\(\int (c+d x+e x^2) (a+b x^3)^2 \, dx\) [321]

Optimal result

Integrand size = 20, antiderivative size = 77 \[ \int \left (c+d x+e x^2\right ) \left (a+b x^3\right )^2 \, dx=a^2 c x+\frac {1}{2} a^2 d x^2+\frac {1}{2} a b c x^4+\frac {2}{5} a b d x^5+\frac {1}{7} b^2 c x^7+\frac {1}{8} b^2 d x^8+\frac {e \left (a+b x^3\right )^3}{9 b} \]

[Out]

a^2*c*x+1/2*a^2*d*x^2+1/2*a*b*c*x^4+2/5*a*b*d*x^5+1/7*b^2*c*x^7+1/8*b^2*d*x^8+1/9*e*(b*x^3+a)^3/b

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {1596, 1864} \[ \int \left (c+d x+e x^2\right ) \left (a+b x^3\right )^2 \, dx=a^2 c x+\frac {1}{2} a^2 d x^2+\frac {1}{2} a b c x^4+\frac {2}{5} a b d x^5+\frac {e \left (a+b x^3\right )^3}{9 b}+\frac {1}{7} b^2 c x^7+\frac {1}{8} b^2 d x^8 \]

[In]

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

[Out]

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

Rule 1596

Int[(Px_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[Coeff[Px, x, n - 1]*((a + b*x^n)^(p + 1)/(b*n*(p +
1))), x] + Int[(Px - Coeff[Px, x, n - 1]*x^(n - 1))*(a + b*x^n)^p, x] /; FreeQ[{a, b}, x] && PolyQ[Px, x] && I
GtQ[p, 1] && IGtQ[n, 1] && NeQ[Coeff[Px, x, n - 1], 0] && NeQ[Px, Coeff[Px, x, n - 1]*x^(n - 1)] &&  !MatchQ[P
x, (Qx_.)*((c_) + (d_.)*x^(m_))^(q_) /; FreeQ[{c, d}, x] && PolyQ[Qx, x] && IGtQ[q, 1] && IGtQ[m, 1] && NeQ[Co
eff[Qx*(a + b*x^n)^p, x, m - 1], 0] && GtQ[m*q, n*p]]

Rule 1864

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^n)^p, x], x] /; FreeQ[
{a, b, n}, x] && PolyQ[Pq, x] && (IGtQ[p, 0] || EqQ[n, 1])

Rubi steps \begin{align*} \text {integral}& = \frac {e \left (a+b x^3\right )^3}{9 b}+\int (c+d x) \left (a+b x^3\right )^2 \, dx \\ & = \frac {e \left (a+b x^3\right )^3}{9 b}+\int \left (a^2 c+a^2 d x+2 a b c x^3+2 a b d x^4+b^2 c x^6+b^2 d x^7\right ) \, dx \\ & = a^2 c x+\frac {1}{2} a^2 d x^2+\frac {1}{2} a b c x^4+\frac {2}{5} a b d x^5+\frac {1}{7} b^2 c x^7+\frac {1}{8} b^2 d x^8+\frac {e \left (a+b x^3\right )^3}{9 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.19 \[ \int \left (c+d x+e x^2\right ) \left (a+b x^3\right )^2 \, dx=a^2 c x+\frac {1}{2} a^2 d x^2+\frac {1}{3} a^2 e x^3+\frac {1}{2} a b c x^4+\frac {2}{5} a b d x^5+\frac {1}{3} a b e x^6+\frac {1}{7} b^2 c x^7+\frac {1}{8} b^2 d x^8+\frac {1}{9} b^2 e x^9 \]

[In]

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

[Out]

a^2*c*x + (a^2*d*x^2)/2 + (a^2*e*x^3)/3 + (a*b*c*x^4)/2 + (2*a*b*d*x^5)/5 + (a*b*e*x^6)/3 + (b^2*c*x^7)/7 + (b
^2*d*x^8)/8 + (b^2*e*x^9)/9

Maple [A] (verified)

Time = 1.54 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00

method	result	size
gosper	\(\frac {1}{9} b^{2} e \,x^{9}+\frac {1}{8} b^{2} d \,x^{8}+\frac {1}{7} b^{2} c \,x^{7}+\frac {1}{3} a e b \,x^{6}+\frac {2}{5} x^{5} d b a +\frac {1}{2} a b c \,x^{4}+\frac {1}{3} a^{2} e \,x^{3}+\frac {1}{2} a^{2} d \,x^{2}+a^{2} c x\)	\(77\)
default	\(\frac {1}{9} b^{2} e \,x^{9}+\frac {1}{8} b^{2} d \,x^{8}+\frac {1}{7} b^{2} c \,x^{7}+\frac {1}{3} a e b \,x^{6}+\frac {2}{5} x^{5} d b a +\frac {1}{2} a b c \,x^{4}+\frac {1}{3} a^{2} e \,x^{3}+\frac {1}{2} a^{2} d \,x^{2}+a^{2} c x\)	\(77\)
norman	\(\frac {1}{9} b^{2} e \,x^{9}+\frac {1}{8} b^{2} d \,x^{8}+\frac {1}{7} b^{2} c \,x^{7}+\frac {1}{3} a e b \,x^{6}+\frac {2}{5} x^{5} d b a +\frac {1}{2} a b c \,x^{4}+\frac {1}{3} a^{2} e \,x^{3}+\frac {1}{2} a^{2} d \,x^{2}+a^{2} c x\)	\(77\)
risch	\(\frac {1}{9} b^{2} e \,x^{9}+\frac {1}{8} b^{2} d \,x^{8}+\frac {1}{7} b^{2} c \,x^{7}+\frac {1}{3} a e b \,x^{6}+\frac {2}{5} x^{5} d b a +\frac {1}{2} a b c \,x^{4}+\frac {1}{3} a^{2} e \,x^{3}+\frac {1}{2} a^{2} d \,x^{2}+a^{2} c x\)	\(77\)
parallelrisch	\(\frac {1}{9} b^{2} e \,x^{9}+\frac {1}{8} b^{2} d \,x^{8}+\frac {1}{7} b^{2} c \,x^{7}+\frac {1}{3} a e b \,x^{6}+\frac {2}{5} x^{5} d b a +\frac {1}{2} a b c \,x^{4}+\frac {1}{3} a^{2} e \,x^{3}+\frac {1}{2} a^{2} d \,x^{2}+a^{2} c x\)	\(77\)

[In]

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

[Out]

1/9*b^2*e*x^9+1/8*b^2*d*x^8+1/7*b^2*c*x^7+1/3*a*e*b*x^6+2/5*x^5*d*b*a+1/2*a*b*c*x^4+1/3*a^2*e*x^3+1/2*a^2*d*x^
2+a^2*c*x

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.99 \[ \int \left (c+d x+e x^2\right ) \left (a+b x^3\right )^2 \, dx=\frac {1}{9} \, b^{2} e x^{9} + \frac {1}{8} \, b^{2} d x^{8} + \frac {1}{7} \, b^{2} c x^{7} + \frac {1}{3} \, a b e x^{6} + \frac {2}{5} \, a b d x^{5} + \frac {1}{2} \, a b c x^{4} + \frac {1}{3} \, a^{2} e x^{3} + \frac {1}{2} \, a^{2} d x^{2} + a^{2} c x \]

[In]

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

[Out]

1/9*b^2*e*x^9 + 1/8*b^2*d*x^8 + 1/7*b^2*c*x^7 + 1/3*a*b*e*x^6 + 2/5*a*b*d*x^5 + 1/2*a*b*c*x^4 + 1/3*a^2*e*x^3
+ 1/2*a^2*d*x^2 + a^2*c*x

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.14 \[ \int \left (c+d x+e x^2\right ) \left (a+b x^3\right )^2 \, dx=a^{2} c x + \frac {a^{2} d x^{2}}{2} + \frac {a^{2} e x^{3}}{3} + \frac {a b c x^{4}}{2} + \frac {2 a b d x^{5}}{5} + \frac {a b e x^{6}}{3} + \frac {b^{2} c x^{7}}{7} + \frac {b^{2} d x^{8}}{8} + \frac {b^{2} e x^{9}}{9} \]

[In]

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

[Out]

a**2*c*x + a**2*d*x**2/2 + a**2*e*x**3/3 + a*b*c*x**4/2 + 2*a*b*d*x**5/5 + a*b*e*x**6/3 + b**2*c*x**7/7 + b**2
*d*x**8/8 + b**2*e*x**9/9

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.99 \[ \int \left (c+d x+e x^2\right ) \left (a+b x^3\right )^2 \, dx=\frac {1}{9} \, b^{2} e x^{9} + \frac {1}{8} \, b^{2} d x^{8} + \frac {1}{7} \, b^{2} c x^{7} + \frac {1}{3} \, a b e x^{6} + \frac {2}{5} \, a b d x^{5} + \frac {1}{2} \, a b c x^{4} + \frac {1}{3} \, a^{2} e x^{3} + \frac {1}{2} \, a^{2} d x^{2} + a^{2} c x \]

[In]

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

[Out]

1/9*b^2*e*x^9 + 1/8*b^2*d*x^8 + 1/7*b^2*c*x^7 + 1/3*a*b*e*x^6 + 2/5*a*b*d*x^5 + 1/2*a*b*c*x^4 + 1/3*a^2*e*x^3
+ 1/2*a^2*d*x^2 + a^2*c*x

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.99 \[ \int \left (c+d x+e x^2\right ) \left (a+b x^3\right )^2 \, dx=\frac {1}{9} \, b^{2} e x^{9} + \frac {1}{8} \, b^{2} d x^{8} + \frac {1}{7} \, b^{2} c x^{7} + \frac {1}{3} \, a b e x^{6} + \frac {2}{5} \, a b d x^{5} + \frac {1}{2} \, a b c x^{4} + \frac {1}{3} \, a^{2} e x^{3} + \frac {1}{2} \, a^{2} d x^{2} + a^{2} c x \]

[In]

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

[Out]

1/9*b^2*e*x^9 + 1/8*b^2*d*x^8 + 1/7*b^2*c*x^7 + 1/3*a*b*e*x^6 + 2/5*a*b*d*x^5 + 1/2*a*b*c*x^4 + 1/3*a^2*e*x^3
+ 1/2*a^2*d*x^2 + a^2*c*x

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.99 \[ \int \left (c+d x+e x^2\right ) \left (a+b x^3\right )^2 \, dx=\frac {e\,a^2\,x^3}{3}+\frac {d\,a^2\,x^2}{2}+c\,a^2\,x+\frac {e\,a\,b\,x^6}{3}+\frac {2\,d\,a\,b\,x^5}{5}+\frac {c\,a\,b\,x^4}{2}+\frac {e\,b^2\,x^9}{9}+\frac {d\,b^2\,x^8}{8}+\frac {c\,b^2\,x^7}{7} \]

[In]

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

[Out]

(a^2*d*x^2)/2 + (b^2*c*x^7)/7 + (a^2*e*x^3)/3 + (b^2*d*x^8)/8 + (b^2*e*x^9)/9 + a^2*c*x + (a*b*c*x^4)/2 + (2*a
*b*d*x^5)/5 + (a*b*e*x^6)/3