Integrand size = 31, antiderivative size = 291 \[ \int \tan ^3(c+d x) (a+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=\frac {\left (2 a^2 B-a A b (3+n)-b^2 B \left (6+5 n+n^2\right )\right ) (a+b \tan (c+d x))^{1+n}}{b^3 d (1+n) (2+n) (3+n)}+\frac {(i A+B) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a-i b}\right ) (a+b \tan (c+d x))^{1+n}}{2 (i a+b) d (1+n)}+\frac {(A+i B) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a+i b}\right ) (a+b \tan (c+d x))^{1+n}}{2 (a+i b) d (1+n)}-\frac {(2 a B-A b (3+n)) \tan (c+d x) (a+b \tan (c+d x))^{1+n}}{b^2 d (2+n) (3+n)}+\frac {B \tan ^2(c+d x) (a+b \tan (c+d x))^{1+n}}{b d (3+n)} \]
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Time = 0.76 (sec) , antiderivative size = 289, normalized size of antiderivative = 0.99, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {3688, 3728, 3711, 3620, 3618, 70} \[ \int \tan ^3(c+d x) (a+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=\frac {\left (2 a^2 B-a A b (n+3)-b^2 B (n+2) (n+3)\right ) (a+b \tan (c+d x))^{n+1}}{b^3 d (n+1) (n+2) (n+3)}-\frac {\tan (c+d x) (2 a B-A b (n+3)) (a+b \tan (c+d x))^{n+1}}{b^2 d (n+2) (n+3)}+\frac {(B+i A) (a+b \tan (c+d x))^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {a+b \tan (c+d x)}{a-i b}\right )}{2 d (n+1) (b+i a)}+\frac {(A+i B) (a+b \tan (c+d x))^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {a+b \tan (c+d x)}{a+i b}\right )}{2 d (n+1) (a+i b)}+\frac {B \tan ^2(c+d x) (a+b \tan (c+d x))^{n+1}}{b d (n+3)} \]
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Rule 70
Rule 3618
Rule 3620
Rule 3688
Rule 3711
Rule 3728
Rubi steps \begin{align*} \text {integral}& = \frac {B \tan ^2(c+d x) (a+b \tan (c+d x))^{1+n}}{b d (3+n)}+\frac {\int \tan (c+d x) (a+b \tan (c+d x))^n \left (-2 a B-b B (3+n) \tan (c+d x)-(2 a B-A b (3+n)) \tan ^2(c+d x)\right ) \, dx}{b (3+n)} \\ & = -\frac {(2 a B-A b (3+n)) \tan (c+d x) (a+b \tan (c+d x))^{1+n}}{b^2 d (2+n) (3+n)}+\frac {B \tan ^2(c+d x) (a+b \tan (c+d x))^{1+n}}{b d (3+n)}+\frac {\int (a+b \tan (c+d x))^n \left (a (2 a B-A b (3+n))-A b^2 (2+n) (3+n) \tan (c+d x)+\left (2 a^2 B-a A b (3+n)-b^2 B (2+n) (3+n)\right ) \tan ^2(c+d x)\right ) \, dx}{b^2 (2+n) (3+n)} \\ & = \frac {\left (2 a^2 B-a A b (3+n)-b^2 B (2+n) (3+n)\right ) (a+b \tan (c+d x))^{1+n}}{b^3 d (1+n) (2+n) (3+n)}-\frac {(2 a B-A b (3+n)) \tan (c+d x) (a+b \tan (c+d x))^{1+n}}{b^2 d (2+n) (3+n)}+\frac {B \tan ^2(c+d x) (a+b \tan (c+d x))^{1+n}}{b d (3+n)}+\frac {\int (a+b \tan (c+d x))^n \left (b^2 B (2+n) (3+n)-A b^2 (2+n) (3+n) \tan (c+d x)\right ) \, dx}{b^2 (2+n) (3+n)} \\ & = \frac {\left (2 a^2 B-a A b (3+n)-b^2 B (2+n) (3+n)\right ) (a+b \tan (c+d x))^{1+n}}{b^3 d (1+n) (2+n) (3+n)}-\frac {(2 a B-A b (3+n)) \tan (c+d x) (a+b \tan (c+d x))^{1+n}}{b^2 d (2+n) (3+n)}+\frac {B \tan ^2(c+d x) (a+b \tan (c+d x))^{1+n}}{b d (3+n)}+\frac {1}{2} (-i A+B) \int (1-i \tan (c+d x)) (a+b \tan (c+d x))^n \, dx+\frac {1}{2} (i A+B) \int (1+i \tan (c+d x)) (a+b \tan (c+d x))^n \, dx \\ & = \frac {\left (2 a^2 B-a A b (3+n)-b^2 B (2+n) (3+n)\right ) (a+b \tan (c+d x))^{1+n}}{b^3 d (1+n) (2+n) (3+n)}-\frac {(2 a B-A b (3+n)) \tan (c+d x) (a+b \tan (c+d x))^{1+n}}{b^2 d (2+n) (3+n)}+\frac {B \tan ^2(c+d x) (a+b \tan (c+d x))^{1+n}}{b d (3+n)}-\frac {(A-i B) \text {Subst}\left (\int \frac {(a-i b x)^n}{-1+x} \, dx,x,i \tan (c+d x)\right )}{2 d}-\frac {(A+i B) \text {Subst}\left (\int \frac {(a+i b x)^n}{-1+x} \, dx,x,-i \tan (c+d x)\right )}{2 d} \\ & = \frac {\left (2 a^2 B-a A b (3+n)-b^2 B (2+n) (3+n)\right ) (a+b \tan (c+d x))^{1+n}}{b^3 d (1+n) (2+n) (3+n)}+\frac {(A-i B) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a-i b}\right ) (a+b \tan (c+d x))^{1+n}}{2 (a-i b) d (1+n)}+\frac {(A+i B) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a+i b}\right ) (a+b \tan (c+d x))^{1+n}}{2 (a+i b) d (1+n)}-\frac {(2 a B-A b (3+n)) \tan (c+d x) (a+b \tan (c+d x))^{1+n}}{b^2 d (2+n) (3+n)}+\frac {B \tan ^2(c+d x) (a+b \tan (c+d x))^{1+n}}{b d (3+n)} \\ \end{align*}
Time = 2.61 (sec) , antiderivative size = 281, normalized size of antiderivative = 0.97 \[ \int \tan ^3(c+d x) (a+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=\frac {(a+b \tan (c+d x))^{1+n} \left (2 (a-i b) (a+i b) \left (2 a^2 B-a A b (3+n)-b^2 B (2+n) (3+n)\right )+(a+i b) b^3 (A-i B) \left (6+5 n+n^2\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a-i b}\right )+(a-i b) b^3 (A+i B) \left (6+5 n+n^2\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a+i b}\right )-2 (a-i b) (a+i b) b (1+n) (2 a B-A b (3+n)) \tan (c+d x)+2 (a-i b) (a+i b) b^2 B (1+n) (2+n) \tan ^2(c+d x)\right )}{2 (a-i b) (a+i b) b^3 d (1+n) (2+n) (3+n)} \]
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\[\int \tan \left (d x +c \right )^{3} \left (a +b \tan \left (d x +c \right )\right )^{n} \left (A +B \tan \left (d x +c \right )\right )d x\]
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\[ \int \tan ^3(c+d x) (a+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{3} \,d x } \]
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\[ \int \tan ^3(c+d x) (a+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=\int \left (A + B \tan {\left (c + d x \right )}\right ) \left (a + b \tan {\left (c + d x \right )}\right )^{n} \tan ^{3}{\left (c + d x \right )}\, dx \]
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\[ \int \tan ^3(c+d x) (a+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{3} \,d x } \]
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\[ \int \tan ^3(c+d x) (a+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{3} \,d x } \]
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Timed out. \[ \int \tan ^3(c+d x) (a+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=\int {\mathrm {tan}\left (c+d\,x\right )}^3\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^n \,d x \]
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