3.1.30 \(\int \frac {(A+B x^2) (b x^2+c x^4)^3}{x^7} \, dx\)

Optimal. Leaf size=60 \[ A b^3 \log (x)+\frac {3}{2} A b^2 c x^2+\frac {3}{4} A b c^2 x^4+\frac {1}{6} A c^3 x^6+\frac {B \left (b+c x^2\right )^4}{8 c} \]

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Rubi [A]  time = 0.05, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1584, 446, 80, 43} \begin {gather*} \frac {3}{2} A b^2 c x^2+A b^3 \log (x)+\frac {3}{4} A b c^2 x^4+\frac {1}{6} A c^3 x^6+\frac {B \left (b+c x^2\right )^4}{8 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^7,x]

[Out]

(3*A*b^2*c*x^2)/2 + (3*A*b*c^2*x^4)/4 + (A*c^3*x^6)/6 + (B*(b + c*x^2)^4)/(8*c) + A*b^3*Log[x]

Rule 43

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 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

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]]

Rule 1584

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rubi steps

\begin {align*} \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^7} \, dx &=\int \frac {\left (A+B x^2\right ) \left (b+c x^2\right )^3}{x} \, dx\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {(A+B x) (b+c x)^3}{x} \, dx,x,x^2\right )\\ &=\frac {B \left (b+c x^2\right )^4}{8 c}+\frac {1}{2} A \operatorname {Subst}\left (\int \frac {(b+c x)^3}{x} \, dx,x,x^2\right )\\ &=\frac {B \left (b+c x^2\right )^4}{8 c}+\frac {1}{2} A \operatorname {Subst}\left (\int \left (3 b^2 c+\frac {b^3}{x}+3 b c^2 x+c^3 x^2\right ) \, dx,x,x^2\right )\\ &=\frac {3}{2} A b^2 c x^2+\frac {3}{4} A b c^2 x^4+\frac {1}{6} A c^3 x^6+\frac {B \left (b+c x^2\right )^4}{8 c}+A b^3 \log (x)\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 71, normalized size = 1.18 \begin {gather*} A b^3 \log (x)+\frac {1}{2} b^2 x^2 (3 A c+b B)+\frac {1}{6} c^2 x^6 (A c+3 b B)+\frac {3}{4} b c x^4 (A c+b B)+\frac {1}{8} B c^3 x^8 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^7,x]

[Out]

(b^2*(b*B + 3*A*c)*x^2)/2 + (3*b*c*(b*B + A*c)*x^4)/4 + (c^2*(3*b*B + A*c)*x^6)/6 + (B*c^3*x^8)/8 + A*b^3*Log[
x]

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^7} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^7,x]

[Out]

IntegrateAlgebraic[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^7, x]

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fricas [A]  time = 0.38, size = 71, normalized size = 1.18 \begin {gather*} \frac {1}{8} \, B c^{3} x^{8} + \frac {1}{6} \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{6} + \frac {3}{4} \, {\left (B b^{2} c + A b c^{2}\right )} x^{4} + A b^{3} \log \relax (x) + \frac {1}{2} \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^7,x, algorithm="fricas")

[Out]

1/8*B*c^3*x^8 + 1/6*(3*B*b*c^2 + A*c^3)*x^6 + 3/4*(B*b^2*c + A*b*c^2)*x^4 + A*b^3*log(x) + 1/2*(B*b^3 + 3*A*b^
2*c)*x^2

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giac [A]  time = 0.15, size = 78, normalized size = 1.30 \begin {gather*} \frac {1}{8} \, B c^{3} x^{8} + \frac {1}{2} \, B b c^{2} x^{6} + \frac {1}{6} \, A c^{3} x^{6} + \frac {3}{4} \, B b^{2} c x^{4} + \frac {3}{4} \, A b c^{2} x^{4} + \frac {1}{2} \, B b^{3} x^{2} + \frac {3}{2} \, A b^{2} c x^{2} + \frac {1}{2} \, A b^{3} \log \left (x^{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^7,x, algorithm="giac")

[Out]

1/8*B*c^3*x^8 + 1/2*B*b*c^2*x^6 + 1/6*A*c^3*x^6 + 3/4*B*b^2*c*x^4 + 3/4*A*b*c^2*x^4 + 1/2*B*b^3*x^2 + 3/2*A*b^
2*c*x^2 + 1/2*A*b^3*log(x^2)

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maple [A]  time = 0.04, size = 76, normalized size = 1.27 \begin {gather*} \frac {B \,c^{3} x^{8}}{8}+\frac {A \,c^{3} x^{6}}{6}+\frac {B b \,c^{2} x^{6}}{2}+\frac {3 A b \,c^{2} x^{4}}{4}+\frac {3 B \,b^{2} c \,x^{4}}{4}+\frac {3 A \,b^{2} c \,x^{2}}{2}+\frac {B \,b^{3} x^{2}}{2}+A \,b^{3} \ln \relax (x ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)*(c*x^4+b*x^2)^3/x^7,x)

[Out]

1/8*B*c^3*x^8+1/6*A*c^3*x^6+1/2*B*x^6*b*c^2+3/4*A*b*c^2*x^4+3/4*B*x^4*b^2*c+3/2*A*b^2*c*x^2+1/2*B*x^2*b^3+A*b^
3*ln(x)

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maxima [A]  time = 1.31, size = 74, normalized size = 1.23 \begin {gather*} \frac {1}{8} \, B c^{3} x^{8} + \frac {1}{6} \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{6} + \frac {3}{4} \, {\left (B b^{2} c + A b c^{2}\right )} x^{4} + \frac {1}{2} \, A b^{3} \log \left (x^{2}\right ) + \frac {1}{2} \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^7,x, algorithm="maxima")

[Out]

1/8*B*c^3*x^8 + 1/6*(3*B*b*c^2 + A*c^3)*x^6 + 3/4*(B*b^2*c + A*b*c^2)*x^4 + 1/2*A*b^3*log(x^2) + 1/2*(B*b^3 +
3*A*b^2*c)*x^2

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mupad [B]  time = 0.03, size = 67, normalized size = 1.12 \begin {gather*} x^2\,\left (\frac {B\,b^3}{2}+\frac {3\,A\,c\,b^2}{2}\right )+x^6\,\left (\frac {A\,c^3}{6}+\frac {B\,b\,c^2}{2}\right )+\frac {B\,c^3\,x^8}{8}+A\,b^3\,\ln \relax (x)+\frac {3\,b\,c\,x^4\,\left (A\,c+B\,b\right )}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^7,x)

[Out]

x^2*((B*b^3)/2 + (3*A*b^2*c)/2) + x^6*((A*c^3)/6 + (B*b*c^2)/2) + (B*c^3*x^8)/8 + A*b^3*log(x) + (3*b*c*x^4*(A
*c + B*b))/4

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sympy [A]  time = 0.18, size = 80, normalized size = 1.33 \begin {gather*} A b^{3} \log {\relax (x )} + \frac {B c^{3} x^{8}}{8} + x^{6} \left (\frac {A c^{3}}{6} + \frac {B b c^{2}}{2}\right ) + x^{4} \left (\frac {3 A b c^{2}}{4} + \frac {3 B b^{2} c}{4}\right ) + x^{2} \left (\frac {3 A b^{2} c}{2} + \frac {B b^{3}}{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)*(c*x**4+b*x**2)**3/x**7,x)

[Out]

A*b**3*log(x) + B*c**3*x**8/8 + x**6*(A*c**3/6 + B*b*c**2/2) + x**4*(3*A*b*c**2/4 + 3*B*b**2*c/4) + x**2*(3*A*
b**2*c/2 + B*b**3/2)

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