Optimal. Leaf size=208 \[ -\frac {2 (-4 B+3 i A) \sqrt {a+i a \tan (e+f x)}}{105 c^3 f \sqrt {c-i c \tan (e+f x)}}-\frac {2 (-4 B+3 i A) \sqrt {a+i a \tan (e+f x)}}{105 c^2 f (c-i c \tan (e+f x))^{3/2}}-\frac {(-4 B+3 i A) \sqrt {a+i a \tan (e+f x)}}{35 c f (c-i c \tan (e+f x))^{5/2}}-\frac {(B+i A) \sqrt {a+i a \tan (e+f x)}}{7 f (c-i c \tan (e+f x))^{7/2}} \]
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Rubi [A] time = 0.28, antiderivative size = 208, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {3588, 78, 45, 37} \[ -\frac {2 (-4 B+3 i A) \sqrt {a+i a \tan (e+f x)}}{105 c^3 f \sqrt {c-i c \tan (e+f x)}}-\frac {2 (-4 B+3 i A) \sqrt {a+i a \tan (e+f x)}}{105 c^2 f (c-i c \tan (e+f x))^{3/2}}-\frac {(-4 B+3 i A) \sqrt {a+i a \tan (e+f x)}}{35 c f (c-i c \tan (e+f x))^{5/2}}-\frac {(B+i A) \sqrt {a+i a \tan (e+f x)}}{7 f (c-i c \tan (e+f x))^{7/2}} \]
Antiderivative was successfully verified.
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Rule 37
Rule 45
Rule 78
Rule 3588
Rubi steps
\begin {align*} \int \frac {\sqrt {a+i a \tan (e+f x)} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{7/2}} \, dx &=\frac {(a c) \operatorname {Subst}\left (\int \frac {A+B x}{\sqrt {a+i a x} (c-i c x)^{9/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {(i A+B) \sqrt {a+i a \tan (e+f x)}}{7 f (c-i c \tan (e+f x))^{7/2}}+\frac {(a (3 A+4 i B)) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+i a x} (c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{7 f}\\ &=-\frac {(i A+B) \sqrt {a+i a \tan (e+f x)}}{7 f (c-i c \tan (e+f x))^{7/2}}-\frac {(3 i A-4 B) \sqrt {a+i a \tan (e+f x)}}{35 c f (c-i c \tan (e+f x))^{5/2}}+\frac {(2 a (3 A+4 i B)) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+i a x} (c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{35 c f}\\ &=-\frac {(i A+B) \sqrt {a+i a \tan (e+f x)}}{7 f (c-i c \tan (e+f x))^{7/2}}-\frac {(3 i A-4 B) \sqrt {a+i a \tan (e+f x)}}{35 c f (c-i c \tan (e+f x))^{5/2}}-\frac {2 (3 i A-4 B) \sqrt {a+i a \tan (e+f x)}}{105 c^2 f (c-i c \tan (e+f x))^{3/2}}+\frac {(2 a (3 A+4 i B)) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+i a x} (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{105 c^2 f}\\ &=-\frac {(i A+B) \sqrt {a+i a \tan (e+f x)}}{7 f (c-i c \tan (e+f x))^{7/2}}-\frac {(3 i A-4 B) \sqrt {a+i a \tan (e+f x)}}{35 c f (c-i c \tan (e+f x))^{5/2}}-\frac {2 (3 i A-4 B) \sqrt {a+i a \tan (e+f x)}}{105 c^2 f (c-i c \tan (e+f x))^{3/2}}-\frac {2 (3 i A-4 B) \sqrt {a+i a \tan (e+f x)}}{105 c^3 f \sqrt {c-i c \tan (e+f x)}}\\ \end {align*}
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Mathematica [A] time = 12.31, size = 136, normalized size = 0.65 \[ \frac {\cos (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)} (\cos (4 (e+f x))+i \sin (4 (e+f x))) (-(3 A+4 i B) (7 \sin (e+f x)+15 \sin (3 (e+f x)))+7 (B-12 i A) \cos (e+f x)+15 (3 B-4 i A) \cos (3 (e+f x)))}{420 c^4 f} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.65, size = 128, normalized size = 0.62 \[ \frac {{\left ({\left (-15 i \, A - 15 \, B\right )} e^{\left (9 i \, f x + 9 i \, e\right )} + {\left (-78 i \, A - 36 \, B\right )} e^{\left (7 i \, f x + 7 i \, e\right )} + {\left (-168 i \, A + 14 \, B\right )} e^{\left (5 i \, f x + 5 i \, e\right )} + {\left (-210 i \, A + 140 \, B\right )} e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (-105 i \, A + 105 \, B\right )} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{840 \, c^{4} f} \]
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 \[ \int \frac {{\left (B \tan \left (f x + e\right ) + A\right )} \sqrt {i \, a \tan \left (f x + e\right ) + a}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {7}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.52, size = 147, normalized size = 0.71 \[ \frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (-1+i \tan \left (f x +e \right )\right )}\, \left (8 i B \left (\tan ^{4}\left (f x +e \right )\right )+30 i A \left (\tan ^{3}\left (f x +e \right )\right )+6 A \left (\tan ^{4}\left (f x +e \right )\right )-84 i B \left (\tan ^{2}\left (f x +e \right )\right )-40 B \left (\tan ^{3}\left (f x +e \right )\right )-75 i A \tan \left (f x +e \right )-63 A \left (\tan ^{2}\left (f x +e \right )\right )+13 i B +65 B \tan \left (f x +e \right )+36 A \right )}{105 f \,c^{4} \left (\tan \left (f x +e \right )+i\right )^{5}} \]
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 [B] time = 10.80, size = 246, normalized size = 1.18 \[ -\frac {\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (A\,105{}\mathrm {i}-105\,B+A\,\cos \left (2\,e+2\,f\,x\right )\,105{}\mathrm {i}+A\,\cos \left (4\,e+4\,f\,x\right )\,63{}\mathrm {i}+A\,\cos \left (6\,e+6\,f\,x\right )\,15{}\mathrm {i}-35\,B\,\cos \left (2\,e+2\,f\,x\right )+21\,B\,\cos \left (4\,e+4\,f\,x\right )+15\,B\,\cos \left (6\,e+6\,f\,x\right )-105\,A\,\sin \left (2\,e+2\,f\,x\right )-63\,A\,\sin \left (4\,e+4\,f\,x\right )-15\,A\,\sin \left (6\,e+6\,f\,x\right )-B\,\sin \left (2\,e+2\,f\,x\right )\,35{}\mathrm {i}+B\,\sin \left (4\,e+4\,f\,x\right )\,21{}\mathrm {i}+B\,\sin \left (6\,e+6\,f\,x\right )\,15{}\mathrm {i}\right )}{840\,c^3\,f\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}} \]
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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