Integrand size = 21, antiderivative size = 62 \[ \int (c+d x)^3 \left (a+b (c+d x)^2\right )^p \, dx=-\frac {a \left (a+b (c+d x)^2\right )^{1+p}}{2 b^2 d (1+p)}+\frac {\left (a+b (c+d x)^2\right )^{2+p}}{2 b^2 d (2+p)} \]
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Time = 0.04 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {379, 272, 45} \[ \int (c+d x)^3 \left (a+b (c+d x)^2\right )^p \, dx=\frac {\left (a+b (c+d x)^2\right )^{p+2}}{2 b^2 d (p+2)}-\frac {a \left (a+b (c+d x)^2\right )^{p+1}}{2 b^2 d (p+1)} \]
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Rule 45
Rule 272
Rule 379
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int x^3 \left (a+b x^2\right )^p \, dx,x,c+d x\right )}{d} \\ & = \frac {\text {Subst}\left (\int x (a+b x)^p \, dx,x,(c+d x)^2\right )}{2 d} \\ & = \frac {\text {Subst}\left (\int \left (-\frac {a (a+b x)^p}{b}+\frac {(a+b x)^{1+p}}{b}\right ) \, dx,x,(c+d x)^2\right )}{2 d} \\ & = -\frac {a \left (a+b (c+d x)^2\right )^{1+p}}{2 b^2 d (1+p)}+\frac {\left (a+b (c+d x)^2\right )^{2+p}}{2 b^2 d (2+p)} \\ \end{align*}
Time = 0.22 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.82 \[ \int (c+d x)^3 \left (a+b (c+d x)^2\right )^p \, dx=\frac {\left (a+b (c+d x)^2\right )^{1+p} \left (-a+b (1+p) (c+d x)^2\right )}{2 b^2 d (1+p) (2+p)} \]
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Time = 4.00 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.47
method | result | size |
gosper | \(-\frac {\left (b \,d^{2} x^{2}+2 b c d x +b \,c^{2}+a \right )^{1+p} \left (-b \,d^{2} p \,x^{2}-2 b c d p x -b \,d^{2} x^{2}-b \,c^{2} p -2 b c d x -b \,c^{2}+a \right )}{2 b^{2} d \left (p^{2}+3 p +2\right )}\) | \(91\) |
risch | \(-\frac {\left (-b^{2} d^{4} p \,x^{4}-4 b^{2} c \,d^{3} p \,x^{3}-d^{4} x^{4} b^{2}-6 b^{2} c^{2} d^{2} p \,x^{2}-4 c \,d^{3} x^{3} b^{2}-4 b^{2} c^{3} d p x -6 b^{2} c^{2} d^{2} x^{2}-a b \,d^{2} p \,x^{2}-b^{2} c^{4} p -4 b^{2} d x \,c^{3}-2 a b c d p x -b^{2} c^{4}-a b \,c^{2} p +a^{2}\right ) \left (a +b \left (d x +c \right )^{2}\right )^{p}}{2 \left (1+p \right ) \left (2+p \right ) b^{2} d}\) | \(178\) |
norman | \(\frac {c \left (2 b \,c^{2} p +2 b \,c^{2}+a p \right ) x \,{\mathrm e}^{p \ln \left (a +b \left (d x +c \right )^{2}\right )}}{b \left (p^{2}+3 p +2\right )}+\frac {d^{3} x^{4} {\mathrm e}^{p \ln \left (a +b \left (d x +c \right )^{2}\right )}}{4+2 p}+\frac {2 c \,d^{2} x^{3} {\mathrm e}^{p \ln \left (a +b \left (d x +c \right )^{2}\right )}}{2+p}-\frac {\left (-b^{2} c^{4} p -b^{2} c^{4}-a b \,c^{2} p +a^{2}\right ) {\mathrm e}^{p \ln \left (a +b \left (d x +c \right )^{2}\right )}}{2 b^{2} \left (p^{2}+3 p +2\right ) d}+\frac {d \left (6 b \,c^{2} p +6 b \,c^{2}+a p \right ) x^{2} {\mathrm e}^{p \ln \left (a +b \left (d x +c \right )^{2}\right )}}{2 b \left (p^{2}+3 p +2\right )}\) | \(220\) |
parallelrisch | \(\frac {x^{4} \left (a +b \left (d x +c \right )^{2}\right )^{p} b^{2} d^{5} p +d^{5} \left (a +b \left (d x +c \right )^{2}\right )^{p} x^{4} b^{2}+4 x^{3} \left (a +b \left (d x +c \right )^{2}\right )^{p} b^{2} c \,d^{4} p +4 c \,d^{4} \left (a +b \left (d x +c \right )^{2}\right )^{p} x^{3} b^{2}+6 x^{2} \left (a +b \left (d x +c \right )^{2}\right )^{p} b^{2} c^{2} d^{3} p +6 x^{2} \left (a +b \left (d x +c \right )^{2}\right )^{p} b^{2} c^{2} d^{3}+4 x \left (a +b \left (d x +c \right )^{2}\right )^{p} b^{2} c^{3} d^{2} p +x^{2} \left (a +b \left (d x +c \right )^{2}\right )^{p} a b \,d^{3} p +4 x \left (a +b \left (d x +c \right )^{2}\right )^{p} b^{2} c^{3} d^{2}+\left (a +b \left (d x +c \right )^{2}\right )^{p} b^{2} c^{4} d p +2 x \left (a +b \left (d x +c \right )^{2}\right )^{p} a b c \,d^{2} p +\left (a +b \left (d x +c \right )^{2}\right )^{p} b^{2} c^{4} d +\left (a +b \left (d x +c \right )^{2}\right )^{p} a b \,c^{2} d p -\left (a +b \left (d x +c \right )^{2}\right )^{p} a^{2} d}{2 \left (2+p \right ) \left (1+p \right ) b^{2} d^{2}}\) | \(353\) |
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Leaf count of result is larger than twice the leaf count of optimal. 183 vs. \(2 (58) = 116\).
Time = 0.24 (sec) , antiderivative size = 183, normalized size of antiderivative = 2.95 \[ \int (c+d x)^3 \left (a+b (c+d x)^2\right )^p \, dx=\frac {{\left (b^{2} c^{4} + {\left (b^{2} d^{4} p + b^{2} d^{4}\right )} x^{4} + 4 \, {\left (b^{2} c d^{3} p + b^{2} c d^{3}\right )} x^{3} + {\left (6 \, b^{2} c^{2} d^{2} + {\left (6 \, b^{2} c^{2} + a b\right )} d^{2} p\right )} x^{2} - a^{2} + {\left (b^{2} c^{4} + a b c^{2}\right )} p + 2 \, {\left (2 \, b^{2} c^{3} d + {\left (2 \, b^{2} c^{3} + a b c\right )} d p\right )} x\right )} {\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a\right )}^{p}}{2 \, {\left (b^{2} d p^{2} + 3 \, b^{2} d p + 2 \, b^{2} d\right )}} \]
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Timed out. \[ \int (c+d x)^3 \left (a+b (c+d x)^2\right )^p \, dx=\text {Timed out} \]
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Leaf count of result is larger than twice the leaf count of optimal. 140 vs. \(2 (58) = 116\).
Time = 0.24 (sec) , antiderivative size = 140, normalized size of antiderivative = 2.26 \[ \int (c+d x)^3 \left (a+b (c+d x)^2\right )^p \, dx=\frac {{\left (b^{2} d^{4} {\left (p + 1\right )} x^{4} + 4 \, b^{2} c d^{3} {\left (p + 1\right )} x^{3} + b^{2} c^{4} {\left (p + 1\right )} + a b c^{2} p + {\left (6 \, b^{2} c^{2} d^{2} {\left (p + 1\right )} + a b d^{2} p\right )} x^{2} - a^{2} + 2 \, {\left (2 \, b^{2} c^{3} d {\left (p + 1\right )} + a b c d p\right )} x\right )} {\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a\right )}^{p}}{2 \, {\left (p^{2} + 3 \, p + 2\right )} b^{2} d} \]
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Time = 0.28 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.60 \[ \int (c+d x)^3 \left (a+b (c+d x)^2\right )^p \, dx=\frac {{\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a\right )}^{2} {\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a\right )}^{p}}{2 \, b^{2} d {\left (p + 2\right )}} - \frac {{\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a\right )}^{p + 1} a}{2 \, b^{2} d {\left (p + 1\right )}} \]
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Time = 5.86 (sec) , antiderivative size = 168, normalized size of antiderivative = 2.71 \[ \int (c+d x)^3 \left (a+b (c+d x)^2\right )^p \, dx={\left (a+b\,{\left (c+d\,x\right )}^2\right )}^p\,\left (\frac {d^3\,x^4\,\left (p+1\right )}{2\,\left (p^2+3\,p+2\right )}+\frac {\left (b\,c^2+a\right )\,\left (b\,c^2-a+b\,c^2\,p\right )}{2\,b^2\,d\,\left (p^2+3\,p+2\right )}+\frac {c\,x\,\left (a\,p+2\,b\,c^2+2\,b\,c^2\,p\right )}{b\,\left (p^2+3\,p+2\right )}+\frac {2\,c\,d^2\,x^3\,\left (p+1\right )}{p^2+3\,p+2}+\frac {d\,x^2\,\left (a\,p+6\,b\,c^2+6\,b\,c^2\,p\right )}{2\,b\,\left (p^2+3\,p+2\right )}\right ) \]
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