Optimal. Leaf size=309 \[ \frac{2 \left (35 a^2 A-77 a b B-45 A b^2\right ) \sqrt{a+b \tan (c+d x)}}{105 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (245 a^2 A b+105 a^3 B-161 a b^2 B-15 A b^3\right ) \sqrt{a+b \tan (c+d x)}}{105 a d \sqrt{\tan (c+d x)}}+\frac{(-b+i a)^{5/2} (-B+i A) \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{2 a (7 a B+10 A b) \sqrt{a+b \tan (c+d x)}}{35 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{(b+i a)^{5/2} (B+i A) \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{2 a A (a+b \tan (c+d x))^{3/2}}{7 d \tan ^{\frac{7}{2}}(c+d x)} \]
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Rubi [A] time = 1.51909, antiderivative size = 309, normalized size of antiderivative = 1., number of steps used = 11, 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} \[ \frac{2 \left (35 a^2 A-77 a b B-45 A b^2\right ) \sqrt{a+b \tan (c+d x)}}{105 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (245 a^2 A b+105 a^3 B-161 a b^2 B-15 A b^3\right ) \sqrt{a+b \tan (c+d x)}}{105 a d \sqrt{\tan (c+d x)}}+\frac{(-b+i a)^{5/2} (-B+i A) \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{2 a (7 a B+10 A b) \sqrt{a+b \tan (c+d x)}}{35 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{(b+i a)^{5/2} (B+i A) \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{2 a A (a+b \tan (c+d x))^{3/2}}{7 d \tan ^{\frac{7}{2}}(c+d x)} \]
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{9}{2}}(c+d x)} \, dx &=-\frac{2 a A (a+b \tan (c+d x))^{3/2}}{7 d \tan ^{\frac{7}{2}}(c+d x)}+\frac{2}{7} \int \frac{\sqrt{a+b \tan (c+d x)} \left (\frac{1}{2} a (10 A b+7 a B)-\frac{7}{2} \left (a^2 A-A b^2-2 a b B\right ) \tan (c+d x)-\frac{1}{2} b (4 a A-7 b B) \tan ^2(c+d x)\right )}{\tan ^{\frac{7}{2}}(c+d x)} \, dx\\ &=-\frac{2 a (10 A b+7 a B) \sqrt{a+b \tan (c+d x)}}{35 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 a A (a+b \tan (c+d x))^{3/2}}{7 d \tan ^{\frac{7}{2}}(c+d x)}+\frac{4}{35} \int \frac{-\frac{1}{4} a \left (35 a^2 A-45 A b^2-77 a b B\right )-\frac{35}{4} \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \tan (c+d x)-\frac{1}{4} b \left (60 a A b+28 a^2 B-35 b^2 B\right ) \tan ^2(c+d x)}{\tan ^{\frac{5}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}} \, dx\\ &=-\frac{2 a (10 A b+7 a B) \sqrt{a+b \tan (c+d x)}}{35 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (35 a^2 A-45 A b^2-77 a b B\right ) \sqrt{a+b \tan (c+d x)}}{105 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 a A (a+b \tan (c+d x))^{3/2}}{7 d \tan ^{\frac{7}{2}}(c+d x)}-\frac{8 \int \frac{\frac{1}{8} a \left (245 a^2 A b-15 A b^3+105 a^3 B-161 a b^2 B\right )-\frac{105}{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{1}{4} a b \left (35 a^2 A-45 A b^2-77 a b B\right ) \tan ^2(c+d x)}{\tan ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}} \, dx}{105 a}\\ &=-\frac{2 a (10 A b+7 a B) \sqrt{a+b \tan (c+d x)}}{35 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (35 a^2 A-45 A b^2-77 a b B\right ) \sqrt{a+b \tan (c+d x)}}{105 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (245 a^2 A b-15 A b^3+105 a^3 B-161 a b^2 B\right ) \sqrt{a+b \tan (c+d x)}}{105 a d \sqrt{\tan (c+d x)}}-\frac{2 a A (a+b \tan (c+d x))^{3/2}}{7 d \tan ^{\frac{7}{2}}(c+d x)}+\frac{16 \int \frac{\frac{105}{16} a^2 \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right )+\frac{105}{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)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx}{105 a^2}\\ &=-\frac{2 a (10 A b+7 a B) \sqrt{a+b \tan (c+d x)}}{35 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (35 a^2 A-45 A b^2-77 a b B\right ) \sqrt{a+b \tan (c+d x)}}{105 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (245 a^2 A b-15 A b^3+105 a^3 B-161 a b^2 B\right ) \sqrt{a+b \tan (c+d x)}}{105 a d \sqrt{\tan (c+d x)}}-\frac{2 a A (a+b \tan (c+d x))^{3/2}}{7 d \tan ^{\frac{7}{2}}(c+d x)}+\frac{1}{2} \left ((a-i b)^3 (A-i B)\right ) \int \frac{1+i \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx+\frac{1}{2} \left ((a+i b)^3 (A+i B)\right ) \int \frac{1-i \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx\\ &=-\frac{2 a (10 A b+7 a B) \sqrt{a+b \tan (c+d x)}}{35 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (35 a^2 A-45 A b^2-77 a b B\right ) \sqrt{a+b \tan (c+d x)}}{105 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (245 a^2 A b-15 A b^3+105 a^3 B-161 a b^2 B\right ) \sqrt{a+b \tan (c+d x)}}{105 a d \sqrt{\tan (c+d x)}}-\frac{2 a A (a+b \tan (c+d x))^{3/2}}{7 d \tan ^{\frac{7}{2}}(c+d x)}+\frac{\left ((a-i b)^3 (A-i 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 ((a+i b)^3 (A+i 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{2 a (10 A b+7 a B) \sqrt{a+b \tan (c+d x)}}{35 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (35 a^2 A-45 A b^2-77 a b B\right ) \sqrt{a+b \tan (c+d x)}}{105 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (245 a^2 A b-15 A b^3+105 a^3 B-161 a b^2 B\right ) \sqrt{a+b \tan (c+d x)}}{105 a d \sqrt{\tan (c+d x)}}-\frac{2 a A (a+b \tan (c+d x))^{3/2}}{7 d \tan ^{\frac{7}{2}}(c+d x)}+\frac{\left ((a-i b)^3 (A-i 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 ((a+i b)^3 (A+i 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{(i a-b)^{5/2} (i A-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} (i A+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 (10 A b+7 a B) \sqrt{a+b \tan (c+d x)}}{35 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (35 a^2 A-45 A b^2-77 a b B\right ) \sqrt{a+b \tan (c+d x)}}{105 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (245 a^2 A b-15 A b^3+105 a^3 B-161 a b^2 B\right ) \sqrt{a+b \tan (c+d x)}}{105 a d \sqrt{\tan (c+d x)}}-\frac{2 a A (a+b \tan (c+d x))^{3/2}}{7 d \tan ^{\frac{7}{2}}(c+d x)}\\ \end{align*}
Mathematica [A] time = 5.9971, size = 385, normalized size = 1.25 \[ \frac{8 \left (245 a^2 A b+105 a^3 B-161 a b^2 B-15 A b^3\right ) \tan ^3(c+d x) \sqrt{a+b \tan (c+d x)}+8 a \left (35 a^2 A-77 a b B-45 A b^2\right ) \tan ^2(c+d x) \sqrt{a+b \tan (c+d x)}-6 a \left (28 a^2 B+60 a A b-35 b^2 B\right ) \tan (c+d x) \sqrt{a+b \tan (c+d x)}-5 a \left (24 a^2 A-49 a b B-28 A b^2\right ) \sqrt{a+b \tan (c+d x)}-35 a b (a B+4 A b) \sqrt{a+b \tan (c+d x)}+420 \sqrt [4]{-1} a \tan ^{\frac{7}{2}}(c+d x) \left (i (-a-i b)^{5/2} (A+i B) \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{-a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )-(a-i b)^{5/2} (B+i A) \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )\right )-210 a b B (a+b \tan (c+d x))^{3/2}}{420 a d \tan ^{\frac{7}{2}}(c+d x)} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.95, size = 2654465, normalized size = 8590.5 \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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