Optimal. Leaf size=387 \[ \frac{(A-i B) (a+b \tan (c+d x))^{n+1} \text{Hypergeometric2F1}\left (1,n+1,n+2,\frac{a+b \tan (c+d x)}{a-i b}\right )}{2 d (n+1) (b+i a)}-\frac{(A+i B) (a+b \tan (c+d x))^{n+1} \text{Hypergeometric2F1}\left (1,n+1,n+2,\frac{a+b \tan (c+d x)}{a+i b}\right )}{2 d (n+1) (-b+i a)}-\frac{\tan ^2(c+d x) (3 a B-A b (n+4)) (a+b \tan (c+d x))^{n+1}}{b^2 d (n+3) (n+4)}-\frac{\left (A b^3 (n+2) (n+3) (n+4)-a \left (b^2 B (n+3) (n+4)-2 a (3 a B-A b (n+4))\right )\right ) (a+b \tan (c+d x))^{n+1}}{b^4 d (n+1) (n+2) (n+3) (n+4)}-\frac{\tan (c+d x) \left (b^2 B (n+3) (n+4)-2 a (3 a B-A b (n+4))\right ) (a+b \tan (c+d x))^{n+1}}{b^3 d (n+2) (n+3) (n+4)}+\frac{B \tan ^3(c+d x) (a+b \tan (c+d x))^{n+1}}{b d (n+4)} \]
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Rubi [A] time = 1.05107, antiderivative size = 385, normalized size of antiderivative = 0.99, number of steps used = 9, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {3607, 3647, 3630, 3539, 3537, 68} \[ -\frac{\left (-2 a^2 A b (n+4)+6 a^3 B-a b^2 B (n+3) (n+4)+A b^3 (n+2) (n+3) (n+4)\right ) (a+b \tan (c+d x))^{n+1}}{b^4 d (n+1) (n+2) (n+3) (n+4)}+\frac{\tan (c+d x) \left (6 a^2 B-2 a A b (n+4)-b^2 B (n+3) (n+4)\right ) (a+b \tan (c+d x))^{n+1}}{b^3 d (n+2) (n+3) (n+4)}-\frac{\tan ^2(c+d x) (3 a B-A b (n+4)) (a+b \tan (c+d x))^{n+1}}{b^2 d (n+3) (n+4)}+\frac{(A-i B) (a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{a+b \tan (c+d x)}{a-i b}\right )}{2 d (n+1) (b+i a)}-\frac{(A+i B) (a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{a+b \tan (c+d x)}{a+i b}\right )}{2 d (n+1) (-b+i a)}+\frac{B \tan ^3(c+d x) (a+b \tan (c+d x))^{n+1}}{b d (n+4)} \]
Antiderivative was successfully verified.
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Rule 3607
Rule 3647
Rule 3630
Rule 3539
Rule 3537
Rule 68
Rubi steps
\begin{align*} \int \tan ^4(c+d x) (a+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx &=\frac{B \tan ^3(c+d x) (a+b \tan (c+d x))^{1+n}}{b d (4+n)}+\frac{\int \tan ^2(c+d x) (a+b \tan (c+d x))^n \left (-3 a B-b B (4+n) \tan (c+d x)-(3 a B-A b (4+n)) \tan ^2(c+d x)\right ) \, dx}{b (4+n)}\\ &=-\frac{(3 a B-A b (4+n)) \tan ^2(c+d x) (a+b \tan (c+d x))^{1+n}}{b^2 d (3+n) (4+n)}+\frac{B \tan ^3(c+d x) (a+b \tan (c+d x))^{1+n}}{b d (4+n)}+\frac{\int \tan (c+d x) (a+b \tan (c+d x))^n \left (2 a (3 a B-A b (4+n))-A b^2 (3+n) (4+n) \tan (c+d x)+\left (6 a^2 B-2 a A b (4+n)-b^2 B (3+n) (4+n)\right ) \tan ^2(c+d x)\right ) \, dx}{b^2 (3+n) (4+n)}\\ &=\frac{\left (6 a^2 B-2 a A b (4+n)-b^2 B (3+n) (4+n)\right ) \tan (c+d x) (a+b \tan (c+d x))^{1+n}}{b^3 d (2+n) (3+n) (4+n)}-\frac{(3 a B-A b (4+n)) \tan ^2(c+d x) (a+b \tan (c+d x))^{1+n}}{b^2 d (3+n) (4+n)}+\frac{B \tan ^3(c+d x) (a+b \tan (c+d x))^{1+n}}{b d (4+n)}+\frac{\int (a+b \tan (c+d x))^n \left (-a \left (6 a^2 B-2 a A b (4+n)-b^2 B (3+n) (4+n)\right )+b^3 B (2+n) (3+n) (4+n) \tan (c+d x)-\left (6 a^3 B-2 a^2 A b (4+n)-a b^2 B (3+n) (4+n)+A b^3 (2+n) (3+n) (4+n)\right ) \tan ^2(c+d x)\right ) \, dx}{b^3 (2+n) (3+n) (4+n)}\\ &=-\frac{\left (6 a^3 B-2 a^2 A b (4+n)-a b^2 B (3+n) (4+n)+A b^3 (2+n) (3+n) (4+n)\right ) (a+b \tan (c+d x))^{1+n}}{b^4 d (1+n) (2+n) (3+n) (4+n)}+\frac{\left (6 a^2 B-2 a A b (4+n)-b^2 B (3+n) (4+n)\right ) \tan (c+d x) (a+b \tan (c+d x))^{1+n}}{b^3 d (2+n) (3+n) (4+n)}-\frac{(3 a B-A b (4+n)) \tan ^2(c+d x) (a+b \tan (c+d x))^{1+n}}{b^2 d (3+n) (4+n)}+\frac{B \tan ^3(c+d x) (a+b \tan (c+d x))^{1+n}}{b d (4+n)}+\frac{\int (a+b \tan (c+d x))^n \left (A b^3 (2+n) (3+n) (4+n)+b^3 B (2+n) (3+n) (4+n) \tan (c+d x)\right ) \, dx}{b^3 (2+n) (3+n) (4+n)}\\ &=-\frac{\left (6 a^3 B-2 a^2 A b (4+n)-a b^2 B (3+n) (4+n)+A b^3 (2+n) (3+n) (4+n)\right ) (a+b \tan (c+d x))^{1+n}}{b^4 d (1+n) (2+n) (3+n) (4+n)}+\frac{\left (6 a^2 B-2 a A b (4+n)-b^2 B (3+n) (4+n)\right ) \tan (c+d x) (a+b \tan (c+d x))^{1+n}}{b^3 d (2+n) (3+n) (4+n)}-\frac{(3 a B-A b (4+n)) \tan ^2(c+d x) (a+b \tan (c+d x))^{1+n}}{b^2 d (3+n) (4+n)}+\frac{B \tan ^3(c+d x) (a+b \tan (c+d x))^{1+n}}{b d (4+n)}+\frac{1}{2} (A-i B) \int (1+i \tan (c+d x)) (a+b \tan (c+d x))^n \, dx+\frac{1}{2} (A+i B) \int (1-i \tan (c+d x)) (a+b \tan (c+d x))^n \, dx\\ &=-\frac{\left (6 a^3 B-2 a^2 A b (4+n)-a b^2 B (3+n) (4+n)+A b^3 (2+n) (3+n) (4+n)\right ) (a+b \tan (c+d x))^{1+n}}{b^4 d (1+n) (2+n) (3+n) (4+n)}+\frac{\left (6 a^2 B-2 a A b (4+n)-b^2 B (3+n) (4+n)\right ) \tan (c+d x) (a+b \tan (c+d x))^{1+n}}{b^3 d (2+n) (3+n) (4+n)}-\frac{(3 a B-A b (4+n)) \tan ^2(c+d x) (a+b \tan (c+d x))^{1+n}}{b^2 d (3+n) (4+n)}+\frac{B \tan ^3(c+d x) (a+b \tan (c+d x))^{1+n}}{b d (4+n)}-\frac{(i A-B) \operatorname{Subst}\left (\int \frac{(a+i b x)^n}{-1+x} \, dx,x,-i \tan (c+d x)\right )}{2 d}+\frac{(i A+B) \operatorname{Subst}\left (\int \frac{(a-i b x)^n}{-1+x} \, dx,x,i \tan (c+d x)\right )}{2 d}\\ &=-\frac{\left (6 a^3 B-2 a^2 A b (4+n)-a b^2 B (3+n) (4+n)+A b^3 (2+n) (3+n) (4+n)\right ) (a+b \tan (c+d x))^{1+n}}{b^4 d (1+n) (2+n) (3+n) (4+n)}-\frac{(i A+B) \, _2F_1\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) \, _2F_1\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{\left (6 a^2 B-2 a A b (4+n)-b^2 B (3+n) (4+n)\right ) \tan (c+d x) (a+b \tan (c+d x))^{1+n}}{b^3 d (2+n) (3+n) (4+n)}-\frac{(3 a B-A b (4+n)) \tan ^2(c+d x) (a+b \tan (c+d x))^{1+n}}{b^2 d (3+n) (4+n)}+\frac{B \tan ^3(c+d x) (a+b \tan (c+d x))^{1+n}}{b d (4+n)}\\ \end{align*}
Mathematica [A] time = 5.78065, size = 384, normalized size = 0.99 \[ \frac{(a+b \tan (c+d x))^{n+1} \left (i \left (b^4 \left (n^3+9 n^2+26 n+24\right ) (-(a+i b)) (A-i B) \text{Hypergeometric2F1}\left (1,n+1,n+2,\frac{a+b \tan (c+d x)}{a-i b}\right )+b^4 \left (n^3+9 n^2+26 n+24\right ) (a-i b) (A+i B) \text{Hypergeometric2F1}\left (1,n+1,n+2,\frac{a+b \tan (c+d x)}{a+i b}\right )+2 i (a-i b) (a+i b) \left (-2 a^2 A b (n+4)+6 a^3 B-a b^2 B (n+3) (n+4)+A b^3 (n+2) (n+3) (n+4)\right )\right )+2 b (n+1) (a-i b) (a+i b) \tan (c+d x) \left (6 a^2 B-2 a A b (n+4)-b^2 B (n+3) (n+4)\right )-2 b^2 (n+1) (n+2) (a-i b) (a+i b) \tan ^2(c+d x) (3 a B-A b (n+4))+2 b^3 B (n+1) (n+2) (n+3) (a-i b) (a+i b) \tan ^3(c+d x)\right )}{2 b^4 d (n+1) (n+2) (n+3) (n+4) (a-i b) (a+i b)} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.362, size = 0, normalized size = 0. \begin{align*} \int \left ( \tan \left ( dx+c \right ) \right ) ^{4} \left ( a+b\tan \left ( dx+c \right ) \right ) ^{n} \left ( A+B\tan \left ( dx+c \right ) \right ) \, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (B \tan \left (d x + c\right )^{5} + A \tan \left (d x + c\right )^{4}\right )}{\left (b \tan \left (d x + c\right ) + a\right )}^{n}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \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 )^{4}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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