3.9 \(\int \frac{\log (c (a+b x^2)^p)}{x^4} \, dx\)

Optimal. Leaf size=60 \[ -\frac{2 b^{3/2} p \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{3 a^{3/2}}-\frac{\log \left (c \left (a+b x^2\right )^p\right )}{3 x^3}-\frac{2 b p}{3 a x} \]

[Out]

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

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Rubi [A]  time = 0.0295252, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {2455, 325, 205} \[ -\frac{2 b^{3/2} p \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{3 a^{3/2}}-\frac{\log \left (c \left (a+b x^2\right )^p\right )}{3 x^3}-\frac{2 b p}{3 a x} \]

Antiderivative was successfully verified.

[In]

Int[Log[c*(a + b*x^2)^p]/x^4,x]

[Out]

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

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

Mathematica [C]  time = 0.0030129, size = 49, normalized size = 0.82 \[ -\frac{\log \left (c \left (a+b x^2\right )^p\right )}{3 x^3}-\frac{2 b p \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};-\frac{b x^2}{a}\right )}{3 a x} \]

Antiderivative was successfully verified.

[In]

Integrate[Log[c*(a + b*x^2)^p]/x^4,x]

[Out]

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

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Maple [C]  time = 0.34, size = 211, normalized size = 3.5 \begin{align*} -{\frac{\ln \left ( \left ( b{x}^{2}+a \right ) ^{p} \right ) }{3\,{x}^{3}}}-{\frac{i\pi \,a{\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{\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 \left ({\it csgn} \left ( ic \left ( b{x}^{2}+a \right ) ^{p} \right ) \right ) ^{3}+i\pi \,a \left ({\it csgn} \left ( ic \left ( b{x}^{2}+a \right ) ^{p} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) -2\,\sum _{{\it \_R}={\it RootOf} \left ({a}^{3}{{\it \_Z}}^{2}+{b}^{3}{p}^{2} \right ) }{\it \_R}\,\ln \left ( \left ( 3\,{a}^{3}{{\it \_R}}^{2}+2\,{b}^{3}{p}^{2} \right ) x+{a}^{2}bp{\it \_R} \right ) a{x}^{3}+4\,{x}^{2}pb+2\,\ln \left ( c \right ) a}{6\,a{x}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(c*(b*x^2+a)^p)/x^4,x)

[Out]

-1/3/x^3*ln((b*x^2+a)^p)-1/6*(I*Pi*a*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)^2-I*Pi*a*csgn(I*(b*x^2+a)^p)*cs
gn(I*c*(b*x^2+a)^p)*csgn(I*c)-I*Pi*a*csgn(I*c*(b*x^2+a)^p)^3+I*Pi*a*csgn(I*c*(b*x^2+a)^p)^2*csgn(I*c)-2*sum(_R
*ln((3*_R^2*a^3+2*b^3*p^2)*x+a^2*b*p*_R),_R=RootOf(_Z^2*a^3+b^3*p^2))*a*x^3+4*x^2*p*b+2*ln(c)*a)/a/x^3

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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^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.85948, size = 311, normalized size = 5.18 \begin{align*} \left [\frac{b p x^{3} \sqrt{-\frac{b}{a}} \log \left (\frac{b x^{2} - 2 \, a x \sqrt{-\frac{b}{a}} - a}{b x^{2} + a}\right ) - 2 \, b p x^{2} - a p \log \left (b x^{2} + a\right ) - a \log \left (c\right )}{3 \, a x^{3}}, -\frac{2 \, b p x^{3} \sqrt{\frac{b}{a}} \arctan \left (x \sqrt{\frac{b}{a}}\right ) + 2 \, b p x^{2} + a p \log \left (b x^{2} + a\right ) + a \log \left (c\right )}{3 \, a x^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(b*x^2+a)^p)/x^4,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 178.799, size = 774, normalized size = 12.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(c*(b*x**2+a)**p)/x**4,x)

[Out]

Piecewise((-log(0**p*c)/(3*x**3), Eq(a, 0) & Eq(b, 0)), (-p*log(b)/(3*x**3) - 2*p*log(x)/(3*x**3) - 2*p/(9*x**
3) - log(c)/(3*x**3), Eq(a, 0)), (-log(a**p*c)/(3*x**3), Eq(b, 0)), (-I*a**(7/2)*p*sqrt(1/b)*log(a + b*x**2)/(
3*I*a**(7/2)*x**3*sqrt(1/b) + 3*I*a**(5/2)*b*x**5*sqrt(1/b)) - I*a**(7/2)*sqrt(1/b)*log(c)/(3*I*a**(7/2)*x**3*
sqrt(1/b) + 3*I*a**(5/2)*b*x**5*sqrt(1/b)) - I*a**(5/2)*p*x**2*sqrt(1/b)*log(a + b*x**2)/(3*I*a**(7/2)*x**3*sq
rt(1/b)/b + 3*I*a**(5/2)*x**5*sqrt(1/b)) - 2*I*a**(5/2)*p*x**2*sqrt(1/b)/(3*I*a**(7/2)*x**3*sqrt(1/b)/b + 3*I*
a**(5/2)*x**5*sqrt(1/b)) - I*a**(5/2)*x**2*sqrt(1/b)*log(c)/(3*I*a**(7/2)*x**3*sqrt(1/b)/b + 3*I*a**(5/2)*x**5
*sqrt(1/b)) - 2*I*a**(3/2)*b*p*x**4*sqrt(1/b)/(3*I*a**(7/2)*x**3*sqrt(1/b)/b + 3*I*a**(5/2)*x**5*sqrt(1/b)) +
a**2*p*x**3*log(a + b*x**2)/(3*I*a**(7/2)*x**3*sqrt(1/b)/b + 3*I*a**(5/2)*x**5*sqrt(1/b)) - 2*a**2*p*x**3*log(
-I*sqrt(a)*sqrt(1/b) + x)/(3*I*a**(7/2)*x**3*sqrt(1/b)/b + 3*I*a**(5/2)*x**5*sqrt(1/b)) + a**2*x**3*log(c)/(3*
I*a**(7/2)*x**3*sqrt(1/b)/b + 3*I*a**(5/2)*x**5*sqrt(1/b)) + a*b*p*x**5*log(a + b*x**2)/(3*I*a**(7/2)*x**3*sqr
t(1/b)/b + 3*I*a**(5/2)*x**5*sqrt(1/b)) - 2*a*b*p*x**5*log(-I*sqrt(a)*sqrt(1/b) + x)/(3*I*a**(7/2)*x**3*sqrt(1
/b)/b + 3*I*a**(5/2)*x**5*sqrt(1/b)) + a*b*x**5*log(c)/(3*I*a**(7/2)*x**3*sqrt(1/b)/b + 3*I*a**(5/2)*x**5*sqrt
(1/b)), True))

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Giac [A]  time = 1.228, size = 78, normalized size = 1.3 \begin{align*} -\frac{2 \, b^{2} p \arctan \left (\frac{b x}{\sqrt{a b}}\right )}{3 \, \sqrt{a b} a} - \frac{p \log \left (b x^{2} + a\right )}{3 \, x^{3}} - \frac{2 \, b p x^{2} + a \log \left (c\right )}{3 \, a x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(b*x^2+a)^p)/x^4,x, algorithm="giac")

[Out]

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