Optimal. Leaf size=434 \[ \frac{b \left (\frac{a n \left (\frac{3 a^2}{b^2}-2 n+5\right )}{b^2}-\frac{\sqrt{-b^2} \left (a^2 b^2 \left (-n^2-2 n+6\right )+3 a^4+b^4 \left (n^2-4 n+3\right )\right )}{b^6}\right ) (a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{a+b \tan (c+d x)}{a-\sqrt{-b^2}}\right )}{16 d (n+1) \left (\frac{a^2}{b^2}+1\right )^2 \left (a-\sqrt{-b^2}\right )}+\frac{b \left (\frac{\sqrt{-b^2} \left (a^2 b^2 \left (-n^2-2 n+6\right )+3 a^4+b^4 \left (n^2-4 n+3\right )\right )}{b^6}+\frac{a n \left (\frac{3 a^2}{b^2}-2 n+5\right )}{b^2}\right ) (a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{a+b \tan (c+d x)}{a+\sqrt{-b^2}}\right )}{16 d (n+1) \left (\frac{a^2}{b^2}+1\right )^2 \left (a+\sqrt{-b^2}\right )}+\frac{\cos ^4(c+d x) (a \tan (c+d x)+b) (a+b \tan (c+d x))^{n+1}}{4 d \left (a^2+b^2\right )}+\frac{b \cos ^2(c+d x) \left (a b \left (\frac{3 a^2}{b^2}-2 n+5\right ) \tan (c+d x)+a^2 (n+1)+b^2 (3-n)\right ) (a+b \tan (c+d x))^{n+1}}{8 d \left (a^2+b^2\right )^2} \]
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Rubi [A] time = 0.686767, antiderivative size = 434, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {3506, 741, 823, 831, 68} \[ \frac{b \left (\frac{a n \left (\frac{3 a^2}{b^2}-2 n+5\right )}{b^2}-\frac{\sqrt{-b^2} \left (a^2 b^2 \left (-n^2-2 n+6\right )+3 a^4+b^4 \left (n^2-4 n+3\right )\right )}{b^6}\right ) (a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{a+b \tan (c+d x)}{a-\sqrt{-b^2}}\right )}{16 d (n+1) \left (\frac{a^2}{b^2}+1\right )^2 \left (a-\sqrt{-b^2}\right )}+\frac{b \left (\frac{\sqrt{-b^2} \left (a^2 b^2 \left (-n^2-2 n+6\right )+3 a^4+b^4 \left (n^2-4 n+3\right )\right )}{b^6}+\frac{a n \left (\frac{3 a^2}{b^2}-2 n+5\right )}{b^2}\right ) (a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{a+b \tan (c+d x)}{a+\sqrt{-b^2}}\right )}{16 d (n+1) \left (\frac{a^2}{b^2}+1\right )^2 \left (a+\sqrt{-b^2}\right )}+\frac{\cos ^4(c+d x) (a \tan (c+d x)+b) (a+b \tan (c+d x))^{n+1}}{4 d \left (a^2+b^2\right )}+\frac{b \cos ^2(c+d x) \left (a b \left (\frac{3 a^2}{b^2}-2 n+5\right ) \tan (c+d x)+a^2 (n+1)+b^2 (3-n)\right ) (a+b \tan (c+d x))^{n+1}}{8 d \left (a^2+b^2\right )^2} \]
Antiderivative was successfully verified.
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Rule 3506
Rule 741
Rule 823
Rule 831
Rule 68
Rubi steps
\begin{align*} \int \cos ^4(c+d x) (a+b \tan (c+d x))^n \, dx &=\frac{\operatorname{Subst}\left (\int \frac{(a+x)^n}{\left (1+\frac{x^2}{b^2}\right )^3} \, dx,x,b \tan (c+d x)\right )}{b d}\\ &=\frac{\cos ^4(c+d x) (b+a \tan (c+d x)) (a+b \tan (c+d x))^{1+n}}{4 \left (a^2+b^2\right ) d}-\frac{b \operatorname{Subst}\left (\int \frac{(a+x)^n \left (-3-\frac{3 a^2}{b^2}+n-\frac{a (2-n) x}{b^2}\right )}{\left (1+\frac{x^2}{b^2}\right )^2} \, dx,x,b \tan (c+d x)\right )}{4 \left (a^2+b^2\right ) d}\\ &=\frac{\cos ^4(c+d x) (b+a \tan (c+d x)) (a+b \tan (c+d x))^{1+n}}{4 \left (a^2+b^2\right ) d}+\frac{b \cos ^2(c+d x) (a+b \tan (c+d x))^{1+n} \left (b^2 (3-n)+a^2 (1+n)+a b \left (5+\frac{3 a^2}{b^2}-2 n\right ) \tan (c+d x)\right )}{8 \left (a^2+b^2\right )^2 d}+\frac{b^5 \operatorname{Subst}\left (\int \frac{(a+x)^n \left (\frac{3 a^4+a^2 b^2 \left (6-2 n-n^2\right )+b^4 \left (3-4 n+n^2\right )}{b^6}-\frac{a \left (5+\frac{3 a^2}{b^2}-2 n\right ) n x}{b^4}\right )}{1+\frac{x^2}{b^2}} \, dx,x,b \tan (c+d x)\right )}{8 \left (a^2+b^2\right )^2 d}\\ &=\frac{\cos ^4(c+d x) (b+a \tan (c+d x)) (a+b \tan (c+d x))^{1+n}}{4 \left (a^2+b^2\right ) d}+\frac{b \cos ^2(c+d x) (a+b \tan (c+d x))^{1+n} \left (b^2 (3-n)+a^2 (1+n)+a b \left (5+\frac{3 a^2}{b^2}-2 n\right ) \tan (c+d x)\right )}{8 \left (a^2+b^2\right )^2 d}+\frac{b^5 \operatorname{Subst}\left (\int \left (\frac{\left (\frac{a \left (5+\frac{3 a^2}{b^2}-2 n\right ) n}{b^2}+\frac{\sqrt{-b^2} \left (3 a^4+a^2 b^2 \left (6-2 n-n^2\right )+b^4 \left (3-4 n+n^2\right )\right )}{b^6}\right ) (a+x)^n}{2 \left (\sqrt{-b^2}-x\right )}+\frac{\left (-\frac{a \left (5+\frac{3 a^2}{b^2}-2 n\right ) n}{b^2}+\frac{\sqrt{-b^2} \left (3 a^4+a^2 b^2 \left (6-2 n-n^2\right )+b^4 \left (3-4 n+n^2\right )\right )}{b^6}\right ) (a+x)^n}{2 \left (\sqrt{-b^2}+x\right )}\right ) \, dx,x,b \tan (c+d x)\right )}{8 \left (a^2+b^2\right )^2 d}\\ &=\frac{\cos ^4(c+d x) (b+a \tan (c+d x)) (a+b \tan (c+d x))^{1+n}}{4 \left (a^2+b^2\right ) d}+\frac{b \cos ^2(c+d x) (a+b \tan (c+d x))^{1+n} \left (b^2 (3-n)+a^2 (1+n)+a b \left (5+\frac{3 a^2}{b^2}-2 n\right ) \tan (c+d x)\right )}{8 \left (a^2+b^2\right )^2 d}-\frac{\left (b^5 \left (\frac{a \left (5+\frac{3 a^2}{b^2}-2 n\right ) n}{b^2}-\frac{\sqrt{-b^2} \left (3 a^4+a^2 b^2 \left (6-2 n-n^2\right )+b^4 \left (3-4 n+n^2\right )\right )}{b^6}\right )\right ) \operatorname{Subst}\left (\int \frac{(a+x)^n}{\sqrt{-b^2}+x} \, dx,x,b \tan (c+d x)\right )}{16 \left (a^2+b^2\right )^2 d}+\frac{\left (b^5 \left (\frac{a \left (5+\frac{3 a^2}{b^2}-2 n\right ) n}{b^2}+\frac{\sqrt{-b^2} \left (3 a^4+a^2 b^2 \left (6-2 n-n^2\right )+b^4 \left (3-4 n+n^2\right )\right )}{b^6}\right )\right ) \operatorname{Subst}\left (\int \frac{(a+x)^n}{\sqrt{-b^2}-x} \, dx,x,b \tan (c+d x)\right )}{16 \left (a^2+b^2\right )^2 d}\\ &=\frac{b^5 \left (\frac{a \left (5+\frac{3 a^2}{b^2}-2 n\right ) n}{b^2}-\frac{\sqrt{-b^2} \left (3 a^4+a^2 b^2 \left (6-2 n-n^2\right )+b^4 \left (3-4 n+n^2\right )\right )}{b^6}\right ) \, _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}}{16 \left (a^2+b^2\right )^2 \left (a-\sqrt{-b^2}\right ) d (1+n)}+\frac{b^5 \left (\frac{a \left (5+\frac{3 a^2}{b^2}-2 n\right ) n}{b^2}+\frac{\sqrt{-b^2} \left (3 a^4+a^2 b^2 \left (6-2 n-n^2\right )+b^4 \left (3-4 n+n^2\right )\right )}{b^6}\right ) \, _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}}{16 \left (a^2+b^2\right )^2 \left (a+\sqrt{-b^2}\right ) d (1+n)}+\frac{\cos ^4(c+d x) (b+a \tan (c+d x)) (a+b \tan (c+d x))^{1+n}}{4 \left (a^2+b^2\right ) d}+\frac{b \cos ^2(c+d x) (a+b \tan (c+d x))^{1+n} \left (b^2 (3-n)+a^2 (1+n)+a b \left (5+\frac{3 a^2}{b^2}-2 n\right ) \tan (c+d x)\right )}{8 \left (a^2+b^2\right )^2 d}\\ \end{align*}
Mathematica [A] time = 4.34034, size = 360, normalized size = 0.83 \[ \frac{(a+b \tan (c+d x))^{n+1} \left (\frac{\frac{\left (\sqrt{-b^2} \left (a^2 b^2 \left (n^2+2 n-6\right )-3 a^4-b^4 \left (n^2-4 n+3\right )\right )+a b^2 n \left (3 a^2+b^2 (5-2 n)\right )\right ) \, _2F_1\left (1,n+1;n+2;\frac{a+b \tan (c+d x)}{a-\sqrt{-b^2}}\right )}{a-\sqrt{-b^2}}+\frac{\left (\sqrt{-b^2} \left (-a^2 b^2 \left (n^2+2 n-6\right )+3 a^4+b^4 \left (n^2-4 n+3\right )\right )+a b^2 n \left (3 a^2+b^2 (5-2 n)\right )\right ) \, _2F_1\left (1,n+1;n+2;\frac{a+b \tan (c+d x)}{a+\sqrt{-b^2}}\right )}{a+\sqrt{-b^2}}}{(n+1) \left (a^2+b^2\right )}-\frac{2 b \cos ^2(c+d x) \left (-a \left (3 a^2+b^2 (5-2 n)\right ) \tan (c+d x)-a^2 b (n+1)+b^3 (n-3)\right )}{a^2+b^2}+4 b \cos ^4(c+d x) (a \tan (c+d x)+b)\right )}{16 b d \left (a^2+b^2\right )} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.328, size = 0, normalized size = 0. \begin{align*} \int \left ( \cos \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} \cos \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} \cos \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} \cos \left (d x + c\right )^{4}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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