Optimal. Leaf size=184 \[ \frac {11 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{4 d}-\frac {2 \sqrt {2} a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {i a^2 \cot (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {a^2 \cot ^2(c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {5 i a \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d} \]
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Rubi [A]
time = 0.39, antiderivative size = 184, normalized size of antiderivative = 1.00, number of steps
used = 9, number of rules used = 9, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {3634, 3677,
3679, 3681, 3561, 212, 3680, 65, 214} \begin {gather*} \frac {11 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{4 d}-\frac {2 \sqrt {2} a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {a^2 \cot ^2(c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {i a^2 \cot (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {5 i a \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d} \end {gather*}
Antiderivative was successfully verified.
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Rule 65
Rule 212
Rule 214
Rule 3561
Rule 3634
Rule 3677
Rule 3679
Rule 3680
Rule 3681
Rubi steps
\begin {align*} \int \cot ^3(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx &=-\frac {a^2 \cot ^2(c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {1}{2} \int \frac {\cot ^2(c+d x) \left (-\frac {7 i a^2}{2}+\frac {9}{2} a^2 \tan (c+d x)\right )}{\sqrt {a+i a \tan (c+d x)}} \, dx\\ &=-\frac {i a^2 \cot (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {a^2 \cot ^2(c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {\int \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)} \left (-\frac {5 i a^3}{2}+\frac {3}{2} a^3 \tan (c+d x)\right ) \, dx}{2 a^2}\\ &=-\frac {i a^2 \cot (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {a^2 \cot ^2(c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {5 i a \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}-\frac {\int \cot (c+d x) \sqrt {a+i a \tan (c+d x)} \left (\frac {11 a^4}{4}+\frac {5}{4} i a^4 \tan (c+d x)\right ) \, dx}{2 a^3}\\ &=-\frac {i a^2 \cot (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {a^2 \cot ^2(c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {5 i a \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}-\frac {11}{8} \int \cot (c+d x) (a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)} \, dx-(2 i a) \int \sqrt {a+i a \tan (c+d x)} \, dx\\ &=-\frac {i a^2 \cot (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {a^2 \cot ^2(c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {5 i a \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}-\frac {\left (11 a^2\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {a+i a x}} \, dx,x,\tan (c+d x)\right )}{8 d}-\frac {\left (4 a^2\right ) \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{d}\\ &=-\frac {2 \sqrt {2} a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {i a^2 \cot (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {a^2 \cot ^2(c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {5 i a \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {(11 i a) \text {Subst}\left (\int \frac {1}{i-\frac {i x^2}{a}} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{4 d}\\ &=\frac {11 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{4 d}-\frac {2 \sqrt {2} a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {i a^2 \cot (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {a^2 \cot ^2(c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {5 i a \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}\\ \end {align*}
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Mathematica [A]
time = 2.08, size = 190, normalized size = 1.03 \begin {gather*} -\frac {a e^{-\frac {1}{2} i (2 c+3 d x)} \sqrt {\frac {a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \left (16 \sinh ^{-1}\left (e^{i (c+d x)}\right )-11 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {2} e^{i (c+d x)}}{\sqrt {1+e^{2 i (c+d x)}}}\right )+\sqrt {1+e^{2 i (c+d x)}} (5 i+2 \cot (c+d x)) \csc (c+d x)\right ) \left (\cos \left (\frac {d x}{2}\right )+i \sin \left (\frac {d x}{2}\right )\right )}{4 \sqrt {2} d} \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. 1141 vs. \(2 (148 ) = 296\).
time = 0.87, size = 1142, normalized size = 6.21
method | result | size |
default | \(\text {Expression too large to display}\) | \(1142\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.49, size = 178, normalized size = 0.97 \begin {gather*} \frac {a^{2} {\left (\frac {8 \, \sqrt {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 )}{\sqrt {a}} - \frac {11 \, \log \left (\frac {\sqrt {i \, a \tan \left (d x + c\right ) + a} - \sqrt {a}}{\sqrt {i \, a \tan \left (d x + c\right ) + a} + \sqrt {a}}\right )}{\sqrt {a}} + \frac {2 \, {\left (5 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} - 3 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} - 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a + a^{2}}\right )}}{8 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice
the leaf count of optimal. 546 vs. \(2 (141) = 282\).
time = 0.46, size = 546, normalized size = 2.97 \begin {gather*} -\frac {16 \, \sqrt {2} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a^{3}}{d^{2}}} \log \left (\frac {4 \, {\left (a^{2} e^{\left (i \, d x + i \, c\right )} + {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a^{3}}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{a}\right ) - 16 \, \sqrt {2} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a^{3}}{d^{2}}} \log \left (\frac {4 \, {\left (a^{2} e^{\left (i \, d x + i \, c\right )} - {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a^{3}}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{a}\right ) - 11 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a^{3}}{d^{2}}} \log \left (16 \, {\left (3 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 2 \, \sqrt {2} {\left (d e^{\left (3 i \, d x + 3 i \, c\right )} + d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a^{3}}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} + a^{2}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) + 11 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a^{3}}{d^{2}}} \log \left (16 \, {\left (3 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 2 \, \sqrt {2} {\left (d e^{\left (3 i \, d x + 3 i \, c\right )} + d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a^{3}}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} + a^{2}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) - 4 \, \sqrt {2} {\left (7 \, a e^{\left (5 i \, d x + 5 i \, c\right )} + 4 \, a e^{\left (3 i \, d x + 3 i \, c\right )} - 3 \, a e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{16 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \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 \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {3}{2}} \cot ^{3}{\left (c + d x \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 [B]
time = 4.07, size = 136, normalized size = 0.74 \begin {gather*} -\frac {\mathrm {atan}\left (\frac {\sqrt {a^3}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{a^2}\right )\,\sqrt {a^3}\,11{}\mathrm {i}}{4\,d}-\frac {5\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{4\,d\,{\mathrm {tan}\left (c+d\,x\right )}^2}+\frac {3\,a\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{4\,d\,{\mathrm {tan}\left (c+d\,x\right )}^2}+\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a^3}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2\,a^2}\right )\,\sqrt {a^3}\,2{}\mathrm {i}}{d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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