\(\int x (a+b x^2)^5 (A+B x^2) \, dx\) [31]

Optimal result

Integrand size = 18, antiderivative size = 42 \[ \int x \left (a+b x^2\right )^5 \left (A+B x^2\right ) \, dx=\frac {(A b-a B) \left (a+b x^2\right )^6}{12 b^2}+\frac {B \left (a+b x^2\right )^7}{14 b^2} \]

[Out]

1/12*(A*b-B*a)*(b*x^2+a)^6/b^2+1/14*B*(b*x^2+a)^7/b^2

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {455, 45} \[ \int x \left (a+b x^2\right )^5 \left (A+B x^2\right ) \, dx=\frac {\left (a+b x^2\right )^6 (A b-a B)}{12 b^2}+\frac {B \left (a+b x^2\right )^7}{14 b^2} \]

[In]

Int[x*(a + b*x^2)^5*(A + B*x^2),x]

[Out]

((A*b - a*B)*(a + b*x^2)^6)/(12*b^2) + (B*(a + b*x^2)^7)/(14*b^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 455

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int (a+b x)^5 (A+B x) \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (\frac {(A b-a B) (a+b x)^5}{b}+\frac {B (a+b x)^6}{b}\right ) \, dx,x,x^2\right ) \\ & = \frac {(A b-a B) \left (a+b x^2\right )^6}{12 b^2}+\frac {B \left (a+b x^2\right )^7}{14 b^2} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(107\) vs. \(2(42)=84\).

Time = 0.02 (sec) , antiderivative size = 107, normalized size of antiderivative = 2.55 \[ \int x \left (a+b x^2\right )^5 \left (A+B x^2\right ) \, dx=\frac {1}{84} x^2 \left (42 a^5 A+21 a^4 (5 A b+a B) x^2+70 a^3 b (2 A b+a B) x^4+105 a^2 b^2 (A b+a B) x^6+42 a b^3 (A b+2 a B) x^8+7 b^4 (A b+5 a B) x^{10}+6 b^5 B x^{12}\right ) \]

[In]

Integrate[x*(a + b*x^2)^5*(A + B*x^2),x]

[Out]

(x^2*(42*a^5*A + 21*a^4*(5*A*b + a*B)*x^2 + 70*a^3*b*(2*A*b + a*B)*x^4 + 105*a^2*b^2*(A*b + a*B)*x^6 + 42*a*b^
3*(A*b + 2*a*B)*x^8 + 7*b^4*(A*b + 5*a*B)*x^10 + 6*b^5*B*x^12))/84

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(119\) vs. \(2(38)=76\).

Time = 2.49 (sec) , antiderivative size = 120, normalized size of antiderivative = 2.86

method	result	size
norman	\(\frac {a^{5} A \,x^{2}}{2}+\left (\frac {5}{4} a^{4} b A +\frac {1}{4} a^{5} B \right ) x^{4}+\left (\frac {5}{3} a^{3} b^{2} A +\frac {5}{6} a^{4} b B \right ) x^{6}+\left (\frac {5}{4} a^{2} b^{3} A +\frac {5}{4} a^{3} b^{2} B \right ) x^{8}+\left (\frac {1}{2} a \,b^{4} A +a^{2} b^{3} B \right ) x^{10}+\left (\frac {1}{12} b^{5} A +\frac {5}{12} a \,b^{4} B \right ) x^{12}+\frac {b^{5} B \,x^{14}}{14}\)	\(120\)
default	\(\frac {b^{5} B \,x^{14}}{14}+\frac {\left (b^{5} A +5 a \,b^{4} B \right ) x^{12}}{12}+\frac {\left (5 a \,b^{4} A +10 a^{2} b^{3} B \right ) x^{10}}{10}+\frac {\left (10 a^{2} b^{3} A +10 a^{3} b^{2} B \right ) x^{8}}{8}+\frac {\left (10 a^{3} b^{2} A +5 a^{4} b B \right ) x^{6}}{6}+\frac {\left (5 a^{4} b A +a^{5} B \right ) x^{4}}{4}+\frac {a^{5} A \,x^{2}}{2}\)	\(124\)
gosper	\(\frac {1}{2} a^{5} A \,x^{2}+\frac {5}{4} x^{4} a^{4} b A +\frac {1}{4} x^{4} a^{5} B +\frac {5}{3} x^{6} a^{3} b^{2} A +\frac {5}{6} x^{6} a^{4} b B +\frac {5}{4} x^{8} a^{2} b^{3} A +\frac {5}{4} x^{8} a^{3} b^{2} B +\frac {1}{2} x^{10} a \,b^{4} A +x^{10} a^{2} b^{3} B +\frac {1}{12} x^{12} b^{5} A +\frac {5}{12} x^{12} a \,b^{4} B +\frac {1}{14} b^{5} B \,x^{14}\)	\(125\)
risch	\(\frac {1}{2} a^{5} A \,x^{2}+\frac {5}{4} x^{4} a^{4} b A +\frac {1}{4} x^{4} a^{5} B +\frac {5}{3} x^{6} a^{3} b^{2} A +\frac {5}{6} x^{6} a^{4} b B +\frac {5}{4} x^{8} a^{2} b^{3} A +\frac {5}{4} x^{8} a^{3} b^{2} B +\frac {1}{2} x^{10} a \,b^{4} A +x^{10} a^{2} b^{3} B +\frac {1}{12} x^{12} b^{5} A +\frac {5}{12} x^{12} a \,b^{4} B +\frac {1}{14} b^{5} B \,x^{14}\)	\(125\)
parallelrisch	\(\frac {1}{2} a^{5} A \,x^{2}+\frac {5}{4} x^{4} a^{4} b A +\frac {1}{4} x^{4} a^{5} B +\frac {5}{3} x^{6} a^{3} b^{2} A +\frac {5}{6} x^{6} a^{4} b B +\frac {5}{4} x^{8} a^{2} b^{3} A +\frac {5}{4} x^{8} a^{3} b^{2} B +\frac {1}{2} x^{10} a \,b^{4} A +x^{10} a^{2} b^{3} B +\frac {1}{12} x^{12} b^{5} A +\frac {5}{12} x^{12} a \,b^{4} B +\frac {1}{14} b^{5} B \,x^{14}\)	\(125\)

[In]

int(x*(b*x^2+a)^5*(B*x^2+A),x,method=_RETURNVERBOSE)

[Out]

1/2*a^5*A*x^2+(5/4*a^4*b*A+1/4*a^5*B)*x^4+(5/3*a^3*b^2*A+5/6*a^4*b*B)*x^6+(5/4*a^2*b^3*A+5/4*a^3*b^2*B)*x^8+(1
/2*a*b^4*A+a^2*b^3*B)*x^10+(1/12*b^5*A+5/12*a*b^4*B)*x^12+1/14*b^5*B*x^14

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 119 vs. \(2 (38) = 76\).

Time = 0.28 (sec) , antiderivative size = 119, normalized size of antiderivative = 2.83 \[ \int x \left (a+b x^2\right )^5 \left (A+B x^2\right ) \, dx=\frac {1}{14} \, B b^{5} x^{14} + \frac {1}{12} \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{12} + \frac {1}{2} \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{10} + \frac {5}{4} \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{8} + \frac {1}{2} \, A a^{5} x^{2} + \frac {5}{6} \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{6} + \frac {1}{4} \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x^{4} \]

[In]

integrate(x*(b*x^2+a)^5*(B*x^2+A),x, algorithm="fricas")

[Out]

1/14*B*b^5*x^14 + 1/12*(5*B*a*b^4 + A*b^5)*x^12 + 1/2*(2*B*a^2*b^3 + A*a*b^4)*x^10 + 5/4*(B*a^3*b^2 + A*a^2*b^
3)*x^8 + 1/2*A*a^5*x^2 + 5/6*(B*a^4*b + 2*A*a^3*b^2)*x^6 + 1/4*(B*a^5 + 5*A*a^4*b)*x^4

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 133 vs. \(2 (36) = 72\).

Time = 0.03 (sec) , antiderivative size = 133, normalized size of antiderivative = 3.17 \[ \int x \left (a+b x^2\right )^5 \left (A+B x^2\right ) \, dx=\frac {A a^{5} x^{2}}{2} + \frac {B b^{5} x^{14}}{14} + x^{12} \left (\frac {A b^{5}}{12} + \frac {5 B a b^{4}}{12}\right ) + x^{10} \left (\frac {A a b^{4}}{2} + B a^{2} b^{3}\right ) + x^{8} \cdot \left (\frac {5 A a^{2} b^{3}}{4} + \frac {5 B a^{3} b^{2}}{4}\right ) + x^{6} \cdot \left (\frac {5 A a^{3} b^{2}}{3} + \frac {5 B a^{4} b}{6}\right ) + x^{4} \cdot \left (\frac {5 A a^{4} b}{4} + \frac {B a^{5}}{4}\right ) \]

[In]

integrate(x*(b*x**2+a)**5*(B*x**2+A),x)

[Out]

A*a**5*x**2/2 + B*b**5*x**14/14 + x**12*(A*b**5/12 + 5*B*a*b**4/12) + x**10*(A*a*b**4/2 + B*a**2*b**3) + x**8*
(5*A*a**2*b**3/4 + 5*B*a**3*b**2/4) + x**6*(5*A*a**3*b**2/3 + 5*B*a**4*b/6) + x**4*(5*A*a**4*b/4 + B*a**5/4)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 119 vs. \(2 (38) = 76\).

Time = 0.18 (sec) , antiderivative size = 119, normalized size of antiderivative = 2.83 \[ \int x \left (a+b x^2\right )^5 \left (A+B x^2\right ) \, dx=\frac {1}{14} \, B b^{5} x^{14} + \frac {1}{12} \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{12} + \frac {1}{2} \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{10} + \frac {5}{4} \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{8} + \frac {1}{2} \, A a^{5} x^{2} + \frac {5}{6} \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{6} + \frac {1}{4} \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x^{4} \]

[In]

integrate(x*(b*x^2+a)^5*(B*x^2+A),x, algorithm="maxima")

[Out]

1/14*B*b^5*x^14 + 1/12*(5*B*a*b^4 + A*b^5)*x^12 + 1/2*(2*B*a^2*b^3 + A*a*b^4)*x^10 + 5/4*(B*a^3*b^2 + A*a^2*b^
3)*x^8 + 1/2*A*a^5*x^2 + 5/6*(B*a^4*b + 2*A*a^3*b^2)*x^6 + 1/4*(B*a^5 + 5*A*a^4*b)*x^4

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 124 vs. \(2 (38) = 76\).

Time = 0.30 (sec) , antiderivative size = 124, normalized size of antiderivative = 2.95 \[ \int x \left (a+b x^2\right )^5 \left (A+B x^2\right ) \, dx=\frac {1}{14} \, B b^{5} x^{14} + \frac {5}{12} \, B a b^{4} x^{12} + \frac {1}{12} \, A b^{5} x^{12} + B a^{2} b^{3} x^{10} + \frac {1}{2} \, A a b^{4} x^{10} + \frac {5}{4} \, B a^{3} b^{2} x^{8} + \frac {5}{4} \, A a^{2} b^{3} x^{8} + \frac {5}{6} \, B a^{4} b x^{6} + \frac {5}{3} \, A a^{3} b^{2} x^{6} + \frac {1}{4} \, B a^{5} x^{4} + \frac {5}{4} \, A a^{4} b x^{4} + \frac {1}{2} \, A a^{5} x^{2} \]

[In]

integrate(x*(b*x^2+a)^5*(B*x^2+A),x, algorithm="giac")

[Out]

1/14*B*b^5*x^14 + 5/12*B*a*b^4*x^12 + 1/12*A*b^5*x^12 + B*a^2*b^3*x^10 + 1/2*A*a*b^4*x^10 + 5/4*B*a^3*b^2*x^8
+ 5/4*A*a^2*b^3*x^8 + 5/6*B*a^4*b*x^6 + 5/3*A*a^3*b^2*x^6 + 1/4*B*a^5*x^4 + 5/4*A*a^4*b*x^4 + 1/2*A*a^5*x^2

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 107, normalized size of antiderivative = 2.55 \[ \int x \left (a+b x^2\right )^5 \left (A+B x^2\right ) \, dx=x^4\,\left (\frac {B\,a^5}{4}+\frac {5\,A\,b\,a^4}{4}\right )+x^{12}\,\left (\frac {A\,b^5}{12}+\frac {5\,B\,a\,b^4}{12}\right )+\frac {A\,a^5\,x^2}{2}+\frac {B\,b^5\,x^{14}}{14}+\frac {5\,a^2\,b^2\,x^8\,\left (A\,b+B\,a\right )}{4}+\frac {5\,a^3\,b\,x^6\,\left (2\,A\,b+B\,a\right )}{6}+\frac {a\,b^3\,x^{10}\,\left (A\,b+2\,B\,a\right )}{2} \]

[In]

int(x*(A + B*x^2)*(a + b*x^2)^5,x)

[Out]

x^4*((B*a^5)/4 + (5*A*a^4*b)/4) + x^12*((A*b^5)/12 + (5*B*a*b^4)/12) + (A*a^5*x^2)/2 + (B*b^5*x^14)/14 + (5*a^
2*b^2*x^8*(A*b + B*a))/4 + (5*a^3*b*x^6*(2*A*b + B*a))/6 + (a*b^3*x^10*(A*b + 2*B*a))/2