Integrand size = 50, antiderivative size = 112 \[ \int \frac {\left (a+b F^{\frac {c \sqrt {d+e x}}{\sqrt {f+g x}}}\right )^2}{d f+(e f+d g) x+e g x^2} \, dx=\frac {4 a b \operatorname {ExpIntegralEi}\left (\frac {c \sqrt {d+e x} \log (F)}{\sqrt {f+g x}}\right )}{e f-d g}+\frac {2 b^2 \operatorname {ExpIntegralEi}\left (\frac {2 c \sqrt {d+e x} \log (F)}{\sqrt {f+g x}}\right )}{e f-d g}+\frac {2 a^2 \log \left (\frac {\sqrt {d+e x}}{\sqrt {f+g x}}\right )}{e f-d g} \]
4*a*b*Ei(c*ln(F)*(e*x+d)^(1/2)/(g*x+f)^(1/2))/(-d*g+e*f)+2*b^2*Ei(2*c*ln(F )*(e*x+d)^(1/2)/(g*x+f)^(1/2))/(-d*g+e*f)+2*a^2*ln((e*x+d)^(1/2)/(g*x+f)^( 1/2))/(-d*g+e*f)
\[ \int \frac {\left (a+b F^{\frac {c \sqrt {d+e x}}{\sqrt {f+g x}}}\right )^2}{d f+(e f+d g) x+e g x^2} \, dx=\int \frac {\left (a+b F^{\frac {c \sqrt {d+e x}}{\sqrt {f+g x}}}\right )^2}{d f+(e f+d g) x+e g x^2} \, dx \]
Time = 0.41 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.82, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.060, Rules used = {2728, 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.
\(\displaystyle \int \frac {\left (a+b F^{\frac {c \sqrt {d+e x}}{\sqrt {f+g x}}}\right )^2}{x (d g+e f)+d f+e g x^2} \, dx\) |
\(\Big \downarrow \) 2728 |
\(\displaystyle \frac {2 \int \frac {\left (b F^{\frac {c \sqrt {d+e x}}{\sqrt {f+g x}}}+a\right )^2 \sqrt {f+g x}}{\sqrt {d+e x}}d\frac {\sqrt {d+e x}}{\sqrt {f+g x}}}{e f-d g}\) |
\(\Big \downarrow \) 2614 |
\(\displaystyle \frac {2 \int \left (\frac {2 a b \sqrt {f+g x} F^{\frac {c \sqrt {d+e x}}{\sqrt {f+g x}}}}{\sqrt {d+e x}}+\frac {b^2 \sqrt {f+g x} F^{\frac {2 c \sqrt {d+e x}}{\sqrt {f+g x}}}}{\sqrt {d+e x}}+\frac {a^2 \sqrt {f+g x}}{\sqrt {d+e x}}\right )d\frac {\sqrt {d+e x}}{\sqrt {f+g x}}}{e f-d g}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 \left (a^2 \log \left (\frac {\sqrt {d+e x}}{\sqrt {f+g x}}\right )+2 a b \operatorname {ExpIntegralEi}\left (\frac {c \sqrt {d+e x} \log (F)}{\sqrt {f+g x}}\right )+b^2 \operatorname {ExpIntegralEi}\left (\frac {2 c \sqrt {d+e x} \log (F)}{\sqrt {f+g x}}\right )\right )}{e f-d g}\) |
(2*(2*a*b*ExpIntegralEi[(c*Sqrt[d + e*x]*Log[F])/Sqrt[f + g*x]] + b^2*ExpI ntegralEi[(2*c*Sqrt[d + e*x]*Log[F])/Sqrt[f + g*x]] + a^2*Log[Sqrt[d + e*x ]/Sqrt[f + g*x]]))/(e*f - d*g)
3.6.45.3.1 Defintions of rubi rules used
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[((a_.) + (b_.)*(F_)^(((c_.)*Sqrt[(d_.) + (e_.)*(x_)])/Sqrt[(f_.) + (g_. )*(x_)]))^(n_.)/((A_.) + (B_.)*(x_) + (C_.)*(x_)^2), x_Symbol] :> Simp[2*e* (g/(C*(e*f - d*g))) Subst[Int[(a + b*F^(c*x))^n/x, x], x, Sqrt[d + e*x]/S qrt[f + g*x]], x] /; FreeQ[{a, b, c, d, e, f, g, A, B, C, F}, x] && EqQ[C*d *f - A*e*g, 0] && EqQ[B*e*g - C*(e*f + d*g), 0] && IGtQ[n, 0]
\[\int \frac {\left (a +b \,F^{\frac {c \sqrt {e x +d}}{\sqrt {g x +f}}}\right )^{2}}{d f +\left (d g +e f \right ) x +e g \,x^{2}}d x\]
\[ \int \frac {\left (a+b F^{\frac {c \sqrt {d+e x}}{\sqrt {f+g x}}}\right )^2}{d f+(e f+d g) x+e g x^2} \, dx=\int { \frac {{\left (F^{\frac {\sqrt {e x + d} c}{\sqrt {g x + f}}} b + a\right )}^{2}}{e g x^{2} + d f + {\left (e f + d g\right )} x} \,d x } \]
integrate((a+b*F^(c*(e*x+d)^(1/2)/(g*x+f)^(1/2)))^2/(d*f+(d*g+e*f)*x+e*g*x ^2),x, algorithm="fricas")
integral((2*F^(sqrt(e*x + d)*c/sqrt(g*x + f))*a*b + F^(2*sqrt(e*x + d)*c/s qrt(g*x + f))*b^2 + a^2)/(e*g*x^2 + d*f + (e*f + d*g)*x), x)
\[ \int \frac {\left (a+b F^{\frac {c \sqrt {d+e x}}{\sqrt {f+g x}}}\right )^2}{d f+(e f+d g) x+e g x^2} \, dx=\int \frac {\left (F^{\frac {c \sqrt {d + e x}}{\sqrt {f + g x}}} b + a\right )^{2}}{\left (d + e x\right ) \left (f + g x\right )}\, dx \]
\[ \int \frac {\left (a+b F^{\frac {c \sqrt {d+e x}}{\sqrt {f+g x}}}\right )^2}{d f+(e f+d g) x+e g x^2} \, dx=\int { \frac {{\left (F^{\frac {\sqrt {e x + d} c}{\sqrt {g x + f}}} b + a\right )}^{2}}{e g x^{2} + d f + {\left (e f + d g\right )} x} \,d x } \]
integrate((a+b*F^(c*(e*x+d)^(1/2)/(g*x+f)^(1/2)))^2/(d*f+(d*g+e*f)*x+e*g*x ^2),x, algorithm="maxima")
a^2*(log(e*x + d)/(e*f - d*g) - log(g*x + f)/(e*f - d*g)) + b^2*integrate( F^(2*sqrt(e*x + d)*c/sqrt(g*x + f))/(e*g*x^2 + d*f + (e*f + d*g)*x), x) + 2*a*b*integrate(F^(sqrt(e*x + d)*c/sqrt(g*x + f))/(e*g*x^2 + d*f + (e*f + d*g)*x), x)
\[ \int \frac {\left (a+b F^{\frac {c \sqrt {d+e x}}{\sqrt {f+g x}}}\right )^2}{d f+(e f+d g) x+e g x^2} \, dx=\int { \frac {{\left (F^{\frac {\sqrt {e x + d} c}{\sqrt {g x + f}}} b + a\right )}^{2}}{e g x^{2} + d f + {\left (e f + d g\right )} x} \,d x } \]
integrate((a+b*F^(c*(e*x+d)^(1/2)/(g*x+f)^(1/2)))^2/(d*f+(d*g+e*f)*x+e*g*x ^2),x, algorithm="giac")
Timed out. \[ \int \frac {\left (a+b F^{\frac {c \sqrt {d+e x}}{\sqrt {f+g x}}}\right )^2}{d f+(e f+d g) x+e g x^2} \, dx=\int \frac {{\left (a+F^{\frac {c\,\sqrt {d+e\,x}}{\sqrt {f+g\,x}}}\,b\right )}^2}{e\,g\,x^2+\left (d\,g+e\,f\right )\,x+d\,f} \,d x \]