Integrand size = 15, antiderivative size = 68 \[ \int \frac {f^{\frac {c}{a+b x}}}{x^2} \, dx=-\frac {b f^{\frac {c}{a+b x}}}{a}-\frac {f^{\frac {c}{a+b x}}}{x}-\frac {b c f^{\frac {c}{a}} \operatorname {ExpIntegralEi}\left (-\frac {b c x \log (f)}{a (a+b x)}\right ) \log (f)}{a^2} \] Output:
-b*f^(c/(b*x+a))/a-f^(c/(b*x+a))/x-b*c*f^(c/a)*Ei(-b*c*x*ln(f)/a/(b*x+a))* ln(f)/a^2
Time = 0.12 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00 \[ \int \frac {f^{\frac {c}{a+b x}}}{x^2} \, dx=-\frac {b f^{\frac {c}{a+b x}}}{a}-\frac {f^{\frac {c}{a+b x}}}{x}-\frac {b c f^{\frac {c}{a}} \operatorname {ExpIntegralEi}\left (-\frac {b c x \log (f)}{a^2+a b x}\right ) \log (f)}{a^2} \] Input:
Integrate[f^(c/(a + b*x))/x^2,x]
Output:
-((b*f^(c/(a + b*x)))/a) - f^(c/(a + b*x))/x - (b*c*f^(c/a)*ExpIntegralEi[ -((b*c*x*Log[f])/(a^2 + a*b*x))]*Log[f])/a^2
Time = 0.94 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.10, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2653, 7293, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {f^{\frac {c}{a+b x}}}{x^2} \, dx\) |
\(\Big \downarrow \) 2653 |
\(\displaystyle -b c \log (f) \int \frac {f^{\frac {c}{a+b x}}}{x (a+b x)^2}dx-\frac {f^{\frac {c}{a+b x}}}{x}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -b c \log (f) \int \left (\frac {f^{\frac {c}{a+b x}}}{a^2 x}-\frac {b f^{\frac {c}{a+b x}}}{a^2 (a+b x)}-\frac {b f^{\frac {c}{a+b x}}}{a (a+b x)^2}\right )dx-\frac {f^{\frac {c}{a+b x}}}{x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -b c \log (f) \left (\frac {f^{\frac {c}{a}} \operatorname {ExpIntegralEi}\left (-\frac {b c x \log (f)}{a (a+b x)}\right )}{a^2}+\frac {f^{\frac {c}{a+b x}}}{a c \log (f)}\right )-\frac {f^{\frac {c}{a+b x}}}{x}\) |
Input:
Int[f^(c/(a + b*x))/x^2,x]
Output:
-(f^(c/(a + b*x))/x) - b*c*((f^(c/a)*ExpIntegralEi[-((b*c*x*Log[f])/(a*(a + b*x)))])/a^2 + f^(c/(a + b*x))/(a*c*Log[f]))*Log[f]
Int[(F_)^((a_.) + (b_.)/((c_.) + (d_.)*(x_)))*((e_.) + (f_.)*(x_))^(m_), x_ Symbol] :> Simp[(e + f*x)^(m + 1)*(F^(a + b/(c + d*x))/(f*(m + 1))), x] + S imp[b*d*(Log[F]/(f*(m + 1))) Int[(e + f*x)^(m + 1)*(F^(a + b/(c + d*x))/( c + d*x)^2), x], x] /; FreeQ[{F, a, b, c, d, e, f}, x] && NeQ[d*e - c*f, 0] && ILtQ[m, -1]
Time = 0.12 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.18
method | result | size |
risch | \(\frac {\ln \left (f \right ) b c \,f^{\frac {c}{b x +a}}}{a^{2} \left (\frac {c \ln \left (f \right )}{b x +a}-\frac {c \ln \left (f \right )}{a}\right )}+\frac {\ln \left (f \right ) b c \,f^{\frac {c}{a}} \operatorname {expIntegral}_{1}\left (-\frac {c \ln \left (f \right )}{b x +a}+\frac {c \ln \left (f \right )}{a}\right )}{a^{2}}\) | \(80\) |
Input:
int(f^(c/(b*x+a))/x^2,x,method=_RETURNVERBOSE)
Output:
1/a^2*ln(f)*b*c*f^(c/(b*x+a))/(c*ln(f)/(b*x+a)-c*ln(f)/a)+1/a^2*ln(f)*b*c* f^(1/a*c)*Ei(1,-c*ln(f)/(b*x+a)+c*ln(f)/a)
Time = 0.09 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.88 \[ \int \frac {f^{\frac {c}{a+b x}}}{x^2} \, dx=-\frac {b c f^{\frac {c}{a}} x {\rm Ei}\left (-\frac {b c x \log \left (f\right )}{a b x + a^{2}}\right ) \log \left (f\right ) + {\left (a b x + a^{2}\right )} f^{\frac {c}{b x + a}}}{a^{2} x} \] Input:
integrate(f^(c/(b*x+a))/x^2,x, algorithm="fricas")
Output:
-(b*c*f^(c/a)*x*Ei(-b*c*x*log(f)/(a*b*x + a^2))*log(f) + (a*b*x + a^2)*f^( c/(b*x + a)))/(a^2*x)
\[ \int \frac {f^{\frac {c}{a+b x}}}{x^2} \, dx=\int \frac {f^{\frac {c}{a + b x}}}{x^{2}}\, dx \] Input:
integrate(f**(c/(b*x+a))/x**2,x)
Output:
Integral(f**(c/(a + b*x))/x**2, x)
\[ \int \frac {f^{\frac {c}{a+b x}}}{x^2} \, dx=\int { \frac {f^{\frac {c}{b x + a}}}{x^{2}} \,d x } \] Input:
integrate(f^(c/(b*x+a))/x^2,x, algorithm="maxima")
Output:
integrate(f^(c/(b*x + a))/x^2, x)
\[ \int \frac {f^{\frac {c}{a+b x}}}{x^2} \, dx=\int { \frac {f^{\frac {c}{b x + a}}}{x^{2}} \,d x } \] Input:
integrate(f^(c/(b*x+a))/x^2,x, algorithm="giac")
Output:
integrate(f^(c/(b*x + a))/x^2, x)
Timed out. \[ \int \frac {f^{\frac {c}{a+b x}}}{x^2} \, dx=\int \frac {f^{\frac {c}{a+b\,x}}}{x^2} \,d x \] Input:
int(f^(c/(a + b*x))/x^2,x)
Output:
int(f^(c/(a + b*x))/x^2, x)
\[ \int \frac {f^{\frac {c}{a+b x}}}{x^2} \, dx =\text {Too large to display} \] Input:
int(f^(c/(b*x+a))/x^2,x)
Output:
( - 4*f**(c/(a + b*x))*log(f)**2*a**3*c**2 - 6*f**(c/(a + b*x))*log(f)**2* a**2*b*c**2*x - 4*f**(c/(a + b*x))*log(f)**2*a*b**2*c**2*x**2 - f**(c/(a + b*x))*log(f)**2*b**3*c**2*x**3 - 4*f**(c/(a + b*x))*log(f)*a**3*b*c*x - 6 *f**(c/(a + b*x))*log(f)*a**2*b**2*c*x**2 - 2*f**(c/(a + b*x))*log(f)*a*b* *3*c*x**3 + 2*f**(c/(a + b*x))*a**4*b*x + 4*f**(c/(a + b*x))*a**3*b**2*x** 2 + 2*f**(c/(a + b*x))*a**2*b**3*x**3 + int(f**(c/(a + b*x))/(log(f)*a**4* c*x**2 + 4*log(f)*a**3*b*c*x**3 + 6*log(f)*a**2*b**2*c*x**4 + 4*log(f)*a*b **3*c*x**5 + log(f)*b**4*c*x**6 + 4*a**5*x**2 + 16*a**4*b*x**3 + 24*a**3*b **2*x**4 + 16*a**2*b**3*x**5 + 4*a*b**4*x**6),x)*log(f)**4*a**6*c**4*x + 2 *int(f**(c/(a + b*x))/(log(f)*a**4*c*x**2 + 4*log(f)*a**3*b*c*x**3 + 6*log (f)*a**2*b**2*c*x**4 + 4*log(f)*a*b**3*c*x**5 + log(f)*b**4*c*x**6 + 4*a** 5*x**2 + 16*a**4*b*x**3 + 24*a**3*b**2*x**4 + 16*a**2*b**3*x**5 + 4*a*b**4 *x**6),x)*log(f)**4*a**5*b*c**4*x**2 + int(f**(c/(a + b*x))/(log(f)*a**4*c *x**2 + 4*log(f)*a**3*b*c*x**3 + 6*log(f)*a**2*b**2*c*x**4 + 4*log(f)*a*b* *3*c*x**5 + log(f)*b**4*c*x**6 + 4*a**5*x**2 + 16*a**4*b*x**3 + 24*a**3*b* *2*x**4 + 16*a**2*b**3*x**5 + 4*a*b**4*x**6),x)*log(f)**4*a**4*b**2*c**4*x **3 + 4*int(f**(c/(a + b*x))/(log(f)*a**4*c*x**2 + 4*log(f)*a**3*b*c*x**3 + 6*log(f)*a**2*b**2*c*x**4 + 4*log(f)*a*b**3*c*x**5 + log(f)*b**4*c*x**6 + 4*a**5*x**2 + 16*a**4*b*x**3 + 24*a**3*b**2*x**4 + 16*a**2*b**3*x**5 + 4 *a*b**4*x**6),x)*log(f)**3*a**7*c**3*x + 8*int(f**(c/(a + b*x))/(log(f)...