Optimal. Leaf size=343 \[ \frac {7 c^{9/2} (-7 B+2 i A) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{a^{5/2} f}+\frac {7 c^4 (-7 B+2 i A) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 a^3 f}+\frac {7 c^3 (-7 B+2 i A) \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{6 a^3 f}+\frac {14 c^2 (-7 B+2 i A) (c-i c \tan (e+f x))^{5/2}}{15 a^2 f \sqrt {a+i a \tan (e+f x)}}-\frac {2 c (-7 B+2 i A) (c-i c \tan (e+f x))^{7/2}}{15 a f (a+i a \tan (e+f x))^{3/2}}+\frac {(-B+i A) (c-i c \tan (e+f x))^{9/2}}{5 f (a+i a \tan (e+f x))^{5/2}} \]
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Rubi [A] time = 0.38, antiderivative size = 343, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {3588, 78, 47, 50, 63, 217, 203} \[ \frac {7 c^{9/2} (-7 B+2 i A) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{a^{5/2} f}+\frac {7 c^4 (-7 B+2 i A) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 a^3 f}+\frac {7 c^3 (-7 B+2 i A) \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{6 a^3 f}+\frac {14 c^2 (-7 B+2 i A) (c-i c \tan (e+f x))^{5/2}}{15 a^2 f \sqrt {a+i a \tan (e+f x)}}-\frac {2 c (-7 B+2 i A) (c-i c \tan (e+f x))^{7/2}}{15 a f (a+i a \tan (e+f x))^{3/2}}+\frac {(-B+i A) (c-i c \tan (e+f x))^{9/2}}{5 f (a+i a \tan (e+f x))^{5/2}} \]
Antiderivative was successfully verified.
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Rule 47
Rule 50
Rule 63
Rule 78
Rule 203
Rule 217
Rule 3588
Rubi steps
\begin {align*} \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{9/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx &=\frac {(a c) \operatorname {Subst}\left (\int \frac {(A+B x) (c-i c x)^{7/2}}{(a+i a x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {(i A-B) (c-i c \tan (e+f x))^{9/2}}{5 f (a+i a \tan (e+f x))^{5/2}}-\frac {((2 A+7 i B) c) \operatorname {Subst}\left (\int \frac {(c-i c x)^{7/2}}{(a+i a x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{5 f}\\ &=-\frac {2 (2 i A-7 B) c (c-i c \tan (e+f x))^{7/2}}{15 a f (a+i a \tan (e+f x))^{3/2}}+\frac {(i A-B) (c-i c \tan (e+f x))^{9/2}}{5 f (a+i a \tan (e+f x))^{5/2}}+\frac {\left (7 (2 A+7 i B) c^2\right ) \operatorname {Subst}\left (\int \frac {(c-i c x)^{5/2}}{(a+i a x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{15 a f}\\ &=\frac {14 (2 i A-7 B) c^2 (c-i c \tan (e+f x))^{5/2}}{15 a^2 f \sqrt {a+i a \tan (e+f x)}}-\frac {2 (2 i A-7 B) c (c-i c \tan (e+f x))^{7/2}}{15 a f (a+i a \tan (e+f x))^{3/2}}+\frac {(i A-B) (c-i c \tan (e+f x))^{9/2}}{5 f (a+i a \tan (e+f x))^{5/2}}-\frac {\left (7 (2 A+7 i B) c^3\right ) \operatorname {Subst}\left (\int \frac {(c-i c x)^{3/2}}{\sqrt {a+i a x}} \, dx,x,\tan (e+f x)\right )}{3 a^2 f}\\ &=\frac {7 (2 i A-7 B) c^3 \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{6 a^3 f}+\frac {14 (2 i A-7 B) c^2 (c-i c \tan (e+f x))^{5/2}}{15 a^2 f \sqrt {a+i a \tan (e+f x)}}-\frac {2 (2 i A-7 B) c (c-i c \tan (e+f x))^{7/2}}{15 a f (a+i a \tan (e+f x))^{3/2}}+\frac {(i A-B) (c-i c \tan (e+f x))^{9/2}}{5 f (a+i a \tan (e+f x))^{5/2}}-\frac {\left (7 (2 A+7 i B) c^4\right ) \operatorname {Subst}\left (\int \frac {\sqrt {c-i c x}}{\sqrt {a+i a x}} \, dx,x,\tan (e+f x)\right )}{2 a^2 f}\\ &=\frac {7 (2 i A-7 B) c^4 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 a^3 f}+\frac {7 (2 i A-7 B) c^3 \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{6 a^3 f}+\frac {14 (2 i A-7 B) c^2 (c-i c \tan (e+f x))^{5/2}}{15 a^2 f \sqrt {a+i a \tan (e+f x)}}-\frac {2 (2 i A-7 B) c (c-i c \tan (e+f x))^{7/2}}{15 a f (a+i a \tan (e+f x))^{3/2}}+\frac {(i A-B) (c-i c \tan (e+f x))^{9/2}}{5 f (a+i a \tan (e+f x))^{5/2}}-\frac {\left (7 (2 A+7 i B) c^5\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+i a x} \sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{2 a^2 f}\\ &=\frac {7 (2 i A-7 B) c^4 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 a^3 f}+\frac {7 (2 i A-7 B) c^3 \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{6 a^3 f}+\frac {14 (2 i A-7 B) c^2 (c-i c \tan (e+f x))^{5/2}}{15 a^2 f \sqrt {a+i a \tan (e+f x)}}-\frac {2 (2 i A-7 B) c (c-i c \tan (e+f x))^{7/2}}{15 a f (a+i a \tan (e+f x))^{3/2}}+\frac {(i A-B) (c-i c \tan (e+f x))^{9/2}}{5 f (a+i a \tan (e+f x))^{5/2}}+\frac {\left (7 (2 i A-7 B) c^5\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {2 c-\frac {c x^2}{a}}} \, dx,x,\sqrt {a+i a \tan (e+f x)}\right )}{a^3 f}\\ &=\frac {7 (2 i A-7 B) c^4 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 a^3 f}+\frac {7 (2 i A-7 B) c^3 \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{6 a^3 f}+\frac {14 (2 i A-7 B) c^2 (c-i c \tan (e+f x))^{5/2}}{15 a^2 f \sqrt {a+i a \tan (e+f x)}}-\frac {2 (2 i A-7 B) c (c-i c \tan (e+f x))^{7/2}}{15 a f (a+i a \tan (e+f x))^{3/2}}+\frac {(i A-B) (c-i c \tan (e+f x))^{9/2}}{5 f (a+i a \tan (e+f x))^{5/2}}+\frac {\left (7 (2 i A-7 B) c^5\right ) \operatorname {Subst}\left (\int \frac {1}{1+\frac {c x^2}{a}} \, dx,x,\frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {c-i c \tan (e+f x)}}\right )}{a^3 f}\\ &=\frac {7 (2 i A-7 B) c^{9/2} \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{a^{5/2} f}+\frac {7 (2 i A-7 B) c^4 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 a^3 f}+\frac {7 (2 i A-7 B) c^3 \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{6 a^3 f}+\frac {14 (2 i A-7 B) c^2 (c-i c \tan (e+f x))^{5/2}}{15 a^2 f \sqrt {a+i a \tan (e+f x)}}-\frac {2 (2 i A-7 B) c (c-i c \tan (e+f x))^{7/2}}{15 a f (a+i a \tan (e+f x))^{3/2}}+\frac {(i A-B) (c-i c \tan (e+f x))^{9/2}}{5 f (a+i a \tan (e+f x))^{5/2}}\\ \end {align*}
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Mathematica [A] time = 15.12, size = 247, normalized size = 0.72 \[ -\frac {\sqrt {2} c^4 e^{-4 i (e+f x)} \sqrt {\frac {c}{1+e^{2 i (e+f x)}}} \left (105 (7 B-2 i A) e^{5 i (e+f x)} \left (1+e^{2 i (e+f x)}\right )^2 \tan ^{-1}\left (e^{i (e+f x)}\right )-2 i A \left (-8 e^{2 i (e+f x)}+56 e^{4 i (e+f x)}+175 e^{6 i (e+f x)}+105 e^{8 i (e+f x)}+6\right )+B \left (-56 e^{2 i (e+f x)}+392 e^{4 i (e+f x)}+1225 e^{6 i (e+f x)}+735 e^{8 i (e+f x)}+12\right )\right )}{15 a^2 f \left (1+e^{2 i (e+f x)}\right )^2 \sqrt {a+i a \tan (e+f x)}} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.73, size = 606, normalized size = 1.77 \[ \frac {15 \, {\left (a^{3} f e^{\left (7 i \, f x + 7 i \, e\right )} + a^{3} f e^{\left (5 i \, f x + 5 i \, e\right )}\right )} \sqrt {\frac {{\left (196 \, A^{2} + 1372 i \, A B - 2401 \, B^{2}\right )} c^{9}}{a^{5} f^{2}}} \log \left (\frac {2 \, {\left ({\left ({\left (-56 i \, A + 196 \, B\right )} c^{4} e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (-56 i \, A + 196 \, B\right )} c^{4} 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}} + 2 \, {\left (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} - a^{3} f\right )} \sqrt {\frac {{\left (196 \, A^{2} + 1372 i \, A B - 2401 \, B^{2}\right )} c^{9}}{a^{5} f^{2}}}\right )}}{{\left (-14 i \, A + 49 \, B\right )} c^{4} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-14 i \, A + 49 \, B\right )} c^{4}}\right ) - 15 \, {\left (a^{3} f e^{\left (7 i \, f x + 7 i \, e\right )} + a^{3} f e^{\left (5 i \, f x + 5 i \, e\right )}\right )} \sqrt {\frac {{\left (196 \, A^{2} + 1372 i \, A B - 2401 \, B^{2}\right )} c^{9}}{a^{5} f^{2}}} \log \left (\frac {2 \, {\left ({\left ({\left (-56 i \, A + 196 \, B\right )} c^{4} e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (-56 i \, A + 196 \, B\right )} c^{4} 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}} - 2 \, {\left (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} - a^{3} f\right )} \sqrt {\frac {{\left (196 \, A^{2} + 1372 i \, A B - 2401 \, B^{2}\right )} c^{9}}{a^{5} f^{2}}}\right )}}{{\left (-14 i \, A + 49 \, B\right )} c^{4} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-14 i \, A + 49 \, B\right )} c^{4}}\right ) + 2 \, {\left ({\left (420 i \, A - 1470 \, B\right )} c^{4} e^{\left (8 i \, f x + 8 i \, e\right )} + {\left (700 i \, A - 2450 \, B\right )} c^{4} e^{\left (6 i \, f x + 6 i \, e\right )} + {\left (224 i \, A - 784 \, B\right )} c^{4} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (-32 i \, A + 112 \, B\right )} c^{4} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (24 i \, A - 24 \, B\right )} c^{4}\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}}}{60 \, {\left (a^{3} f e^{\left (7 i \, f x + 7 i \, e\right )} + a^{3} f e^{\left (5 i \, f x + 5 i \, e\right )}\right )}} \]
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 \[ \int \frac {{\left (B \tan \left (f x + e\right ) + A\right )} {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {9}{2}}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.53, size = 899, normalized size = 2.62 \[ \text {result too large to display} \]
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 \[ \int \frac {\left (A+B\,\mathrm {tan}\left (e+f\,x\right )\right )\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{9/2}}{{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \]
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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