3.1.50 \(\int \frac {x (A+B x^2)}{b x^2+c x^4} \, dx\) [50]

Optimal. Leaf size=34 \[ \frac {A \log (x)}{b}+\frac {(b B-A c) \log \left (b+c x^2\right )}{2 b c} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 34, 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, 78} \begin {gather*} \frac {(b B-A c) \log \left (b+c x^2\right )}{2 b c}+\frac {A \log (x)}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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

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Mathematica [A]
time = 0.01, size = 34, normalized size = 1.00 \begin {gather*} \frac {A \log (x)}{b}+\frac {(b B-A c) \log \left (b+c x^2\right )}{2 b c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.38, size = 33, normalized size = 0.97

method	result	size
default	\(-\frac {\left (A c -B b \right ) \ln \left (c \,x^{2}+b \right )}{2 b c}+\frac {A \ln \left (x \right )}{b}\)	\(33\)
norman	\(-\frac {\left (A c -B b \right ) \ln \left (c \,x^{2}+b \right )}{2 b c}+\frac {A \ln \left (x \right )}{b}\)	\(33\)
risch	\(-\frac {\ln \left (c \,x^{2}+b \right ) A}{2 b}+\frac {\ln \left (c \,x^{2}+b \right ) B}{2 c}+\frac {A \ln \left (x \right )}{b}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.29, size = 35, normalized size = 1.03 \begin {gather*} \frac {A \log \left (x^{2}\right )}{2 \, b} + \frac {{\left (B b - A c\right )} \log \left (c x^{2} + b\right )}{2 \, b c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 2.19, size = 32, normalized size = 0.94 \begin {gather*} \frac {2 \, A c \log \left (x\right ) + {\left (B b - A c\right )} \log \left (c x^{2} + b\right )}{2 \, b c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.38, size = 26, normalized size = 0.76 \begin {gather*} \frac {A \log {\left (x \right )}}{b} + \frac {\left (- A c + B b\right ) \log {\left (\frac {b}{c} + x^{2} \right )}}{2 b c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.64, size = 34, normalized size = 1.00 \begin {gather*} \frac {A \log \left ({\left | x \right |}\right )}{b} + \frac {{\left (B b - A c\right )} \log \left ({\left | c x^{2} + b \right |}\right )}{2 \, b c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.08, size = 32, normalized size = 0.94 \begin {gather*} \frac {A\,\ln \left (x\right )}{b}-\frac {\ln \left (c\,x^2+b\right )\,\left (A\,c-B\,b\right )}{2\,b\,c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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