Optimal. Leaf size=315 \[ -\frac {((7-5 i) A-2 i B) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^3 d}+\frac {((7-5 i) A-2 i B) \tan ^{-1}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{16 \sqrt {2} a^3 d}-\frac {((7+5 i) A-2 i B) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{32 \sqrt {2} a^3 d}+\frac {((7+5 i) A-2 i B) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{32 \sqrt {2} a^3 d}+\frac {5 A \sqrt {\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {(4 A+i B) \sqrt {\tan (c+d x)}}{12 a d (a+i a \tan (c+d x))^2}+\frac {(A+i B) \sqrt {\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3} \]
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Rubi [A] time = 0.64, antiderivative size = 315, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 8, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {3596, 3534, 1168, 1162, 617, 204, 1165, 628} \[ -\frac {((7-5 i) A-2 i B) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^3 d}+\frac {((7-5 i) A-2 i B) \tan ^{-1}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{16 \sqrt {2} a^3 d}-\frac {((7+5 i) A-2 i B) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{32 \sqrt {2} a^3 d}+\frac {((7+5 i) A-2 i B) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{32 \sqrt {2} a^3 d}+\frac {5 A \sqrt {\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {(4 A+i B) \sqrt {\tan (c+d x)}}{12 a d (a+i a \tan (c+d x))^2}+\frac {(A+i B) \sqrt {\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3} \]
Antiderivative was successfully verified.
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Rule 204
Rule 617
Rule 628
Rule 1162
Rule 1165
Rule 1168
Rule 3534
Rule 3596
Rubi steps
\begin {align*} \int \frac {A+B \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3} \, dx &=\frac {(A+i B) \sqrt {\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3}+\frac {\int \frac {\frac {1}{2} a (11 A-i B)-\frac {5}{2} a (i A-B) \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2} \, dx}{6 a^2}\\ &=\frac {(A+i B) \sqrt {\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3}+\frac {(4 A+i B) \sqrt {\tan (c+d x)}}{12 a d (a+i a \tan (c+d x))^2}+\frac {\int \frac {3 a^2 (6 A-i B)-3 a^2 (4 i A-B) \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+i a \tan (c+d x))} \, dx}{24 a^4}\\ &=\frac {(A+i B) \sqrt {\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3}+\frac {(4 A+i B) \sqrt {\tan (c+d x)}}{12 a d (a+i a \tan (c+d x))^2}+\frac {5 A \sqrt {\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {\int \frac {3 a^3 (7 A-2 i B)-15 i a^3 A \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx}{48 a^6}\\ &=\frac {(A+i B) \sqrt {\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3}+\frac {(4 A+i B) \sqrt {\tan (c+d x)}}{12 a d (a+i a \tan (c+d x))^2}+\frac {5 A \sqrt {\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {\operatorname {Subst}\left (\int \frac {3 a^3 (7 A-2 i B)-15 i a^3 A x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{24 a^6 d}\\ &=\frac {(A+i B) \sqrt {\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3}+\frac {(4 A+i B) \sqrt {\tan (c+d x)}}{12 a d (a+i a \tan (c+d x))^2}+\frac {5 A \sqrt {\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {((7-5 i) A-2 i B) \operatorname {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{16 a^3 d}+\frac {((7+5 i) A-2 i B) \operatorname {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{16 a^3 d}\\ &=\frac {(A+i B) \sqrt {\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3}+\frac {(4 A+i B) \sqrt {\tan (c+d x)}}{12 a d (a+i a \tan (c+d x))^2}+\frac {5 A \sqrt {\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {((7-5 i) A-2 i B) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{32 a^3 d}+\frac {((7-5 i) A-2 i B) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{32 a^3 d}-\frac {((7+5 i) A-2 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^3 d}-\frac {((7+5 i) A-2 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^3 d}\\ &=-\frac {((7+5 i) A-2 i B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^3 d}+\frac {((7+5 i) A-2 i B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^3 d}+\frac {(A+i B) \sqrt {\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3}+\frac {(4 A+i B) \sqrt {\tan (c+d x)}}{12 a d (a+i a \tan (c+d x))^2}+\frac {5 A \sqrt {\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {((7-5 i) A-2 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^3 d}-\frac {((7-5 i) A-2 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^3 d}\\ &=-\frac {((7-5 i) A-2 i B) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^3 d}+\frac {((7-5 i) A-2 i B) \tan ^{-1}\left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^3 d}-\frac {((7+5 i) A-2 i B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^3 d}+\frac {((7+5 i) A-2 i B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^3 d}+\frac {(A+i B) \sqrt {\tan (c+d x)}}{6 d (a+i a \tan (c+d x))^3}+\frac {(4 A+i B) \sqrt {\tan (c+d x)}}{12 a d (a+i a \tan (c+d x))^2}+\frac {5 A \sqrt {\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}\\ \end {align*}
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Mathematica [A] time = 3.41, size = 258, normalized size = 0.82 \[ \frac {\sec ^2(c+d x) (\cos (d x)+i \sin (d x))^3 (A+B \tan (c+d x)) \left (\frac {4}{3} \sin (c+d x) (\cos (3 d x)-i \sin (3 d x)) ((-B+19 i A) \sin (2 (c+d x))+3 (7 A+i B) \cos (2 (c+d x))+6 A+3 i B)+(-\sin (3 c)+i \cos (3 c)) \sqrt {\sin (2 (c+d x))} \sec (c+d x) \left ((2 B+(5+7 i) A) \sin ^{-1}(\cos (c+d x)-\sin (c+d x))-(1+i) ((1+6 i) A+(1-i) B) \log \left (\sin (c+d x)+\sqrt {\sin (2 (c+d x))}+\cos (c+d x)\right )\right )\right )}{32 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3 (A \cos (c+d x)+B \sin (c+d x))} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.60, size = 682, normalized size = 2.17 \[ \frac {{\left (3 \, a^{3} d \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{6} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (\frac {{\left ({\left (16 i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + 16 i \, a^{3} 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^{6} d^{2}}} + 16 \, {\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 )}}{8 \, {\left (i \, A + B\right )}}\right ) - 3 \, a^{3} d \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{6} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (\frac {{\left ({\left (-16 i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} - 16 i \, a^{3} 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^{6} d^{2}}} + 16 \, {\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 )}}{8 \, {\left (i \, A + B\right )}}\right ) + 3 \, a^{3} d \sqrt {\frac {36 i \, A^{2} + 12 \, A B - i \, B^{2}}{a^{6} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (\frac {{\left ({\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} 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 {36 i \, A^{2} + 12 \, A B - i \, B^{2}}{a^{6} d^{2}}} + 6 i \, A + B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{3} d}\right ) - 3 \, a^{3} d \sqrt {\frac {36 i \, A^{2} + 12 \, A B - i \, B^{2}}{a^{6} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (-\frac {{\left ({\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} 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 {36 i \, A^{2} + 12 \, A B - i \, B^{2}}{a^{6} d^{2}}} - 6 i \, A - B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{3} d}\right ) + 2 \, {\left (2 \, {\left (10 \, A + i \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (26 \, A + 5 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (7 \, A + 4 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + A + i \, 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}}\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{96 \, a^{3} d} \]
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 \[ \int \frac {B \tan \left (d x + c\right ) + A}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} \sqrt {\tan \left (d x + c\right )}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.46, size = 323, normalized size = 1.03 \[ \frac {i \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right ) A}{4 d \,a^{3} \left (\sqrt {2}+i \sqrt {2}\right )}+\frac {\arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right ) B}{4 d \,a^{3} \left (\sqrt {2}+i \sqrt {2}\right )}-\frac {5 i A \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {19 \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right ) A}{12 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {i \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right ) B}{12 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {9 i A \left (\sqrt {\tan }\left (d x +c \right )\right )}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {B \left (\sqrt {\tan }\left (d x +c \right )\right )}{4 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {3 i \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right ) A}{2 d \,a^{3} \left (\sqrt {2}-i \sqrt {2}\right )}-\frac {\arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right ) B}{4 d \,a^{3} \left (\sqrt {2}-i \sqrt {2}\right )} \]
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 = 6.67, size = 308, normalized size = 0.98 \[ \frac {\frac {9\,A\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{8\,a^3\,d}-\frac {5\,A\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}}{8\,a^3\,d}+\frac {A\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,19{}\mathrm {i}}{12\,a^3\,d}}{-{\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,3{}\mathrm {i}+1}+\frac {-\frac {B\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}}{12\,a^3\,d}+\frac {B\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,1{}\mathrm {i}}{4\,a^3\,d}}{-{\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,3{}\mathrm {i}+1}-\frac {{\left (-1\right )}^{1/4}\,B\,\mathrm {atan}\left ({\left (-1\right )}^{1/4}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{8\,a^3\,d}-\frac {{\left (-1\right )}^{1/4}\,B\,\mathrm {atanh}\left ({\left (-1\right )}^{1/4}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{8\,a^3\,d}-\mathrm {atan}\left (\frac {8\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {A^2\,9{}\mathrm {i}}{64\,a^6\,d^2}}}{3\,A}\right )\,\sqrt {\frac {A^2\,9{}\mathrm {i}}{64\,a^6\,d^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {16\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {A^2\,1{}\mathrm {i}}{256\,a^6\,d^2}}}{A}\right )\,\sqrt {-\frac {A^2\,1{}\mathrm {i}}{256\,a^6\,d^2}}\,2{}\mathrm {i} \]
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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