\(\int (d+e x)^2 (a d e+(c d^2+a e^2) x+c d e x^2) \, dx\) [1830]

Optimal result

Integrand size = 33, antiderivative size = 39 \[ \int (d+e x)^2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right ) \, dx=\frac {1}{4} \left (a-\frac {c d^2}{e^2}\right ) (d+e x)^4+\frac {c d (d+e x)^5}{5 e^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {640, 45} \[ \int (d+e x)^2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right ) \, dx=\frac {1}{4} (d+e x)^4 \left (a-\frac {c d^2}{e^2}\right )+\frac {c d (d+e x)^5}{5 e^2} \]

[In]

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

[Out]

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

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 (a e+c d x) (d+e x)^3 \, dx \\ & = \int \left (\frac {\left (-c d^2+a e^2\right ) (d+e x)^3}{e}+\frac {c d (d+e x)^4}{e}\right ) \, dx \\ & = \frac {1}{4} \left (a-\frac {c d^2}{e^2}\right ) (d+e x)^4+\frac {c d (d+e x)^5}{5 e^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.87 \[ \int (d+e x)^2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right ) \, dx=\frac {1}{20} x \left (5 a e \left (4 d^3+6 d^2 e x+4 d e^2 x^2+e^3 x^3\right )+c d x \left (10 d^3+20 d^2 e x+15 d e^2 x^2+4 e^3 x^3\right )\right ) \]

[In]

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

[Out]

(x*(5*a*e*(4*d^3 + 6*d^2*e*x + 4*d*e^2*x^2 + e^3*x^3) + c*d*x*(10*d^3 + 20*d^2*e*x + 15*d*e^2*x^2 + 4*e^3*x^3)
))/20

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(75\) vs. \(2(35)=70\).

Time = 2.18 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.95

method	result	size
norman	\(\frac {d \,e^{3} c \,x^{5}}{5}+\left (\frac {1}{4} e^{4} a +\frac {3}{4} d^{2} e^{2} c \right ) x^{4}+\left (a d \,e^{3}+c \,d^{3} e \right ) x^{3}+\left (\frac {3}{2} a \,d^{2} e^{2}+\frac {1}{2} c \,d^{4}\right ) x^{2}+a \,d^{3} e x\)	\(76\)
risch	\(\frac {1}{5} d \,e^{3} c \,x^{5}+\frac {1}{4} x^{4} e^{4} a +\frac {3}{4} x^{4} d^{2} e^{2} c +x^{3} a d \,e^{3}+c \,d^{3} e \,x^{3}+\frac {3}{2} x^{2} a \,d^{2} e^{2}+\frac {1}{2} x^{2} c \,d^{4}+a \,d^{3} e x\)	\(79\)
parallelrisch	\(\frac {1}{5} d \,e^{3} c \,x^{5}+\frac {1}{4} x^{4} e^{4} a +\frac {3}{4} x^{4} d^{2} e^{2} c +x^{3} a d \,e^{3}+c \,d^{3} e \,x^{3}+\frac {3}{2} x^{2} a \,d^{2} e^{2}+\frac {1}{2} x^{2} c \,d^{4}+a \,d^{3} e x\)	\(79\)
gosper	\(\frac {x \left (4 d \,e^{3} c \,x^{4}+5 a \,e^{4} x^{3}+15 x^{3} d^{2} e^{2} c +20 a d \,e^{3} x^{2}+20 c \,d^{3} e \,x^{2}+30 a \,d^{2} e^{2} x +10 x c \,d^{4}+20 a \,d^{3} e \right )}{20}\)	\(80\)
default	\(\frac {d \,e^{3} c \,x^{5}}{5}+\frac {\left (2 d^{2} e^{2} c +e^{2} \left (e^{2} a +c \,d^{2}\right )\right ) x^{4}}{4}+\frac {\left (c \,d^{3} e +2 d e \left (e^{2} a +c \,d^{2}\right )+a d \,e^{3}\right ) x^{3}}{3}+\frac {\left (d^{2} \left (e^{2} a +c \,d^{2}\right )+2 a \,d^{2} e^{2}\right ) x^{2}}{2}+a \,d^{3} e x\)	\(112\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 75 vs. \(2 (35) = 70\).

Time = 0.27 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.92 \[ \int (d+e x)^2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right ) \, dx=\frac {1}{5} \, c d e^{3} x^{5} + a d^{3} e x + \frac {1}{4} \, {\left (3 \, c d^{2} e^{2} + a e^{4}\right )} x^{4} + {\left (c d^{3} e + a d e^{3}\right )} x^{3} + \frac {1}{2} \, {\left (c d^{4} + 3 \, a d^{2} e^{2}\right )} x^{2} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 80 vs. \(2 (34) = 68\).

Time = 0.03 (sec) , antiderivative size = 80, normalized size of antiderivative = 2.05 \[ \int (d+e x)^2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right ) \, dx=a d^{3} e x + \frac {c d e^{3} x^{5}}{5} + x^{4} \left (\frac {a e^{4}}{4} + \frac {3 c d^{2} e^{2}}{4}\right ) + x^{3} \left (a d e^{3} + c d^{3} e\right ) + x^{2} \cdot \left (\frac {3 a d^{2} e^{2}}{2} + \frac {c d^{4}}{2}\right ) \]

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 75 vs. \(2 (35) = 70\).

Time = 0.19 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.92 \[ \int (d+e x)^2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right ) \, dx=\frac {1}{5} \, c d e^{3} x^{5} + a d^{3} e x + \frac {1}{4} \, {\left (3 \, c d^{2} e^{2} + a e^{4}\right )} x^{4} + {\left (c d^{3} e + a d e^{3}\right )} x^{3} + \frac {1}{2} \, {\left (c d^{4} + 3 \, a d^{2} e^{2}\right )} x^{2} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 78 vs. \(2 (35) = 70\).

Time = 0.28 (sec) , antiderivative size = 78, normalized size of antiderivative = 2.00 \[ \int (d+e x)^2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right ) \, dx=\frac {1}{5} \, c d e^{3} x^{5} + \frac {3}{4} \, c d^{2} e^{2} x^{4} + \frac {1}{4} \, a e^{4} x^{4} + c d^{3} e x^{3} + a d e^{3} x^{3} + \frac {1}{2} \, c d^{4} x^{2} + \frac {3}{2} \, a d^{2} e^{2} x^{2} + a d^{3} e x \]

[In]

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

[Out]

1/5*c*d*e^3*x^5 + 3/4*c*d^2*e^2*x^4 + 1/4*a*e^4*x^4 + c*d^3*e*x^3 + a*d*e^3*x^3 + 1/2*c*d^4*x^2 + 3/2*a*d^2*e^
2*x^2 + a*d^3*e*x

Mupad [B] (verification not implemented)

Time = 9.60 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.92 \[ \int (d+e x)^2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right ) \, dx=x^2\,\left (\frac {c\,d^4}{2}+\frac {3\,a\,d^2\,e^2}{2}\right )+x^4\,\left (\frac {3\,c\,d^2\,e^2}{4}+\frac {a\,e^4}{4}\right )+x^3\,\left (c\,d^3\,e+a\,d\,e^3\right )+a\,d^3\,e\,x+\frac {c\,d\,e^3\,x^5}{5} \]

[In]

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

[Out]

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