Optimal. Leaf size=178 \[ -\frac{2 (5 B+i A) \sqrt{a+i a \tan (c+d x)}}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 (13 A-5 i B) \sqrt{a+i a \tan (c+d x)}}{15 d \sqrt{\tan (c+d x)}}-\frac{(1+i) \sqrt{a} (A-i B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{2 A \sqrt{a+i a \tan (c+d x)}}{5 d \tan ^{\frac{5}{2}}(c+d x)} \]
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Rubi [A] time = 0.535027, antiderivative size = 178, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {3598, 12, 3544, 205} \[ -\frac{2 (5 B+i A) \sqrt{a+i a \tan (c+d x)}}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 (13 A-5 i B) \sqrt{a+i a \tan (c+d x)}}{15 d \sqrt{\tan (c+d x)}}-\frac{(1+i) \sqrt{a} (A-i B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{2 A \sqrt{a+i a \tan (c+d x)}}{5 d \tan ^{\frac{5}{2}}(c+d x)} \]
Antiderivative was successfully verified.
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Rule 3598
Rule 12
Rule 3544
Rule 205
Rubi steps
\begin{align*} \int \frac{\sqrt{a+i a \tan (c+d x)} (A+B \tan (c+d x))}{\tan ^{\frac{7}{2}}(c+d x)} \, dx &=-\frac{2 A \sqrt{a+i a \tan (c+d x)}}{5 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2 \int \frac{\sqrt{a+i a \tan (c+d x)} \left (\frac{1}{2} a (i A+5 B)-2 a A \tan (c+d x)\right )}{\tan ^{\frac{5}{2}}(c+d x)} \, dx}{5 a}\\ &=-\frac{2 A \sqrt{a+i a \tan (c+d x)}}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 (i A+5 B) \sqrt{a+i a \tan (c+d x)}}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{4 \int \frac{\sqrt{a+i a \tan (c+d x)} \left (-\frac{1}{4} a^2 (13 A-5 i B)-\frac{1}{2} a^2 (i A+5 B) \tan (c+d x)\right )}{\tan ^{\frac{3}{2}}(c+d x)} \, dx}{15 a^2}\\ &=-\frac{2 A \sqrt{a+i a \tan (c+d x)}}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 (i A+5 B) \sqrt{a+i a \tan (c+d x)}}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 (13 A-5 i B) \sqrt{a+i a \tan (c+d x)}}{15 d \sqrt{\tan (c+d x)}}+\frac{8 \int -\frac{15 a^3 (i A+B) \sqrt{a+i a \tan (c+d x)}}{8 \sqrt{\tan (c+d x)}} \, dx}{15 a^3}\\ &=-\frac{2 A \sqrt{a+i a \tan (c+d x)}}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 (i A+5 B) \sqrt{a+i a \tan (c+d x)}}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 (13 A-5 i B) \sqrt{a+i a \tan (c+d x)}}{15 d \sqrt{\tan (c+d x)}}+(-i A-B) \int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx\\ &=-\frac{2 A \sqrt{a+i a \tan (c+d x)}}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 (i A+5 B) \sqrt{a+i a \tan (c+d x)}}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 (13 A-5 i B) \sqrt{a+i a \tan (c+d x)}}{15 d \sqrt{\tan (c+d x)}}-\frac{\left (2 a^2 (A-i B)\right ) \operatorname{Subst}\left (\int \frac{1}{-i a-2 a^2 x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}\\ &=-\frac{(1+i) \sqrt{a} (A-i B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{2 A \sqrt{a+i a \tan (c+d x)}}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 (i A+5 B) \sqrt{a+i a \tan (c+d x)}}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 (13 A-5 i B) \sqrt{a+i a \tan (c+d x)}}{15 d \sqrt{\tan (c+d x)}}\\ \end{align*}
Mathematica [A] time = 7.41365, size = 211, normalized size = 1.19 \[ -\frac{(A-i B) e^{-i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right ) \sqrt{a+i a \tan (c+d x)}}{d \sqrt{-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}}}-\frac{\csc ^2(c+d x) \sqrt{a+i a \tan (c+d x)} ((5 B+i A) \sin (2 (c+d x))+(16 A-5 i B) \cos (2 (c+d x))-10 A+5 i B)}{15 d \sqrt{\tan (c+d x)}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.048, size = 630, normalized size = 3.5 \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 [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 [B] time = 1.82182, size = 1482, normalized size = 8.33 \begin{align*} \text{result too large to display} \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 [A] time = 1.44699, size = 265, normalized size = 1.49 \begin{align*} \frac{\left (i + 1\right ) \, \sqrt{-2 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a + 2 \, a^{2}}{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{5} +{\left (-\left (2 i - 2\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{4} + \left (2 i - 2\right ) \, a^{5}\right )} \sqrt{-2 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a + 2 \, a^{2}} \sqrt{i \, a \tan \left (d x + c\right ) + a} B}{{\left (-2 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{5} a + 12 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} a^{2} - 28 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a^{3} + 32 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{4} - 18 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{5} + 4 i \, a^{6}\right )} d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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