\(\int (d+e x) (b x+c x^2)^3 \, dx\) [247]

Optimal result

Integrand size = 17, antiderivative size = 75 \[ \int (d+e x) \left (b x+c x^2\right )^3 \, dx=\frac {1}{4} b^3 d x^4+\frac {1}{5} b^2 (3 c d+b e) x^5+\frac {1}{2} b c (c d+b e) x^6+\frac {1}{7} c^2 (c d+3 b e) x^7+\frac {1}{8} c^3 e x^8 \]

[Out]

1/4*b^3*d*x^4+1/5*b^2*(b*e+3*c*d)*x^5+1/2*b*c*(b*e+c*d)*x^6+1/7*c^2*(3*b*e+c*d)*x^7+1/8*c^3*e*x^8

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {645} \[ \int (d+e x) \left (b x+c x^2\right )^3 \, dx=\frac {1}{4} b^3 d x^4+\frac {1}{5} b^2 x^5 (b e+3 c d)+\frac {1}{7} c^2 x^7 (3 b e+c d)+\frac {1}{2} b c x^6 (b e+c d)+\frac {1}{8} c^3 e x^8 \]

[In]

Int[(d + e*x)*(b*x + c*x^2)^3,x]

[Out]

(b^3*d*x^4)/4 + (b^2*(3*c*d + b*e)*x^5)/5 + (b*c*(c*d + b*e)*x^6)/2 + (c^2*(c*d + 3*b*e)*x^7)/7 + (c^3*e*x^8)/
8

Rule 645

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

Rubi steps \begin{align*} \text {integral}& = \int \left (b^3 d x^3+b^2 (3 c d+b e) x^4+3 b c (c d+b e) x^5+c^2 (c d+3 b e) x^6+c^3 e x^7\right ) \, dx \\ & = \frac {1}{4} b^3 d x^4+\frac {1}{5} b^2 (3 c d+b e) x^5+\frac {1}{2} b c (c d+b e) x^6+\frac {1}{7} c^2 (c d+3 b e) x^7+\frac {1}{8} c^3 e x^8 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00 \[ \int (d+e x) \left (b x+c x^2\right )^3 \, dx=\frac {1}{4} b^3 d x^4+\frac {1}{5} b^2 (3 c d+b e) x^5+\frac {1}{2} b c (c d+b e) x^6+\frac {1}{7} c^2 (c d+3 b e) x^7+\frac {1}{8} c^3 e x^8 \]

[In]

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

[Out]

(b^3*d*x^4)/4 + (b^2*(3*c*d + b*e)*x^5)/5 + (b*c*(c*d + b*e)*x^6)/2 + (c^2*(c*d + 3*b*e)*x^7)/7 + (c^3*e*x^8)/
8

Maple [A] (verified)

Time = 1.97 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00

method	result	size
norman	\(\frac {c^{3} e \,x^{8}}{8}+\left (\frac {3}{7} b \,c^{2} e +\frac {1}{7} c^{3} d \right ) x^{7}+\left (\frac {1}{2} b^{2} c e +\frac {1}{2} b \,c^{2} d \right ) x^{6}+\left (\frac {1}{5} b^{3} e +\frac {3}{5} b^{2} c d \right ) x^{5}+\frac {d \,x^{4} b^{3}}{4}\)	\(75\)
gosper	\(\frac {x^{4} \left (35 c^{3} x^{4} e +120 b \,c^{2} x^{3} e +40 c^{3} d \,x^{3}+140 b^{2} c e \,x^{2}+140 b \,c^{2} d \,x^{2}+56 b^{3} e x +168 x \,b^{2} c d +70 b^{3} d \right )}{280}\)	\(76\)
default	\(\frac {c^{3} e \,x^{8}}{8}+\frac {\left (3 b \,c^{2} e +c^{3} d \right ) x^{7}}{7}+\frac {\left (3 b^{2} c e +3 b \,c^{2} d \right ) x^{6}}{6}+\frac {\left (b^{3} e +3 b^{2} c d \right ) x^{5}}{5}+\frac {d \,x^{4} b^{3}}{4}\)	\(76\)
risch	\(\frac {1}{8} c^{3} e \,x^{8}+\frac {3}{7} x^{7} b \,c^{2} e +\frac {1}{7} c^{3} d \,x^{7}+\frac {1}{2} x^{6} b^{2} c e +\frac {1}{2} x^{6} b \,c^{2} d +\frac {1}{5} b^{3} e \,x^{5}+\frac {3}{5} b^{2} c d \,x^{5}+\frac {1}{4} d \,x^{4} b^{3}\)	\(78\)
parallelrisch	\(\frac {1}{8} c^{3} e \,x^{8}+\frac {3}{7} x^{7} b \,c^{2} e +\frac {1}{7} c^{3} d \,x^{7}+\frac {1}{2} x^{6} b^{2} c e +\frac {1}{2} x^{6} b \,c^{2} d +\frac {1}{5} b^{3} e \,x^{5}+\frac {3}{5} b^{2} c d \,x^{5}+\frac {1}{4} d \,x^{4} b^{3}\)	\(78\)

[In]

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

[Out]

1/8*c^3*e*x^8+(3/7*b*c^2*e+1/7*c^3*d)*x^7+(1/2*b^2*c*e+1/2*b*c^2*d)*x^6+(1/5*b^3*e+3/5*b^2*c*d)*x^5+1/4*d*x^4*
b^3

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.97 \[ \int (d+e x) \left (b x+c x^2\right )^3 \, dx=\frac {1}{8} \, c^{3} e x^{8} + \frac {1}{4} \, b^{3} d x^{4} + \frac {1}{7} \, {\left (c^{3} d + 3 \, b c^{2} e\right )} x^{7} + \frac {1}{2} \, {\left (b c^{2} d + b^{2} c e\right )} x^{6} + \frac {1}{5} \, {\left (3 \, b^{2} c d + b^{3} e\right )} x^{5} \]

[In]

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

[Out]

1/8*c^3*e*x^8 + 1/4*b^3*d*x^4 + 1/7*(c^3*d + 3*b*c^2*e)*x^7 + 1/2*(b*c^2*d + b^2*c*e)*x^6 + 1/5*(3*b^2*c*d + b
^3*e)*x^5

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.07 \[ \int (d+e x) \left (b x+c x^2\right )^3 \, dx=\frac {b^{3} d x^{4}}{4} + \frac {c^{3} e x^{8}}{8} + x^{7} \cdot \left (\frac {3 b c^{2} e}{7} + \frac {c^{3} d}{7}\right ) + x^{6} \left (\frac {b^{2} c e}{2} + \frac {b c^{2} d}{2}\right ) + x^{5} \left (\frac {b^{3} e}{5} + \frac {3 b^{2} c d}{5}\right ) \]

[In]

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

[Out]

b**3*d*x**4/4 + c**3*e*x**8/8 + x**7*(3*b*c**2*e/7 + c**3*d/7) + x**6*(b**2*c*e/2 + b*c**2*d/2) + x**5*(b**3*e
/5 + 3*b**2*c*d/5)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.97 \[ \int (d+e x) \left (b x+c x^2\right )^3 \, dx=\frac {1}{8} \, c^{3} e x^{8} + \frac {1}{4} \, b^{3} d x^{4} + \frac {1}{7} \, {\left (c^{3} d + 3 \, b c^{2} e\right )} x^{7} + \frac {1}{2} \, {\left (b c^{2} d + b^{2} c e\right )} x^{6} + \frac {1}{5} \, {\left (3 \, b^{2} c d + b^{3} e\right )} x^{5} \]

[In]

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

[Out]

1/8*c^3*e*x^8 + 1/4*b^3*d*x^4 + 1/7*(c^3*d + 3*b*c^2*e)*x^7 + 1/2*(b*c^2*d + b^2*c*e)*x^6 + 1/5*(3*b^2*c*d + b
^3*e)*x^5

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.03 \[ \int (d+e x) \left (b x+c x^2\right )^3 \, dx=\frac {1}{8} \, c^{3} e x^{8} + \frac {1}{7} \, c^{3} d x^{7} + \frac {3}{7} \, b c^{2} e x^{7} + \frac {1}{2} \, b c^{2} d x^{6} + \frac {1}{2} \, b^{2} c e x^{6} + \frac {3}{5} \, b^{2} c d x^{5} + \frac {1}{5} \, b^{3} e x^{5} + \frac {1}{4} \, b^{3} d x^{4} \]

[In]

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

[Out]

1/8*c^3*e*x^8 + 1/7*c^3*d*x^7 + 3/7*b*c^2*e*x^7 + 1/2*b*c^2*d*x^6 + 1/2*b^2*c*e*x^6 + 3/5*b^2*c*d*x^5 + 1/5*b^
3*e*x^5 + 1/4*b^3*d*x^4

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.92 \[ \int (d+e x) \left (b x+c x^2\right )^3 \, dx=x^5\,\left (\frac {e\,b^3}{5}+\frac {3\,c\,d\,b^2}{5}\right )+x^7\,\left (\frac {d\,c^3}{7}+\frac {3\,b\,e\,c^2}{7}\right )+\frac {b^3\,d\,x^4}{4}+\frac {c^3\,e\,x^8}{8}+\frac {b\,c\,x^6\,\left (b\,e+c\,d\right )}{2} \]

[In]

int((b*x + c*x^2)^3*(d + e*x),x)

[Out]

x^5*((b^3*e)/5 + (3*b^2*c*d)/5) + x^7*((c^3*d)/7 + (3*b*c^2*e)/7) + (b^3*d*x^4)/4 + (c^3*e*x^8)/8 + (b*c*x^6*(
b*e + c*d))/2