Optimal. Leaf size=227 \[ \frac {3 a^{5/2} (3 B+2 i A) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{\sqrt {c} f}-\frac {3 a^2 (3 B+2 i A) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 c f}-\frac {a (3 B+2 i A) (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 c f}-\frac {(B+i A) (a+i a \tan (e+f x))^{5/2}}{f \sqrt {c-i c \tan (e+f x)}} \]
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Rubi [A] time = 0.30, antiderivative size = 227, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {3588, 78, 50, 63, 217, 203} \[ \frac {3 a^{5/2} (3 B+2 i A) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{\sqrt {c} f}-\frac {3 a^2 (3 B+2 i A) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 c f}-\frac {a (3 B+2 i A) (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 c f}-\frac {(B+i A) (a+i a \tan (e+f x))^{5/2}}{f \sqrt {c-i c \tan (e+f x)}} \]
Antiderivative was successfully verified.
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Rule 50
Rule 63
Rule 78
Rule 203
Rule 217
Rule 3588
Rubi steps
\begin {align*} \int \frac {(a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x))}{\sqrt {c-i c \tan (e+f x)}} \, dx &=\frac {(a c) \operatorname {Subst}\left (\int \frac {(a+i a x)^{3/2} (A+B x)}{(c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {(i A+B) (a+i a \tan (e+f x))^{5/2}}{f \sqrt {c-i c \tan (e+f x)}}-\frac {(a (2 A-3 i B)) \operatorname {Subst}\left (\int \frac {(a+i a x)^{3/2}}{\sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {(i A+B) (a+i a \tan (e+f x))^{5/2}}{f \sqrt {c-i c \tan (e+f x)}}-\frac {a (2 i A+3 B) (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 c f}-\frac {\left (3 a^2 (2 A-3 i B)\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a+i a x}}{\sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{2 f}\\ &=-\frac {(i A+B) (a+i a \tan (e+f x))^{5/2}}{f \sqrt {c-i c \tan (e+f x)}}-\frac {3 a^2 (2 i A+3 B) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 c f}-\frac {a (2 i A+3 B) (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 c f}-\frac {\left (3 a^3 (2 A-3 i B)\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 f}\\ &=-\frac {(i A+B) (a+i a \tan (e+f x))^{5/2}}{f \sqrt {c-i c \tan (e+f x)}}-\frac {3 a^2 (2 i A+3 B) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 c f}-\frac {a (2 i A+3 B) (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 c f}+\frac {\left (3 a^2 (2 i A+3 B)\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 )}{f}\\ &=-\frac {(i A+B) (a+i a \tan (e+f x))^{5/2}}{f \sqrt {c-i c \tan (e+f x)}}-\frac {3 a^2 (2 i A+3 B) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 c f}-\frac {a (2 i A+3 B) (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 c f}+\frac {\left (3 a^2 (2 i A+3 B)\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 )}{f}\\ &=\frac {3 a^{5/2} (2 i A+3 B) \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 )}{\sqrt {c} f}-\frac {(i A+B) (a+i a \tan (e+f x))^{5/2}}{f \sqrt {c-i c \tan (e+f x)}}-\frac {3 a^2 (2 i A+3 B) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 c f}-\frac {a (2 i A+3 B) (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 c f}\\ \end {align*}
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Mathematica [A] time = 11.25, size = 239, normalized size = 1.05 \[ \frac {(a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x)) \left (\frac {3 (3 B+2 i A) e^{-3 i (e+f x)} \sqrt {\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \tan ^{-1}\left (e^{i (e+f x)}\right )}{\sqrt {\frac {c}{1+e^{2 i (e+f x)}}}}-\frac {(\tan (e+f x)+i) \sqrt {\sec (e+f x)} \sqrt {c-i c \tan (e+f x)} ((-5 B-2 i A) \sin (2 (e+f x))+(10 A-13 i B) \cos (2 (e+f x))+5 (2 A-3 i B))}{4 c}\right )}{f \sec ^{\frac {7}{2}}(e+f x) (A \cos (e+f x)+B \sin (e+f x))} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.82, size = 527, normalized size = 2.32 \[ -\frac {\sqrt {\frac {{\left (36 \, A^{2} - 108 i \, A B - 81 \, B^{2}\right )} a^{5}}{c f^{2}}} {\left (c f e^{\left (2 i \, f x + 2 i \, e\right )} + c f\right )} \log \left (\frac {2 \, {\left ({\left ({\left (24 i \, A + 36 \, B\right )} a^{2} e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (24 i \, A + 36 \, B\right )} a^{2} 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 \, \sqrt {\frac {{\left (36 \, A^{2} - 108 i \, A B - 81 \, B^{2}\right )} a^{5}}{c f^{2}}} {\left (c f e^{\left (2 i \, f x + 2 i \, e\right )} - c f\right )}\right )}}{{\left (6 i \, A + 9 \, B\right )} a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (6 i \, A + 9 \, B\right )} a^{2}}\right ) - \sqrt {\frac {{\left (36 \, A^{2} - 108 i \, A B - 81 \, B^{2}\right )} a^{5}}{c f^{2}}} {\left (c f e^{\left (2 i \, f x + 2 i \, e\right )} + c f\right )} \log \left (\frac {2 \, {\left ({\left ({\left (24 i \, A + 36 \, B\right )} a^{2} e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (24 i \, A + 36 \, B\right )} a^{2} 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 \, \sqrt {\frac {{\left (36 \, A^{2} - 108 i \, A B - 81 \, B^{2}\right )} a^{5}}{c f^{2}}} {\left (c f e^{\left (2 i \, f x + 2 i \, e\right )} - c f\right )}\right )}}{{\left (6 i \, A + 9 \, B\right )} a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (6 i \, A + 9 \, B\right )} a^{2}}\right ) - 2 \, {\left ({\left (-8 i \, A - 8 \, B\right )} a^{2} e^{\left (5 i \, f x + 5 i \, e\right )} + {\left (-20 i \, A - 30 \, B\right )} a^{2} e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (-12 i \, A - 18 \, B\right )} a^{2} 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}}}{4 \, {\left (c f e^{\left (2 i \, f x + 2 i \, e\right )} + c f\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 \[ \int \frac {{\left (B \tan \left (f x + e\right ) + A\right )} {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {5}{2}}}{\sqrt {-i \, c \tan \left (f x + e\right ) + c}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.57, size = 565, normalized size = 2.49 \[ \frac {i \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (-1+i \tan \left (f x +e \right )\right )}\, a^{2} \left (6 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 ^{2}\left (f x +e \right )\right ) a c +18 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 ) \tan \left (f x +e \right ) a c +4 i B \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\, \left (\tan ^{2}\left (f x +e \right )\right )+9 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 -B \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\, \left (\tan ^{3}\left (f x +e \right )\right )-6 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 ) a c -12 i A \sqrt {c a}\, \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \tan \left (f x +e \right )-12 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 -2 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}-14 i B \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}-9 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 -19 B \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\, \tan \left (f x +e \right )+10 A \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\right )}{2 f c \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \left (\tan \left (f x +e \right )+i\right )^{2} \sqrt {c a}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 1.10, size = 996, normalized size = 4.39 \[ \text {result too large to display} \]
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 \[ \int \frac {\left (A+B\,\mathrm {tan}\left (e+f\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}} \,d x \]
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 \[ \int \frac {\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {5}{2}} \left (A + B \tan {\left (e + f x \right )}\right )}{\sqrt {- i c \left (\tan {\left (e + f x \right )} + i\right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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