\(\int \frac {\log (c (a+b x^2)^p)}{x^5} \, dx\) [10]

Optimal result

Integrand size = 16, antiderivative size = 64 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^5} \, dx=-\frac {b p}{4 a x^2}-\frac {b^2 p \log (x)}{2 a^2}+\frac {b^2 p \log \left (a+b x^2\right )}{4 a^2}-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{4 x^4} \]

[Out]

-1/4*b*p/a/x^2-1/2*b^2*p*ln(x)/a^2+1/4*b^2*p*ln(b*x^2+a)/a^2-1/4*ln(c*(b*x^2+a)^p)/x^4

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {2504, 2442, 46} \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^5} \, dx=\frac {b^2 p \log \left (a+b x^2\right )}{4 a^2}-\frac {b^2 p \log (x)}{2 a^2}-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{4 x^4}-\frac {b p}{4 a x^2} \]

[In]

Int[Log[c*(a + b*x^2)^p]/x^5,x]

[Out]

-1/4*(b*p)/(a*x^2) - (b^2*p*Log[x])/(2*a^2) + (b^2*p*Log[a + b*x^2])/(4*a^2) - Log[c*(a + b*x^2)^p]/(4*x^4)

Rule 46

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x
)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 2442

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[(f + g*
x)^(q + 1)*((a + b*Log[c*(d + e*x)^n])/(g*(q + 1))), x] - Dist[b*e*(n/(g*(q + 1))), Int[(f + g*x)^(q + 1)/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && NeQ[q, -1]

Rule 2504

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(a + b*Log[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p,
q}, x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) &&  !(EqQ[q, 1] && ILtQ[n, 0] &&
IGtQ[m, 0])

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {\log \left (c (a+b x)^p\right )}{x^3} \, dx,x,x^2\right ) \\ & = -\frac {\log \left (c \left (a+b x^2\right )^p\right )}{4 x^4}+\frac {1}{4} (b p) \text {Subst}\left (\int \frac {1}{x^2 (a+b x)} \, dx,x,x^2\right ) \\ & = -\frac {\log \left (c \left (a+b x^2\right )^p\right )}{4 x^4}+\frac {1}{4} (b p) \text {Subst}\left (\int \left (\frac {1}{a x^2}-\frac {b}{a^2 x}+\frac {b^2}{a^2 (a+b x)}\right ) \, dx,x,x^2\right ) \\ & = -\frac {b p}{4 a x^2}-\frac {b^2 p \log (x)}{2 a^2}+\frac {b^2 p \log \left (a+b x^2\right )}{4 a^2}-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{4 x^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.88 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^5} \, dx=\frac {1}{4} b p \left (-\frac {1}{a x^2}-\frac {2 b \log (x)}{a^2}+\frac {b \log \left (a+b x^2\right )}{a^2}\right )-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{4 x^4} \]

[In]

Integrate[Log[c*(a + b*x^2)^p]/x^5,x]

[Out]

(b*p*(-(1/(a*x^2)) - (2*b*Log[x])/a^2 + (b*Log[a + b*x^2])/a^2))/4 - Log[c*(a + b*x^2)^p]/(4*x^4)

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.84

method	result	size
parts	\(-\frac {\ln \left (c \left (b \,x^{2}+a \right )^{p}\right )}{4 x^{4}}+\frac {p b \left (-\frac {1}{2 a \,x^{2}}-\frac {b \ln \left (x \right )}{a^{2}}+\frac {b \ln \left (b \,x^{2}+a \right )}{2 a^{2}}\right )}{2}\)	\(54\)
parallelrisch	\(-\frac {2 b^{2} p^{2} \ln \left (x \right ) x^{4}-x^{4} \ln \left (c \left (b \,x^{2}+a \right )^{p}\right ) b^{2} p -b^{2} p^{2} x^{4}+a b \,p^{2} x^{2}+\ln \left (c \left (b \,x^{2}+a \right )^{p}\right ) a^{2} p}{4 x^{4} a^{2} p}\)	\(84\)
risch	\(-\frac {\ln \left (\left (b \,x^{2}+a \right )^{p}\right )}{4 x^{4}}-\frac {4 b^{2} p \ln \left (x \right ) x^{4}-2 b^{2} p \ln \left (-b \,x^{2}-a \right ) x^{4}+i \pi \,a^{2} \operatorname {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{2}-i \pi \,a^{2} \operatorname {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \right )-i \pi \,a^{2} {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{3}+i \pi \,a^{2} {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )+2 a b p \,x^{2}+2 \ln \left (c \right ) a^{2}}{8 a^{2} x^{4}}\)	\(198\)

[In]

int(ln(c*(b*x^2+a)^p)/x^5,x,method=_RETURNVERBOSE)

[Out]

-1/4*ln(c*(b*x^2+a)^p)/x^4+1/2*p*b*(-1/2/a/x^2-1/a^2*b*ln(x)+1/2*b/a^2*ln(b*x^2+a))

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.91 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^5} \, dx=-\frac {2 \, b^{2} p x^{4} \log \left (x\right ) + a b p x^{2} + a^{2} \log \left (c\right ) - {\left (b^{2} p x^{4} - a^{2} p\right )} \log \left (b x^{2} + a\right )}{4 \, a^{2} x^{4}} \]

[In]

integrate(log(c*(b*x^2+a)^p)/x^5,x, algorithm="fricas")

[Out]

-1/4*(2*b^2*p*x^4*log(x) + a*b*p*x^2 + a^2*log(c) - (b^2*p*x^4 - a^2*p)*log(b*x^2 + a))/(a^2*x^4)

Sympy [A] (verification not implemented)

Time = 2.63 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.30 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^5} \, dx=\begin {cases} - \frac {\log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{4 x^{4}} - \frac {b p}{4 a x^{2}} - \frac {b^{2} p \log {\left (x \right )}}{2 a^{2}} + \frac {b^{2} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{4 a^{2}} & \text {for}\: a \neq 0 \\- \frac {p}{8 x^{4}} - \frac {\log {\left (c \left (b x^{2}\right )^{p} \right )}}{4 x^{4}} & \text {otherwise} \end {cases} \]

[In]

integrate(ln(c*(b*x**2+a)**p)/x**5,x)

[Out]

Piecewise((-log(c*(a + b*x**2)**p)/(4*x**4) - b*p/(4*a*x**2) - b**2*p*log(x)/(2*a**2) + b**2*log(c*(a + b*x**2
)**p)/(4*a**2), Ne(a, 0)), (-p/(8*x**4) - log(c*(b*x**2)**p)/(4*x**4), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.84 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^5} \, dx=\frac {1}{4} \, b p {\left (\frac {b \log \left (b x^{2} + a\right )}{a^{2}} - \frac {b \log \left (x^{2}\right )}{a^{2}} - \frac {1}{a x^{2}}\right )} - \frac {\log \left ({\left (b x^{2} + a\right )}^{p} c\right )}{4 \, x^{4}} \]

[In]

integrate(log(c*(b*x^2+a)^p)/x^5,x, algorithm="maxima")

[Out]

1/4*b*p*(b*log(b*x^2 + a)/a^2 - b*log(x^2)/a^2 - 1/(a*x^2)) - 1/4*log((b*x^2 + a)^p*c)/x^4

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 132 vs. \(2 (56) = 112\).

Time = 0.31 (sec) , antiderivative size = 132, normalized size of antiderivative = 2.06 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^5} \, dx=-\frac {\frac {b^{3} p \log \left (b x^{2} + a\right )}{{\left (b x^{2} + a\right )}^{2} - 2 \, {\left (b x^{2} + a\right )} a + a^{2}} - \frac {b^{3} p \log \left (b x^{2} + a\right )}{a^{2}} + \frac {b^{3} p \log \left (b x^{2}\right )}{a^{2}} + \frac {{\left (b x^{2} + a\right )} b^{3} p - a b^{3} p + a b^{3} \log \left (c\right )}{{\left (b x^{2} + a\right )}^{2} a - 2 \, {\left (b x^{2} + a\right )} a^{2} + a^{3}}}{4 \, b} \]

[In]

integrate(log(c*(b*x^2+a)^p)/x^5,x, algorithm="giac")

[Out]

-1/4*(b^3*p*log(b*x^2 + a)/((b*x^2 + a)^2 - 2*(b*x^2 + a)*a + a^2) - b^3*p*log(b*x^2 + a)/a^2 + b^3*p*log(b*x^
2)/a^2 + ((b*x^2 + a)*b^3*p - a*b^3*p + a*b^3*log(c))/((b*x^2 + a)^2*a - 2*(b*x^2 + a)*a^2 + a^3))/b

Mupad [B] (verification not implemented)

Time = 1.24 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.88 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^5} \, dx=\frac {b^2\,p\,\ln \left (b\,x^2+a\right )}{4\,a^2}-\frac {\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}{4\,x^4}-\frac {b^2\,p\,\ln \left (x\right )}{2\,a^2}-\frac {b\,p}{4\,a\,x^2} \]

[In]

int(log(c*(a + b*x^2)^p)/x^5,x)

[Out]

(b^2*p*log(a + b*x^2))/(4*a^2) - log(c*(a + b*x^2)^p)/(4*x^4) - (b^2*p*log(x))/(2*a^2) - (b*p)/(4*a*x^2)