Integrand size = 28, antiderivative size = 54 \[ \int \frac {q+p x}{\sqrt {b+a x} \left (f+\sqrt {b+a x}\right )} \, dx=\frac {p x}{a}-\frac {2 f p \sqrt {b+a x}}{a^2}-\frac {2 \left (b p-f^2 p-a q\right ) \log \left (f+\sqrt {b+a x}\right )}{a^2} \]
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Time = 0.29 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {711} \[ \int \frac {q+p x}{\sqrt {b+a x} \left (f+\sqrt {b+a x}\right )} \, dx=-\frac {2 \left (-a q+b p+f^2 (-p)\right ) \log \left (\sqrt {a x+b}+f\right )}{a^2}-\frac {2 f p \sqrt {a x+b}}{a^2}+\frac {p x}{a} \]
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Rule 711
Rubi steps \begin{align*} \text {integral}& = \frac {2 \text {Subst}\left (\int \frac {-b p+a q+p x^2}{f+x} \, dx,x,\sqrt {b+a x}\right )}{a^2} \\ & = \frac {2 \text {Subst}\left (\int \left (-f p+p x+\frac {-b p+f^2 p+a q}{f+x}\right ) \, dx,x,\sqrt {b+a x}\right )}{a^2} \\ & = \frac {p x}{a}-\frac {2 f p \sqrt {b+a x}}{a^2}-\frac {2 \left (b p-f^2 p-a q\right ) \log \left (f+\sqrt {b+a x}\right )}{a^2} \\ \end{align*}
Time = 0.05 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.94 \[ \int \frac {q+p x}{\sqrt {b+a x} \left (f+\sqrt {b+a x}\right )} \, dx=\frac {p \left (b+a x-2 f \sqrt {b+a x}\right )+2 \left (-b p+f^2 p+a q\right ) \log \left (f+\sqrt {b+a x}\right )}{a^2} \]
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Time = 1.01 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.93
method | result | size |
derivativedivides | \(\frac {-2 f p \sqrt {a x +b}+p \left (a x +b \right )+2 \left (f^{2} p +a q -b p \right ) \ln \left (f +\sqrt {a x +b}\right )}{a^{2}}\) | \(50\) |
default | \(\frac {-2 f p \sqrt {a x +b}+p \left (a x +b \right )+2 \left (f^{2} p +a q -b p \right ) \ln \left (f +\sqrt {a x +b}\right )}{a^{2}}\) | \(50\) |
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none
Time = 0.26 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.83 \[ \int \frac {q+p x}{\sqrt {b+a x} \left (f+\sqrt {b+a x}\right )} \, dx=\frac {a p x - 2 \, \sqrt {a x + b} f p + 2 \, {\left ({\left (f^{2} - b\right )} p + a q\right )} \log \left (f + \sqrt {a x + b}\right )}{a^{2}} \]
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Time = 1.28 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.35 \[ \int \frac {q+p x}{\sqrt {b+a x} \left (f+\sqrt {b+a x}\right )} \, dx=\begin {cases} \frac {2 \left (- \frac {f p \sqrt {a x + b}}{a} + \frac {p \left (a x + b\right )}{2 a} - \frac {\left (- a q + b p - f^{2} p\right ) \log {\left (f + \sqrt {a x + b} \right )}}{a}\right )}{a} & \text {for}\: a \neq 0 \\\frac {\frac {p x^{2}}{2} + q x}{\sqrt {b} \left (\sqrt {b} + f\right )} & \text {otherwise} \end {cases} \]
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none
Time = 0.19 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.07 \[ \int \frac {q+p x}{\sqrt {b+a x} \left (f+\sqrt {b+a x}\right )} \, dx=\frac {\frac {2 \, {\left ({\left (f^{2} - b\right )} p + a q\right )} \log \left (f + \sqrt {a x + b}\right )}{a} - \frac {2 \, \sqrt {a x + b} f p - {\left (a x + b\right )} p}{a}}{a} \]
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Time = 0.42 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.13 \[ \int \frac {q+p x}{\sqrt {b+a x} \left (f+\sqrt {b+a x}\right )} \, dx=\frac {2 \, {\left (f^{2} p - b p + a q\right )} \log \left ({\left | f + \sqrt {a x + b} \right |}\right )}{a^{2}} - \frac {2 \, \sqrt {a x + b} a^{2} f p - {\left (a x + b\right )} a^{2} p}{a^{4}} \]
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Time = 18.26 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.93 \[ \int \frac {q+p x}{\sqrt {b+a x} \left (f+\sqrt {b+a x}\right )} \, dx=\frac {\ln \left (f+\sqrt {b+a\,x}\right )\,\left (2\,p\,f^2+2\,a\,q-2\,b\,p\right )}{a^2}+\frac {p\,x}{a}-\frac {2\,f\,p\,\sqrt {b+a\,x}}{a^2} \]
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