Optimal. Leaf size=219 \[ -\frac {63 i \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{128 \sqrt {2} \sqrt {a} d}+\frac {63 i a^2}{160 d (a+i a \tan (c+d x))^{5/2}}-\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{5/2}}-\frac {9 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}+\frac {21 i a}{64 d (a+i a \tan (c+d x))^{3/2}}+\frac {63 i}{128 d \sqrt {a+i a \tan (c+d x)}} \]
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Rubi [A]
time = 0.10, antiderivative size = 219, normalized size of antiderivative = 1.00, number of steps
used = 8, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {3568, 44, 53,
65, 212} \begin {gather*} -\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} \end {gather*}
Antiderivative was successfully verified.
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Rule 44
Rule 53
Rule 65
Rule 212
Rule 3568
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 ) \text {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 ) \text {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 ) \text {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 ) \text {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) \text {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) \text {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) \text {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*}
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Mathematica [A]
time = 0.87, size = 152, normalized size = 0.69 \begin {gather*} -\frac {i e^{-4 i (c+d x)} \left (\sqrt {1+e^{2 i (c+d x)}} \left (-8-56 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)}\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)}} \end {gather*}
Antiderivative was successfully verified.
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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice
the leaf count of optimal. 361 vs. \(2 (174 ) = 348\).
time = 0.98, size = 362, normalized size = 1.65
method | result | size |
default | \(-\frac {\sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (-512 i \left (\cos ^{6}\left (d x +c \right )\right )-512 \sin \left (d x +c \right ) \left (\cos ^{5}\left (d x +c \right )\right )+315 i \cos \left (d x +c \right ) \arctan \left (\frac {\left (-i \cos \left (d x +c \right )+\sin \left (d x +c \right )+i\right ) \sqrt {2}}{2 \sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}-96 i \left (\cos ^{4}\left (d x +c \right )\right )+315 \sin \left (d x +c \right ) \arctan \left (\frac {\left (-i \cos \left (d x +c \right )+\sin \left (d x +c \right )+i\right ) \sqrt {2}}{2 \sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}+315 i \arctan \left (\frac {\left (-i \cos \left (d x +c \right )+\sin \left (d x +c \right )+i\right ) \sqrt {2}}{2 \sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}-672 \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )-420 i \left (\cos ^{2}\left (d x +c \right )\right )-1260 \sin \left (d x +c \right ) \cos \left (d x +c \right )\right )}{2560 d a}\) | \(362\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.50, size = 192, normalized size = 0.88 \begin {gather*} \frac {i \, {\left (315 \, \sqrt {2} \sqrt {a} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + \frac {4 \, {\left (315 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} a - 1050 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a^{2} + 672 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{3} + 192 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{4} + 128 \, a^{5}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {9}{2}} - 4 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a + 4 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{2}}\right )}}{2560 \, a d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.39, size = 293, normalized size = 1.34 \begin {gather*} \frac {{\left (-315 i \, \sqrt {\frac {1}{2}} a d \sqrt {\frac {1}{a d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a d^{2}}} + a 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 (5 i \, d x + 5 i \, c\right )} \log \left (-4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a d^{2}}} - a 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 )}\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{1280 \, a d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\cos ^{4}{\left (c + d x \right )}}{\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\cos \left (c+d\,x\right )}^4}{\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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