Integrand size = 35, antiderivative size = 270 \[ \int \tan ^5(d+e x) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)} \, dx=-\frac {\sqrt {a-b+c} \text {arctanh}\left (\frac {2 a-b+(b-2 c) \tan ^2(d+e x)}{2 \sqrt {a-b+c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 e}+\frac {\left (b^3+2 b^2 c-4 b (a-2 c) c-8 c^2 (a+2 c)\right ) \text {arctanh}\left (\frac {b+2 c \tan ^2(d+e x)}{2 \sqrt {c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{32 c^{5/2} e}-\frac {\left ((b-2 c) (b+4 c)+2 c (b+2 c) \tan ^2(d+e x)\right ) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{16 c^2 e}+\frac {\left (a+b \tan ^2(d+e x)+c \tan ^4(d+e x)\right )^{3/2}}{6 c e} \]
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Time = 0.66 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {3781, 1265, 1667, 828, 857, 635, 212, 738} \[ \int \tan ^5(d+e x) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)} \, dx=\frac {\left (-4 b c (a-2 c)-8 c^2 (a+2 c)+b^3+2 b^2 c\right ) \text {arctanh}\left (\frac {b+2 c \tan ^2(d+e x)}{2 \sqrt {c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{32 c^{5/2} e}-\frac {\sqrt {a-b+c} \text {arctanh}\left (\frac {2 a+(b-2 c) \tan ^2(d+e x)-b}{2 \sqrt {a-b+c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 e}-\frac {\left (2 c (b+2 c) \tan ^2(d+e x)+(b-2 c) (b+4 c)\right ) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{16 c^2 e}+\frac {\left (a+b \tan ^2(d+e x)+c \tan ^4(d+e x)\right )^{3/2}}{6 c e} \]
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Rule 212
Rule 635
Rule 738
Rule 828
Rule 857
Rule 1265
Rule 1667
Rule 3781
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x^5 \sqrt {a+b x^2+c x^4}}{1+x^2} \, dx,x,\tan (d+e x)\right )}{e} \\ & = \frac {\text {Subst}\left (\int \frac {x^2 \sqrt {a+b x+c x^2}}{1+x} \, dx,x,\tan ^2(d+e x)\right )}{2 e} \\ & = \frac {\left (a+b \tan ^2(d+e x)+c \tan ^4(d+e x)\right )^{3/2}}{6 c e}+\frac {\text {Subst}\left (\int \frac {\left (-\frac {3 b}{2}-\frac {3}{2} (b+2 c) x\right ) \sqrt {a+b x+c x^2}}{1+x} \, dx,x,\tan ^2(d+e x)\right )}{6 c e} \\ & = -\frac {\left ((b-2 c) (b+4 c)+2 c (b+2 c) \tan ^2(d+e x)\right ) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{16 c^2 e}+\frac {\left (a+b \tan ^2(d+e x)+c \tan ^4(d+e x)\right )^{3/2}}{6 c e}-\frac {\text {Subst}\left (\int \frac {-\frac {3}{4} (b-2 c) \left (b^2-4 a c+4 b c\right )-\frac {3}{4} \left (b^3+2 b^2 c-4 b (a-2 c) c-8 c^2 (a+2 c)\right ) x}{(1+x) \sqrt {a+b x+c x^2}} \, dx,x,\tan ^2(d+e x)\right )}{24 c^2 e} \\ & = -\frac {\left ((b-2 c) (b+4 c)+2 c (b+2 c) \tan ^2(d+e x)\right ) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{16 c^2 e}+\frac {\left (a+b \tan ^2(d+e x)+c \tan ^4(d+e x)\right )^{3/2}}{6 c e}+\frac {(a-b+c) \text {Subst}\left (\int \frac {1}{(1+x) \sqrt {a+b x+c x^2}} \, dx,x,\tan ^2(d+e x)\right )}{2 e}+\frac {\left (b^3+2 b^2 c-4 b (a-2 c) c-8 c^2 (a+2 c)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x+c x^2}} \, dx,x,\tan ^2(d+e x)\right )}{32 c^2 e} \\ & = -\frac {\left ((b-2 c) (b+4 c)+2 c (b+2 c) \tan ^2(d+e x)\right ) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{16 c^2 e}+\frac {\left (a+b \tan ^2(d+e x)+c \tan ^4(d+e x)\right )^{3/2}}{6 c e}-\frac {(a-b+c) \text {Subst}\left (\int \frac {1}{4 a-4 b+4 c-x^2} \, dx,x,\frac {2 a-b-(-b+2 c) \tan ^2(d+e x)}{\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{e}+\frac {\left (b^3+2 b^2 c-4 b (a-2 c) c-8 c^2 (a+2 c)\right ) \text {Subst}\left (\int \frac {1}{4 c-x^2} \, dx,x,\frac {b+2 c \tan ^2(d+e x)}{\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{16 c^2 e} \\ & = -\frac {\sqrt {a-b+c} \text {arctanh}\left (\frac {2 a-b+(b-2 c) \tan ^2(d+e x)}{2 \sqrt {a-b+c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 e}+\frac {\left (b^3+2 b^2 c-4 b (a-2 c) c-8 c^2 (a+2 c)\right ) \text {arctanh}\left (\frac {b+2 c \tan ^2(d+e x)}{2 \sqrt {c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{32 c^{5/2} e}-\frac {\left ((b-2 c) (b+4 c)+2 c (b+2 c) \tan ^2(d+e x)\right ) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{16 c^2 e}+\frac {\left (a+b \tan ^2(d+e x)+c \tan ^4(d+e x)\right )^{3/2}}{6 c e} \\ \end{align*}
Time = 6.06 (sec) , antiderivative size = 290, normalized size of antiderivative = 1.07 \[ \int \tan ^5(d+e x) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)} \, dx=\frac {-48 c^{5/2} \sqrt {a-b+c} \text {arctanh}\left (\frac {2 a-b+(b-2 c) \tan ^2(d+e x)}{2 \sqrt {a-b+c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )+3 \left (b^3+2 b^2 c-4 b (a-2 c) c-8 c^2 (a+2 c)\right ) \text {arctanh}\left (\frac {b+2 c \tan ^2(d+e x)}{2 \sqrt {c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )+\frac {1}{4} \sqrt {c} \left (-9 b^2+24 a c-16 b c+84 c^2-4 \left (3 b^2+6 b c-8 c (a+2 c)\right ) \cos (2 (d+e x))+\left (-3 b^2+8 a c-8 b c+44 c^2\right ) \cos (4 (d+e x))\right ) \sec ^4(d+e x) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{96 c^{5/2} e} \]
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Time = 0.29 (sec) , antiderivative size = 455, normalized size of antiderivative = 1.69
method | result | size |
derivativedivides | \(\frac {\frac {\left (a +b \tan \left (e x +d \right )^{2}+c \tan \left (e x +d \right )^{4}\right )^{\frac {3}{2}}}{6 c}-\frac {b \left (\frac {\left (b +2 c \tan \left (e x +d \right )^{2}\right ) \sqrt {a +b \tan \left (e x +d \right )^{2}+c \tan \left (e x +d \right )^{4}}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c \tan \left (e x +d \right )^{2}}{\sqrt {c}}+\sqrt {a +b \tan \left (e x +d \right )^{2}+c \tan \left (e x +d \right )^{4}}\right )}{8 c^{\frac {3}{2}}}\right )}{4 c}-\frac {\left (b +2 c \tan \left (e x +d \right )^{2}\right ) \sqrt {a +b \tan \left (e x +d \right )^{2}+c \tan \left (e x +d \right )^{4}}}{8 c}-\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c \tan \left (e x +d \right )^{2}}{\sqrt {c}}+\sqrt {a +b \tan \left (e x +d \right )^{2}+c \tan \left (e x +d \right )^{4}}\right )}{16 c^{\frac {3}{2}}}+\frac {\sqrt {c \left (1+\tan \left (e x +d \right )^{2}\right )^{2}+\left (b -2 c \right ) \left (1+\tan \left (e x +d \right )^{2}\right )+a -b +c}}{2}+\frac {\left (b -2 c \right ) \ln \left (\frac {\frac {b}{2}-c +\left (1+\tan \left (e x +d \right )^{2}\right ) c}{\sqrt {c}}+\sqrt {c \left (1+\tan \left (e x +d \right )^{2}\right )^{2}+\left (b -2 c \right ) \left (1+\tan \left (e x +d \right )^{2}\right )+a -b +c}\right )}{4 \sqrt {c}}-\frac {\sqrt {a -b +c}\, \ln \left (\frac {2 a -2 b +2 c +\left (b -2 c \right ) \left (1+\tan \left (e x +d \right )^{2}\right )+2 \sqrt {a -b +c}\, \sqrt {c \left (1+\tan \left (e x +d \right )^{2}\right )^{2}+\left (b -2 c \right ) \left (1+\tan \left (e x +d \right )^{2}\right )+a -b +c}}{1+\tan \left (e x +d \right )^{2}}\right )}{2}}{e}\) | \(455\) |
default | \(\frac {\frac {\left (a +b \tan \left (e x +d \right )^{2}+c \tan \left (e x +d \right )^{4}\right )^{\frac {3}{2}}}{6 c}-\frac {b \left (\frac {\left (b +2 c \tan \left (e x +d \right )^{2}\right ) \sqrt {a +b \tan \left (e x +d \right )^{2}+c \tan \left (e x +d \right )^{4}}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c \tan \left (e x +d \right )^{2}}{\sqrt {c}}+\sqrt {a +b \tan \left (e x +d \right )^{2}+c \tan \left (e x +d \right )^{4}}\right )}{8 c^{\frac {3}{2}}}\right )}{4 c}-\frac {\left (b +2 c \tan \left (e x +d \right )^{2}\right ) \sqrt {a +b \tan \left (e x +d \right )^{2}+c \tan \left (e x +d \right )^{4}}}{8 c}-\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c \tan \left (e x +d \right )^{2}}{\sqrt {c}}+\sqrt {a +b \tan \left (e x +d \right )^{2}+c \tan \left (e x +d \right )^{4}}\right )}{16 c^{\frac {3}{2}}}+\frac {\sqrt {c \left (1+\tan \left (e x +d \right )^{2}\right )^{2}+\left (b -2 c \right ) \left (1+\tan \left (e x +d \right )^{2}\right )+a -b +c}}{2}+\frac {\left (b -2 c \right ) \ln \left (\frac {\frac {b}{2}-c +\left (1+\tan \left (e x +d \right )^{2}\right ) c}{\sqrt {c}}+\sqrt {c \left (1+\tan \left (e x +d \right )^{2}\right )^{2}+\left (b -2 c \right ) \left (1+\tan \left (e x +d \right )^{2}\right )+a -b +c}\right )}{4 \sqrt {c}}-\frac {\sqrt {a -b +c}\, \ln \left (\frac {2 a -2 b +2 c +\left (b -2 c \right ) \left (1+\tan \left (e x +d \right )^{2}\right )+2 \sqrt {a -b +c}\, \sqrt {c \left (1+\tan \left (e x +d \right )^{2}\right )^{2}+\left (b -2 c \right ) \left (1+\tan \left (e x +d \right )^{2}\right )+a -b +c}}{1+\tan \left (e x +d \right )^{2}}\right )}{2}}{e}\) | \(455\) |
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Time = 2.89 (sec) , antiderivative size = 1405, normalized size of antiderivative = 5.20 \[ \int \tan ^5(d+e x) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)} \, dx=\text {Too large to display} \]
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\[ \int \tan ^5(d+e x) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)} \, dx=\int \sqrt {a + b \tan ^{2}{\left (d + e x \right )} + c \tan ^{4}{\left (d + e x \right )}} \tan ^{5}{\left (d + e x \right )}\, dx \]
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\[ \int \tan ^5(d+e x) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)} \, dx=\int { \sqrt {c \tan \left (e x + d\right )^{4} + b \tan \left (e x + d\right )^{2} + a} \tan \left (e x + d\right )^{5} \,d x } \]
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Timed out. \[ \int \tan ^5(d+e x) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)} \, dx=\text {Timed out} \]
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Timed out. \[ \int \tan ^5(d+e x) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)} \, dx=\int {\mathrm {tan}\left (d+e\,x\right )}^5\,\sqrt {c\,{\mathrm {tan}\left (d+e\,x\right )}^4+b\,{\mathrm {tan}\left (d+e\,x\right )}^2+a} \,d x \]
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