Integrand size = 31, antiderivative size = 205 \[ \int \frac {(a+b \tan (c+d x)) (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=-\frac {(b (A-B)+a (A+B)) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {(b (A-B)+a (A+B)) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {(a (A-B)-b (A+B)) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}+\frac {(a (A-B)-b (A+B)) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}+\frac {2 b B \sqrt {\tan (c+d x)}}{d} \]
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Time = 0.23 (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 = {3673, 3615, 1182, 1176, 631, 210, 1179, 642} \[ \int \frac {(a+b \tan (c+d x)) (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=-\frac {(a (A+B)+b (A-B)) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {(a (A+B)+b (A-B)) \arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} d}-\frac {(a (A-B)-b (A+B)) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d}+\frac {(a (A-B)-b (A+B)) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d}+\frac {2 b B \sqrt {\tan (c+d x)}}{d} \]
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Rule 210
Rule 631
Rule 642
Rule 1176
Rule 1179
Rule 1182
Rule 3615
Rule 3673
Rubi steps \begin{align*} \text {integral}& = \frac {2 b B \sqrt {\tan (c+d x)}}{d}+\int \frac {a A-b B+(A b+a B) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx \\ & = \frac {2 b B \sqrt {\tan (c+d x)}}{d}+\frac {2 \text {Subst}\left (\int \frac {a A-b B+(A b+a B) x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d} \\ & = \frac {2 b B \sqrt {\tan (c+d x)}}{d}+\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 b B \sqrt {\tan (c+d x)}}{d}+\frac {(b (A-B)+a (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)) \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 {\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 {\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)) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}+\frac {(a (A-B)-b (A+B)) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}+\frac {2 b B \sqrt {\tan (c+d x)}}{d}+\frac {(b (A-B)+a (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 {(b (A-B)+a (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 {(b (A-B)+a (A+B)) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {(b (A-B)+a (A+B)) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {(a (A-B)-b (A+B)) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}+\frac {(a (A-B)-b (A+B)) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}+\frac {2 b B \sqrt {\tan (c+d x)}}{d} \\ \end{align*}
Result contains complex when optimal does not.
Time = 0.18 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.46 \[ \int \frac {(a+b \tan (c+d x)) (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=-\frac {\sqrt [4]{-1} (a-i b) (A-i B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )+\sqrt [4]{-1} (a+i b) (A+i B) \text {arctanh}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )-2 b B \sqrt {\tan (c+d x)}}{d} \]
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Time = 0.04 (sec) , antiderivative size = 203, normalized size of antiderivative = 0.99
method | result | size |
derivativedivides | \(\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right ) B b +\frac {\left (a A -B b \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (A b +B a \right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{d}\) | \(203\) |
default | \(\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right ) B b +\frac {\left (a A -B b \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (A b +B a \right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{d}\) | \(203\) |
parts | \(\frac {\left (A b +B a \right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4 d}+\frac {B b \left (2 \left (\sqrt {\tan }\left (d x +c \right )\right )-\frac {\sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}+\frac {a A \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4 d}\) | \(289\) |
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Leaf count of result is larger than twice the leaf count of optimal. 2216 vs. \(2 (175) = 350\).
Time = 0.38 (sec) , antiderivative size = 2216, normalized size of antiderivative = 10.81 \[ \int \frac {(a+b \tan (c+d x)) (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=\text {Too large to display} \]
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\[ \int \frac {(a+b \tan (c+d x)) (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=\int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \left (a + b \tan {\left (c + d x \right )}\right )}{\sqrt {\tan {\left (c + d x \right )}}}\, dx \]
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Time = 0.37 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.85 \[ \int \frac {(a+b \tan (c+d x)) (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=\frac {2 \, \sqrt {2} {\left ({\left (A + B\right )} a + {\left (A - B\right )} b\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} {\left ({\left (A + B\right )} a + {\left (A - B\right )} b\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + \sqrt {2} {\left ({\left (A - B\right )} a - {\left (A + B\right )} b\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - \sqrt {2} {\left ({\left (A - B\right )} a - {\left (A + B\right )} b\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + 8 \, B b \sqrt {\tan \left (d x + c\right )}}{4 \, d} \]
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Timed out. \[ \int \frac {(a+b \tan (c+d x)) (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=\text {Timed out} \]
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Time = 10.38 (sec) , antiderivative size = 1420, normalized size of antiderivative = 6.93 \[ \int \frac {(a+b \tan (c+d x)) (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=2\,\mathrm {atanh}\left (\frac {32\,A^2\,a^2\,\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}}}{\frac {16\,A^3\,b^3}{d}+\frac {16\,A\,a\,\sqrt {-A^4\,a^4\,d^4+2\,A^4\,a^2\,b^2\,d^4-A^4\,b^4\,d^4}}{d^3}-\frac {16\,A^3\,a^2\,b}{d}}-\frac {32\,A^2\,b^2\,\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}}}{\frac {16\,A^3\,b^3}{d}+\frac {16\,A\,a\,\sqrt {-A^4\,a^4\,d^4+2\,A^4\,a^2\,b^2\,d^4-A^4\,b^4\,d^4}}{d^3}-\frac {16\,A^3\,a^2\,b}{d}}\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\,A^2\,a^2\,\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}}}{\frac {16\,A\,a\,\sqrt {-A^4\,a^4\,d^4+2\,A^4\,a^2\,b^2\,d^4-A^4\,b^4\,d^4}}{d^3}-\frac {16\,A^3\,b^3}{d}+\frac {16\,A^3\,a^2\,b}{d}}-\frac {32\,A^2\,b^2\,\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}}}{\frac {16\,A\,a\,\sqrt {-A^4\,a^4\,d^4+2\,A^4\,a^2\,b^2\,d^4-A^4\,b^4\,d^4}}{d^3}-\frac {16\,A^3\,b^3}{d}+\frac {16\,A^3\,a^2\,b}{d}}\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 {B^2\,a\,b}{2\,d^2}-\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 {16\,B^3\,a^3}{d}+\frac {16\,B\,b\,\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\,b^2}{d}}-\frac {32\,B^2\,b^2\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {B^2\,a\,b}{2\,d^2}-\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 {16\,B^3\,a^3}{d}+\frac {16\,B\,b\,\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\,b^2}{d}}\right )\,\sqrt {\frac {B^2\,a\,b}{2\,d^2}-\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}}+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\,b\,\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^3}{d}+\frac {16\,B^3\,a\,b^2}{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\,b\,\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^3}{d}+\frac {16\,B^3\,a\,b^2}{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\,B\,b\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{d} \]
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