\(\int \frac {(d+e x)^5}{(a d e+(c d^2+a e^2) x+c d e x^2)^3} \, dx\) [1889]

Optimal result

Integrand size = 35, antiderivative size = 85 \[ \int \frac {(d+e x)^5}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=-\frac {\left (c d^2-a e^2\right )^2}{2 c^3 d^3 (a e+c d x)^2}-\frac {2 e \left (c d^2-a e^2\right )}{c^3 d^3 (a e+c d x)}+\frac {e^2 \log (a e+c d x)}{c^3 d^3} \]

[Out]

-1/2*(-a*e^2+c*d^2)^2/c^3/d^3/(c*d*x+a*e)^2-2*e*(-a*e^2+c*d^2)/c^3/d^3/(c*d*x+a*e)+e^2*ln(c*d*x+a*e)/c^3/d^3

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {640, 45} \[ \int \frac {(d+e x)^5}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\frac {e^2 \log (a e+c d x)}{c^3 d^3}-\frac {2 e \left (c d^2-a e^2\right )}{c^3 d^3 (a e+c d x)}-\frac {\left (c d^2-a e^2\right )^2}{2 c^3 d^3 (a e+c d x)^2} \]

[In]

Int[(d + e*x)^5/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^3,x]

[Out]

-1/2*(c*d^2 - a*e^2)^2/(c^3*d^3*(a*e + c*d*x)^2) - (2*e*(c*d^2 - a*e^2))/(c^3*d^3*(a*e + c*d*x)) + (e^2*Log[a*
e + c*d*x])/(c^3*d^3)

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 640

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

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.76 \[ \int \frac {(d+e x)^5}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\frac {-\frac {\left (c d^2-a e^2\right ) \left (3 a e^2+c d (d+4 e x)\right )}{(a e+c d x)^2}+2 e^2 \log (a e+c d x)}{2 c^3 d^3} \]

[In]

Integrate[(d + e*x)^5/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^3,x]

[Out]

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

Maple [A] (verified)

Time = 2.60 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.06

method	result	size
risch	\(\frac {\frac {2 e \left (e^{2} a -c \,d^{2}\right ) x}{c^{2} d^{2}}+\frac {3 a^{2} e^{4}-2 a c \,d^{2} e^{2}-c^{2} d^{4}}{2 c^{3} d^{3}}}{\left (c d x +a e \right )^{2}}+\frac {e^{2} \ln \left (c d x +a e \right )}{c^{3} d^{3}}\)	\(90\)
default	\(-\frac {a^{2} e^{4}-2 a c \,d^{2} e^{2}+c^{2} d^{4}}{2 c^{3} d^{3} \left (c d x +a e \right )^{2}}+\frac {e^{2} \ln \left (c d x +a e \right )}{c^{3} d^{3}}+\frac {2 e \left (e^{2} a -c \,d^{2}\right )}{c^{3} d^{3} \left (c d x +a e \right )}\)	\(95\)
parallelrisch	\(\frac {2 \ln \left (c d x +a e \right ) x^{2} c^{2} d^{2} e^{2}+4 \ln \left (c d x +a e \right ) x a c d \,e^{3}+2 \ln \left (c d x +a e \right ) a^{2} e^{4}+4 x a c d \,e^{3}-4 x \,c^{2} d^{3} e +3 a^{2} e^{4}-2 a c \,d^{2} e^{2}-c^{2} d^{4}}{2 c^{3} d^{3} \left (c d x +a e \right )^{2}}\)	\(123\)
norman	\(\frac {\frac {\left (3 a^{2} e^{6}-3 c^{2} d^{4} e^{2}\right ) x}{c^{3} d^{2} e}+\frac {3 a^{2} e^{4}-2 a c \,d^{2} e^{2}-c^{2} d^{4}}{2 c^{3} d}+\frac {\left (3 a^{2} e^{8}+6 a c \,d^{2} e^{6}-9 d^{4} e^{4} c^{2}\right ) x^{2}}{2 c^{3} d^{3} e^{2}}+\frac {2 \left (a \,e^{6}-c \,d^{2} e^{4}\right ) x^{3}}{c^{2} d^{2} e}}{\left (c d x +a e \right )^{2} \left (e x +d \right )^{2}}+\frac {e^{2} \ln \left (c d x +a e \right )}{c^{3} d^{3}}\)	\(179\)

[In]

int((e*x+d)^5/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.48 \[ \int \frac {(d+e x)^5}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=-\frac {c^{2} d^{4} + 2 \, a c d^{2} e^{2} - 3 \, a^{2} e^{4} + 4 \, {\left (c^{2} d^{3} e - a c d e^{3}\right )} x - 2 \, {\left (c^{2} d^{2} e^{2} x^{2} + 2 \, a c d e^{3} x + a^{2} e^{4}\right )} \log \left (c d x + a e\right )}{2 \, {\left (c^{5} d^{5} x^{2} + 2 \, a c^{4} d^{4} e x + a^{2} c^{3} d^{3} e^{2}\right )}} \]

[In]

integrate((e*x+d)^5/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.28 \[ \int \frac {(d+e x)^5}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\frac {3 a^{2} e^{4} - 2 a c d^{2} e^{2} - c^{2} d^{4} + x \left (4 a c d e^{3} - 4 c^{2} d^{3} e\right )}{2 a^{2} c^{3} d^{3} e^{2} + 4 a c^{4} d^{4} e x + 2 c^{5} d^{5} x^{2}} + \frac {e^{2} \log {\left (a e + c d x \right )}}{c^{3} d^{3}} \]

[In]

integrate((e*x+d)**5/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**3,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.24 \[ \int \frac {(d+e x)^5}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=-\frac {c^{2} d^{4} + 2 \, a c d^{2} e^{2} - 3 \, a^{2} e^{4} + 4 \, {\left (c^{2} d^{3} e - a c d e^{3}\right )} x}{2 \, {\left (c^{5} d^{5} x^{2} + 2 \, a c^{4} d^{4} e x + a^{2} c^{3} d^{3} e^{2}\right )}} + \frac {e^{2} \log \left (c d x + a e\right )}{c^{3} d^{3}} \]

[In]

integrate((e*x+d)^5/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.05 \[ \int \frac {(d+e x)^5}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\frac {e^{2} \log \left ({\left | c d x + a e \right |}\right )}{c^{3} d^{3}} - \frac {4 \, {\left (c d^{2} e - a e^{3}\right )} x + \frac {c^{2} d^{4} + 2 \, a c d^{2} e^{2} - 3 \, a^{2} e^{4}}{c d}}{2 \, {\left (c d x + a e\right )}^{2} c^{2} d^{2}} \]

[In]

integrate((e*x+d)^5/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.25 \[ \int \frac {(d+e x)^5}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\frac {e^2\,\ln \left (a\,e+c\,d\,x\right )}{c^3\,d^3}-\frac {\frac {-3\,a^2\,e^4+2\,a\,c\,d^2\,e^2+c^2\,d^4}{2\,c^3\,d^3}-\frac {2\,e\,x\,\left (a\,e^2-c\,d^2\right )}{c^2\,d^2}}{a^2\,e^2+2\,a\,c\,d\,e\,x+c^2\,d^2\,x^2} \]

[In]

int((d + e*x)^5/(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^3,x)

[Out]

(e^2*log(a*e + c*d*x))/(c^3*d^3) - ((c^2*d^4 - 3*a^2*e^4 + 2*a*c*d^2*e^2)/(2*c^3*d^3) - (2*e*x*(a*e^2 - c*d^2)
)/(c^2*d^2))/(a^2*e^2 + c^2*d^2*x^2 + 2*a*c*d*e*x)