Optimal. Leaf size=284 \[ \frac {2 c^{7/2} (-6 B+i A) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{a^{5/2} f}+\frac {c^3 (-6 B+i A) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{a^3 f}+\frac {2 c^2 (-6 B+i A) (c-i c \tan (e+f x))^{3/2}}{3 a^2 f \sqrt {a+i a \tan (e+f x)}}-\frac {2 c (-6 B+i A) (c-i c \tan (e+f x))^{5/2}}{15 a f (a+i a \tan (e+f x))^{3/2}}+\frac {(-B+i A) (c-i c \tan (e+f x))^{7/2}}{5 f (a+i a \tan (e+f x))^{5/2}} \]
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Rubi [A] time = 0.34, antiderivative size = 284, normalized size of antiderivative = 1.00, number of steps used = 8, 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} \[ \frac {2 c^{7/2} (-6 B+i A) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{a^{5/2} f}+\frac {c^3 (-6 B+i A) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{a^3 f}+\frac {2 c^2 (-6 B+i A) (c-i c \tan (e+f x))^{3/2}}{3 a^2 f \sqrt {a+i a \tan (e+f x)}}-\frac {2 c (-6 B+i A) (c-i c \tan (e+f x))^{5/2}}{15 a f (a+i a \tan (e+f x))^{3/2}}+\frac {(-B+i A) (c-i c \tan (e+f x))^{7/2}}{5 f (a+i a \tan (e+f x))^{5/2}} \]
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))^{7/2}}{(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)^{5/2}}{(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))^{7/2}}{5 f (a+i a \tan (e+f x))^{5/2}}-\frac {((A+6 i B) c) \operatorname {Subst}\left (\int \frac {(c-i c x)^{5/2}}{(a+i a x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{5 f}\\ &=-\frac {2 (i A-6 B) c (c-i c \tan (e+f x))^{5/2}}{15 a f (a+i a \tan (e+f x))^{3/2}}+\frac {(i A-B) (c-i c \tan (e+f x))^{7/2}}{5 f (a+i a \tan (e+f x))^{5/2}}+\frac {\left ((A+6 i B) c^2\right ) \operatorname {Subst}\left (\int \frac {(c-i c x)^{3/2}}{(a+i a x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{3 a f}\\ &=\frac {2 (i A-6 B) c^2 (c-i c \tan (e+f x))^{3/2}}{3 a^2 f \sqrt {a+i a \tan (e+f x)}}-\frac {2 (i A-6 B) c (c-i c \tan (e+f x))^{5/2}}{15 a f (a+i a \tan (e+f x))^{3/2}}+\frac {(i A-B) (c-i c \tan (e+f x))^{7/2}}{5 f (a+i a \tan (e+f x))^{5/2}}-\frac {\left ((A+6 i B) c^3\right ) \operatorname {Subst}\left (\int \frac {\sqrt {c-i c x}}{\sqrt {a+i a x}} \, dx,x,\tan (e+f x)\right )}{a^2 f}\\ &=\frac {(i A-6 B) c^3 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{a^3 f}+\frac {2 (i A-6 B) c^2 (c-i c \tan (e+f x))^{3/2}}{3 a^2 f \sqrt {a+i a \tan (e+f x)}}-\frac {2 (i A-6 B) c (c-i c \tan (e+f x))^{5/2}}{15 a f (a+i a \tan (e+f x))^{3/2}}+\frac {(i A-B) (c-i c \tan (e+f x))^{7/2}}{5 f (a+i a \tan (e+f x))^{5/2}}-\frac {\left ((A+6 i B) c^4\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+i a x} \sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{a^2 f}\\ &=\frac {(i A-6 B) c^3 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{a^3 f}+\frac {2 (i A-6 B) c^2 (c-i c \tan (e+f x))^{3/2}}{3 a^2 f \sqrt {a+i a \tan (e+f x)}}-\frac {2 (i A-6 B) c (c-i c \tan (e+f x))^{5/2}}{15 a f (a+i a \tan (e+f x))^{3/2}}+\frac {(i A-B) (c-i c \tan (e+f x))^{7/2}}{5 f (a+i a \tan (e+f x))^{5/2}}+\frac {\left (2 (i A-6 B) c^4\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 {(i A-6 B) c^3 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{a^3 f}+\frac {2 (i A-6 B) c^2 (c-i c \tan (e+f x))^{3/2}}{3 a^2 f \sqrt {a+i a \tan (e+f x)}}-\frac {2 (i A-6 B) c (c-i c \tan (e+f x))^{5/2}}{15 a f (a+i a \tan (e+f x))^{3/2}}+\frac {(i A-B) (c-i c \tan (e+f x))^{7/2}}{5 f (a+i a \tan (e+f x))^{5/2}}+\frac {\left (2 (i A-6 B) c^4\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 {2 (i A-6 B) c^{7/2} \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 {(i A-6 B) c^3 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{a^3 f}+\frac {2 (i A-6 B) c^2 (c-i c \tan (e+f x))^{3/2}}{3 a^2 f \sqrt {a+i a \tan (e+f x)}}-\frac {2 (i A-6 B) c (c-i c \tan (e+f x))^{5/2}}{15 a f (a+i a \tan (e+f x))^{3/2}}+\frac {(i A-B) (c-i c \tan (e+f x))^{7/2}}{5 f (a+i a \tan (e+f x))^{5/2}}\\ \end {align*}
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Mathematica [A] time = 14.25, size = 205, normalized size = 0.72 \[ \frac {2 \sqrt {2} c^2 e^{-4 i (e+f x)} \left (\frac {c}{1+e^{2 i (e+f x)}}\right )^{3/2} \left (15 i (A+6 i B) e^{5 i (e+f x)} \left (1+e^{2 i (e+f x)}\right ) \tan ^{-1}\left (e^{i (e+f x)}\right )+i A \left (-2 e^{2 i (e+f x)}+10 e^{4 i (e+f x)}+15 e^{6 i (e+f x)}+3\right )-3 B \left (-4 e^{2 i (e+f x)}+20 e^{4 i (e+f x)}+30 e^{6 i (e+f x)}+1\right )\right )}{15 a^2 f \sqrt {a+i a \tan (e+f x)}} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.71, size = 537, normalized size = 1.89 \[ \frac {{\left (15 \, a^{3} \sqrt {\frac {{\left (4 \, A^{2} + 48 i \, A B - 144 \, B^{2}\right )} c^{7}}{a^{5} f^{2}}} f e^{\left (5 i \, f x + 5 i \, e\right )} \log \left (\frac {2 \, {\left ({\left ({\left (-4 i \, A + 24 \, B\right )} c^{3} e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (-4 i \, A + 24 \, B\right )} c^{3} 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}} + {\left (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} - a^{3} f\right )} \sqrt {\frac {{\left (4 \, A^{2} + 48 i \, A B - 144 \, B^{2}\right )} c^{7}}{a^{5} f^{2}}}\right )}}{{\left (-i \, A + 6 \, B\right )} c^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-i \, A + 6 \, B\right )} c^{3}}\right ) - 15 \, a^{3} \sqrt {\frac {{\left (4 \, A^{2} + 48 i \, A B - 144 \, B^{2}\right )} c^{7}}{a^{5} f^{2}}} f e^{\left (5 i \, f x + 5 i \, e\right )} \log \left (\frac {2 \, {\left ({\left ({\left (-4 i \, A + 24 \, B\right )} c^{3} e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (-4 i \, A + 24 \, B\right )} c^{3} 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}} - {\left (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} - a^{3} f\right )} \sqrt {\frac {{\left (4 \, A^{2} + 48 i \, A B - 144 \, B^{2}\right )} c^{7}}{a^{5} f^{2}}}\right )}}{{\left (-i \, A + 6 \, B\right )} c^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-i \, A + 6 \, B\right )} c^{3}}\right ) + 2 \, {\left ({\left (60 i \, A - 360 \, B\right )} c^{3} e^{\left (6 i \, f x + 6 i \, e\right )} + {\left (40 i \, A - 240 \, B\right )} c^{3} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (-8 i \, A + 48 \, B\right )} c^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (12 i \, A - 12 \, B\right )} c^{3}\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}}\right )} e^{\left (-5 i \, f x - 5 i \, e\right )}}{60 \, a^{3} 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 \[ \int \frac {{\left (B \tan \left (f x + e\right ) + A\right )} {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {7}{2}}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.43, size = 835, normalized size = 2.94 \[ \frac {\sqrt {-c \left (-1+i \tan \left (f x +e \right )\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, c^{3} \left (-60 i A \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) \tan \left (f x +e \right ) a c +60 i A \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) \left (\tan ^{3}\left (f x +e \right )\right ) a c -15 A \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) \left (\tan ^{4}\left (f x +e \right )\right ) a c +246 i B \left (\tan ^{3}\left (f x +e \right )\right ) \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}-90 i B \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) a c -360 B \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) \left (\tan ^{3}\left (f x +e \right )\right ) a c -15 B \left (\tan ^{4}\left (f x +e \right )\right ) \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}-94 i A \left (\tan ^{2}\left (f x +e \right )\right ) \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}-474 i B \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\, \tan \left (f x +e \right )+90 A \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) \left (\tan ^{2}\left (f x +e \right )\right ) a c +46 A \left (\tan ^{3}\left (f x +e \right )\right ) \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}-90 i B \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) \left (\tan ^{4}\left (f x +e \right )\right ) a c +540 i B \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) \left (\tan ^{2}\left (f x +e \right )\right ) a c +360 B \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) \tan \left (f x +e \right ) a c +564 B \left (\tan ^{2}\left (f x +e \right )\right ) \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}+26 i A \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}-15 A \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) a c -74 A \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\, \tan \left (f x +e \right )-141 B \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\right )}{15 f \,a^{3} \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \left (-\tan \left (f x +e \right )+i\right )^{4} \sqrt {c a}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 1.30, size = 1032, normalized size = 3.63 \[ \text {result too large to display} \]
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 \[ \int \frac {\left (A+B\,\mathrm {tan}\left (e+f\,x\right )\right )\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{7/2}}{{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \]
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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