Optimal. Leaf size=300 \[ -\frac{(-b+i a)^{5/2} (-B+i A) \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{2 a (a B+2 A b) \sqrt{\cot (c+d x)} \sqrt{a+b \tan (c+d x)}}{d}-\frac{(b+i a)^{5/2} (B+i A) \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{2 a A \cot ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d}+\frac{2 b^{5/2} B \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d} \]
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Rubi [A] time = 2.28855, antiderivative size = 300, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 11, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.314, Rules used = {4241, 3605, 3645, 3655, 6725, 63, 217, 206, 93, 205, 208} \[ -\frac{(-b+i a)^{5/2} (-B+i A) \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{2 a (a B+2 A b) \sqrt{\cot (c+d x)} \sqrt{a+b \tan (c+d x)}}{d}-\frac{(b+i a)^{5/2} (B+i A) \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{2 a A \cot ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d}+\frac{2 b^{5/2} B \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d} \]
Antiderivative was successfully verified.
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Rule 4241
Rule 3605
Rule 3645
Rule 3655
Rule 6725
Rule 63
Rule 217
Rule 206
Rule 93
Rule 205
Rule 208
Rubi steps
\begin{align*} \int \cot ^{\frac{5}{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{5}{2}}(c+d x)} \, dx\\ &=-\frac{2 a A \cot ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d}+\frac{1}{3} \left (2 \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{\sqrt{a+b \tan (c+d x)} \left (\frac{3}{2} a (2 A b+a B)-\frac{3}{2} \left (a^2 A-A b^2-2 a b B\right ) \tan (c+d x)+\frac{3}{2} b^2 B \tan ^2(c+d x)\right )}{\tan ^{\frac{3}{2}}(c+d x)} \, dx\\ &=-\frac{2 a (2 A b+a B) \sqrt{\cot (c+d x)} \sqrt{a+b \tan (c+d x)}}{d}-\frac{2 a A \cot ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d}+\frac{1}{3} \left (4 \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{-\frac{3}{4} a \left (a^2 A-3 A b^2-3 a b B\right )-\frac{3}{4} \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \tan (c+d x)+\frac{3}{4} b^3 B \tan ^2(c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx\\ &=-\frac{2 a (2 A b+a B) \sqrt{\cot (c+d x)} \sqrt{a+b \tan (c+d x)}}{d}-\frac{2 a A \cot ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d}+\frac{\left (4 \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{-\frac{3}{4} a \left (a^2 A-3 A b^2-3 a b B\right )-\frac{3}{4} \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) x+\frac{3}{4} b^3 B x^2}{\sqrt{x} \sqrt{a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{3 d}\\ &=-\frac{2 a (2 A b+a B) \sqrt{\cot (c+d x)} \sqrt{a+b \tan (c+d x)}}{d}-\frac{2 a A \cot ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d}+\frac{\left (4 \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \left (\frac{3 b^3 B}{4 \sqrt{x} \sqrt{a+b x}}-\frac{3 \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B+\left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) x\right )}{4 \sqrt{x} \sqrt{a+b x} \left (1+x^2\right )}\right ) \, dx,x,\tan (c+d x)\right )}{3 d}\\ &=-\frac{2 a (2 A b+a B) \sqrt{\cot (c+d x)} \sqrt{a+b \tan (c+d x)}}{d}-\frac{2 a A \cot ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d}-\frac{\left (\sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{a^3 A-3 a A b^2-3 a^2 b B+b^3 B+\left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) x}{\sqrt{x} \sqrt{a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{d}+\frac{\left (b^3 B \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac{2 a (2 A b+a B) \sqrt{\cot (c+d x)} \sqrt{a+b \tan (c+d x)}}{d}-\frac{2 a A \cot ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d}-\frac{\left (\sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \left (\frac{-3 a^2 A b+A b^3-a^3 B+3 a b^2 B+i \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right )}{2 (i-x) \sqrt{x} \sqrt{a+b x}}+\frac{3 a^2 A b-A b^3+a^3 B-3 a b^2 B+i \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right )}{2 \sqrt{x} (i+x) \sqrt{a+b x}}\right ) \, dx,x,\tan (c+d x)\right )}{d}+\frac{\left (2 b^3 B \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}\\ &=-\frac{2 a (2 A b+a B) \sqrt{\cot (c+d x)} \sqrt{a+b \tan (c+d x)}}{d}-\frac{2 a A \cot ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d}+\frac{\left (2 b^3 B \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{1-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 (i A+B) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} (i+x) \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}-\frac{\left (\left (-3 a^2 A b+A b^3-a^3 B+3 a b^2 B+i \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right )\right ) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{(i-x) \sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=\frac{2 b^{5/2} B \tanh ^{-1}\left (\frac{\sqrt{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 a (2 A b+a B) \sqrt{\cot (c+d x)} \sqrt{a+b \tan (c+d x)}}{d}-\frac{2 a A \cot ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d}-\frac{\left ((a-i b)^3 (i A+B) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{i-(-a+i b) x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{\left (\left (-3 a^2 A b+A b^3-a^3 B+3 a b^2 B+i \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right )\right ) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{i-(a+i 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{2 b^{5/2} B \tanh ^{-1}\left (\frac{\sqrt{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 a (2 A b+a B) \sqrt{\cot (c+d x)} \sqrt{a+b \tan (c+d x)}}{d}-\frac{2 a A \cot ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d}\\ \end{align*}
Mathematica [C] time = 40.15, size = 130606, normalized size = 435.35 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
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Maple [C] time = 2.002, size = 46754, normalized size = 155.9 \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{5}{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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