Integrand size = 16, antiderivative size = 118 \[ \int \frac {1}{\left (a+a \tan ^2(c+d x)\right )^{7/2}} \, dx=\frac {\tan (c+d x)}{7 d \left (a \sec ^2(c+d x)\right )^{7/2}}+\frac {6 \tan (c+d x)}{35 a d \left (a \sec ^2(c+d x)\right )^{5/2}}+\frac {8 \tan (c+d x)}{35 a^2 d \left (a \sec ^2(c+d x)\right )^{3/2}}+\frac {16 \tan (c+d x)}{35 a^3 d \sqrt {a \sec ^2(c+d x)}} \]
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Time = 0.06 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3738, 4207, 198, 197} \[ \int \frac {1}{\left (a+a \tan ^2(c+d x)\right )^{7/2}} \, dx=\frac {16 \tan (c+d x)}{35 a^3 d \sqrt {a \sec ^2(c+d x)}}+\frac {8 \tan (c+d x)}{35 a^2 d \left (a \sec ^2(c+d x)\right )^{3/2}}+\frac {6 \tan (c+d x)}{35 a d \left (a \sec ^2(c+d x)\right )^{5/2}}+\frac {\tan (c+d x)}{7 d \left (a \sec ^2(c+d x)\right )^{7/2}} \]
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Rule 197
Rule 198
Rule 3738
Rule 4207
Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{\left (a \sec ^2(c+d x)\right )^{7/2}} \, dx \\ & = \frac {a \text {Subst}\left (\int \frac {1}{\left (a+a x^2\right )^{9/2}} \, dx,x,\tan (c+d x)\right )}{d} \\ & = \frac {\tan (c+d x)}{7 d \left (a \sec ^2(c+d x)\right )^{7/2}}+\frac {6 \text {Subst}\left (\int \frac {1}{\left (a+a x^2\right )^{7/2}} \, dx,x,\tan (c+d x)\right )}{7 d} \\ & = \frac {\tan (c+d x)}{7 d \left (a \sec ^2(c+d x)\right )^{7/2}}+\frac {6 \tan (c+d x)}{35 a d \left (a \sec ^2(c+d x)\right )^{5/2}}+\frac {24 \text {Subst}\left (\int \frac {1}{\left (a+a x^2\right )^{5/2}} \, dx,x,\tan (c+d x)\right )}{35 a d} \\ & = \frac {\tan (c+d x)}{7 d \left (a \sec ^2(c+d x)\right )^{7/2}}+\frac {6 \tan (c+d x)}{35 a d \left (a \sec ^2(c+d x)\right )^{5/2}}+\frac {8 \tan (c+d x)}{35 a^2 d \left (a \sec ^2(c+d x)\right )^{3/2}}+\frac {16 \text {Subst}\left (\int \frac {1}{\left (a+a x^2\right )^{3/2}} \, dx,x,\tan (c+d x)\right )}{35 a^2 d} \\ & = \frac {\tan (c+d x)}{7 d \left (a \sec ^2(c+d x)\right )^{7/2}}+\frac {6 \tan (c+d x)}{35 a d \left (a \sec ^2(c+d x)\right )^{5/2}}+\frac {8 \tan (c+d x)}{35 a^2 d \left (a \sec ^2(c+d x)\right )^{3/2}}+\frac {16 \tan (c+d x)}{35 a^3 d \sqrt {a \sec ^2(c+d x)}} \\ \end{align*}
Time = 0.20 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.53 \[ \int \frac {1}{\left (a+a \tan ^2(c+d x)\right )^{7/2}} \, dx=\frac {\left (35-35 \sin ^2(c+d x)+21 \sin ^4(c+d x)-5 \sin ^6(c+d x)\right ) \tan (c+d x)}{35 a^3 d \sqrt {a \sec ^2(c+d x)}} \]
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Time = 0.07 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.01
method | result | size |
derivativedivides | \(\frac {a \left (\frac {\tan \left (d x +c \right )}{7 a \left (a +a \tan \left (d x +c \right )^{2}\right )^{\frac {7}{2}}}+\frac {\frac {6 \tan \left (d x +c \right )}{35 a \left (a +a \tan \left (d x +c \right )^{2}\right )^{\frac {5}{2}}}+\frac {6 \left (\frac {4 \tan \left (d x +c \right )}{15 a \left (a +a \tan \left (d x +c \right )^{2}\right )^{\frac {3}{2}}}+\frac {8 \tan \left (d x +c \right )}{15 a^{2} \sqrt {a +a \tan \left (d x +c \right )^{2}}}\right )}{7 a}}{a}\right )}{d}\) | \(119\) |
default | \(\frac {a \left (\frac {\tan \left (d x +c \right )}{7 a \left (a +a \tan \left (d x +c \right )^{2}\right )^{\frac {7}{2}}}+\frac {\frac {6 \tan \left (d x +c \right )}{35 a \left (a +a \tan \left (d x +c \right )^{2}\right )^{\frac {5}{2}}}+\frac {6 \left (\frac {4 \tan \left (d x +c \right )}{15 a \left (a +a \tan \left (d x +c \right )^{2}\right )^{\frac {3}{2}}}+\frac {8 \tan \left (d x +c \right )}{15 a^{2} \sqrt {a +a \tan \left (d x +c \right )^{2}}}\right )}{7 a}}{a}\right )}{d}\) | \(119\) |
risch | \(-\frac {i {\mathrm e}^{8 i \left (d x +c \right )}}{896 d \sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) a^{3}}-\frac {35 i {\mathrm e}^{2 i \left (d x +c \right )}}{128 d \sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) a^{3}}+\frac {35 i}{128 a^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, d}+\frac {7 i {\mathrm e}^{-2 i \left (d x +c \right )}}{128 d \sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) a^{3}}-\frac {11 i \cos \left (6 d x +6 c \right )}{1120 a^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, d}+\frac {27 \sin \left (6 d x +6 c \right )}{2240 a^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, d}-\frac {7 i \cos \left (4 d x +4 c \right )}{160 a^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, d}+\frac {21 \sin \left (4 d x +4 c \right )}{320 a^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, d}\) | \(447\) |
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Time = 0.25 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\left (a+a \tan ^2(c+d x)\right )^{7/2}} \, dx=\frac {{\left (16 \, \tan \left (d x + c\right )^{7} + 56 \, \tan \left (d x + c\right )^{5} + 70 \, \tan \left (d x + c\right )^{3} + 35 \, \tan \left (d x + c\right )\right )} \sqrt {a \tan \left (d x + c\right )^{2} + a}}{35 \, {\left (a^{4} d \tan \left (d x + c\right )^{8} + 4 \, a^{4} d \tan \left (d x + c\right )^{6} + 6 \, a^{4} d \tan \left (d x + c\right )^{4} + 4 \, a^{4} d \tan \left (d x + c\right )^{2} + a^{4} d\right )}} \]
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\[ \int \frac {1}{\left (a+a \tan ^2(c+d x)\right )^{7/2}} \, dx=\int \frac {1}{\left (a \tan ^{2}{\left (c + d x \right )} + a\right )^{\frac {7}{2}}}\, dx \]
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Time = 0.40 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.42 \[ \int \frac {1}{\left (a+a \tan ^2(c+d x)\right )^{7/2}} \, dx=\frac {5 \, \sin \left (7 \, d x + 7 \, c\right ) + 49 \, \sin \left (5 \, d x + 5 \, c\right ) + 245 \, \sin \left (3 \, d x + 3 \, c\right ) + 1225 \, \sin \left (d x + c\right )}{2240 \, a^{\frac {7}{2}} d} \]
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Leaf count of result is larger than twice the leaf count of optimal. 2158 vs. \(2 (102) = 204\).
Time = 2.85 (sec) , antiderivative size = 2158, normalized size of antiderivative = 18.29 \[ \int \frac {1}{\left (a+a \tan ^2(c+d x)\right )^{7/2}} \, dx=\text {Too large to display} \]
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Time = 11.20 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.36 \[ \int \frac {1}{\left (a+a \tan ^2(c+d x)\right )^{7/2}} \, dx=\frac {16\,\mathrm {tan}\left (c+d\,x\right )\,\sqrt {a\,{\mathrm {tan}\left (c+d\,x\right )}^2+a}}{35\,a^4\,d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )}+\frac {8\,\mathrm {tan}\left (c+d\,x\right )\,\sqrt {a\,{\mathrm {tan}\left (c+d\,x\right )}^2+a}}{35\,a^4\,d\,{\left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )}^2}+\frac {6\,\mathrm {tan}\left (c+d\,x\right )\,\sqrt {a\,{\mathrm {tan}\left (c+d\,x\right )}^2+a}}{35\,a^4\,d\,{\left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )}^3}+\frac {\mathrm {tan}\left (c+d\,x\right )\,\sqrt {a\,{\mathrm {tan}\left (c+d\,x\right )}^2+a}}{7\,a^4\,d\,{\left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )}^4} \]
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