Optimal. Leaf size=62 \[ \frac {2 \left (a+b \sqrt {c+d x}\right )^{p+2}}{b^2 d (p+2)}-\frac {2 a \left (a+b \sqrt {c+d x}\right )^{p+1}}{b^2 d (p+1)} \]
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Rubi [A] time = 0.04, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {247, 190, 43} \begin {gather*} \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)} \end {gather*}
Antiderivative was successfully verified.
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Rule 43
Rule 190
Rule 247
Rubi steps
\begin {align*} \int \left (a+b \sqrt {c+d x}\right )^p \, dx &=\frac {\operatorname {Subst}\left (\int \left (a+b \sqrt {x}\right )^p \, dx,x,c+d x\right )}{d}\\ &=\frac {2 \operatorname {Subst}\left (\int x (a+b x)^p \, dx,x,\sqrt {c+d x}\right )}{d}\\ &=\frac {2 \operatorname {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*}
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Mathematica [A] time = 0.04, size = 53, normalized size = 0.85 \begin {gather*} \frac {2 \left (a+b \sqrt {c+d x}\right )^{p+1} \left (b (p+1) \sqrt {c+d x}-a\right )}{b^2 d (p+1) (p+2)} \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [F] time = 0.12, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a+b \sqrt {c+d x}\right )^p \, dx \end {gather*}
Verification is not applicable to the result.
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fricas [A] time = 0.99, size = 81, normalized size = 1.31 \begin {gather*} \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} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.44, size = 129, normalized size = 2.08 \begin {gather*} \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} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.01, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a +\sqrt {d x +c}\, b \right )^{p}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.93, size = 60, normalized size = 0.97 \begin {gather*} \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} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.58, size = 146, normalized size = 2.35 \begin {gather*} \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 . \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \sqrt {c + d x}\right )^{p}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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