3.42 \(\int \frac {(a+b x^3)^5 (A+B x^3)}{x^{10}} \, dx\)

Optimal. Leaf size=114 \[ -\frac {a^5 A}{9 x^9}-\frac {a^4 (a B+5 A b)}{6 x^6}-\frac {5 a^3 b (a B+2 A b)}{3 x^3}+10 a^2 b^2 \log (x) (a B+A b)+\frac {1}{6} b^4 x^6 (5 a B+A b)+\frac {5}{3} a b^3 x^3 (2 a B+A b)+\frac {1}{9} b^5 B x^9 \]

[Out]

-1/9*a^5*A/x^9-1/6*a^4*(5*A*b+B*a)/x^6-5/3*a^3*b*(2*A*b+B*a)/x^3+5/3*a*b^3*(A*b+2*B*a)*x^3+1/6*b^4*(A*b+5*B*a)
*x^6+1/9*b^5*B*x^9+10*a^2*b^2*(A*b+B*a)*ln(x)

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Rubi [A]  time = 0.10, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {446, 76} \[ 10 a^2 b^2 \log (x) (a B+A b)-\frac {5 a^3 b (a B+2 A b)}{3 x^3}-\frac {a^4 (a B+5 A b)}{6 x^6}-\frac {a^5 A}{9 x^9}+\frac {1}{6} b^4 x^6 (5 a B+A b)+\frac {5}{3} a b^3 x^3 (2 a B+A b)+\frac {1}{9} b^5 B x^9 \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x^3)^5*(A + B*x^3))/x^10,x]

[Out]

-(a^5*A)/(9*x^9) - (a^4*(5*A*b + a*B))/(6*x^6) - (5*a^3*b*(2*A*b + a*B))/(3*x^3) + (5*a*b^3*(A*b + 2*a*B)*x^3)
/3 + (b^4*(A*b + 5*a*B)*x^6)/6 + (b^5*B*x^9)/9 + 10*a^2*b^2*(A*b + a*B)*Log[x]

Rule 76

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])

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(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] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

\begin {align*} \int \frac {\left (a+b x^3\right )^5 \left (A+B x^3\right )}{x^{10}} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {(a+b x)^5 (A+B x)}{x^4} \, dx,x,x^3\right )\\ &=\frac {1}{3} \operatorname {Subst}\left (\int \left (5 a b^3 (A b+2 a B)+\frac {a^5 A}{x^4}+\frac {a^4 (5 A b+a B)}{x^3}+\frac {5 a^3 b (2 A b+a B)}{x^2}+\frac {10 a^2 b^2 (A b+a B)}{x}+b^4 (A b+5 a B) x+b^5 B x^2\right ) \, dx,x,x^3\right )\\ &=-\frac {a^5 A}{9 x^9}-\frac {a^4 (5 A b+a B)}{6 x^6}-\frac {5 a^3 b (2 A b+a B)}{3 x^3}+\frac {5}{3} a b^3 (A b+2 a B) x^3+\frac {1}{6} b^4 (A b+5 a B) x^6+\frac {1}{9} b^5 B x^9+10 a^2 b^2 (A b+a B) \log (x)\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 106, normalized size = 0.93 \[ \frac {1}{18} \left (-\frac {2 a^5 A}{x^9}-\frac {3 a^4 (a B+5 A b)}{x^6}-\frac {30 a^3 b (a B+2 A b)}{x^3}+180 a^2 b^2 \log (x) (a B+A b)+3 b^4 x^6 (5 a B+A b)+30 a b^3 x^3 (2 a B+A b)+2 b^5 B x^9\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^3)^5*(A + B*x^3))/x^10,x]

[Out]

((-2*a^5*A)/x^9 - (3*a^4*(5*A*b + a*B))/x^6 - (30*a^3*b*(2*A*b + a*B))/x^3 + 30*a*b^3*(A*b + 2*a*B)*x^3 + 3*b^
4*(A*b + 5*a*B)*x^6 + 2*b^5*B*x^9 + 180*a^2*b^2*(A*b + a*B)*Log[x])/18

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fricas [A]  time = 0.89, size = 123, normalized size = 1.08 \[ \frac {2 \, B b^{5} x^{18} + 3 \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{15} + 30 \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{12} + 180 \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{9} \log \relax (x) - 30 \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{6} - 2 \, A a^{5} - 3 \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x^{3}}{18 \, x^{9}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^5*(B*x^3+A)/x^10,x, algorithm="fricas")

[Out]

1/18*(2*B*b^5*x^18 + 3*(5*B*a*b^4 + A*b^5)*x^15 + 30*(2*B*a^2*b^3 + A*a*b^4)*x^12 + 180*(B*a^3*b^2 + A*a^2*b^3
)*x^9*log(x) - 30*(B*a^4*b + 2*A*a^3*b^2)*x^6 - 2*A*a^5 - 3*(B*a^5 + 5*A*a^4*b)*x^3)/x^9

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giac [A]  time = 0.15, size = 150, normalized size = 1.32 \[ \frac {1}{9} \, B b^{5} x^{9} + \frac {5}{6} \, B a b^{4} x^{6} + \frac {1}{6} \, A b^{5} x^{6} + \frac {10}{3} \, B a^{2} b^{3} x^{3} + \frac {5}{3} \, A a b^{4} x^{3} + 10 \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} \log \left ({\left | x \right |}\right ) - \frac {110 \, B a^{3} b^{2} x^{9} + 110 \, A a^{2} b^{3} x^{9} + 30 \, B a^{4} b x^{6} + 60 \, A a^{3} b^{2} x^{6} + 3 \, B a^{5} x^{3} + 15 \, A a^{4} b x^{3} + 2 \, A a^{5}}{18 \, x^{9}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^5*(B*x^3+A)/x^10,x, algorithm="giac")

[Out]

1/9*B*b^5*x^9 + 5/6*B*a*b^4*x^6 + 1/6*A*b^5*x^6 + 10/3*B*a^2*b^3*x^3 + 5/3*A*a*b^4*x^3 + 10*(B*a^3*b^2 + A*a^2
*b^3)*log(abs(x)) - 1/18*(110*B*a^3*b^2*x^9 + 110*A*a^2*b^3*x^9 + 30*B*a^4*b*x^6 + 60*A*a^3*b^2*x^6 + 3*B*a^5*
x^3 + 15*A*a^4*b*x^3 + 2*A*a^5)/x^9

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maple [A]  time = 0.05, size = 124, normalized size = 1.09 \[ \frac {B \,b^{5} x^{9}}{9}+\frac {A \,b^{5} x^{6}}{6}+\frac {5 B a \,b^{4} x^{6}}{6}+\frac {5 A a \,b^{4} x^{3}}{3}+\frac {10 B \,a^{2} b^{3} x^{3}}{3}+10 A \,a^{2} b^{3} \ln \relax (x )+10 B \,a^{3} b^{2} \ln \relax (x )-\frac {10 A \,a^{3} b^{2}}{3 x^{3}}-\frac {5 B \,a^{4} b}{3 x^{3}}-\frac {5 A \,a^{4} b}{6 x^{6}}-\frac {B \,a^{5}}{6 x^{6}}-\frac {A \,a^{5}}{9 x^{9}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)^5*(B*x^3+A)/x^10,x)

[Out]

1/9*B*b^5*x^9+1/6*A*x^6*b^5+5/6*B*x^6*a*b^4+5/3*A*x^3*a*b^4+10/3*B*x^3*a^2*b^3-10/3*a^3*b^2/x^3*A-5/3*a^4*b/x^
3*B-1/9*a^5*A/x^9-5/6*a^4/x^6*A*b-1/6*a^5/x^6*B+10*A*ln(x)*a^2*b^3+10*B*ln(x)*a^3*b^2

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maxima [A]  time = 0.47, size = 123, normalized size = 1.08 \[ \frac {1}{9} \, B b^{5} x^{9} + \frac {1}{6} \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{6} + \frac {5}{3} \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{3} + \frac {10}{3} \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} \log \left (x^{3}\right ) - \frac {30 \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{6} + 2 \, A a^{5} + 3 \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x^{3}}{18 \, x^{9}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^5*(B*x^3+A)/x^10,x, algorithm="maxima")

[Out]

1/9*B*b^5*x^9 + 1/6*(5*B*a*b^4 + A*b^5)*x^6 + 5/3*(2*B*a^2*b^3 + A*a*b^4)*x^3 + 10/3*(B*a^3*b^2 + A*a^2*b^3)*l
og(x^3) - 1/18*(30*(B*a^4*b + 2*A*a^3*b^2)*x^6 + 2*A*a^5 + 3*(B*a^5 + 5*A*a^4*b)*x^3)/x^9

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mupad [B]  time = 0.05, size = 118, normalized size = 1.04 \[ x^6\,\left (\frac {A\,b^5}{6}+\frac {5\,B\,a\,b^4}{6}\right )-\frac {\frac {A\,a^5}{9}+x^6\,\left (\frac {5\,B\,a^4\,b}{3}+\frac {10\,A\,a^3\,b^2}{3}\right )+x^3\,\left (\frac {B\,a^5}{6}+\frac {5\,A\,b\,a^4}{6}\right )}{x^9}+\ln \relax (x)\,\left (10\,B\,a^3\,b^2+10\,A\,a^2\,b^3\right )+\frac {B\,b^5\,x^9}{9}+\frac {5\,a\,b^3\,x^3\,\left (A\,b+2\,B\,a\right )}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x^3)*(a + b*x^3)^5)/x^10,x)

[Out]

x^6*((A*b^5)/6 + (5*B*a*b^4)/6) - ((A*a^5)/9 + x^6*((10*A*a^3*b^2)/3 + (5*B*a^4*b)/3) + x^3*((B*a^5)/6 + (5*A*
a^4*b)/6))/x^9 + log(x)*(10*A*a^2*b^3 + 10*B*a^3*b^2) + (B*b^5*x^9)/9 + (5*a*b^3*x^3*(A*b + 2*B*a))/3

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sympy [A]  time = 2.24, size = 129, normalized size = 1.13 \[ \frac {B b^{5} x^{9}}{9} + 10 a^{2} b^{2} \left (A b + B a\right ) \log {\relax (x )} + x^{6} \left (\frac {A b^{5}}{6} + \frac {5 B a b^{4}}{6}\right ) + x^{3} \left (\frac {5 A a b^{4}}{3} + \frac {10 B a^{2} b^{3}}{3}\right ) + \frac {- 2 A a^{5} + x^{6} \left (- 60 A a^{3} b^{2} - 30 B a^{4} b\right ) + x^{3} \left (- 15 A a^{4} b - 3 B a^{5}\right )}{18 x^{9}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a)**5*(B*x**3+A)/x**10,x)

[Out]

B*b**5*x**9/9 + 10*a**2*b**2*(A*b + B*a)*log(x) + x**6*(A*b**5/6 + 5*B*a*b**4/6) + x**3*(5*A*a*b**4/3 + 10*B*a
**2*b**3/3) + (-2*A*a**5 + x**6*(-60*A*a**3*b**2 - 30*B*a**4*b) + x**3*(-15*A*a**4*b - 3*B*a**5))/(18*x**9)

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