e+f x}} \, dx=\text {Exception raised: ValueError} \]
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 = 283, normalized size of antiderivative = 1.54 \[ \int \frac {(a+b x)^3}{(c+d x) \sqrt {e+f x}} \, dx=-\frac {2 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \arctan \left (\frac {\sqrt {f x + e} d}{\sqrt {-d^{2} e + c d f}}\right )}{\sqrt {-d^{2} e + c d f} d^{3}} + \frac {2 \, {\left (3 \, {\left (f x + e\right )}^{\frac {5}{2}} b^{3} d^{4} f^{12} - 10 \, {\left (f x + e\right )}^{\frac {3}{2}} b^{3} d^{4} e f^{12} + 15 \, \sqrt {f x + e} b^{3} d^{4} e^{2} f^{12} - 5 \, {\left (f x + e\right )}^{\frac {3}{2}} b^{3} c d^{3} f^{13} + 15 \, {\left (f x + e\right )}^{\frac {3}{2}} a b^{2} d^{4} f^{13} + 15 \, \sqrt {f x + e} b^{3} c d^{3} e f^{13} - 45 \, \sqrt {f x + e} a b^{2} d^{4} e f^{13} + 15 \, \sqrt {f x + e} b^{3} c^{2} d^{2} f^{14} - 45 \, \sqrt {f x + e} a b^{2} c d^{3} f^{14} + 45 \, \sqrt {f x + e} a^{2} b d^{4} f^{14}\right )}}{15 \, d^{5} f^{15}} \]
-2*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*arctan(sqrt(f*x + e )*d/sqrt(-d^2*e + c*d*f))/(sqrt(-d^2*e + c*d*f)*d^3) + 2/15*(3*(f*x + e)^( 5/2)*b^3*d^4*f^12 - 10*(f*x + e)^(3/2)*b^3*d^4*e*f^12 + 15*sqrt(f*x + e)*b ^3*d^4*e^2*f^12 - 5*(f*x + e)^(3/2)*b^3*c*d^3*f^13 + 15*(f*x + e)^(3/2)*a* b^2*d^4*f^13 + 15*sqrt(f*x + e)*b^3*c*d^3*e*f^13 - 45*sqrt(f*x + e)*a*b^2* d^4*e*f^13 + 15*sqrt(f*x + e)*b^3*c^2*d^2*f^14 - 45*sqrt(f*x + e)*a*b^2*c* d^3*f^14 + 45*sqrt(f*x + e)*a^2*b*d^4*f^14)/(d^5*f^15)
Time = 0.11 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.43 \[ \int \frac {(a+b x)^3}{(c+d x) \sqrt {e+f x}} \, dx=\sqrt {e+f\,x}\,\left (\frac {\left (\frac {6\,b^3\,e-6\,a\,b^2\,f}{d\,f^3}+\frac {2\,b^3\,\left (c\,f^4-d\,e\,f^3\right )}{d^2\,f^6}\right )\,\left (c\,f^4-d\,e\,f^3\right )}{d\,f^3}+\frac {6\,b\,{\left (a\,f-b\,e\right )}^2}{d\,f^3}\right )-{\left (e+f\,x\right )}^{3/2}\,\left (\frac {6\,b^3\,e-6\,a\,b^2\,f}{3\,d\,f^3}+\frac {2\,b^3\,\left (c\,f^4-d\,e\,f^3\right )}{3\,d^2\,f^6}\right )+\frac {2\,b^3\,{\left (e+f\,x\right )}^{5/2}}{5\,d\,f^3}+\frac {2\,\mathrm {atan}\left (\frac {\sqrt {d}\,\sqrt {e+f\,x}\,{\left (a\,d-b\,c\right )}^3}{\sqrt {c\,f-d\,e}\,\left (a^3\,d^3-3\,a^2\,b\,c\,d^2+3\,a\,b^2\,c^2\,d-b^3\,c^3\right )}\right )\,{\left (a\,d-b\,c\right )}^3}{d^{7/2}\,\sqrt {c\,f-d\,e}} \]
(e + f*x)^(1/2)*((((6*b^3*e - 6*a*b^2*f)/(d*f^3) + (2*b^3*(c*f^4 - d*e*f^3 ))/(d^2*f^6))*(c*f^4 - d*e*f^3))/(d*f^3) + (6*b*(a*f - b*e)^2)/(d*f^3)) - (e + f*x)^(3/2)*((6*b^3*e - 6*a*b^2*f)/(3*d*f^3) + (2*b^3*(c*f^4 - d*e*f^3 ))/(3*d^2*f^6)) + (2*b^3*(e + f*x)^(5/2))/(5*d*f^3) + (2*atan((d^(1/2)*(e + f*x)^(1/2)*(a*d - b*c)^3)/((c*f - d*e)^(1/2)*(a^3*d^3 - b^3*c^3 + 3*a*b^ 2*c^2*d - 3*a^2*b*c*d^2)))*(a*d - b*c)^3)/(d^(7/2)*(c*f - d*e)^(1/2))