Optimal. Leaf size=168 \[ -\frac {(A-i B) (a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {a+b \tan (c+d x)}{a-i b}\right )}{2 d (n+1) (a-i b)}-\frac {(A+i B) (a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {a+b \tan (c+d x)}{a+i b}\right )}{2 d (n+1) (a+i b)}+\frac {B (a+b \tan (c+d x))^{n+1}}{b d (n+1)} \]
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Rubi [A] time = 0.18, antiderivative size = 168, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {3592, 3539, 3537, 68} \[ -\frac {(A-i B) (a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {a+b \tan (c+d x)}{a-i b}\right )}{2 d (n+1) (a-i b)}-\frac {(A+i B) (a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {a+b \tan (c+d x)}{a+i b}\right )}{2 d (n+1) (a+i b)}+\frac {B (a+b \tan (c+d x))^{n+1}}{b d (n+1)} \]
Antiderivative was successfully verified.
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Rule 68
Rule 3537
Rule 3539
Rule 3592
Rubi steps
\begin {align*} \int \tan (c+d x) (a+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx &=\frac {B (a+b \tan (c+d x))^{1+n}}{b d (1+n)}+\int (-B+A \tan (c+d x)) (a+b \tan (c+d x))^n \, dx\\ &=\frac {B (a+b \tan (c+d x))^{1+n}}{b d (1+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 {B (a+b \tan (c+d x))^{1+n}}{b d (1+n)}+\frac {(A-i B) \operatorname {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) \operatorname {Subst}\left (\int \frac {(a+i b x)^n}{-1+x} \, dx,x,-i \tan (c+d x)\right )}{2 d}\\ &=\frac {B (a+b \tan (c+d x))^{1+n}}{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 (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 (a+i b) d (1+n)}\\ \end {align*}
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Mathematica [A] time = 0.26, size = 125, normalized size = 0.74 \[ \frac {(a+b \tan (c+d x))^{n+1} \left (-\frac {(A-i B) \, _2F_1\left (1,n+1;n+2;\frac {a+b \tan (c+d x)}{a-i b}\right )}{a-i b}-\frac {(A+i B) \, _2F_1\left (1,n+1;n+2;\frac {a+b \tan (c+d x)}{a+i b}\right )}{a+i b}+\frac {2 B}{b}\right )}{2 d (n+1)} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.60, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (B \tan \left (d x + c\right )^{2} + A \tan \left (d x + c\right )\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{n}, x\right ) \]
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 \[ \int {\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.94, size = 0, normalized size = 0.00 \[ \int \tan \left (d x +c \right ) \left (a +b \tan \left (d x +c \right )\right )^{n} \left (A +B \tan \left (d x +c \right )\right )\, dx \]
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 \[ \int {\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \mathrm {tan}\left (c+d\,x\right )\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^n \,d x \]
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 \[ \int \left (A + B \tan {\left (c + d x \right )}\right ) \left (a + b \tan {\left (c + d x \right )}\right )^{n} \tan {\left (c + d x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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