Optimal. Leaf size=114 \[ -\frac {2 A \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} d}+\frac {(A-i B) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} \sqrt {a} d}+\frac {A+i B}{d \sqrt {a+i a \tan (c+d x)}} \]
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Rubi [A]
time = 0.24, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps
used = 7, number of rules used = 7, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.206, Rules used = {3677, 3681,
3561, 212, 3680, 65, 214} \begin {gather*} \frac {A+i B}{d \sqrt {a+i a \tan (c+d x)}}+\frac {(A-i B) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} \sqrt {a} d}-\frac {2 A \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} d} \end {gather*}
Antiderivative was successfully verified.
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Rule 65
Rule 212
Rule 214
Rule 3561
Rule 3677
Rule 3680
Rule 3681
Rubi steps
\begin {align*} \int \frac {\cot (c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx &=\frac {A+i B}{d \sqrt {a+i a \tan (c+d x)}}+\frac {\int \cot (c+d x) \sqrt {a+i a \tan (c+d x)} \left (a A-\frac {1}{2} a (i A-B) \tan (c+d x)\right ) \, dx}{a^2}\\ &=\frac {A+i B}{d \sqrt {a+i a \tan (c+d x)}}+\frac {A \int \cot (c+d x) (a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)} \, dx}{a^2}+\frac {(i A+B) \int \sqrt {a+i a \tan (c+d x)} \, dx}{2 a}\\ &=\frac {A+i B}{d \sqrt {a+i a \tan (c+d x)}}+\frac {A \text {Subst}\left (\int \frac {1}{x \sqrt {a+i a x}} \, dx,x,\tan (c+d x)\right )}{d}+\frac {(A-i 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-i B) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} \sqrt {a} d}+\frac {A+i B}{d \sqrt {a+i a \tan (c+d x)}}-\frac {(2 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 )}{a d}\\ &=-\frac {2 A \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} d}+\frac {(A-i B) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} \sqrt {a} d}+\frac {A+i B}{d \sqrt {a+i a \tan (c+d x)}}\\ \end {align*}
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Mathematica [A]
time = 2.49, size = 208, normalized size = 1.82 \begin {gather*} \frac {\left ((A+i B) \sqrt {1+e^{2 i (c+d x)}}+(A-i B) e^{i (c+d x)} \sinh ^{-1}\left (e^{i (c+d x)}\right )-2 \sqrt {2} A e^{i (c+d x)} \tanh ^{-1}\left (\frac {\sqrt {2} e^{i (c+d x)}}{\sqrt {1+e^{2 i (c+d x)}}}\right )\right ) \sqrt {\sec (c+d x)}}{2 d \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+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)}}} \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. 947 vs. \(2 (93 ) = 186\).
time = 0.94, size = 948, normalized size = 8.32
method | result | size |
default | \(\text {Expression too large to display}\) | \(948\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.56, size = 133, normalized size = 1.17 \begin {gather*} -\frac {\frac {\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 )}{\sqrt {a}} - \frac {4 \, A \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 {4 \, {\left (A + i \, B\right )}}{\sqrt {i \, a \tan \left (d x + c\right ) + a}}}{4 \, 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. 575 vs. \(2 (88) = 176\).
time = 2.09, size = 575, normalized size = 5.04 \begin {gather*} -\frac {{\left (\sqrt {2} a d \sqrt {\frac {A^{2} - 2 i \, A B - B^{2}}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} \log \left (-\frac {4 \, {\left ({\left (-i \, A - B\right )} a e^{\left (i \, d x + i \, c\right )} + {\left (i \, a d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {A^{2} - 2 i \, A B - B^{2}}{a d^{2}}}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - \sqrt {2} a d \sqrt {\frac {A^{2} - 2 i \, A B - B^{2}}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} \log \left (-\frac {4 \, {\left ({\left (-i \, A - B\right )} a e^{\left (i \, d x + i \, c\right )} + {\left (-i \, a d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {A^{2} - 2 i \, A B - B^{2}}{a d^{2}}}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) + 2 \, a d \sqrt {\frac {A^{2}}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} \log \left (\frac {16 \, {\left (3 \, A a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + A a^{2} + 2 \, \sqrt {2} {\left (a^{2} d e^{\left (3 i \, d x + 3 i \, c\right )} + a^{2} d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {A^{2}}{a d^{2}}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{A}\right ) - 2 \, a d \sqrt {\frac {A^{2}}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} \log \left (\frac {16 \, {\left (3 \, A a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + A a^{2} - 2 \, \sqrt {2} {\left (a^{2} d e^{\left (3 i \, d x + 3 i \, c\right )} + a^{2} d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {A^{2}}{a d^{2}}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{A}\right ) - 2 \, \sqrt {2} {\left ({\left (A + i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + A + i \, B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{4 \, 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 {\left (A + B \tan {\left (c + d x \right )}\right ) \cot {\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 [B]
time = 7.14, size = 515, normalized size = 4.52 \begin {gather*} \frac {A+B\,1{}\mathrm {i}}{d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}-\frac {2\,A\,\mathrm {atanh}\left (\frac {28\,A^3\,a^{3/2}\,d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{28\,d\,A^3\,a^2+8{}\mathrm {i}\,d\,A^2\,B\,a^2+4\,d\,A\,B^2\,a^2}+\frac {4\,A\,B^2\,a^{3/2}\,d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{28\,d\,A^3\,a^2+8{}\mathrm {i}\,d\,A^2\,B\,a^2+4\,d\,A\,B^2\,a^2}+\frac {A^2\,B\,a^{3/2}\,d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,8{}\mathrm {i}}{28\,d\,A^3\,a^2+8{}\mathrm {i}\,d\,A^2\,B\,a^2+4\,d\,A\,B^2\,a^2}\right )}{\sqrt {a}\,d}+\frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,A^3\,{\left (-a\right )}^{3/2}\,d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,7{}\mathrm {i}}{2\,\left (7\,d\,A^3\,a^2-5{}\mathrm {i}\,d\,A^2\,B\,a^2+3\,d\,A\,B^2\,a^2-1{}\mathrm {i}\,d\,B^3\,a^2\right )}+\frac {\sqrt {2}\,B^3\,{\left (-a\right )}^{3/2}\,d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\left (7\,d\,A^3\,a^2-5{}\mathrm {i}\,d\,A^2\,B\,a^2+3\,d\,A\,B^2\,a^2-1{}\mathrm {i}\,d\,B^3\,a^2\right )}+\frac {\sqrt {2}\,A\,B^2\,{\left (-a\right )}^{3/2}\,d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,3{}\mathrm {i}}{2\,\left (7\,d\,A^3\,a^2-5{}\mathrm {i}\,d\,A^2\,B\,a^2+3\,d\,A\,B^2\,a^2-1{}\mathrm {i}\,d\,B^3\,a^2\right )}+\frac {5\,\sqrt {2}\,A^2\,B\,{\left (-a\right )}^{3/2}\,d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\left (7\,d\,A^3\,a^2-5{}\mathrm {i}\,d\,A^2\,B\,a^2+3\,d\,A\,B^2\,a^2-1{}\mathrm {i}\,d\,B^3\,a^2\right )}\right )\,\left (B+A\,1{}\mathrm {i}\right )}{2\,\sqrt {-a}\,d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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