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

Optimal. Leaf size=75 \[ \frac {1}{3} A b^3 x^3+\frac {1}{5} b^2 x^5 (3 A c+b B)+\frac {1}{9} c^2 x^9 (A c+3 b B)+\frac {3}{7} b c x^7 (A c+b B)+\frac {1}{11} B c^3 x^{11} \]

[Out]

1/3*A*b^3*x^3+1/5*b^2*(3*A*c+B*b)*x^5+3/7*b*c*(A*c+B*b)*x^7+1/9*c^2*(A*c+3*B*b)*x^9+1/11*B*c^3*x^11

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Rubi [A]  time = 0.06, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {1584, 448} \[ \frac {1}{5} b^2 x^5 (3 A c+b B)+\frac {1}{3} A b^3 x^3+\frac {1}{9} c^2 x^9 (A c+3 b B)+\frac {3}{7} b c x^7 (A c+b B)+\frac {1}{11} B c^3 x^{11} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(A*b^3*x^3)/3 + (b^2*(b*B + 3*A*c)*x^5)/5 + (3*b*c*(b*B + A*c)*x^7)/7 + (c^2*(3*b*B + A*c)*x^9)/9 + (B*c^3*x^1
1)/11

Rule 448

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandI
ntegrand[(e*x)^m*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[p, 0] && IGtQ[q, 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^4} \, dx &=\int x^2 \left (A+B x^2\right ) \left (b+c x^2\right )^3 \, dx\\ &=\int \left (A b^3 x^2+b^2 (b B+3 A c) x^4+3 b c (b B+A c) x^6+c^2 (3 b B+A c) x^8+B c^3 x^{10}\right ) \, dx\\ &=\frac {1}{3} A b^3 x^3+\frac {1}{5} b^2 (b B+3 A c) x^5+\frac {3}{7} b c (b B+A c) x^7+\frac {1}{9} c^2 (3 b B+A c) x^9+\frac {1}{11} B c^3 x^{11}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 75, normalized size = 1.00 \[ \frac {1}{3} A b^3 x^3+\frac {1}{5} b^2 x^5 (3 A c+b B)+\frac {1}{9} c^2 x^9 (A c+3 b B)+\frac {3}{7} b c x^7 (A c+b B)+\frac {1}{11} B c^3 x^{11} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(A*b^3*x^3)/3 + (b^2*(b*B + 3*A*c)*x^5)/5 + (3*b*c*(b*B + A*c)*x^7)/7 + (c^2*(3*b*B + A*c)*x^9)/9 + (B*c^3*x^1
1)/11

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fricas [A]  time = 0.83, size = 73, normalized size = 0.97 \[ \frac {1}{11} \, B c^{3} x^{11} + \frac {1}{9} \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{9} + \frac {3}{7} \, {\left (B b^{2} c + A b c^{2}\right )} x^{7} + \frac {1}{3} \, A b^{3} x^{3} + \frac {1}{5} \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x^{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/11*B*c^3*x^11 + 1/9*(3*B*b*c^2 + A*c^3)*x^9 + 3/7*(B*b^2*c + A*b*c^2)*x^7 + 1/3*A*b^3*x^3 + 1/5*(B*b^3 + 3*A
*b^2*c)*x^5

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giac [A]  time = 0.15, size = 77, normalized size = 1.03 \[ \frac {1}{11} \, B c^{3} x^{11} + \frac {1}{3} \, B b c^{2} x^{9} + \frac {1}{9} \, A c^{3} x^{9} + \frac {3}{7} \, B b^{2} c x^{7} + \frac {3}{7} \, A b c^{2} x^{7} + \frac {1}{5} \, B b^{3} x^{5} + \frac {3}{5} \, A b^{2} c x^{5} + \frac {1}{3} \, A b^{3} x^{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/11*B*c^3*x^11 + 1/3*B*b*c^2*x^9 + 1/9*A*c^3*x^9 + 3/7*B*b^2*c*x^7 + 3/7*A*b*c^2*x^7 + 1/5*B*b^3*x^5 + 3/5*A*
b^2*c*x^5 + 1/3*A*b^3*x^3

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maple [A]  time = 0.04, size = 76, normalized size = 1.01 \[ \frac {B \,c^{3} x^{11}}{11}+\frac {\left (A \,c^{3}+3 B b \,c^{2}\right ) x^{9}}{9}+\frac {A \,b^{3} x^{3}}{3}+\frac {\left (3 A b \,c^{2}+3 B c \,b^{2}\right ) x^{7}}{7}+\frac {\left (3 A c \,b^{2}+B \,b^{3}\right ) x^{5}}{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/11*B*c^3*x^11+1/9*(A*c^3+3*B*b*c^2)*x^9+1/7*(3*A*b*c^2+3*B*b^2*c)*x^7+1/5*(3*A*b^2*c+B*b^3)*x^5+1/3*A*b^3*x^
3

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maxima [A]  time = 1.37, size = 73, normalized size = 0.97 \[ \frac {1}{11} \, B c^{3} x^{11} + \frac {1}{9} \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{9} + \frac {3}{7} \, {\left (B b^{2} c + A b c^{2}\right )} x^{7} + \frac {1}{3} \, A b^{3} x^{3} + \frac {1}{5} \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x^{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/11*B*c^3*x^11 + 1/9*(3*B*b*c^2 + A*c^3)*x^9 + 3/7*(B*b^2*c + A*b*c^2)*x^7 + 1/3*A*b^3*x^3 + 1/5*(B*b^3 + 3*A
*b^2*c)*x^5

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mupad [B]  time = 0.03, size = 69, normalized size = 0.92 \[ x^5\,\left (\frac {B\,b^3}{5}+\frac {3\,A\,c\,b^2}{5}\right )+x^9\,\left (\frac {A\,c^3}{9}+\frac {B\,b\,c^2}{3}\right )+\frac {A\,b^3\,x^3}{3}+\frac {B\,c^3\,x^{11}}{11}+\frac {3\,b\,c\,x^7\,\left (A\,c+B\,b\right )}{7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^5*((B*b^3)/5 + (3*A*b^2*c)/5) + x^9*((A*c^3)/9 + (B*b*c^2)/3) + (A*b^3*x^3)/3 + (B*c^3*x^11)/11 + (3*b*c*x^7
*(A*c + B*b))/7

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sympy [A]  time = 0.09, size = 82, normalized size = 1.09 \[ \frac {A b^{3} x^{3}}{3} + \frac {B c^{3} x^{11}}{11} + x^{9} \left (\frac {A c^{3}}{9} + \frac {B b c^{2}}{3}\right ) + x^{7} \left (\frac {3 A b c^{2}}{7} + \frac {3 B b^{2} c}{7}\right ) + x^{5} \left (\frac {3 A b^{2} c}{5} + \frac {B b^{3}}{5}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*b**3*x**3/3 + B*c**3*x**11/11 + x**9*(A*c**3/9 + B*b*c**2/3) + x**7*(3*A*b*c**2/7 + 3*B*b**2*c/7) + x**5*(3*
A*b**2*c/5 + B*b**3/5)

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