Optimal. Leaf size=307 \[ \frac {(2 A+(5-7 i) B) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^3 d}-\frac {(2 A+(5-7 i) B) \tan ^{-1}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{16 \sqrt {2} a^3 d}-\frac {(2 A-(5+7 i) B) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{32 \sqrt {2} a^3 d}+\frac {(2 A-(5+7 i) B) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{32 \sqrt {2} a^3 d}+\frac {5 B \sqrt {\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {(-B+i A) \tan ^{\frac {5}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(A+4 i B) \tan ^{\frac {3}{2}}(c+d x)}{12 a d (a+i a \tan (c+d x))^2} \]
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Rubi [A] time = 0.64, antiderivative size = 307, 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 = {3595, 3534, 1168, 1162, 617, 204, 1165, 628} \[ \frac {(2 A+(5-7 i) B) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^3 d}-\frac {(2 A+(5-7 i) B) \tan ^{-1}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{16 \sqrt {2} a^3 d}-\frac {(2 A-(5+7 i) B) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{32 \sqrt {2} a^3 d}+\frac {(2 A-(5+7 i) B) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{32 \sqrt {2} a^3 d}+\frac {5 B \sqrt {\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {(-B+i A) \tan ^{\frac {5}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(A+4 i B) \tan ^{\frac {3}{2}}(c+d x)}{12 a d (a+i a \tan (c+d x))^2} \]
Antiderivative was successfully verified.
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Rule 204
Rule 617
Rule 628
Rule 1162
Rule 1165
Rule 1168
Rule 3534
Rule 3595
Rubi steps
\begin {align*} \int \frac {\tan ^{\frac {5}{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 {5}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {\int \frac {\tan ^{\frac {3}{2}}(c+d x) \left (\frac {5}{2} a (i A-B)-\frac {1}{2} a (A-11 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 {5}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(A+4 i B) \tan ^{\frac {3}{2}}(c+d x)}{12 a d (a+i a \tan (c+d x))^2}+\frac {\int \frac {\sqrt {\tan (c+d x)} \left (-3 a^2 (A+4 i B)-3 a^2 (i A+6 B) \tan (c+d x)\right )}{a+i a \tan (c+d x)} \, dx}{24 a^4}\\ &=\frac {(i A-B) \tan ^{\frac {5}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(A+4 i B) \tan ^{\frac {3}{2}}(c+d x)}{12 a d (a+i a \tan (c+d x))^2}+\frac {5 B \sqrt {\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {\int \frac {15 a^3 B+3 a^3 (2 A-7 i B) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx}{48 a^6}\\ &=\frac {(i A-B) \tan ^{\frac {5}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(A+4 i B) \tan ^{\frac {3}{2}}(c+d x)}{12 a d (a+i a \tan (c+d x))^2}+\frac {5 B \sqrt {\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {\operatorname {Subst}\left (\int \frac {15 a^3 B+3 a^3 (2 A-7 i B) x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{24 a^6 d}\\ &=\frac {(i A-B) \tan ^{\frac {5}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(A+4 i B) \tan ^{\frac {3}{2}}(c+d x)}{12 a d (a+i a \tan (c+d x))^2}+\frac {5 B \sqrt {\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {(2 A-(5+7 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 {(2 A+(5-7 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 {(i A-B) \tan ^{\frac {5}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(A+4 i B) \tan ^{\frac {3}{2}}(c+d x)}{12 a d (a+i a \tan (c+d x))^2}+\frac {5 B \sqrt {\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {(2 A-(5+7 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 {(2 A-(5+7 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 {(2 A+(5-7 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 {(2 A+(5-7 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 {(2 A-(5+7 i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^3 d}+\frac {(2 A-(5+7 i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^3 d}+\frac {(i A-B) \tan ^{\frac {5}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(A+4 i B) \tan ^{\frac {3}{2}}(c+d x)}{12 a d (a+i a \tan (c+d x))^2}+\frac {5 B \sqrt {\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {(2 A+(5-7 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 {(2 A+(5-7 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 {(2 A+(5-7 i) B) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^3 d}-\frac {(2 A+(5-7 i) B) \tan ^{-1}\left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^3 d}-\frac {(2 A-(5+7 i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^3 d}+\frac {(2 A-(5+7 i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^3 d}+\frac {(i A-B) \tan ^{\frac {5}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(A+4 i B) \tan ^{\frac {3}{2}}(c+d x)}{12 a d (a+i a \tan (c+d x))^2}+\frac {5 B \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 = 2.80, size = 254, normalized size = 0.83 \[ \frac {\sec ^2(c+d x) (\cos (d x)+i \sin (d x))^3 (A+B \tan (c+d x)) \left (\frac {4}{3} \sin (c+d x) (\cos (3 d x)-i \sin (3 d x)) ((A+19 i B) \sin (2 (c+d x))+3 (7 B-i A) \cos (2 (c+d x))+3 i A-6 B)+(\cos (3 c)+i \sin (3 c)) \sqrt {\sin (2 (c+d x))} \sec (c+d x) \left ((2 A+(5-7 i) B) \sin ^{-1}(\cos (c+d x)-\sin (c+d x))+(1-i) ((1+i) A+(1-6 i) B) \log \left (\sin (c+d x)+\sqrt {\sin (2 (c+d x))}+\cos (c+d x)\right )\right )\right )}{32 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3 (A \cos (c+d x)+B \sin (c+d x))} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.69, size = 676, normalized size = 2.20 \[ \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 {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 \, 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 {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 \, a^{3} d \sqrt {\frac {-i \, A^{2} - 12 \, A B + 36 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 {-i \, A^{2} - 12 \, A B + 36 i \, B^{2}}{a^{6} d^{2}}} + A - 6 i \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{3} d}\right ) - 3 \, a^{3} d \sqrt {\frac {-i \, A^{2} - 12 \, A B + 36 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 {-i \, A^{2} - 12 \, A B + 36 i \, B^{2}}{a^{6} d^{2}}} - A + 6 i \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{3} d}\right ) + 2 \, {\left ({\left (-2 i \, A + 20 \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (i \, A + 14 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (2 i \, A - 5 \, 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}}\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 \[ \int \frac {{\left (B \tan \left (d x + c\right ) + A\right )} \tan \left (d x + c\right )^{\frac {5}{2}}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.44, size = 323, normalized size = 1.05 \[ -\frac {\arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right ) A}{4 d \,a^{3} \left (\sqrt {2}+i \sqrt {2}\right )}+\frac {i \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right ) B}{4 d \,a^{3} \left (\sqrt {2}+i \sqrt {2}\right )}-\frac {9 i B \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {A \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{4 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {19 B \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{12 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {i \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right ) A}{12 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {5 i B \left (\sqrt {\tan }\left (d x +c \right )\right )}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {\arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right ) A}{4 d \,a^{3} \left (\sqrt {2}-i \sqrt {2}\right )}+\frac {3 i \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right ) B}{2 d \,a^{3} \left (\sqrt {2}-i \sqrt {2}\right )} \]
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 = 8.17, size = 308, normalized size = 1.00 \[ \frac {\frac {5\,B\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{8\,a^3\,d}-\frac {9\,B\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}}{8\,a^3\,d}+\frac {B\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,19{}\mathrm {i}}{12\,a^3\,d}}{-{\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,3{}\mathrm {i}+1}+\frac {\frac {A\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}}{12\,a^3\,d}+\frac {A\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}\,1{}\mathrm {i}}{4\,a^3\,d}}{-{\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,3{}\mathrm {i}+1}-\frac {{\left (-1\right )}^{1/4}\,A\,\mathrm {atan}\left ({\left (-1\right )}^{1/4}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{8\,a^3\,d}+\frac {{\left (-1\right )}^{1/4}\,A\,\mathrm {atanh}\left ({\left (-1\right )}^{1/4}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{8\,a^3\,d}+\mathrm {atan}\left (\frac {8\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {B^2\,9{}\mathrm {i}}{64\,a^6\,d^2}}}{3\,B}\right )\,\sqrt {\frac {B^2\,9{}\mathrm {i}}{64\,a^6\,d^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {16\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {B^2\,1{}\mathrm {i}}{256\,a^6\,d^2}}}{B}\right )\,\sqrt {-\frac {B^2\,1{}\mathrm {i}}{256\,a^6\,d^2}}\,2{}\mathrm {i} \]
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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