Integrand size = 16, antiderivative size = 44 \[ \int x \left (a+b \sec \left (c+d x^2\right )\right )^2 \, dx=\frac {a^2 x^2}{2}+\frac {a b \text {arctanh}\left (\sin \left (c+d x^2\right )\right )}{d}+\frac {b^2 \tan \left (c+d x^2\right )}{2 d} \]
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Time = 0.07 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {4289, 3858, 3855, 3852, 8} \[ \int x \left (a+b \sec \left (c+d x^2\right )\right )^2 \, dx=\frac {a^2 x^2}{2}+\frac {a b \text {arctanh}\left (\sin \left (c+d x^2\right )\right )}{d}+\frac {b^2 \tan \left (c+d x^2\right )}{2 d} \]
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Rule 8
Rule 3852
Rule 3855
Rule 3858
Rule 4289
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int (a+b \sec (c+d x))^2 \, dx,x,x^2\right ) \\ & = \frac {a^2 x^2}{2}+(a b) \text {Subst}\left (\int \sec (c+d x) \, dx,x,x^2\right )+\frac {1}{2} b^2 \text {Subst}\left (\int \sec ^2(c+d x) \, dx,x,x^2\right ) \\ & = \frac {a^2 x^2}{2}+\frac {a b \text {arctanh}\left (\sin \left (c+d x^2\right )\right )}{d}-\frac {b^2 \text {Subst}\left (\int 1 \, dx,x,-\tan \left (c+d x^2\right )\right )}{2 d} \\ & = \frac {a^2 x^2}{2}+\frac {a b \text {arctanh}\left (\sin \left (c+d x^2\right )\right )}{d}+\frac {b^2 \tan \left (c+d x^2\right )}{2 d} \\ \end{align*}
Time = 0.25 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.93 \[ \int x \left (a+b \sec \left (c+d x^2\right )\right )^2 \, dx=\frac {a^2 d x^2+2 a b \text {arctanh}\left (\sin \left (c+d x^2\right )\right )+b^2 \tan \left (c+d x^2\right )}{2 d} \]
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Time = 0.39 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.14
method | result | size |
parts | \(\frac {a^{2} x^{2}}{2}+\frac {b^{2} \tan \left (d \,x^{2}+c \right )}{2 d}+\frac {b a \ln \left (\sec \left (d \,x^{2}+c \right )+\tan \left (d \,x^{2}+c \right )\right )}{d}\) | \(50\) |
derivativedivides | \(\frac {a^{2} \left (d \,x^{2}+c \right )+2 b a \ln \left (\sec \left (d \,x^{2}+c \right )+\tan \left (d \,x^{2}+c \right )\right )+b^{2} \tan \left (d \,x^{2}+c \right )}{2 d}\) | \(52\) |
default | \(\frac {a^{2} \left (d \,x^{2}+c \right )+2 b a \ln \left (\sec \left (d \,x^{2}+c \right )+\tan \left (d \,x^{2}+c \right )\right )+b^{2} \tan \left (d \,x^{2}+c \right )}{2 d}\) | \(52\) |
risch | \(\frac {a^{2} x^{2}}{2}+\frac {i b^{2}}{d \left (1+{\mathrm e}^{2 i \left (d \,x^{2}+c \right )}\right )}+\frac {b a \ln \left ({\mathrm e}^{i \left (d \,x^{2}+c \right )}+i\right )}{d}-\frac {b a \ln \left ({\mathrm e}^{i \left (d \,x^{2}+c \right )}-i\right )}{d}\) | \(77\) |
parallelrisch | \(\frac {a^{2} d \,x^{2} \cos \left (d \,x^{2}+c \right )-2 b a \ln \left (\tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )-1\right ) \cos \left (d \,x^{2}+c \right )+2 b a \ln \left (\tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )+1\right ) \cos \left (d \,x^{2}+c \right )+b^{2} \sin \left (d \,x^{2}+c \right )}{2 d \cos \left (d \,x^{2}+c \right )}\) | \(97\) |
norman | \(\frac {-\frac {a^{2} x^{2}}{2}+\frac {a^{2} x^{2} \tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )^{2}}{2}-\frac {b^{2} \tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )}{d}}{\tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )^{2}-1}+\frac {b a \ln \left (\tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {b a \ln \left (\tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )-1\right )}{d}\) | \(110\) |
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Leaf count of result is larger than twice the leaf count of optimal. 91 vs. \(2 (40) = 80\).
Time = 0.27 (sec) , antiderivative size = 91, normalized size of antiderivative = 2.07 \[ \int x \left (a+b \sec \left (c+d x^2\right )\right )^2 \, dx=\frac {a^{2} d x^{2} \cos \left (d x^{2} + c\right ) + a b \cos \left (d x^{2} + c\right ) \log \left (\sin \left (d x^{2} + c\right ) + 1\right ) - a b \cos \left (d x^{2} + c\right ) \log \left (-\sin \left (d x^{2} + c\right ) + 1\right ) + b^{2} \sin \left (d x^{2} + c\right )}{2 \, d \cos \left (d x^{2} + c\right )} \]
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\[ \int x \left (a+b \sec \left (c+d x^2\right )\right )^2 \, dx=\int x \left (a + b \sec {\left (c + d x^{2} \right )}\right )^{2}\, dx \]
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Leaf count of result is larger than twice the leaf count of optimal. 96 vs. \(2 (40) = 80\).
Time = 0.23 (sec) , antiderivative size = 96, normalized size of antiderivative = 2.18 \[ \int x \left (a+b \sec \left (c+d x^2\right )\right )^2 \, dx=\frac {1}{2} \, a^{2} x^{2} + \frac {a b \log \left (\sec \left (d x^{2} + c\right ) + \tan \left (d x^{2} + c\right )\right )}{d} + \frac {b^{2} \sin \left (2 \, d x^{2} + 2 \, c\right )}{d \cos \left (2 \, d x^{2} + 2 \, c\right )^{2} + d \sin \left (2 \, d x^{2} + 2 \, c\right )^{2} + 2 \, d \cos \left (2 \, d x^{2} + 2 \, c\right ) + d} \]
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Leaf count of result is larger than twice the leaf count of optimal. 88 vs. \(2 (40) = 80\).
Time = 0.32 (sec) , antiderivative size = 88, normalized size of antiderivative = 2.00 \[ \int x \left (a+b \sec \left (c+d x^2\right )\right )^2 \, dx=\frac {{\left (d x^{2} + c\right )} a^{2} + 2 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x^{2} + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 2 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x^{2} + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, b^{2} \tan \left (\frac {1}{2} \, d x^{2} + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x^{2} + \frac {1}{2} \, c\right )^{2} - 1}}{2 \, d} \]
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Time = 13.78 (sec) , antiderivative size = 100, normalized size of antiderivative = 2.27 \[ \int x \left (a+b \sec \left (c+d x^2\right )\right )^2 \, dx=\frac {a^2\,x^2}{2}+\frac {b^2\,1{}\mathrm {i}}{d\,\left ({\mathrm {e}}^{2{}\mathrm {i}\,d\,x^2+c\,2{}\mathrm {i}}+1\right )}+\frac {a\,b\,\ln \left (-a\,b\,x\,4{}\mathrm {i}-4\,a\,b\,x\,{\mathrm {e}}^{d\,x^2\,1{}\mathrm {i}}\,{\mathrm {e}}^{c\,1{}\mathrm {i}}\right )}{d}-\frac {a\,b\,\ln \left (a\,b\,x\,4{}\mathrm {i}-4\,a\,b\,x\,{\mathrm {e}}^{d\,x^2\,1{}\mathrm {i}}\,{\mathrm {e}}^{c\,1{}\mathrm {i}}\right )}{d} \]
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