Optimal. Leaf size=297 \[ \frac{\left (2 a^2-b^2 (n+2) (n+3)\right ) (a+b \tan (c+d x))^{n+1}}{b^3 d (n+1) (n+2) (n+3)}-\frac{\sqrt{-b^2} (a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{a+b \tan (c+d x)}{a-\sqrt{-b^2}}\right )}{2 b d (n+1) \left (a-\sqrt{-b^2}\right )}+\frac{\sqrt{-b^2} (a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{a+b \tan (c+d x)}{a+\sqrt{-b^2}}\right )}{2 b d (n+1) \left (a+\sqrt{-b^2}\right )}-\frac{2 a \tan (c+d x) (a+b \tan (c+d x))^{n+1}}{b^2 d (n+2) (n+3)}+\frac{\tan ^2(c+d x) (a+b \tan (c+d x))^{n+1}}{b d (n+3)} \]
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Rubi [A] time = 0.519569, antiderivative size = 297, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {3566, 3647, 3631, 3485, 712, 68} \[ \frac{\left (2 a^2-b^2 (n+2) (n+3)\right ) (a+b \tan (c+d x))^{n+1}}{b^3 d (n+1) (n+2) (n+3)}-\frac{\sqrt{-b^2} (a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{a+b \tan (c+d x)}{a-\sqrt{-b^2}}\right )}{2 b d (n+1) \left (a-\sqrt{-b^2}\right )}+\frac{\sqrt{-b^2} (a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{a+b \tan (c+d x)}{a+\sqrt{-b^2}}\right )}{2 b d (n+1) \left (a+\sqrt{-b^2}\right )}-\frac{2 a \tan (c+d x) (a+b \tan (c+d x))^{n+1}}{b^2 d (n+2) (n+3)}+\frac{\tan ^2(c+d x) (a+b \tan (c+d x))^{n+1}}{b d (n+3)} \]
Antiderivative was successfully verified.
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Rule 3566
Rule 3647
Rule 3631
Rule 3485
Rule 712
Rule 68
Rubi steps
\begin{align*} \int \tan ^4(c+d x) (a+b \tan (c+d x))^n \, dx &=\frac{\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 (3+n) \tan (c+d x)-2 a \tan ^2(c+d x)\right ) \, dx}{b (3+n)}\\ &=-\frac{2 a \tan (c+d x) (a+b \tan (c+d x))^{1+n}}{b^2 d (2+n) (3+n)}+\frac{\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 (2 a^2+\left (2 a^2-b^2 (2+n) (3+n)\right ) \tan ^2(c+d x)\right ) \, dx}{b^2 (2+n) (3+n)}\\ &=\frac{\left (2 a^2-b^2 (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 \tan (c+d x) (a+b \tan (c+d x))^{1+n}}{b^2 d (2+n) (3+n)}+\frac{\tan ^2(c+d x) (a+b \tan (c+d x))^{1+n}}{b d (3+n)}+\int (a+b \tan (c+d x))^n \, dx\\ &=\frac{\left (2 a^2-b^2 (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 \tan (c+d x) (a+b \tan (c+d x))^{1+n}}{b^2 d (2+n) (3+n)}+\frac{\tan ^2(c+d x) (a+b \tan (c+d x))^{1+n}}{b d (3+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{\left (2 a^2-b^2 (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 \tan (c+d x) (a+b \tan (c+d x))^{1+n}}{b^2 d (2+n) (3+n)}+\frac{\tan ^2(c+d x) (a+b \tan (c+d x))^{1+n}}{b d (3+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{\left (2 a^2-b^2 (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 \tan (c+d x) (a+b \tan (c+d x))^{1+n}}{b^2 d (2+n) (3+n)}+\frac{\tan ^2(c+d x) (a+b \tan (c+d x))^{1+n}}{b d (3+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{\left (2 a^2-b^2 (2+n) (3+n)\right ) (a+b \tan (c+d x))^{1+n}}{b^3 d (1+n) (2+n) (3+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 \, _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{2 a \tan (c+d x) (a+b \tan (c+d x))^{1+n}}{b^2 d (2+n) (3+n)}+\frac{\tan ^2(c+d x) (a+b \tan (c+d x))^{1+n}}{b d (3+n)}\\ \end{align*}
Mathematica [C] time = 2.09864, size = 249, normalized size = 0.84 \[ -\frac{(a+b \tan (c+d x))^{n+1} \left (2 (-b+i a) (b+i a) \left (2 a^2-b^2 \left (n^2+5 n+6\right )\right )+i b^3 (n+2) (n+3) (a+i b) \, _2F_1\left (1,n+1;n+2;\frac{a+b \tan (c+d x)}{a-i b}\right )-b^3 (n+2) (n+3) (b+i a) \, _2F_1\left (1,n+1;n+2;\frac{a+b \tan (c+d x)}{a+i b}\right )-2 b^2 (n+1) (n+2) (a-i b) (a+i b) \tan ^2(c+d x)+4 a b (n+1) (a-i b) (a+i b) \tan (c+d x)\right )}{2 b^3 d (n+1) (n+2) (n+3) (a-i b) (a+i b)} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.227, 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}\, 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} \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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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} \tan \left (d x + c\right )^{4}, 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} \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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