Integrand size = 18, antiderivative size = 261 \[ \int \frac {x^3}{a+b \sec \left (c+d x^2\right )} \, dx=\frac {x^4}{4 a}+\frac {i b x^2 \log \left (1+\frac {a e^{i \left (c+d x^2\right )}}{b-\sqrt {-a^2+b^2}}\right )}{2 a \sqrt {-a^2+b^2} d}-\frac {i b x^2 \log \left (1+\frac {a e^{i \left (c+d x^2\right )}}{b+\sqrt {-a^2+b^2}}\right )}{2 a \sqrt {-a^2+b^2} d}+\frac {b \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (c+d x^2\right )}}{b-\sqrt {-a^2+b^2}}\right )}{2 a \sqrt {-a^2+b^2} d^2}-\frac {b \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (c+d x^2\right )}}{b+\sqrt {-a^2+b^2}}\right )}{2 a \sqrt {-a^2+b^2} d^2} \]
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Time = 0.66 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {4289, 4276, 3402, 2296, 2221, 2317, 2438} \[ \int \frac {x^3}{a+b \sec \left (c+d x^2\right )} \, dx=\frac {b \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (d x^2+c\right )}}{b-\sqrt {b^2-a^2}}\right )}{2 a d^2 \sqrt {b^2-a^2}}-\frac {b \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (d x^2+c\right )}}{b+\sqrt {b^2-a^2}}\right )}{2 a d^2 \sqrt {b^2-a^2}}+\frac {i b x^2 \log \left (1+\frac {a e^{i \left (c+d x^2\right )}}{b-\sqrt {b^2-a^2}}\right )}{2 a d \sqrt {b^2-a^2}}-\frac {i b x^2 \log \left (1+\frac {a e^{i \left (c+d x^2\right )}}{\sqrt {b^2-a^2}+b}\right )}{2 a d \sqrt {b^2-a^2}}+\frac {x^4}{4 a} \]
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Rule 2221
Rule 2296
Rule 2317
Rule 2438
Rule 3402
Rule 4276
Rule 4289
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {x}{a+b \sec (c+d x)} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (\frac {x}{a}-\frac {b x}{a (b+a \cos (c+d x))}\right ) \, dx,x,x^2\right ) \\ & = \frac {x^4}{4 a}-\frac {b \text {Subst}\left (\int \frac {x}{b+a \cos (c+d x)} \, dx,x,x^2\right )}{2 a} \\ & = \frac {x^4}{4 a}-\frac {b \text {Subst}\left (\int \frac {e^{i (c+d x)} x}{a+2 b e^{i (c+d x)}+a e^{2 i (c+d x)}} \, dx,x,x^2\right )}{a} \\ & = \frac {x^4}{4 a}-\frac {b \text {Subst}\left (\int \frac {e^{i (c+d x)} x}{2 b-2 \sqrt {-a^2+b^2}+2 a e^{i (c+d x)}} \, dx,x,x^2\right )}{\sqrt {-a^2+b^2}}+\frac {b \text {Subst}\left (\int \frac {e^{i (c+d x)} x}{2 b+2 \sqrt {-a^2+b^2}+2 a e^{i (c+d x)}} \, dx,x,x^2\right )}{\sqrt {-a^2+b^2}} \\ & = \frac {x^4}{4 a}+\frac {i b x^2 \log \left (1+\frac {a e^{i \left (c+d x^2\right )}}{b-\sqrt {-a^2+b^2}}\right )}{2 a \sqrt {-a^2+b^2} d}-\frac {i b x^2 \log \left (1+\frac {a e^{i \left (c+d x^2\right )}}{b+\sqrt {-a^2+b^2}}\right )}{2 a \sqrt {-a^2+b^2} d}-\frac {(i b) \text {Subst}\left (\int \log \left (1+\frac {2 a e^{i (c+d x)}}{2 b-2 \sqrt {-a^2+b^2}}\right ) \, dx,x,x^2\right )}{2 a \sqrt {-a^2+b^2} d}+\frac {(i b) \text {Subst}\left (\int \log \left (1+\frac {2 a e^{i (c+d x)}}{2 b+2 \sqrt {-a^2+b^2}}\right ) \, dx,x,x^2\right )}{2 a \sqrt {-a^2+b^2} d} \\ & = \frac {x^4}{4 a}+\frac {i b x^2 \log \left (1+\frac {a e^{i \left (c+d x^2\right )}}{b-\sqrt {-a^2+b^2}}\right )}{2 a \sqrt {-a^2+b^2} d}-\frac {i b x^2 \log \left (1+\frac {a e^{i \left (c+d x^2\right )}}{b+\sqrt {-a^2+b^2}}\right )}{2 a \sqrt {-a^2+b^2} d}-\frac {b \text {Subst}\left (\int \frac {\log \left (1+\frac {2 a x}{2 b-2 \sqrt {-a^2+b^2}}\right )}{x} \, dx,x,e^{i \left (c+d x^2\right )}\right )}{2 a \sqrt {-a^2+b^2} d^2}+\frac {b \text {Subst}\left (\int \frac {\log \left (1+\frac {2 a x}{2 b+2 \sqrt {-a^2+b^2}}\right )}{x} \, dx,x,e^{i \left (c+d x^2\right )}\right )}{2 a \sqrt {-a^2+b^2} d^2} \\ & = \frac {x^4}{4 a}+\frac {i b x^2 \log \left (1+\frac {a e^{i \left (c+d x^2\right )}}{b-\sqrt {-a^2+b^2}}\right )}{2 a \sqrt {-a^2+b^2} d}-\frac {i b x^2 \log \left (1+\frac {a e^{i \left (c+d x^2\right )}}{b+\sqrt {-a^2+b^2}}\right )}{2 a \sqrt {-a^2+b^2} d}+\frac {b \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (c+d x^2\right )}}{b-\sqrt {-a^2+b^2}}\right )}{2 a \sqrt {-a^2+b^2} d^2}-\frac {b \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (c+d x^2\right )}}{b+\sqrt {-a^2+b^2}}\right )}{2 a \sqrt {-a^2+b^2} d^2} \\ \end{align*}
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(845\) vs. \(2(261)=522\).
Time = 1.61 (sec) , antiderivative size = 845, normalized size of antiderivative = 3.24 \[ \int \frac {x^3}{a+b \sec \left (c+d x^2\right )} \, dx=\frac {\left (b+a \cos \left (c+d x^2\right )\right ) \left (x^4-\frac {2 b \left (2 \left (c+d x^2\right ) \text {arctanh}\left (\frac {(a+b) \cot \left (\frac {1}{2} \left (c+d x^2\right )\right )}{\sqrt {a^2-b^2}}\right )-2 \left (c+\arccos \left (-\frac {b}{a}\right )\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} \left (c+d x^2\right )\right )}{\sqrt {a^2-b^2}}\right )+\left (\arccos \left (-\frac {b}{a}\right )-2 i \text {arctanh}\left (\frac {(a+b) \cot \left (\frac {1}{2} \left (c+d x^2\right )\right )}{\sqrt {a^2-b^2}}\right )+2 i \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} \left (c+d x^2\right )\right )}{\sqrt {a^2-b^2}}\right )\right ) \log \left (\frac {\sqrt {a^2-b^2} e^{-\frac {1}{2} i \left (c+d x^2\right )}}{\sqrt {2} \sqrt {a} \sqrt {b+a \cos \left (c+d x^2\right )}}\right )+\left (\arccos \left (-\frac {b}{a}\right )+2 i \left (\text {arctanh}\left (\frac {(a+b) \cot \left (\frac {1}{2} \left (c+d x^2\right )\right )}{\sqrt {a^2-b^2}}\right )-\text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} \left (c+d x^2\right )\right )}{\sqrt {a^2-b^2}}\right )\right )\right ) \log \left (\frac {\sqrt {a^2-b^2} e^{\frac {1}{2} i \left (c+d x^2\right )}}{\sqrt {2} \sqrt {a} \sqrt {b+a \cos \left (c+d x^2\right )}}\right )-\left (\arccos \left (-\frac {b}{a}\right )-2 i \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} \left (c+d x^2\right )\right )}{\sqrt {a^2-b^2}}\right )\right ) \log \left (\frac {(a+b) \left (a-b-i \sqrt {a^2-b^2}\right ) \left (1+i \tan \left (\frac {1}{2} \left (c+d x^2\right )\right )\right )}{a \left (a+b+\sqrt {a^2-b^2} \tan \left (\frac {1}{2} \left (c+d x^2\right )\right )\right )}\right )-\left (\arccos \left (-\frac {b}{a}\right )+2 i \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} \left (c+d x^2\right )\right )}{\sqrt {a^2-b^2}}\right )\right ) \log \left (\frac {(a+b) \left (-i a+i b+\sqrt {a^2-b^2}\right ) \left (i+\tan \left (\frac {1}{2} \left (c+d x^2\right )\right )\right )}{a \left (a+b+\sqrt {a^2-b^2} \tan \left (\frac {1}{2} \left (c+d x^2\right )\right )\right )}\right )+i \left (\operatorname {PolyLog}\left (2,\frac {\left (b-i \sqrt {a^2-b^2}\right ) \left (a+b-\sqrt {a^2-b^2} \tan \left (\frac {1}{2} \left (c+d x^2\right )\right )\right )}{a \left (a+b+\sqrt {a^2-b^2} \tan \left (\frac {1}{2} \left (c+d x^2\right )\right )\right )}\right )-\operatorname {PolyLog}\left (2,\frac {\left (b+i \sqrt {a^2-b^2}\right ) \left (a+b-\sqrt {a^2-b^2} \tan \left (\frac {1}{2} \left (c+d x^2\right )\right )\right )}{a \left (a+b+\sqrt {a^2-b^2} \tan \left (\frac {1}{2} \left (c+d x^2\right )\right )\right )}\right )\right )\right )}{\sqrt {a^2-b^2} d^2}\right ) \sec \left (c+d x^2\right )}{4 a \left (a+b \sec \left (c+d x^2\right )\right )} \]
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\[\int \frac {x^{3}}{a +b \sec \left (d \,x^{2}+c \right )}d x\]
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1060 vs. \(2 (221) = 442\).
Time = 0.40 (sec) , antiderivative size = 1060, normalized size of antiderivative = 4.06 \[ \int \frac {x^3}{a+b \sec \left (c+d x^2\right )} \, dx=\text {Too large to display} \]
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\[ \int \frac {x^3}{a+b \sec \left (c+d x^2\right )} \, dx=\int \frac {x^{3}}{a + b \sec {\left (c + d x^{2} \right )}}\, dx \]
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\[ \int \frac {x^3}{a+b \sec \left (c+d x^2\right )} \, dx=\int { \frac {x^{3}}{b \sec \left (d x^{2} + c\right ) + a} \,d x } \]
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\[ \int \frac {x^3}{a+b \sec \left (c+d x^2\right )} \, dx=\int { \frac {x^{3}}{b \sec \left (d x^{2} + c\right ) + a} \,d x } \]
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Timed out. \[ \int \frac {x^3}{a+b \sec \left (c+d x^2\right )} \, dx=\int \frac {x^3}{a+\frac {b}{\cos \left (d\,x^2+c\right )}} \,d x \]
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