Optimal. Leaf size=259 \[ \frac{2 \left (15 a^2 A-20 a b B-3 A b^2\right ) \sqrt{a+b \tan (c+d x)}}{15 a d \sqrt{\tan (c+d x)}}+\frac{(a+i b)^2 (-B+i A) \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d \sqrt{-b+i a}}-\frac{2 (5 a B+6 A b) \sqrt{a+b \tan (c+d x)}}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{(b+i a)^{3/2} (B+i A) \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{2 a A \sqrt{a+b \tan (c+d x)}}{5 d \tan ^{\frac{5}{2}}(c+d x)} \]
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Rubi [A] time = 1.15287, antiderivative size = 259, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3605, 3649, 3616, 3615, 93, 203, 206} \[ \frac{2 \left (15 a^2 A-20 a b B-3 A b^2\right ) \sqrt{a+b \tan (c+d x)}}{15 a d \sqrt{\tan (c+d x)}}+\frac{(a+i b)^2 (-B+i A) \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d \sqrt{-b+i a}}-\frac{2 (5 a B+6 A b) \sqrt{a+b \tan (c+d x)}}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{(b+i a)^{3/2} (B+i A) \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{2 a A \sqrt{a+b \tan (c+d x)}}{5 d \tan ^{\frac{5}{2}}(c+d x)} \]
Antiderivative was successfully verified.
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Rule 3605
Rule 3649
Rule 3616
Rule 3615
Rule 93
Rule 203
Rule 206
Rubi steps
\begin{align*} \int \frac{(a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac{7}{2}}(c+d x)} \, dx &=-\frac{2 a A \sqrt{a+b \tan (c+d x)}}{5 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2}{5} \int \frac{\frac{1}{2} a (6 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 (4 a A-5 b B) \tan ^2(c+d x)}{\tan ^{\frac{5}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}} \, dx\\ &=-\frac{2 a A \sqrt{a+b \tan (c+d x)}}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 (6 A b+5 a B) \sqrt{a+b \tan (c+d x)}}{15 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{4 \int \frac{\frac{1}{4} a \left (15 a^2 A-3 A b^2-20 a b B\right )+\frac{15}{4} a \left (2 a A b+a^2 B-b^2 B\right ) \tan (c+d x)+\frac{1}{2} a b (6 A b+5 a B) \tan ^2(c+d x)}{\tan ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}} \, dx}{15 a}\\ &=-\frac{2 a A \sqrt{a+b \tan (c+d x)}}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 (6 A b+5 a B) \sqrt{a+b \tan (c+d x)}}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (15 a^2 A-3 A b^2-20 a b B\right ) \sqrt{a+b \tan (c+d x)}}{15 a d \sqrt{\tan (c+d x)}}+\frac{8 \int \frac{-\frac{15}{8} a^2 \left (2 a A b+a^2 B-b^2 B\right )+\frac{15}{8} a^2 \left (a^2 A-A b^2-2 a b B\right ) \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx}{15 a^2}\\ &=-\frac{2 a A \sqrt{a+b \tan (c+d x)}}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 (6 A b+5 a B) \sqrt{a+b \tan (c+d x)}}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (15 a^2 A-3 A b^2-20 a b B\right ) \sqrt{a+b \tan (c+d x)}}{15 a d \sqrt{\tan (c+d x)}}+\frac{1}{2} \left ((a+i b)^2 (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{1}{2} \left ((a-i b)^2 (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{2 a A \sqrt{a+b \tan (c+d x)}}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 (6 A b+5 a B) \sqrt{a+b \tan (c+d x)}}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (15 a^2 A-3 A b^2-20 a b B\right ) \sqrt{a+b \tan (c+d x)}}{15 a d \sqrt{\tan (c+d x)}}+\frac{\left ((a+i b)^2 (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 ((a-i b)^2 (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{2 a A \sqrt{a+b \tan (c+d x)}}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 (6 A b+5 a B) \sqrt{a+b \tan (c+d x)}}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (15 a^2 A-3 A b^2-20 a b B\right ) \sqrt{a+b \tan (c+d x)}}{15 a d \sqrt{\tan (c+d x)}}+\frac{\left ((a+i b)^2 (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 ((a-i b)^2 (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{(a+i b)^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 )}{\sqrt{i a-b} d}+\frac{(i a+b)^{3/2} (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 A \sqrt{a+b \tan (c+d x)}}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 (6 A b+5 a B) \sqrt{a+b \tan (c+d x)}}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (15 a^2 A-3 A b^2-20 a b B\right ) \sqrt{a+b \tan (c+d x)}}{15 a d \sqrt{\tan (c+d x)}}\\ \end{align*}
Mathematica [A] time = 2.2734, size = 286, normalized size = 1.1 \[ \frac{4 \left (15 a^2 A-20 a b B-3 A b^2\right ) \tan ^2(c+d x) \sqrt{a+b \tan (c+d x)}-4 a (5 a B+6 A b) \tan (c+d x) \sqrt{a+b \tan (c+d x)}-3 a (4 a A-5 b B) \sqrt{a+b \tan (c+d x)}+30 \sqrt [4]{-1} a \tan ^{\frac{5}{2}}(c+d x) \left ((-a-i b)^{3/2} (A+i B) \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{-a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )-(a-i b)^{3/2} (A-i B) \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )\right )-15 a b B \sqrt{a+b \tan (c+d x)}}{30 a d \tan ^{\frac{5}{2}}(c+d x)} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.875, size = 2398570, normalized size = 9260.9 \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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