Optimal. Leaf size=326 \[ \frac {2 \left (a^2 A-2 a b B-A b^2\right ) \sqrt {\cot (c+d x)}}{d}+\frac {\left (a^2 (A+B)+2 a b (A-B)-b^2 (A+B)\right ) \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d}-\frac {\left (a^2 (A+B)+2 a b (A-B)-b^2 (A+B)\right ) \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d}+\frac {\left (a^2 (A-B)-2 a b (A+B)-b^2 (A-B)\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}-\frac {\left (a^2 (A-B)-2 a b (A+B)-b^2 (A-B)\right ) \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} d}-\frac {2 a (5 a B+7 A b) \cot ^{\frac {3}{2}}(c+d x)}{15 d}-\frac {2 a A \cot ^{\frac {3}{2}}(c+d x) (a \cot (c+d x)+b)}{5 d} \]
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Rubi [A] time = 0.61, antiderivative size = 326, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 11, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3581, 3607, 3630, 3528, 3534, 1168, 1162, 617, 204, 1165, 628} \[ \frac {2 \left (a^2 A-2 a b B-A b^2\right ) \sqrt {\cot (c+d x)}}{d}+\frac {\left (a^2 (A+B)+2 a b (A-B)-b^2 (A+B)\right ) \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d}-\frac {\left (a^2 (A+B)+2 a b (A-B)-b^2 (A+B)\right ) \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d}+\frac {\left (a^2 (A-B)-2 a b (A+B)-b^2 (A-B)\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}-\frac {\left (a^2 (A-B)-2 a b (A+B)-b^2 (A-B)\right ) \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} d}-\frac {2 a (5 a B+7 A b) \cot ^{\frac {3}{2}}(c+d x)}{15 d}-\frac {2 a A \cot ^{\frac {3}{2}}(c+d x) (a \cot (c+d x)+b)}{5 d} \]
Antiderivative was successfully verified.
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Rule 204
Rule 617
Rule 628
Rule 1162
Rule 1165
Rule 1168
Rule 3528
Rule 3534
Rule 3581
Rule 3607
Rule 3630
Rubi steps
\begin {align*} \int \cot ^{\frac {7}{2}}(c+d x) (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx &=\int \sqrt {\cot (c+d x)} (b+a \cot (c+d x))^2 (B+A \cot (c+d x)) \, dx\\ &=-\frac {2 a A \cot ^{\frac {3}{2}}(c+d x) (b+a \cot (c+d x))}{5 d}-\frac {2}{5} \int \sqrt {\cot (c+d x)} \left (\frac {1}{2} b (3 a A-5 b B)+\frac {5}{2} \left (a^2 A-A b^2-2 a b B\right ) \cot (c+d x)-\frac {1}{2} a (7 A b+5 a B) \cot ^2(c+d x)\right ) \, dx\\ &=-\frac {2 a (7 A b+5 a B) \cot ^{\frac {3}{2}}(c+d x)}{15 d}-\frac {2 a A \cot ^{\frac {3}{2}}(c+d x) (b+a \cot (c+d x))}{5 d}-\frac {2}{5} \int \sqrt {\cot (c+d x)} \left (\frac {5}{2} \left (2 a A b+a^2 B-b^2 B\right )+\frac {5}{2} \left (a^2 A-A b^2-2 a b B\right ) \cot (c+d x)\right ) \, dx\\ &=\frac {2 \left (a^2 A-A b^2-2 a b B\right ) \sqrt {\cot (c+d x)}}{d}-\frac {2 a (7 A b+5 a B) \cot ^{\frac {3}{2}}(c+d x)}{15 d}-\frac {2 a A \cot ^{\frac {3}{2}}(c+d x) (b+a \cot (c+d x))}{5 d}-\frac {2}{5} \int \frac {-\frac {5}{2} \left (a^2 A-A b^2-2 a b B\right )+\frac {5}{2} \left (2 a A b+a^2 B-b^2 B\right ) \cot (c+d x)}{\sqrt {\cot (c+d x)}} \, dx\\ &=\frac {2 \left (a^2 A-A b^2-2 a b B\right ) \sqrt {\cot (c+d x)}}{d}-\frac {2 a (7 A b+5 a B) \cot ^{\frac {3}{2}}(c+d x)}{15 d}-\frac {2 a A \cot ^{\frac {3}{2}}(c+d x) (b+a \cot (c+d x))}{5 d}-\frac {4 \operatorname {Subst}\left (\int \frac {\frac {5}{2} \left (a^2 A-A b^2-2 a b B\right )-\frac {5}{2} \left (2 a A b+a^2 B-b^2 B\right ) x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{5 d}\\ &=\frac {2 \left (a^2 A-A b^2-2 a b B\right ) \sqrt {\cot (c+d x)}}{d}-\frac {2 a (7 A b+5 a B) \cot ^{\frac {3}{2}}(c+d x)}{15 d}-\frac {2 a A \cot ^{\frac {3}{2}}(c+d x) (b+a \cot (c+d x))}{5 d}-\frac {\left (a^2 (A-B)-b^2 (A-B)-2 a b (A+B)\right ) \operatorname {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{d}-\frac {\left (2 a b (A-B)+a^2 (A+B)-b^2 (A+B)\right ) \operatorname {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{d}\\ &=\frac {2 \left (a^2 A-A b^2-2 a b B\right ) \sqrt {\cot (c+d x)}}{d}-\frac {2 a (7 A b+5 a B) \cot ^{\frac {3}{2}}(c+d x)}{15 d}-\frac {2 a A \cot ^{\frac {3}{2}}(c+d x) (b+a \cot (c+d x))}{5 d}-\frac {\left (a^2 (A-B)-b^2 (A-B)-2 a b (A+B)\right ) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{2 d}-\frac {\left (a^2 (A-B)-b^2 (A-B)-2 a b (A+B)\right ) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{2 d}+\frac {\left (2 a b (A-B)+a^2 (A+B)-b^2 (A+B)\right ) \operatorname {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{2 \sqrt {2} d}+\frac {\left (2 a b (A-B)+a^2 (A+B)-b^2 (A+B)\right ) \operatorname {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{2 \sqrt {2} d}\\ &=\frac {2 \left (a^2 A-A b^2-2 a b B\right ) \sqrt {\cot (c+d x)}}{d}-\frac {2 a (7 A b+5 a B) \cot ^{\frac {3}{2}}(c+d x)}{15 d}-\frac {2 a A \cot ^{\frac {3}{2}}(c+d x) (b+a \cot (c+d x))}{5 d}+\frac {\left (2 a b (A-B)+a^2 (A+B)-b^2 (A+B)\right ) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} d}-\frac {\left (2 a b (A-B)+a^2 (A+B)-b^2 (A+B)\right ) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} d}-\frac {\left (a^2 (A-B)-b^2 (A-B)-2 a b (A+B)\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}+\frac {\left (a^2 (A-B)-b^2 (A-B)-2 a b (A+B)\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}\\ &=\frac {\left (a^2 (A-B)-b^2 (A-B)-2 a b (A+B)\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}-\frac {\left (a^2 (A-B)-b^2 (A-B)-2 a b (A+B)\right ) \tan ^{-1}\left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}+\frac {2 \left (a^2 A-A b^2-2 a b B\right ) \sqrt {\cot (c+d x)}}{d}-\frac {2 a (7 A b+5 a B) \cot ^{\frac {3}{2}}(c+d x)}{15 d}-\frac {2 a A \cot ^{\frac {3}{2}}(c+d x) (b+a \cot (c+d x))}{5 d}+\frac {\left (2 a b (A-B)+a^2 (A+B)-b^2 (A+B)\right ) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} d}-\frac {\left (2 a b (A-B)+a^2 (A+B)-b^2 (A+B)\right ) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} d}\\ \end {align*}
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Mathematica [A] time = 1.86, size = 255, normalized size = 0.78 \[ -\frac {\sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (30 \sqrt {2} \left (a^2 (A-B)-2 a b (A+B)+b^2 (B-A)\right ) \left (\tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )-\tan ^{-1}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )\right )-\frac {120 \left (a^2 A-2 a b B-A b^2\right )}{\sqrt {\tan (c+d x)}}-15 \sqrt {2} \left (a^2 (A+B)+2 a b (A-B)-b^2 (A+B)\right ) \left (\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )-\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )\right )+\frac {24 a^2 A}{\tan ^{\frac {5}{2}}(c+d x)}+\frac {40 a (a B+2 A b)}{\tan ^{\frac {3}{2}}(c+d x)}\right )}{60 d} \]
Antiderivative was successfully verified.
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fricas [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{2} \cot \left (d x + c\right )^{\frac {7}{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 2.11, size = 13170, normalized size = 40.40 \[ \text {output too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 1.97, size = 279, normalized size = 0.86 \[ -\frac {30 \, \sqrt {2} {\left ({\left (A - B\right )} a^{2} - 2 \, {\left (A + B\right )} a b - {\left (A - B\right )} b^{2}\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + 30 \, \sqrt {2} {\left ({\left (A - B\right )} a^{2} - 2 \, {\left (A + B\right )} a b - {\left (A - B\right )} b^{2}\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + 15 \, \sqrt {2} {\left ({\left (A + B\right )} a^{2} + 2 \, {\left (A - B\right )} a b - {\left (A + B\right )} b^{2}\right )} \log \left (\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) - 15 \, \sqrt {2} {\left ({\left (A + B\right )} a^{2} + 2 \, {\left (A - B\right )} a b - {\left (A + B\right )} b^{2}\right )} \log \left (-\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) + \frac {24 \, A a^{2}}{\tan \left (d x + c\right )^{\frac {5}{2}}} - \frac {120 \, {\left (A a^{2} - 2 \, B a b - A b^{2}\right )}}{\sqrt {\tan \left (d x + c\right )}} + \frac {40 \, {\left (B a^{2} + 2 \, A a b\right )}}{\tan \left (d x + c\right )^{\frac {3}{2}}}}{60 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int {\mathrm {cot}\left (c+d\,x\right )}^{7/2}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^2 \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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