Optimal. Leaf size=378 \[ \frac{2 \left (231 a^2 A b+105 a^3 B-135 a b^2 B-5 A b^3\right ) \sqrt{a+b \tan (c+d x)}}{315 a d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (21 a^2 A-45 a b B-25 A b^2\right ) \sqrt{a+b \tan (c+d x)}}{105 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 \left (-483 a^2 A b^2+315 a^4 A-735 a^3 b B+45 a b^3 B-10 A b^4\right ) \sqrt{a+b \tan (c+d x)}}{315 a^2 d \sqrt{\tan (c+d x)}}+\frac{(-b+i a)^{5/2} (A+i B) \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{2 a (3 a B+4 A b) \sqrt{a+b \tan (c+d x)}}{21 d \tan ^{\frac{7}{2}}(c+d x)}-\frac{(b+i a)^{5/2} (A-i B) \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{2 a A (a+b \tan (c+d x))^{3/2}}{9 d \tan ^{\frac{9}{2}}(c+d x)} \]
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Rubi [A] time = 1.91132, antiderivative size = 378, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 8, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.229, Rules used = {3605, 3645, 3649, 3616, 3615, 93, 203, 206} \[ \frac{2 \left (231 a^2 A b+105 a^3 B-135 a b^2 B-5 A b^3\right ) \sqrt{a+b \tan (c+d x)}}{315 a d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (21 a^2 A-45 a b B-25 A b^2\right ) \sqrt{a+b \tan (c+d x)}}{105 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 \left (-483 a^2 A b^2+315 a^4 A-735 a^3 b B+45 a b^3 B-10 A b^4\right ) \sqrt{a+b \tan (c+d x)}}{315 a^2 d \sqrt{\tan (c+d x)}}+\frac{(-b+i a)^{5/2} (A+i B) \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{2 a (3 a B+4 A b) \sqrt{a+b \tan (c+d x)}}{21 d \tan ^{\frac{7}{2}}(c+d x)}-\frac{(b+i a)^{5/2} (A-i B) \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{2 a A (a+b \tan (c+d x))^{3/2}}{9 d \tan ^{\frac{9}{2}}(c+d x)} \]
Antiderivative was successfully verified.
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Rule 3605
Rule 3645
Rule 3649
Rule 3616
Rule 3615
Rule 93
Rule 203
Rule 206
Rubi steps
\begin{align*} \int \frac{(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac{11}{2}}(c+d x)} \, dx &=-\frac{2 a A (a+b \tan (c+d x))^{3/2}}{9 d \tan ^{\frac{9}{2}}(c+d x)}+\frac{2}{9} \int \frac{\sqrt{a+b \tan (c+d x)} \left (\frac{3}{2} a (4 A b+3 a B)-\frac{9}{2} \left (a^2 A-A b^2-2 a b B\right ) \tan (c+d x)-\frac{3}{2} b (2 a A-3 b B) \tan ^2(c+d x)\right )}{\tan ^{\frac{9}{2}}(c+d x)} \, dx\\ &=-\frac{2 a (4 A b+3 a B) \sqrt{a+b \tan (c+d x)}}{21 d \tan ^{\frac{7}{2}}(c+d x)}-\frac{2 a A (a+b \tan (c+d x))^{3/2}}{9 d \tan ^{\frac{9}{2}}(c+d x)}+\frac{4}{63} \int \frac{-\frac{3}{4} a \left (21 a^2 A-25 A b^2-45 a b 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)-\frac{3}{4} b \left (38 a A b+18 a^2 B-21 b^2 B\right ) \tan ^2(c+d x)}{\tan ^{\frac{7}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}} \, dx\\ &=-\frac{2 a (4 A b+3 a B) \sqrt{a+b \tan (c+d x)}}{21 d \tan ^{\frac{7}{2}}(c+d x)}+\frac{2 \left (21 a^2 A-25 A b^2-45 a b B\right ) \sqrt{a+b \tan (c+d x)}}{105 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 a A (a+b \tan (c+d x))^{3/2}}{9 d \tan ^{\frac{9}{2}}(c+d x)}-\frac{8 \int \frac{\frac{3}{8} a \left (231 a^2 A b-5 A b^3+105 a^3 B-135 a b^2 B\right )-\frac{315}{8} a \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) \tan (c+d x)-\frac{3}{2} a b \left (21 a^2 A-25 A b^2-45 a b B\right ) \tan ^2(c+d x)}{\tan ^{\frac{5}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}} \, dx}{315 a}\\ &=-\frac{2 a (4 A b+3 a B) \sqrt{a+b \tan (c+d x)}}{21 d \tan ^{\frac{7}{2}}(c+d x)}+\frac{2 \left (21 a^2 A-25 A b^2-45 a b B\right ) \sqrt{a+b \tan (c+d x)}}{105 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (231 a^2 A b-5 A b^3+105 a^3 B-135 a b^2 B\right ) \sqrt{a+b \tan (c+d x)}}{315 a d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 a A (a+b \tan (c+d x))^{3/2}}{9 d \tan ^{\frac{9}{2}}(c+d x)}+\frac{16 \int \frac{\frac{3}{16} a \left (315 a^4 A-483 a^2 A b^2-10 A b^4-735 a^3 b B+45 a b^3 B\right )+\frac{945}{16} a^2 \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \tan (c+d x)+\frac{3}{8} a b \left (231 a^2 A b-5 A b^3+105 a^3 B-135 a b^2 B\right ) \tan ^2(c+d x)}{\tan ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}} \, dx}{945 a^2}\\ &=-\frac{2 a (4 A b+3 a B) \sqrt{a+b \tan (c+d x)}}{21 d \tan ^{\frac{7}{2}}(c+d x)}+\frac{2 \left (21 a^2 A-25 A b^2-45 a b B\right ) \sqrt{a+b \tan (c+d x)}}{105 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (231 a^2 A b-5 A b^3+105 a^3 B-135 a b^2 B\right ) \sqrt{a+b \tan (c+d x)}}{315 a d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 \left (315 a^4 A-483 a^2 A b^2-10 A b^4-735 a^3 b B+45 a b^3 B\right ) \sqrt{a+b \tan (c+d x)}}{315 a^2 d \sqrt{\tan (c+d x)}}-\frac{2 a A (a+b \tan (c+d x))^{3/2}}{9 d \tan ^{\frac{9}{2}}(c+d x)}-\frac{32 \int \frac{-\frac{945}{32} a^3 \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right )+\frac{945}{32} a^3 \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)} \sqrt{a+b \tan (c+d x)}} \, dx}{945 a^3}\\ &=-\frac{2 a (4 A b+3 a B) \sqrt{a+b \tan (c+d x)}}{21 d \tan ^{\frac{7}{2}}(c+d x)}+\frac{2 \left (21 a^2 A-25 A b^2-45 a b B\right ) \sqrt{a+b \tan (c+d x)}}{105 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (231 a^2 A b-5 A b^3+105 a^3 B-135 a b^2 B\right ) \sqrt{a+b \tan (c+d x)}}{315 a d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 \left (315 a^4 A-483 a^2 A b^2-10 A b^4-735 a^3 b B+45 a b^3 B\right ) \sqrt{a+b \tan (c+d x)}}{315 a^2 d \sqrt{\tan (c+d x)}}-\frac{2 a A (a+b \tan (c+d x))^{3/2}}{9 d \tan ^{\frac{9}{2}}(c+d x)}+\frac{1}{2} \left ((a-i b)^3 (i A+B)\right ) \int \frac{1+i \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx-\frac{\left (16 \left (-\frac{945}{32} a^3 \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right )+\frac{945}{32} i a^3 \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right )\right )\right ) \int \frac{1-i \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx}{945 a^3}\\ &=-\frac{2 a (4 A b+3 a B) \sqrt{a+b \tan (c+d x)}}{21 d \tan ^{\frac{7}{2}}(c+d x)}+\frac{2 \left (21 a^2 A-25 A b^2-45 a b B\right ) \sqrt{a+b \tan (c+d x)}}{105 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (231 a^2 A b-5 A b^3+105 a^3 B-135 a b^2 B\right ) \sqrt{a+b \tan (c+d x)}}{315 a d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 \left (315 a^4 A-483 a^2 A b^2-10 A b^4-735 a^3 b B+45 a b^3 B\right ) \sqrt{a+b \tan (c+d x)}}{315 a^2 d \sqrt{\tan (c+d x)}}-\frac{2 a A (a+b \tan (c+d x))^{3/2}}{9 d \tan ^{\frac{9}{2}}(c+d x)}+\frac{\left ((a-i b)^3 (i A+B)\right ) \operatorname{Subst}\left (\int \frac{1}{(1-i x) \sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}-\frac{\left (16 \left (-\frac{945}{32} a^3 \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right )+\frac{945}{32} i a^3 \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right )\right )\right ) \operatorname{Subst}\left (\int \frac{1}{(1+i x) \sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{945 a^3 d}\\ &=-\frac{2 a (4 A b+3 a B) \sqrt{a+b \tan (c+d x)}}{21 d \tan ^{\frac{7}{2}}(c+d x)}+\frac{2 \left (21 a^2 A-25 A b^2-45 a b B\right ) \sqrt{a+b \tan (c+d x)}}{105 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (231 a^2 A b-5 A b^3+105 a^3 B-135 a b^2 B\right ) \sqrt{a+b \tan (c+d x)}}{315 a d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 \left (315 a^4 A-483 a^2 A b^2-10 A b^4-735 a^3 b B+45 a b^3 B\right ) \sqrt{a+b \tan (c+d x)}}{315 a^2 d \sqrt{\tan (c+d x)}}-\frac{2 a A (a+b \tan (c+d x))^{3/2}}{9 d \tan ^{\frac{9}{2}}(c+d x)}+\frac{\left ((a-i b)^3 (i A+B)\right ) \operatorname{Subst}\left (\int \frac{1}{1-(i a+b) x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{\left (32 \left (-\frac{945}{32} a^3 \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right )+\frac{945}{32} i a^3 \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right )\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1-(-i a+b) x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{945 a^3 d}\\ &=\frac{(i a-b)^{5/2} (A+i B) \tan ^{-1}\left (\frac{\sqrt{i a-b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{(i a+b)^{5/2} (A-i B) \tanh ^{-1}\left (\frac{\sqrt{i a+b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{2 a (4 A b+3 a B) \sqrt{a+b \tan (c+d x)}}{21 d \tan ^{\frac{7}{2}}(c+d x)}+\frac{2 \left (21 a^2 A-25 A b^2-45 a b B\right ) \sqrt{a+b \tan (c+d x)}}{105 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (231 a^2 A b-5 A b^3+105 a^3 B-135 a b^2 B\right ) \sqrt{a+b \tan (c+d x)}}{315 a d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 \left (315 a^4 A-483 a^2 A b^2-10 A b^4-735 a^3 b B+45 a b^3 B\right ) \sqrt{a+b \tan (c+d x)}}{315 a^2 d \sqrt{\tan (c+d x)}}-\frac{2 a A (a+b \tan (c+d x))^{3/2}}{9 d \tan ^{\frac{9}{2}}(c+d x)}\\ \end{align*}
Mathematica [A] time = 6.86106, size = 543, normalized size = 1.44 \[ -\frac{b B (a+b \tan (c+d x))^{3/2}}{3 d \tan ^{\frac{9}{2}}(c+d x)}+\frac{1}{3} \left (-\frac{3 b (a B+2 A b) \sqrt{a+b \tan (c+d x)}}{8 d \tan ^{\frac{9}{2}}(c+d x)}+\frac{1}{4} \left (-\frac{\left (16 a^2 A-33 a b B-18 A b^2\right ) \sqrt{a+b \tan (c+d x)}}{6 d \tan ^{\frac{9}{2}}(c+d x)}-\frac{2 \left (\frac{6 a \left (18 a^2 B+38 a A b-21 b^2 B\right ) \sqrt{a+b \tan (c+d x)}}{7 d \tan ^{\frac{7}{2}}(c+d x)}-\frac{2 \left (\frac{18 a^2 \left (21 a^2 A-45 a b B-25 A b^2\right ) \sqrt{a+b \tan (c+d x)}}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 \left (-\frac{3 a^2 \left (231 a^2 A b+105 a^3 B-135 a b^2 B-5 A b^3\right ) \sqrt{a+b \tan (c+d x)}}{d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 \left (-\frac{9 a^2 \left (-483 a^2 A b^2+315 a^4 A-735 a^3 b B+45 a b^3 B-10 A b^4\right ) \sqrt{a+b \tan (c+d x)}}{2 d \sqrt{\tan (c+d x)}}+\frac{2835 a^4 \left (\sqrt [4]{-1} (-a-i b)^{5/2} (A+i B) \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{-a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )+\sqrt [4]{-1} (a-i b)^{5/2} (A-i B) \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )\right )}{4 d}\right )}{3 a}\right )}{5 a}\right )}{7 a}\right )}{9 a}\right )\right ) \]
Antiderivative was successfully verified.
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Maple [B] time = 0.967, size = 2656820, normalized size = 7028.6 \begin{align*} \text{output too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [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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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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