\(\int \frac {(d+e x)^5}{(b x+c x^2)^2} \, dx\) [267]

Optimal result

Integrand size = 19, antiderivative size = 118 \[ \int \frac {(d+e x)^5}{\left (b x+c x^2\right )^2} \, dx=-\frac {d^5}{b^2 x}+\frac {e^4 (5 c d-2 b e) x}{c^3}+\frac {e^5 x^2}{2 c^2}-\frac {(c d-b e)^5}{b^2 c^4 (b+c x)}-\frac {d^4 (2 c d-5 b e) \log (x)}{b^3}+\frac {(c d-b e)^4 (2 c d+3 b e) \log (b+c x)}{b^3 c^4} \]

[Out]

-d^5/b^2/x+e^4*(-2*b*e+5*c*d)*x/c^3+1/2*e^5*x^2/c^2-(-b*e+c*d)^5/b^2/c^4/(c*x+b)-d^4*(-5*b*e+2*c*d)*ln(x)/b^3+
(-b*e+c*d)^4*(3*b*e+2*c*d)*ln(c*x+b)/b^3/c^4

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {712} \[ \int \frac {(d+e x)^5}{\left (b x+c x^2\right )^2} \, dx=\frac {(c d-b e)^4 (3 b e+2 c d) \log (b+c x)}{b^3 c^4}-\frac {d^4 \log (x) (2 c d-5 b e)}{b^3}-\frac {(c d-b e)^5}{b^2 c^4 (b+c x)}-\frac {d^5}{b^2 x}+\frac {e^4 x (5 c d-2 b e)}{c^3}+\frac {e^5 x^2}{2 c^2} \]

[In]

Int[(d + e*x)^5/(b*x + c*x^2)^2,x]

[Out]

-(d^5/(b^2*x)) + (e^4*(5*c*d - 2*b*e)*x)/c^3 + (e^5*x^2)/(2*c^2) - (c*d - b*e)^5/(b^2*c^4*(b + c*x)) - (d^4*(2
*c*d - 5*b*e)*Log[x])/b^3 + ((c*d - b*e)^4*(2*c*d + 3*b*e)*Log[b + c*x])/(b^3*c^4)

Rule 712

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*
e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {e^4 (5 c d-2 b e)}{c^3}+\frac {d^5}{b^2 x^2}+\frac {d^4 (-2 c d+5 b e)}{b^3 x}+\frac {e^5 x}{c^2}-\frac {(-c d+b e)^5}{b^2 c^3 (b+c x)^2}+\frac {(-c d+b e)^4 (2 c d+3 b e)}{b^3 c^3 (b+c x)}\right ) \, dx \\ & = -\frac {d^5}{b^2 x}+\frac {e^4 (5 c d-2 b e) x}{c^3}+\frac {e^5 x^2}{2 c^2}-\frac {(c d-b e)^5}{b^2 c^4 (b+c x)}-\frac {d^4 (2 c d-5 b e) \log (x)}{b^3}+\frac {(c d-b e)^4 (2 c d+3 b e) \log (b+c x)}{b^3 c^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.98 \[ \int \frac {(d+e x)^5}{\left (b x+c x^2\right )^2} \, dx=-\frac {d^5}{b^2 x}+\frac {e^4 (5 c d-2 b e) x}{c^3}+\frac {e^5 x^2}{2 c^2}+\frac {(-c d+b e)^5}{b^2 c^4 (b+c x)}+\frac {d^4 (-2 c d+5 b e) \log (x)}{b^3}+\frac {(c d-b e)^4 (2 c d+3 b e) \log (b+c x)}{b^3 c^4} \]

[In]

Integrate[(d + e*x)^5/(b*x + c*x^2)^2,x]

[Out]

-(d^5/(b^2*x)) + (e^4*(5*c*d - 2*b*e)*x)/c^3 + (e^5*x^2)/(2*c^2) + (-(c*d) + b*e)^5/(b^2*c^4*(b + c*x)) + (d^4
*(-2*c*d + 5*b*e)*Log[x])/b^3 + ((c*d - b*e)^4*(2*c*d + 3*b*e)*Log[b + c*x])/(b^3*c^4)

Maple [A] (verified)

Time = 1.91 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.69

method	result	size
default	\(-\frac {e^{4} \left (-\frac {1}{2} c e \,x^{2}+2 b e x -5 c d x \right )}{c^{3}}-\frac {d^{5}}{b^{2} x}+\frac {d^{4} \left (5 b e -2 c d \right ) \ln \left (x \right )}{b^{3}}+\frac {\left (3 b^{5} e^{5}-10 b^{4} c d \,e^{4}+10 b^{3} c^{2} d^{2} e^{3}-5 b \,c^{4} d^{4} e +2 c^{5} d^{5}\right ) \ln \left (c x +b \right )}{c^{4} b^{3}}-\frac {-b^{5} e^{5}+5 b^{4} c d \,e^{4}-10 b^{3} c^{2} d^{2} e^{3}+10 b^{2} c^{3} d^{3} e^{2}-5 b \,c^{4} d^{4} e +c^{5} d^{5}}{b^{2} c^{4} \left (c x +b \right )}\)	\(200\)
norman	\(\frac {-\frac {d^{5}}{b}+\frac {e^{5} x^{4}}{2 c}-\frac {e^{4} \left (3 b e -10 c d \right ) x^{3}}{2 c^{2}}-\frac {\left (3 b^{5} e^{5}-10 b^{4} c d \,e^{4}+10 b^{3} c^{2} d^{2} e^{3}-10 b^{2} c^{3} d^{3} e^{2}+5 b \,c^{4} d^{4} e -2 c^{5} d^{5}\right ) x^{2}}{b^{3} c^{3}}}{x \left (c x +b \right )}+\frac {\left (3 b^{5} e^{5}-10 b^{4} c d \,e^{4}+10 b^{3} c^{2} d^{2} e^{3}-5 b \,c^{4} d^{4} e +2 c^{5} d^{5}\right ) \ln \left (c x +b \right )}{c^{4} b^{3}}+\frac {d^{4} \left (5 b e -2 c d \right ) \ln \left (x \right )}{b^{3}}\)	\(211\)
risch	\(\frac {e^{5} x^{2}}{2 c^{2}}-\frac {2 e^{5} b x}{c^{3}}+\frac {5 e^{4} d x}{c^{2}}+\frac {\frac {\left (b^{5} e^{5}-5 b^{4} c d \,e^{4}+10 b^{3} c^{2} d^{2} e^{3}-10 b^{2} c^{3} d^{3} e^{2}+5 b \,c^{4} d^{4} e -2 c^{5} d^{5}\right ) x}{b^{2} c}-\frac {c^{3} d^{5}}{b}}{c^{3} x \left (c x +b \right )}+\frac {5 d^{4} \ln \left (x \right ) e}{b^{2}}-\frac {2 d^{5} \ln \left (x \right ) c}{b^{3}}+\frac {3 b^{2} \ln \left (-c x -b \right ) e^{5}}{c^{4}}-\frac {10 b \ln \left (-c x -b \right ) d \,e^{4}}{c^{3}}+\frac {10 \ln \left (-c x -b \right ) d^{2} e^{3}}{c^{2}}-\frac {5 \ln \left (-c x -b \right ) d^{4} e}{b^{2}}+\frac {2 c \ln \left (-c x -b \right ) d^{5}}{b^{3}}\)	\(248\)
parallelrisch	\(\frac {20 \ln \left (c x +b \right ) x \,b^{4} c^{2} d^{2} e^{3}-2 c^{4} b^{2} d^{5}-10 \ln \left (c x +b \right ) x \,b^{2} c^{4} d^{4} e +10 \ln \left (x \right ) x^{2} b \,c^{5} d^{4} e -20 \ln \left (c x +b \right ) x^{2} b^{4} c^{2} d \,e^{4}+20 \ln \left (c x +b \right ) x^{2} b^{3} c^{3} d^{2} e^{3}-10 \ln \left (c x +b \right ) x^{2} b \,c^{5} d^{4} e +10 \ln \left (x \right ) x \,b^{2} c^{4} d^{4} e -20 \ln \left (c x +b \right ) x \,b^{5} c d \,e^{4}+6 \ln \left (c x +b \right ) x^{2} b^{5} c \,e^{5}-4 \ln \left (x \right ) x b \,c^{5} d^{5}+4 x^{2} c^{6} d^{5}+x^{4} b^{3} c^{3} e^{5}-3 x^{3} b^{4} c^{2} e^{5}-6 x^{2} b^{5} c \,e^{5}-4 \ln \left (x \right ) x^{2} c^{6} d^{5}+4 \ln \left (c x +b \right ) x^{2} c^{6} d^{5}+6 \ln \left (c x +b \right ) x \,b^{6} e^{5}+10 x^{3} b^{3} c^{3} d \,e^{4}+20 x^{2} b^{4} c^{2} d \,e^{4}-20 x^{2} b^{3} c^{3} d^{2} e^{3}+20 x^{2} b^{2} c^{4} d^{3} e^{2}-10 x^{2} b \,c^{5} d^{4} e +4 \ln \left (c x +b \right ) x b \,c^{5} d^{5}}{2 b^{3} c^{4} x \left (c x +b \right )}\)	\(399\)

[In]

int((e*x+d)^5/(c*x^2+b*x)^2,x,method=_RETURNVERBOSE)

[Out]

-e^4/c^3*(-1/2*c*e*x^2+2*b*e*x-5*c*d*x)-d^5/b^2/x+d^4*(5*b*e-2*c*d)/b^3*ln(x)+1/c^4*(3*b^5*e^5-10*b^4*c*d*e^4+
10*b^3*c^2*d^2*e^3-5*b*c^4*d^4*e+2*c^5*d^5)/b^3*ln(c*x+b)-(-b^5*e^5+5*b^4*c*d*e^4-10*b^3*c^2*d^2*e^3+10*b^2*c^
3*d^3*e^2-5*b*c^4*d^4*e+c^5*d^5)/b^2/c^4/(c*x+b)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 349 vs. \(2 (116) = 232\).

Time = 0.28 (sec) , antiderivative size = 349, normalized size of antiderivative = 2.96 \[ \int \frac {(d+e x)^5}{\left (b x+c x^2\right )^2} \, dx=\frac {b^{3} c^{3} e^{5} x^{4} - 2 \, b^{2} c^{4} d^{5} + {\left (10 \, b^{3} c^{3} d e^{4} - 3 \, b^{4} c^{2} e^{5}\right )} x^{3} + 2 \, {\left (5 \, b^{4} c^{2} d e^{4} - 2 \, b^{5} c e^{5}\right )} x^{2} - 2 \, {\left (2 \, b c^{5} d^{5} - 5 \, b^{2} c^{4} d^{4} e + 10 \, b^{3} c^{3} d^{3} e^{2} - 10 \, b^{4} c^{2} d^{2} e^{3} + 5 \, b^{5} c d e^{4} - b^{6} e^{5}\right )} x + 2 \, {\left ({\left (2 \, c^{6} d^{5} - 5 \, b c^{5} d^{4} e + 10 \, b^{3} c^{3} d^{2} e^{3} - 10 \, b^{4} c^{2} d e^{4} + 3 \, b^{5} c e^{5}\right )} x^{2} + {\left (2 \, b c^{5} d^{5} - 5 \, b^{2} c^{4} d^{4} e + 10 \, b^{4} c^{2} d^{2} e^{3} - 10 \, b^{5} c d e^{4} + 3 \, b^{6} e^{5}\right )} x\right )} \log \left (c x + b\right ) - 2 \, {\left ({\left (2 \, c^{6} d^{5} - 5 \, b c^{5} d^{4} e\right )} x^{2} + {\left (2 \, b c^{5} d^{5} - 5 \, b^{2} c^{4} d^{4} e\right )} x\right )} \log \left (x\right )}{2 \, {\left (b^{3} c^{5} x^{2} + b^{4} c^{4} x\right )}} \]

[In]

integrate((e*x+d)^5/(c*x^2+b*x)^2,x, algorithm="fricas")

[Out]

1/2*(b^3*c^3*e^5*x^4 - 2*b^2*c^4*d^5 + (10*b^3*c^3*d*e^4 - 3*b^4*c^2*e^5)*x^3 + 2*(5*b^4*c^2*d*e^4 - 2*b^5*c*e
^5)*x^2 - 2*(2*b*c^5*d^5 - 5*b^2*c^4*d^4*e + 10*b^3*c^3*d^3*e^2 - 10*b^4*c^2*d^2*e^3 + 5*b^5*c*d*e^4 - b^6*e^5
)*x + 2*((2*c^6*d^5 - 5*b*c^5*d^4*e + 10*b^3*c^3*d^2*e^3 - 10*b^4*c^2*d*e^4 + 3*b^5*c*e^5)*x^2 + (2*b*c^5*d^5
- 5*b^2*c^4*d^4*e + 10*b^4*c^2*d^2*e^3 - 10*b^5*c*d*e^4 + 3*b^6*e^5)*x)*log(c*x + b) - 2*((2*c^6*d^5 - 5*b*c^5
*d^4*e)*x^2 + (2*b*c^5*d^5 - 5*b^2*c^4*d^4*e)*x)*log(x))/(b^3*c^5*x^2 + b^4*c^4*x)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 381 vs. \(2 (110) = 220\).

Time = 2.15 (sec) , antiderivative size = 381, normalized size of antiderivative = 3.23 \[ \int \frac {(d+e x)^5}{\left (b x+c x^2\right )^2} \, dx=x \left (- \frac {2 b e^{5}}{c^{3}} + \frac {5 d e^{4}}{c^{2}}\right ) + \frac {- b c^{4} d^{5} + x \left (b^{5} e^{5} - 5 b^{4} c d e^{4} + 10 b^{3} c^{2} d^{2} e^{3} - 10 b^{2} c^{3} d^{3} e^{2} + 5 b c^{4} d^{4} e - 2 c^{5} d^{5}\right )}{b^{3} c^{4} x + b^{2} c^{5} x^{2}} + \frac {e^{5} x^{2}}{2 c^{2}} + \frac {d^{4} \cdot \left (5 b e - 2 c d\right ) \log {\left (x + \frac {- 5 b^{2} c^{3} d^{4} e + 2 b c^{4} d^{5} + b c^{3} d^{4} \cdot \left (5 b e - 2 c d\right )}{3 b^{5} e^{5} - 10 b^{4} c d e^{4} + 10 b^{3} c^{2} d^{2} e^{3} - 10 b c^{4} d^{4} e + 4 c^{5} d^{5}} \right )}}{b^{3}} + \frac {\left (b e - c d\right )^{4} \cdot \left (3 b e + 2 c d\right ) \log {\left (x + \frac {- 5 b^{2} c^{3} d^{4} e + 2 b c^{4} d^{5} + \frac {b \left (b e - c d\right )^{4} \cdot \left (3 b e + 2 c d\right )}{c}}{3 b^{5} e^{5} - 10 b^{4} c d e^{4} + 10 b^{3} c^{2} d^{2} e^{3} - 10 b c^{4} d^{4} e + 4 c^{5} d^{5}} \right )}}{b^{3} c^{4}} \]

[In]

integrate((e*x+d)**5/(c*x**2+b*x)**2,x)

[Out]

x*(-2*b*e**5/c**3 + 5*d*e**4/c**2) + (-b*c**4*d**5 + x*(b**5*e**5 - 5*b**4*c*d*e**4 + 10*b**3*c**2*d**2*e**3 -
10*b**2*c**3*d**3*e**2 + 5*b*c**4*d**4*e - 2*c**5*d**5))/(b**3*c**4*x + b**2*c**5*x**2) + e**5*x**2/(2*c**2)
+ d**4*(5*b*e - 2*c*d)*log(x + (-5*b**2*c**3*d**4*e + 2*b*c**4*d**5 + b*c**3*d**4*(5*b*e - 2*c*d))/(3*b**5*e**
5 - 10*b**4*c*d*e**4 + 10*b**3*c**2*d**2*e**3 - 10*b*c**4*d**4*e + 4*c**5*d**5))/b**3 + (b*e - c*d)**4*(3*b*e
+ 2*c*d)*log(x + (-5*b**2*c**3*d**4*e + 2*b*c**4*d**5 + b*(b*e - c*d)**4*(3*b*e + 2*c*d)/c)/(3*b**5*e**5 - 10*
b**4*c*d*e**4 + 10*b**3*c**2*d**2*e**3 - 10*b*c**4*d**4*e + 4*c**5*d**5))/(b**3*c**4)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.83 \[ \int \frac {(d+e x)^5}{\left (b x+c x^2\right )^2} \, dx=-\frac {b c^{4} d^{5} + {\left (2 \, c^{5} d^{5} - 5 \, b c^{4} d^{4} e + 10 \, b^{2} c^{3} d^{3} e^{2} - 10 \, b^{3} c^{2} d^{2} e^{3} + 5 \, b^{4} c d e^{4} - b^{5} e^{5}\right )} x}{b^{2} c^{5} x^{2} + b^{3} c^{4} x} - \frac {{\left (2 \, c d^{5} - 5 \, b d^{4} e\right )} \log \left (x\right )}{b^{3}} + \frac {c e^{5} x^{2} + 2 \, {\left (5 \, c d e^{4} - 2 \, b e^{5}\right )} x}{2 \, c^{3}} + \frac {{\left (2 \, c^{5} d^{5} - 5 \, b c^{4} d^{4} e + 10 \, b^{3} c^{2} d^{2} e^{3} - 10 \, b^{4} c d e^{4} + 3 \, b^{5} e^{5}\right )} \log \left (c x + b\right )}{b^{3} c^{4}} \]

[In]

integrate((e*x+d)^5/(c*x^2+b*x)^2,x, algorithm="maxima")

[Out]

-(b*c^4*d^5 + (2*c^5*d^5 - 5*b*c^4*d^4*e + 10*b^2*c^3*d^3*e^2 - 10*b^3*c^2*d^2*e^3 + 5*b^4*c*d*e^4 - b^5*e^5)*
x)/(b^2*c^5*x^2 + b^3*c^4*x) - (2*c*d^5 - 5*b*d^4*e)*log(x)/b^3 + 1/2*(c*e^5*x^2 + 2*(5*c*d*e^4 - 2*b*e^5)*x)/
c^3 + (2*c^5*d^5 - 5*b*c^4*d^4*e + 10*b^3*c^2*d^2*e^3 - 10*b^4*c*d*e^4 + 3*b^5*e^5)*log(c*x + b)/(b^3*c^4)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.83 \[ \int \frac {(d+e x)^5}{\left (b x+c x^2\right )^2} \, dx=-\frac {{\left (2 \, c d^{5} - 5 \, b d^{4} e\right )} \log \left ({\left | x \right |}\right )}{b^{3}} + \frac {c^{2} e^{5} x^{2} + 10 \, c^{2} d e^{4} x - 4 \, b c e^{5} x}{2 \, c^{4}} + \frac {{\left (2 \, c^{5} d^{5} - 5 \, b c^{4} d^{4} e + 10 \, b^{3} c^{2} d^{2} e^{3} - 10 \, b^{4} c d e^{4} + 3 \, b^{5} e^{5}\right )} \log \left ({\left | c x + b \right |}\right )}{b^{3} c^{4}} - \frac {b c^{4} d^{5} + {\left (2 \, c^{5} d^{5} - 5 \, b c^{4} d^{4} e + 10 \, b^{2} c^{3} d^{3} e^{2} - 10 \, b^{3} c^{2} d^{2} e^{3} + 5 \, b^{4} c d e^{4} - b^{5} e^{5}\right )} x}{{\left (c x + b\right )} b^{2} c^{4} x} \]

[In]

integrate((e*x+d)^5/(c*x^2+b*x)^2,x, algorithm="giac")

[Out]

-(2*c*d^5 - 5*b*d^4*e)*log(abs(x))/b^3 + 1/2*(c^2*e^5*x^2 + 10*c^2*d*e^4*x - 4*b*c*e^5*x)/c^4 + (2*c^5*d^5 - 5
*b*c^4*d^4*e + 10*b^3*c^2*d^2*e^3 - 10*b^4*c*d*e^4 + 3*b^5*e^5)*log(abs(c*x + b))/(b^3*c^4) - (b*c^4*d^5 + (2*
c^5*d^5 - 5*b*c^4*d^4*e + 10*b^2*c^3*d^3*e^2 - 10*b^3*c^2*d^2*e^3 + 5*b^4*c*d*e^4 - b^5*e^5)*x)/((c*x + b)*b^2
*c^4*x)

Mupad [B] (verification not implemented)

Time = 9.68 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.85 \[ \int \frac {(d+e x)^5}{\left (b x+c x^2\right )^2} \, dx=\frac {e^5\,x^2}{2\,c^2}-\frac {\frac {c^3\,d^5}{b}-\frac {x\,\left (b^5\,e^5-5\,b^4\,c\,d\,e^4+10\,b^3\,c^2\,d^2\,e^3-10\,b^2\,c^3\,d^3\,e^2+5\,b\,c^4\,d^4\,e-2\,c^5\,d^5\right )}{b^2\,c}}{c^4\,x^2+b\,c^3\,x}-x\,\left (\frac {2\,b\,e^5}{c^3}-\frac {5\,d\,e^4}{c^2}\right )+\frac {\ln \left (b+c\,x\right )\,\left (3\,b^5\,e^5-10\,b^4\,c\,d\,e^4+10\,b^3\,c^2\,d^2\,e^3-5\,b\,c^4\,d^4\,e+2\,c^5\,d^5\right )}{b^3\,c^4}+\frac {d^4\,\ln \left (x\right )\,\left (5\,b\,e-2\,c\,d\right )}{b^3} \]

[In]

int((d + e*x)^5/(b*x + c*x^2)^2,x)

[Out]

(e^5*x^2)/(2*c^2) - ((c^3*d^5)/b - (x*(b^5*e^5 - 2*c^5*d^5 - 10*b^2*c^3*d^3*e^2 + 10*b^3*c^2*d^2*e^3 + 5*b*c^4
*d^4*e - 5*b^4*c*d*e^4))/(b^2*c))/(c^4*x^2 + b*c^3*x) - x*((2*b*e^5)/c^3 - (5*d*e^4)/c^2) + (log(b + c*x)*(3*b
^5*e^5 + 2*c^5*d^5 + 10*b^3*c^2*d^2*e^3 - 5*b*c^4*d^4*e - 10*b^4*c*d*e^4))/(b^3*c^4) + (d^4*log(x)*(5*b*e - 2*
c*d))/b^3