Optimal. Leaf size=282 \[ \frac {\left (a^2 A+2 a b B-A b^2\right ) \tan ^{m+1}(c+d x) \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\tan ^2(c+d x)\right )}{d (m+1) \left (a^2+b^2\right )^2}-\frac {\left (a^2 (-B)+2 a A b+b^2 B\right ) \tan ^{m+2}(c+d x) \, _2F_1\left (1,\frac {m+2}{2};\frac {m+4}{2};-\tan ^2(c+d x)\right )}{d (m+2) \left (a^2+b^2\right )^2}+\frac {b (A b-a B) \tan ^{m+1}(c+d x)}{a d \left (a^2+b^2\right ) (a+b \tan (c+d x))}+\frac {b \left (-\left (a^3 (B-B m)\right )+a^2 A b (2-m)+a b^2 B (m+1)-A b^3 m\right ) \tan ^{m+1}(c+d x) \, _2F_1\left (1,m+1;m+2;-\frac {b \tan (c+d x)}{a}\right )}{a^2 d (m+1) \left (a^2+b^2\right )^2} \]
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Rubi [A] time = 0.70, antiderivative size = 282, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {3609, 3653, 3538, 3476, 364, 3634, 64} \[ \frac {\left (a^2 A+2 a b B-A b^2\right ) \tan ^{m+1}(c+d x) \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\tan ^2(c+d x)\right )}{d (m+1) \left (a^2+b^2\right )^2}+\frac {b \left (a^2 A b (2-m)+a^3 (-(B-B m))+a b^2 B (m+1)-A b^3 m\right ) \tan ^{m+1}(c+d x) \, _2F_1\left (1,m+1;m+2;-\frac {b \tan (c+d x)}{a}\right )}{a^2 d (m+1) \left (a^2+b^2\right )^2}-\frac {\left (a^2 (-B)+2 a A b+b^2 B\right ) \tan ^{m+2}(c+d x) \, _2F_1\left (1,\frac {m+2}{2};\frac {m+4}{2};-\tan ^2(c+d x)\right )}{d (m+2) \left (a^2+b^2\right )^2}+\frac {b (A b-a B) \tan ^{m+1}(c+d x)}{a d \left (a^2+b^2\right ) (a+b \tan (c+d x))} \]
Antiderivative was successfully verified.
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Rule 64
Rule 364
Rule 3476
Rule 3538
Rule 3609
Rule 3634
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))^2} \, dx &=\frac {b (A b-a B) \tan ^{1+m}(c+d x)}{a \left (a^2+b^2\right ) d (a+b \tan (c+d x))}+\frac {\int \frac {\tan ^m(c+d x) \left (a^2 A-A b^2 m+a b B (1+m)-a (A b-a B) \tan (c+d x)-b (A b-a B) m \tan ^2(c+d x)\right )}{a+b \tan (c+d x)} \, dx}{a \left (a^2+b^2\right )}\\ &=\frac {b (A b-a B) \tan ^{1+m}(c+d x)}{a \left (a^2+b^2\right ) d (a+b \tan (c+d x))}+\frac {\int \tan ^m(c+d x) \left (a \left (a^2 A-A b^2+2 a b B\right )-a \left (2 a A b-a^2 B+b^2 B\right ) \tan (c+d x)\right ) \, dx}{a \left (a^2+b^2\right )^2}+\frac {\left (a^2 b (A b-a B)-a^2 b (A b-a B) m+b^2 \left (a^2 A-A b^2 m+a b B (1+m)\right )\right ) \int \frac {\tan ^m(c+d x) \left (1+\tan ^2(c+d x)\right )}{a+b \tan (c+d x)} \, dx}{a \left (a^2+b^2\right )^2}\\ &=\frac {b (A b-a B) \tan ^{1+m}(c+d x)}{a \left (a^2+b^2\right ) d (a+b \tan (c+d x))}+\frac {\left (a^2 A-A b^2+2 a b B\right ) \int \tan ^m(c+d x) \, dx}{\left (a^2+b^2\right )^2}-\frac {\left (2 a A b-a^2 B+b^2 B\right ) \int \tan ^{1+m}(c+d x) \, dx}{\left (a^2+b^2\right )^2}+\frac {\left (a^2 b (A b-a B)-a^2 b (A b-a B) m+b^2 \left (a^2 A-A b^2 m+a b B (1+m)\right )\right ) \operatorname {Subst}\left (\int \frac {x^m}{a+b x} \, dx,x,\tan (c+d x)\right )}{a \left (a^2+b^2\right )^2 d}\\ &=-\frac {b \left (a^3 B (1-m)-a^2 A b (2-m)+A b^3 m-a b^2 B (1+m)\right ) \, _2F_1\left (1,1+m;2+m;-\frac {b \tan (c+d x)}{a}\right ) \tan ^{1+m}(c+d x)}{a^2 \left (a^2+b^2\right )^2 d (1+m)}+\frac {b (A b-a B) \tan ^{1+m}(c+d x)}{a \left (a^2+b^2\right ) d (a+b \tan (c+d x))}+\frac {\left (a^2 A-A b^2+2 a b 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 )^2 d}-\frac {\left (2 a A b-a^2 B+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 )^2 d}\\ &=\frac {\left (a^2 A-A b^2+2 a b 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 )^2 d (1+m)}-\frac {b \left (a^3 B (1-m)-a^2 A b (2-m)+A b^3 m-a b^2 B (1+m)\right ) \, _2F_1\left (1,1+m;2+m;-\frac {b \tan (c+d x)}{a}\right ) \tan ^{1+m}(c+d x)}{a^2 \left (a^2+b^2\right )^2 d (1+m)}-\frac {\left (2 a A b-a^2 B+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 )^2 d (2+m)}+\frac {b (A b-a B) \tan ^{1+m}(c+d x)}{a \left (a^2+b^2\right ) d (a+b \tan (c+d x))}\\ \end {align*}
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Mathematica [A] time = 3.07, size = 239, normalized size = 0.85 \[ \frac {\tan ^{m+1}(c+d x) \left (\frac {a \left (\frac {\left (a^2 A+2 a b B-A b^2\right ) \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\tan ^2(c+d x)\right )}{m+1}+\frac {\left (a^2 B-2 a A b-b^2 B\right ) \tan (c+d x) \, _2F_1\left (1,\frac {m+2}{2};\frac {m+4}{2};-\tan ^2(c+d x)\right )}{m+2}\right )}{a^2+b^2}+\frac {b \left (a^3 B (m-1)-a^2 A b (m-2)+a b^2 B (m+1)-A b^3 m\right ) \, _2F_1\left (1,m+1;m+2;-\frac {b \tan (c+d x)}{a}\right )}{a (m+1) \left (a^2+b^2\right )}+\frac {b (A b-a B)}{a+b \tan (c+d x)}\right )}{a d \left (a^2+b^2\right )} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.58, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B \tan \left (d x + c\right ) + A\right )} \tan \left (d x + c\right )^{m}}{b^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) + a^{2}}, x\right ) \]
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 \[ \int \frac {{\left (B \tan \left (d x + c\right ) + A\right )} \tan \left (d x + c\right )^{m}}{{\left (b \tan \left (d x + c\right ) + a\right )}^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 3.94, size = 0, normalized size = 0.00 \[ \int \frac {\left (\tan ^{m}\left (d x +c \right )\right ) \left (A +B \tan \left (d x +c \right )\right )}{\left (a +b \tan \left (d x +c \right )\right )^{2}}\, dx \]
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 \[ \int \frac {{\left (B \tan \left (d x + c\right ) + A\right )} \tan \left (d x + c\right )^{m}}{{\left (b \tan \left (d x + c\right ) + a\right )}^{2}}\,{d x} \]
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 \[ \int \frac {{\mathrm {tan}\left (c+d\,x\right )}^m\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )}{{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^2} \,d x \]
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 \[ \int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \tan ^{m}{\left (c + d x \right )}}{\left (a + b \tan {\left (c + d x \right )}\right )^{2}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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