Optimal. Leaf size=349 \[ \frac{2 \left (35 a^2 A-77 a b B-45 A b^2\right ) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{105 d}+\frac{2 \left (245 a^2 A b+105 a^3 B-161 a b^2 B-15 A b^3\right ) \sqrt{\cot (c+d x)} \sqrt{a+b \tan (c+d x)}}{105 a d}-\frac{2 a (7 a B+10 A b) \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{35 d}+\frac{(-b+i a)^{5/2} (-B+i A) \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}+\frac{(b+i a)^{5/2} (B+i A) \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{2 a A \cot ^{\frac{7}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{7 d} \]
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Rubi [A] time = 1.65267, antiderivative size = 349, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 9, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.257, Rules used = {4241, 3605, 3645, 3649, 3616, 3615, 93, 203, 206} \[ \frac{2 \left (35 a^2 A-77 a b B-45 A b^2\right ) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{105 d}+\frac{2 \left (245 a^2 A b+105 a^3 B-161 a b^2 B-15 A b^3\right ) \sqrt{\cot (c+d x)} \sqrt{a+b \tan (c+d x)}}{105 a d}-\frac{2 a (7 a B+10 A b) \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{35 d}+\frac{(-b+i a)^{5/2} (-B+i A) \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}+\frac{(b+i a)^{5/2} (B+i A) \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{2 a A \cot ^{\frac{7}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{7 d} \]
Antiderivative was successfully verified.
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Rule 4241
Rule 3605
Rule 3645
Rule 3649
Rule 3616
Rule 3615
Rule 93
Rule 203
Rule 206
Rubi steps
\begin{align*} \int \cot ^{\frac{9}{2}}(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx &=\left (\sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac{9}{2}}(c+d x)} \, dx\\ &=-\frac{2 a A \cot ^{\frac{7}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{7 d}+\frac{1}{7} \left (2 \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{\sqrt{a+b \tan (c+d x)} \left (\frac{1}{2} a (10 A b+7 a B)-\frac{7}{2} \left (a^2 A-A b^2-2 a b B\right ) \tan (c+d x)-\frac{1}{2} b (4 a A-7 b B) \tan ^2(c+d x)\right )}{\tan ^{\frac{7}{2}}(c+d x)} \, dx\\ &=-\frac{2 a (10 A b+7 a B) \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{35 d}-\frac{2 a A \cot ^{\frac{7}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{7 d}+\frac{1}{35} \left (4 \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{-\frac{1}{4} a \left (35 a^2 A-45 A b^2-77 a b B\right )-\frac{35}{4} \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \tan (c+d x)-\frac{1}{4} b \left (60 a A b+28 a^2 B-35 b^2 B\right ) \tan ^2(c+d x)}{\tan ^{\frac{5}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}} \, dx\\ &=\frac{2 \left (35 a^2 A-45 A b^2-77 a b B\right ) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{105 d}-\frac{2 a (10 A b+7 a B) \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{35 d}-\frac{2 a A \cot ^{\frac{7}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{7 d}-\frac{\left (8 \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{\frac{1}{8} a \left (245 a^2 A b-15 A b^3+105 a^3 B-161 a b^2 B\right )-\frac{105}{8} a \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) \tan (c+d x)-\frac{1}{4} a b \left (35 a^2 A-45 A b^2-77 a b B\right ) \tan ^2(c+d x)}{\tan ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}} \, dx}{105 a}\\ &=\frac{2 \left (245 a^2 A b-15 A b^3+105 a^3 B-161 a b^2 B\right ) \sqrt{\cot (c+d x)} \sqrt{a+b \tan (c+d x)}}{105 a d}+\frac{2 \left (35 a^2 A-45 A b^2-77 a b B\right ) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{105 d}-\frac{2 a (10 A b+7 a B) \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{35 d}-\frac{2 a A \cot ^{\frac{7}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{7 d}+\frac{\left (16 \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{\frac{105}{16} a^2 \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right )+\frac{105}{16} a^2 \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx}{105 a^2}\\ &=\frac{2 \left (245 a^2 A b-15 A b^3+105 a^3 B-161 a b^2 B\right ) \sqrt{\cot (c+d x)} \sqrt{a+b \tan (c+d x)}}{105 a d}+\frac{2 \left (35 a^2 A-45 A b^2-77 a b B\right ) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{105 d}-\frac{2 a (10 A b+7 a B) \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{35 d}-\frac{2 a A \cot ^{\frac{7}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{7 d}+\frac{1}{2} \left ((a-i b)^3 (A-i B) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{1+i \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx+\frac{1}{2} \left ((a+i b)^3 (A+i B) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{1-i \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx\\ &=\frac{2 \left (245 a^2 A b-15 A b^3+105 a^3 B-161 a b^2 B\right ) \sqrt{\cot (c+d x)} \sqrt{a+b \tan (c+d x)}}{105 a d}+\frac{2 \left (35 a^2 A-45 A b^2-77 a b B\right ) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{105 d}-\frac{2 a (10 A b+7 a B) \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{35 d}-\frac{2 a A \cot ^{\frac{7}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{7 d}+\frac{\left ((a-i b)^3 (A-i B) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{(1-i x) \sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac{\left ((a+i b)^3 (A+i B) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{(1+i x) \sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=\frac{2 \left (245 a^2 A b-15 A b^3+105 a^3 B-161 a b^2 B\right ) \sqrt{\cot (c+d x)} \sqrt{a+b \tan (c+d x)}}{105 a d}+\frac{2 \left (35 a^2 A-45 A b^2-77 a b B\right ) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{105 d}-\frac{2 a (10 A b+7 a B) \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{35 d}-\frac{2 a A \cot ^{\frac{7}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{7 d}+\frac{\left ((a-i b)^3 (A-i B) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{1-(i a+b) x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}+\frac{\left ((a+i b)^3 (A+i B) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{1-(-i a+b) x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}\\ &=\frac{(i a-b)^{5/2} (i A-B) \tan ^{-1}\left (\frac{\sqrt{i a-b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right ) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}}{d}+\frac{(i a+b)^{5/2} (i A+B) \tanh ^{-1}\left (\frac{\sqrt{i a+b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right ) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}}{d}+\frac{2 \left (245 a^2 A b-15 A b^3+105 a^3 B-161 a b^2 B\right ) \sqrt{\cot (c+d x)} \sqrt{a+b \tan (c+d x)}}{105 a d}+\frac{2 \left (35 a^2 A-45 A b^2-77 a b B\right ) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{105 d}-\frac{2 a (10 A b+7 a B) \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{35 d}-\frac{2 a A \cot ^{\frac{7}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{7 d}\\ \end{align*}
Mathematica [A] time = 5.17812, size = 386, normalized size = 1.11 \[ -\frac{\cot ^{\frac{7}{2}}(c+d x) \left (6 a \left (28 a^2 B+60 a A b-35 b^2 B\right ) \tan (c+d x) \sqrt{a+b \tan (c+d x)}+5 a \left (24 a^2 A-49 a b B-28 A b^2\right ) \sqrt{a+b \tan (c+d x)}-4 \tan ^2(c+d x) \left (2 \left (245 a^2 A b+105 a^3 B-161 a b^2 B-15 A b^3\right ) \tan (c+d x) \sqrt{a+b \tan (c+d x)}+2 a \left (35 a^2 A-77 a b B-45 A b^2\right ) \sqrt{a+b \tan (c+d x)}+105 \sqrt [4]{-1} a \tan ^{\frac{3}{2}}(c+d x) \left (i (-a-i b)^{5/2} (A+i B) \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{-a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )-(a-i b)^{5/2} (B+i A) \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )\right )\right )+35 a b (a B+4 A b) \sqrt{a+b \tan (c+d x)}+210 a b B (a+b \tan (c+d x))^{3/2}\right )}{420 a d} \]
Antiderivative was successfully verified.
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Maple [C] time = 2.56, size = 67683, normalized size = 193.9 \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 [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 )}^{\frac{5}{2}} \cot \left (d x + c\right )^{\frac{9}{2}}\,{d x} \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(-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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