Integrand size = 23, antiderivative size = 147 \[ \int \frac {d+e \left (F^{c (a+b x)}\right )^n}{(f+g x)^3} \, dx=-\frac {d}{2 g (f+g x)^2}-\frac {e \left (F^{a c+b c x}\right )^n}{2 g (f+g x)^2}-\frac {b c e \left (F^{a c+b c x}\right )^n n \log (F)}{2 g^2 (f+g x)}+\frac {b^2 c^2 e F^{c \left (a-\frac {b f}{g}\right ) n-c n (a+b x)} \left (F^{a c+b c x}\right )^n n^2 \operatorname {ExpIntegralEi}\left (\frac {b c n (f+g x) \log (F)}{g}\right ) \log ^2(F)}{2 g^3} \] Output:
-1/2*d/g/(g*x+f)^2-1/2*e*(F^(b*c*x+a*c))^n/g/(g*x+f)^2-1/2*b*c*e*(F^(b*c*x +a*c))^n*n*ln(F)/g^2/(g*x+f)+1/2*b^2*c^2*e*F^(c*(a-b*f/g)*n-c*n*(b*x+a))*( F^(b*c*x+a*c))^n*n^2*Ei(b*c*n*(g*x+f)*ln(F)/g)*ln(F)^2/g^3
Time = 0.64 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.76 \[ \int \frac {d+e \left (F^{c (a+b x)}\right )^n}{(f+g x)^3} \, dx=-\frac {d g^2-b^2 c^2 e F^{-\frac {b c n (f+g x)}{g}} \left (F^{c (a+b x)}\right )^n n^2 (f+g x)^2 \operatorname {ExpIntegralEi}\left (\frac {b c n (f+g x) \log (F)}{g}\right ) \log ^2(F)+e \left (F^{c (a+b x)}\right )^n g (g+b c n (f+g x) \log (F))}{2 g^3 (f+g x)^2} \] Input:
Integrate[(d + e*(F^(c*(a + b*x)))^n)/(f + g*x)^3,x]
Output:
-1/2*(d*g^2 - (b^2*c^2*e*(F^(c*(a + b*x)))^n*n^2*(f + g*x)^2*ExpIntegralEi [(b*c*n*(f + g*x)*Log[F])/g]*Log[F]^2)/F^((b*c*n*(f + g*x))/g) + e*(F^(c*( a + b*x)))^n*g*(g + b*c*n*(f + g*x)*Log[F]))/(g^3*(f + g*x)^2)
Time = 0.69 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2614, 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.
\(\displaystyle \int \frac {e \left (F^{c (a+b x)}\right )^n+d}{(f+g x)^3} \, dx\) |
\(\Big \downarrow \) 2614 |
\(\displaystyle \int \left (\frac {e \left (F^{a c+b c x}\right )^n}{(f+g x)^3}+\frac {d}{(f+g x)^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {b^2 c^2 e n^2 \log ^2(F) \left (F^{a c+b c x}\right )^n F^{c n \left (a-\frac {b f}{g}\right )-c n (a+b x)} \operatorname {ExpIntegralEi}\left (\frac {b c n (f+g x) \log (F)}{g}\right )}{2 g^3}-\frac {b c e n \log (F) \left (F^{a c+b c x}\right )^n}{2 g^2 (f+g x)}-\frac {e \left (F^{a c+b c x}\right )^n}{2 g (f+g x)^2}-\frac {d}{2 g (f+g x)^2}\) |
Input:
Int[(d + e*(F^(c*(a + b*x)))^n)/(f + g*x)^3,x]
Output:
-1/2*d/(g*(f + g*x)^2) - (e*(F^(a*c + b*c*x))^n)/(2*g*(f + g*x)^2) - (b*c* e*(F^(a*c + b*c*x))^n*n*Log[F])/(2*g^2*(f + g*x)) + (b^2*c^2*e*F^(c*(a - ( b*f)/g)*n - c*n*(a + b*x))*(F^(a*c + b*c*x))^n*n^2*ExpIntegralEi[(b*c*n*(f + g*x)*Log[F])/g]*Log[F]^2)/(2*g^3)
Int[((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.))^(p_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c + d*x)^m, (a + b*(F ^(g*(e + f*x)))^n)^p, x], x] /; FreeQ[{F, a, b, c, d, e, f, g, m, n}, x] && IGtQ[p, 0]
\[\int \frac {d +e \left (F^{c \left (b x +a \right )}\right )^{n}}{\left (g x +f \right )^{3}}d x\]
Input:
int((d+e*(F^(c*(b*x+a)))^n)/(g*x+f)^3,x)
Output:
int((d+e*(F^(c*(b*x+a)))^n)/(g*x+f)^3,x)
Time = 0.10 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.10 \[ \int \frac {d+e \left (F^{c (a+b x)}\right )^n}{(f+g x)^3} \, dx=-\frac {d g^{2} - \frac {{\left (b^{2} c^{2} e g^{2} n^{2} x^{2} + 2 \, b^{2} c^{2} e f g n^{2} x + b^{2} c^{2} e f^{2} n^{2}\right )} {\rm Ei}\left (\frac {{\left (b c g n x + b c f n\right )} \log \left (F\right )}{g}\right ) \log \left (F\right )^{2}}{F^{\frac {{\left (b c f - a c g\right )} n}{g}}} + {\left (e g^{2} + {\left (b c e g^{2} n x + b c e f g n\right )} \log \left (F\right )\right )} F^{b c n x + a c n}}{2 \, {\left (g^{5} x^{2} + 2 \, f g^{4} x + f^{2} g^{3}\right )}} \] Input:
integrate((d+e*(F^((b*x+a)*c))^n)/(g*x+f)^3,x, algorithm="fricas")
Output:
-1/2*(d*g^2 - (b^2*c^2*e*g^2*n^2*x^2 + 2*b^2*c^2*e*f*g*n^2*x + b^2*c^2*e*f ^2*n^2)*Ei((b*c*g*n*x + b*c*f*n)*log(F)/g)*log(F)^2/F^((b*c*f - a*c*g)*n/g ) + (e*g^2 + (b*c*e*g^2*n*x + b*c*e*f*g*n)*log(F))*F^(b*c*n*x + a*c*n))/(g ^5*x^2 + 2*f*g^4*x + f^2*g^3)
\[ \int \frac {d+e \left (F^{c (a+b x)}\right )^n}{(f+g x)^3} \, dx=\int \frac {d + e \left (F^{a c + b c x}\right )^{n}}{\left (f + g x\right )^{3}}\, dx \] Input:
integrate((d+e*(F**((b*x+a)*c))**n)/(g*x+f)**3,x)
Output:
Integral((d + e*(F**(a*c + b*c*x))**n)/(f + g*x)**3, x)
\[ \int \frac {d+e \left (F^{c (a+b x)}\right )^n}{(f+g x)^3} \, dx=\int { \frac {{\left (F^{{\left (b x + a\right )} c}\right )}^{n} e + d}{{\left (g x + f\right )}^{3}} \,d x } \] Input:
integrate((d+e*(F^((b*x+a)*c))^n)/(g*x+f)^3,x, algorithm="maxima")
Output:
F^(a*c*n)*e*integrate(F^(b*c*n*x)/(g^3*x^3 + 3*f*g^2*x^2 + 3*f^2*g*x + f^3 ), x) - 1/2*d/(g^3*x^2 + 2*f*g^2*x + f^2*g)
\[ \int \frac {d+e \left (F^{c (a+b x)}\right )^n}{(f+g x)^3} \, dx=\int { \frac {{\left (F^{{\left (b x + a\right )} c}\right )}^{n} e + d}{{\left (g x + f\right )}^{3}} \,d x } \] Input:
integrate((d+e*(F^((b*x+a)*c))^n)/(g*x+f)^3,x, algorithm="giac")
Output:
integrate(((F^((b*x + a)*c))^n*e + d)/(g*x + f)^3, x)
Timed out. \[ \int \frac {d+e \left (F^{c (a+b x)}\right )^n}{(f+g x)^3} \, dx=\int \frac {d+e\,{\left (F^{c\,\left (a+b\,x\right )}\right )}^n}{{\left (f+g\,x\right )}^3} \,d x \] Input:
int((d + e*(F^(c*(a + b*x)))^n)/(f + g*x)^3,x)
Output:
int((d + e*(F^(c*(a + b*x)))^n)/(f + g*x)^3, x)
\[ \int \frac {d+e \left (F^{c (a+b x)}\right )^n}{(f+g x)^3} \, dx=\frac {2 f^{a c n} \left (\int \frac {f^{b c n x}}{g^{3} x^{3}+3 f \,g^{2} x^{2}+3 f^{2} g x +f^{3}}d x \right ) e \,f^{2} g +4 f^{a c n} \left (\int \frac {f^{b c n x}}{g^{3} x^{3}+3 f \,g^{2} x^{2}+3 f^{2} g x +f^{3}}d x \right ) e f \,g^{2} x +2 f^{a c n} \left (\int \frac {f^{b c n x}}{g^{3} x^{3}+3 f \,g^{2} x^{2}+3 f^{2} g x +f^{3}}d x \right ) e \,g^{3} x^{2}-d}{2 g \left (g^{2} x^{2}+2 f g x +f^{2}\right )} \] Input:
int((d+e*(F^((b*x+a)*c))^n)/(g*x+f)^3,x)
Output:
(2*f**(a*c*n)*int(f**(b*c*n*x)/(f**3 + 3*f**2*g*x + 3*f*g**2*x**2 + g**3*x **3),x)*e*f**2*g + 4*f**(a*c*n)*int(f**(b*c*n*x)/(f**3 + 3*f**2*g*x + 3*f* g**2*x**2 + g**3*x**3),x)*e*f*g**2*x + 2*f**(a*c*n)*int(f**(b*c*n*x)/(f**3 + 3*f**2*g*x + 3*f*g**2*x**2 + g**3*x**3),x)*e*g**3*x**2 - d)/(2*g*(f**2 + 2*f*g*x + g**2*x**2))