Optimal. Leaf size=218 \[ \frac {2 \sqrt [4]{-1} \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac {7 \sqrt {\tan (c+d x)}}{4 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {\tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {i \tan ^{\frac {3}{2}}(c+d x)}{2 a d (a+i a \tan (c+d x))^{3/2}} \]
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Rubi [A] time = 0.71, antiderivative size = 218, 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 = {3558, 3595, 3601, 3544, 205, 3599, 63, 217, 203} \[ \frac {2 \sqrt [4]{-1} \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac {7 \sqrt {\tan (c+d x)}}{4 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}-\frac {\tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {i \tan ^{\frac {3}{2}}(c+d x)}{2 a d (a+i a \tan (c+d x))^{3/2}} \]
Antiderivative was successfully verified.
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Rule 63
Rule 203
Rule 205
Rule 217
Rule 3544
Rule 3558
Rule 3595
Rule 3599
Rule 3601
Rubi steps
\begin {align*} \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx &=-\frac {\tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\int \frac {\tan ^{\frac {3}{2}}(c+d x) \left (-\frac {5 a}{2}+5 i a \tan (c+d x)\right )}{(a+i a \tan (c+d x))^{3/2}} \, dx}{5 a^2}\\ &=-\frac {\tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {i \tan ^{\frac {3}{2}}(c+d x)}{2 a d (a+i a \tan (c+d x))^{3/2}}+\frac {\int \frac {\sqrt {\tan (c+d x)} \left (-\frac {45 i a^2}{4}-15 a^2 \tan (c+d x)\right )}{\sqrt {a+i a \tan (c+d x)}} \, dx}{15 a^4}\\ &=-\frac {\tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {i \tan ^{\frac {3}{2}}(c+d x)}{2 a d (a+i a \tan (c+d x))^{3/2}}+\frac {7 \sqrt {\tan (c+d x)}}{4 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {\int \frac {\sqrt {a+i a \tan (c+d x)} \left (\frac {105 a^3}{8}-15 i a^3 \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx}{15 a^6}\\ &=-\frac {\tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {i \tan ^{\frac {3}{2}}(c+d x)}{2 a d (a+i a \tan (c+d x))^{3/2}}+\frac {7 \sqrt {\tan (c+d x)}}{4 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {\int \frac {(a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx}{a^4}+\frac {\int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx}{8 a^3}\\ &=-\frac {\tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {i \tan ^{\frac {3}{2}}(c+d x)}{2 a d (a+i a \tan (c+d x))^{3/2}}+\frac {7 \sqrt {\tan (c+d x)}}{4 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {\operatorname {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {a+i a x}} \, dx,x,\tan (c+d x)\right )}{a^2 d}-\frac {i \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 )}{4 a d}\\ &=\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}-\frac {\tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {i \tan ^{\frac {3}{2}}(c+d x)}{2 a d (a+i a \tan (c+d x))^{3/2}}+\frac {7 \sqrt {\tan (c+d x)}}{4 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {2 \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+i a x^2}} \, dx,x,\sqrt {\tan (c+d x)}\right )}{a^2 d}\\ &=\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}-\frac {\tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {i \tan ^{\frac {3}{2}}(c+d x)}{2 a d (a+i a \tan (c+d x))^{3/2}}+\frac {7 \sqrt {\tan (c+d x)}}{4 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {2 \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 )}{a^2 d}\\ &=\frac {2 \sqrt [4]{-1} \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}-\frac {\tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {i \tan ^{\frac {3}{2}}(c+d x)}{2 a d (a+i a \tan (c+d x))^{3/2}}+\frac {7 \sqrt {\tan (c+d x)}}{4 a^2 d \sqrt {a+i a \tan (c+d x)}}\\ \end {align*}
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Mathematica [A] time = 2.73, size = 241, normalized size = 1.11 \[ \frac {i e^{-6 i (c+d x)} \sqrt {\frac {a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \left (-8 e^{2 i (c+d x)}+48 e^{4 i (c+d x)}-41 e^{6 i (c+d x)}-5 e^{5 i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac {e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )+40 \sqrt {2} e^{5 i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac {\sqrt {2} e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )+1\right )}{20 \sqrt {2} a^3 d \sqrt {\tan (c+d x)}} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.74, size = 621, normalized size = 2.85 \[ \frac {{\left (10 \, a^{3} d \sqrt {-\frac {i}{8 \, a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (a^{3} d \sqrt {-\frac {i}{8 \, a^{5} d^{2}}} e^{\left (i \, d x + i \, c\right )} + \frac {1}{4} \, \sqrt {2} \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}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}\right ) - 10 \, a^{3} d \sqrt {-\frac {i}{8 \, a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (-a^{3} d \sqrt {-\frac {i}{8 \, a^{5} d^{2}}} e^{\left (i \, d x + i \, c\right )} + \frac {1}{4} \, \sqrt {2} \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}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}\right ) - 10 \, a^{3} d \sqrt {-\frac {4 i}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (\frac {52 \, {\left (4 \, \sqrt {2} \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}} {\left (e^{\left (3 i \, d x + 3 i \, c\right )} + e^{\left (i \, d x + i \, c\right )}\right )} + {\left (3 \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} - a^{3} d\right )} \sqrt {-\frac {4 i}{a^{5} d^{2}}}\right )}}{605 \, {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}}\right ) + 10 \, a^{3} d \sqrt {-\frac {4 i}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (\frac {52 \, {\left (4 \, \sqrt {2} \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}} {\left (e^{\left (3 i \, d x + 3 i \, c\right )} + e^{\left (i \, d x + i \, c\right )}\right )} - {\left (3 \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} - a^{3} d\right )} \sqrt {-\frac {4 i}{a^{5} d^{2}}}\right )}}{605 \, {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}}\right ) + \sqrt {2} \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}} {\left (41 \, e^{\left (6 i \, d x + 6 i \, c\right )} + 34 \, e^{\left (4 i \, d x + 4 i \, c\right )} - 6 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{40 \, a^{3} d} \]
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 \[ \int \frac {\tan \left (d x + c\right )^{\frac {7}{2}}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.19, size = 963, normalized size = 4.42 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]
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 \[ \int \frac {{\mathrm {tan}\left (c+d\,x\right )}^{7/2}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \]
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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