Optimal. Leaf size=393 \[ \frac {\left (\frac {1}{16}+\frac {i}{16}\right ) ((29+i) A+(1+76 i) B) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^3 d}-\frac {\left (\frac {1}{16}+\frac {i}{16}\right ) ((29+i) A+(1+76 i) B) \tan ^{-1}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a^3 d}-\frac {3 (-5 B+2 i A) \tan ^{\frac {5}{2}}(c+d x)}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {7 (4 A+11 i B) \tan ^{\frac {3}{2}}(c+d x)}{24 a^3 d}+\frac {15 (-5 B+2 i A) \sqrt {\tan (c+d x)}}{8 a^3 d}-\frac {((28-30 i) A+(75+77 i) B) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{32 \sqrt {2} a^3 d}-\frac {\left (\frac {1}{32}+\frac {i}{32}\right ) ((1+29 i) A-(76+i) B) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a^3 d}+\frac {(-B+i A) \tan ^{\frac {9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(A+2 i B) \tan ^{\frac {7}{2}}(c+d x)}{4 a d (a+i a \tan (c+d x))^2} \]
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Rubi [A] time = 0.83, antiderivative size = 393, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 9, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3595, 3528, 3534, 1168, 1162, 617, 204, 1165, 628} \[ -\frac {3 (-5 B+2 i A) \tan ^{\frac {5}{2}}(c+d x)}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {7 (4 A+11 i B) \tan ^{\frac {3}{2}}(c+d x)}{24 a^3 d}+\frac {\left (\frac {1}{16}+\frac {i}{16}\right ) ((29+i) A+(1+76 i) B) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^3 d}-\frac {\left (\frac {1}{16}+\frac {i}{16}\right ) ((29+i) A+(1+76 i) B) \tan ^{-1}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a^3 d}+\frac {15 (-5 B+2 i A) \sqrt {\tan (c+d x)}}{8 a^3 d}-\frac {((28-30 i) A+(75+77 i) B) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{32 \sqrt {2} a^3 d}-\frac {\left (\frac {1}{32}+\frac {i}{32}\right ) ((1+29 i) A-(76+i) B) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a^3 d}+\frac {(-B+i A) \tan ^{\frac {9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(A+2 i B) \tan ^{\frac {7}{2}}(c+d x)}{4 a d (a+i a \tan (c+d x))^2} \]
Antiderivative was successfully verified.
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Rule 204
Rule 617
Rule 628
Rule 1162
Rule 1165
Rule 1168
Rule 3528
Rule 3534
Rule 3595
Rubi steps
\begin {align*} \int \frac {\tan ^{\frac {9}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx &=\frac {(i A-B) \tan ^{\frac {9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {\int \frac {\tan ^{\frac {7}{2}}(c+d x) \left (\frac {9}{2} a (i A-B)+\frac {3}{2} a (A+5 i B) \tan (c+d x)\right )}{(a+i a \tan (c+d x))^2} \, dx}{6 a^2}\\ &=\frac {(i A-B) \tan ^{\frac {9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(A+2 i B) \tan ^{\frac {7}{2}}(c+d x)}{4 a d (a+i a \tan (c+d x))^2}+\frac {\int \frac {\tan ^{\frac {5}{2}}(c+d x) \left (-21 a^2 (A+2 i B)+3 a^2 (5 i A-16 B) \tan (c+d x)\right )}{a+i a \tan (c+d x)} \, dx}{24 a^4}\\ &=\frac {(i A-B) \tan ^{\frac {9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(A+2 i B) \tan ^{\frac {7}{2}}(c+d x)}{4 a d (a+i a \tan (c+d x))^2}-\frac {3 (2 i A-5 B) \tan ^{\frac {5}{2}}(c+d x)}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {\int \tan ^{\frac {3}{2}}(c+d x) \left (-45 a^3 (2 i A-5 B)-21 a^3 (4 A+11 i B) \tan (c+d x)\right ) \, dx}{48 a^6}\\ &=\frac {7 (4 A+11 i B) \tan ^{\frac {3}{2}}(c+d x)}{24 a^3 d}+\frac {(i A-B) \tan ^{\frac {9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(A+2 i B) \tan ^{\frac {7}{2}}(c+d x)}{4 a d (a+i a \tan (c+d x))^2}-\frac {3 (2 i A-5 B) \tan ^{\frac {5}{2}}(c+d x)}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {\int \sqrt {\tan (c+d x)} \left (21 a^3 (4 A+11 i B)-45 a^3 (2 i A-5 B) \tan (c+d x)\right ) \, dx}{48 a^6}\\ &=\frac {15 (2 i A-5 B) \sqrt {\tan (c+d x)}}{8 a^3 d}+\frac {7 (4 A+11 i B) \tan ^{\frac {3}{2}}(c+d x)}{24 a^3 d}+\frac {(i A-B) \tan ^{\frac {9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(A+2 i B) \tan ^{\frac {7}{2}}(c+d x)}{4 a d (a+i a \tan (c+d x))^2}-\frac {3 (2 i A-5 B) \tan ^{\frac {5}{2}}(c+d x)}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {\int \frac {45 a^3 (2 i A-5 B)+21 a^3 (4 A+11 i B) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx}{48 a^6}\\ &=\frac {15 (2 i A-5 B) \sqrt {\tan (c+d x)}}{8 a^3 d}+\frac {7 (4 A+11 i B) \tan ^{\frac {3}{2}}(c+d x)}{24 a^3 d}+\frac {(i A-B) \tan ^{\frac {9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(A+2 i B) \tan ^{\frac {7}{2}}(c+d x)}{4 a d (a+i a \tan (c+d x))^2}-\frac {3 (2 i A-5 B) \tan ^{\frac {5}{2}}(c+d x)}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {\operatorname {Subst}\left (\int \frac {45 a^3 (2 i A-5 B)+21 a^3 (4 A+11 i B) x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{24 a^6 d}\\ &=\frac {15 (2 i A-5 B) \sqrt {\tan (c+d x)}}{8 a^3 d}+\frac {7 (4 A+11 i B) \tan ^{\frac {3}{2}}(c+d x)}{24 a^3 d}+\frac {(i A-B) \tan ^{\frac {9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(A+2 i B) \tan ^{\frac {7}{2}}(c+d x)}{4 a d (a+i a \tan (c+d x))^2}-\frac {3 (2 i A-5 B) \tan ^{\frac {5}{2}}(c+d x)}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {\left (\left (\frac {1}{16}+\frac {i}{16}\right ) ((29+i) A+(1+76 i) B)\right ) \operatorname {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{a^3 d}+\frac {((28-30 i) A+(75+77 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 {15 (2 i A-5 B) \sqrt {\tan (c+d x)}}{8 a^3 d}+\frac {7 (4 A+11 i B) \tan ^{\frac {3}{2}}(c+d x)}{24 a^3 d}+\frac {(i A-B) \tan ^{\frac {9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(A+2 i B) \tan ^{\frac {7}{2}}(c+d x)}{4 a d (a+i a \tan (c+d x))^2}-\frac {3 (2 i A-5 B) \tan ^{\frac {5}{2}}(c+d x)}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {\left (\left (\frac {1}{32}+\frac {i}{32}\right ) ((29+i) A+(1+76 i) B)\right ) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{a^3 d}-\frac {\left (\left (\frac {1}{32}+\frac {i}{32}\right ) ((29+i) A+(1+76 i) B)\right ) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{a^3 d}-\frac {((28-30 i) A+(75+77 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 {((28-30 i) A+(75+77 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 {((28-30 i) A+(75+77 i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^3 d}+\frac {((28-30 i) A+(75+77 i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^3 d}+\frac {15 (2 i A-5 B) \sqrt {\tan (c+d x)}}{8 a^3 d}+\frac {7 (4 A+11 i B) \tan ^{\frac {3}{2}}(c+d x)}{24 a^3 d}+\frac {(i A-B) \tan ^{\frac {9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(A+2 i B) \tan ^{\frac {7}{2}}(c+d x)}{4 a d (a+i a \tan (c+d x))^2}-\frac {3 (2 i A-5 B) \tan ^{\frac {5}{2}}(c+d x)}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}--\frac {\left (\left (\frac {1}{16}+\frac {i}{16}\right ) ((29+i) A+(1+76 i) B)\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^3 d}-\frac {\left (\left (\frac {1}{16}+\frac {i}{16}\right ) ((29+i) A+(1+76 i) B)\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^3 d}\\ &=\frac {\left (\frac {1}{16}+\frac {i}{16}\right ) ((29+i) A+(1+76 i) B) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^3 d}-\frac {\left (\frac {1}{16}+\frac {i}{16}\right ) ((29+i) A+(1+76 i) B) \tan ^{-1}\left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^3 d}-\frac {((28-30 i) A+(75+77 i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^3 d}+\frac {((28-30 i) A+(75+77 i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^3 d}+\frac {15 (2 i A-5 B) \sqrt {\tan (c+d x)}}{8 a^3 d}+\frac {7 (4 A+11 i B) \tan ^{\frac {3}{2}}(c+d x)}{24 a^3 d}+\frac {(i A-B) \tan ^{\frac {9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(A+2 i B) \tan ^{\frac {7}{2}}(c+d x)}{4 a d (a+i a \tan (c+d x))^2}-\frac {3 (2 i A-5 B) \tan ^{\frac {5}{2}}(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 = 5.37, size = 300, normalized size = 0.76 \[ \frac {\sec ^3(c+d x) (\cos (d x)+i \sin (d x))^3 (A+B \tan (c+d x)) \left (3 (-\sin (3 c)+i \cos (3 c)) \sqrt {\sin (2 (c+d x))} \left (((30-28 i) A+(77+75 i) B) \sin ^{-1}(\cos (c+d x)-\sin (c+d x))+(1+i) ((1-76 i) B-(29-i) A) \log \left (\sin (c+d x)+\sqrt {\sin (2 (c+d x))}+\cos (c+d x)\right )\right )+\tan (c+d x) (\sin (3 d x)+i \cos (3 d x)) (2 (90 A+241 i B) \cos (2 (c+d x))+(147 A+349 i B) \cos (4 (c+d x))+194 i A \sin (2 (c+d x))+145 i A \sin (4 (c+d x))+33 A-502 B \sin (2 (c+d x))-347 B \sin (4 (c+d x))+69 i B)\right )}{96 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3 (A \cos (c+d x)+B \sin (c+d x))} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.68, size = 777, normalized size = 1.98 \[ -\frac {3 \, {\left (a^{3} d e^{\left (8 i \, d x + 8 i \, c\right )} + a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )}\right )} \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{6} d^{2}}} \log \left (\frac {2 \, {\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 {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{6} 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 ) - 3 \, {\left (a^{3} d e^{\left (8 i \, d x + 8 i \, c\right )} + a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )}\right )} \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{6} d^{2}}} \log \left (-\frac {2 \, {\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 {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{6} 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 ) - 3 \, {\left (a^{3} d e^{\left (8 i \, d x + 8 i \, c\right )} + a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )}\right )} \sqrt {\frac {-841 i \, A^{2} + 4408 \, A B + 5776 i \, B^{2}}{a^{6} d^{2}}} \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 {-841 i \, A^{2} + 4408 \, A B + 5776 i \, B^{2}}{a^{6} d^{2}}} + 29 \, A + 76 i \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{3} d}\right ) + 3 \, {\left (a^{3} d e^{\left (8 i \, d x + 8 i \, c\right )} + a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )}\right )} \sqrt {\frac {-841 i \, A^{2} + 4408 \, A B + 5776 i \, B^{2}}{a^{6} d^{2}}} \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 {-841 i \, A^{2} + 4408 \, A B + 5776 i \, B^{2}}{a^{6} d^{2}}} - 29 \, A - 76 i \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{3} d}\right ) - 2 \, {\left ({\left (146 i \, A - 348 \, B\right )} e^{\left (8 i \, d x + 8 i \, c\right )} + {\left (187 i \, A - 492 \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (33 i \, A - 69 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (-7 i \, A + 10 \, 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}}}{96 \, {\left (a^{3} d e^{\left (8 i \, d x + 8 i \, c\right )} + a^{3} d e^{\left (6 i \, d x + 6 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 \[ \int \frac {{\left (B \tan \left (d x + c\right ) + A\right )} \tan \left (d x + c\right )^{\frac {9}{2}}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.45, size = 404, normalized size = 1.03 \[ \frac {2 i B \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3 d \,a^{3}}-\frac {6 B \left (\sqrt {\tan }\left (d x +c \right )\right )}{d \,a^{3}}+\frac {2 i A \left (\sqrt {\tan }\left (d x +c \right )\right )}{d \,a^{3}}+\frac {\arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right ) A}{4 d \,a^{3} \left (\sqrt {2}+i \sqrt {2}\right )}-\frac {i \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right ) B}{4 d \,a^{3} \left (\sqrt {2}+i \sqrt {2}\right )}+\frac {35 i B \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {5 A \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{2 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {91 B \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{12 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {49 i \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right ) A}{12 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {7 A \left (\sqrt {\tan }\left (d x +c \right )\right )}{4 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {27 i B \left (\sqrt {\tan }\left (d x +c \right )\right )}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {29 \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right ) A}{4 d \,a^{3} \left (\sqrt {2}-i \sqrt {2}\right )}-\frac {19 i \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right ) B}{d \,a^{3} \left (\sqrt {2}-i \sqrt {2}\right )} \]
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 = 10.22, size = 431, normalized size = 1.10 \[ \mathrm {atan}\left (\frac {a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {A^2\,1{}\mathrm {i}}{256\,a^6\,d^2}}\,16{}\mathrm {i}}{A}\right )\,\sqrt {\frac {A^2\,1{}\mathrm {i}}{256\,a^6\,d^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (\frac {a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {A^2\,841{}\mathrm {i}}{256\,a^6\,d^2}}\,16{}\mathrm {i}}{29\,A}\right )\,\sqrt {-\frac {A^2\,841{}\mathrm {i}}{256\,a^6\,d^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (\frac {16\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {B^2\,1{}\mathrm {i}}{256\,a^6\,d^2}}}{B}\right )\,\sqrt {-\frac {B^2\,1{}\mathrm {i}}{256\,a^6\,d^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (\frac {4\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {B^2\,361{}\mathrm {i}}{16\,a^6\,d^2}}}{19\,B}\right )\,\sqrt {\frac {B^2\,361{}\mathrm {i}}{16\,a^6\,d^2}}\,2{}\mathrm {i}-\frac {\frac {49\,A\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}}{12\,a^3\,d}-\frac {A\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,7{}\mathrm {i}}{4\,a^3\,d}+\frac {A\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}\,5{}\mathrm {i}}{2\,a^3\,d}}{-{\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,3{}\mathrm {i}+1}-\frac {\frac {27\,B\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{8\,a^3\,d}-\frac {35\,B\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}}{8\,a^3\,d}+\frac {B\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,91{}\mathrm {i}}{12\,a^3\,d}}{-{\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,3{}\mathrm {i}+1}+\frac {A\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,2{}\mathrm {i}}{a^3\,d}-\frac {6\,B\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{a^3\,d}+\frac {B\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,2{}\mathrm {i}}{3\,a^3\,d} \]
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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