Optimal. Leaf size=62 \[ -\frac {2 a (i A+B)}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac {2 a B}{3 c f (c-i c \tan (e+f x))^{3/2}} \]
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Rubi [A]
time = 0.08, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps
used = 3, number of rules used = 2, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.049, Rules used = {3669, 45}
\begin {gather*} \frac {2 a B}{3 c f (c-i c \tan (e+f x))^{3/2}}-\frac {2 a (B+i A)}{5 f (c-i c \tan (e+f x))^{5/2}} \end {gather*}
Antiderivative was successfully verified.
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Rule 45
Rule 3669
Rubi steps
\begin {align*} \int \frac {(a+i a \tan (e+f x)) (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{5/2}} \, dx &=\frac {(a c) \text {Subst}\left (\int \frac {A+B x}{(c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {(a c) \text {Subst}\left (\int \left (\frac {A-i B}{(c-i c x)^{7/2}}+\frac {i B}{c (c-i c x)^{5/2}}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {2 a (i A+B)}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac {2 a B}{3 c f (c-i c \tan (e+f x))^{3/2}}\\ \end {align*}
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Mathematica [A]
time = 2.19, size = 100, normalized size = 1.61 \begin {gather*} \frac {2 a \cos ^2(e+f x) (\cos (f x)-i \sin (f x)) ((-3 i A+2 B) \cos (e+f x)-5 i B \sin (e+f x)) (\cos (3 e+4 f x)+i \sin (3 e+4 f x)) \sqrt {c-i c \tan (e+f x)}}{15 c^3 f} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.24, size = 53, normalized size = 0.85
method | result | size |
derivativedivides | \(\frac {2 i a \left (-\frac {c \left (-i B +A \right )}{5 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}-\frac {i B}{3 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}\right )}{f c}\) | \(53\) |
default | \(\frac {2 i a \left (-\frac {c \left (-i B +A \right )}{5 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}-\frac {i B}{3 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}\right )}{f c}\) | \(53\) |
risch | \(-\frac {a \left (3 i A \,{\mathrm e}^{4 i \left (f x +e \right )}+3 B \,{\mathrm e}^{4 i \left (f x +e \right )}+6 i A \,{\mathrm e}^{2 i \left (f x +e \right )}-4 B \,{\mathrm e}^{2 i \left (f x +e \right )}+3 i A -7 B \right ) \sqrt {2}}{60 c^{2} \sqrt {\frac {c}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}\) | \(88\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.28, size = 48, normalized size = 0.77 \begin {gather*} -\frac {2 i \, {\left (5 i \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} B a + 3 \, {\left (A - i \, B\right )} a c\right )}}{15 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} c f} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 2.60, size = 96, normalized size = 1.55 \begin {gather*} -\frac {\sqrt {2} {\left (3 \, {\left (i \, A + B\right )} a e^{\left (6 i \, f x + 6 i \, e\right )} - {\left (-9 i \, A + B\right )} a e^{\left (4 i \, f x + 4 i \, e\right )} - {\left (-9 i \, A + 11 \, B\right )} a e^{\left (2 i \, f x + 2 i \, e\right )} - {\left (-3 i \, A + 7 \, B\right )} a\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{60 \, c^{3} 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*} i a \left (\int \left (- \frac {i A}{- c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - 2 i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\right )\, dx + \int \frac {A \tan {\left (e + f x \right )}}{- c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - 2 i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx + \int \frac {B \tan ^{2}{\left (e + f x \right )}}{- c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - 2 i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx + \int \left (- \frac {i B \tan {\left (e + f x \right )}}{- c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - 2 i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\right )\, dx\right ) \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 = 10.95, size = 232, normalized size = 3.74 \begin {gather*} \frac {a\,\sqrt {-\frac {c\,\left (-2\,{\cos \left (e+f\,x\right )}^2+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{2\,{\cos \left (e+f\,x\right )}^2}}\,\left (7\,B+11\,B\,\left (2\,{\cos \left (e+f\,x\right )}^2-1\right )+9\,A\,\sin \left (2\,e+2\,f\,x\right )+9\,A\,\sin \left (4\,e+4\,f\,x\right )+3\,A\,\sin \left (6\,e+6\,f\,x\right )+B\,\left (2\,{\cos \left (2\,e+2\,f\,x\right )}^2-1\right )-3\,B\,\left (2\,{\cos \left (3\,e+3\,f\,x\right )}^2-1\right )-A\,3{}\mathrm {i}-A\,\left (2\,{\cos \left (e+f\,x\right )}^2-1\right )\,9{}\mathrm {i}+B\,\sin \left (2\,e+2\,f\,x\right )\,11{}\mathrm {i}+B\,\sin \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}-B\,\sin \left (6\,e+6\,f\,x\right )\,3{}\mathrm {i}-A\,\left (2\,{\cos \left (2\,e+2\,f\,x\right )}^2-1\right )\,9{}\mathrm {i}-A\,\left (2\,{\cos \left (3\,e+3\,f\,x\right )}^2-1\right )\,3{}\mathrm {i}\right )}{60\,c^3\,f} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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