Optimal. Leaf size=264 \[ \frac{2 a (A b-a B) \tan ^2(c+d x)}{b d \left (a^2+b^2\right ) \sqrt{a+b \tan (c+d x)}}-\frac{2 \left (-4 a^2 B+3 a A b-b^2 B\right ) \tan (c+d x) \sqrt{a+b \tan (c+d x)}}{3 b^2 d \left (a^2+b^2\right )}+\frac{2 \left (6 a^2 A b-8 a^3 B-5 a b^2 B+3 A b^3\right ) \sqrt{a+b \tan (c+d x)}}{3 b^3 d \left (a^2+b^2\right )}+\frac{(A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )}{d (a-i b)^{3/2}}+\frac{(A+i B) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{d (a+i b)^{3/2}} \]
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Rubi [A] time = 0.723921, antiderivative size = 264, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.212, Rules used = {3605, 3647, 3630, 3539, 3537, 63, 208} \[ \frac{2 a (A b-a B) \tan ^2(c+d x)}{b d \left (a^2+b^2\right ) \sqrt{a+b \tan (c+d x)}}-\frac{2 \left (-4 a^2 B+3 a A b-b^2 B\right ) \tan (c+d x) \sqrt{a+b \tan (c+d x)}}{3 b^2 d \left (a^2+b^2\right )}+\frac{2 \left (6 a^2 A b-8 a^3 B-5 a b^2 B+3 A b^3\right ) \sqrt{a+b \tan (c+d x)}}{3 b^3 d \left (a^2+b^2\right )}+\frac{(A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )}{d (a-i b)^{3/2}}+\frac{(A+i B) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{d (a+i b)^{3/2}} \]
Antiderivative was successfully verified.
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Rule 3605
Rule 3647
Rule 3630
Rule 3539
Rule 3537
Rule 63
Rule 208
Rubi steps
\begin{align*} \int \frac{\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx &=\frac{2 a (A b-a B) \tan ^2(c+d x)}{b \left (a^2+b^2\right ) d \sqrt{a+b \tan (c+d x)}}+\frac{2 \int \frac{\tan (c+d x) \left (-2 a (A b-a B)+\frac{1}{2} b (A b-a B) \tan (c+d x)-\frac{1}{2} \left (3 a A b-4 a^2 B-b^2 B\right ) \tan ^2(c+d x)\right )}{\sqrt{a+b \tan (c+d x)}} \, dx}{b \left (a^2+b^2\right )}\\ &=\frac{2 a (A b-a B) \tan ^2(c+d x)}{b \left (a^2+b^2\right ) d \sqrt{a+b \tan (c+d x)}}-\frac{2 \left (3 a A b-4 a^2 B-b^2 B\right ) \tan (c+d x) \sqrt{a+b \tan (c+d x)}}{3 b^2 \left (a^2+b^2\right ) d}+\frac{4 \int \frac{\frac{1}{2} a \left (3 a A b-4 a^2 B-b^2 B\right )-\frac{3}{4} b^2 (a A+b B) \tan (c+d x)+\frac{1}{4} \left (6 a^2 A b+3 A b^3-8 a^3 B-5 a b^2 B\right ) \tan ^2(c+d x)}{\sqrt{a+b \tan (c+d x)}} \, dx}{3 b^2 \left (a^2+b^2\right )}\\ &=\frac{2 a (A b-a B) \tan ^2(c+d x)}{b \left (a^2+b^2\right ) d \sqrt{a+b \tan (c+d x)}}+\frac{2 \left (6 a^2 A b+3 A b^3-8 a^3 B-5 a b^2 B\right ) \sqrt{a+b \tan (c+d x)}}{3 b^3 \left (a^2+b^2\right ) d}-\frac{2 \left (3 a A b-4 a^2 B-b^2 B\right ) \tan (c+d x) \sqrt{a+b \tan (c+d x)}}{3 b^2 \left (a^2+b^2\right ) d}+\frac{4 \int \frac{-\frac{3}{4} b^2 (A b-a B)-\frac{3}{4} b^2 (a A+b B) \tan (c+d x)}{\sqrt{a+b \tan (c+d x)}} \, dx}{3 b^2 \left (a^2+b^2\right )}\\ &=\frac{2 a (A b-a B) \tan ^2(c+d x)}{b \left (a^2+b^2\right ) d \sqrt{a+b \tan (c+d x)}}+\frac{2 \left (6 a^2 A b+3 A b^3-8 a^3 B-5 a b^2 B\right ) \sqrt{a+b \tan (c+d x)}}{3 b^3 \left (a^2+b^2\right ) d}-\frac{2 \left (3 a A b-4 a^2 B-b^2 B\right ) \tan (c+d x) \sqrt{a+b \tan (c+d x)}}{3 b^2 \left (a^2+b^2\right ) d}-\frac{((i a+b) (A+i B)) \int \frac{1-i \tan (c+d x)}{\sqrt{a+b \tan (c+d x)}} \, dx}{2 \left (a^2+b^2\right )}+\frac{(i A+B) \int \frac{1+i \tan (c+d x)}{\sqrt{a+b \tan (c+d x)}} \, dx}{2 (a-i b)}\\ &=\frac{2 a (A b-a B) \tan ^2(c+d x)}{b \left (a^2+b^2\right ) d \sqrt{a+b \tan (c+d x)}}+\frac{2 \left (6 a^2 A b+3 A b^3-8 a^3 B-5 a b^2 B\right ) \sqrt{a+b \tan (c+d x)}}{3 b^3 \left (a^2+b^2\right ) d}-\frac{2 \left (3 a A b-4 a^2 B-b^2 B\right ) \tan (c+d x) \sqrt{a+b \tan (c+d x)}}{3 b^2 \left (a^2+b^2\right ) 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) 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) d}\\ &=\frac{2 a (A b-a B) \tan ^2(c+d x)}{b \left (a^2+b^2\right ) d \sqrt{a+b \tan (c+d x)}}+\frac{2 \left (6 a^2 A b+3 A b^3-8 a^3 B-5 a b^2 B\right ) \sqrt{a+b \tan (c+d x)}}{3 b^3 \left (a^2+b^2\right ) d}-\frac{2 \left (3 a A b-4 a^2 B-b^2 B\right ) \tan (c+d x) \sqrt{a+b \tan (c+d x)}}{3 b^2 \left (a^2+b^2\right ) d}+\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) 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) b d}\\ &=\frac{(A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )}{(a-i b)^{3/2} d}+\frac{(A+i B) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{(a+i b)^{3/2} d}+\frac{2 a (A b-a B) \tan ^2(c+d x)}{b \left (a^2+b^2\right ) d \sqrt{a+b \tan (c+d x)}}+\frac{2 \left (6 a^2 A b+3 A b^3-8 a^3 B-5 a b^2 B\right ) \sqrt{a+b \tan (c+d x)}}{3 b^3 \left (a^2+b^2\right ) d}-\frac{2 \left (3 a A b-4 a^2 B-b^2 B\right ) \tan (c+d x) \sqrt{a+b \tan (c+d x)}}{3 b^2 \left (a^2+b^2\right ) d}\\ \end{align*}
Mathematica [C] time = 3.29794, size = 300, normalized size = 1.14 \[ \frac{\frac{3 i (a A+b B) \left ((a+i b) \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},\frac{a+b \tan (c+d x)}{a-i b}\right )-(a-i b) \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},\frac{a+b \tan (c+d x)}{a+i b}\right )\right )}{\left (a^2+b^2\right ) \sqrt{a+b \tan (c+d x)}}+\frac{2 \left (-8 a^2 B+6 a A b+3 b^2 B\right )}{b^2 \sqrt{a+b \tan (c+d x)}}+\frac{2 (3 A b-4 a B) \tan (c+d x)}{b \sqrt{a+b \tan (c+d x)}}+3 i A \left (\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )}{\sqrt{a-i b}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{\sqrt{a+i b}}\right )+\frac{2 B \tan ^2(c+d x)}{\sqrt{a+b \tan (c+d x)}}}{3 b d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.14, size = 8025, normalized size = 30.4 \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(-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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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] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + B \tan{\left (c + d x \right )}\right ) \tan ^{3}{\left (c + d x \right )}}{\left (a + b \tan{\left (c + d x \right )}\right )^{\frac{3}{2}}}\, dx \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 )} \tan \left (d x + c\right )^{3}}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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