Optimal. Leaf size=160 \[ -\frac {a^{3/2} \sqrt {c} (B+2 i A) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{f}+\frac {a (B+2 i A) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 f}+\frac {B (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 f} \]
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Rubi [A] time = 0.26, antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {3588, 80, 50, 63, 217, 203} \[ -\frac {a^{3/2} \sqrt {c} (B+2 i A) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{f}+\frac {a (B+2 i A) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 f}+\frac {B (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 f} \]
Antiderivative was successfully verified.
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Rule 50
Rule 63
Rule 80
Rule 203
Rule 217
Rule 3588
Rubi steps
\begin {align*} \int (a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} \, dx &=\frac {(a c) \operatorname {Subst}\left (\int \frac {\sqrt {a+i a x} (A+B x)}{\sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {B (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 f}+\frac {(a (2 A-i B) c) \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 {a (2 i A+B) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 f}+\frac {B (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 f}+\frac {\left (a^2 (2 A-i B) c\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 {a (2 i A+B) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 f}+\frac {B (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 f}-\frac {(a (2 i A+B) c) \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 {a (2 i A+B) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 f}+\frac {B (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 f}-\frac {(a (2 i A+B) c) \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 {a^{3/2} (2 i A+B) \sqrt {c} \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 )}{f}+\frac {a (2 i A+B) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 f}+\frac {B (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 f}\\ \end {align*}
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Mathematica [A] time = 6.80, size = 220, normalized size = 1.38 \[ \frac {(a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x)) \left (\frac {\cos (e) (\tan (e)+i) \sqrt {\sec (e+f x)} \sqrt {c-i c \tan (e+f x)} (2 A+B \tan (e+f x)-2 i B)}{2 \cos (f x)+2 i \sin (f x)}-\frac {i c (2 A-i B) e^{-2 i (e+f x)} \sqrt {\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \tan ^{-1}\left (e^{i (e+f x)}\right )}{\sqrt {\frac {c}{1+e^{2 i (e+f x)}}}}\right )}{f \sec ^{\frac {5}{2}}(e+f x) (A \cos (e+f x)+B \sin (e+f x))} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.72, size = 458, normalized size = 2.86 \[ \frac {\sqrt {\frac {{\left (4 \, A^{2} - 4 i \, A B - B^{2}\right )} a^{3} c}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {2 \, {\left ({\left ({\left (8 i \, A + 4 \, B\right )} a e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (8 i \, A + 4 \, B\right )} a e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} + 2 \, \sqrt {\frac {{\left (4 \, A^{2} - 4 i \, A B - B^{2}\right )} a^{3} c}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} - f\right )}\right )}}{{\left (2 i \, A + B\right )} a e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (2 i \, A + B\right )} a}\right ) - \sqrt {\frac {{\left (4 \, A^{2} - 4 i \, A B - B^{2}\right )} a^{3} c}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {2 \, {\left ({\left ({\left (8 i \, A + 4 \, B\right )} a e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (8 i \, A + 4 \, B\right )} a e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - 2 \, \sqrt {\frac {{\left (4 \, A^{2} - 4 i \, A B - B^{2}\right )} a^{3} c}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} - f\right )}\right )}}{{\left (2 i \, A + B\right )} a e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (2 i \, A + B\right )} a}\right ) + 2 \, {\left ({\left (4 i \, A + 6 \, B\right )} a e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (4 i \, A + 2 \, B\right )} a e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{4 \, {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 23.46, size = 565, normalized size = 3.53 \[ \frac {-3 i \, B a^{\frac {3}{2}} \sqrt {c} e^{\left (8 i \, f x + 8 i \, e\right )} \log \left (e^{\left (i \, f x + i \, e\right )} + i\right ) - 12 i \, B a^{\frac {3}{2}} \sqrt {c} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (e^{\left (i \, f x + i \, e\right )} + i\right ) - 18 i \, B a^{\frac {3}{2}} \sqrt {c} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (e^{\left (i \, f x + i \, e\right )} + i\right ) - 12 i \, B a^{\frac {3}{2}} \sqrt {c} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (e^{\left (i \, f x + i \, e\right )} + i\right ) + 3 i \, B a^{\frac {3}{2}} \sqrt {c} e^{\left (8 i \, f x + 8 i \, e\right )} \log \left (e^{\left (i \, f x + i \, e\right )} - i\right ) + 12 i \, B a^{\frac {3}{2}} \sqrt {c} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (e^{\left (i \, f x + i \, e\right )} - i\right ) + 18 i \, B a^{\frac {3}{2}} \sqrt {c} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (e^{\left (i \, f x + i \, e\right )} - i\right ) + 12 i \, B a^{\frac {3}{2}} \sqrt {c} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (e^{\left (i \, f x + i \, e\right )} - i\right ) + 10 \, B a^{\frac {3}{2}} \sqrt {c} e^{\left (7 i \, f x + 7 i \, e\right )} + 26 \, B a^{\frac {3}{2}} \sqrt {c} e^{\left (5 i \, f x + 5 i \, e\right )} + 22 \, B a^{\frac {3}{2}} \sqrt {c} e^{\left (3 i \, f x + 3 i \, e\right )} + 6 \, B a^{\frac {3}{2}} \sqrt {c} e^{\left (i \, f x + i \, e\right )} - 3 i \, B a^{\frac {3}{2}} \sqrt {c} \log \left (e^{\left (i \, f x + i \, e\right )} + i\right ) + 3 i \, B a^{\frac {3}{2}} \sqrt {c} \log \left (e^{\left (i \, f x + i \, e\right )} - i\right )}{8 \, {\left (f e^{\left (8 i \, f x + 8 i \, e\right )} + 4 \, f e^{\left (6 i \, f x + 6 i \, e\right )} + 6 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 4 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} - \frac {i \, {\left ({\left (8 \, A a^{\frac {3}{2}} \sqrt {c} - i \, B a^{\frac {3}{2}} \sqrt {c}\right )} \arctan \left (e^{\left (i \, f x + i \, e\right )}\right ) - \frac {8 \, A a^{\frac {3}{2}} \sqrt {c} e^{\left (3 i \, f x + 3 i \, e\right )} - 7 i \, B a^{\frac {3}{2}} \sqrt {c} e^{\left (3 i \, f x + 3 i \, e\right )} + 8 \, A a^{\frac {3}{2}} \sqrt {c} e^{\left (i \, f x + i \, e\right )} - i \, B a^{\frac {3}{2}} \sqrt {c} e^{\left (i \, f x + i \, e\right )}}{{\left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}^{2}}\right )}}{4 \, f} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.61, size = 223, normalized size = 1.39 \[ \frac {\sqrt {-c \left (-1+i \tan \left (f x +e \right )\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, a \left (-i B \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) a c +i B \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\, \tan \left (f x +e \right )+2 i A \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}+2 A \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) a c +2 B \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\right )}{2 f \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.96, size = 779, normalized size = 4.87 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \left (A+B\,\mathrm {tan}\left (e+f\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}} \,d x \]
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 \[ \int \left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {3}{2}} \sqrt {- i c \left (\tan {\left (e + f x \right )} + i\right )} \left (A + B \tan {\left (e + f x \right )}\right )\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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