Optimal. Leaf size=224 \[ \frac{2 b (A b-a B)}{3 a d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac{2 b \left (3 a^2 A b-2 a^3 B+A b^3\right )}{a^2 d \left (a^2+b^2\right )^2 \sqrt{a+b \tan (c+d x)}}-\frac{2 A \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a}}\right )}{a^{5/2} d}+\frac{(A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )}{d (a-i b)^{5/2}}+\frac{(A+i B) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{d (a+i b)^{5/2}} \]
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Rubi [A] time = 0.918239, antiderivative size = 224, 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 = {3609, 3649, 3653, 3539, 3537, 63, 208, 3634} \[ \frac{2 b (A b-a B)}{3 a d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac{2 b \left (3 a^2 A b-2 a^3 B+A b^3\right )}{a^2 d \left (a^2+b^2\right )^2 \sqrt{a+b \tan (c+d x)}}-\frac{2 A \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a}}\right )}{a^{5/2} d}+\frac{(A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )}{d (a-i b)^{5/2}}+\frac{(A+i B) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{d (a+i b)^{5/2}} \]
Antiderivative was successfully verified.
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Rule 3609
Rule 3649
Rule 3653
Rule 3539
Rule 3537
Rule 63
Rule 208
Rule 3634
Rubi steps
\begin{align*} \int \frac{\cot (c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx &=\frac{2 b (A b-a B)}{3 a \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac{2 \int \frac{\cot (c+d x) \left (\frac{3}{2} A \left (a^2+b^2\right )-\frac{3}{2} a (A b-a B) \tan (c+d x)+\frac{3}{2} b (A b-a B) \tan ^2(c+d x)\right )}{(a+b \tan (c+d x))^{3/2}} \, dx}{3 a \left (a^2+b^2\right )}\\ &=\frac{2 b (A b-a B)}{3 a \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac{2 b \left (3 a^2 A b+A b^3-2 a^3 B\right )}{a^2 \left (a^2+b^2\right )^2 d \sqrt{a+b \tan (c+d x)}}+\frac{4 \int \frac{\cot (c+d x) \left (\frac{3}{4} A \left (a^2+b^2\right )^2-\frac{3}{4} a^2 \left (2 a A b-a^2 B+b^2 B\right ) \tan (c+d x)+\frac{3}{4} b \left (3 a^2 A b+A b^3-2 a^3 B\right ) \tan ^2(c+d x)\right )}{\sqrt{a+b \tan (c+d x)}} \, dx}{3 a^2 \left (a^2+b^2\right )^2}\\ &=\frac{2 b (A b-a B)}{3 a \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac{2 b \left (3 a^2 A b+A b^3-2 a^3 B\right )}{a^2 \left (a^2+b^2\right )^2 d \sqrt{a+b \tan (c+d x)}}+\frac{A \int \frac{\cot (c+d x) \left (1+\tan ^2(c+d x)\right )}{\sqrt{a+b \tan (c+d x)}} \, dx}{a^2}+\frac{4 \int \frac{-\frac{3}{4} a^2 \left (2 a A b-a^2 B+b^2 B\right )-\frac{3}{4} a^2 \left (a^2 A-A b^2+2 a b B\right ) \tan (c+d x)}{\sqrt{a+b \tan (c+d x)}} \, dx}{3 a^2 \left (a^2+b^2\right )^2}\\ &=\frac{2 b (A b-a B)}{3 a \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac{2 b \left (3 a^2 A b+A b^3-2 a^3 B\right )}{a^2 \left (a^2+b^2\right )^2 d \sqrt{a+b \tan (c+d x)}}-\frac{(i A-B) \int \frac{1-i \tan (c+d x)}{\sqrt{a+b \tan (c+d x)}} \, dx}{2 (a+i b)^2}+\frac{(i A+B) \int \frac{1+i \tan (c+d x)}{\sqrt{a+b \tan (c+d x)}} \, dx}{2 (a-i b)^2}+\frac{A \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{a^2 d}\\ &=\frac{2 b (A b-a B)}{3 a \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac{2 b \left (3 a^2 A b+A b^3-2 a^3 B\right )}{a^2 \left (a^2+b^2\right )^2 d \sqrt{a+b \tan (c+d x)}}+\frac{(2 A) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b \tan (c+d x)}\right )}{a^2 b d}-\frac{(A-i B) \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{a-i b x}} \, dx,x,i \tan (c+d x)\right )}{2 (a-i b)^2 d}-\frac{(A+i B) \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{a+i b x}} \, dx,x,-i \tan (c+d x)\right )}{2 (a+i b)^2 d}\\ &=-\frac{2 A \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a}}\right )}{a^{5/2} d}+\frac{2 b (A b-a B)}{3 a \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac{2 b \left (3 a^2 A b+A b^3-2 a^3 B\right )}{a^2 \left (a^2+b^2\right )^2 d \sqrt{a+b \tan (c+d x)}}+\frac{(i (A+i B)) \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 )}{(a+i b)^2 b d}-\frac{(i A+B) \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 )}{(a-i b)^2 b d}\\ &=-\frac{2 A \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a}}\right )}{a^{5/2} d}+\frac{(A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )}{(a-i b)^{5/2} d}+\frac{(A+i B) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{(a+i b)^{5/2} d}+\frac{2 b (A b-a B)}{3 a \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac{2 b \left (3 a^2 A b+A b^3-2 a^3 B\right )}{a^2 \left (a^2+b^2\right )^2 d \sqrt{a+b \tan (c+d x)}}\\ \end{align*}
Mathematica [A] time = 4.89794, size = 242, normalized size = 1.08 \[ \frac{2 \left (\frac{3 b \left (3 a^2 A b-2 a^3 B+A b^3\right )}{a \left (a^2+b^2\right ) \sqrt{a+b \tan (c+d x)}}-\frac{3 A \left (a^2+b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a}}\right )}{a^{3/2}}+\frac{b (A b-a B)}{(a+b \tan (c+d x))^{3/2}}+\frac{3 a (a+i b) (A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )}{2 (a-i b)^{3/2}}+\frac{3 a (a-i b) (A+i B) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{2 (a+i b)^{3/2}}\right )}{3 a d \left (a^2+b^2\right )} \]
Antiderivative was successfully verified.
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Maple [C] time = 5.095, size = 185586, normalized size = 828.5 \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] 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 )}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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