Integrand size = 20, antiderivative size = 66 \[ \int \frac {f^{b x+c x^2}}{(b+2 c x)^3} \, dx=-\frac {f^{b x+c x^2}}{4 c (b+2 c x)^2}+\frac {f^{-\frac {b^2}{4 c}} \operatorname {ExpIntegralEi}\left (\frac {(b+2 c x)^2 \log (f)}{4 c}\right ) \log (f)}{16 c^2} \] Output:
-1/4*f^(c*x^2+b*x)/c/(2*c*x+b)^2+1/16*Ei(1/4*(2*c*x+b)^2*ln(f)/c)*ln(f)/c^ 2/(f^(1/4*b^2/c))
Time = 0.22 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.17 \[ \int \frac {f^{b x+c x^2}}{(b+2 c x)^3} \, dx=\frac {f^{-\frac {b^2}{4 c}} \left (-4 c f^{\frac {(b+2 c x)^2}{4 c}}+(b+2 c x)^2 \operatorname {ExpIntegralEi}\left (\frac {(b+2 c x)^2 \log (f)}{4 c}\right ) \log (f)\right )}{16 c^2 (b+2 c x)^2} \] Input:
Integrate[f^(b*x + c*x^2)/(b + 2*c*x)^3,x]
Output:
(-4*c*f^((b + 2*c*x)^2/(4*c)) + (b + 2*c*x)^2*ExpIntegralEi[((b + 2*c*x)^2 *Log[f])/(4*c)]*Log[f])/(16*c^2*f^(b^2/(4*c))*(b + 2*c*x)^2)
Time = 0.38 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2669, 2668}
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.
\(\displaystyle \int \frac {f^{b x+c x^2}}{(b+2 c x)^3} \, dx\) |
\(\Big \downarrow \) 2669 |
\(\displaystyle \frac {\log (f) \int \frac {f^{c x^2+b x}}{b+2 c x}dx}{4 c}-\frac {f^{b x+c x^2}}{4 c (b+2 c x)^2}\) |
\(\Big \downarrow \) 2668 |
\(\displaystyle \frac {\log (f) f^{-\frac {b^2}{4 c}} \operatorname {ExpIntegralEi}\left (\frac {(b+2 c x)^2 \log (f)}{4 c}\right )}{16 c^2}-\frac {f^{b x+c x^2}}{4 c (b+2 c x)^2}\) |
Input:
Int[f^(b*x + c*x^2)/(b + 2*c*x)^3,x]
Output:
-1/4*f^(b*x + c*x^2)/(c*(b + 2*c*x)^2) + (ExpIntegralEi[((b + 2*c*x)^2*Log [f])/(4*c)]*Log[f])/(16*c^2*f^(b^2/(4*c)))
Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)/((d_.) + (e_.)*(x_)), x_Symbol ] :> Simp[(1/(2*e))*F^(a - b^2/(4*c))*ExpIntegralEi[(b + 2*c*x)^2*(Log[F]/( 4*c))], x] /; FreeQ[{F, a, b, c, d, e}, x] && EqQ[b*e - 2*c*d, 0]
Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_.) + (e_.)*(x_))^(m_), x_S ymbol] :> Simp[(d + e*x)^(m + 1)*(F^(a + b*x + c*x^2)/(e*(m + 1))), x] - Si mp[2*c*(Log[F]/(e^2*(m + 1))) Int[(d + e*x)^(m + 2)*F^(a + b*x + c*x^2), x], x] /; FreeQ[{F, a, b, c, d, e}, x] && EqQ[b*e - 2*c*d, 0] && LtQ[m, -1]
Time = 0.12 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.12
method | result | size |
risch | \(-\frac {f^{\frac {\left (2 c x +b \right )^{2}}{4 c}} f^{-\frac {b^{2}}{4 c}}}{4 c \left (2 c x +b \right )^{2}}-\frac {\ln \left (f \right ) f^{-\frac {b^{2}}{4 c}} \operatorname {expIntegral}_{1}\left (-\frac {\left (2 c x +b \right )^{2} \ln \left (f \right )}{4 c}\right )}{16 c^{2}}\) | \(74\) |
Input:
int(f^(c*x^2+b*x)/(2*c*x+b)^3,x,method=_RETURNVERBOSE)
Output:
-1/4/c/(2*c*x+b)^2*f^(1/4*(2*c*x+b)^2/c)*f^(-1/4*b^2/c)-1/16/c^2*ln(f)*f^( -1/4*b^2/c)*Ei(1,-1/4*(2*c*x+b)^2*ln(f)/c)
Time = 0.08 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.52 \[ \int \frac {f^{b x+c x^2}}{(b+2 c x)^3} \, dx=-\frac {4 \, c f^{c x^{2} + b x} - \frac {{\left (4 \, c^{2} x^{2} + 4 \, b c x + b^{2}\right )} {\rm Ei}\left (\frac {{\left (4 \, c^{2} x^{2} + 4 \, b c x + b^{2}\right )} \log \left (f\right )}{4 \, c}\right ) \log \left (f\right )}{f^{\frac {b^{2}}{4 \, c}}}}{16 \, {\left (4 \, c^{4} x^{2} + 4 \, b c^{3} x + b^{2} c^{2}\right )}} \] Input:
integrate(f^(c*x^2+b*x)/(2*c*x+b)^3,x, algorithm="fricas")
Output:
-1/16*(4*c*f^(c*x^2 + b*x) - (4*c^2*x^2 + 4*b*c*x + b^2)*Ei(1/4*(4*c^2*x^2 + 4*b*c*x + b^2)*log(f)/c)*log(f)/f^(1/4*b^2/c))/(4*c^4*x^2 + 4*b*c^3*x + b^2*c^2)
\[ \int \frac {f^{b x+c x^2}}{(b+2 c x)^3} \, dx=\int \frac {f^{b x + c x^{2}}}{\left (b + 2 c x\right )^{3}}\, dx \] Input:
integrate(f**(c*x**2+b*x)/(2*c*x+b)**3,x)
Output:
Integral(f**(b*x + c*x**2)/(b + 2*c*x)**3, x)
\[ \int \frac {f^{b x+c x^2}}{(b+2 c x)^3} \, dx=\int { \frac {f^{c x^{2} + b x}}{{\left (2 \, c x + b\right )}^{3}} \,d x } \] Input:
integrate(f^(c*x^2+b*x)/(2*c*x+b)^3,x, algorithm="maxima")
Output:
integrate(f^(c*x^2 + b*x)/(2*c*x + b)^3, x)
\[ \int \frac {f^{b x+c x^2}}{(b+2 c x)^3} \, dx=\int { \frac {f^{c x^{2} + b x}}{{\left (2 \, c x + b\right )}^{3}} \,d x } \] Input:
integrate(f^(c*x^2+b*x)/(2*c*x+b)^3,x, algorithm="giac")
Output:
integrate(f^(c*x^2 + b*x)/(2*c*x + b)^3, x)
Timed out. \[ \int \frac {f^{b x+c x^2}}{(b+2 c x)^3} \, dx=\int \frac {f^{c\,x^2+b\,x}}{{\left (b+2\,c\,x\right )}^3} \,d x \] Input:
int(f^(b*x + c*x^2)/(b + 2*c*x)^3,x)
Output:
int(f^(b*x + c*x^2)/(b + 2*c*x)^3, x)
\[ \int \frac {f^{b x+c x^2}}{(b+2 c x)^3} \, dx=\text {too large to display} \] Input:
int(f^(c*x^2+b*x)/(2*c*x+b)^3,x)
Output:
( - f**(b*x + c*x**2) + 2*int(f**(b*x + c*x**2)/(log(f)*b**5 + 6*log(f)*b* *4*c*x + 12*log(f)*b**3*c**2*x**2 + 8*log(f)*b**2*c**3*x**3 - 4*b**3*c - 2 4*b**2*c**2*x - 48*b*c**3*x**2 - 32*c**4*x**3),x)*log(f)**2*b**6 + 8*int(f **(b*x + c*x**2)/(log(f)*b**5 + 6*log(f)*b**4*c*x + 12*log(f)*b**3*c**2*x* *2 + 8*log(f)*b**2*c**3*x**3 - 4*b**3*c - 24*b**2*c**2*x - 48*b*c**3*x**2 - 32*c**4*x**3),x)*log(f)**2*b**5*c*x + 8*int(f**(b*x + c*x**2)/(log(f)*b* *5 + 6*log(f)*b**4*c*x + 12*log(f)*b**3*c**2*x**2 + 8*log(f)*b**2*c**3*x** 3 - 4*b**3*c - 24*b**2*c**2*x - 48*b*c**3*x**2 - 32*c**4*x**3),x)*log(f)** 2*b**4*c**2*x**2 - 16*int(f**(b*x + c*x**2)/(log(f)*b**5 + 6*log(f)*b**4*c *x + 12*log(f)*b**3*c**2*x**2 + 8*log(f)*b**2*c**3*x**3 - 4*b**3*c - 24*b* *2*c**2*x - 48*b*c**3*x**2 - 32*c**4*x**3),x)*log(f)*b**4*c - 64*int(f**(b *x + c*x**2)/(log(f)*b**5 + 6*log(f)*b**4*c*x + 12*log(f)*b**3*c**2*x**2 + 8*log(f)*b**2*c**3*x**3 - 4*b**3*c - 24*b**2*c**2*x - 48*b*c**3*x**2 - 32 *c**4*x**3),x)*log(f)*b**3*c**2*x - 64*int(f**(b*x + c*x**2)/(log(f)*b**5 + 6*log(f)*b**4*c*x + 12*log(f)*b**3*c**2*x**2 + 8*log(f)*b**2*c**3*x**3 - 4*b**3*c - 24*b**2*c**2*x - 48*b*c**3*x**2 - 32*c**4*x**3),x)*log(f)*b**2 *c**3*x**2 + 32*int(f**(b*x + c*x**2)/(log(f)*b**5 + 6*log(f)*b**4*c*x + 1 2*log(f)*b**3*c**2*x**2 + 8*log(f)*b**2*c**3*x**3 - 4*b**3*c - 24*b**2*c** 2*x - 48*b*c**3*x**2 - 32*c**4*x**3),x)*b**2*c**2 + 128*int(f**(b*x + c*x* *2)/(log(f)*b**5 + 6*log(f)*b**4*c*x + 12*log(f)*b**3*c**2*x**2 + 8*log...