3.1.14 \(\int \frac {(A+B x^2) (b x^2+c x^4)^2}{x^2} \, dx\) [14]

Optimal. Leaf size=55 \[ \frac {1}{3} A b^2 x^3+\frac {1}{5} b (b B+2 A c) x^5+\frac {1}{7} c (2 b B+A c) x^7+\frac {1}{9} B c^2 x^9 \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 55, 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 = {1598, 459} \begin {gather*} \frac {1}{3} A b^2 x^3+\frac {1}{7} c x^7 (A c+2 b B)+\frac {1}{5} b x^5 (2 A c+b B)+\frac {1}{9} B c^2 x^9 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 459

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 1598

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 )^2}{x^2} \, dx &=\int x^2 \left (A+B x^2\right ) \left (b+c x^2\right )^2 \, dx\\ &=\int \left (A b^2 x^2+b (b B+2 A c) x^4+c (2 b B+A c) x^6+B c^2 x^8\right ) \, dx\\ &=\frac {1}{3} A b^2 x^3+\frac {1}{5} b (b B+2 A c) x^5+\frac {1}{7} c (2 b B+A c) x^7+\frac {1}{9} B c^2 x^9\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 55, normalized size = 1.00 \begin {gather*} \frac {1}{3} A b^2 x^3+\frac {1}{5} b (b B+2 A c) x^5+\frac {1}{7} c (2 b B+A c) x^7+\frac {1}{9} B c^2 x^9 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.42, size = 52, normalized size = 0.95

method	result	size
default	\(\frac {B \,c^{2} x^{9}}{9}+\frac {\left (A \,c^{2}+2 b B c \right ) x^{7}}{7}+\frac {\left (2 A b c +b^{2} B \right ) x^{5}}{5}+\frac {A \,b^{2} x^{3}}{3}\)	\(52\)
risch	\(\frac {1}{9} B \,c^{2} x^{9}+\frac {1}{7} x^{7} A \,c^{2}+\frac {2}{7} x^{7} b B c +\frac {2}{5} x^{5} A b c +\frac {1}{5} b^{2} B \,x^{5}+\frac {1}{3} A \,b^{2} x^{3}\)	\(54\)
gosper	\(\frac {x^{3} \left (35 B \,c^{2} x^{6}+45 A \,c^{2} x^{4}+90 x^{4} b B c +126 A b c \,x^{2}+63 b^{2} B \,x^{2}+105 b^{2} A \right )}{315}\)	\(56\)
norman	\(\frac {\left (\frac {1}{7} A \,c^{2}+\frac {2}{7} b B c \right ) x^{8}+\left (\frac {2}{5} A b c +\frac {1}{5} b^{2} B \right ) x^{6}+\frac {A \,b^{2} x^{4}}{3}+\frac {B \,c^{2} x^{10}}{9}}{x}\)	\(56\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)*(c*x^4+b*x^2)^2/x^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.27, size = 51, normalized size = 0.93 \begin {gather*} \frac {1}{9} \, B c^{2} x^{9} + \frac {1}{7} \, {\left (2 \, B b c + A c^{2}\right )} x^{7} + \frac {1}{3} \, A b^{2} x^{3} + \frac {1}{5} \, {\left (B b^{2} + 2 \, A b c\right )} x^{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 1.25, size = 51, normalized size = 0.93 \begin {gather*} \frac {1}{9} \, B c^{2} x^{9} + \frac {1}{7} \, {\left (2 \, B b c + A c^{2}\right )} x^{7} + \frac {1}{3} \, A b^{2} x^{3} + \frac {1}{5} \, {\left (B b^{2} + 2 \, A b c\right )} x^{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.01, size = 56, normalized size = 1.02 \begin {gather*} \frac {A b^{2} x^{3}}{3} + \frac {B c^{2} x^{9}}{9} + x^{7} \left (\frac {A c^{2}}{7} + \frac {2 B b c}{7}\right ) + x^{5} \cdot \left (\frac {2 A b c}{5} + \frac {B b^{2}}{5}\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)**2/x**2,x)

[Out]

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

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Giac [A]
time = 1.14, size = 53, normalized size = 0.96 \begin {gather*} \frac {1}{9} \, B c^{2} x^{9} + \frac {2}{7} \, B b c x^{7} + \frac {1}{7} \, A c^{2} x^{7} + \frac {1}{5} \, B b^{2} x^{5} + \frac {2}{5} \, A b c x^{5} + \frac {1}{3} \, A b^{2} x^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.05, size = 51, normalized size = 0.93 \begin {gather*} x^5\,\left (\frac {B\,b^2}{5}+\frac {2\,A\,c\,b}{5}\right )+x^7\,\left (\frac {A\,c^2}{7}+\frac {2\,B\,b\,c}{7}\right )+\frac {A\,b^2\,x^3}{3}+\frac {B\,c^2\,x^9}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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