\(\int \frac {(a+b x) (a^2+2 a b x+b^2 x^2)}{(d+e x)^3} \, dx\) [1903]

Optimal result

Integrand size = 29, antiderivative size = 78 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^3} \, dx=\frac {b^3 x}{e^3}+\frac {(b d-a e)^3}{2 e^4 (d+e x)^2}-\frac {3 b (b d-a e)^2}{e^4 (d+e x)}-\frac {3 b^2 (b d-a e) \log (d+e x)}{e^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {27, 45} \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^3} \, dx=-\frac {3 b^2 (b d-a e) \log (d+e x)}{e^4}-\frac {3 b (b d-a e)^2}{e^4 (d+e x)}+\frac {(b d-a e)^3}{2 e^4 (d+e x)^2}+\frac {b^3 x}{e^3} \]

[In]

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

[Out]

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

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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])

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.46 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^3} \, dx=\frac {-a^3 e^3-3 a^2 b e^2 (d+2 e x)+3 a b^2 d e (3 d+4 e x)+b^3 \left (-5 d^3-4 d^2 e x+4 d e^2 x^2+2 e^3 x^3\right )-6 b^2 (b d-a e) (d+e x)^2 \log (d+e x)}{2 e^4 (d+e x)^2} \]

[In]

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

[Out]

(-(a^3*e^3) - 3*a^2*b*e^2*(d + 2*e*x) + 3*a*b^2*d*e*(3*d + 4*e*x) + b^3*(-5*d^3 - 4*d^2*e*x + 4*d*e^2*x^2 + 2*
e^3*x^3) - 6*b^2*(b*d - a*e)*(d + e*x)^2*Log[d + e*x])/(2*e^4*(d + e*x)^2)

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.46

method	result	size
default	\(\frac {b^{3} x}{e^{3}}-\frac {3 b \left (e^{2} a^{2}-2 a b d e +b^{2} d^{2}\right )}{e^{4} \left (e x +d \right )}+\frac {3 b^{2} \left (a e -b d \right ) \ln \left (e x +d \right )}{e^{4}}-\frac {a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}}{2 e^{4} \left (e x +d \right )^{2}}\)	\(114\)
norman	\(\frac {\frac {b^{3} x^{3}}{e}-\frac {a^{3} e^{3}+3 a^{2} b d \,e^{2}-9 a \,b^{2} d^{2} e +9 b^{3} d^{3}}{2 e^{4}}-\frac {\left (3 a^{2} b \,e^{2}-6 a \,b^{2} d e +6 b^{3} d^{2}\right ) x}{e^{3}}}{\left (e x +d \right )^{2}}+\frac {3 b^{2} \left (a e -b d \right ) \ln \left (e x +d \right )}{e^{4}}\)	\(116\)
risch	\(\frac {b^{3} x}{e^{3}}+\frac {\left (-3 a^{2} b \,e^{2}+6 a \,b^{2} d e -3 b^{3} d^{2}\right ) x -\frac {a^{3} e^{3}+3 a^{2} b d \,e^{2}-9 a \,b^{2} d^{2} e +5 b^{3} d^{3}}{2 e}}{e^{3} \left (e x +d \right )^{2}}+\frac {3 b^{2} \ln \left (e x +d \right ) a}{e^{3}}-\frac {3 b^{3} \ln \left (e x +d \right ) d}{e^{4}}\)	\(121\)
parallelrisch	\(\frac {6 \ln \left (e x +d \right ) x^{2} a \,b^{2} e^{3}-6 \ln \left (e x +d \right ) x^{2} b^{3} d \,e^{2}+2 b^{3} x^{3} e^{3}+12 \ln \left (e x +d \right ) x a \,b^{2} d \,e^{2}-12 \ln \left (e x +d \right ) x \,b^{3} d^{2} e +6 \ln \left (e x +d \right ) a \,b^{2} d^{2} e -6 \ln \left (e x +d \right ) b^{3} d^{3}-6 x \,a^{2} b \,e^{3}+12 x a \,b^{2} d \,e^{2}-12 x \,b^{3} d^{2} e -a^{3} e^{3}-3 a^{2} b d \,e^{2}+9 a \,b^{2} d^{2} e -9 b^{3} d^{3}}{2 e^{4} \left (e x +d \right )^{2}}\)	\(191\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 188 vs. \(2 (76) = 152\).

Time = 0.36 (sec) , antiderivative size = 188, normalized size of antiderivative = 2.41 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^3} \, dx=\frac {2 \, b^{3} e^{3} x^{3} + 4 \, b^{3} d e^{2} x^{2} - 5 \, b^{3} d^{3} + 9 \, a b^{2} d^{2} e - 3 \, a^{2} b d e^{2} - a^{3} e^{3} - 2 \, {\left (2 \, b^{3} d^{2} e - 6 \, a b^{2} d e^{2} + 3 \, a^{2} b e^{3}\right )} x - 6 \, {\left (b^{3} d^{3} - a b^{2} d^{2} e + {\left (b^{3} d e^{2} - a b^{2} e^{3}\right )} x^{2} + 2 \, {\left (b^{3} d^{2} e - a b^{2} d e^{2}\right )} x\right )} \log \left (e x + d\right )}{2 \, {\left (e^{6} x^{2} + 2 \, d e^{5} x + d^{2} e^{4}\right )}} \]

[In]

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

[Out]

1/2*(2*b^3*e^3*x^3 + 4*b^3*d*e^2*x^2 - 5*b^3*d^3 + 9*a*b^2*d^2*e - 3*a^2*b*d*e^2 - a^3*e^3 - 2*(2*b^3*d^2*e -
6*a*b^2*d*e^2 + 3*a^2*b*e^3)*x - 6*(b^3*d^3 - a*b^2*d^2*e + (b^3*d*e^2 - a*b^2*e^3)*x^2 + 2*(b^3*d^2*e - a*b^2
*d*e^2)*x)*log(e*x + d))/(e^6*x^2 + 2*d*e^5*x + d^2*e^4)

Sympy [A] (verification not implemented)

Time = 0.44 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.64 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^3} \, dx=\frac {b^{3} x}{e^{3}} + \frac {3 b^{2} \left (a e - b d\right ) \log {\left (d + e x \right )}}{e^{4}} + \frac {- a^{3} e^{3} - 3 a^{2} b d e^{2} + 9 a b^{2} d^{2} e - 5 b^{3} d^{3} + x \left (- 6 a^{2} b e^{3} + 12 a b^{2} d e^{2} - 6 b^{3} d^{2} e\right )}{2 d^{2} e^{4} + 4 d e^{5} x + 2 e^{6} x^{2}} \]

[In]

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

[Out]

b**3*x/e**3 + 3*b**2*(a*e - b*d)*log(d + e*x)/e**4 + (-a**3*e**3 - 3*a**2*b*d*e**2 + 9*a*b**2*d**2*e - 5*b**3*
d**3 + x*(-6*a**2*b*e**3 + 12*a*b**2*d*e**2 - 6*b**3*d**2*e))/(2*d**2*e**4 + 4*d*e**5*x + 2*e**6*x**2)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.60 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^3} \, dx=\frac {b^{3} x}{e^{3}} - \frac {5 \, b^{3} d^{3} - 9 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} + a^{3} e^{3} + 6 \, {\left (b^{3} d^{2} e - 2 \, a b^{2} d e^{2} + a^{2} b e^{3}\right )} x}{2 \, {\left (e^{6} x^{2} + 2 \, d e^{5} x + d^{2} e^{4}\right )}} - \frac {3 \, {\left (b^{3} d - a b^{2} e\right )} \log \left (e x + d\right )}{e^{4}} \]

[In]

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

[Out]

b^3*x/e^3 - 1/2*(5*b^3*d^3 - 9*a*b^2*d^2*e + 3*a^2*b*d*e^2 + a^3*e^3 + 6*(b^3*d^2*e - 2*a*b^2*d*e^2 + a^2*b*e^
3)*x)/(e^6*x^2 + 2*d*e^5*x + d^2*e^4) - 3*(b^3*d - a*b^2*e)*log(e*x + d)/e^4

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.44 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^3} \, dx=\frac {b^{3} x}{e^{3}} - \frac {3 \, {\left (b^{3} d - a b^{2} e\right )} \log \left ({\left | e x + d \right |}\right )}{e^{4}} - \frac {5 \, b^{3} d^{3} - 9 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} + a^{3} e^{3} + 6 \, {\left (b^{3} d^{2} e - 2 \, a b^{2} d e^{2} + a^{2} b e^{3}\right )} x}{2 \, {\left (e x + d\right )}^{2} e^{4}} \]

[In]

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

[Out]

b^3*x/e^3 - 3*(b^3*d - a*b^2*e)*log(abs(e*x + d))/e^4 - 1/2*(5*b^3*d^3 - 9*a*b^2*d^2*e + 3*a^2*b*d*e^2 + a^3*e
^3 + 6*(b^3*d^2*e - 2*a*b^2*d*e^2 + a^2*b*e^3)*x)/((e*x + d)^2*e^4)

Mupad [B] (verification not implemented)

Time = 10.85 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.67 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^3} \, dx=\frac {b^3\,x}{e^3}-\frac {\ln \left (d+e\,x\right )\,\left (3\,b^3\,d-3\,a\,b^2\,e\right )}{e^4}-\frac {\frac {a^3\,e^3+3\,a^2\,b\,d\,e^2-9\,a\,b^2\,d^2\,e+5\,b^3\,d^3}{2\,e}+x\,\left (3\,a^2\,b\,e^2-6\,a\,b^2\,d\,e+3\,b^3\,d^2\right )}{d^2\,e^3+2\,d\,e^4\,x+e^5\,x^2} \]

[In]

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

[Out]

(b^3*x)/e^3 - (log(d + e*x)*(3*b^3*d - 3*a*b^2*e))/e^4 - ((a^3*e^3 + 5*b^3*d^3 - 9*a*b^2*d^2*e + 3*a^2*b*d*e^2
)/(2*e) + x*(3*b^3*d^2 + 3*a^2*b*e^2 - 6*a*b^2*d*e))/(d^2*e^3 + e^5*x^2 + 2*d*e^4*x)