Optimal. Leaf size=326 \[ -\frac{\left (a^2 (A-B)-2 a b (A+B)-b^2 (A-B)\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}+\frac{\left (a^2 (A-B)-2 a b (A+B)-b^2 (A-B)\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} d}+\frac{2 \left (a^2 B+2 a A b-b^2 B\right ) \sqrt{\tan (c+d x)}}{d}+\frac{\left (a^2 (A+B)+2 a b (A-B)-b^2 (A+B)\right ) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{\left (a^2 (A+B)+2 a b (A-B)-b^2 (A+B)\right ) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}+\frac{2 b (7 a B+5 A b) \tan ^{\frac{3}{2}}(c+d x)}{15 d}+\frac{2 b B \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d} \]
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Rubi [A] time = 0.519279, antiderivative size = 326, 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 = {3607, 3630, 3528, 3534, 1168, 1162, 617, 204, 1165, 628} \[ -\frac{\left (a^2 (A-B)-2 a b (A+B)-b^2 (A-B)\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}+\frac{\left (a^2 (A-B)-2 a b (A+B)-b^2 (A-B)\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} d}+\frac{2 \left (a^2 B+2 a A b-b^2 B\right ) \sqrt{\tan (c+d x)}}{d}+\frac{\left (a^2 (A+B)+2 a b (A-B)-b^2 (A+B)\right ) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{\left (a^2 (A+B)+2 a b (A-B)-b^2 (A+B)\right ) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}+\frac{2 b (7 a B+5 A b) \tan ^{\frac{3}{2}}(c+d x)}{15 d}+\frac{2 b B \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d} \]
Antiderivative was successfully verified.
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Rule 3607
Rule 3630
Rule 3528
Rule 3534
Rule 1168
Rule 1162
Rule 617
Rule 204
Rule 1165
Rule 628
Rubi steps
\begin{align*} \int \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx &=\frac{2 b B \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}+\frac{2}{5} \int \sqrt{\tan (c+d x)} \left (\frac{1}{2} a (5 a A-3 b B)+\frac{5}{2} \left (2 a A b+a^2 B-b^2 B\right ) \tan (c+d x)+\frac{1}{2} b (5 A b+7 a B) \tan ^2(c+d x)\right ) \, dx\\ &=\frac{2 b (5 A b+7 a B) \tan ^{\frac{3}{2}}(c+d x)}{15 d}+\frac{2 b B \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}+\frac{2}{5} \int \sqrt{\tan (c+d x)} \left (\frac{5}{2} \left (a^2 A-A b^2-2 a b B\right )+\frac{5}{2} \left (2 a A b+a^2 B-b^2 B\right ) \tan (c+d x)\right ) \, dx\\ &=\frac{2 \left (2 a A b+a^2 B-b^2 B\right ) \sqrt{\tan (c+d x)}}{d}+\frac{2 b (5 A b+7 a B) \tan ^{\frac{3}{2}}(c+d x)}{15 d}+\frac{2 b B \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}+\frac{2}{5} \int \frac{-\frac{5}{2} \left (2 a A b+a^2 B-b^2 B\right )+\frac{5}{2} \left (a^2 A-A b^2-2 a b B\right ) \tan (c+d x)}{\sqrt{\tan (c+d x)}} \, dx\\ &=\frac{2 \left (2 a A b+a^2 B-b^2 B\right ) \sqrt{\tan (c+d x)}}{d}+\frac{2 b (5 A b+7 a B) \tan ^{\frac{3}{2}}(c+d x)}{15 d}+\frac{2 b B \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}+\frac{4 \operatorname{Subst}\left (\int \frac{-\frac{5}{2} \left (2 a A b+a^2 B-b^2 B\right )+\frac{5}{2} \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 )}{5 d}\\ &=\frac{2 \left (2 a A b+a^2 B-b^2 B\right ) \sqrt{\tan (c+d x)}}{d}+\frac{2 b (5 A b+7 a B) \tan ^{\frac{3}{2}}(c+d x)}{15 d}+\frac{2 b B \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))}{5 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 )}{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 \left (2 a A b+a^2 B-b^2 B\right ) \sqrt{\tan (c+d x)}}{d}+\frac{2 b (5 A b+7 a B) \tan ^{\frac{3}{2}}(c+d x)}{15 d}+\frac{2 b B \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))}{5 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 (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 \left (2 a A b+a^2 B-b^2 B\right ) \sqrt{\tan (c+d x)}}{d}+\frac{2 b (5 A b+7 a B) \tan ^{\frac{3}{2}}(c+d x)}{15 d}+\frac{2 b B \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))}{5 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 ) \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 \left (2 a A b+a^2 B-b^2 B\right ) \sqrt{\tan (c+d x)}}{d}+\frac{2 b (5 A b+7 a B) \tan ^{\frac{3}{2}}(c+d x)}{15 d}+\frac{2 b B \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}\\ \end{align*}
Mathematica [C] time = 1.19863, size = 151, normalized size = 0.46 \[ \frac{2 \sqrt{\tan (c+d x)} \left (15 \left (a^2 B+2 a A b-b^2 B\right )+5 b (2 a B+A b) \tan (c+d x)+3 b^2 B \tan ^2(c+d x)\right )+15 \sqrt [4]{-1} (a-i b)^2 (B+i A) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )-15 (-1)^{3/4} (a+i b)^2 (A+i B) \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{15 d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.022, size = 762, normalized size = 2.3 \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 = 1.71455, size = 371, normalized size = 1.14 \begin{align*} \frac{24 \, B b^{2} \tan \left (d x + c\right )^{\frac{5}{2}} + 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 ) + 40 \,{\left (2 \, B a b + A b^{2}\right )} \tan \left (d x + c\right )^{\frac{3}{2}} + 120 \,{\left (B a^{2} + 2 \, A a b - B b^{2}\right )} \sqrt{\tan \left (d x + c\right )}}{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 \left (A + B \tan{\left (c + d x \right )}\right ) \left (a + b \tan{\left (c + d x \right )}\right )^{2} \sqrt{\tan{\left (c + d x \right )}}\, dx \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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