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

Optimal. Leaf size=28 \[ \frac {1}{3} x^3 (A c+b B)+A b x+\frac {1}{5} B c x^5 \]

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Rubi [A]  time = 0.02, antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1584, 373} \begin {gather*} \frac {1}{3} x^3 (A c+b B)+A b x+\frac {1}{5} B c x^5 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 373

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n
)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, 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 )}{x^2} \, dx &=\int \left (A+B x^2\right ) \left (b+c x^2\right ) \, dx\\ &=\int \left (A b+(b B+A c) x^2+B c x^4\right ) \, dx\\ &=A b x+\frac {1}{3} (b B+A c) x^3+\frac {1}{5} B c x^5\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 28, normalized size = 1.00 \begin {gather*} \frac {1}{3} x^3 (A c+b B)+A b x+\frac {1}{5} B c x^5 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.03, size = 32, normalized size = 1.14 \begin {gather*} A b x+\frac {1}{3} A c x^3+\frac {1}{3} b B x^3+\frac {1}{5} B c x^5 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.38, size = 24, normalized size = 0.86 \begin {gather*} \frac {1}{5} \, B c x^{5} + \frac {1}{3} \, {\left (B b + A c\right )} x^{3} + A b x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.15, size = 26, normalized size = 0.93 \begin {gather*} \frac {1}{5} \, B c x^{5} + \frac {1}{3} \, B b x^{3} + \frac {1}{3} \, A c x^{3} + A b x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 25, normalized size = 0.89 \begin {gather*} \frac {B c \,x^{5}}{5}+A b x +\frac {\left (A c +b B \right ) x^{3}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.41, size = 24, normalized size = 0.86 \begin {gather*} \frac {1}{5} \, B c x^{5} + \frac {1}{3} \, {\left (B b + A c\right )} x^{3} + A b x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.06, size = 25, normalized size = 0.89 \begin {gather*} \frac {B\,c\,x^5}{5}+\left (\frac {A\,c}{3}+\frac {B\,b}{3}\right )\,x^3+A\,b\,x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.07, size = 26, normalized size = 0.93 \begin {gather*} A b x + \frac {B c x^{5}}{5} + x^{3} \left (\frac {A c}{3} + \frac {B b}{3}\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)/x**2,x)

[Out]

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

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