3.1.17 \(\int \frac {(a+b x^3)^2 (A+B x^3)}{x^4} \, dx\)

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

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Rubi [A]  time = 0.05, 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} \begin {gather*} -\frac {a^2 A}{3 x^3}+\frac {1}{3} b x^3 (2 a B+A b)+a \log (x) (a B+2 A b)+\frac {1}{6} b^2 B x^6 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x^3)^2*(A + B*x^3))/x^4,x]

[Out]

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

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Mathematica [A]  time = 0.03, size = 49, normalized size = 0.96 \begin {gather*} \frac {1}{6} \left (-\frac {2 a^2 A}{x^3}+2 b x^3 (2 a B+A b)+6 a \log (x) (a B+2 A b)+b^2 B x^6\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^3)^2*(A + B*x^3))/x^4,x]

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a+b x^3\right )^2 \left (A+B x^3\right )}{x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[((a + b*x^3)^2*(A + B*x^3))/x^4,x]

[Out]

IntegrateAlgebraic[((a + b*x^3)^2*(A + B*x^3))/x^4, x]

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fricas [A]  time = 0.64, size = 54, normalized size = 1.06 \begin {gather*} \frac {B b^{2} x^{9} + 2 \, {\left (2 \, B a b + A b^{2}\right )} x^{6} + 6 \, {\left (B a^{2} + 2 \, A a b\right )} x^{3} \log \relax (x) - 2 \, A a^{2}}{6 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^2*(B*x^3+A)/x^4,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.15, size = 69, normalized size = 1.35 \begin {gather*} \frac {1}{6} \, B b^{2} x^{6} + \frac {2}{3} \, B a b x^{3} + \frac {1}{3} \, A b^{2} x^{3} + {\left (B a^{2} + 2 \, A a b\right )} \log \left ({\left | x \right |}\right ) - \frac {B a^{2} x^{3} + 2 \, A a b x^{3} + A a^{2}}{3 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^2*(B*x^3+A)/x^4,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 51, normalized size = 1.00 \begin {gather*} \frac {B \,b^{2} x^{6}}{6}+\frac {A \,b^{2} x^{3}}{3}+\frac {2 B a b \,x^{3}}{3}+2 A a b \ln \relax (x )+B \,a^{2} \ln \relax (x )-\frac {A \,a^{2}}{3 x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)^2*(B*x^3+A)/x^4,x)

[Out]

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

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maxima [A]  time = 0.48, size = 52, normalized size = 1.02 \begin {gather*} \frac {1}{6} \, B b^{2} x^{6} + \frac {1}{3} \, {\left (2 \, B a b + A b^{2}\right )} x^{3} + \frac {1}{3} \, {\left (B a^{2} + 2 \, A a b\right )} \log \left (x^{3}\right ) - \frac {A a^{2}}{3 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^2*(B*x^3+A)/x^4,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.04, size = 49, normalized size = 0.96 \begin {gather*} x^3\,\left (\frac {A\,b^2}{3}+\frac {2\,B\,a\,b}{3}\right )+\ln \relax (x)\,\left (B\,a^2+2\,A\,b\,a\right )-\frac {A\,a^2}{3\,x^3}+\frac {B\,b^2\,x^6}{6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x^3)*(a + b*x^3)^2)/x^4,x)

[Out]

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

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sympy [A]  time = 0.27, size = 51, normalized size = 1.00 \begin {gather*} - \frac {A a^{2}}{3 x^{3}} + \frac {B b^{2} x^{6}}{6} + a \left (2 A b + B a\right ) \log {\relax (x )} + x^{3} \left (\frac {A b^{2}}{3} + \frac {2 B a b}{3}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a)**2*(B*x**3+A)/x**4,x)

[Out]

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

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