Optimal. Leaf size=245 \[ -\frac{b n (a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{b \tan (c+d x)}{a}+1\right )}{a^2 d (n+1)}-\frac{b (a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{a+b \tan (c+d x)}{a-\sqrt{-b^2}}\right )}{2 \sqrt{-b^2} d (n+1) \left (a-\sqrt{-b^2}\right )}+\frac{b (a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{a+b \tan (c+d x)}{a+\sqrt{-b^2}}\right )}{2 \sqrt{-b^2} d (n+1) \left (a+\sqrt{-b^2}\right )}-\frac{\cot (c+d x) (a+b \tan (c+d x))^{n+1}}{a d} \]
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Rubi [A] time = 0.361871, antiderivative size = 245, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.381, Rules used = {3569, 3653, 12, 3485, 712, 68, 3634, 65} \[ -\frac{b n (a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{b \tan (c+d x)}{a}+1\right )}{a^2 d (n+1)}-\frac{b (a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{a+b \tan (c+d x)}{a-\sqrt{-b^2}}\right )}{2 \sqrt{-b^2} d (n+1) \left (a-\sqrt{-b^2}\right )}+\frac{b (a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{a+b \tan (c+d x)}{a+\sqrt{-b^2}}\right )}{2 \sqrt{-b^2} d (n+1) \left (a+\sqrt{-b^2}\right )}-\frac{\cot (c+d x) (a+b \tan (c+d x))^{n+1}}{a d} \]
Antiderivative was successfully verified.
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Rule 3569
Rule 3653
Rule 12
Rule 3485
Rule 712
Rule 68
Rule 3634
Rule 65
Rubi steps
\begin{align*} \int \cot ^2(c+d x) (a+b \tan (c+d x))^n \, dx &=-\frac{\cot (c+d x) (a+b \tan (c+d x))^{1+n}}{a d}-\frac{\int \cot (c+d x) (a+b \tan (c+d x))^n \left (-b n+a \tan (c+d x)-b n \tan ^2(c+d x)\right ) \, dx}{a}\\ &=-\frac{\cot (c+d x) (a+b \tan (c+d x))^{1+n}}{a d}-\frac{\int a (a+b \tan (c+d x))^n \, dx}{a}+\frac{(b n) \int \cot (c+d x) (a+b \tan (c+d x))^n \left (1+\tan ^2(c+d x)\right ) \, dx}{a}\\ &=-\frac{\cot (c+d x) (a+b \tan (c+d x))^{1+n}}{a d}+\frac{(b n) \operatorname{Subst}\left (\int \frac{(a+b x)^n}{x} \, dx,x,\tan (c+d x)\right )}{a d}-\int (a+b \tan (c+d x))^n \, dx\\ &=-\frac{\cot (c+d x) (a+b \tan (c+d x))^{1+n}}{a d}-\frac{b n \, _2F_1\left (1,1+n;2+n;1+\frac{b \tan (c+d x)}{a}\right ) (a+b \tan (c+d x))^{1+n}}{a^2 d (1+n)}-\frac{b \operatorname{Subst}\left (\int \frac{(a+x)^n}{b^2+x^2} \, dx,x,b \tan (c+d x)\right )}{d}\\ &=-\frac{\cot (c+d x) (a+b \tan (c+d x))^{1+n}}{a d}-\frac{b n \, _2F_1\left (1,1+n;2+n;1+\frac{b \tan (c+d x)}{a}\right ) (a+b \tan (c+d x))^{1+n}}{a^2 d (1+n)}-\frac{b \operatorname{Subst}\left (\int \left (\frac{\sqrt{-b^2} (a+x)^n}{2 b^2 \left (\sqrt{-b^2}-x\right )}+\frac{\sqrt{-b^2} (a+x)^n}{2 b^2 \left (\sqrt{-b^2}+x\right )}\right ) \, dx,x,b \tan (c+d x)\right )}{d}\\ &=-\frac{\cot (c+d x) (a+b \tan (c+d x))^{1+n}}{a d}-\frac{b n \, _2F_1\left (1,1+n;2+n;1+\frac{b \tan (c+d x)}{a}\right ) (a+b \tan (c+d x))^{1+n}}{a^2 d (1+n)}+\frac{b \operatorname{Subst}\left (\int \frac{(a+x)^n}{\sqrt{-b^2}-x} \, dx,x,b \tan (c+d x)\right )}{2 \sqrt{-b^2} d}+\frac{b \operatorname{Subst}\left (\int \frac{(a+x)^n}{\sqrt{-b^2}+x} \, dx,x,b \tan (c+d x)\right )}{2 \sqrt{-b^2} d}\\ &=-\frac{\cot (c+d x) (a+b \tan (c+d x))^{1+n}}{a d}-\frac{b \, _2F_1\left (1,1+n;2+n;\frac{a+b \tan (c+d x)}{a-\sqrt{-b^2}}\right ) (a+b \tan (c+d x))^{1+n}}{2 \sqrt{-b^2} \left (a-\sqrt{-b^2}\right ) d (1+n)}+\frac{b \, _2F_1\left (1,1+n;2+n;\frac{a+b \tan (c+d x)}{a+\sqrt{-b^2}}\right ) (a+b \tan (c+d x))^{1+n}}{2 \sqrt{-b^2} \left (a+\sqrt{-b^2}\right ) d (1+n)}-\frac{b n \, _2F_1\left (1,1+n;2+n;1+\frac{b \tan (c+d x)}{a}\right ) (a+b \tan (c+d x))^{1+n}}{a^2 d (1+n)}\\ \end{align*}
Mathematica [C] time = 0.728011, size = 190, normalized size = 0.78 \[ -\frac{\tan (c+d x) (a \cot (c+d x)+b) (a+b \tan (c+d x))^n \left (a^2 (b-i a) \, _2F_1\left (1,n+1;n+2;\frac{a+b \tan (c+d x)}{a-i b}\right )+(a-i b) \left (i a^2 \, _2F_1\left (1,n+1;n+2;\frac{a+b \tan (c+d x)}{a+i b}\right )+2 (a+i b) \left (b n \, _2F_1\left (1,n+1;n+2;\frac{b \tan (c+d x)}{a}+1\right )+a (n+1) \cot (c+d x)\right )\right )\right )}{2 a^2 d (n+1) (a-i b) (a+i b)} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.22, size = 0, normalized size = 0. \begin{align*} \int \left ( \cot \left ( dx+c \right ) \right ) ^{2} \left ( a+b\tan \left ( dx+c \right ) \right ) ^{n}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \tan \left (d x + c\right ) + a\right )}^{n} \cot \left (d x + c\right )^{2}\,{d x} \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 ) + a\right )}^{n} \cot \left (d x + c\right )^{2}, 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 )}^{n} \cot \left (d x + c\right )^{2}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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