Optimal. Leaf size=210 \[ \frac {\sqrt {a} (9 i A+14 B) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{8 d}-\frac {\sqrt {2} \sqrt {a} (i A+B) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {(7 A-2 i B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{8 d}-\frac {(i A+6 B) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d}-\frac {A \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d} \]
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Rubi [A]
time = 0.50, antiderivative size = 210, normalized size of antiderivative = 1.00, number of steps
used = 9, number of rules used = 7, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {3679, 3681,
3561, 212, 3680, 65, 214} \begin {gather*} \frac {\sqrt {a} (14 B+9 i A) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{8 d}-\frac {\sqrt {2} \sqrt {a} (B+i A) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {(6 B+i A) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d}+\frac {(7 A-2 i B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{8 d}-\frac {A \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d} \end {gather*}
Antiderivative was successfully verified.
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Rule 65
Rule 212
Rule 214
Rule 3561
Rule 3679
Rule 3680
Rule 3681
Rubi steps
\begin {align*} \int \cot ^4(c+d x) \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx &=-\frac {A \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}+\frac {\int \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)} \left (\frac {1}{2} a (i A+6 B)-\frac {5}{2} a A \tan (c+d x)\right ) \, dx}{3 a}\\ &=-\frac {(i A+6 B) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d}-\frac {A \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}+\frac {\int \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)} \left (-\frac {3}{4} a^2 (7 A-2 i B)-\frac {3}{4} a^2 (i A+6 B) \tan (c+d x)\right ) \, dx}{6 a^2}\\ &=\frac {(7 A-2 i B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{8 d}-\frac {(i A+6 B) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d}-\frac {A \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}+\frac {\int \cot (c+d x) \sqrt {a+i a \tan (c+d x)} \left (-\frac {3}{8} a^3 (9 i A+14 B)+\frac {3}{8} a^3 (7 A-2 i B) \tan (c+d x)\right ) \, dx}{6 a^3}\\ &=\frac {(7 A-2 i B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{8 d}-\frac {(i A+6 B) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d}-\frac {A \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}+(A-i B) \int \sqrt {a+i a \tan (c+d x)} \, dx-\frac {(9 i A+14 B) \int \cot (c+d x) (a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)} \, dx}{16 a}\\ &=\frac {(7 A-2 i B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{8 d}-\frac {(i A+6 B) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d}-\frac {A \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}-\frac {(2 a (i A+B)) \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{d}-\frac {(a (9 i A+14 B)) \text {Subst}\left (\int \frac {1}{x \sqrt {a+i a x}} \, dx,x,\tan (c+d x)\right )}{16 d}\\ &=-\frac {\sqrt {2} \sqrt {a} (i A+B) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {(7 A-2 i B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{8 d}-\frac {(i A+6 B) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d}-\frac {A \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}-\frac {(9 A-14 i B) \text {Subst}\left (\int \frac {1}{i-\frac {i x^2}{a}} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{8 d}\\ &=\frac {\sqrt {a} (9 i A+14 B) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{8 d}-\frac {\sqrt {2} \sqrt {a} (i A+B) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {(7 A-2 i B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{8 d}-\frac {(i A+6 B) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d}-\frac {A \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\\ \end {align*}
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Mathematica [A]
time = 4.41, size = 414, normalized size = 1.97 \begin {gather*} \frac {\left (-\frac {2 i \left (32 \sqrt {2} (A-i B) \sinh ^{-1}\left (e^{i (c+d x)}\right )+(9 A-14 i B) \left (\log \left (\left (-1+e^{i (c+d x)}\right )^2\right )-\log \left (\left (1+e^{i (c+d x)}\right )^2\right )+\log \left (3+3 e^{2 i (c+d x)}+2 \sqrt {2} \sqrt {1+e^{2 i (c+d x)}}-2 e^{i (c+d x)} \left (1+\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}\right )\right )-\log \left (3+3 e^{2 i (c+d x)}+2 \sqrt {2} \sqrt {1+e^{2 i (c+d x)}}+2 e^{i (c+d x)} \left (1+\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}\right )\right )\right )\right )}{\sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}}}-\frac {4 \csc ^3(c+d x) (-13 A+6 i B+(29 A-6 i B) \cos (2 (c+d x))+2 (i A+6 B) \sin (2 (c+d x)))}{3 \sqrt {\sec (c+d x)}}\right ) \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x))}{64 d \sec ^{\frac {3}{2}}(c+d x) (A \cos (c+d x)+B \sin (c+d x))} \end {gather*}
Warning: Unable to verify antiderivative.
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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. 1782 vs. \(2 (172 ) = 344\).
time = 0.51, size = 1783, normalized size = 8.49
method | result | size |
default | \(\text {Expression too large to display}\) | \(1783\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.58, size = 249, normalized size = 1.19 \begin {gather*} \frac {i \, a^{3} {\left (\frac {2 \, {\left (3 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} {\left (7 \, A - 2 i \, B\right )} - 40 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} A a + 3 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} {\left (9 \, A + 2 i \, B\right )} a^{2}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a^{2} - 3 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{3} + 3 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{4} - a^{5}} + \frac {24 \, \sqrt {2} {\left (A - i \, B\right )} \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 )}{a^{\frac {5}{2}}} - \frac {3 \, {\left (9 \, A - 14 i \, B\right )} \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 )}{a^{\frac {5}{2}}}\right )}}{48 \, 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. 823 vs. \(2 (163) = 326\).
time = 1.91, size = 823, normalized size = 3.92 \begin {gather*} \frac {48 \, \sqrt {2} {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {-\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} a}{d^{2}}} \log \left (-\frac {4 \, {\left ({\left (-i \, A - B\right )} a e^{\left (i \, d x + i \, c\right )} + {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {-\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} a}{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 )}}{i \, A + B}\right ) - 48 \, \sqrt {2} {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {-\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} a}{d^{2}}} \log \left (-\frac {4 \, {\left ({\left (-i \, A - B\right )} a e^{\left (i \, d x + i \, c\right )} - {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {-\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} a}{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 )}}{i \, A + B}\right ) - 3 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {-\frac {{\left (81 \, A^{2} - 252 i \, A B - 196 \, B^{2}\right )} a}{d^{2}}} \log \left (-\frac {16 \, {\left (3 \, {\left (-9 i \, A - 14 \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (-9 i \, A - 14 \, B\right )} a^{2} + 2 \, \sqrt {2} {\left (a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {-\frac {{\left (81 \, A^{2} - 252 i \, A B - 196 \, B^{2}\right )} a}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{9 i \, A + 14 \, B}\right ) + 3 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {-\frac {{\left (81 \, A^{2} - 252 i \, A B - 196 \, B^{2}\right )} a}{d^{2}}} \log \left (-\frac {16 \, {\left (3 \, {\left (-9 i \, A - 14 \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (-9 i \, A - 14 \, B\right )} a^{2} - 2 \, \sqrt {2} {\left (a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {-\frac {{\left (81 \, A^{2} - 252 i \, A B - 196 \, B^{2}\right )} a}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{9 i \, A + 14 \, B}\right ) + 4 \, \sqrt {2} {\left ({\left (31 i \, A + 18 \, B\right )} e^{\left (7 i \, d x + 7 i \, c\right )} + {\left (5 i \, A + 6 \, B\right )} e^{\left (5 i \, d x + 5 i \, c\right )} + {\left (i \, A - 18 \, B\right )} e^{\left (3 i \, d x + 3 i \, c\right )} - 3 \, {\left (-9 i \, A + 2 \, B\right )} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{96 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, 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 \sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )} \left (A + B \tan {\left (c + d x \right )}\right ) \cot ^{4}{\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 = 7.09, size = 735, normalized size = 3.50 \begin {gather*} -\frac {\frac {\left (9\,A\,a^3+B\,a^3\,2{}\mathrm {i}\right )\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{8\,d}+\frac {\left (7\,A\,a-B\,a\,2{}\mathrm {i}\right )\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}\,1{}\mathrm {i}}{8\,d}-\frac {A\,a^2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,5{}\mathrm {i}}{3\,d}}{3\,a\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2-3\,a^2\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )-{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3+a^3}-\frac {\mathrm {atan}\left (\frac {47\,\sqrt {32}\,A^3\,a^{9/2}\,d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{8\,\left (47{}\mathrm {i}\,d\,A^3\,a^5+51\,d\,A^2\,B\,a^5+64{}\mathrm {i}\,d\,A\,B^2\,a^5+68\,d\,B^3\,a^5\right )}-\frac {\sqrt {32}\,B^3\,a^{9/2}\,d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,17{}\mathrm {i}}{2\,\left (47{}\mathrm {i}\,d\,A^3\,a^5+51\,d\,A^2\,B\,a^5+64{}\mathrm {i}\,d\,A\,B^2\,a^5+68\,d\,B^3\,a^5\right )}+\frac {8\,\sqrt {32}\,A\,B^2\,a^{9/2}\,d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{47{}\mathrm {i}\,d\,A^3\,a^5+51\,d\,A^2\,B\,a^5+64{}\mathrm {i}\,d\,A\,B^2\,a^5+68\,d\,B^3\,a^5}-\frac {\sqrt {32}\,A^2\,B\,a^{9/2}\,d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,51{}\mathrm {i}}{8\,\left (47{}\mathrm {i}\,d\,A^3\,a^5+51\,d\,A^2\,B\,a^5+64{}\mathrm {i}\,d\,A\,B^2\,a^5+68\,d\,B^3\,a^5\right )}\right )\,\left (B+A\,1{}\mathrm {i}\right )\,\sqrt {\frac {a}{32}}\,8{}\mathrm {i}}{d}+\frac {\sqrt {a}\,\mathrm {atan}\left (\frac {423\,A^3\,a^{9/2}\,d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{8\,\left (\frac {423{}\mathrm {i}\,d\,A^3\,a^5}{8}+\frac {347\,d\,A^2\,B\,a^5}{4}+\frac {139{}\mathrm {i}\,d\,A\,B^2\,a^5}{2}+119\,d\,B^3\,a^5\right )}-\frac {B^3\,a^{9/2}\,d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,119{}\mathrm {i}}{\frac {423{}\mathrm {i}\,d\,A^3\,a^5}{8}+\frac {347\,d\,A^2\,B\,a^5}{4}+\frac {139{}\mathrm {i}\,d\,A\,B^2\,a^5}{2}+119\,d\,B^3\,a^5}+\frac {139\,A\,B^2\,a^{9/2}\,d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\left (\frac {423{}\mathrm {i}\,d\,A^3\,a^5}{8}+\frac {347\,d\,A^2\,B\,a^5}{4}+\frac {139{}\mathrm {i}\,d\,A\,B^2\,a^5}{2}+119\,d\,B^3\,a^5\right )}-\frac {A^2\,B\,a^{9/2}\,d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,347{}\mathrm {i}}{4\,\left (\frac {423{}\mathrm {i}\,d\,A^3\,a^5}{8}+\frac {347\,d\,A^2\,B\,a^5}{4}+\frac {139{}\mathrm {i}\,d\,A\,B^2\,a^5}{2}+119\,d\,B^3\,a^5\right )}\right )\,\left (14\,B+A\,9{}\mathrm {i}\right )\,1{}\mathrm {i}}{8\,d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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