Optimal. Leaf size=219 \[ -\frac{i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{5/2}}-\frac{9 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}+\frac{63 i a^2}{160 d (a+i a \tan (c+d x))^{5/2}}+\frac{21 i a}{64 d (a+i a \tan (c+d x))^{3/2}}+\frac{63 i}{128 d \sqrt{a+i a \tan (c+d x)}}-\frac{63 i \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{128 \sqrt{2} \sqrt{a} d} \]
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Rubi [A] time = 0.12402, antiderivative size = 219, 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} \[ -\frac{i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{5/2}}-\frac{9 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}+\frac{63 i a^2}{160 d (a+i a \tan (c+d x))^{5/2}}+\frac{21 i a}{64 d (a+i a \tan (c+d x))^{3/2}}+\frac{63 i}{128 d \sqrt{a+i a \tan (c+d x)}}-\frac{63 i \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{128 \sqrt{2} \sqrt{a} d} \]
Antiderivative was successfully verified.
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Rule 3487
Rule 51
Rule 63
Rule 206
Rubi steps
\begin{align*} \int \frac{\cos ^4(c+d x)}{\sqrt{a+i a \tan (c+d x)}} \, dx &=-\frac{\left (i a^5\right ) \operatorname{Subst}\left (\int \frac{1}{(a-x)^3 (a+x)^{7/2}} \, dx,x,i a \tan (c+d x)\right )}{d}\\ &=-\frac{i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{5/2}}-\frac{\left (9 i a^4\right ) \operatorname{Subst}\left (\int \frac{1}{(a-x)^2 (a+x)^{7/2}} \, dx,x,i a \tan (c+d x)\right )}{8 d}\\ &=-\frac{i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{5/2}}-\frac{9 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}-\frac{\left (63 i a^3\right ) \operatorname{Subst}\left (\int \frac{1}{(a-x) (a+x)^{7/2}} \, dx,x,i a \tan (c+d x)\right )}{32 d}\\ &=\frac{63 i a^2}{160 d (a+i a \tan (c+d x))^{5/2}}-\frac{i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{5/2}}-\frac{9 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}-\frac{\left (63 i a^2\right ) \operatorname{Subst}\left (\int \frac{1}{(a-x) (a+x)^{5/2}} \, dx,x,i a \tan (c+d x)\right )}{64 d}\\ &=\frac{63 i a^2}{160 d (a+i a \tan (c+d x))^{5/2}}-\frac{i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{5/2}}-\frac{9 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}+\frac{21 i a}{64 d (a+i a \tan (c+d x))^{3/2}}-\frac{(63 i a) \operatorname{Subst}\left (\int \frac{1}{(a-x) (a+x)^{3/2}} \, dx,x,i a \tan (c+d x)\right )}{128 d}\\ &=\frac{63 i a^2}{160 d (a+i a \tan (c+d x))^{5/2}}-\frac{i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{5/2}}-\frac{9 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}+\frac{21 i a}{64 d (a+i a \tan (c+d x))^{3/2}}+\frac{63 i}{128 d \sqrt{a+i a \tan (c+d x)}}-\frac{(63 i) \operatorname{Subst}\left (\int \frac{1}{(a-x) \sqrt{a+x}} \, dx,x,i a \tan (c+d x)\right )}{256 d}\\ &=\frac{63 i a^2}{160 d (a+i a \tan (c+d x))^{5/2}}-\frac{i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{5/2}}-\frac{9 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}+\frac{21 i a}{64 d (a+i a \tan (c+d x))^{3/2}}+\frac{63 i}{128 d \sqrt{a+i a \tan (c+d x)}}-\frac{(63 i) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\sqrt{a+i a \tan (c+d x)}\right )}{128 d}\\ &=-\frac{63 i \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{128 \sqrt{2} \sqrt{a} d}+\frac{63 i a^2}{160 d (a+i a \tan (c+d x))^{5/2}}-\frac{i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{5/2}}-\frac{9 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}+\frac{21 i a}{64 d (a+i a \tan (c+d x))^{3/2}}+\frac{63 i}{128 d \sqrt{a+i a \tan (c+d x)}}\\ \end{align*}
Mathematica [A] time = 0.627008, size = 152, normalized size = 0.69 \[ -\frac{i e^{-4 i (c+d x)} \left (\sqrt{1+e^{2 i (c+d x)}} \left (-56 e^{2 i (c+d x)}-288 e^{4 i (c+d x)}+85 e^{6 i (c+d x)}+10 e^{8 i (c+d x)}-8\right )+315 e^{5 i (c+d x)} \sinh ^{-1}\left (e^{i (c+d x)}\right )\right )}{640 d \sqrt{1+e^{2 i (c+d x)}} \sqrt{a+i a \tan (c+d x)}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.361, size = 368, normalized size = 1.7 \begin{align*}{\frac{1}{2560\,ad}\sqrt{{\frac{a \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) \right ) }{\cos \left ( dx+c \right ) }}} \left ( 512\,i \left ( \cos \left ( dx+c \right ) \right ) ^{6}+512\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}\sin \left ( dx+c \right ) +315\,i\sqrt{2}\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\arctan \left ({\frac{\sqrt{2} \left ( i\cos \left ( dx+c \right ) -i-\sin \left ( dx+c \right ) \right ) }{2\,\sin \left ( dx+c \right ) }{\frac{1}{\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}}}} \right ) \cos \left ( dx+c \right ) +96\,i \left ( \cos \left ( dx+c \right ) \right ) ^{4}+315\,i\sqrt{2}\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\arctan \left ({\frac{\sqrt{2} \left ( i\cos \left ( dx+c \right ) -i-\sin \left ( dx+c \right ) \right ) }{2\,\sin \left ( dx+c \right ) }{\frac{1}{\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}}}} \right ) +315\,\sqrt{2}\arctan \left ( 1/2\,{\frac{\sqrt{2} \left ( i\cos \left ( dx+c \right ) -i-\sin \left ( dx+c \right ) \right ) }{\sin \left ( dx+c \right ) }{\frac{1}{\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}}}} \right ) \sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\sin \left ( dx+c \right ) +672\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) +420\,i \left ( \cos \left ( dx+c \right ) \right ) ^{2}+1260\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) \right ) } \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 = 2.57692, size = 922, normalized size = 4.21 \begin{align*} \frac{{\left (-315 i \, \sqrt{\frac{1}{2}} a d \sqrt{\frac{1}{a d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left ({\left (2 \, \sqrt{\frac{1}{2}} a d \sqrt{\frac{1}{a d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + 315 i \, \sqrt{\frac{1}{2}} a d \sqrt{\frac{1}{a d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (-{\left (2 \, \sqrt{\frac{1}{2}} a d \sqrt{\frac{1}{a d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (-10 i \, e^{\left (10 i \, d x + 10 i \, c\right )} - 95 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 203 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 344 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 64 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 8 i\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{1280 \, a 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] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (d x + c\right )^{4}}{\sqrt{i \, a \tan \left (d x + c\right ) + a}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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