Integrand size = 15, antiderivative size = 62 \[ \int \left (a+b \sqrt {c+d x}\right )^p \, dx=-\frac {2 a \left (a+b \sqrt {c+d x}\right )^{1+p}}{b^2 d (1+p)}+\frac {2 \left (a+b \sqrt {c+d x}\right )^{2+p}}{b^2 d (2+p)} \]
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Time = 0.03 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {253, 196, 45} \[ \int \left (a+b \sqrt {c+d x}\right )^p \, dx=\frac {2 \left (a+b \sqrt {c+d x}\right )^{p+2}}{b^2 d (p+2)}-\frac {2 a \left (a+b \sqrt {c+d x}\right )^{p+1}}{b^2 d (p+1)} \]
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Rule 45
Rule 196
Rule 253
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \left (a+b \sqrt {x}\right )^p \, dx,x,c+d x\right )}{d} \\ & = \frac {2 \text {Subst}\left (\int x (a+b x)^p \, dx,x,\sqrt {c+d x}\right )}{d} \\ & = \frac {2 \text {Subst}\left (\int \left (-\frac {a (a+b x)^p}{b}+\frac {(a+b x)^{1+p}}{b}\right ) \, dx,x,\sqrt {c+d x}\right )}{d} \\ & = -\frac {2 a \left (a+b \sqrt {c+d x}\right )^{1+p}}{b^2 d (1+p)}+\frac {2 \left (a+b \sqrt {c+d x}\right )^{2+p}}{b^2 d (2+p)} \\ \end{align*}
Time = 0.09 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.85 \[ \int \left (a+b \sqrt {c+d x}\right )^p \, dx=\frac {2 \left (a+b \sqrt {c+d x}\right )^{1+p} \left (-a+b (1+p) \sqrt {c+d x}\right )}{b^2 d (1+p) (2+p)} \]
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\[\int \left (a +b \sqrt {d x +c}\right )^{p}d x\]
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none
Time = 0.29 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.31 \[ \int \left (a+b \sqrt {c+d x}\right )^p \, dx=\frac {2 \, {\left (b^{2} c p + \sqrt {d x + c} a b p + b^{2} c - a^{2} + {\left (b^{2} d p + b^{2} d\right )} x\right )} {\left (\sqrt {d x + c} b + a\right )}^{p}}{b^{2} d p^{2} + 3 \, b^{2} d p + 2 \, b^{2} d} \]
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\[ \int \left (a+b \sqrt {c+d x}\right )^p \, dx=\int \left (a + b \sqrt {c + d x}\right )^{p}\, dx \]
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none
Time = 0.20 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.97 \[ \int \left (a+b \sqrt {c+d x}\right )^p \, dx=\frac {2 \, {\left ({\left (d x + c\right )} b^{2} {\left (p + 1\right )} + \sqrt {d x + c} a b p - a^{2}\right )} {\left (\sqrt {d x + c} b + a\right )}^{p}}{{\left (p^{2} + 3 \, p + 2\right )} b^{2} d} \]
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Leaf count of result is larger than twice the leaf count of optimal. 129 vs. \(2 (58) = 116\).
Time = 0.57 (sec) , antiderivative size = 129, normalized size of antiderivative = 2.08 \[ \int \left (a+b \sqrt {c+d x}\right )^p \, dx=\frac {2 \, {\left ({\left (\sqrt {d x + c} b + a\right )}^{2} {\left (\sqrt {d x + c} b + a\right )}^{p} p - {\left (\sqrt {d x + c} b + a\right )} {\left (\sqrt {d x + c} b + a\right )}^{p} a p + {\left (\sqrt {d x + c} b + a\right )}^{2} {\left (\sqrt {d x + c} b + a\right )}^{p} - 2 \, {\left (\sqrt {d x + c} b + a\right )} {\left (\sqrt {d x + c} b + a\right )}^{p} a\right )}}{{\left (p^{2} + 3 \, p + 2\right )} b^{2} d} \]
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Time = 17.52 (sec) , antiderivative size = 146, normalized size of antiderivative = 2.35 \[ \int \left (a+b \sqrt {c+d x}\right )^p \, dx=\left \{\begin {array}{cl} -\frac {2\,a\,\ln \left (a+b\,\sqrt {c+d\,x}\right )-2\,b\,\sqrt {c+d\,x}}{b^2\,d} & \text {\ if\ \ }p=-1\\ \frac {2\,\left (\ln \left (a+b\,\sqrt {c+d\,x}\right )+\frac {a}{a+b\,\sqrt {c+d\,x}}\right )}{b^2\,d} & \text {\ if\ \ }p=-2\\ \frac {4\,{\left (a+b\,\sqrt {c+d\,x}\right )}^{p+2}}{b^2\,d\,\left (2\,p+4\right )}-\frac {4\,a\,{\left (a+b\,\sqrt {c+d\,x}\right )}^{p+1}}{b^2\,d\,\left (2\,p+2\right )} & \text {\ if\ \ }p\neq -1\wedge p\neq -2 \end {array}\right . \]
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