3.61 \(\int \frac {\log (e (f (a+b x)^p (c+d x)^q)^r)}{x \log ^2(i (j (h x)^t)^u)} \, dx\)

Optimal. Leaf size=42 \[ \text {Int}\left (\frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{x \log ^2\left (i \left (j (h x)^t\right )^u\right )},x\right ) \]

[Out]

CannotIntegrate(ln(e*(f*(b*x+a)^p*(d*x+c)^q)^r)/x/ln(i*(j*(h*x)^t)^u)^2,x)

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Rubi [A]  time = 0.43, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \[ \int \frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{x \log ^2\left (i \left (j (h x)^t\right )^u\right )} \, dx \]

Verification is Not applicable to the result.

[In]

Int[Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]/(x*Log[i*(j*(h*x)^t)^u]^2),x]

[Out]

Defer[Int][Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]/(x*Log[i*(j*(h*x)^t)^u]^2), x]

Rubi steps

\begin {align*} \int \frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{x \log ^2\left (61 \left (j (h x)^t\right )^u\right )} \, dx &=\int \frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{x \log ^2\left (61 \left (j (h x)^t\right )^u\right )} \, dx\\ \end {align*}

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Mathematica [A]  time = 2.55, size = 0, normalized size = 0.00 \[ \int \frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{x \log ^2\left (i \left (j (h x)^t\right )^u\right )} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]/(x*Log[i*(j*(h*x)^t)^u]^2),x]

[Out]

Integrate[Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]/(x*Log[i*(j*(h*x)^t)^u]^2), x]

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fricas [A]  time = 0.97, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\log \left (\left ({\left (b x + a\right )}^{p} {\left (d x + c\right )}^{q} f\right )^{r} e\right )}{x \log \left (\left (\left (h x\right )^{t} j\right )^{u} i\right )^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(e*(f*(b*x+a)^p*(d*x+c)^q)^r)/x/log(i*(j*(h*x)^t)^u)^2,x, algorithm="fricas")

[Out]

integral(log(((b*x + a)^p*(d*x + c)^q*f)^r*e)/(x*log(((h*x)^t*j)^u*i)^2), x)

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giac [A]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\log \left (\left ({\left (b x + a\right )}^{p} {\left (d x + c\right )}^{q} f\right )^{r} e\right )}{x \log \left (\left (\left (h x\right )^{t} j\right )^{u} i\right )^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(e*(f*(b*x+a)^p*(d*x+c)^q)^r)/x/log(i*(j*(h*x)^t)^u)^2,x, algorithm="giac")

[Out]

integrate(log(((b*x + a)^p*(d*x + c)^q*f)^r*e)/(x*log(((h*x)^t*j)^u*i)^2), x)

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maple [A]  time = 1.72, size = 0, normalized size = 0.00 \[ \int \frac {\ln \left (e \left (f \left (b x +a \right )^{p} \left (d x +c \right )^{q}\right )^{r}\right )}{x \ln \left (i \left (j \left (h x \right )^{t}\right )^{u}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(e*(f*(b*x+a)^p*(d*x+c)^q)^r)/x/ln(i*(j*(h*x)^t)^u)^2,x)

[Out]

int(ln(e*(f*(b*x+a)^p*(d*x+c)^q)^r)/x/ln(i*(j*(h*x)^t)^u)^2,x)

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maxima [A]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {r \log \relax (f) + \log \left ({\left ({\left (b x + a\right )}^{p}\right )}^{r}\right ) + \log \left ({\left ({\left (d x + c\right )}^{q}\right )}^{r}\right ) + \log \relax (e)}{t^{2} u^{2} \log \relax (h) + t u^{2} \log \relax (j) + t u \log \relax (i) + t u \log \left ({\left (x^{t}\right )}^{u}\right )} + \int \frac {b c p r + a d q r + {\left (p r + q r\right )} b d x}{{\left (t^{2} u^{2} \log \relax (h) + t u^{2} \log \relax (j) + t u \log \relax (i)\right )} b d x^{2} + {\left (t^{2} u^{2} \log \relax (h) + t u^{2} \log \relax (j) + t u \log \relax (i)\right )} a c + {\left ({\left (t^{2} u^{2} \log \relax (h) + t u^{2} \log \relax (j) + t u \log \relax (i)\right )} b c + {\left (t^{2} u^{2} \log \relax (h) + t u^{2} \log \relax (j) + t u \log \relax (i)\right )} a d\right )} x + {\left (b d t u x^{2} + a c t u + {\left (b c t u + a d t u\right )} x\right )} \log \left ({\left (x^{t}\right )}^{u}\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(e*(f*(b*x+a)^p*(d*x+c)^q)^r)/x/log(i*(j*(h*x)^t)^u)^2,x, algorithm="maxima")

[Out]

-(r*log(f) + log(((b*x + a)^p)^r) + log(((d*x + c)^q)^r) + log(e))/(t^2*u^2*log(h) + t*u^2*log(j) + t*u*log(i)
+ t*u*log((x^t)^u)) + integrate((b*c*p*r + a*d*q*r + (p*r + q*r)*b*d*x)/((t^2*u^2*log(h) + t*u^2*log(j) + t*u
*log(i))*b*d*x^2 + (t^2*u^2*log(h) + t*u^2*log(j) + t*u*log(i))*a*c + ((t^2*u^2*log(h) + t*u^2*log(j) + t*u*lo
g(i))*b*c + (t^2*u^2*log(h) + t*u^2*log(j) + t*u*log(i))*a*d)*x + (b*d*t*u*x^2 + a*c*t*u + (b*c*t*u + a*d*t*u)
*x)*log((x^t)^u)), x)

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mupad [A]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\ln \left (e\,{\left (f\,{\left (a+b\,x\right )}^p\,{\left (c+d\,x\right )}^q\right )}^r\right )}{x\,{\ln \left (i\,{\left (j\,{\left (h\,x\right )}^t\right )}^u\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(e*(f*(a + b*x)^p*(c + d*x)^q)^r)/(x*log(i*(j*(h*x)^t)^u)^2),x)

[Out]

int(log(e*(f*(a + b*x)^p*(c + d*x)^q)^r)/(x*log(i*(j*(h*x)^t)^u)^2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(e*(f*(b*x+a)**p*(d*x+c)**q)**r)/x/ln(i*(j*(h*x)**t)**u)**2,x)

[Out]

Timed out

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