Optimal. Leaf size=380 \[ \frac{2 a \left (5 a^2 A-15 a b B-14 A b^2\right ) \sqrt{\cot (c+d x)}}{5 d}+\frac{\left (3 a^2 b (A-B)+a^3 (A+B)-3 a b^2 (A+B)-b^3 (A-B)\right ) \log \left (\cot (c+d x)-\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{\left (3 a^2 b (A-B)+a^3 (A+B)-3 a b^2 (A+B)-b^3 (A-B)\right ) \log \left (\cot (c+d x)+\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{2 \sqrt{2} d}+\frac{\left (-3 a^2 b (A+B)+a^3 (A-B)-3 a b^2 (A-B)+b^3 (A+B)\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} d}-\frac{\left (-3 a^2 b (A+B)+a^3 (A-B)-3 a b^2 (A-B)+b^3 (A+B)\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{\sqrt{2} d}-\frac{2 a^2 (5 a B+9 A b) \cot ^{\frac{3}{2}}(c+d x)}{15 d}-\frac{2 a A \sqrt{\cot (c+d x)} (a \cot (c+d x)+b)^2}{5 d} \]
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Rubi [A] time = 0.769157, antiderivative size = 380, normalized size of antiderivative = 1., 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, 3637, 3630, 3534, 1168, 1162, 617, 204, 1165, 628} \[ \frac{2 a \left (5 a^2 A-15 a b B-14 A b^2\right ) \sqrt{\cot (c+d x)}}{5 d}+\frac{\left (3 a^2 b (A-B)+a^3 (A+B)-3 a b^2 (A+B)-b^3 (A-B)\right ) \log \left (\cot (c+d x)-\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{\left (3 a^2 b (A-B)+a^3 (A+B)-3 a b^2 (A+B)-b^3 (A-B)\right ) \log \left (\cot (c+d x)+\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{2 \sqrt{2} d}+\frac{\left (-3 a^2 b (A+B)+a^3 (A-B)-3 a b^2 (A-B)+b^3 (A+B)\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} d}-\frac{\left (-3 a^2 b (A+B)+a^3 (A-B)-3 a b^2 (A-B)+b^3 (A+B)\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{\sqrt{2} d}-\frac{2 a^2 (5 a B+9 A b) \cot ^{\frac{3}{2}}(c+d x)}{15 d}-\frac{2 a A \sqrt{\cot (c+d x)} (a \cot (c+d x)+b)^2}{5 d} \]
Antiderivative was successfully verified.
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Rule 3581
Rule 3607
Rule 3637
Rule 3630
Rule 3534
Rule 1168
Rule 1162
Rule 617
Rule 204
Rule 1165
Rule 628
Rubi steps
\begin{align*} \int \cot ^{\frac{7}{2}}(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx &=\int \frac{(b+a \cot (c+d x))^3 (B+A \cot (c+d x))}{\sqrt{\cot (c+d x)}} \, dx\\ &=-\frac{2 a A \sqrt{\cot (c+d x)} (b+a \cot (c+d x))^2}{5 d}-\frac{2}{5} \int \frac{(b+a \cot (c+d x)) \left (\frac{1}{2} b (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 (9 A b+5 a B) \cot ^2(c+d x)\right )}{\sqrt{\cot (c+d x)}} \, dx\\ &=-\frac{2 a^2 (9 A b+5 a B) \cot ^{\frac{3}{2}}(c+d x)}{15 d}-\frac{2 a A \sqrt{\cot (c+d x)} (b+a \cot (c+d x))^2}{5 d}-\frac{4}{15} \int \frac{\frac{3}{4} b^2 (a A-5 b B)+\frac{15}{4} \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \cot (c+d x)+\frac{3}{4} a \left (5 a^2 A-14 A b^2-15 a b B\right ) \cot ^2(c+d x)}{\sqrt{\cot (c+d x)}} \, dx\\ &=\frac{2 a \left (5 a^2 A-14 A b^2-15 a b B\right ) \sqrt{\cot (c+d x)}}{5 d}-\frac{2 a^2 (9 A b+5 a B) \cot ^{\frac{3}{2}}(c+d x)}{15 d}-\frac{2 a A \sqrt{\cot (c+d x)} (b+a \cot (c+d x))^2}{5 d}-\frac{4}{15} \int \frac{-\frac{15}{4} \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right )+\frac{15}{4} \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \cot (c+d x)}{\sqrt{\cot (c+d x)}} \, dx\\ &=\frac{2 a \left (5 a^2 A-14 A b^2-15 a b B\right ) \sqrt{\cot (c+d x)}}{5 d}-\frac{2 a^2 (9 A b+5 a B) \cot ^{\frac{3}{2}}(c+d x)}{15 d}-\frac{2 a A \sqrt{\cot (c+d x)} (b+a \cot (c+d x))^2}{5 d}-\frac{8 \operatorname{Subst}\left (\int \frac{\frac{15}{4} \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right )-\frac{15}{4} \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) x^2}{1+x^4} \, dx,x,\sqrt{\cot (c+d x)}\right )}{15 d}\\ &=\frac{2 a \left (5 a^2 A-14 A b^2-15 a b B\right ) \sqrt{\cot (c+d x)}}{5 d}-\frac{2 a^2 (9 A b+5 a B) \cot ^{\frac{3}{2}}(c+d x)}{15 d}-\frac{2 a A \sqrt{\cot (c+d x)} (b+a \cot (c+d x))^2}{5 d}-\frac{\left (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a 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{\left (a^3 (A-B)-3 a b^2 (A-B)-3 a^2 b (A+B)+b^3 (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 a \left (5 a^2 A-14 A b^2-15 a b B\right ) \sqrt{\cot (c+d x)}}{5 d}-\frac{2 a^2 (9 A b+5 a B) \cot ^{\frac{3}{2}}(c+d x)}{15 d}-\frac{2 a A \sqrt{\cot (c+d x)} (b+a \cot (c+d x))^2}{5 d}+\frac{\left (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a 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 (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a 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 (a^3 (A-B)-3 a b^2 (A-B)-3 a^2 b (A+B)+b^3 (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^3 (A-B)-3 a b^2 (A-B)-3 a^2 b (A+B)+b^3 (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{2 a \left (5 a^2 A-14 A b^2-15 a b B\right ) \sqrt{\cot (c+d x)}}{5 d}-\frac{2 a^2 (9 A b+5 a B) \cot ^{\frac{3}{2}}(c+d x)}{15 d}-\frac{2 a A \sqrt{\cot (c+d x)} (b+a \cot (c+d x))^2}{5 d}+\frac{\left (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a 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 (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a 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^3 (A-B)-3 a b^2 (A-B)-3 a^2 b (A+B)+b^3 (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^3 (A-B)-3 a b^2 (A-B)-3 a^2 b (A+B)+b^3 (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^3 (A-B)-3 a b^2 (A-B)-3 a^2 b (A+B)+b^3 (A+B)\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} d}-\frac{\left (a^3 (A-B)-3 a b^2 (A-B)-3 a^2 b (A+B)+b^3 (A+B)\right ) \tan ^{-1}\left (1+\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} d}+\frac{2 a \left (5 a^2 A-14 A b^2-15 a b B\right ) \sqrt{\cot (c+d x)}}{5 d}-\frac{2 a^2 (9 A b+5 a B) \cot ^{\frac{3}{2}}(c+d x)}{15 d}-\frac{2 a A \sqrt{\cot (c+d x)} (b+a \cot (c+d x))^2}{5 d}+\frac{\left (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a 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 (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a 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*}
Mathematica [A] time = 2.33852, size = 286, normalized size = 0.75 \[ \frac{2 \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \left (-\frac{\left (-3 a^2 b (A+B)+a^3 (A-B)+3 a b^2 (B-A)+b^3 (A+B)\right ) \left (\tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )-\tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )\right )}{2 \sqrt{2}}+\frac{a \left (a^2 A-3 a b B-3 A b^2\right )}{\sqrt{\tan (c+d x)}}+\frac{\left (3 a^2 b (A-B)+a^3 (A+B)-3 a b^2 (A+B)+b^3 (B-A)\right ) \left (\log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )-\log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )\right )}{4 \sqrt{2}}-\frac{a^2 (a B+3 A b)}{3 \tan ^{\frac{3}{2}}(c+d x)}-\frac{a^3 A}{5 \tan ^{\frac{5}{2}}(c+d x)}\right )}{d} \]
Antiderivative was successfully verified.
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Maple [C] time = 0.796, size = 17628, normalized size = 46.4 \begin{align*} \text{output too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.70487, size = 446, normalized size = 1.17 \begin{align*} -\frac{30 \, \sqrt{2}{\left ({\left (A - B\right )} a^{3} - 3 \,{\left (A + B\right )} a^{2} b - 3 \,{\left (A - B\right )} a b^{2} +{\left (A + B\right )} b^{3}\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^{3} - 3 \,{\left (A + B\right )} a^{2} b - 3 \,{\left (A - B\right )} a b^{2} +{\left (A + B\right )} b^{3}\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^{3} + 3 \,{\left (A - B\right )} a^{2} b - 3 \,{\left (A + B\right )} a b^{2} -{\left (A - B\right )} b^{3}\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^{3} + 3 \,{\left (A - B\right )} a^{2} b - 3 \,{\left (A + B\right )} a b^{2} -{\left (A - B\right )} b^{3}\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^{3}}{\tan \left (d x + c\right )^{\frac{5}{2}}} - \frac{120 \,{\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2}\right )}}{\sqrt{\tan \left (d x + c\right )}} + \frac{40 \,{\left (B a^{3} + 3 \, A a^{2} b\right )}}{\tan \left (d x + c\right )^{\frac{3}{2}}}}{60 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \tan \left (d x + c\right ) + A\right )}{\left (b \tan \left (d x + c\right ) + a\right )}^{3} \cot \left (d x + c\right )^{\frac{7}{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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