3.11 \(\int \frac{\log (c (a+b x^2)^p)}{x^6} \, dx\)

Optimal. Leaf size=74 \[ \frac{2 b^2 p}{5 a^2 x}+\frac{2 b^{5/2} p \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{5 a^{5/2}}-\frac{\log \left (c \left (a+b x^2\right )^p\right )}{5 x^5}-\frac{2 b p}{15 a x^3} \]

[Out]

(-2*b*p)/(15*a*x^3) + (2*b^2*p)/(5*a^2*x) + (2*b^(5/2)*p*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(5*a^(5/2)) - Log[c*(a +
b*x^2)^p]/(5*x^5)

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Rubi [A]  time = 0.0365117, antiderivative size = 74, 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 = {2455, 325, 205} \[ \frac{2 b^2 p}{5 a^2 x}+\frac{2 b^{5/2} p \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{5 a^{5/2}}-\frac{\log \left (c \left (a+b x^2\right )^p\right )}{5 x^5}-\frac{2 b p}{15 a x^3} \]

Antiderivative was successfully verified.

[In]

Int[Log[c*(a + b*x^2)^p]/x^6,x]

[Out]

(-2*b*p)/(15*a*x^3) + (2*b^2*p)/(5*a^2*x) + (2*b^(5/2)*p*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(5*a^(5/2)) - Log[c*(a +
b*x^2)^p]/(5*x^5)

Rule 2455

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

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{\log \left (c \left (a+b x^2\right )^p\right )}{x^6} \, dx &=-\frac{\log \left (c \left (a+b x^2\right )^p\right )}{5 x^5}+\frac{1}{5} (2 b p) \int \frac{1}{x^4 \left (a+b x^2\right )} \, dx\\ &=-\frac{2 b p}{15 a x^3}-\frac{\log \left (c \left (a+b x^2\right )^p\right )}{5 x^5}-\frac{\left (2 b^2 p\right ) \int \frac{1}{x^2 \left (a+b x^2\right )} \, dx}{5 a}\\ &=-\frac{2 b p}{15 a x^3}+\frac{2 b^2 p}{5 a^2 x}-\frac{\log \left (c \left (a+b x^2\right )^p\right )}{5 x^5}+\frac{\left (2 b^3 p\right ) \int \frac{1}{a+b x^2} \, dx}{5 a^2}\\ &=-\frac{2 b p}{15 a x^3}+\frac{2 b^2 p}{5 a^2 x}+\frac{2 b^{5/2} p \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{5 a^{5/2}}-\frac{\log \left (c \left (a+b x^2\right )^p\right )}{5 x^5}\\ \end{align*}

Mathematica [C]  time = 0.0027562, size = 49, normalized size = 0.66 \[ -\frac{\log \left (c \left (a+b x^2\right )^p\right )}{5 x^5}-\frac{2 b p \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};-\frac{b x^2}{a}\right )}{15 a x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Log[c*(a + b*x^2)^p]/x^6,x]

[Out]

(-2*b*p*Hypergeometric2F1[-3/2, 1, -1/2, -((b*x^2)/a)])/(15*a*x^3) - Log[c*(a + b*x^2)^p]/(5*x^5)

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Maple [C]  time = 0.362, size = 235, normalized size = 3.2 \begin{align*} -{\frac{\ln \left ( \left ( b{x}^{2}+a \right ) ^{p} \right ) }{5\,{x}^{5}}}-{\frac{1}{30\,{a}^{3}{x}^{5}} \left ( -6\,\sqrt{-ab}p{b}^{2}\ln \left ( -bx-\sqrt{-ab} \right ){x}^{5}+6\,\sqrt{-ab}p{b}^{2}\ln \left ( -bx+\sqrt{-ab} \right ){x}^{5}+3\,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}-3\,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 ) -3\,i\pi \,{a}^{3} \left ({\it csgn} \left ( ic \left ( b{x}^{2}+a \right ) ^{p} \right ) \right ) ^{3}+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 ) -12\,a{b}^{2}p{x}^{4}+4\,{a}^{2}bp{x}^{2}+6\,\ln \left ( c \right ){a}^{3} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(c*(b*x^2+a)^p)/x^6,x)

[Out]

-1/5/x^5*ln((b*x^2+a)^p)-1/30*(-6*(-a*b)^(1/2)*p*b^2*ln(-b*x-(-a*b)^(1/2))*x^5+6*(-a*b)^(1/2)*p*b^2*ln(-b*x+(-
a*b)^(1/2))*x^5+3*I*Pi*a^3*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)^2-3*I*Pi*a^3*csgn(I*(b*x^2+a)^p)*csgn(I*c
*(b*x^2+a)^p)*csgn(I*c)-3*I*Pi*a^3*csgn(I*c*(b*x^2+a)^p)^3+3*I*Pi*a^3*csgn(I*c*(b*x^2+a)^p)^2*csgn(I*c)-12*a*b
^2*p*x^4+4*a^2*b*p*x^2+6*ln(c)*a^3)/a^3/x^5

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(b*x^2+a)^p)/x^6,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.07726, size = 390, normalized size = 5.27 \begin{align*} \left [\frac{3 \, b^{2} p x^{5} \sqrt{-\frac{b}{a}} \log \left (\frac{b x^{2} + 2 \, a x \sqrt{-\frac{b}{a}} - a}{b x^{2} + a}\right ) + 6 \, b^{2} p x^{4} - 2 \, a b p x^{2} - 3 \, a^{2} p \log \left (b x^{2} + a\right ) - 3 \, a^{2} \log \left (c\right )}{15 \, a^{2} x^{5}}, \frac{6 \, b^{2} p x^{5} \sqrt{\frac{b}{a}} \arctan \left (x \sqrt{\frac{b}{a}}\right ) + 6 \, b^{2} p x^{4} - 2 \, a b p x^{2} - 3 \, a^{2} p \log \left (b x^{2} + a\right ) - 3 \, a^{2} \log \left (c\right )}{15 \, a^{2} x^{5}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(b*x^2+a)^p)/x^6,x, algorithm="fricas")

[Out]

[1/15*(3*b^2*p*x^5*sqrt(-b/a)*log((b*x^2 + 2*a*x*sqrt(-b/a) - a)/(b*x^2 + a)) + 6*b^2*p*x^4 - 2*a*b*p*x^2 - 3*
a^2*p*log(b*x^2 + a) - 3*a^2*log(c))/(a^2*x^5), 1/15*(6*b^2*p*x^5*sqrt(b/a)*arctan(x*sqrt(b/a)) + 6*b^2*p*x^4
- 2*a*b*p*x^2 - 3*a^2*p*log(b*x^2 + a) - 3*a^2*log(c))/(a^2*x^5)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(c*(b*x**2+a)**p)/x**6,x)

[Out]

Timed out

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Giac [A]  time = 1.28366, size = 96, normalized size = 1.3 \begin{align*} \frac{2 \, b^{3} p \arctan \left (\frac{b x}{\sqrt{a b}}\right )}{5 \, \sqrt{a b} a^{2}} - \frac{p \log \left (b x^{2} + a\right )}{5 \, x^{5}} + \frac{6 \, b^{2} p x^{4} - 2 \, a b p x^{2} - 3 \, a^{2} \log \left (c\right )}{15 \, a^{2} x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(b*x^2+a)^p)/x^6,x, algorithm="giac")

[Out]

2/5*b^3*p*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^2) - 1/5*p*log(b*x^2 + a)/x^5 + 1/15*(6*b^2*p*x^4 - 2*a*b*p*x^2 -
3*a^2*log(c))/(a^2*x^5)