Optimal. Leaf size=438 \[ \frac {b (A b-a B) \tan ^{m+1}(c+d x)}{2 a d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}+\frac {\left (a^3 A+3 a^2 b B-3 a A b^2-b^3 B\right ) \tan ^{m+1}(c+d x) \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\tan ^2(c+d x)\right )}{d (m+1) \left (a^2+b^2\right )^3}-\frac {\left (a^3 (-B)+3 a^2 A b+3 a b^2 B-A b^3\right ) \tan ^{m+2}(c+d x) \, _2F_1\left (1,\frac {m+2}{2};\frac {m+4}{2};-\tan ^2(c+d x)\right )}{d (m+2) \left (a^2+b^2\right )^3}+\frac {b \left (a^3 (-B) (3-m)+a^2 A b (5-m)+a b^2 B (m+1)+A b^3 (1-m)\right ) \tan ^{m+1}(c+d x)}{2 a^2 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}-\frac {b \left (a^5 B \left (m^2-3 m+2\right )-a^4 A b \left (m^2-5 m+6\right )-2 a^3 b^2 B \left (-m^2+m+3\right )+2 a^2 A b^3 \left (-m^2+3 m+1\right )+a b^4 B m (m+1)+A b^5 (1-m) m\right ) \tan ^{m+1}(c+d x) \, _2F_1\left (1,m+1;m+2;-\frac {b \tan (c+d x)}{a}\right )}{2 a^3 d (m+1) \left (a^2+b^2\right )^3} \]
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Rubi [A] time = 1.28, antiderivative size = 438, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.258, Rules used = {3609, 3649, 3653, 3538, 3476, 364, 3634, 64} \[ -\frac {b \left (2 a^2 A b^3 \left (-m^2+3 m+1\right )-a^4 A b \left (m^2-5 m+6\right )-2 a^3 b^2 B \left (-m^2+m+3\right )+a^5 B \left (m^2-3 m+2\right )+a b^4 B m (m+1)+A b^5 (1-m) m\right ) \tan ^{m+1}(c+d x) \, _2F_1\left (1,m+1;m+2;-\frac {b \tan (c+d x)}{a}\right )}{2 a^3 d (m+1) \left (a^2+b^2\right )^3}+\frac {\left (a^3 A+3 a^2 b B-3 a A b^2-b^3 B\right ) \tan ^{m+1}(c+d x) \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\tan ^2(c+d x)\right )}{d (m+1) \left (a^2+b^2\right )^3}-\frac {\left (3 a^2 A b+a^3 (-B)+3 a b^2 B-A b^3\right ) \tan ^{m+2}(c+d x) \, _2F_1\left (1,\frac {m+2}{2};\frac {m+4}{2};-\tan ^2(c+d x)\right )}{d (m+2) \left (a^2+b^2\right )^3}+\frac {b \left (a^2 A b (5-m)+a^3 (-B) (3-m)+a b^2 B (m+1)+A b^3 (1-m)\right ) \tan ^{m+1}(c+d x)}{2 a^2 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}+\frac {b (A b-a B) \tan ^{m+1}(c+d x)}{2 a d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2} \]
Antiderivative was successfully verified.
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Rule 64
Rule 364
Rule 3476
Rule 3538
Rule 3609
Rule 3634
Rule 3649
Rule 3653
Rubi steps
\begin {align*} \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^3} \, dx &=\frac {b (A b-a B) \tan ^{1+m}(c+d x)}{2 a \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac {\int \frac {\tan ^m(c+d x) \left (2 a^2 A+A b^2 (1-m)+a b B (1+m)-2 a (A b-a B) \tan (c+d x)+b (A b-a B) (1-m) \tan ^2(c+d x)\right )}{(a+b \tan (c+d x))^2} \, dx}{2 a \left (a^2+b^2\right )}\\ &=\frac {b (A b-a B) \tan ^{1+m}(c+d x)}{2 a \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac {b \left (A b^3 (1-m)-a^3 B (3-m)+a^2 A b (5-m)+a b^2 B (1+m)\right ) \tan ^{1+m}(c+d x)}{2 a^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac {\int \frac {\tan ^m(c+d x) \left (-a^2 b (A b-a B) (3-m) (1+m)+\left (a^2-b^2 m\right ) \left (2 a^2 A+A b^2 (1-m)+a b B (1+m)\right )-2 a^2 \left (2 a A b-a^2 B+b^2 B\right ) \tan (c+d x)-b m \left (A b^3 (1-m)-a^3 B (3-m)+a^2 A b (5-m)+a b^2 B (1+m)\right ) \tan ^2(c+d x)\right )}{a+b \tan (c+d x)} \, dx}{2 a^2 \left (a^2+b^2\right )^2}\\ &=\frac {b (A b-a B) \tan ^{1+m}(c+d x)}{2 a \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac {b \left (A b^3 (1-m)-a^3 B (3-m)+a^2 A b (5-m)+a b^2 B (1+m)\right ) \tan ^{1+m}(c+d x)}{2 a^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac {\int \tan ^m(c+d x) \left (2 a^2 \left (a^3 A-3 a A b^2+3 a^2 b B-b^3 B\right )-2 a^2 \left (3 a^2 A b-A b^3-a^3 B+3 a b^2 B\right ) \tan (c+d x)\right ) \, dx}{2 a^2 \left (a^2+b^2\right )^3}-\frac {\left (b \left (A b^5 (1-m) m+a b^4 B m (1+m)-2 a^3 b^2 B \left (3+m-m^2\right )+2 a^2 A b^3 \left (1+3 m-m^2\right )-a^4 A b \left (6-5 m+m^2\right )+a^5 B \left (2-3 m+m^2\right )\right )\right ) \int \frac {\tan ^m(c+d x) \left (1+\tan ^2(c+d x)\right )}{a+b \tan (c+d x)} \, dx}{2 a^2 \left (a^2+b^2\right )^3}\\ &=\frac {b (A b-a B) \tan ^{1+m}(c+d x)}{2 a \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac {b \left (A b^3 (1-m)-a^3 B (3-m)+a^2 A b (5-m)+a b^2 B (1+m)\right ) \tan ^{1+m}(c+d x)}{2 a^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}-\frac {\left (3 a^2 A b-A b^3-a^3 B+3 a b^2 B\right ) \int \tan ^{1+m}(c+d x) \, dx}{\left (a^2+b^2\right )^3}+\frac {\left (a^3 A-3 a A b^2+3 a^2 b B-b^3 B\right ) \int \tan ^m(c+d x) \, dx}{\left (a^2+b^2\right )^3}-\frac {\left (b \left (A b^5 (1-m) m+a b^4 B m (1+m)-2 a^3 b^2 B \left (3+m-m^2\right )+2 a^2 A b^3 \left (1+3 m-m^2\right )-a^4 A b \left (6-5 m+m^2\right )+a^5 B \left (2-3 m+m^2\right )\right )\right ) \operatorname {Subst}\left (\int \frac {x^m}{a+b x} \, dx,x,\tan (c+d x)\right )}{2 a^2 \left (a^2+b^2\right )^3 d}\\ &=-\frac {b \left (A b^5 (1-m) m+a b^4 B m (1+m)-2 a^3 b^2 B \left (3+m-m^2\right )+2 a^2 A b^3 \left (1+3 m-m^2\right )-a^4 A b \left (6-5 m+m^2\right )+a^5 B \left (2-3 m+m^2\right )\right ) \, _2F_1\left (1,1+m;2+m;-\frac {b \tan (c+d x)}{a}\right ) \tan ^{1+m}(c+d x)}{2 a^3 \left (a^2+b^2\right )^3 d (1+m)}+\frac {b (A b-a B) \tan ^{1+m}(c+d x)}{2 a \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac {b \left (A b^3 (1-m)-a^3 B (3-m)+a^2 A b (5-m)+a b^2 B (1+m)\right ) \tan ^{1+m}(c+d x)}{2 a^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}-\frac {\left (3 a^2 A b-A b^3-a^3 B+3 a b^2 B\right ) \operatorname {Subst}\left (\int \frac {x^{1+m}}{1+x^2} \, dx,x,\tan (c+d x)\right )}{\left (a^2+b^2\right )^3 d}+\frac {\left (a^3 A-3 a A b^2+3 a^2 b B-b^3 B\right ) \operatorname {Subst}\left (\int \frac {x^m}{1+x^2} \, dx,x,\tan (c+d x)\right )}{\left (a^2+b^2\right )^3 d}\\ &=\frac {\left (a^3 A-3 a A b^2+3 a^2 b B-b^3 B\right ) \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\tan ^2(c+d x)\right ) \tan ^{1+m}(c+d x)}{\left (a^2+b^2\right )^3 d (1+m)}-\frac {b \left (A b^5 (1-m) m+a b^4 B m (1+m)-2 a^3 b^2 B \left (3+m-m^2\right )+2 a^2 A b^3 \left (1+3 m-m^2\right )-a^4 A b \left (6-5 m+m^2\right )+a^5 B \left (2-3 m+m^2\right )\right ) \, _2F_1\left (1,1+m;2+m;-\frac {b \tan (c+d x)}{a}\right ) \tan ^{1+m}(c+d x)}{2 a^3 \left (a^2+b^2\right )^3 d (1+m)}-\frac {\left (3 a^2 A b-A b^3-a^3 B+3 a b^2 B\right ) \, _2F_1\left (1,\frac {2+m}{2};\frac {4+m}{2};-\tan ^2(c+d x)\right ) \tan ^{2+m}(c+d x)}{\left (a^2+b^2\right )^3 d (2+m)}+\frac {b (A b-a B) \tan ^{1+m}(c+d x)}{2 a \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac {b \left (A b^3 (1-m)-a^3 B (3-m)+a^2 A b (5-m)+a b^2 B (1+m)\right ) \tan ^{1+m}(c+d x)}{2 a^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}\\ \end {align*}
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Mathematica [A] time = 6.25, size = 534, normalized size = 1.22 \[ \frac {b (A b-a B) \tan ^{m+1}(c+d x)}{2 a d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}+\frac {\frac {\left (b^2 \left (2 a^2 A+a b B (m+1)+A b^2 (1-m)\right )-a (-a b (1-m) (A b-a B)-2 a b (A b-a B))\right ) \tan ^{m+1}(c+d x)}{a d \left (a^2+b^2\right ) (a+b \tan (c+d x))}+\frac {\frac {\left (b^2 \left (\left (a^2-b^2 m\right ) \left (2 a^2 A+a b B (m+1)+A b^2 (1-m)\right )-a^2 b (3-m) (m+1) (A b-a B)\right )+2 a^3 b \left (a^2 (-B)+2 a A b+b^2 B\right )-a^2 b m \left (a^3 (-B) (3-m)+a^2 A b (5-m)+a b^2 B (m+1)+A b^3 (1-m)\right )\right ) \tan ^{m+1}(c+d x) \, _2F_1\left (1,m+1;m+2;-\frac {b \tan (c+d x)}{a}\right )}{a d (m+1) \left (a^2+b^2\right )}+\frac {\frac {2 a^2 \left (a^3 A+3 a^2 b B-3 a A b^2-b^3 B\right ) \tan ^{m+1}(c+d x) \, _2F_1\left (1,\frac {m+1}{2};\frac {m+1}{2}+1;-\tan ^2(c+d x)\right )}{d (m+1)}-\frac {2 a^2 \left (a^3 (-B)+3 a^2 A b+3 a b^2 B-A b^3\right ) \tan ^{m+2}(c+d x) \, _2F_1\left (1,\frac {m+2}{2};\frac {m+2}{2}+1;-\tan ^2(c+d x)\right )}{d (m+2)}}{a^2+b^2}}{a \left (a^2+b^2\right )}}{2 a \left (a^2+b^2\right )} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.63, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B \tan \left (d x + c\right ) + A\right )} \tan \left (d x + c\right )^{m}}{b^{3} \tan \left (d x + c\right )^{3} + 3 \, a b^{2} \tan \left (d x + c\right )^{2} + 3 \, a^{2} b \tan \left (d x + c\right ) + a^{3}}, x\right ) \]
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 \[ \int \frac {{\left (B \tan \left (d x + c\right ) + A\right )} \tan \left (d x + c\right )^{m}}{{\left (b \tan \left (d x + c\right ) + a\right )}^{3}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 4.17, size = 0, normalized size = 0.00 \[ \int \frac {\left (\tan ^{m}\left (d x +c \right )\right ) \left (A +B \tan \left (d x +c \right )\right )}{\left (a +b \tan \left (d x +c \right )\right )^{3}}\, dx \]
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 \[ \int \frac {{\left (B \tan \left (d x + c\right ) + A\right )} \tan \left (d x + c\right )^{m}}{{\left (b \tan \left (d x + c\right ) + a\right )}^{3}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\mathrm {tan}\left (c+d\,x\right )}^m\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )}{{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^3} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \tan ^{m}{\left (c + d x \right )}}{\left (a + b \tan {\left (c + d x \right )}\right )^{3}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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