\(\int \frac {1}{\log ^3(c (a+b x^2)^p)} \, dx\) [121]

Optimal result

Integrand size = 14, antiderivative size = 14 \[ \int \frac {1}{\log ^3\left (c \left (a+b x^2\right )^p\right )} \, dx=\text {Int}\left (\frac {1}{\log ^3\left (c \left (a+b x^2\right )^p\right )},x\right ) \]

[Out]

Unintegrable(1/ln(c*(b*x^2+a)^p)^3,x)

Rubi [N/A]

Not integrable

Time = 0.00 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 0, number of rules used = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \[ \int \frac {1}{\log ^3\left (c \left (a+b x^2\right )^p\right )} \, dx=\int \frac {1}{\log ^3\left (c \left (a+b x^2\right )^p\right )} \, dx \]

[In]

Int[Log[c*(a + b*x^2)^p]^(-3),x]

[Out]

Defer[Int][Log[c*(a + b*x^2)^p]^(-3), x]

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

Mathematica [N/A]

Not integrable

Time = 0.23 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.14 \[ \int \frac {1}{\log ^3\left (c \left (a+b x^2\right )^p\right )} \, dx=\int \frac {1}{\log ^3\left (c \left (a+b x^2\right )^p\right )} \, dx \]

[In]

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

[Out]

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

Maple [N/A]

Not integrable

Time = 0.02 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00

\[\int \frac {1}{{\ln \left (c \left (b \,x^{2}+a \right )^{p}\right )}^{3}}d x\]

[In]

int(1/ln(c*(b*x^2+a)^p)^3,x)

[Out]

int(1/ln(c*(b*x^2+a)^p)^3,x)

Fricas [N/A]

Not integrable

Time = 0.30 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.14 \[ \int \frac {1}{\log ^3\left (c \left (a+b x^2\right )^p\right )} \, dx=\int { \frac {1}{\log \left ({\left (b x^{2} + a\right )}^{p} c\right )^{3}} \,d x } \]

[In]

integrate(1/log(c*(b*x^2+a)^p)^3,x, algorithm="fricas")

[Out]

integral(log((b*x^2 + a)^p*c)^(-3), x)

Sympy [N/A]

Not integrable

Time = 2.67 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.07 \[ \int \frac {1}{\log ^3\left (c \left (a+b x^2\right )^p\right )} \, dx=\int \frac {1}{\log {\left (c \left (a + b x^{2}\right )^{p} \right )}^{3}}\, dx \]

[In]

integrate(1/ln(c*(b*x**2+a)**p)**3,x)

[Out]

Integral(log(c*(a + b*x**2)**p)**(-3), x)

Maxima [N/A]

Not integrable

Time = 0.23 (sec) , antiderivative size = 164, normalized size of antiderivative = 11.71 \[ \int \frac {1}{\log ^3\left (c \left (a+b x^2\right )^p\right )} \, dx=\int { \frac {1}{\log \left ({\left (b x^{2} + a\right )}^{p} c\right )^{3}} \,d x } \]

[In]

integrate(1/log(c*(b*x^2+a)^p)^3,x, algorithm="maxima")

[Out]

-1/8*(b^2*(2*p + log(c))*x^4 + 2*a*b*p*x^2 - a^2*log(c) + (b^2*p*x^4 - a^2*p)*log(b*x^2 + a))/(b^2*p^4*x^3*log
(b*x^2 + a)^2 + 2*b^2*p^3*x^3*log(b*x^2 + a)*log(c) + b^2*p^2*x^3*log(c)^2) + integrate(1/8*(b^2*x^4 + 3*a^2)/
(b^2*p^3*x^4*log(b*x^2 + a) + b^2*p^2*x^4*log(c)), x)

Giac [N/A]

Not integrable

Time = 0.30 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.14 \[ \int \frac {1}{\log ^3\left (c \left (a+b x^2\right )^p\right )} \, dx=\int { \frac {1}{\log \left ({\left (b x^{2} + a\right )}^{p} c\right )^{3}} \,d x } \]

[In]

integrate(1/log(c*(b*x^2+a)^p)^3,x, algorithm="giac")

[Out]

integrate(log((b*x^2 + a)^p*c)^(-3), x)

Mupad [N/A]

Not integrable

Time = 1.22 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.14 \[ \int \frac {1}{\log ^3\left (c \left (a+b x^2\right )^p\right )} \, dx=\int \frac {1}{{\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}^3} \,d x \]

[In]

int(1/log(c*(a + b*x^2)^p)^3,x)

[Out]

int(1/log(c*(a + b*x^2)^p)^3, x)