Optimal. Leaf size=252 \[ -\frac {(-1)^{3/4} a^{5/2} (46 A-45 i B) \text {ArcTan}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{8 d}-\frac {(4+4 i) a^{5/2} (A-i B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {a^2 (18 i A+19 B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{8 d}-\frac {a^2 (2 A-3 i B) \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {i a B \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d} \]
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Rubi [A]
time = 0.60, antiderivative size = 252, normalized size of antiderivative = 1.00, number of steps
used = 10, number of rules used = 9, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.237, Rules used = {3675, 3678,
3682, 3625, 211, 3680, 65, 223, 209} \begin {gather*} -\frac {(-1)^{3/4} a^{5/2} (46 A-45 i B) \text {ArcTan}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{8 d}-\frac {(4+4 i) a^{5/2} (A-i B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {a^2 (2 A-3 i B) \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {a^2 (19 B+18 i A) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{8 d}+\frac {i a B \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d} \end {gather*}
Antiderivative was successfully verified.
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Rule 65
Rule 209
Rule 211
Rule 223
Rule 3625
Rule 3675
Rule 3678
Rule 3680
Rule 3682
Rubi steps
\begin {align*} \int \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx &=\frac {i a B \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac {1}{3} \int \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2} \left (\frac {3}{2} a (2 A-i B)+\frac {3}{2} a (2 i A+3 B) \tan (c+d x)\right ) \, dx\\ &=-\frac {a^2 (2 A-3 i B) \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {i a B \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac {1}{6} \int \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)} \left (\frac {3}{4} a^2 (14 A-13 i B)+\frac {3}{4} a^2 (18 i A+19 B) \tan (c+d x)\right ) \, dx\\ &=\frac {a^2 (18 i A+19 B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{8 d}-\frac {a^2 (2 A-3 i B) \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {i a B \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac {\int \frac {\sqrt {a+i a \tan (c+d x)} \left (-\frac {3}{8} a^3 (18 i A+19 B)+\frac {3}{8} a^3 (46 A-45 i B) \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx}{6 a}\\ &=\frac {a^2 (18 i A+19 B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{8 d}-\frac {a^2 (2 A-3 i B) \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {i a B \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}-\left (4 a^2 (i A+B)\right ) \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx+\frac {1}{16} (a (46 i A+45 B)) \int \frac {(a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx\\ &=\frac {a^2 (18 i A+19 B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{8 d}-\frac {a^2 (2 A-3 i B) \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {i a B \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}-\frac {\left (8 a^4 (A-i B)\right ) \text {Subst}\left (\int \frac {1}{-i a-2 a^2 x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {\left (a^3 (46 i A+45 B)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {a+i a x}} \, dx,x,\tan (c+d x)\right )}{16 d}\\ &=-\frac {(4+4 i) a^{5/2} (A-i B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {a^2 (18 i A+19 B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{8 d}-\frac {a^2 (2 A-3 i B) \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {i a B \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac {\left (a^3 (46 i A+45 B)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+i a x^2}} \, dx,x,\sqrt {\tan (c+d x)}\right )}{8 d}\\ &=-\frac {(4+4 i) a^{5/2} (A-i B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {a^2 (18 i A+19 B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{8 d}-\frac {a^2 (2 A-3 i B) \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {i a B \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac {\left (a^3 (46 i A+45 B)\right ) \text {Subst}\left (\int \frac {1}{1-i a x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{8 d}\\ &=-\frac {\sqrt [4]{-1} a^{5/2} (46 i A+45 B) \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{8 d}-\frac {(4+4 i) a^{5/2} (A-i B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {a^2 (18 i A+19 B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{8 d}-\frac {a^2 (2 A-3 i B) \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {i a B \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}\\ \end {align*}
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Mathematica [A]
time = 6.51, size = 426, normalized size = 1.69 \begin {gather*} \frac {\left (\frac {\sqrt {2} e^{-2 i (c+d x)} \sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \left (-256 i (A-i B) \log \left (e^{i (c+d x)}+\sqrt {-1+e^{2 i (c+d x)}}\right )+\sqrt {2} (46 i A+45 B) \left (\log \left (1-3 e^{2 i (c+d x)}-2 \sqrt {2} e^{i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}}\right )-\log \left (1-3 e^{2 i (c+d x)}+2 \sqrt {2} e^{i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}}\right )\right )\right )}{\sqrt {-1+e^{2 i (c+d x)}} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}}}+\frac {4 \sec ^{\frac {5}{2}}(c+d x) (\cos (2 c)-i \sin (2 c)) (54 i A+49 B+(54 i A+65 B) \cos (2 (c+d x))+(-12 A+26 i B) \sin (2 (c+d x))) \sqrt {\tan (c+d x)}}{3 \cos (2 d x)+3 i \sin (2 d x)}\right ) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{64 d \sec ^{\frac {7}{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. 652 vs. \(2 (202 ) = 404\).
time = 0.12, size = 653, normalized size = 2.59
method | result | size |
derivativedivides | \(-\frac {\left (\sqrt {\tan }\left (d x +c \right )\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a^{2} \left (16 B \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, \left (\tan ^{2}\left (d x +c \right )\right )-52 i B \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, \tan \left (d x +c \right )+54 i A \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a -108 i A \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}+24 A \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, \tan \left (d x +c \right )-48 i \sqrt {i a}\, \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a +57 B \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a -114 B \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}+96 i \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a -48 \sqrt {i a}\, \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a -96 \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \right )}{48 d \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}}\) | \(653\) |
default | \(-\frac {\left (\sqrt {\tan }\left (d x +c \right )\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a^{2} \left (16 B \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, \left (\tan ^{2}\left (d x +c \right )\right )-52 i B \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, \tan \left (d x +c \right )+54 i A \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a -108 i A \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}+24 A \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, \tan \left (d x +c \right )-48 i \sqrt {i a}\, \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a +57 B \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a -114 B \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}+96 i \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a -48 \sqrt {i a}\, \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a -96 \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \right )}{48 d \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}}\) | \(653\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \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. 932 vs. \(2 (188) = 376\).
time = 0.79, size = 932, normalized size = 3.70 \begin {gather*} \frac {96 \, \sqrt {2} \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{5}}{d^{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 )} \log \left (\frac {{\left (\sqrt {2} \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{5}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )} + \sqrt {2} {\left ({\left (-i \, A - B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (-i \, A - B\right )} a^{2}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{{\left (-i \, A - B\right )} a^{2}}\right ) - 96 \, \sqrt {2} \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{5}}{d^{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 )} \log \left (-\frac {{\left (\sqrt {2} \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{5}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )} - \sqrt {2} {\left ({\left (-i \, A - B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (-i \, A - B\right )} a^{2}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{{\left (-i \, A - B\right )} a^{2}}\right ) + 2 \, \sqrt {2} {\left ({\left (66 i \, A + 91 \, B\right )} a^{2} e^{\left (5 i \, d x + 5 i \, c\right )} - 2 \, {\left (-54 i \, A - 49 \, B\right )} a^{2} e^{\left (3 i \, d x + 3 i \, c\right )} - 3 \, {\left (-14 i \, A - 13 \, B\right )} a^{2} 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 {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} + 3 \, \sqrt {\frac {{\left (2116 i \, A^{2} + 4140 \, A B - 2025 i \, B^{2}\right )} a^{5}}{d^{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 )} \log \left (\frac {{\left (\sqrt {2} {\left ({\left (46 i \, A + 45 \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (46 i \, A + 45 \, B\right )} a^{2}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} + 2 \, \sqrt {\frac {{\left (2116 i \, A^{2} + 4140 \, A B - 2025 i \, B^{2}\right )} a^{5}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{{\left (46 i \, A + 45 \, B\right )} a^{2}}\right ) - 3 \, \sqrt {\frac {{\left (2116 i \, A^{2} + 4140 \, A B - 2025 i \, B^{2}\right )} a^{5}}{d^{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 )} \log \left (\frac {{\left (\sqrt {2} {\left ({\left (46 i \, A + 45 \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (46 i \, A + 45 \, B\right )} a^{2}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} - 2 \, \sqrt {\frac {{\left (2116 i \, A^{2} + 4140 \, A B - 2025 i \, B^{2}\right )} a^{5}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{{\left (46 i \, A + 45 \, B\right )} a^{2}}\right )}{48 \, {\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(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \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 \sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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