Optimal. Leaf size=318 \[ \frac {((25+21 i) A-(9-5 i) B) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^2 d}-\frac {\left (\frac {1}{16}-\frac {i}{16}\right ) ((2+23 i) A-(7+2 i) B) \tan ^{-1}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a^2 d}-\frac {5 (5 A+i B)}{8 a^2 d \sqrt {\tan (c+d x)}}+\frac {7 A+3 i B}{8 a^2 d (1+i \tan (c+d x)) \sqrt {\tan (c+d x)}}-\frac {\left (\frac {1}{32}-\frac {i}{32}\right ) ((23+2 i) A+(2+7 i) B) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a^2 d}+\frac {\left (\frac {1}{32}-\frac {i}{32}\right ) ((23+2 i) A+(2+7 i) B) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a^2 d}+\frac {A+i B}{4 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2} \]
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Rubi [A] time = 0.58, antiderivative size = 318, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3596, 3529, 3534, 1168, 1162, 617, 204, 1165, 628} \[ \frac {((25+21 i) A-(9-5 i) B) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^2 d}-\frac {\left (\frac {1}{16}-\frac {i}{16}\right ) ((2+23 i) A-(7+2 i) B) \tan ^{-1}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a^2 d}-\frac {5 (5 A+i B)}{8 a^2 d \sqrt {\tan (c+d x)}}+\frac {7 A+3 i B}{8 a^2 d (1+i \tan (c+d x)) \sqrt {\tan (c+d x)}}-\frac {\left (\frac {1}{32}-\frac {i}{32}\right ) ((23+2 i) A+(2+7 i) B) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a^2 d}+\frac {\left (\frac {1}{32}-\frac {i}{32}\right ) ((23+2 i) A+(2+7 i) B) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a^2 d}+\frac {A+i B}{4 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2} \]
Antiderivative was successfully verified.
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Rule 204
Rule 617
Rule 628
Rule 1162
Rule 1165
Rule 1168
Rule 3529
Rule 3534
Rule 3596
Rubi steps
\begin {align*} \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2} \, dx &=\frac {A+i B}{4 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2}+\frac {\int \frac {\frac {1}{2} a (9 A+i B)-\frac {5}{2} a (i A-B) \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))} \, dx}{4 a^2}\\ &=\frac {7 A+3 i B}{8 a^2 d (1+i \tan (c+d x)) \sqrt {\tan (c+d x)}}+\frac {A+i B}{4 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2}+\frac {\int \frac {\frac {5}{2} a^2 (5 A+i B)-\frac {3}{2} a^2 (7 i A-3 B) \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x)} \, dx}{8 a^4}\\ &=-\frac {5 (5 A+i B)}{8 a^2 d \sqrt {\tan (c+d x)}}+\frac {7 A+3 i B}{8 a^2 d (1+i \tan (c+d x)) \sqrt {\tan (c+d x)}}+\frac {A+i B}{4 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2}+\frac {\int \frac {-\frac {3}{2} a^2 (7 i A-3 B)-\frac {5}{2} a^2 (5 A+i B) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx}{8 a^4}\\ &=-\frac {5 (5 A+i B)}{8 a^2 d \sqrt {\tan (c+d x)}}+\frac {7 A+3 i B}{8 a^2 d (1+i \tan (c+d x)) \sqrt {\tan (c+d x)}}+\frac {A+i B}{4 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2}+\frac {\operatorname {Subst}\left (\int \frac {-\frac {3}{2} a^2 (7 i A-3 B)-\frac {5}{2} a^2 (5 A+i B) x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{4 a^4 d}\\ &=-\frac {5 (5 A+i B)}{8 a^2 d \sqrt {\tan (c+d x)}}+\frac {7 A+3 i B}{8 a^2 d (1+i \tan (c+d x)) \sqrt {\tan (c+d x)}}+\frac {A+i B}{4 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2}-\frac {((25+21 i) A-(9-5 i) B) \operatorname {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{16 a^2 d}+\frac {((25-21 i) A+(9+5 i) B) \operatorname {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{16 a^2 d}\\ &=-\frac {5 (5 A+i B)}{8 a^2 d \sqrt {\tan (c+d x)}}+\frac {7 A+3 i B}{8 a^2 d (1+i \tan (c+d x)) \sqrt {\tan (c+d x)}}+\frac {A+i B}{4 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2}-\frac {((25+21 i) A-(9-5 i) B) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{32 a^2 d}-\frac {((25+21 i) A-(9-5 i) B) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{32 a^2 d}-\frac {((25-21 i) A+(9+5 i) B) \operatorname {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{32 \sqrt {2} a^2 d}-\frac {((25-21 i) A+(9+5 i) B) \operatorname {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{32 \sqrt {2} a^2 d}\\ &=-\frac {((25-21 i) A+(9+5 i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^2 d}+\frac {((25-21 i) A+(9+5 i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^2 d}-\frac {5 (5 A+i B)}{8 a^2 d \sqrt {\tan (c+d x)}}+\frac {7 A+3 i B}{8 a^2 d (1+i \tan (c+d x)) \sqrt {\tan (c+d x)}}+\frac {A+i B}{4 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2}-\frac {((25+21 i) A-(9-5 i) B) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^2 d}+\frac {((25+21 i) A-(9-5 i) B) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^2 d}\\ &=\frac {((25+21 i) A-(9-5 i) B) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^2 d}-\frac {((25+21 i) A-(9-5 i) B) \tan ^{-1}\left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^2 d}-\frac {((25-21 i) A+(9+5 i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^2 d}+\frac {((25-21 i) A+(9+5 i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^2 d}-\frac {5 (5 A+i B)}{8 a^2 d \sqrt {\tan (c+d x)}}+\frac {7 A+3 i B}{8 a^2 d (1+i \tan (c+d x)) \sqrt {\tan (c+d x)}}+\frac {A+i B}{4 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2}\\ \end {align*}
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Mathematica [A] time = 2.63, size = 250, normalized size = 0.79 \[ \frac {\sec (c+d x) (\cos (d x)+i \sin (d x))^2 (A+B \tan (c+d x)) \left ((-2 \cos (2 d x)+2 i \sin (2 d x)) ((-7 B+43 i A) \sin (2 (c+d x))+(41 A+5 i B) \cos (2 (c+d x))-9 A-5 i B)+(-\sin (2 c)+i \cos (2 c)) \sqrt {\sin (2 (c+d x))} \sec (c+d x) \left (((21-25 i) A+(5+9 i) B) \sin ^{-1}(\cos (c+d x)-\sin (c+d x))-(1+i) ((23+2 i) A+(2+7 i) B) \log \left (\sin (c+d x)+\sqrt {\sin (2 (c+d x))}+\cos (c+d x)\right )\right )\right )}{32 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2 (A \cos (c+d x)+B \sin (c+d x))} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.72, size = 762, normalized size = 2.40 \[ \frac {2 \, {\left (a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} - a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{4} d^{2}}} \log \left (\frac {2 \, {\left ({\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{4} d^{2}}} + {\left (A - i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{i \, A + B}\right ) - 2 \, {\left (a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} - a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{4} d^{2}}} \log \left (-\frac {2 \, {\left ({\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{4} d^{2}}} - {\left (A - i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{i \, A + B}\right ) + {\left (a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} - a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \sqrt {\frac {-529 i \, A^{2} + 322 \, A B + 49 i \, B^{2}}{a^{4} d^{2}}} \log \left (\frac {{\left ({\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-529 i \, A^{2} + 322 \, A B + 49 i \, B^{2}}{a^{4} d^{2}}} + 23 \, A + 7 i \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{2} d}\right ) - {\left (a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} - a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \sqrt {\frac {-529 i \, A^{2} + 322 \, A B + 49 i \, B^{2}}{a^{4} d^{2}}} \log \left (-\frac {{\left ({\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-529 i \, A^{2} + 322 \, A B + 49 i \, B^{2}}{a^{4} d^{2}}} - 23 \, A - 7 i \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{2} d}\right ) + 2 \, {\left ({\left (-42 i \, A + 6 \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (-33 i \, A + B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (10 i \, A - 6 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A - B\right )} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{32 \, {\left (a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} - a^{2} d e^{\left (4 i \, d x + 4 i \, c\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 \[ \int \frac {B \tan \left (d x + c\right ) + A}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} \tan \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.36, size = 311, normalized size = 0.98 \[ -\frac {\arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right ) A}{2 d \,a^{2} \left (\sqrt {2}+i \sqrt {2}\right )}+\frac {i \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right ) B}{2 d \,a^{2} \left (\sqrt {2}+i \sqrt {2}\right )}-\frac {2 A}{d \,a^{2} \sqrt {\tan \left (d x +c \right )}}-\frac {9 \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right ) A}{8 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {5 i \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right ) B}{8 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {11 i \left (\sqrt {\tan }\left (d x +c \right )\right ) A}{8 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {7 \left (\sqrt {\tan }\left (d x +c \right )\right ) B}{8 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {7 i \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right ) B}{4 d \,a^{2} \left (\sqrt {2}-i \sqrt {2}\right )}-\frac {23 \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right ) A}{4 d \,a^{2} \left (\sqrt {2}-i \sqrt {2}\right )} \]
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 = 9.83, size = 338, normalized size = 1.06 \[ -\frac {-\frac {5\,B\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}}{8\,a^2\,d}+\frac {B\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,7{}\mathrm {i}}{8\,a^2\,d}}{{\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}+2\,\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}}+2\,\mathrm {atanh}\left (\frac {8\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {A^2\,1{}\mathrm {i}}{64\,a^4\,d^2}}}{A}\right )\,\sqrt {\frac {A^2\,1{}\mathrm {i}}{64\,a^4\,d^2}}+2\,\mathrm {atanh}\left (\frac {16\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {A^2\,529{}\mathrm {i}}{256\,a^4\,d^2}}}{23\,A}\right )\,\sqrt {-\frac {A^2\,529{}\mathrm {i}}{256\,a^4\,d^2}}+\mathrm {atan}\left (\frac {8\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {B^2\,1{}\mathrm {i}}{64\,a^4\,d^2}}}{B}\right )\,\sqrt {-\frac {B^2\,1{}\mathrm {i}}{64\,a^4\,d^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (\frac {16\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {B^2\,49{}\mathrm {i}}{256\,a^4\,d^2}}}{7\,B}\right )\,\sqrt {\frac {B^2\,49{}\mathrm {i}}{256\,a^4\,d^2}}\,2{}\mathrm {i}-\frac {\frac {43\,A\,\mathrm {tan}\left (c+d\,x\right )}{8\,a^2\,d}-\frac {A\,2{}\mathrm {i}}{a^2\,d}+\frac {A\,{\mathrm {tan}\left (c+d\,x\right )}^2\,25{}\mathrm {i}}{8\,a^2\,d}}{2\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}-\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,1{}\mathrm {i}+{\mathrm {tan}\left (c+d\,x\right )}^{5/2}\,1{}\mathrm {i}} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]
Verification of antiderivative is not currently implemented for this CAS.
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