Optimal. Leaf size=296 \[ \frac{2 \left (15 a^2 A+10 a b B-8 A b^2\right ) \sqrt{\cot (c+d x)} \sqrt{a+b \tan (c+d x)}}{15 a^3 d}+\frac{2 (4 A b-5 a B) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{15 a^2 d}+\frac{(-B+i A) \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d \sqrt{-b+i a}}-\frac{(B+i A) \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d \sqrt{b+i a}}-\frac{2 A \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{5 a d} \]
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Rubi [A] time = 1.15661, antiderivative size = 296, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.229, Rules used = {4241, 3609, 3649, 3616, 3615, 93, 203, 206} \[ \frac{2 \left (15 a^2 A+10 a b B-8 A b^2\right ) \sqrt{\cot (c+d x)} \sqrt{a+b \tan (c+d x)}}{15 a^3 d}+\frac{2 (4 A b-5 a B) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{15 a^2 d}+\frac{(-B+i A) \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d \sqrt{-b+i a}}-\frac{(B+i A) \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d \sqrt{b+i a}}-\frac{2 A \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{5 a d} \]
Antiderivative was successfully verified.
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Rule 4241
Rule 3609
Rule 3649
Rule 3616
Rule 3615
Rule 93
Rule 203
Rule 206
Rubi steps
\begin{align*} \int \frac{\cot ^{\frac{7}{2}}(c+d x) (A+B \tan (c+d x))}{\sqrt{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)}{\tan ^{\frac{7}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}} \, dx\\ &=-\frac{2 A \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{5 a d}-\frac{\left (2 \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{\frac{1}{2} (4 A b-5 a B)+\frac{5}{2} a A \tan (c+d x)+2 A b \tan ^2(c+d x)}{\tan ^{\frac{5}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}} \, dx}{5 a}\\ &=\frac{2 (4 A b-5 a B) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{15 a^2 d}-\frac{2 A \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{5 a d}+\frac{\left (4 \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{\frac{1}{4} \left (-15 a^2 A+8 A b^2-10 a b B\right )-\frac{15}{4} a^2 B \tan (c+d x)+\frac{1}{2} b (4 A b-5 a B) \tan ^2(c+d x)}{\tan ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}} \, dx}{15 a^2}\\ &=\frac{2 \left (15 a^2 A-8 A b^2+10 a b B\right ) \sqrt{\cot (c+d x)} \sqrt{a+b \tan (c+d x)}}{15 a^3 d}+\frac{2 (4 A b-5 a B) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{15 a^2 d}-\frac{2 A \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{5 a d}-\frac{\left (8 \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{\frac{15 a^3 B}{8}-\frac{15}{8} a^3 A \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx}{15 a^3}\\ &=\frac{2 \left (15 a^2 A-8 A b^2+10 a b B\right ) \sqrt{\cot (c+d x)} \sqrt{a+b \tan (c+d x)}}{15 a^3 d}+\frac{2 (4 A b-5 a B) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{15 a^2 d}-\frac{2 A \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{5 a d}+\frac{1}{2} \left ((i A-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 ((i A+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 (15 a^2 A-8 A b^2+10 a b B\right ) \sqrt{\cot (c+d x)} \sqrt{a+b \tan (c+d x)}}{15 a^3 d}+\frac{2 (4 A b-5 a B) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{15 a^2 d}-\frac{2 A \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{5 a d}+\frac{\left ((i A-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 ((i A+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 (15 a^2 A-8 A b^2+10 a b B\right ) \sqrt{\cot (c+d x)} \sqrt{a+b \tan (c+d x)}}{15 a^3 d}+\frac{2 (4 A b-5 a B) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{15 a^2 d}-\frac{2 A \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{5 a d}+\frac{\left ((i A-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 ((i A+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) \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)}}{\sqrt{i a-b} d}-\frac{(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)}}{\sqrt{i a+b} d}+\frac{2 \left (15 a^2 A-8 A b^2+10 a b B\right ) \sqrt{\cot (c+d x)} \sqrt{a+b \tan (c+d x)}}{15 a^3 d}+\frac{2 (4 A b-5 a B) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{15 a^2 d}-\frac{2 A \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{5 a d}\\ \end{align*}
Mathematica [A] time = 5.91789, size = 244, normalized size = 0.82 \[ \frac{\sqrt{\cot (c+d x)} \left (-\frac{2 \sqrt{a+b \tan (c+d x)} \left (3 a^2 A \cot ^2(c+d x)-15 a^2 A+a (5 a B-4 A b) \cot (c+d x)-10 a b B+8 A b^2\right )}{a^3}+\frac{15 \sqrt [4]{-1} (A+i B) \sqrt{\tan (c+d x)} \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{-a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{\sqrt{-a-i b}}-\frac{15 \sqrt [4]{-1} (A-i B) \sqrt{\tan (c+d x)} \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{\sqrt{a-i b}}\right )}{15 d} \]
Antiderivative was successfully verified.
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Maple [C] time = 1.385, size = 28811, normalized size = 97.3 \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 \frac{{\left (B \tan \left (d x + c\right ) + A\right )} \cot \left (d x + c\right )^{\frac{7}{2}}}{\sqrt{b \tan \left (d x + c\right ) + a}}\,{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] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \tan \left (d x + c\right ) + A\right )} \cot \left (d x + c\right )^{\frac{7}{2}}}{\sqrt{b \tan \left (d x + c\right ) + a}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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