3.12 \(\int \frac{\log (c (a+b x^2)^p)}{x^7} \, dx\)

Optimal. Leaf size=78 \[ \frac{b^2 p}{6 a^2 x^2}-\frac{b^3 p \log \left (a+b x^2\right )}{6 a^3}+\frac{b^3 p \log (x)}{3 a^3}-\frac{\log \left (c \left (a+b x^2\right )^p\right )}{6 x^6}-\frac{b p}{12 a x^4} \]

[Out]

-(b*p)/(12*a*x^4) + (b^2*p)/(6*a^2*x^2) + (b^3*p*Log[x])/(3*a^3) - (b^3*p*Log[a + b*x^2])/(6*a^3) - Log[c*(a +
b*x^2)^p]/(6*x^6)

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Rubi [A]  time = 0.0601039, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {2454, 2395, 44} \[ \frac{b^2 p}{6 a^2 x^2}-\frac{b^3 p \log \left (a+b x^2\right )}{6 a^3}+\frac{b^3 p \log (x)}{3 a^3}-\frac{\log \left (c \left (a+b x^2\right )^p\right )}{6 x^6}-\frac{b p}{12 a x^4} \]

Antiderivative was successfully verified.

[In]

Int[Log[c*(a + b*x^2)^p]/x^7,x]

[Out]

-(b*p)/(12*a*x^4) + (b^2*p)/(6*a^2*x^2) + (b^3*p*Log[x])/(3*a^3) - (b^3*p*Log[a + b*x^2])/(6*a^3) - Log[c*(a +
b*x^2)^p]/(6*x^6)

Rule 2454

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])

Rule 2395

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 44

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] && L
tQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{\log \left (c \left (a+b x^2\right )^p\right )}{x^7} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{\log \left (c (a+b x)^p\right )}{x^4} \, dx,x,x^2\right )\\ &=-\frac{\log \left (c \left (a+b x^2\right )^p\right )}{6 x^6}+\frac{1}{6} (b p) \operatorname{Subst}\left (\int \frac{1}{x^3 (a+b x)} \, dx,x,x^2\right )\\ &=-\frac{\log \left (c \left (a+b x^2\right )^p\right )}{6 x^6}+\frac{1}{6} (b p) \operatorname{Subst}\left (\int \left (\frac{1}{a x^3}-\frac{b}{a^2 x^2}+\frac{b^2}{a^3 x}-\frac{b^3}{a^3 (a+b x)}\right ) \, dx,x,x^2\right )\\ &=-\frac{b p}{12 a x^4}+\frac{b^2 p}{6 a^2 x^2}+\frac{b^3 p \log (x)}{3 a^3}-\frac{b^3 p \log \left (a+b x^2\right )}{6 a^3}-\frac{\log \left (c \left (a+b x^2\right )^p\right )}{6 x^6}\\ \end{align*}

Mathematica [A]  time = 0.0648393, size = 68, normalized size = 0.87 \[ -\frac{\frac{b p x^2 \left (2 b^2 x^4 \log \left (a+b x^2\right )+a \left (a-2 b x^2\right )-4 b^2 x^4 \log (x)\right )}{a^3}+2 \log \left (c \left (a+b x^2\right )^p\right )}{12 x^6} \]

Antiderivative was successfully verified.

[In]

Integrate[Log[c*(a + b*x^2)^p]/x^7,x]

[Out]

-((b*p*x^2*(a*(a - 2*b*x^2) - 4*b^2*x^4*Log[x] + 2*b^2*x^4*Log[a + b*x^2]))/a^3 + 2*Log[c*(a + b*x^2)^p])/(12*
x^6)

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Maple [C]  time = 0.286, size = 206, normalized size = 2.6 \begin{align*} -{\frac{\ln \left ( \left ( b{x}^{2}+a \right ) ^{p} \right ) }{6\,{x}^{6}}}-{\frac{-4\,{b}^{3}p\ln \left ( x \right ){x}^{6}+2\,{b}^{3}p\ln \left ( b{x}^{2}+a \right ){x}^{6}+i\pi \,{a}^{3}{\it csgn} \left ( i \left ( b{x}^{2}+a \right ) ^{p} \right ) \left ({\it csgn} \left ( ic \left ( b{x}^{2}+a \right ) ^{p} \right ) \right ) ^{2}-i\pi \,{a}^{3}{\it csgn} \left ( i \left ( b{x}^{2}+a \right ) ^{p} \right ){\it csgn} \left ( ic \left ( b{x}^{2}+a \right ) ^{p} \right ){\it csgn} \left ( ic \right ) -i\pi \,{a}^{3} \left ({\it csgn} \left ( ic \left ( b{x}^{2}+a \right ) ^{p} \right ) \right ) ^{3}+i\pi \,{a}^{3} \left ({\it csgn} \left ( ic \left ( b{x}^{2}+a \right ) ^{p} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) -2\,a{b}^{2}p{x}^{4}+{a}^{2}bp{x}^{2}+2\,\ln \left ( c \right ){a}^{3}}{12\,{a}^{3}{x}^{6}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(c*(b*x^2+a)^p)/x^7,x)

[Out]

-1/6/x^6*ln((b*x^2+a)^p)-1/12*(-4*b^3*p*ln(x)*x^6+2*b^3*p*ln(b*x^2+a)*x^6+I*Pi*a^3*csgn(I*(b*x^2+a)^p)*csgn(I*
c*(b*x^2+a)^p)^2-I*Pi*a^3*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)*csgn(I*c)-I*Pi*a^3*csgn(I*c*(b*x^2+a)^p)^3
+I*Pi*a^3*csgn(I*c*(b*x^2+a)^p)^2*csgn(I*c)-2*a*b^2*p*x^4+a^2*b*p*x^2+2*ln(c)*a^3)/a^3/x^6

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Maxima [A]  time = 1.04477, size = 93, normalized size = 1.19 \begin{align*} -\frac{1}{12} \, b p{\left (\frac{2 \, b^{2} \log \left (b x^{2} + a\right )}{a^{3}} - \frac{2 \, b^{2} \log \left (x^{2}\right )}{a^{3}} - \frac{2 \, b x^{2} - a}{a^{2} x^{4}}\right )} - \frac{\log \left ({\left (b x^{2} + a\right )}^{p} c\right )}{6 \, x^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(b*x^2+a)^p)/x^7,x, algorithm="maxima")

[Out]

-1/12*b*p*(2*b^2*log(b*x^2 + a)/a^3 - 2*b^2*log(x^2)/a^3 - (2*b*x^2 - a)/(a^2*x^4)) - 1/6*log((b*x^2 + a)^p*c)
/x^6

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Fricas [A]  time = 2.06221, size = 163, normalized size = 2.09 \begin{align*} \frac{4 \, b^{3} p x^{6} \log \left (x\right ) + 2 \, a b^{2} p x^{4} - a^{2} b p x^{2} - 2 \, a^{3} \log \left (c\right ) - 2 \,{\left (b^{3} p x^{6} + a^{3} p\right )} \log \left (b x^{2} + a\right )}{12 \, a^{3} x^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(b*x^2+a)^p)/x^7,x, algorithm="fricas")

[Out]

1/12*(4*b^3*p*x^6*log(x) + 2*a*b^2*p*x^4 - a^2*b*p*x^2 - 2*a^3*log(c) - 2*(b^3*p*x^6 + a^3*p)*log(b*x^2 + a))/
(a^3*x^6)

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Sympy [A]  time = 54.4347, size = 116, normalized size = 1.49 \begin{align*} \begin{cases} - \frac{p \log{\left (a + b x^{2} \right )}}{6 x^{6}} - \frac{\log{\left (c \right )}}{6 x^{6}} - \frac{b p}{12 a x^{4}} + \frac{b^{2} p}{6 a^{2} x^{2}} + \frac{b^{3} p \log{\left (x \right )}}{3 a^{3}} - \frac{b^{3} p \log{\left (a + b x^{2} \right )}}{6 a^{3}} & \text{for}\: a \neq 0 \\- \frac{p \log{\left (b \right )}}{6 x^{6}} - \frac{p \log{\left (x \right )}}{3 x^{6}} - \frac{p}{18 x^{6}} - \frac{\log{\left (c \right )}}{6 x^{6}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(c*(b*x**2+a)**p)/x**7,x)

[Out]

Piecewise((-p*log(a + b*x**2)/(6*x**6) - log(c)/(6*x**6) - b*p/(12*a*x**4) + b**2*p/(6*a**2*x**2) + b**3*p*log
(x)/(3*a**3) - b**3*p*log(a + b*x**2)/(6*a**3), Ne(a, 0)), (-p*log(b)/(6*x**6) - p*log(x)/(3*x**6) - p/(18*x**
6) - log(c)/(6*x**6), True))

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Giac [B]  time = 1.17609, size = 258, normalized size = 3.31 \begin{align*} -\frac{\frac{2 \, b^{4} p \log \left (b x^{2} + a\right )}{{\left (b x^{2} + a\right )}^{3} - 3 \,{\left (b x^{2} + a\right )}^{2} a + 3 \,{\left (b x^{2} + a\right )} a^{2} - a^{3}} + \frac{2 \, b^{4} p \log \left (b x^{2} + a\right )}{a^{3}} - \frac{2 \, b^{4} p \log \left (b x^{2}\right )}{a^{3}} - \frac{2 \,{\left (b x^{2} + a\right )}^{2} b^{4} p - 5 \,{\left (b x^{2} + a\right )} a b^{4} p + 3 \, a^{2} b^{4} p - 2 \, a^{2} b^{4} \log \left (c\right )}{{\left (b x^{2} + a\right )}^{3} a^{2} - 3 \,{\left (b x^{2} + a\right )}^{2} a^{3} + 3 \,{\left (b x^{2} + a\right )} a^{4} - a^{5}}}{12 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(b*x^2+a)^p)/x^7,x, algorithm="giac")

[Out]

-1/12*(2*b^4*p*log(b*x^2 + a)/((b*x^2 + a)^3 - 3*(b*x^2 + a)^2*a + 3*(b*x^2 + a)*a^2 - a^3) + 2*b^4*p*log(b*x^
2 + a)/a^3 - 2*b^4*p*log(b*x^2)/a^3 - (2*(b*x^2 + a)^2*b^4*p - 5*(b*x^2 + a)*a*b^4*p + 3*a^2*b^4*p - 2*a^2*b^4
*log(c))/((b*x^2 + a)^3*a^2 - 3*(b*x^2 + a)^2*a^3 + 3*(b*x^2 + a)*a^4 - a^5))/b