Optimal. Leaf size=316 \[ \frac {4 a^3 (A-i B) \sqrt {1+i \tan (c+d x)} \tan ^{m+1}(c+d x) F_1\left (m+1;\frac {1}{2},1;m+2;-i \tan (c+d x),i \tan (c+d x)\right )}{d (m+1) \sqrt {a+i a \tan (c+d x)}}+\frac {2 a^2 \left (2 B \left (4 m^2+17 m+19\right )+i A \left (8 m^2+34 m+35\right )\right ) \sqrt {a+i a \tan (c+d x)} \tan ^m(c+d x) (-i \tan (c+d x))^{-m} \, _2F_1\left (\frac {1}{2},-m;\frac {3}{2};i \tan (c+d x)+1\right )}{d (2 m+3) (2 m+5)}+\frac {2 a^2 (-A (2 m+5)+2 i B (m+4)) \sqrt {a+i a \tan (c+d x)} \tan ^{m+1}(c+d x)}{d (2 m+3) (2 m+5)}+\frac {2 i a B (a+i a \tan (c+d x))^{3/2} \tan ^{m+1}(c+d x)}{d (2 m+5)} \]
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Rubi [A] time = 0.99, antiderivative size = 316, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {3594, 3601, 3564, 135, 133, 3599, 67, 65} \[ \frac {4 a^3 (A-i B) \sqrt {1+i \tan (c+d x)} \tan ^{m+1}(c+d x) F_1\left (m+1;\frac {1}{2},1;m+2;-i \tan (c+d x),i \tan (c+d x)\right )}{d (m+1) \sqrt {a+i a \tan (c+d x)}}+\frac {2 a^2 \left (2 B \left (4 m^2+17 m+19\right )+i A \left (8 m^2+34 m+35\right )\right ) \sqrt {a+i a \tan (c+d x)} \tan ^m(c+d x) (-i \tan (c+d x))^{-m} \, _2F_1\left (\frac {1}{2},-m;\frac {3}{2};i \tan (c+d x)+1\right )}{d (2 m+3) (2 m+5)}+\frac {2 a^2 (-A (2 m+5)+2 i B (m+4)) \sqrt {a+i a \tan (c+d x)} \tan ^{m+1}(c+d x)}{d (2 m+3) (2 m+5)}+\frac {2 i a B (a+i a \tan (c+d x))^{3/2} \tan ^{m+1}(c+d x)}{d (2 m+5)} \]
Antiderivative was successfully verified.
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Rule 65
Rule 67
Rule 133
Rule 135
Rule 3564
Rule 3594
Rule 3599
Rule 3601
Rubi steps
\begin {align*} \int \tan ^m(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx &=\frac {2 i a B \tan ^{1+m}(c+d x) (a+i a \tan (c+d x))^{3/2}}{d (5+2 m)}+\frac {2 \int \tan ^m(c+d x) (a+i a \tan (c+d x))^{3/2} \left (-\frac {1}{2} a \left (2 i B (1+m)-2 A \left (\frac {5}{2}+m\right )\right )+\frac {1}{2} a (2 B (4+m)+i A (5+2 m)) \tan (c+d x)\right ) \, dx}{5+2 m}\\ &=\frac {2 a^2 (2 i B (4+m)-A (5+2 m)) \tan ^{1+m}(c+d x) \sqrt {a+i a \tan (c+d x)}}{d (3+2 m) (5+2 m)}+\frac {2 i a B \tan ^{1+m}(c+d x) (a+i a \tan (c+d x))^{3/2}}{d (5+2 m)}+\frac {4 \int \tan ^m(c+d x) \sqrt {a+i a \tan (c+d x)} \left (-\frac {1}{4} a^2 \left (2 i B \left (11+15 m+4 m^2\right )-A \left (25+30 m+8 m^2\right )\right )+\frac {1}{4} a^2 \left (2 B \left (19+17 m+4 m^2\right )+i A \left (35+34 m+8 m^2\right )\right ) \tan (c+d x)\right ) \, dx}{15+16 m+4 m^2}\\ &=\frac {2 a^2 (2 i B (4+m)-A (5+2 m)) \tan ^{1+m}(c+d x) \sqrt {a+i a \tan (c+d x)}}{d (3+2 m) (5+2 m)}+\frac {2 i a B \tan ^{1+m}(c+d x) (a+i a \tan (c+d x))^{3/2}}{d (5+2 m)}+\left (4 a^2 (A-i B)\right ) \int \tan ^m(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx+\frac {\left (a \left (2 i B \left (19+17 m+4 m^2\right )-A \left (35+34 m+8 m^2\right )\right )\right ) \int \tan ^m(c+d x) (a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)} \, dx}{15+16 m+4 m^2}\\ &=\frac {2 a^2 (2 i B (4+m)-A (5+2 m)) \tan ^{1+m}(c+d x) \sqrt {a+i a \tan (c+d x)}}{d (3+2 m) (5+2 m)}+\frac {2 i a B \tan ^{1+m}(c+d x) (a+i a \tan (c+d x))^{3/2}}{d (5+2 m)}+\frac {\left (4 a^4 (i A+B)\right ) \operatorname {Subst}\left (\int \frac {\left (-\frac {i x}{a}\right )^m}{\sqrt {a+x} \left (-a^2+a x\right )} \, dx,x,i a \tan (c+d x)\right )}{d}+\frac {\left (a^3 \left (2 i B \left (19+17 m+4 m^2\right )-A \left (35+34 m+8 m^2\right )\right )\right ) \operatorname {Subst}\left (\int \frac {x^m}{\sqrt {a+i a x}} \, dx,x,\tan (c+d x)\right )}{d \left (15+16 m+4 m^2\right )}\\ &=\frac {2 a^2 (2 i B (4+m)-A (5+2 m)) \tan ^{1+m}(c+d x) \sqrt {a+i a \tan (c+d x)}}{d (3+2 m) (5+2 m)}+\frac {2 i a B \tan ^{1+m}(c+d x) (a+i a \tan (c+d x))^{3/2}}{d (5+2 m)}+\frac {\left (a^3 \left (2 i B \left (19+17 m+4 m^2\right )-A \left (35+34 m+8 m^2\right )\right ) (-i \tan (c+d x))^{-m} \tan ^m(c+d x)\right ) \operatorname {Subst}\left (\int \frac {(-i x)^m}{\sqrt {a+i a x}} \, dx,x,\tan (c+d x)\right )}{d \left (15+16 m+4 m^2\right )}+\frac {\left (4 a^4 (i A+B) \sqrt {1+i \tan (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {\left (-\frac {i x}{a}\right )^m}{\sqrt {1+\frac {x}{a}} \left (-a^2+a x\right )} \, dx,x,i a \tan (c+d x)\right )}{d \sqrt {a+i a \tan (c+d x)}}\\ &=\frac {4 a^3 (A-i B) F_1\left (1+m;\frac {1}{2},1;2+m;-i \tan (c+d x),i \tan (c+d x)\right ) \sqrt {1+i \tan (c+d x)} \tan ^{1+m}(c+d x)}{d (1+m) \sqrt {a+i a \tan (c+d x)}}+\frac {2 a^2 \left (2 B \left (19+17 m+4 m^2\right )+i A \left (35+34 m+8 m^2\right )\right ) \, _2F_1\left (\frac {1}{2},-m;\frac {3}{2};1+i \tan (c+d x)\right ) (-i \tan (c+d x))^{-m} \tan ^m(c+d x) \sqrt {a+i a \tan (c+d x)}}{d \left (15+16 m+4 m^2\right )}+\frac {2 a^2 (2 i B (4+m)-A (5+2 m)) \tan ^{1+m}(c+d x) \sqrt {a+i a \tan (c+d x)}}{d (3+2 m) (5+2 m)}+\frac {2 i a B \tan ^{1+m}(c+d x) (a+i a \tan (c+d x))^{3/2}}{d (5+2 m)}\\ \end {align*}
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Mathematica [F] time = 7.60, size = 0, normalized size = 0.00 \[ \int \tan ^m(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx \]
Verification is Not applicable to the result.
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fricas [F] time = 0.65, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {4 \, \sqrt {2} {\left ({\left (A - i \, B\right )} a^{2} e^{\left (7 i \, d x + 7 i \, c\right )} + {\left (A + i \, B\right )} a^{2} e^{\left (5 i \, d x + 5 i \, c\right )}\right )} \left (\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{m} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [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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maple [F] time = 3.79, size = 0, normalized size = 0.00 \[ \int \left (\tan ^{m}\left (d x +c \right )\right ) \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}} \left (A +B \tan \left (d x +c \right )\right )\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \tan \left (d x + c\right )^{m}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int {\mathrm {tan}\left (c+d\,x\right )}^m\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2} \,d x \]
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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