3.28 \(\int \frac {(A+B x^2) (b x^2+c x^4)^3}{x^5} \, dx\)

Optimal. Leaf size=42 \[ \frac {B \left (b+c x^2\right )^5}{10 c^2}-\frac {\left (b+c x^2\right )^4 (b B-A c)}{8 c^2} \]

[Out]

-1/8*(-A*c+B*b)*(c*x^2+b)^4/c^2+1/10*B*(c*x^2+b)^5/c^2

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Rubi [A]  time = 0.07, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1584, 444, 43} \[ \frac {B \left (b+c x^2\right )^5}{10 c^2}-\frac {\left (b+c x^2\right )^4 (b B-A c)}{8 c^2} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^5,x]

[Out]

-((b*B - A*c)*(b + c*x^2)^4)/(8*c^2) + (B*(b + c*x^2)^5)/(10*c^2)

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 444

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]

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^5} \, dx &=\int x \left (A+B x^2\right ) \left (b+c x^2\right )^3 \, dx\\ &=\frac {1}{2} \operatorname {Subst}\left (\int (A+B x) (b+c x)^3 \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \left (\frac {(-b B+A c) (b+c x)^3}{c}+\frac {B (b+c x)^4}{c}\right ) \, dx,x,x^2\right )\\ &=-\frac {(b B-A c) \left (b+c x^2\right )^4}{8 c^2}+\frac {B \left (b+c x^2\right )^5}{10 c^2}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 69, normalized size = 1.64 \[ \frac {1}{40} x^2 \left (20 A b^3+10 b^2 x^2 (3 A c+b B)+5 c^2 x^6 (A c+3 b B)+20 b c x^4 (A c+b B)+4 B c^3 x^8\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^5,x]

[Out]

(x^2*(20*A*b^3 + 10*b^2*(b*B + 3*A*c)*x^2 + 20*b*c*(b*B + A*c)*x^4 + 5*c^2*(3*b*B + A*c)*x^6 + 4*B*c^3*x^8))/4
0

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fricas [A]  time = 0.88, size = 73, normalized size = 1.74 \[ \frac {1}{10} \, B c^{3} x^{10} + \frac {1}{8} \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{8} + \frac {1}{2} \, {\left (B b^{2} c + A b c^{2}\right )} x^{6} + \frac {1}{2} \, A b^{3} x^{2} + \frac {1}{4} \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x^{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^5,x, algorithm="fricas")

[Out]

1/10*B*c^3*x^10 + 1/8*(3*B*b*c^2 + A*c^3)*x^8 + 1/2*(B*b^2*c + A*b*c^2)*x^6 + 1/2*A*b^3*x^2 + 1/4*(B*b^3 + 3*A
*b^2*c)*x^4

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giac [B]  time = 0.15, size = 77, normalized size = 1.83 \[ \frac {1}{10} \, B c^{3} x^{10} + \frac {3}{8} \, B b c^{2} x^{8} + \frac {1}{8} \, A c^{3} x^{8} + \frac {1}{2} \, B b^{2} c x^{6} + \frac {1}{2} \, A b c^{2} x^{6} + \frac {1}{4} \, B b^{3} x^{4} + \frac {3}{4} \, A b^{2} c x^{4} + \frac {1}{2} \, A b^{3} x^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^5,x, algorithm="giac")

[Out]

1/10*B*c^3*x^10 + 3/8*B*b*c^2*x^8 + 1/8*A*c^3*x^8 + 1/2*B*b^2*c*x^6 + 1/2*A*b*c^2*x^6 + 1/4*B*b^3*x^4 + 3/4*A*
b^2*c*x^4 + 1/2*A*b^3*x^2

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maple [A]  time = 0.04, size = 76, normalized size = 1.81 \[ \frac {B \,c^{3} x^{10}}{10}+\frac {\left (A \,c^{3}+3 B b \,c^{2}\right ) x^{8}}{8}+\frac {A \,b^{3} x^{2}}{2}+\frac {\left (3 A b \,c^{2}+3 B c \,b^{2}\right ) x^{6}}{6}+\frac {\left (3 A c \,b^{2}+B \,b^{3}\right ) x^{4}}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)*(c*x^4+b*x^2)^3/x^5,x)

[Out]

1/10*B*c^3*x^10+1/8*(A*c^3+3*B*b*c^2)*x^8+1/6*(3*A*b*c^2+3*B*b^2*c)*x^6+1/4*(3*A*b^2*c+B*b^3)*x^4+1/2*A*b^3*x^
2

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maxima [A]  time = 1.44, size = 73, normalized size = 1.74 \[ \frac {1}{10} \, B c^{3} x^{10} + \frac {1}{8} \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{8} + \frac {1}{2} \, {\left (B b^{2} c + A b c^{2}\right )} x^{6} + \frac {1}{2} \, A b^{3} x^{2} + \frac {1}{4} \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x^{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^5,x, algorithm="maxima")

[Out]

1/10*B*c^3*x^10 + 1/8*(3*B*b*c^2 + A*c^3)*x^8 + 1/2*(B*b^2*c + A*b*c^2)*x^6 + 1/2*A*b^3*x^2 + 1/4*(B*b^3 + 3*A
*b^2*c)*x^4

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mupad [B]  time = 0.03, size = 69, normalized size = 1.64 \[ x^4\,\left (\frac {B\,b^3}{4}+\frac {3\,A\,c\,b^2}{4}\right )+x^8\,\left (\frac {A\,c^3}{8}+\frac {3\,B\,b\,c^2}{8}\right )+\frac {A\,b^3\,x^2}{2}+\frac {B\,c^3\,x^{10}}{10}+\frac {b\,c\,x^6\,\left (A\,c+B\,b\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^5,x)

[Out]

x^4*((B*b^3)/4 + (3*A*b^2*c)/4) + x^8*((A*c^3)/8 + (3*B*b*c^2)/8) + (A*b^3*x^2)/2 + (B*c^3*x^10)/10 + (b*c*x^6
*(A*c + B*b))/2

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sympy [B]  time = 0.09, size = 80, normalized size = 1.90 \[ \frac {A b^{3} x^{2}}{2} + \frac {B c^{3} x^{10}}{10} + x^{8} \left (\frac {A c^{3}}{8} + \frac {3 B b c^{2}}{8}\right ) + x^{6} \left (\frac {A b c^{2}}{2} + \frac {B b^{2} c}{2}\right ) + x^{4} \left (\frac {3 A b^{2} c}{4} + \frac {B b^{3}}{4}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)*(c*x**4+b*x**2)**3/x**5,x)

[Out]

A*b**3*x**2/2 + B*c**3*x**10/10 + x**8*(A*c**3/8 + 3*B*b*c**2/8) + x**6*(A*b*c**2/2 + B*b**2*c/2) + x**4*(3*A*
b**2*c/4 + B*b**3/4)

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