Integrand size = 50, antiderivative size = 112 \[ \int \frac {\left (a+b F^{\frac {c \sqrt {d+e x}}{\sqrt {f+g x}}}\right )^2}{d f+(e f+d g) x+e g x^2} \, dx=\frac {4 a b \operatorname {ExpIntegralEi}\left (\frac {c \sqrt {d+e x} \log (F)}{\sqrt {f+g x}}\right )}{e f-d g}+\frac {2 b^2 \operatorname {ExpIntegralEi}\left (\frac {2 c \sqrt {d+e x} \log (F)}{\sqrt {f+g x}}\right )}{e f-d g}+\frac {2 a^2 \log \left (\frac {\sqrt {d+e x}}{\sqrt {f+g x}}\right )}{e f-d g} \]
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Time = 0.15 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.060, Rules used = {2328, 2214, 2209} \[ \int \frac {\left (a+b F^{\frac {c \sqrt {d+e x}}{\sqrt {f+g x}}}\right )^2}{d f+(e f+d g) x+e g x^2} \, dx=\frac {2 a^2 \log \left (\frac {\sqrt {d+e x}}{\sqrt {f+g x}}\right )}{e f-d g}+\frac {4 a b \operatorname {ExpIntegralEi}\left (\frac {c \sqrt {d+e x} \log (F)}{\sqrt {f+g x}}\right )}{e f-d g}+\frac {2 b^2 \operatorname {ExpIntegralEi}\left (\frac {2 c \sqrt {d+e x} \log (F)}{\sqrt {f+g x}}\right )}{e f-d g} \]
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Rule 2209
Rule 2214
Rule 2328
Rubi steps \begin{align*} \text {integral}& = \frac {2 \text {Subst}\left (\int \frac {\left (a+b F^{c x}\right )^2}{x} \, dx,x,\frac {\sqrt {d+e x}}{\sqrt {f+g x}}\right )}{e f-d g} \\ & = \frac {2 \text {Subst}\left (\int \left (\frac {a^2}{x}+\frac {2 a b F^{c x}}{x}+\frac {b^2 F^{2 c x}}{x}\right ) \, dx,x,\frac {\sqrt {d+e x}}{\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}+\frac {(4 a b) \text {Subst}\left (\int \frac {F^{c x}}{x} \, dx,x,\frac {\sqrt {d+e x}}{\sqrt {f+g x}}\right )}{e f-d g}+\frac {\left (2 b^2\right ) \text {Subst}\left (\int \frac {F^{2 c x}}{x} \, dx,x,\frac {\sqrt {d+e x}}{\sqrt {f+g x}}\right )}{e f-d g} \\ & = \frac {4 a b \text {Ei}\left (\frac {c \sqrt {d+e x} \log (F)}{\sqrt {f+g x}}\right )}{e f-d g}+\frac {2 b^2 \text {Ei}\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} \\ \end{align*}
\[ \int \frac {\left (a+b F^{\frac {c \sqrt {d+e x}}{\sqrt {f+g x}}}\right )^2}{d f+(e f+d g) x+e g x^2} \, dx=\int \frac {\left (a+b F^{\frac {c \sqrt {d+e x}}{\sqrt {f+g x}}}\right )^2}{d f+(e f+d g) x+e g x^2} \, dx \]
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\[\int \frac {\left (a +b \,F^{\frac {c \sqrt {e x +d}}{\sqrt {g x +f}}}\right )^{2}}{d f +\left (d g +e f \right ) x +e g \,x^{2}}d x\]
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\[ \int \frac {\left (a+b F^{\frac {c \sqrt {d+e x}}{\sqrt {f+g x}}}\right )^2}{d f+(e f+d g) x+e g x^2} \, dx=\int { \frac {{\left (F^{\frac {\sqrt {e x + d} c}{\sqrt {g x + f}}} b + a\right )}^{2}}{e g x^{2} + d f + {\left (e f + d g\right )} x} \,d x } \]
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\[ \int \frac {\left (a+b F^{\frac {c \sqrt {d+e x}}{\sqrt {f+g x}}}\right )^2}{d f+(e f+d g) x+e g x^2} \, dx=\int \frac {\left (F^{\frac {c \sqrt {d + e x}}{\sqrt {f + g x}}} b + a\right )^{2}}{\left (d + e x\right ) \left (f + g x\right )}\, dx \]
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\[ \int \frac {\left (a+b F^{\frac {c \sqrt {d+e x}}{\sqrt {f+g x}}}\right )^2}{d f+(e f+d g) x+e g x^2} \, dx=\int { \frac {{\left (F^{\frac {\sqrt {e x + d} c}{\sqrt {g x + f}}} b + a\right )}^{2}}{e g x^{2} + d f + {\left (e f + d g\right )} x} \,d x } \]
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\[ \int \frac {\left (a+b F^{\frac {c \sqrt {d+e x}}{\sqrt {f+g x}}}\right )^2}{d f+(e f+d g) x+e g x^2} \, dx=\int { \frac {{\left (F^{\frac {\sqrt {e x + d} c}{\sqrt {g x + f}}} b + a\right )}^{2}}{e g x^{2} + d f + {\left (e f + d g\right )} x} \,d x } \]
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Timed out. \[ \int \frac {\left (a+b F^{\frac {c \sqrt {d+e x}}{\sqrt {f+g x}}}\right )^2}{d f+(e f+d g) x+e g x^2} \, dx=\int \frac {{\left (a+F^{\frac {c\,\sqrt {d+e\,x}}{\sqrt {f+g\,x}}}\,b\right )}^2}{e\,g\,x^2+\left (d\,g+e\,f\right )\,x+d\,f} \,d x \]
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