Optimal. Leaf size=463 \[ \frac{2 b \left (22 a^2 B+27 a A b-9 b^2 B\right ) \tan ^{\frac{5}{2}}(c+d x)}{45 d}+\frac{2 \left (3 a^2 A b+a^3 B-3 a b^2 B-A b^3\right ) \tan ^{\frac{3}{2}}(c+d x)}{3 d}+\frac{\left (3 a^2 b (A-B)+a^3 (A+B)-3 a b^2 (A+B)-b^3 (A-B)\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}-\frac{\left (3 a^2 b (A-B)+a^3 (A+B)-3 a b^2 (A+B)-b^3 (A-B)\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} d}+\frac{2 \left (a^3 A-3 a^2 b B-3 a A b^2+b^3 B\right ) \sqrt{\tan (c+d x)}}{d}+\frac{\left (-3 a^2 b (A+B)+a^3 (A-B)-3 a b^2 (A-B)+b^3 (A+B)\right ) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{\left (-3 a^2 b (A+B)+a^3 (A-B)-3 a b^2 (A-B)+b^3 (A+B)\right ) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}+\frac{2 b^2 (13 a B+9 A b) \tan ^{\frac{7}{2}}(c+d x)}{63 d}+\frac{2 b B \tan ^{\frac{5}{2}}(c+d x) (a+b \tan (c+d x))^2}{9 d} \]
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Rubi [A] time = 0.91742, antiderivative size = 463, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 11, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3607, 3637, 3630, 3528, 3534, 1168, 1162, 617, 204, 1165, 628} \[ \frac{2 b \left (22 a^2 B+27 a A b-9 b^2 B\right ) \tan ^{\frac{5}{2}}(c+d x)}{45 d}+\frac{2 \left (3 a^2 A b+a^3 B-3 a b^2 B-A b^3\right ) \tan ^{\frac{3}{2}}(c+d x)}{3 d}+\frac{\left (3 a^2 b (A-B)+a^3 (A+B)-3 a b^2 (A+B)-b^3 (A-B)\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}-\frac{\left (3 a^2 b (A-B)+a^3 (A+B)-3 a b^2 (A+B)-b^3 (A-B)\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} d}+\frac{2 \left (a^3 A-3 a^2 b B-3 a A b^2+b^3 B\right ) \sqrt{\tan (c+d x)}}{d}+\frac{\left (-3 a^2 b (A+B)+a^3 (A-B)-3 a b^2 (A-B)+b^3 (A+B)\right ) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{\left (-3 a^2 b (A+B)+a^3 (A-B)-3 a b^2 (A-B)+b^3 (A+B)\right ) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}+\frac{2 b^2 (13 a B+9 A b) \tan ^{\frac{7}{2}}(c+d x)}{63 d}+\frac{2 b B \tan ^{\frac{5}{2}}(c+d x) (a+b \tan (c+d x))^2}{9 d} \]
Antiderivative was successfully verified.
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Rule 3607
Rule 3637
Rule 3630
Rule 3528
Rule 3534
Rule 1168
Rule 1162
Rule 617
Rule 204
Rule 1165
Rule 628
Rubi steps
\begin{align*} \int \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx &=\frac{2 b B \tan ^{\frac{5}{2}}(c+d x) (a+b \tan (c+d x))^2}{9 d}+\frac{2}{9} \int \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x)) \left (\frac{1}{2} a (9 a A-5 b B)+\frac{9}{2} \left (2 a A b+a^2 B-b^2 B\right ) \tan (c+d x)+\frac{1}{2} b (9 A b+13 a B) \tan ^2(c+d x)\right ) \, dx\\ &=\frac{2 b^2 (9 A b+13 a B) \tan ^{\frac{7}{2}}(c+d x)}{63 d}+\frac{2 b B \tan ^{\frac{5}{2}}(c+d x) (a+b \tan (c+d x))^2}{9 d}-\frac{4}{63} \int \tan ^{\frac{3}{2}}(c+d x) \left (-\frac{7}{4} a^2 (9 a A-5 b B)-\frac{63}{4} \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \tan (c+d x)-\frac{7}{4} b \left (27 a A b+22 a^2 B-9 b^2 B\right ) \tan ^2(c+d x)\right ) \, dx\\ &=\frac{2 b \left (27 a A b+22 a^2 B-9 b^2 B\right ) \tan ^{\frac{5}{2}}(c+d x)}{45 d}+\frac{2 b^2 (9 A b+13 a B) \tan ^{\frac{7}{2}}(c+d x)}{63 d}+\frac{2 b B \tan ^{\frac{5}{2}}(c+d x) (a+b \tan (c+d x))^2}{9 d}-\frac{4}{63} \int \tan ^{\frac{3}{2}}(c+d x) \left (-\frac{63}{4} \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right )-\frac{63}{4} \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \tan (c+d x)\right ) \, dx\\ &=\frac{2 \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \tan ^{\frac{3}{2}}(c+d x)}{3 d}+\frac{2 b \left (27 a A b+22 a^2 B-9 b^2 B\right ) \tan ^{\frac{5}{2}}(c+d x)}{45 d}+\frac{2 b^2 (9 A b+13 a B) \tan ^{\frac{7}{2}}(c+d x)}{63 d}+\frac{2 b B \tan ^{\frac{5}{2}}(c+d x) (a+b \tan (c+d x))^2}{9 d}-\frac{4}{63} \int \sqrt{\tan (c+d x)} \left (\frac{63}{4} \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right )-\frac{63}{4} \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) \tan (c+d x)\right ) \, dx\\ &=\frac{2 \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) \sqrt{\tan (c+d x)}}{d}+\frac{2 \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \tan ^{\frac{3}{2}}(c+d x)}{3 d}+\frac{2 b \left (27 a A b+22 a^2 B-9 b^2 B\right ) \tan ^{\frac{5}{2}}(c+d x)}{45 d}+\frac{2 b^2 (9 A b+13 a B) \tan ^{\frac{7}{2}}(c+d x)}{63 d}+\frac{2 b B \tan ^{\frac{5}{2}}(c+d x) (a+b \tan (c+d x))^2}{9 d}-\frac{4}{63} \int \frac{\frac{63}{4} \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right )+\frac{63}{4} \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \tan (c+d x)}{\sqrt{\tan (c+d x)}} \, dx\\ &=\frac{2 \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) \sqrt{\tan (c+d x)}}{d}+\frac{2 \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \tan ^{\frac{3}{2}}(c+d x)}{3 d}+\frac{2 b \left (27 a A b+22 a^2 B-9 b^2 B\right ) \tan ^{\frac{5}{2}}(c+d x)}{45 d}+\frac{2 b^2 (9 A b+13 a B) \tan ^{\frac{7}{2}}(c+d x)}{63 d}+\frac{2 b B \tan ^{\frac{5}{2}}(c+d x) (a+b \tan (c+d x))^2}{9 d}-\frac{8 \operatorname{Subst}\left (\int \frac{\frac{63}{4} \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right )+\frac{63}{4} \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{63 d}\\ &=\frac{2 \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) \sqrt{\tan (c+d x)}}{d}+\frac{2 \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \tan ^{\frac{3}{2}}(c+d x)}{3 d}+\frac{2 b \left (27 a A b+22 a^2 B-9 b^2 B\right ) \tan ^{\frac{5}{2}}(c+d x)}{45 d}+\frac{2 b^2 (9 A b+13 a B) \tan ^{\frac{7}{2}}(c+d x)}{63 d}+\frac{2 b B \tan ^{\frac{5}{2}}(c+d x) (a+b \tan (c+d x))^2}{9 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{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 \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) \sqrt{\tan (c+d x)}}{d}+\frac{2 \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \tan ^{\frac{3}{2}}(c+d x)}{3 d}+\frac{2 b \left (27 a A b+22 a^2 B-9 b^2 B\right ) \tan ^{\frac{5}{2}}(c+d x)}{45 d}+\frac{2 b^2 (9 A b+13 a B) \tan ^{\frac{7}{2}}(c+d x)}{63 d}+\frac{2 b B \tan ^{\frac{5}{2}}(c+d x) (a+b \tan (c+d x))^2}{9 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{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 ) \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{\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{\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 ) \log \left (1-\sqrt{2} \sqrt{\tan (c+d x)}+\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 ) \log \left (1+\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt{2} d}+\frac{2 \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) \sqrt{\tan (c+d x)}}{d}+\frac{2 \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \tan ^{\frac{3}{2}}(c+d x)}{3 d}+\frac{2 b \left (27 a A b+22 a^2 B-9 b^2 B\right ) \tan ^{\frac{5}{2}}(c+d x)}{45 d}+\frac{2 b^2 (9 A b+13 a B) \tan ^{\frac{7}{2}}(c+d x)}{63 d}+\frac{2 b B \tan ^{\frac{5}{2}}(c+d x) (a+b \tan (c+d x))^2}{9 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{1}{-1-x^2} \, dx,x,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 ) \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 (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a b^2 (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 ) \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 ) \log \left (1-\sqrt{2} \sqrt{\tan (c+d x)}+\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 ) \log \left (1+\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt{2} d}+\frac{2 \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) \sqrt{\tan (c+d x)}}{d}+\frac{2 \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \tan ^{\frac{3}{2}}(c+d x)}{3 d}+\frac{2 b \left (27 a A b+22 a^2 B-9 b^2 B\right ) \tan ^{\frac{5}{2}}(c+d x)}{45 d}+\frac{2 b^2 (9 A b+13 a B) \tan ^{\frac{7}{2}}(c+d x)}{63 d}+\frac{2 b B \tan ^{\frac{5}{2}}(c+d x) (a+b \tan (c+d x))^2}{9 d}\\ \end{align*}
Mathematica [C] time = 3.58508, size = 221, normalized size = 0.48 \[ \frac{2 \left (7 b \left (22 a^2 B+27 a A b-9 b^2 B\right ) \tan ^{\frac{5}{2}}(c+d x)+5 b^2 (13 a B+9 A b) \tan ^{\frac{7}{2}}(c+d x)+\frac{105}{2} (a-i b)^3 (B+i A) \left (-3 (-1)^{3/4} \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )+\sqrt{\tan (c+d x)} (\tan (c+d x)-3 i)\right )+\frac{105}{2} (a+i b)^3 (B-i A) \left (3 (-1)^{3/4} \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )+\sqrt{\tan (c+d x)} (\tan (c+d x)+3 i)\right )+35 b B \tan ^{\frac{5}{2}}(c+d x) (a+b \tan (c+d x))^2\right )}{315 d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.026, size = 1147, normalized size = 2.5 \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 = 1.71492, size = 537, normalized size = 1.16 \begin{align*} \frac{280 \, B b^{3} \tan \left (d x + c\right )^{\frac{9}{2}} + 360 \,{\left (3 \, B a b^{2} + A b^{3}\right )} \tan \left (d x + c\right )^{\frac{7}{2}} + 504 \,{\left (3 \, B a^{2} b + 3 \, A a b^{2} - B b^{3}\right )} \tan \left (d x + c\right )^{\frac{5}{2}} - 630 \, \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 ) - 630 \, \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 ) - 315 \, \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 ) + 315 \, \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 ) + 840 \,{\left (B a^{3} + 3 \, A a^{2} b - 3 \, B a b^{2} - A b^{3}\right )} \tan \left (d x + c\right )^{\frac{3}{2}} + 2520 \,{\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2} + B b^{3}\right )} \sqrt{\tan \left (d x + c\right )}}{1260 \, 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(-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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