Optimal. Leaf size=254 \[ -\frac {i \sqrt {c-i d} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{4 \sqrt {2} a^{5/2} f}+\frac {i \sqrt {c+d \tan (e+f x)}}{5 f (a+i a \tan (e+f x))^{5/2}}+\frac {(5 i c-3 d) \sqrt {c+d \tan (e+f x)}}{30 a (c+i d) f (a+i a \tan (e+f x))^{3/2}}-\frac {\left (20 c d-i \left (15 c^2+3 d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{60 a^2 (c+i d)^2 f \sqrt {a+i a \tan (e+f x)}} \]
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Rubi [A]
time = 0.55, antiderivative size = 254, normalized size of antiderivative = 1.00, number of steps
used = 6, number of rules used = 5, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {3638, 3677, 12,
3625, 214} \begin {gather*} -\frac {i \sqrt {c-i d} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{4 \sqrt {2} a^{5/2} f}-\frac {\left (20 c d-i \left (15 c^2+3 d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{60 a^2 f (c+i d)^2 \sqrt {a+i a \tan (e+f x)}}+\frac {(-3 d+5 i c) \sqrt {c+d \tan (e+f x)}}{30 a f (c+i d) (a+i a \tan (e+f x))^{3/2}}+\frac {i \sqrt {c+d \tan (e+f x)}}{5 f (a+i a \tan (e+f x))^{5/2}} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 214
Rule 3625
Rule 3638
Rule 3677
Rubi steps
\begin {align*} \int \frac {\sqrt {c+d \tan (e+f x)}}{(a+i a \tan (e+f x))^{5/2}} \, dx &=\frac {i \sqrt {c+d \tan (e+f x)}}{5 f (a+i a \tan (e+f x))^{5/2}}-\frac {\int \frac {-a (5 c-i d)-4 a d \tan (e+f x)}{(a+i a \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}} \, dx}{10 a^2}\\ &=\frac {i \sqrt {c+d \tan (e+f x)}}{5 f (a+i a \tan (e+f x))^{5/2}}+\frac {(5 i c-3 d) \sqrt {c+d \tan (e+f x)}}{30 a (c+i d) f (a+i a \tan (e+f x))^{3/2}}+\frac {\int \frac {-\frac {1}{2} a^2 \left (10 c d-i \left (15 c^2+9 d^2\right )\right )+a^2 (5 i c-3 d) d \tan (e+f x)}{\sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}} \, dx}{30 a^4 (i c-d)}\\ &=\frac {i \sqrt {c+d \tan (e+f x)}}{5 f (a+i a \tan (e+f x))^{5/2}}+\frac {(5 i c-3 d) \sqrt {c+d \tan (e+f x)}}{30 a (c+i d) f (a+i a \tan (e+f x))^{3/2}}+\frac {\left (15 i c^2-20 c d+3 i d^2\right ) \sqrt {c+d \tan (e+f x)}}{60 a^2 (c+i d)^2 f \sqrt {a+i a \tan (e+f x)}}+\frac {\int \frac {15 a^3 (c-i d) (c+i d)^2 \sqrt {a+i a \tan (e+f x)}}{4 \sqrt {c+d \tan (e+f x)}} \, dx}{30 a^6 (c+i d)^2}\\ &=\frac {i \sqrt {c+d \tan (e+f x)}}{5 f (a+i a \tan (e+f x))^{5/2}}+\frac {(5 i c-3 d) \sqrt {c+d \tan (e+f x)}}{30 a (c+i d) f (a+i a \tan (e+f x))^{3/2}}+\frac {\left (15 i c^2-20 c d+3 i d^2\right ) \sqrt {c+d \tan (e+f x)}}{60 a^2 (c+i d)^2 f \sqrt {a+i a \tan (e+f x)}}+\frac {(c-i d) \int \frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}} \, dx}{8 a^3}\\ &=\frac {i \sqrt {c+d \tan (e+f x)}}{5 f (a+i a \tan (e+f x))^{5/2}}+\frac {(5 i c-3 d) \sqrt {c+d \tan (e+f x)}}{30 a (c+i d) f (a+i a \tan (e+f x))^{3/2}}+\frac {\left (15 i c^2-20 c d+3 i d^2\right ) \sqrt {c+d \tan (e+f x)}}{60 a^2 (c+i d)^2 f \sqrt {a+i a \tan (e+f x)}}-\frac {(i c+d) \text {Subst}\left (\int \frac {1}{a c-i a d-2 a^2 x^2} \, dx,x,\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {a+i a \tan (e+f x)}}\right )}{4 a f}\\ &=-\frac {i \sqrt {c-i d} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{4 \sqrt {2} a^{5/2} f}+\frac {i \sqrt {c+d \tan (e+f x)}}{5 f (a+i a \tan (e+f x))^{5/2}}+\frac {(5 i c-3 d) \sqrt {c+d \tan (e+f x)}}{30 a (c+i d) f (a+i a \tan (e+f x))^{3/2}}+\frac {\left (15 i c^2-20 c d+3 i d^2\right ) \sqrt {c+d \tan (e+f x)}}{60 a^2 (c+i d)^2 f \sqrt {a+i a \tan (e+f x)}}\\ \end {align*}
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Mathematica [A]
time = 5.82, size = 302, normalized size = 1.19 \begin {gather*} \frac {\sec ^{\frac {5}{2}}(e+f x) \left (-i \sqrt {2} \sqrt {c-i d} e^{2 i (e+f x)} \sqrt {\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \sqrt {1+e^{2 i (e+f x)}} \log \left (2 \left (\sqrt {c-i d} e^{i (e+f x)}+\sqrt {1+e^{2 i (e+f x)}} \sqrt {c-\frac {i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}\right )\right )+\frac {2 i \left (11 c^2+20 i c d-9 d^2+\left (26 c^2+40 i c d-6 d^2\right ) \cos (2 (e+f x))+4 i c (5 c+7 i d) \sin (2 (e+f x))\right ) \sqrt {c+d \tan (e+f x)}}{15 (c+i d)^2 \sqrt {\sec (e+f x)}}\right )}{8 f (a+i a \tan (e+f x))^{5/2}} \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. 2225 vs. \(2 (206 ) = 412\).
time = 0.75, size = 2226, normalized size = 8.76
method | result | size |
derivativedivides | \(\text {Expression too large to display}\) | \(2226\) |
default | \(\text {Expression too large to display}\) | \(2226\) |
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 \begin {gather*} \text {Exception raised: RuntimeError} \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. 521 vs. \(2 (199) = 398\).
time = 0.86, size = 521, normalized size = 2.05 \begin {gather*} \frac {{\left (15 \, \sqrt {\frac {1}{2}} {\left (a^{3} c^{2} + 2 i \, a^{3} c d - a^{3} d^{2}\right )} f \sqrt {-\frac {c - i \, d}{a^{5} f^{2}}} e^{\left (5 i \, f x + 5 i \, e\right )} \log \left (2 i \, \sqrt {\frac {1}{2}} a^{3} f \sqrt {-\frac {c - i \, d}{a^{5} f^{2}}} e^{\left (i \, f x + i \, e\right )} + \sqrt {2} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}\right ) - 15 \, \sqrt {\frac {1}{2}} {\left (a^{3} c^{2} + 2 i \, a^{3} c d - a^{3} d^{2}\right )} f \sqrt {-\frac {c - i \, d}{a^{5} f^{2}}} e^{\left (5 i \, f x + 5 i \, e\right )} \log \left (-2 i \, \sqrt {\frac {1}{2}} a^{3} f \sqrt {-\frac {c - i \, d}{a^{5} f^{2}}} e^{\left (i \, f x + i \, e\right )} + \sqrt {2} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}\right ) - \sqrt {2} {\left (-3 i \, c^{2} + 6 \, c d + 3 i \, d^{2} + {\left (-23 i \, c^{2} + 34 \, c d + 3 i \, d^{2}\right )} e^{\left (6 i \, f x + 6 i \, e\right )} - 2 \, {\left (17 i \, c^{2} - 27 \, c d - 6 i \, d^{2}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} - 2 \, {\left (7 i \, c^{2} - 13 \, c d - 6 i \, d^{2}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-5 i \, f x - 5 i \, e\right )}}{120 \, {\left (a^{3} c^{2} + 2 i \, a^{3} c d - a^{3} d^{2}\right )} f} \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 {\sqrt {c + d \tan {\left (e + f x \right )}}}{\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {5}{2}}}\, dx \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 \frac {\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}}{{\left (a+a\,\mathrm {tan}\left (e+f\,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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