Integrand size = 20, antiderivative size = 71 \[ \int x^m \left (a+b x^3\right )^2 \left (A+B x^3\right ) \, dx=\frac {a^2 A x^{1+m}}{1+m}+\frac {a (2 A b+a B) x^{4+m}}{4+m}+\frac {b (A b+2 a B) x^{7+m}}{7+m}+\frac {b^2 B x^{10+m}}{10+m} \] Output:
a^2*A*x^(1+m)/(1+m)+a*(2*A*b+B*a)*x^(4+m)/(4+m)+b*(A*b+2*B*a)*x^(7+m)/(7+m )+b^2*B*x^(10+m)/(10+m)
Time = 0.15 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.93 \[ \int x^m \left (a+b x^3\right )^2 \left (A+B x^3\right ) \, dx=x^{1+m} \left (\frac {a^2 A}{1+m}+\frac {a (2 A b+a B) x^3}{4+m}+\frac {b (A b+2 a B) x^6}{7+m}+\frac {b^2 B x^9}{10+m}\right ) \] Input:
Integrate[x^m*(a + b*x^3)^2*(A + B*x^3),x]
Output:
x^(1 + m)*((a^2*A)/(1 + m) + (a*(2*A*b + a*B)*x^3)/(4 + m) + (b*(A*b + 2*a *B)*x^6)/(7 + m) + (b^2*B*x^9)/(10 + m))
Time = 0.37 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {950, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x^m \left (a+b x^3\right )^2 \left (A+B x^3\right ) \, dx\) |
\(\Big \downarrow \) 950 |
\(\displaystyle \int \left (a^2 A x^m+a x^{m+3} (a B+2 A b)+b x^{m+6} (2 a B+A b)+b^2 B x^{m+9}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^2 A x^{m+1}}{m+1}+\frac {a x^{m+4} (a B+2 A b)}{m+4}+\frac {b x^{m+7} (2 a B+A b)}{m+7}+\frac {b^2 B x^{m+10}}{m+10}\) |
Input:
Int[x^m*(a + b*x^3)^2*(A + B*x^3),x]
Output:
(a^2*A*x^(1 + m))/(1 + m) + (a*(2*A*b + a*B)*x^(4 + m))/(4 + m) + (b*(A*b + 2*a*B)*x^(7 + m))/(7 + m) + (b^2*B*x^(10 + m))/(10 + m)
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n _))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(a + b*x^n)^p*(c + d*x^ n)^q, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && IGt Q[p, 0] && IGtQ[q, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(260\) vs. \(2(71)=142\).
Time = 0.69 (sec) , antiderivative size = 261, normalized size of antiderivative = 3.68
method | result | size |
risch | \(\frac {x \left (B \,b^{2} m^{3} x^{9}+12 B \,b^{2} m^{2} x^{9}+39 m \,x^{9} B \,b^{2}+A \,b^{2} m^{3} x^{6}+2 B a b \,m^{3} x^{6}+28 b^{2} B \,x^{9}+15 A \,b^{2} m^{2} x^{6}+30 B a b \,m^{2} x^{6}+54 A \,b^{2} x^{6} m +108 B a b \,x^{6} m +2 A a b \,m^{3} x^{3}+40 A \,b^{2} x^{6}+B \,a^{2} m^{3} x^{3}+80 B a b \,x^{6}+36 A a b \,m^{2} x^{3}+18 B \,a^{2} m^{2} x^{3}+174 a A b \,x^{3} m +87 B \,a^{2} x^{3} m +A \,a^{2} m^{3}+140 a A b \,x^{3}+70 B \,a^{2} x^{3}+21 A \,a^{2} m^{2}+138 a^{2} A m +280 a^{2} A \right ) x^{m}}{\left (10+m \right ) \left (7+m \right ) \left (4+m \right ) \left (1+m \right )}\) | \(261\) |
orering | \(\frac {x \left (B \,b^{2} m^{3} x^{9}+12 B \,b^{2} m^{2} x^{9}+39 m \,x^{9} B \,b^{2}+A \,b^{2} m^{3} x^{6}+2 B a b \,m^{3} x^{6}+28 b^{2} B \,x^{9}+15 A \,b^{2} m^{2} x^{6}+30 B a b \,m^{2} x^{6}+54 A \,b^{2} x^{6} m +108 B a b \,x^{6} m +2 A a b \,m^{3} x^{3}+40 A \,b^{2} x^{6}+B \,a^{2} m^{3} x^{3}+80 B a b \,x^{6}+36 A a b \,m^{2} x^{3}+18 B \,a^{2} m^{2} x^{3}+174 a A b \,x^{3} m +87 B \,a^{2} x^{3} m +A \,a^{2} m^{3}+140 a A b \,x^{3}+70 B \,a^{2} x^{3}+21 A \,a^{2} m^{2}+138 a^{2} A m +280 a^{2} A \right ) x^{m}}{\left (10+m \right ) \left (7+m \right ) \left (4+m \right ) \left (1+m \right )}\) | \(261\) |
gosper | \(\frac {x^{1+m} \left (B \,b^{2} m^{3} x^{9}+12 B \,b^{2} m^{2} x^{9}+39 m \,x^{9} B \,b^{2}+A \,b^{2} m^{3} x^{6}+2 B a b \,m^{3} x^{6}+28 b^{2} B \,x^{9}+15 A \,b^{2} m^{2} x^{6}+30 B a b \,m^{2} x^{6}+54 A \,b^{2} x^{6} m +108 B a b \,x^{6} m +2 A a b \,m^{3} x^{3}+40 A \,b^{2} x^{6}+B \,a^{2} m^{3} x^{3}+80 B a b \,x^{6}+36 A a b \,m^{2} x^{3}+18 B \,a^{2} m^{2} x^{3}+174 a A b \,x^{3} m +87 B \,a^{2} x^{3} m +A \,a^{2} m^{3}+140 a A b \,x^{3}+70 B \,a^{2} x^{3}+21 A \,a^{2} m^{2}+138 a^{2} A m +280 a^{2} A \right )}{\left (1+m \right ) \left (4+m \right ) \left (7+m \right ) \left (10+m \right )}\) | \(262\) |
parallelrisch | \(\frac {80 B \,x^{7} x^{m} a b +B \,x^{4} x^{m} a^{2} m^{3}+28 B \,x^{10} x^{m} b^{2}+40 A \,x^{7} x^{m} b^{2}+70 B \,x^{4} x^{m} a^{2}+280 A x \,x^{m} a^{2}+39 B \,x^{10} x^{m} b^{2} m +A \,x^{7} x^{m} b^{2} m^{3}+15 A \,x^{7} x^{m} b^{2} m^{2}+54 A \,x^{7} x^{m} b^{2} m +2 B \,x^{7} x^{m} a b \,m^{3}+30 B \,x^{7} x^{m} a b \,m^{2}+108 B \,x^{7} x^{m} a b m +2 A \,x^{4} x^{m} a b \,m^{3}+36 A \,x^{4} x^{m} a b \,m^{2}+174 A \,x^{4} x^{m} a b m +18 B \,x^{4} x^{m} a^{2} m^{2}+B \,x^{10} x^{m} b^{2} m^{3}+12 B \,x^{10} x^{m} b^{2} m^{2}+87 B \,x^{4} x^{m} a^{2} m +140 A \,x^{4} x^{m} a b +A x \,x^{m} a^{2} m^{3}+21 A x \,x^{m} a^{2} m^{2}+138 A x \,x^{m} a^{2} m}{\left (10+m \right ) \left (7+m \right ) \left (4+m \right ) \left (1+m \right )}\) | \(333\) |
Input:
int(x^m*(b*x^3+a)^2*(B*x^3+A),x,method=_RETURNVERBOSE)
Output:
x*(B*b^2*m^3*x^9+12*B*b^2*m^2*x^9+39*B*b^2*m*x^9+A*b^2*m^3*x^6+2*B*a*b*m^3 *x^6+28*B*b^2*x^9+15*A*b^2*m^2*x^6+30*B*a*b*m^2*x^6+54*A*b^2*m*x^6+108*B*a *b*m*x^6+2*A*a*b*m^3*x^3+40*A*b^2*x^6+B*a^2*m^3*x^3+80*B*a*b*x^6+36*A*a*b* m^2*x^3+18*B*a^2*m^2*x^3+174*A*a*b*m*x^3+87*B*a^2*m*x^3+A*a^2*m^3+140*A*a* b*x^3+70*B*a^2*x^3+21*A*a^2*m^2+138*A*a^2*m+280*A*a^2)*x^m/(10+m)/(7+m)/(4 +m)/(1+m)
Leaf count of result is larger than twice the leaf count of optimal. 215 vs. \(2 (71) = 142\).
Time = 0.13 (sec) , antiderivative size = 215, normalized size of antiderivative = 3.03 \[ \int x^m \left (a+b x^3\right )^2 \left (A+B x^3\right ) \, dx=\frac {{\left ({\left (B b^{2} m^{3} + 12 \, B b^{2} m^{2} + 39 \, B b^{2} m + 28 \, B b^{2}\right )} x^{10} + {\left ({\left (2 \, B a b + A b^{2}\right )} m^{3} + 80 \, B a b + 40 \, A b^{2} + 15 \, {\left (2 \, B a b + A b^{2}\right )} m^{2} + 54 \, {\left (2 \, B a b + A b^{2}\right )} m\right )} x^{7} + {\left ({\left (B a^{2} + 2 \, A a b\right )} m^{3} + 70 \, B a^{2} + 140 \, A a b + 18 \, {\left (B a^{2} + 2 \, A a b\right )} m^{2} + 87 \, {\left (B a^{2} + 2 \, A a b\right )} m\right )} x^{4} + {\left (A a^{2} m^{3} + 21 \, A a^{2} m^{2} + 138 \, A a^{2} m + 280 \, A a^{2}\right )} x\right )} x^{m}}{m^{4} + 22 \, m^{3} + 159 \, m^{2} + 418 \, m + 280} \] Input:
integrate(x^m*(b*x^3+a)^2*(B*x^3+A),x, algorithm="fricas")
Output:
((B*b^2*m^3 + 12*B*b^2*m^2 + 39*B*b^2*m + 28*B*b^2)*x^10 + ((2*B*a*b + A*b ^2)*m^3 + 80*B*a*b + 40*A*b^2 + 15*(2*B*a*b + A*b^2)*m^2 + 54*(2*B*a*b + A *b^2)*m)*x^7 + ((B*a^2 + 2*A*a*b)*m^3 + 70*B*a^2 + 140*A*a*b + 18*(B*a^2 + 2*A*a*b)*m^2 + 87*(B*a^2 + 2*A*a*b)*m)*x^4 + (A*a^2*m^3 + 21*A*a^2*m^2 + 138*A*a^2*m + 280*A*a^2)*x)*x^m/(m^4 + 22*m^3 + 159*m^2 + 418*m + 280)
Leaf count of result is larger than twice the leaf count of optimal. 1057 vs. \(2 (63) = 126\).
Time = 0.62 (sec) , antiderivative size = 1057, normalized size of antiderivative = 14.89 \[ \int x^m \left (a+b x^3\right )^2 \left (A+B x^3\right ) \, dx =\text {Too large to display} \] Input:
integrate(x**m*(b*x**3+a)**2*(B*x**3+A),x)
Output:
Piecewise((-A*a**2/(9*x**9) - A*a*b/(3*x**6) - A*b**2/(3*x**3) - B*a**2/(6 *x**6) - 2*B*a*b/(3*x**3) + B*b**2*log(x), Eq(m, -10)), (-A*a**2/(6*x**6) - 2*A*a*b/(3*x**3) + A*b**2*log(x) - B*a**2/(3*x**3) + 2*B*a*b*log(x) + B* b**2*x**3/3, Eq(m, -7)), (-A*a**2/(3*x**3) + 2*A*a*b*log(x) + A*b**2*x**3/ 3 + B*a**2*log(x) + 2*B*a*b*x**3/3 + B*b**2*x**6/6, Eq(m, -4)), (A*a**2*lo g(x) + 2*A*a*b*x**3/3 + A*b**2*x**6/6 + B*a**2*x**3/3 + B*a*b*x**6/3 + B*b **2*x**9/9, Eq(m, -1)), (A*a**2*m**3*x*x**m/(m**4 + 22*m**3 + 159*m**2 + 4 18*m + 280) + 21*A*a**2*m**2*x*x**m/(m**4 + 22*m**3 + 159*m**2 + 418*m + 2 80) + 138*A*a**2*m*x*x**m/(m**4 + 22*m**3 + 159*m**2 + 418*m + 280) + 280* A*a**2*x*x**m/(m**4 + 22*m**3 + 159*m**2 + 418*m + 280) + 2*A*a*b*m**3*x** 4*x**m/(m**4 + 22*m**3 + 159*m**2 + 418*m + 280) + 36*A*a*b*m**2*x**4*x**m /(m**4 + 22*m**3 + 159*m**2 + 418*m + 280) + 174*A*a*b*m*x**4*x**m/(m**4 + 22*m**3 + 159*m**2 + 418*m + 280) + 140*A*a*b*x**4*x**m/(m**4 + 22*m**3 + 159*m**2 + 418*m + 280) + A*b**2*m**3*x**7*x**m/(m**4 + 22*m**3 + 159*m** 2 + 418*m + 280) + 15*A*b**2*m**2*x**7*x**m/(m**4 + 22*m**3 + 159*m**2 + 4 18*m + 280) + 54*A*b**2*m*x**7*x**m/(m**4 + 22*m**3 + 159*m**2 + 418*m + 2 80) + 40*A*b**2*x**7*x**m/(m**4 + 22*m**3 + 159*m**2 + 418*m + 280) + B*a* *2*m**3*x**4*x**m/(m**4 + 22*m**3 + 159*m**2 + 418*m + 280) + 18*B*a**2*m* *2*x**4*x**m/(m**4 + 22*m**3 + 159*m**2 + 418*m + 280) + 87*B*a**2*m*x**4* x**m/(m**4 + 22*m**3 + 159*m**2 + 418*m + 280) + 70*B*a**2*x**4*x**m/(m...
Time = 0.03 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.28 \[ \int x^m \left (a+b x^3\right )^2 \left (A+B x^3\right ) \, dx=\frac {B b^{2} x^{m + 10}}{m + 10} + \frac {2 \, B a b x^{m + 7}}{m + 7} + \frac {A b^{2} x^{m + 7}}{m + 7} + \frac {B a^{2} x^{m + 4}}{m + 4} + \frac {2 \, A a b x^{m + 4}}{m + 4} + \frac {A a^{2} x^{m + 1}}{m + 1} \] Input:
integrate(x^m*(b*x^3+a)^2*(B*x^3+A),x, algorithm="maxima")
Output:
B*b^2*x^(m + 10)/(m + 10) + 2*B*a*b*x^(m + 7)/(m + 7) + A*b^2*x^(m + 7)/(m + 7) + B*a^2*x^(m + 4)/(m + 4) + 2*A*a*b*x^(m + 4)/(m + 4) + A*a^2*x^(m + 1)/(m + 1)
Leaf count of result is larger than twice the leaf count of optimal. 332 vs. \(2 (71) = 142\).
Time = 0.13 (sec) , antiderivative size = 332, normalized size of antiderivative = 4.68 \[ \int x^m \left (a+b x^3\right )^2 \left (A+B x^3\right ) \, dx=\frac {B b^{2} m^{3} x^{10} x^{m} + 12 \, B b^{2} m^{2} x^{10} x^{m} + 39 \, B b^{2} m x^{10} x^{m} + 2 \, B a b m^{3} x^{7} x^{m} + A b^{2} m^{3} x^{7} x^{m} + 28 \, B b^{2} x^{10} x^{m} + 30 \, B a b m^{2} x^{7} x^{m} + 15 \, A b^{2} m^{2} x^{7} x^{m} + 108 \, B a b m x^{7} x^{m} + 54 \, A b^{2} m x^{7} x^{m} + B a^{2} m^{3} x^{4} x^{m} + 2 \, A a b m^{3} x^{4} x^{m} + 80 \, B a b x^{7} x^{m} + 40 \, A b^{2} x^{7} x^{m} + 18 \, B a^{2} m^{2} x^{4} x^{m} + 36 \, A a b m^{2} x^{4} x^{m} + 87 \, B a^{2} m x^{4} x^{m} + 174 \, A a b m x^{4} x^{m} + A a^{2} m^{3} x x^{m} + 70 \, B a^{2} x^{4} x^{m} + 140 \, A a b x^{4} x^{m} + 21 \, A a^{2} m^{2} x x^{m} + 138 \, A a^{2} m x x^{m} + 280 \, A a^{2} x x^{m}}{m^{4} + 22 \, m^{3} + 159 \, m^{2} + 418 \, m + 280} \] Input:
integrate(x^m*(b*x^3+a)^2*(B*x^3+A),x, algorithm="giac")
Output:
(B*b^2*m^3*x^10*x^m + 12*B*b^2*m^2*x^10*x^m + 39*B*b^2*m*x^10*x^m + 2*B*a* b*m^3*x^7*x^m + A*b^2*m^3*x^7*x^m + 28*B*b^2*x^10*x^m + 30*B*a*b*m^2*x^7*x ^m + 15*A*b^2*m^2*x^7*x^m + 108*B*a*b*m*x^7*x^m + 54*A*b^2*m*x^7*x^m + B*a ^2*m^3*x^4*x^m + 2*A*a*b*m^3*x^4*x^m + 80*B*a*b*x^7*x^m + 40*A*b^2*x^7*x^m + 18*B*a^2*m^2*x^4*x^m + 36*A*a*b*m^2*x^4*x^m + 87*B*a^2*m*x^4*x^m + 174* A*a*b*m*x^4*x^m + A*a^2*m^3*x*x^m + 70*B*a^2*x^4*x^m + 140*A*a*b*x^4*x^m + 21*A*a^2*m^2*x*x^m + 138*A*a^2*m*x*x^m + 280*A*a^2*x*x^m)/(m^4 + 22*m^3 + 159*m^2 + 418*m + 280)
Time = 0.97 (sec) , antiderivative size = 177, normalized size of antiderivative = 2.49 \[ \int x^m \left (a+b x^3\right )^2 \left (A+B x^3\right ) \, dx=x^m\,\left (\frac {B\,b^2\,x^{10}\,\left (m^3+12\,m^2+39\,m+28\right )}{m^4+22\,m^3+159\,m^2+418\,m+280}+\frac {A\,a^2\,x\,\left (m^3+21\,m^2+138\,m+280\right )}{m^4+22\,m^3+159\,m^2+418\,m+280}+\frac {a\,x^4\,\left (2\,A\,b+B\,a\right )\,\left (m^3+18\,m^2+87\,m+70\right )}{m^4+22\,m^3+159\,m^2+418\,m+280}+\frac {b\,x^7\,\left (A\,b+2\,B\,a\right )\,\left (m^3+15\,m^2+54\,m+40\right )}{m^4+22\,m^3+159\,m^2+418\,m+280}\right ) \] Input:
int(x^m*(A + B*x^3)*(a + b*x^3)^2,x)
Output:
x^m*((B*b^2*x^10*(39*m + 12*m^2 + m^3 + 28))/(418*m + 159*m^2 + 22*m^3 + m ^4 + 280) + (A*a^2*x*(138*m + 21*m^2 + m^3 + 280))/(418*m + 159*m^2 + 22*m ^3 + m^4 + 280) + (a*x^4*(2*A*b + B*a)*(87*m + 18*m^2 + m^3 + 70))/(418*m + 159*m^2 + 22*m^3 + m^4 + 280) + (b*x^7*(A*b + 2*B*a)*(54*m + 15*m^2 + m^ 3 + 40))/(418*m + 159*m^2 + 22*m^3 + m^4 + 280))
Time = 0.23 (sec) , antiderivative size = 176, normalized size of antiderivative = 2.48 \[ \int x^m \left (a+b x^3\right )^2 \left (A+B x^3\right ) \, dx=\frac {x^{m} x \left (b^{3} m^{3} x^{9}+12 b^{3} m^{2} x^{9}+39 b^{3} m \,x^{9}+3 a \,b^{2} m^{3} x^{6}+28 b^{3} x^{9}+45 a \,b^{2} m^{2} x^{6}+162 a \,b^{2} m \,x^{6}+3 a^{2} b \,m^{3} x^{3}+120 a \,b^{2} x^{6}+54 a^{2} b \,m^{2} x^{3}+261 a^{2} b m \,x^{3}+a^{3} m^{3}+210 a^{2} b \,x^{3}+21 a^{3} m^{2}+138 a^{3} m +280 a^{3}\right )}{m^{4}+22 m^{3}+159 m^{2}+418 m +280} \] Input:
int(x^m*(b*x^3+a)^2*(B*x^3+A),x)
Output:
(x**m*x*(a**3*m**3 + 21*a**3*m**2 + 138*a**3*m + 280*a**3 + 3*a**2*b*m**3* x**3 + 54*a**2*b*m**2*x**3 + 261*a**2*b*m*x**3 + 210*a**2*b*x**3 + 3*a*b** 2*m**3*x**6 + 45*a*b**2*m**2*x**6 + 162*a*b**2*m*x**6 + 120*a*b**2*x**6 + b**3*m**3*x**9 + 12*b**3*m**2*x**9 + 39*b**3*m*x**9 + 28*b**3*x**9))/(m**4 + 22*m**3 + 159*m**2 + 418*m + 280)