Optimal. Leaf size=380 \[ -\frac{\left (-3 a^2 b (A+B)+a^3 (A-B)-3 a b^2 (A-B)+b^3 (A+B)\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}+\frac{\left (-3 a^2 b (A+B)+a^3 (A-B)-3 a b^2 (A-B)+b^3 (A+B)\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} d}+\frac{2 a \left (5 a^2 A-15 a b B-14 A b^2\right )}{5 d \sqrt{\tan (c+d x)}}+\frac{\left (3 a^2 b (A-B)+a^3 (A+B)-3 a b^2 (A+B)-b^3 (A-B)\right ) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{\left (3 a^2 b (A-B)+a^3 (A+B)-3 a b^2 (A+B)-b^3 (A-B)\right ) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{2 a^2 (5 a B+9 A b)}{15 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 a A (a+b \tan (c+d x))^2}{5 d \tan ^{\frac{5}{2}}(c+d x)} \]
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Rubi [A] time = 0.659335, antiderivative size = 380, 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 = {3605, 3635, 3628, 3534, 1168, 1162, 617, 204, 1165, 628} \[ -\frac{\left (-3 a^2 b (A+B)+a^3 (A-B)-3 a b^2 (A-B)+b^3 (A+B)\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}+\frac{\left (-3 a^2 b (A+B)+a^3 (A-B)-3 a b^2 (A-B)+b^3 (A+B)\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} d}+\frac{2 a \left (5 a^2 A-15 a b B-14 A b^2\right )}{5 d \sqrt{\tan (c+d x)}}+\frac{\left (3 a^2 b (A-B)+a^3 (A+B)-3 a b^2 (A+B)-b^3 (A-B)\right ) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{\left (3 a^2 b (A-B)+a^3 (A+B)-3 a b^2 (A+B)-b^3 (A-B)\right ) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{2 a^2 (5 a B+9 A b)}{15 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 a A (a+b \tan (c+d x))^2}{5 d \tan ^{\frac{5}{2}}(c+d x)} \]
Antiderivative was successfully verified.
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Rule 3605
Rule 3635
Rule 3628
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))^3 (A+B \tan (c+d x))}{\tan ^{\frac{7}{2}}(c+d x)} \, dx &=-\frac{2 a A (a+b \tan (c+d x))^2}{5 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2}{5} \int \frac{(a+b \tan (c+d x)) \left (\frac{1}{2} a (9 A b+5 a B)-\frac{5}{2} \left (a^2 A-A b^2-2 a b B\right ) \tan (c+d x)-\frac{1}{2} b (a A-5 b B) \tan ^2(c+d x)\right )}{\tan ^{\frac{5}{2}}(c+d x)} \, dx\\ &=-\frac{2 a^2 (9 A b+5 a B)}{15 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 a A (a+b \tan (c+d x))^2}{5 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2}{5} \int \frac{-\frac{1}{2} a \left (5 a^2 A-14 A b^2-15 a b B\right )-\frac{5}{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}{2} b^2 (a A-5 b B) \tan ^2(c+d x)}{\tan ^{\frac{3}{2}}(c+d x)} \, dx\\ &=-\frac{2 a^2 (9 A b+5 a B)}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 a \left (5 a^2 A-14 A b^2-15 a b B\right )}{5 d \sqrt{\tan (c+d x)}}-\frac{2 a A (a+b \tan (c+d x))^2}{5 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2}{5} \int \frac{-\frac{5}{2} \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right )+\frac{5}{2} \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) \tan (c+d x)}{\sqrt{\tan (c+d x)}} \, dx\\ &=-\frac{2 a^2 (9 A b+5 a B)}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 a \left (5 a^2 A-14 A b^2-15 a b B\right )}{5 d \sqrt{\tan (c+d x)}}-\frac{2 a A (a+b \tan (c+d x))^2}{5 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{4 \operatorname{Subst}\left (\int \frac{-\frac{5}{2} \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right )+\frac{5}{2} \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{5 d}\\ &=-\frac{2 a^2 (9 A b+5 a B)}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 a \left (5 a^2 A-14 A b^2-15 a b B\right )}{5 d \sqrt{\tan (c+d x)}}-\frac{2 a A (a+b \tan (c+d x))^2}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{\left (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a 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{\left (a^3 (A-B)-3 a b^2 (A-B)-3 a^2 b (A+B)+b^3 (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 (9 A b+5 a B)}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 a \left (5 a^2 A-14 A b^2-15 a b B\right )}{5 d \sqrt{\tan (c+d x)}}-\frac{2 a A (a+b \tan (c+d x))^2}{5 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{\left (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a 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 (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a 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 (a^3 (A-B)-3 a b^2 (A-B)-3 a^2 b (A+B)+b^3 (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^3 (A-B)-3 a b^2 (A-B)-3 a^2 b (A+B)+b^3 (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 (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a 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 (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a 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 (9 A b+5 a B)}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 a \left (5 a^2 A-14 A b^2-15 a b B\right )}{5 d \sqrt{\tan (c+d x)}}-\frac{2 a A (a+b \tan (c+d x))^2}{5 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{\left (a^3 (A-B)-3 a b^2 (A-B)-3 a^2 b (A+B)+b^3 (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^3 (A-B)-3 a b^2 (A-B)-3 a^2 b (A+B)+b^3 (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^3 (A-B)-3 a b^2 (A-B)-3 a^2 b (A+B)+b^3 (A+B)\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}+\frac{\left (a^3 (A-B)-3 a b^2 (A-B)-3 a^2 b (A+B)+b^3 (A+B)\right ) \tan ^{-1}\left (1+\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}+\frac{\left (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a 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 (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a 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 (9 A b+5 a B)}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 a \left (5 a^2 A-14 A b^2-15 a b B\right )}{5 d \sqrt{\tan (c+d x)}}-\frac{2 a A (a+b \tan (c+d x))^2}{5 d \tan ^{\frac{5}{2}}(c+d x)}\\ \end{align*}
Mathematica [C] time = 1.34563, size = 166, normalized size = 0.44 \[ -\frac{2 \left (3 \left (a^3 A-3 a^2 b B-3 a A b^2+b^3 B\right ) \text{Hypergeometric2F1}\left (-\frac{5}{4},1,-\frac{1}{4},-\tan ^2(c+d x)\right )+5 \left (3 a^2 A b+a^3 B-3 a b^2 B-A b^3\right ) \tan (c+d x) \text{Hypergeometric2F1}\left (-\frac{3}{4},1,\frac{1}{4},-\tan ^2(c+d x)\right )+b \left (9 a^2 B+5 b (3 a B+A b) \tan (c+d x)+9 a A b+15 b^2 B \tan ^2(c+d x)-3 b^2 B\right )\right )}{15 d \tan ^{\frac{5}{2}}(c+d x)} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.028, size = 1007, normalized size = 2.7 \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 = 1.75453, size = 441, normalized size = 1.16 \begin{align*} \frac{30 \, \sqrt{2}{\left ({\left (A - B\right )} a^{3} - 3 \,{\left (A + B\right )} a^{2} b - 3 \,{\left (A - B\right )} a b^{2} +{\left (A + B\right )} b^{3}\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^{3} - 3 \,{\left (A + B\right )} a^{2} b - 3 \,{\left (A - B\right )} a b^{2} +{\left (A + B\right )} b^{3}\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^{3} + 3 \,{\left (A - B\right )} a^{2} b - 3 \,{\left (A + B\right )} a b^{2} -{\left (A - B\right )} b^{3}\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^{3} + 3 \,{\left (A - B\right )} a^{2} b - 3 \,{\left (A + B\right )} a b^{2} -{\left (A - B\right )} b^{3}\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^{3} - 15 \,{\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2}\right )} \tan \left (d x + c\right )^{2} + 5 \,{\left (B a^{3} + 3 \, A a^{2} 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 )^{3}}{\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 = 1.69691, size = 660, normalized size = 1.74 \begin{align*} \frac{{\left (\sqrt{2} A a^{3} - \sqrt{2} B a^{3} - 3 \, \sqrt{2} A a^{2} b - 3 \, \sqrt{2} B a^{2} b - 3 \, \sqrt{2} A a b^{2} + 3 \, \sqrt{2} B a b^{2} + \sqrt{2} A b^{3} + \sqrt{2} B b^{3}\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^{3} - \sqrt{2} B a^{3} - 3 \, \sqrt{2} A a^{2} b - 3 \, \sqrt{2} B a^{2} b - 3 \, \sqrt{2} A a b^{2} + 3 \, \sqrt{2} B a b^{2} + \sqrt{2} A b^{3} + \sqrt{2} B b^{3}\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^{3} + \sqrt{2} B a^{3} + 3 \, \sqrt{2} A a^{2} b - 3 \, \sqrt{2} B a^{2} b - 3 \, \sqrt{2} A a b^{2} - 3 \, \sqrt{2} B a b^{2} - \sqrt{2} A b^{3} + \sqrt{2} B b^{3}\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^{3} + \sqrt{2} B a^{3} + 3 \, \sqrt{2} A a^{2} b - 3 \, \sqrt{2} B a^{2} b - 3 \, \sqrt{2} A a b^{2} - 3 \, \sqrt{2} B a b^{2} - \sqrt{2} A b^{3} + \sqrt{2} B b^{3}\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^{3} \tan \left (d x + c\right )^{2} - 45 \, B a^{2} b \tan \left (d x + c\right )^{2} - 45 \, A a b^{2} \tan \left (d x + c\right )^{2} - 5 \, B a^{3} \tan \left (d x + c\right ) - 15 \, A a^{2} b \tan \left (d x + c\right ) - 3 \, A a^{3}\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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