Integrand size = 21, antiderivative size = 22 \[ \int \sec ^2(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {(a+b \tan (c+d x))^4}{4 b d} \]
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Time = 0.05 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {3587, 32} \[ \int \sec ^2(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {(a+b \tan (c+d x))^4}{4 b d} \]
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Rule 32
Rule 3587
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int (a+x)^3 \, dx,x,b \tan (c+d x)\right )}{b d} \\ & = \frac {(a+b \tan (c+d x))^4}{4 b d} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(57\) vs. \(2(22)=44\).
Time = 0.18 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.59 \[ \int \sec ^2(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {\tan (c+d x) \left (4 a^3+6 a^2 b \tan (c+d x)+4 a b^2 \tan ^2(c+d x)+b^3 \tan ^3(c+d x)\right )}{4 d} \]
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Leaf count of result is larger than twice the leaf count of optimal. \(71\) vs. \(2(20)=40\).
Time = 4.50 (sec) , antiderivative size = 72, normalized size of antiderivative = 3.27
method | result | size |
derivativedivides | \(\frac {\frac {b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{4 \cos \left (d x +c \right )^{4}}+\frac {a \,b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{\cos \left (d x +c \right )^{3}}+\frac {3 a^{2} b}{2 \cos \left (d x +c \right )^{2}}+a^{3} \tan \left (d x +c \right )}{d}\) | \(72\) |
default | \(\frac {\frac {b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{4 \cos \left (d x +c \right )^{4}}+\frac {a \,b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{\cos \left (d x +c \right )^{3}}+\frac {3 a^{2} b}{2 \cos \left (d x +c \right )^{2}}+a^{3} \tan \left (d x +c \right )}{d}\) | \(72\) |
risch | \(-\frac {2 \left (-i a^{3} {\mathrm e}^{6 i \left (d x +c \right )}+3 i a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-3 a^{2} b \,{\mathrm e}^{6 i \left (d x +c \right )}+b^{3} {\mathrm e}^{6 i \left (d x +c \right )}-3 i a^{3} {\mathrm e}^{4 i \left (d x +c \right )}+3 i a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-6 a^{2} b \,{\mathrm e}^{4 i \left (d x +c \right )}-3 i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}+i a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-3 a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}+b^{3} {\mathrm e}^{2 i \left (d x +c \right )}-i a^{3}+i a \,b^{2}\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}\) | \(197\) |
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Leaf count of result is larger than twice the leaf count of optimal. 78 vs. \(2 (20) = 40\).
Time = 0.24 (sec) , antiderivative size = 78, normalized size of antiderivative = 3.55 \[ \int \sec ^2(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {b^{3} + 2 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left (a b^{2} \cos \left (d x + c\right ) + {\left (a^{3} - a b^{2}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{4}} \]
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\[ \int \sec ^2(c+d x) (a+b \tan (c+d x))^3 \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right )^{3} \sec ^{2}{\left (c + d x \right )}\, dx \]
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none
Time = 0.24 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \sec ^2(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {{\left (b \tan \left (d x + c\right ) + a\right )}^{4}}{4 \, b d} \]
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Leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (20) = 40\).
Time = 0.74 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.59 \[ \int \sec ^2(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {b^{3} \tan \left (d x + c\right )^{4} + 4 \, a b^{2} \tan \left (d x + c\right )^{3} + 6 \, a^{2} b \tan \left (d x + c\right )^{2} + 4 \, a^{3} \tan \left (d x + c\right )}{4 \, d} \]
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Time = 4.41 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.50 \[ \int \sec ^2(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {a^3\,\mathrm {tan}\left (c+d\,x\right )+\frac {3\,a^2\,b\,{\mathrm {tan}\left (c+d\,x\right )}^2}{2}+a\,b^2\,{\mathrm {tan}\left (c+d\,x\right )}^3+\frac {b^3\,{\mathrm {tan}\left (c+d\,x\right )}^4}{4}}{d} \]
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