Optimal. Leaf size=391 \[ -\frac{\left (a^2 (-(A+B))+2 a b (A-B)+b^2 (A+B)\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d \left (a^2+b^2\right )^2}+\frac{\left (a^2 (-(A+B))+2 a b (A-B)+b^2 (A+B)\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} d \left (a^2+b^2\right )^2}+\frac{\sqrt{a} \left (a^2 A b+a^3 B+5 a b^2 B-3 A b^3\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{b^{3/2} d \left (a^2+b^2\right )^2}+\frac{a (A b-a B) \sqrt{\tan (c+d x)}}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}+\frac{\left (a^2 (A-B)+2 a b (A+B)-b^2 (A-B)\right ) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )^2}-\frac{\left (a^2 (A-B)+2 a b (A+B)-b^2 (A-B)\right ) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )^2} \]
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Rubi [A] time = 0.788364, antiderivative size = 391, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 12, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {3605, 3653, 3534, 1168, 1162, 617, 204, 1165, 628, 3634, 63, 205} \[ -\frac{\left (a^2 (-(A+B))+2 a b (A-B)+b^2 (A+B)\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d \left (a^2+b^2\right )^2}+\frac{\left (a^2 (-(A+B))+2 a b (A-B)+b^2 (A+B)\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} d \left (a^2+b^2\right )^2}+\frac{\sqrt{a} \left (a^2 A b+a^3 B+5 a b^2 B-3 A b^3\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{b^{3/2} d \left (a^2+b^2\right )^2}+\frac{a (A b-a B) \sqrt{\tan (c+d x)}}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}+\frac{\left (a^2 (A-B)+2 a b (A+B)-b^2 (A-B)\right ) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )^2}-\frac{\left (a^2 (A-B)+2 a b (A+B)-b^2 (A-B)\right ) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d \left (a^2+b^2\right )^2} \]
Antiderivative was successfully verified.
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Rule 3605
Rule 3653
Rule 3534
Rule 1168
Rule 1162
Rule 617
Rule 204
Rule 1165
Rule 628
Rule 3634
Rule 63
Rule 205
Rubi steps
\begin{align*} \int \frac{\tan ^{\frac{3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx &=\frac{a (A b-a B) \sqrt{\tan (c+d x)}}{b \left (a^2+b^2\right ) d (a+b \tan (c+d x))}+\frac{\int \frac{-\frac{1}{2} a (A b-a B)+b (A b-a B) \tan (c+d x)+\frac{1}{2} \left (a A b+a^2 B+2 b^2 B\right ) \tan ^2(c+d x)}{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))} \, dx}{b \left (a^2+b^2\right )}\\ &=\frac{a (A b-a B) \sqrt{\tan (c+d x)}}{b \left (a^2+b^2\right ) d (a+b \tan (c+d x))}+\frac{\int \frac{-b \left (a^2 A-A b^2+2 a b B\right )+b \left (2 a A b-a^2 B+b^2 B\right ) \tan (c+d x)}{\sqrt{\tan (c+d x)}} \, dx}{b \left (a^2+b^2\right )^2}+\frac{\left (a \left (a^2 A b-3 A b^3+a^3 B+5 a b^2 B\right )\right ) \int \frac{1+\tan ^2(c+d x)}{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))} \, dx}{2 b \left (a^2+b^2\right )^2}\\ &=\frac{a (A b-a B) \sqrt{\tan (c+d x)}}{b \left (a^2+b^2\right ) d (a+b \tan (c+d x))}+\frac{2 \operatorname{Subst}\left (\int \frac{-b \left (a^2 A-A b^2+2 a b B\right )+b \left (2 a A b-a^2 B+b^2 B\right ) x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{b \left (a^2+b^2\right )^2 d}+\frac{\left (a \left (a^2 A b-3 A b^3+a^3 B+5 a b^2 B\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} (a+b x)} \, dx,x,\tan (c+d x)\right )}{2 b \left (a^2+b^2\right )^2 d}\\ &=\frac{a (A b-a B) \sqrt{\tan (c+d x)}}{b \left (a^2+b^2\right ) d (a+b \tan (c+d x))}+\frac{\left (a \left (a^2 A b-3 A b^3+a^3 B+5 a b^2 B\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{b \left (a^2+b^2\right )^2 d}-\frac{\left (a^2 (A-B)-b^2 (A-B)+2 a b (A+B)\right ) \operatorname{Subst}\left (\int \frac{1-x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{\left (a^2+b^2\right )^2 d}+\frac{\left (2 a b (A-B)-a^2 (A+B)+b^2 (A+B)\right ) \operatorname{Subst}\left (\int \frac{1+x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{\left (a^2+b^2\right )^2 d}\\ &=\frac{\sqrt{a} \left (a^2 A b-3 A b^3+a^3 B+5 a b^2 B\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{b^{3/2} \left (a^2+b^2\right )^2 d}+\frac{a (A b-a B) \sqrt{\tan (c+d x)}}{b \left (a^2+b^2\right ) d (a+b \tan (c+d x))}+\frac{\left (a^2 (A-B)-b^2 (A-B)+2 a b (A+B)\right ) \operatorname{Subst}\left (\int \frac{\sqrt{2}+2 x}{-1-\sqrt{2} x-x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{2 \sqrt{2} \left (a^2+b^2\right )^2 d}+\frac{\left (a^2 (A-B)-b^2 (A-B)+2 a b (A+B)\right ) \operatorname{Subst}\left (\int \frac{\sqrt{2}-2 x}{-1+\sqrt{2} x-x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{2 \sqrt{2} \left (a^2+b^2\right )^2 d}+\frac{\left (2 a b (A-B)-a^2 (A+B)+b^2 (A+B)\right ) \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2} x+x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{2 \left (a^2+b^2\right )^2 d}+\frac{\left (2 a b (A-B)-a^2 (A+B)+b^2 (A+B)\right ) \operatorname{Subst}\left (\int \frac{1}{1+\sqrt{2} x+x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{2 \left (a^2+b^2\right )^2 d}\\ &=\frac{\sqrt{a} \left (a^2 A b-3 A b^3+a^3 B+5 a b^2 B\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{b^{3/2} \left (a^2+b^2\right )^2 d}+\frac{\left (a^2 (A-B)-b^2 (A-B)+2 a b (A+B)\right ) \log \left (1-\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt{2} \left (a^2+b^2\right )^2 d}-\frac{\left (a^2 (A-B)-b^2 (A-B)+2 a b (A+B)\right ) \log \left (1+\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt{2} \left (a^2+b^2\right )^2 d}+\frac{a (A b-a B) \sqrt{\tan (c+d x)}}{b \left (a^2+b^2\right ) d (a+b \tan (c+d x))}+\frac{\left (2 a b (A-B)-a^2 (A+B)+b^2 (A+B)\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} \left (a^2+b^2\right )^2 d}-\frac{\left (2 a b (A-B)-a^2 (A+B)+b^2 (A+B)\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} \left (a^2+b^2\right )^2 d}\\ &=-\frac{\left (2 a b (A-B)-a^2 (A+B)+b^2 (A+B)\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} \left (a^2+b^2\right )^2 d}+\frac{\left (2 a b (A-B)-a^2 (A+B)+b^2 (A+B)\right ) \tan ^{-1}\left (1+\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} \left (a^2+b^2\right )^2 d}+\frac{\sqrt{a} \left (a^2 A b-3 A b^3+a^3 B+5 a b^2 B\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{b^{3/2} \left (a^2+b^2\right )^2 d}+\frac{\left (a^2 (A-B)-b^2 (A-B)+2 a b (A+B)\right ) \log \left (1-\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt{2} \left (a^2+b^2\right )^2 d}-\frac{\left (a^2 (A-B)-b^2 (A-B)+2 a b (A+B)\right ) \log \left (1+\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt{2} \left (a^2+b^2\right )^2 d}+\frac{a (A b-a B) \sqrt{\tan (c+d x)}}{b \left (a^2+b^2\right ) d (a+b \tan (c+d x))}\\ \end{align*}
Mathematica [C] time = 1.77828, size = 230, normalized size = 0.59 \[ \frac{\frac{\left (a^2 B+a A b+2 b^2 B\right ) \sqrt{\tan (c+d x)}}{\left (a^2+b^2\right ) (a+b \tan (c+d x))}+\frac{\sqrt{a} \left (a^2 A b+a^3 B+5 a b^2 B-3 A b^3\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )+\sqrt [4]{-1} b^{3/2} \left ((a+i b)^2 (A-i B) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )+(a-i b)^2 (A+i B) \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )\right )}{\sqrt{b} \left (a^2+b^2\right )^2}-\frac{2 B \sqrt{\tan (c+d x)}}{a+b \tan (c+d x)}}{b d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.056, size = 1136, normalized size = 2.9 \begin{align*} \text{result 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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