3.20 \(\int \frac {(a+b x^3)^2 (A+B x^3)}{x^7} \, dx\)

Optimal. Leaf size=51 \[ -\frac {a^2 A}{6 x^6}-\frac {a (a B+2 A b)}{3 x^3}+b \log (x) (2 a B+A b)+\frac {1}{3} b^2 B x^3 \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 51, 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} \[ -\frac {a^2 A}{6 x^6}-\frac {a (a B+2 A b)}{3 x^3}+b \log (x) (2 a B+A b)+\frac {1}{3} b^2 B x^3 \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x^3)^2*(A + B*x^3))/x^7,x]

[Out]

-(a^2*A)/(6*x^6) - (a*(2*A*b + a*B))/(3*x^3) + (b^2*B*x^3)/3 + b*(A*b + 2*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 )^2 \left (A+B x^3\right )}{x^7} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {(a+b x)^2 (A+B x)}{x^3} \, dx,x,x^3\right )\\ &=\frac {1}{3} \operatorname {Subst}\left (\int \left (b^2 B+\frac {a^2 A}{x^3}+\frac {a (2 A b+a B)}{x^2}+\frac {b (A b+2 a B)}{x}\right ) \, dx,x,x^3\right )\\ &=-\frac {a^2 A}{6 x^6}-\frac {a (2 A b+a B)}{3 x^3}+\frac {1}{3} b^2 B x^3+b (A b+2 a B) \log (x)\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 51, normalized size = 1.00 \[ \frac {1}{6} \left (-\frac {a^2 \left (A+2 B x^3\right )}{x^6}+6 b \log (x) (2 a B+A b)-\frac {4 a A b}{x^3}+2 b^2 B x^3\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^3)^2*(A + B*x^3))/x^7,x]

[Out]

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

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fricas [A]  time = 1.00, size = 55, normalized size = 1.08 \[ \frac {2 \, B b^{2} x^{9} + 6 \, {\left (2 \, B a b + A b^{2}\right )} x^{6} \log \relax (x) - 2 \, {\left (B a^{2} + 2 \, A a b\right )} x^{3} - A a^{2}}{6 \, x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^2*(B*x^3+A)/x^7,x, algorithm="fricas")

[Out]

1/6*(2*B*b^2*x^9 + 6*(2*B*a*b + A*b^2)*x^6*log(x) - 2*(B*a^2 + 2*A*a*b)*x^3 - A*a^2)/x^6

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giac [A]  time = 0.21, size = 70, normalized size = 1.37 \[ \frac {1}{3} \, B b^{2} x^{3} + {\left (2 \, B a b + A b^{2}\right )} \log \left ({\left | x \right |}\right ) - \frac {6 \, B a b x^{6} + 3 \, A b^{2} x^{6} + 2 \, B a^{2} x^{3} + 4 \, A a b x^{3} + A a^{2}}{6 \, x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^2*(B*x^3+A)/x^7,x, algorithm="giac")

[Out]

1/3*B*b^2*x^3 + (2*B*a*b + A*b^2)*log(abs(x)) - 1/6*(6*B*a*b*x^6 + 3*A*b^2*x^6 + 2*B*a^2*x^3 + 4*A*a*b*x^3 + A
*a^2)/x^6

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maple [A]  time = 0.07, size = 51, normalized size = 1.00 \[ \frac {B \,b^{2} x^{3}}{3}+A \,b^{2} \ln \relax (x )+2 B a b \ln \relax (x )-\frac {2 A a b}{3 x^{3}}-\frac {B \,a^{2}}{3 x^{3}}-\frac {A \,a^{2}}{6 x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)^2*(B*x^3+A)/x^7,x)

[Out]

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

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maxima [A]  time = 0.67, size = 54, normalized size = 1.06 \[ \frac {1}{3} \, B b^{2} x^{3} + \frac {1}{3} \, {\left (2 \, B a b + A b^{2}\right )} \log \left (x^{3}\right ) - \frac {2 \, {\left (B a^{2} + 2 \, A a b\right )} x^{3} + A a^{2}}{6 \, x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^2*(B*x^3+A)/x^7,x, algorithm="maxima")

[Out]

1/3*B*b^2*x^3 + 1/3*(2*B*a*b + A*b^2)*log(x^3) - 1/6*(2*(B*a^2 + 2*A*a*b)*x^3 + A*a^2)/x^6

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mupad [B]  time = 2.36, size = 52, normalized size = 1.02 \[ \ln \relax (x)\,\left (A\,b^2+2\,B\,a\,b\right )-\frac {x^3\,\left (\frac {B\,a^2}{3}+\frac {2\,A\,b\,a}{3}\right )+\frac {A\,a^2}{6}}{x^6}+\frac {B\,b^2\,x^3}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x^3)*(a + b*x^3)^2)/x^7,x)

[Out]

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

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sympy [A]  time = 0.76, size = 51, normalized size = 1.00 \[ \frac {B b^{2} x^{3}}{3} + b \left (A b + 2 B a\right ) \log {\relax (x )} + \frac {- A a^{2} + x^{3} \left (- 4 A a b - 2 B a^{2}\right )}{6 x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a)**2*(B*x**3+A)/x**7,x)

[Out]

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

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