Optimal. Leaf size=219 \[ -\frac {45 \sqrt [4]{-1} a^{5/2} \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{8 d}-\frac {(4-4 i) a^{5/2} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}+\frac {13 i a^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d}+\frac {19 a^2 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{8 d} \]
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Rubi [A] time = 0.70, antiderivative size = 219, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {3556, 3597, 3601, 3544, 205, 3599, 63, 217, 203} \[ -\frac {45 \sqrt [4]{-1} a^{5/2} \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{8 d}-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}+\frac {13 i a^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d}+\frac {19 a^2 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{8 d}-\frac {(4-4 i) a^{5/2} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d} \]
Antiderivative was successfully verified.
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Rule 63
Rule 203
Rule 205
Rule 217
Rule 3544
Rule 3556
Rule 3597
Rule 3599
Rule 3601
Rubi steps
\begin {align*} \int \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx &=-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}+\frac {1}{3} a \int \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)} \left (\frac {11 a}{2}+\frac {13}{2} i a \tan (c+d x)\right ) \, dx\\ &=\frac {13 i a^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d}-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}+\frac {1}{6} \int \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)} \left (-\frac {39 i a^2}{4}+\frac {57}{4} a^2 \tan (c+d x)\right ) \, dx\\ &=\frac {19 a^2 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{8 d}+\frac {13 i a^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d}-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}+\frac {\int \frac {\sqrt {a+i a \tan (c+d x)} \left (-\frac {57 a^3}{8}-\frac {135}{8} i a^3 \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx}{6 a}\\ &=\frac {19 a^2 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{8 d}+\frac {13 i a^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d}-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}+\frac {1}{16} (45 a) \int \frac {(a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx-\left (4 a^2\right ) \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx\\ &=\frac {19 a^2 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{8 d}+\frac {13 i a^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d}-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}+\frac {\left (45 a^3\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {a+i a x}} \, dx,x,\tan (c+d x)\right )}{16 d}+\frac {\left (8 i a^4\right ) \operatorname {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 {(4-4 i) a^{5/2} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {19 a^2 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{8 d}+\frac {13 i a^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d}-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}+\frac {\left (45 a^3\right ) \operatorname {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} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {19 a^2 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{8 d}+\frac {13 i a^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d}-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}+\frac {\left (45 a^3\right ) \operatorname {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 {45 \sqrt [4]{-1} a^{5/2} \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} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {19 a^2 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{8 d}+\frac {13 i a^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d}-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\\ \end {align*}
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Mathematica [A] time = 3.09, size = 258, normalized size = 1.18 \[ \frac {i a^2 e^{-i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}} \sqrt {\frac {a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \left (64 \tanh ^{-1}\left (\frac {e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )-45 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {2} e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )\right )}{8 \sqrt {2} d \sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}}}+\frac {a^2 \sqrt {\tan (c+d x)} \sec ^2(c+d x) \sqrt {a+i a \tan (c+d x)} (26 i \sin (2 (c+d x))+65 \cos (2 (c+d x))+49)}{48 d} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.61, size = 694, normalized size = 3.17 \[ \frac {\sqrt {2} {\left (91 \, a^{2} e^{\left (5 i \, d x + 5 i \, c\right )} + 98 \, a^{2} e^{\left (3 i \, d x + 3 i \, c\right )} + 39 \, 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}} + 12 \, \sqrt {-\frac {2025 i \, a^{5}}{64 \, 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 (45 \, \sqrt {2} {\left (a^{2} e^{\left (2 i \, d x + 2 i \, c\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}} + 16 \, \sqrt {-\frac {2025 i \, a^{5}}{64 \, d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{45 \, a^{2}}\right ) - 12 \, \sqrt {-\frac {2025 i \, a^{5}}{64 \, 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 (45 \, \sqrt {2} {\left (a^{2} e^{\left (2 i \, d x + 2 i \, c\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}} - 16 \, \sqrt {-\frac {2025 i \, a^{5}}{64 \, d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{45 \, a^{2}}\right ) - 12 \, \sqrt {-\frac {32 i \, 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 (4 \, \sqrt {2} {\left (a^{2} e^{\left (2 i \, d x + 2 i \, c\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}} + \sqrt {-\frac {32 i \, a^{5}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{4 \, a^{2}}\right ) + 12 \, \sqrt {-\frac {32 i \, 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 (4 \, \sqrt {2} {\left (a^{2} e^{\left (2 i \, d x + 2 i \, c\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}} - \sqrt {-\frac {32 i \, a^{5}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{4 \, a^{2}}\right )}{24 \, {\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 )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [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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maple [B] time = 0.20, size = 450, normalized size = 2.05 \[ -\frac {\left (\sqrt {\tan }\left (d x +c \right )\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a^{2} \left (48 i \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, a -52 i \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, \tan \left (d x +c \right )+16 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, \left (\tan ^{2}\left (d x +c \right )\right )+192 i \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a -48 \sqrt {i a}\, \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a -135 \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a -114 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\right )}{48 d \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}} \]
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 \[ \int {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \tan \left (d x + c\right )^{\frac {3}{2}}\,{d x} \]
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 \[ \int {\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2} \,d x \]
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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