Optimal. Leaf size=182 \[ -\frac{2 a^{5/2} A \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a}}\right )}{d}+\frac{2 b (2 a B+A b) \sqrt{a+b \tan (c+d x)}}{d}+\frac{(a-i b)^{5/2} (A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )}{d}+\frac{(a+i b)^{5/2} (A+i B) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{d}+\frac{2 b B (a+b \tan (c+d x))^{3/2}}{3 d} \]
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Rubi [A] time = 0.84733, antiderivative size = 182, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 8, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.258, Rules used = {3607, 3647, 3653, 3539, 3537, 63, 208, 3634} \[ -\frac{2 a^{5/2} A \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a}}\right )}{d}+\frac{2 b (2 a B+A b) \sqrt{a+b \tan (c+d x)}}{d}+\frac{(a-i b)^{5/2} (A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )}{d}+\frac{(a+i b)^{5/2} (A+i B) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{d}+\frac{2 b B (a+b \tan (c+d x))^{3/2}}{3 d} \]
Antiderivative was successfully verified.
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Rule 3607
Rule 3647
Rule 3653
Rule 3539
Rule 3537
Rule 63
Rule 208
Rule 3634
Rubi steps
\begin{align*} \int \cot (c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx &=\frac{2 b B (a+b \tan (c+d x))^{3/2}}{3 d}+\frac{2}{3} \int \cot (c+d x) \sqrt{a+b \tan (c+d x)} \left (\frac{3 a^2 A}{2}+\frac{3}{2} \left (2 a A b+a^2 B-b^2 B\right ) \tan (c+d x)+\frac{3}{2} b (A b+2 a B) \tan ^2(c+d x)\right ) \, dx\\ &=\frac{2 b (A b+2 a B) \sqrt{a+b \tan (c+d x)}}{d}+\frac{2 b B (a+b \tan (c+d x))^{3/2}}{3 d}+\frac{4}{3} \int \frac{\cot (c+d x) \left (\frac{3 a^3 A}{4}+\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 \left (3 a A b+3 a^2 B-b^2 B\right ) \tan ^2(c+d x)\right )}{\sqrt{a+b \tan (c+d x)}} \, dx\\ &=\frac{2 b (A b+2 a B) \sqrt{a+b \tan (c+d x)}}{d}+\frac{2 b B (a+b \tan (c+d x))^{3/2}}{3 d}+\frac{4}{3} \int \frac{\frac{3}{4} \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right )-\frac{3}{4} \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) \tan (c+d x)}{\sqrt{a+b \tan (c+d x)}} \, dx+\left (a^3 A\right ) \int \frac{\cot (c+d x) \left (1+\tan ^2(c+d x)\right )}{\sqrt{a+b \tan (c+d x)}} \, dx\\ &=\frac{2 b (A b+2 a B) \sqrt{a+b \tan (c+d x)}}{d}+\frac{2 b B (a+b \tan (c+d x))^{3/2}}{3 d}+\frac{1}{2} \left ((a-i b)^3 (i A+B)\right ) \int \frac{1+i \tan (c+d x)}{\sqrt{a+b \tan (c+d x)}} \, dx+\frac{1}{3} \left (2 \left (\frac{3}{4} \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right )-\frac{3}{4} i \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right )\right )\right ) \int \frac{1-i \tan (c+d x)}{\sqrt{a+b \tan (c+d x)}} \, dx+\frac{\left (a^3 A\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{2 b (A b+2 a B) \sqrt{a+b \tan (c+d x)}}{d}+\frac{2 b B (a+b \tan (c+d x))^{3/2}}{3 d}+\frac{\left (2 a^3 A\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b \tan (c+d x)}\right )}{b d}-\frac{\left ((a-i b)^3 (A-i B)\right ) \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{a-i b x}} \, dx,x,i \tan (c+d x)\right )}{2 d}-\frac{\left ((a+i b)^3 (A+i B)\right ) \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{a+i b x}} \, dx,x,-i \tan (c+d x)\right )}{2 d}\\ &=-\frac{2 a^{5/2} A \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a}}\right )}{d}+\frac{2 b (A b+2 a B) \sqrt{a+b \tan (c+d x)}}{d}+\frac{2 b B (a+b \tan (c+d x))^{3/2}}{3 d}+\frac{\left (i (a+i b)^3 (A+i B)\right ) \operatorname{Subst}\left (\int \frac{1}{-1+\frac{i a}{b}-\frac{i x^2}{b}} \, dx,x,\sqrt{a+b \tan (c+d x)}\right )}{b d}-\frac{\left ((a-i b)^3 (i A+B)\right ) \operatorname{Subst}\left (\int \frac{1}{-1-\frac{i a}{b}+\frac{i x^2}{b}} \, dx,x,\sqrt{a+b \tan (c+d x)}\right )}{b d}\\ &=-\frac{2 a^{5/2} A \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a}}\right )}{d}+\frac{(a-i b)^{5/2} (A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )}{d}+\frac{(a+i b)^{5/2} (A+i B) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{d}+\frac{2 b (A b+2 a B) \sqrt{a+b \tan (c+d x)}}{d}+\frac{2 b B (a+b \tan (c+d x))^{3/2}}{3 d}\\ \end{align*}
Mathematica [A] time = 1.10937, size = 177, normalized size = 0.97 \[ \frac{2 \left (-3 a^{5/2} A \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a}}\right )+3 b (2 a B+A b) \sqrt{a+b \tan (c+d x)}+\frac{3}{2} (a-i b)^{5/2} (A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )+\frac{3}{2} (a+i b)^{5/2} (A+i B) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )+b B (a+b \tan (c+d x))^{3/2}\right )}{3 d} \]
Antiderivative was successfully verified.
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Maple [C] time = 3.712, size = 55566, normalized size = 305.3 \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(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \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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