Integrand size = 21, antiderivative size = 84 \[ \int \frac {f^{a+b x+c x^2}}{(b+2 c x)^2} \, dx=-\frac {f^{a+b x+c x^2}}{2 c (b+2 c x)}+\frac {f^{a-\frac {b^2}{4 c}} \sqrt {\pi } \text {erfi}\left (\frac {(b+2 c x) \sqrt {\log (f)}}{2 \sqrt {c}}\right ) \sqrt {\log (f)}}{4 c^{3/2}} \] Output:
-1/2*f^(c*x^2+b*x+a)/c/(2*c*x+b)+1/4*f^(a-1/4*b^2/c)*Pi^(1/2)*erfi(1/2*(2* c*x+b)*ln(f)^(1/2)/c^(1/2))*ln(f)^(1/2)/c^(3/2)
Time = 0.74 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.14 \[ \int \frac {f^{a+b x+c x^2}}{(b+2 c x)^2} \, dx=\frac {f^{a-\frac {b^2}{4 c}} \left (-2 \sqrt {c} f^{\frac {(b+2 c x)^2}{4 c}}+\sqrt {\pi } (b+2 c x) \text {erfi}\left (\frac {(b+2 c x) \sqrt {\log (f)}}{2 \sqrt {c}}\right ) \sqrt {\log (f)}\right )}{4 c^{3/2} (b+2 c x)} \] Input:
Integrate[f^(a + b*x + c*x^2)/(b + 2*c*x)^2,x]
Output:
(f^(a - b^2/(4*c))*(-2*Sqrt[c]*f^((b + 2*c*x)^2/(4*c)) + Sqrt[Pi]*(b + 2*c *x)*Erfi[((b + 2*c*x)*Sqrt[Log[f]])/(2*Sqrt[c])]*Sqrt[Log[f]]))/(4*c^(3/2) *(b + 2*c*x))
Time = 0.41 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2669, 2664, 2633}
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.
\(\displaystyle \int \frac {f^{a+b x+c x^2}}{(b+2 c x)^2} \, dx\) |
\(\Big \downarrow \) 2669 |
\(\displaystyle \frac {\log (f) \int f^{c x^2+b x+a}dx}{2 c}-\frac {f^{a+b x+c x^2}}{2 c (b+2 c x)}\) |
\(\Big \downarrow \) 2664 |
\(\displaystyle \frac {\log (f) f^{a-\frac {b^2}{4 c}} \int f^{\frac {(b+2 c x)^2}{4 c}}dx}{2 c}-\frac {f^{a+b x+c x^2}}{2 c (b+2 c x)}\) |
\(\Big \downarrow \) 2633 |
\(\displaystyle \frac {\sqrt {\pi } \sqrt {\log (f)} f^{a-\frac {b^2}{4 c}} \text {erfi}\left (\frac {\sqrt {\log (f)} (b+2 c x)}{2 \sqrt {c}}\right )}{4 c^{3/2}}-\frac {f^{a+b x+c x^2}}{2 c (b+2 c x)}\) |
Input:
Int[f^(a + b*x + c*x^2)/(b + 2*c*x)^2,x]
Output:
-1/2*f^(a + b*x + c*x^2)/(c*(b + 2*c*x)) + (f^(a - b^2/(4*c))*Sqrt[Pi]*Erf i[((b + 2*c*x)*Sqrt[Log[f]])/(2*Sqrt[c])]*Sqrt[Log[f]])/(4*c^(3/2))
Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt [Pi]*(Erfi[(c + d*x)*Rt[b*Log[F], 2]]/(2*d*Rt[b*Log[F], 2])), x] /; FreeQ[{ F, a, b, c, d}, x] && PosQ[b]
Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[F^(a - b^2/ (4*c)) Int[F^((b + 2*c*x)^2/(4*c)), x], x] /; FreeQ[{F, a, b, c}, x]
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 = 0.11 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.20
method | result | size |
risch | \(-\frac {f^{\frac {\left (2 c x +b \right )^{2}}{4 c}} f^{\frac {4 a c -b^{2}}{4 c}}}{2 c \left (2 c x +b \right )}+\frac {\ln \left (f \right ) \sqrt {\pi }\, f^{\frac {4 a c -b^{2}}{4 c}} \operatorname {erf}\left (\frac {\sqrt {-\frac {\ln \left (f \right )}{c}}\, \left (2 c x +b \right )}{2}\right )}{4 c^{2} \sqrt {-\frac {\ln \left (f \right )}{c}}}\) | \(101\) |
Input:
int(f^(c*x^2+b*x+a)/(2*c*x+b)^2,x,method=_RETURNVERBOSE)
Output:
-1/2/c/(2*c*x+b)*f^(1/4*(2*c*x+b)^2/c)*f^(1/4*(4*a*c-b^2)/c)+1/4/c^2*ln(f) *Pi^(1/2)*f^(1/4*(4*a*c-b^2)/c)/(-ln(f)/c)^(1/2)*erf(1/2*(-ln(f)/c)^(1/2)* (2*c*x+b))
Time = 0.07 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.01 \[ \int \frac {f^{a+b x+c x^2}}{(b+2 c x)^2} \, dx=-\frac {2 \, c f^{c x^{2} + b x + a} + \frac {\sqrt {\pi } {\left (2 \, c x + b\right )} \sqrt {-c \log \left (f\right )} \operatorname {erf}\left (\frac {{\left (2 \, c x + b\right )} \sqrt {-c \log \left (f\right )}}{2 \, c}\right )}{f^{\frac {b^{2} - 4 \, a c}{4 \, c}}}}{4 \, {\left (2 \, c^{3} x + b c^{2}\right )}} \] Input:
integrate(f^(c*x^2+b*x+a)/(2*c*x+b)^2,x, algorithm="fricas")
Output:
-1/4*(2*c*f^(c*x^2 + b*x + a) + sqrt(pi)*(2*c*x + b)*sqrt(-c*log(f))*erf(1 /2*(2*c*x + b)*sqrt(-c*log(f))/c)/f^(1/4*(b^2 - 4*a*c)/c))/(2*c^3*x + b*c^ 2)
\[ \int \frac {f^{a+b x+c x^2}}{(b+2 c x)^2} \, dx=\int \frac {f^{a + b x + c x^{2}}}{\left (b + 2 c x\right )^{2}}\, dx \] Input:
integrate(f**(c*x**2+b*x+a)/(2*c*x+b)**2,x)
Output:
Integral(f**(a + b*x + c*x**2)/(b + 2*c*x)**2, x)
\[ \int \frac {f^{a+b x+c x^2}}{(b+2 c x)^2} \, dx=\int { \frac {f^{c x^{2} + b x + a}}{{\left (2 \, c x + b\right )}^{2}} \,d x } \] Input:
integrate(f^(c*x^2+b*x+a)/(2*c*x+b)^2,x, algorithm="maxima")
Output:
integrate(f^(c*x^2 + b*x + a)/(2*c*x + b)^2, x)
\[ \int \frac {f^{a+b x+c x^2}}{(b+2 c x)^2} \, dx=\int { \frac {f^{c x^{2} + b x + a}}{{\left (2 \, c x + b\right )}^{2}} \,d x } \] Input:
integrate(f^(c*x^2+b*x+a)/(2*c*x+b)^2,x, algorithm="giac")
Output:
integrate(f^(c*x^2 + b*x + a)/(2*c*x + b)^2, x)
Time = 0.40 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.90 \[ \int \frac {f^{a+b x+c x^2}}{(b+2 c x)^2} \, dx=\frac {f^{a-\frac {b^2}{4\,c}}\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {\frac {b\,\ln \left (f\right )}{2}+c\,x\,\ln \left (f\right )}{\sqrt {c\,\ln \left (f\right )}}\right )\,\ln \left (f\right )}{4\,c\,\sqrt {c\,\ln \left (f\right )}}-\frac {f^a\,f^{c\,x^2}\,f^{b\,x}}{2\,c\,\left (b+2\,c\,x\right )} \] Input:
int(f^(a + b*x + c*x^2)/(b + 2*c*x)^2,x)
Output:
(f^(a - b^2/(4*c))*pi^(1/2)*erfi(((b*log(f))/2 + c*x*log(f))/(c*log(f))^(1 /2))*log(f))/(4*c*(c*log(f))^(1/2)) - (f^a*f^(c*x^2)*f^(b*x))/(2*c*(b + 2* c*x))
\[ \int \frac {f^{a+b x+c x^2}}{(b+2 c x)^2} \, dx=f^{a} \left (\int \frac {f^{c \,x^{2}+b x}}{4 c^{2} x^{2}+4 b c x +b^{2}}d x \right ) \] Input:
int(f^(c*x^2+b*x+a)/(2*c*x+b)^2,x)
Output:
f**a*int(f**(b*x + c*x**2)/(b**2 + 4*b*c*x + 4*c**2*x**2),x)