Optimal. Leaf size=220 \[ \frac{\sqrt{a} \left (8 a^2 A-20 a b B-15 A b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a}}\right )}{4 d}-\frac{(a-i b)^{5/2} (A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )}{d}-\frac{(a+i b)^{5/2} (A+i B) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{d}-\frac{a (4 a B+7 A b) \cot (c+d x) \sqrt{a+b \tan (c+d x)}}{4 d}-\frac{a A \cot ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{2 d} \]
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Rubi [A] time = 0.927036, antiderivative size = 220, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 8, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.242, Rules used = {3605, 3645, 3653, 3539, 3537, 63, 208, 3634} \[ \frac{\sqrt{a} \left (8 a^2 A-20 a b B-15 A b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a}}\right )}{4 d}-\frac{(a-i b)^{5/2} (A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )}{d}-\frac{(a+i b)^{5/2} (A+i B) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{d}-\frac{a (4 a B+7 A b) \cot (c+d x) \sqrt{a+b \tan (c+d x)}}{4 d}-\frac{a A \cot ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{2 d} \]
Antiderivative was successfully verified.
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Rule 3605
Rule 3645
Rule 3653
Rule 3539
Rule 3537
Rule 63
Rule 208
Rule 3634
Rubi steps
\begin{align*} \int \cot ^3(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx &=-\frac{a A \cot ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{2 d}+\frac{1}{2} \int \cot ^2(c+d x) \sqrt{a+b \tan (c+d x)} \left (\frac{1}{2} a (7 A b+4 a B)-2 \left (a^2 A-A b^2-2 a b B\right ) \tan (c+d x)-\frac{1}{2} b (a A-4 b B) \tan ^2(c+d x)\right ) \, dx\\ &=-\frac{a (7 A b+4 a B) \cot (c+d x) \sqrt{a+b \tan (c+d x)}}{4 d}-\frac{a A \cot ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{2 d}+\frac{1}{2} \int \frac{\cot (c+d x) \left (-\frac{1}{4} a \left (8 a^2 A-15 A b^2-20 a b B\right )-2 \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 (9 a A b+4 a^2 B-8 b^2 B\right ) \tan ^2(c+d x)\right )}{\sqrt{a+b \tan (c+d x)}} \, dx\\ &=-\frac{a (7 A b+4 a B) \cot (c+d x) \sqrt{a+b \tan (c+d x)}}{4 d}-\frac{a A \cot ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{2 d}+\frac{1}{2} \int \frac{-2 \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right )+2 \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-\frac{1}{8} \left (a \left (8 a^2 A-15 A b^2-20 a b B\right )\right ) \int \frac{\cot (c+d x) \left (1+\tan ^2(c+d x)\right )}{\sqrt{a+b \tan (c+d x)}} \, dx\\ &=-\frac{a (7 A b+4 a B) \cot (c+d x) \sqrt{a+b \tan (c+d x)}}{4 d}-\frac{a A \cot ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{2 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}{4} \left (-2 \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right )+2 i \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right )\right ) \int \frac{1-i \tan (c+d x)}{\sqrt{a+b \tan (c+d x)}} \, dx-\frac{\left (a \left (8 a^2 A-15 A b^2-20 a b B\right )\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{8 d}\\ &=-\frac{a (7 A b+4 a B) \cot (c+d x) \sqrt{a+b \tan (c+d x)}}{4 d}-\frac{a A \cot ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{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{\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 \left (8 a^2 A-15 A b^2-20 a b B\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b \tan (c+d x)}\right )}{4 b d}\\ &=\frac{\sqrt{a} \left (8 a^2 A-15 A b^2-20 a b B\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a}}\right )}{4 d}-\frac{a (7 A b+4 a B) \cot (c+d x) \sqrt{a+b \tan (c+d x)}}{4 d}-\frac{a A \cot ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{2 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{\sqrt{a} \left (8 a^2 A-15 A b^2-20 a b B\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a}}\right )}{4 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{a (7 A b+4 a B) \cot (c+d x) \sqrt{a+b \tan (c+d x)}}{4 d}-\frac{a A \cot ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{2 d}\\ \end{align*}
Mathematica [B] time = 2.34958, size = 448, normalized size = 2.04 \[ -\frac{-\sqrt{a} \left (8 a^2 A-20 a b B-15 A b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a}}\right )+4 a^2 A \sqrt{a+i b} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )+2 a^2 A \cot ^2(c+d x) \sqrt{a+b \tan (c+d x)}+4 i a^2 B \sqrt{a+i b} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )+4 a^2 B \cot (c+d x) \sqrt{a+b \tan (c+d x)}-4 A b^2 \sqrt{a+i b} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )+4 (a-i b)^{5/2} (A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )+8 i a A b \sqrt{a+i b} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )+9 a A b \cot (c+d x) \sqrt{a+b \tan (c+d x)}-4 i b^2 B \sqrt{a+i b} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )-8 a b B \sqrt{a+i b} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{4 d} \]
Antiderivative was successfully verified.
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Maple [C] time = 2.441, size = 128221, normalized size = 582.8 \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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