Optimal. Leaf size=317 \[ -\frac{\left (a^2 (A-B)-2 a b (A+B)-b^2 (A-B)\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}+\frac{\left (a^2 (A-B)-2 a b (A+B)-b^2 (A-B)\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} d}+\frac{2 \left (a^2 A-2 a b B-A b^2\right )}{d \sqrt{\tan (c+d x)}}+\frac{\left (a^2 (A+B)+2 a b (A-B)-b^2 (A+B)\right ) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{\left (a^2 (A+B)+2 a b (A-B)-b^2 (A+B)\right ) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{2 a^2 A}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 a (a B+2 A b)}{3 d \tan ^{\frac{3}{2}}(c+d x)} \]
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Rubi [A] time = 0.445564, antiderivative size = 317, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 10, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.303, Rules used = {3604, 3628, 3529, 3534, 1168, 1162, 617, 204, 1165, 628} \[ -\frac{\left (a^2 (A-B)-2 a b (A+B)-b^2 (A-B)\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}+\frac{\left (a^2 (A-B)-2 a b (A+B)-b^2 (A-B)\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} d}+\frac{2 \left (a^2 A-2 a b B-A b^2\right )}{d \sqrt{\tan (c+d x)}}+\frac{\left (a^2 (A+B)+2 a b (A-B)-b^2 (A+B)\right ) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{\left (a^2 (A+B)+2 a b (A-B)-b^2 (A+B)\right ) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{2 a^2 A}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 a (a B+2 A b)}{3 d \tan ^{\frac{3}{2}}(c+d x)} \]
Antiderivative was successfully verified.
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Rule 3604
Rule 3628
Rule 3529
Rule 3534
Rule 1168
Rule 1162
Rule 617
Rule 204
Rule 1165
Rule 628
Rubi steps
\begin{align*} \int \frac{(a+b \tan (c+d x))^2 (A+B \tan (c+d x))}{\tan ^{\frac{7}{2}}(c+d x)} \, dx &=-\frac{2 a^2 A}{5 d \tan ^{\frac{5}{2}}(c+d x)}+\int \frac{a (2 A b+a B)-\left (a^2 A-A b^2-2 a b B\right ) \tan (c+d x)+b^2 B \tan ^2(c+d x)}{\tan ^{\frac{5}{2}}(c+d x)} \, dx\\ &=-\frac{2 a^2 A}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 a (2 A b+a B)}{3 d \tan ^{\frac{3}{2}}(c+d x)}+\int \frac{-a^2 A+A b^2+2 a b B+\left (b^2 B-a (2 A b+a B)\right ) \tan (c+d x)}{\tan ^{\frac{3}{2}}(c+d x)} \, dx\\ &=-\frac{2 a^2 A}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 a (2 A b+a B)}{3 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (a^2 A-A b^2-2 a b B\right )}{d \sqrt{\tan (c+d x)}}+\int \frac{b^2 B-a (2 A b+a B)+\left (a^2 A-A b^2-2 a b B\right ) \tan (c+d x)}{\sqrt{\tan (c+d x)}} \, dx\\ &=-\frac{2 a^2 A}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 a (2 A b+a B)}{3 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (a^2 A-A b^2-2 a b B\right )}{d \sqrt{\tan (c+d x)}}+\frac{2 \operatorname{Subst}\left (\int \frac{b^2 B-a (2 A b+a B)+\left (a^2 A-A b^2-2 a b B\right ) x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}\\ &=-\frac{2 a^2 A}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 a (2 A b+a B)}{3 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (a^2 A-A b^2-2 a b B\right )}{d \sqrt{\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{1+x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{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 )}{d}\\ &=-\frac{2 a^2 A}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 a (2 A b+a B)}{3 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (a^2 A-A b^2-2 a b B\right )}{d \sqrt{\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{1}{1-\sqrt{2} x+x^2} \, dx,x,\sqrt{\tan (c+d x)}\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}{1+\sqrt{2} x+x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{2 d}+\frac{\left (2 a b (A-B)+a^2 (A+B)-b^2 (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} d}+\frac{\left (2 a b (A-B)+a^2 (A+B)-b^2 (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} d}\\ &=\frac{\left (2 a b (A-B)+a^2 (A+B)-b^2 (A+B)\right ) \log \left (1-\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt{2} d}-\frac{\left (2 a b (A-B)+a^2 (A+B)-b^2 (A+B)\right ) \log \left (1+\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt{2} d}-\frac{2 a^2 A}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 a (2 A b+a B)}{3 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (a^2 A-A b^2-2 a b B\right )}{d \sqrt{\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{1}{-1-x^2} \, dx,x,1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}-\frac{\left (a^2 (A-B)-b^2 (A-B)-2 a b (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} d}\\ &=-\frac{\left (a^2 (A-B)-b^2 (A-B)-2 a b (A+B)\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}+\frac{\left (a^2 (A-B)-b^2 (A-B)-2 a b (A+B)\right ) \tan ^{-1}\left (1+\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}+\frac{\left (2 a b (A-B)+a^2 (A+B)-b^2 (A+B)\right ) \log \left (1-\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt{2} d}-\frac{\left (2 a b (A-B)+a^2 (A+B)-b^2 (A+B)\right ) \log \left (1+\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt{2} d}-\frac{2 a^2 A}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 a (2 A b+a B)}{3 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (a^2 A-A b^2-2 a b B\right )}{d \sqrt{\tan (c+d x)}}\\ \end{align*}
Mathematica [C] time = 0.597168, size = 120, normalized size = 0.38 \[ \frac{2 \left (\left (-3 a^2 A+6 a b B+3 A b^2\right ) \text{Hypergeometric2F1}\left (-\frac{5}{4},1,-\frac{1}{4},-\tan ^2(c+d x)\right )-5 \left (a^2 B+2 a A b-b^2 B\right ) \tan (c+d x) \text{Hypergeometric2F1}\left (-\frac{3}{4},1,\frac{1}{4},-\tan ^2(c+d x)\right )-b (6 a B+3 A b+5 b B \tan (c+d x))\right )}{15 d \tan ^{\frac{5}{2}}(c+d x)} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.029, size = 762, normalized size = 2.4 \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 [A] time = 1.82011, size = 373, normalized size = 1.18 \begin{align*} \frac{30 \, \sqrt{2}{\left ({\left (A - B\right )} a^{2} - 2 \,{\left (A + B\right )} a b -{\left (A - B\right )} b^{2}\right )} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} + 2 \, \sqrt{\tan \left (d x + c\right )}\right )}\right ) + 30 \, \sqrt{2}{\left ({\left (A - B\right )} a^{2} - 2 \,{\left (A + B\right )} a b -{\left (A - B\right )} b^{2}\right )} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} - 2 \, \sqrt{\tan \left (d x + c\right )}\right )}\right ) - 15 \, \sqrt{2}{\left ({\left (A + B\right )} a^{2} + 2 \,{\left (A - B\right )} a b -{\left (A + B\right )} b^{2}\right )} \log \left (\sqrt{2} \sqrt{\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + 15 \, \sqrt{2}{\left ({\left (A + B\right )} a^{2} + 2 \,{\left (A - B\right )} a b -{\left (A + B\right )} b^{2}\right )} \log \left (-\sqrt{2} \sqrt{\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - \frac{8 \,{\left (3 \, A a^{2} - 15 \,{\left (A a^{2} - 2 \, B a b - A b^{2}\right )} \tan \left (d x + c\right )^{2} + 5 \,{\left (B a^{2} + 2 \, A a b\right )} \tan \left (d x + c\right )\right )}}{\tan \left (d x + c\right )^{\frac{5}{2}}}}{60 \, d} \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 ) \left (a + b \tan{\left (c + d x \right )}\right )^{2}}{\tan ^{\frac{7}{2}}{\left (c + d x \right )}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.58846, size = 529, normalized size = 1.67 \begin{align*} \frac{{\left (\sqrt{2} A a^{2} - \sqrt{2} B a^{2} - 2 \, \sqrt{2} A a b - 2 \, \sqrt{2} B a b - \sqrt{2} A b^{2} + \sqrt{2} B b^{2}\right )} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} + 2 \, \sqrt{\tan \left (d x + c\right )}\right )}\right )}{2 \, d} + \frac{{\left (\sqrt{2} A a^{2} - \sqrt{2} B a^{2} - 2 \, \sqrt{2} A a b - 2 \, \sqrt{2} B a b - \sqrt{2} A b^{2} + \sqrt{2} B b^{2}\right )} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} - 2 \, \sqrt{\tan \left (d x + c\right )}\right )}\right )}{2 \, d} - \frac{{\left (\sqrt{2} A a^{2} + \sqrt{2} B a^{2} + 2 \, \sqrt{2} A a b - 2 \, \sqrt{2} B a b - \sqrt{2} A b^{2} - \sqrt{2} B b^{2}\right )} \log \left (\sqrt{2} \sqrt{\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )}{4 \, d} + \frac{{\left (\sqrt{2} A a^{2} + \sqrt{2} B a^{2} + 2 \, \sqrt{2} A a b - 2 \, \sqrt{2} B a b - \sqrt{2} A b^{2} - \sqrt{2} B b^{2}\right )} \log \left (-\sqrt{2} \sqrt{\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )}{4 \, d} + \frac{2 \,{\left (15 \, A a^{2} \tan \left (d x + c\right )^{2} - 30 \, B a b \tan \left (d x + c\right )^{2} - 15 \, A b^{2} \tan \left (d x + c\right )^{2} - 5 \, B a^{2} \tan \left (d x + c\right ) - 10 \, A a b \tan \left (d x + c\right ) - 3 \, A a^{2}\right )}}{15 \, d \tan \left (d x + c\right )^{\frac{5}{2}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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