Optimal. Leaf size=239 \[ -\frac{i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{3/2}}-\frac{3 i a^5}{16 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{3/2}}-\frac{21 i a^4}{64 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{3/2}}+\frac{35 i a^3}{128 d (a+i a \tan (c+d x))^{3/2}}+\frac{105 i a^2}{256 d \sqrt{a+i a \tan (c+d x)}}-\frac{105 i a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{256 \sqrt{2} d} \]
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Rubi [A] time = 0.134344, antiderivative size = 239, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {3487, 51, 63, 206} \[ -\frac{i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{3/2}}-\frac{3 i a^5}{16 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{3/2}}-\frac{21 i a^4}{64 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{3/2}}+\frac{35 i a^3}{128 d (a+i a \tan (c+d x))^{3/2}}+\frac{105 i a^2}{256 d \sqrt{a+i a \tan (c+d x)}}-\frac{105 i a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{256 \sqrt{2} d} \]
Antiderivative was successfully verified.
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Rule 3487
Rule 51
Rule 63
Rule 206
Rubi steps
\begin{align*} \int \cos ^6(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx &=-\frac{\left (i a^7\right ) \operatorname{Subst}\left (\int \frac{1}{(a-x)^4 (a+x)^{5/2}} \, dx,x,i a \tan (c+d x)\right )}{d}\\ &=-\frac{i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{3/2}}-\frac{\left (3 i a^6\right ) \operatorname{Subst}\left (\int \frac{1}{(a-x)^3 (a+x)^{5/2}} \, dx,x,i a \tan (c+d x)\right )}{4 d}\\ &=-\frac{i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{3/2}}-\frac{3 i a^5}{16 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{3/2}}-\frac{\left (21 i a^5\right ) \operatorname{Subst}\left (\int \frac{1}{(a-x)^2 (a+x)^{5/2}} \, dx,x,i a \tan (c+d x)\right )}{32 d}\\ &=-\frac{i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{3/2}}-\frac{3 i a^5}{16 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{3/2}}-\frac{21 i a^4}{64 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{3/2}}-\frac{\left (105 i a^4\right ) \operatorname{Subst}\left (\int \frac{1}{(a-x) (a+x)^{5/2}} \, dx,x,i a \tan (c+d x)\right )}{128 d}\\ &=\frac{35 i a^3}{128 d (a+i a \tan (c+d x))^{3/2}}-\frac{i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{3/2}}-\frac{3 i a^5}{16 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{3/2}}-\frac{21 i a^4}{64 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{3/2}}-\frac{\left (105 i a^3\right ) \operatorname{Subst}\left (\int \frac{1}{(a-x) (a+x)^{3/2}} \, dx,x,i a \tan (c+d x)\right )}{256 d}\\ &=\frac{35 i a^3}{128 d (a+i a \tan (c+d x))^{3/2}}-\frac{i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{3/2}}-\frac{3 i a^5}{16 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{3/2}}-\frac{21 i a^4}{64 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{3/2}}+\frac{105 i a^2}{256 d \sqrt{a+i a \tan (c+d x)}}-\frac{\left (105 i a^2\right ) \operatorname{Subst}\left (\int \frac{1}{(a-x) \sqrt{a+x}} \, dx,x,i a \tan (c+d x)\right )}{512 d}\\ &=\frac{35 i a^3}{128 d (a+i a \tan (c+d x))^{3/2}}-\frac{i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{3/2}}-\frac{3 i a^5}{16 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{3/2}}-\frac{21 i a^4}{64 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{3/2}}+\frac{105 i a^2}{256 d \sqrt{a+i a \tan (c+d x)}}-\frac{\left (105 i a^2\right ) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\sqrt{a+i a \tan (c+d x)}\right )}{256 d}\\ &=-\frac{105 i a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{256 \sqrt{2} d}+\frac{35 i a^3}{128 d (a+i a \tan (c+d x))^{3/2}}-\frac{i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{3/2}}-\frac{3 i a^5}{16 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{3/2}}-\frac{21 i a^4}{64 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{3/2}}+\frac{105 i a^2}{256 d \sqrt{a+i a \tan (c+d x)}}\\ \end{align*}
Mathematica [A] time = 0.981902, size = 169, normalized size = 0.71 \[ \frac{a e^{-4 i (c+d x)} \cos ^2(c+d x) (\tan (c+d x)-i) \left (\sqrt{1+e^{2 i (c+d x)}} \left (-208 e^{2 i (c+d x)}+165 e^{4 i (c+d x)}+50 e^{6 i (c+d x)}+8 e^{8 i (c+d x)}-16\right )+315 e^{3 i (c+d x)} \sinh ^{-1}\left (e^{i (c+d x)}\right )\right ) \sqrt{a+i a \tan (c+d x)}}{768 d \sqrt{1+e^{2 i (c+d x)}}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.365, size = 1086, normalized size = 4.5 \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(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.76755, size = 959, normalized size = 4.01 \begin{align*} \frac{{\left (315 \, \sqrt{\frac{1}{2}} \sqrt{-\frac{a^{3}}{d^{2}}} d e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (\frac{{\left (210 i \, \sqrt{\frac{1}{2}} \sqrt{-\frac{a^{3}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )} + 105 \, \sqrt{2}{\left (a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{105 \, a}\right ) - 315 \, \sqrt{\frac{1}{2}} \sqrt{-\frac{a^{3}}{d^{2}}} d e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (\frac{{\left (-210 i \, \sqrt{\frac{1}{2}} \sqrt{-\frac{a^{3}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )} + 105 \, \sqrt{2}{\left (a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{105 \, a}\right ) + \sqrt{2}{\left (-8 i \, a e^{\left (10 i \, d x + 10 i \, c\right )} - 58 i \, a e^{\left (8 i \, d x + 8 i \, c\right )} - 215 i \, a e^{\left (6 i \, d x + 6 i \, c\right )} + 43 i \, a e^{\left (4 i \, d x + 4 i \, c\right )} + 224 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} + 16 i \, a\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{1536 \, d} \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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