Optimal. Leaf size=93 \[ -\frac {i a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} d}-\frac {i a^2 \sqrt {a+i a \tan (c+d x)}}{2 d (a-i a \tan (c+d x))} \]
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Rubi [A] time = 0.08, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3487, 51, 63, 206} \[ -\frac {i a^2 \sqrt {a+i a \tan (c+d x)}}{2 d (a-i a \tan (c+d x))}-\frac {i a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} d} \]
Antiderivative was successfully verified.
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Rule 51
Rule 63
Rule 206
Rule 3487
Rubi steps
\begin {align*} \int \cos ^2(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx &=-\frac {\left (i a^3\right ) \operatorname {Subst}\left (\int \frac {1}{(a-x)^2 \sqrt {a+x}} \, dx,x,i a \tan (c+d x)\right )}{d}\\ &=-\frac {i a^2 \sqrt {a+i a \tan (c+d x)}}{2 d (a-i a \tan (c+d x))}-\frac {\left (i a^2\right ) \operatorname {Subst}\left (\int \frac {1}{(a-x) \sqrt {a+x}} \, dx,x,i a \tan (c+d x)\right )}{4 d}\\ &=-\frac {i a^2 \sqrt {a+i a \tan (c+d x)}}{2 d (a-i a \tan (c+d x))}-\frac {\left (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 )}{2 d}\\ &=-\frac {i a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} d}-\frac {i a^2 \sqrt {a+i a \tan (c+d x)}}{2 d (a-i a \tan (c+d x))}\\ \end {align*}
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Mathematica [A] time = 0.64, size = 97, normalized size = 1.04 \[ -\frac {i a e^{-i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \left (e^{i (c+d x)} \sqrt {1+e^{2 i (c+d x)}}+\sinh ^{-1}\left (e^{i (c+d x)}\right )\right ) \sqrt {a+i a \tan (c+d x)}}{4 d} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.46, size = 241, normalized size = 2.59 \[ \frac {\sqrt {\frac {1}{2}} \sqrt {-\frac {a^{3}}{d^{2}}} d \log \left (\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (4 i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + 4 i \, d\right )} \sqrt {-\frac {a^{3}}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} + 4 \, a^{2} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{a}\right ) - \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{3}}{d^{2}}} d \log \left (\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-4 i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - 4 i \, d\right )} \sqrt {-\frac {a^{3}}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} + 4 \, a^{2} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{a}\right ) + \sqrt {2} {\left (-i \, a e^{\left (3 i \, d x + 3 i \, c\right )} - i \, a e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{4 \, d} \]
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 \[ \int {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cos \left (d x + c\right )^{2}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 1.27, size = 398, normalized size = 4.28 \[ -\frac {\left (-i \sin \left (d x +c \right ) \cos \left (d x +c \right ) \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \sqrt {2}-\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right ) \sqrt {2}-i \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \sqrt {2}\, \sin \left (d x +c \right )-\sqrt {2}\, \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \sin \left (d x +c \right )+8 i \left (\cos ^{4}\left (d x +c \right )\right )-8 \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )-4 i \left (\cos ^{3}\left (d x +c \right )\right )+4 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-4 i \left (\cos ^{2}\left (d x +c \right )\right )\right ) \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, a}{8 d \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )-1\right ) \cos \left (d x +c \right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.80, size = 98, normalized size = 1.05 \[ \frac {i \, {\left (\sqrt {2} a^{\frac {5}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + \frac {16 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{3}}{4 i \, a \tan \left (d x + c\right ) - 4 \, a}\right )}}{8 \, a d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int {\cos \left (c+d\,x\right )}^2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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