Integrand size = 31, antiderivative size = 205 \[ \int \frac {(a+b \tan (c+d x)) (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\frac {(a (A-B)-b (A+B)) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {(a (A-B)-b (A+B)) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {(b (A-B)+a (A+B)) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}+\frac {(b (A-B)+a (A+B)) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {2 a A}{d \sqrt {\tan (c+d x)}} \]
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Time = 0.33 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.258, Rules used = {3672, 3615, 1182, 1176, 631, 210, 1179, 642} \[ \int \frac {(a+b \tan (c+d x)) (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\frac {(a (A-B)-b (A+B)) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {(a (A-B)-b (A+B)) \arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} d}-\frac {(a (A+B)+b (A-B)) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d}+\frac {(a (A+B)+b (A-B)) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d}-\frac {2 a A}{d \sqrt {\tan (c+d x)}} \]
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Rule 210
Rule 631
Rule 642
Rule 1176
Rule 1179
Rule 1182
Rule 3615
Rule 3672
Rubi steps \begin{align*} \text {integral}& = -\frac {2 a A}{d \sqrt {\tan (c+d x)}}+\int \frac {A b+a B-(a A-b B) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx \\ & = -\frac {2 a A}{d \sqrt {\tan (c+d x)}}+\frac {2 \text {Subst}\left (\int \frac {A b+a B+(-a A+b B) x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d} \\ & = -\frac {2 a A}{d \sqrt {\tan (c+d x)}}+\frac {(b (A-B)+a (A+B)) \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}-\frac {(a (A-B)-b (A+B)) \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d} \\ & = -\frac {2 a A}{d \sqrt {\tan (c+d x)}}-\frac {(b (A-B)+a (A+B)) \text {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 \sqrt {2} d}-\frac {(b (A-B)+a (A+B)) \text {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 \sqrt {2} d}-\frac {(a (A-B)-b (A+B)) \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 d}-\frac {(a (A-B)-b (A+B)) \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 d} \\ & = -\frac {(b (A-B)+a (A+B)) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}+\frac {(b (A-B)+a (A+B)) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {2 a A}{d \sqrt {\tan (c+d x)}}-\frac {(a (A-B)-b (A+B)) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {(a (A-B)-b (A+B)) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d} \\ & = \frac {(a (A-B)-b (A+B)) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {(a (A-B)-b (A+B)) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {(b (A-B)+a (A+B)) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}+\frac {(b (A-B)+a (A+B)) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {2 a A}{d \sqrt {\tan (c+d x)}} \\ \end{align*}
Time = 0.68 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.77 \[ \int \frac {(a+b \tan (c+d x)) (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {-2 \sqrt {2} (a (A-B)-b (A+B)) \left (\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )-\arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )\right )+\sqrt {2} (b (A-B)+a (A+B)) \left (\log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )-\log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )\right )+\frac {8 a A}{\sqrt {\tan (c+d x)}}}{4 d} \]
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Time = 0.04 (sec) , antiderivative size = 203, normalized size of antiderivative = 0.99
method | result | size |
derivativedivides | \(\frac {-\frac {2 a A}{\sqrt {\tan \left (d x +c \right )}}+\frac {\left (A b +B a \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (-a A +B b \right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{d}\) | \(203\) |
default | \(\frac {-\frac {2 a A}{\sqrt {\tan \left (d x +c \right )}}+\frac {\left (A b +B a \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (-a A +B b \right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{d}\) | \(203\) |
parts | \(\frac {\left (A b +B a \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4 d}+\frac {a A \left (-\frac {\sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}-\frac {2}{\sqrt {\tan \left (d x +c \right )}}\right )}{d}+\frac {B b \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4 d}\) | \(289\) |
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Leaf count of result is larger than twice the leaf count of optimal. 2244 vs. \(2 (175) = 350\).
Time = 0.39 (sec) , antiderivative size = 2244, normalized size of antiderivative = 10.95 \[ \int \frac {(a+b \tan (c+d x)) (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\text {Too large to display} \]
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\[ \int \frac {(a+b \tan (c+d x)) (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \left (a + b \tan {\left (c + d x \right )}\right )}{\tan ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \]
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Time = 0.31 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.85 \[ \int \frac {(a+b \tan (c+d x)) (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {2 \, \sqrt {2} {\left ({\left (A - B\right )} a - {\left (A + B\right )} b\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} {\left ({\left (A - B\right )} a - {\left (A + B\right )} b\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - \sqrt {2} {\left ({\left (A + B\right )} a + {\left (A - B\right )} b\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + \sqrt {2} {\left ({\left (A + B\right )} a + {\left (A - B\right )} b\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + \frac {8 \, A a}{\sqrt {\tan \left (d x + c\right )}}}{4 \, d} \]
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Timed out. \[ \int \frac {(a+b \tan (c+d x)) (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\text {Timed out} \]
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Time = 10.79 (sec) , antiderivative size = 1420, normalized size of antiderivative = 6.93 \[ \int \frac {(a+b \tan (c+d x)) (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=2\,\mathrm {atanh}\left (\frac {32\,A^2\,a^2\,d^3\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {A^2\,a\,b}{2\,d^2}-\frac {\sqrt {-A^4\,a^4\,d^4+2\,A^4\,a^2\,b^2\,d^4-A^4\,b^4\,d^4}}{4\,d^4}}}{16\,A\,b\,\sqrt {-A^4\,a^4\,d^4+2\,A^4\,a^2\,b^2\,d^4-A^4\,b^4\,d^4}+16\,A^3\,a^3\,d^2-16\,A^3\,a\,b^2\,d^2}-\frac {32\,A^2\,b^2\,d^3\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {A^2\,a\,b}{2\,d^2}-\frac {\sqrt {-A^4\,a^4\,d^4+2\,A^4\,a^2\,b^2\,d^4-A^4\,b^4\,d^4}}{4\,d^4}}}{16\,A\,b\,\sqrt {-A^4\,a^4\,d^4+2\,A^4\,a^2\,b^2\,d^4-A^4\,b^4\,d^4}+16\,A^3\,a^3\,d^2-16\,A^3\,a\,b^2\,d^2}\right )\,\sqrt {\frac {A^2\,a\,b}{2\,d^2}-\frac {\sqrt {-A^4\,a^4\,d^4+2\,A^4\,a^2\,b^2\,d^4-A^4\,b^4\,d^4}}{4\,d^4}}-2\,\mathrm {atanh}\left (\frac {32\,A^2\,a^2\,d^3\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {\sqrt {-A^4\,a^4\,d^4+2\,A^4\,a^2\,b^2\,d^4-A^4\,b^4\,d^4}}{4\,d^4}+\frac {A^2\,a\,b}{2\,d^2}}}{16\,A\,b\,\sqrt {-A^4\,a^4\,d^4+2\,A^4\,a^2\,b^2\,d^4-A^4\,b^4\,d^4}-16\,A^3\,a^3\,d^2+16\,A^3\,a\,b^2\,d^2}-\frac {32\,A^2\,b^2\,d^3\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {\sqrt {-A^4\,a^4\,d^4+2\,A^4\,a^2\,b^2\,d^4-A^4\,b^4\,d^4}}{4\,d^4}+\frac {A^2\,a\,b}{2\,d^2}}}{16\,A\,b\,\sqrt {-A^4\,a^4\,d^4+2\,A^4\,a^2\,b^2\,d^4-A^4\,b^4\,d^4}-16\,A^3\,a^3\,d^2+16\,A^3\,a\,b^2\,d^2}\right )\,\sqrt {\frac {\sqrt {-A^4\,a^4\,d^4+2\,A^4\,a^2\,b^2\,d^4-A^4\,b^4\,d^4}}{4\,d^4}+\frac {A^2\,a\,b}{2\,d^2}}-2\,\mathrm {atanh}\left (\frac {32\,B^2\,a^2\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {\sqrt {-B^4\,a^4\,d^4+2\,B^4\,a^2\,b^2\,d^4-B^4\,b^4\,d^4}}{4\,d^4}-\frac {B^2\,a\,b}{2\,d^2}}}{\frac {16\,B\,a\,\sqrt {-B^4\,a^4\,d^4+2\,B^4\,a^2\,b^2\,d^4-B^4\,b^4\,d^4}}{d^3}-\frac {16\,B^3\,b^3}{d}+\frac {16\,B^3\,a^2\,b}{d}}-\frac {32\,B^2\,b^2\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {\sqrt {-B^4\,a^4\,d^4+2\,B^4\,a^2\,b^2\,d^4-B^4\,b^4\,d^4}}{4\,d^4}-\frac {B^2\,a\,b}{2\,d^2}}}{\frac {16\,B\,a\,\sqrt {-B^4\,a^4\,d^4+2\,B^4\,a^2\,b^2\,d^4-B^4\,b^4\,d^4}}{d^3}-\frac {16\,B^3\,b^3}{d}+\frac {16\,B^3\,a^2\,b}{d}}\right )\,\sqrt {-\frac {\sqrt {-B^4\,a^4\,d^4+2\,B^4\,a^2\,b^2\,d^4-B^4\,b^4\,d^4}}{4\,d^4}-\frac {B^2\,a\,b}{2\,d^2}}+2\,\mathrm {atanh}\left (\frac {32\,B^2\,a^2\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {\sqrt {-B^4\,a^4\,d^4+2\,B^4\,a^2\,b^2\,d^4-B^4\,b^4\,d^4}}{4\,d^4}-\frac {B^2\,a\,b}{2\,d^2}}}{\frac {16\,B^3\,b^3}{d}+\frac {16\,B\,a\,\sqrt {-B^4\,a^4\,d^4+2\,B^4\,a^2\,b^2\,d^4-B^4\,b^4\,d^4}}{d^3}-\frac {16\,B^3\,a^2\,b}{d}}-\frac {32\,B^2\,b^2\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {\sqrt {-B^4\,a^4\,d^4+2\,B^4\,a^2\,b^2\,d^4-B^4\,b^4\,d^4}}{4\,d^4}-\frac {B^2\,a\,b}{2\,d^2}}}{\frac {16\,B^3\,b^3}{d}+\frac {16\,B\,a\,\sqrt {-B^4\,a^4\,d^4+2\,B^4\,a^2\,b^2\,d^4-B^4\,b^4\,d^4}}{d^3}-\frac {16\,B^3\,a^2\,b}{d}}\right )\,\sqrt {\frac {\sqrt {-B^4\,a^4\,d^4+2\,B^4\,a^2\,b^2\,d^4-B^4\,b^4\,d^4}}{4\,d^4}-\frac {B^2\,a\,b}{2\,d^2}}-\frac {2\,A\,a}{d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}} \]
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