3.1.8 \(\int \frac {(A+B x^2) (b x^2+c x^4)}{x^5} \, dx\) [8]

Optimal. Leaf size=29 \[ -\frac {A b}{2 x^2}+\frac {1}{2} B c x^2+(b B+A c) \log (x) \]

[Out]

-1/2*A*b/x^2+1/2*B*c*x^2+(A*c+B*b)*ln(x)

________________________________________________________________________________________

Rubi [A]
time = 0.02, antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1598, 457, 77} \begin {gather*} \log (x) (A c+b B)-\frac {A b}{2 x^2}+\frac {1}{2} B c x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(A*b)/x^2 + (B*c*x^2)/2 + (b*B + A*c)*Log[x]

Rule 77

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && N
eQ[b*e + a*f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && Rational
Q[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])

Rule 457

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

________________________________________________________________________________________

Mathematica [A]
time = 0.01, size = 29, normalized size = 1.00 \begin {gather*} -\frac {A b}{2 x^2}+\frac {1}{2} B c x^2+(b B+A c) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(A*b)/x^2 + (B*c*x^2)/2 + (b*B + A*c)*Log[x]

________________________________________________________________________________________

Maple [A]
time = 0.03, size = 26, normalized size = 0.90

method	result	size
default	\(-\frac {A b}{2 x^{2}}+\frac {B c \,x^{2}}{2}+\left (A c +B b \right ) \ln \left (x \right )\)	\(26\)
risch	\(-\frac {A b}{2 x^{2}}+\frac {B c \,x^{2}}{2}+A \ln \left (x \right ) c +b B \ln \left (x \right )\)	\(26\)
norman	\(\frac {-\frac {1}{2} A b \,x^{2}+\frac {1}{2} B c \,x^{6}}{x^{4}}+\left (A c +B b \right ) \ln \left (x \right )\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*A*b/x^2+1/2*B*c*x^2+(A*c+B*b)*ln(x)

________________________________________________________________________________________

Maxima [A]
time = 0.27, size = 28, normalized size = 0.97 \begin {gather*} \frac {1}{2} \, B c x^{2} + \frac {1}{2} \, {\left (B b + A c\right )} \log \left (x^{2}\right ) - \frac {A b}{2 \, 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)/x^5,x, algorithm="maxima")

[Out]

1/2*B*c*x^2 + 1/2*(B*b + A*c)*log(x^2) - 1/2*A*b/x^2

________________________________________________________________________________________

Fricas [A]
time = 1.33, size = 30, normalized size = 1.03 \begin {gather*} \frac {B c x^{4} + 2 \, {\left (B b + A c\right )} x^{2} \log \left (x\right ) - A b}{2 \, 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)/x^5,x, algorithm="fricas")

[Out]

1/2*(B*c*x^4 + 2*(B*b + A*c)*x^2*log(x) - A*b)/x^2

________________________________________________________________________________________

Sympy [A]
time = 0.08, size = 26, normalized size = 0.90 \begin {gather*} - \frac {A b}{2 x^{2}} + \frac {B c x^{2}}{2} + \left (A c + B b\right ) \log {\left (x \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**5,x)

[Out]

-A*b/(2*x**2) + B*c*x**2/2 + (A*c + B*b)*log(x)

________________________________________________________________________________________

Giac [A]
time = 0.63, size = 42, normalized size = 1.45 \begin {gather*} \frac {1}{2} \, B c x^{2} + \frac {1}{2} \, {\left (B b + A c\right )} \log \left (x^{2}\right ) - \frac {B b x^{2} + A c x^{2} + A b}{2 \, 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)/x^5,x, algorithm="giac")

[Out]

1/2*B*c*x^2 + 1/2*(B*b + A*c)*log(x^2) - 1/2*(B*b*x^2 + A*c*x^2 + A*b)/x^2

________________________________________________________________________________________

Mupad [B]
time = 0.04, size = 25, normalized size = 0.86 \begin {gather*} \ln \left (x\right )\,\left (A\,c+B\,b\right )-\frac {A\,b}{2\,x^2}+\frac {B\,c\,x^2}{2} \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^5,x)

[Out]

log(x)*(A*c + B*b) - (A*b)/(2*x^2) + (B*c*x^2)/2

________________________________________________________________________________________