\(\int x^3 (a+b x)^2 (A+B x) \, dx\) [92]

Optimal result

Integrand size = 16, antiderivative size = 55 \[ \int x^3 (a+b x)^2 (A+B x) \, dx=\frac {1}{4} a^2 A x^4+\frac {1}{5} a (2 A b+a B) x^5+\frac {1}{6} b (A b+2 a B) x^6+\frac {1}{7} b^2 B x^7 \]

[Out]

1/4*a^2*A*x^4+1/5*a*(2*A*b+B*a)*x^5+1/6*b*(A*b+2*B*a)*x^6+1/7*b^2*B*x^7

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {77} \[ \int x^3 (a+b x)^2 (A+B x) \, dx=\frac {1}{4} a^2 A x^4+\frac {1}{6} b x^6 (2 a B+A b)+\frac {1}{5} a x^5 (a B+2 A b)+\frac {1}{7} b^2 B x^7 \]

[In]

Int[x^3*(a + b*x)^2*(A + B*x),x]

[Out]

(a^2*A*x^4)/4 + (a*(2*A*b + a*B)*x^5)/5 + (b*(A*b + 2*a*B)*x^6)/6 + (b^2*B*x^7)/7

Rule 77

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && N
eQ[b*e + a*f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && Rational
Q[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00 \[ \int x^3 (a+b x)^2 (A+B x) \, dx=\frac {1}{4} a^2 A x^4+\frac {1}{5} a (2 A b+a B) x^5+\frac {1}{6} b (A b+2 a B) x^6+\frac {1}{7} b^2 B x^7 \]

[In]

Integrate[x^3*(a + b*x)^2*(A + B*x),x]

[Out]

(a^2*A*x^4)/4 + (a*(2*A*b + a*B)*x^5)/5 + (b*(A*b + 2*a*B)*x^6)/6 + (b^2*B*x^7)/7

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.95

method	result	size
default	\(\frac {b^{2} B \,x^{7}}{7}+\frac {\left (b^{2} A +2 a b B \right ) x^{6}}{6}+\frac {\left (2 a b A +a^{2} B \right ) x^{5}}{5}+\frac {a^{2} A \,x^{4}}{4}\)	\(52\)
norman	\(\frac {b^{2} B \,x^{7}}{7}+\left (\frac {1}{6} b^{2} A +\frac {1}{3} a b B \right ) x^{6}+\left (\frac {2}{5} a b A +\frac {1}{5} a^{2} B \right ) x^{5}+\frac {a^{2} A \,x^{4}}{4}\)	\(52\)
gosper	\(\frac {1}{7} b^{2} B \,x^{7}+\frac {1}{6} x^{6} b^{2} A +\frac {1}{3} x^{6} a b B +\frac {2}{5} x^{5} a b A +\frac {1}{5} x^{5} a^{2} B +\frac {1}{4} a^{2} A \,x^{4}\)	\(54\)
risch	\(\frac {1}{7} b^{2} B \,x^{7}+\frac {1}{6} x^{6} b^{2} A +\frac {1}{3} x^{6} a b B +\frac {2}{5} x^{5} a b A +\frac {1}{5} x^{5} a^{2} B +\frac {1}{4} a^{2} A \,x^{4}\)	\(54\)
parallelrisch	\(\frac {1}{7} b^{2} B \,x^{7}+\frac {1}{6} x^{6} b^{2} A +\frac {1}{3} x^{6} a b B +\frac {2}{5} x^{5} a b A +\frac {1}{5} x^{5} a^{2} B +\frac {1}{4} a^{2} A \,x^{4}\)	\(54\)

[In]

int(x^3*(b*x+a)^2*(B*x+A),x,method=_RETURNVERBOSE)

[Out]

1/7*b^2*B*x^7+1/6*(A*b^2+2*B*a*b)*x^6+1/5*(2*A*a*b+B*a^2)*x^5+1/4*a^2*A*x^4

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.93 \[ \int x^3 (a+b x)^2 (A+B x) \, dx=\frac {1}{7} \, B b^{2} x^{7} + \frac {1}{4} \, A a^{2} x^{4} + \frac {1}{6} \, {\left (2 \, B a b + A b^{2}\right )} x^{6} + \frac {1}{5} \, {\left (B a^{2} + 2 \, A a b\right )} x^{5} \]

[In]

integrate(x^3*(b*x+a)^2*(B*x+A),x, algorithm="fricas")

[Out]

1/7*B*b^2*x^7 + 1/4*A*a^2*x^4 + 1/6*(2*B*a*b + A*b^2)*x^6 + 1/5*(B*a^2 + 2*A*a*b)*x^5

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.98 \[ \int x^3 (a+b x)^2 (A+B x) \, dx=\frac {A a^{2} x^{4}}{4} + \frac {B b^{2} x^{7}}{7} + x^{6} \left (\frac {A b^{2}}{6} + \frac {B a b}{3}\right ) + x^{5} \cdot \left (\frac {2 A a b}{5} + \frac {B a^{2}}{5}\right ) \]

[In]

integrate(x**3*(b*x+a)**2*(B*x+A),x)

[Out]

A*a**2*x**4/4 + B*b**2*x**7/7 + x**6*(A*b**2/6 + B*a*b/3) + x**5*(2*A*a*b/5 + B*a**2/5)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.93 \[ \int x^3 (a+b x)^2 (A+B x) \, dx=\frac {1}{7} \, B b^{2} x^{7} + \frac {1}{4} \, A a^{2} x^{4} + \frac {1}{6} \, {\left (2 \, B a b + A b^{2}\right )} x^{6} + \frac {1}{5} \, {\left (B a^{2} + 2 \, A a b\right )} x^{5} \]

[In]

integrate(x^3*(b*x+a)^2*(B*x+A),x, algorithm="maxima")

[Out]

1/7*B*b^2*x^7 + 1/4*A*a^2*x^4 + 1/6*(2*B*a*b + A*b^2)*x^6 + 1/5*(B*a^2 + 2*A*a*b)*x^5

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.96 \[ \int x^3 (a+b x)^2 (A+B x) \, dx=\frac {1}{7} \, B b^{2} x^{7} + \frac {1}{3} \, B a b x^{6} + \frac {1}{6} \, A b^{2} x^{6} + \frac {1}{5} \, B a^{2} x^{5} + \frac {2}{5} \, A a b x^{5} + \frac {1}{4} \, A a^{2} x^{4} \]

[In]

integrate(x^3*(b*x+a)^2*(B*x+A),x, algorithm="giac")

[Out]

1/7*B*b^2*x^7 + 1/3*B*a*b*x^6 + 1/6*A*b^2*x^6 + 1/5*B*a^2*x^5 + 2/5*A*a*b*x^5 + 1/4*A*a^2*x^4

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.93 \[ \int x^3 (a+b x)^2 (A+B x) \, dx=x^5\,\left (\frac {B\,a^2}{5}+\frac {2\,A\,b\,a}{5}\right )+x^6\,\left (\frac {A\,b^2}{6}+\frac {B\,a\,b}{3}\right )+\frac {A\,a^2\,x^4}{4}+\frac {B\,b^2\,x^7}{7} \]

[In]

int(x^3*(A + B*x)*(a + b*x)^2,x)

[Out]

x^5*((B*a^2)/5 + (2*A*a*b)/5) + x^6*((A*b^2)/6 + (B*a*b)/3) + (A*a^2*x^4)/4 + (B*b^2*x^7)/7