Optimal. Leaf size=159 \[ \frac{\cos (c+d x) \sqrt{1-\frac{a+b \tan (c+d x)}{a-\sqrt{-b^2}}} \sqrt{1-\frac{a+b \tan (c+d x)}{a+\sqrt{-b^2}}} (a+b \tan (c+d x))^{n+1} F_1\left (n+1;\frac{1}{2},\frac{1}{2};n+2;\frac{a+b \tan (c+d x)}{a-\sqrt{-b^2}},\frac{a+b \tan (c+d x)}{a+\sqrt{-b^2}}\right )}{b d (n+1)} \]
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Rubi [A] time = 0.113323, antiderivative size = 159, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {3512, 760, 133} \[ \frac{\cos (c+d x) \sqrt{1-\frac{a+b \tan (c+d x)}{a-\sqrt{-b^2}}} \sqrt{1-\frac{a+b \tan (c+d x)}{a+\sqrt{-b^2}}} (a+b \tan (c+d x))^{n+1} F_1\left (n+1;\frac{1}{2},\frac{1}{2};n+2;\frac{a+b \tan (c+d x)}{a-\sqrt{-b^2}},\frac{a+b \tan (c+d x)}{a+\sqrt{-b^2}}\right )}{b d (n+1)} \]
Antiderivative was successfully verified.
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Rule 3512
Rule 760
Rule 133
Rubi steps
\begin{align*} \int \sec (c+d x) (a+b \tan (c+d x))^n \, dx &=\frac{\sec (c+d x) \operatorname{Subst}\left (\int \frac{(a+x)^n}{\sqrt{1+\frac{x^2}{b^2}}} \, dx,x,b \tan (c+d x)\right )}{b d \sqrt{\sec ^2(c+d x)}}\\ &=\frac{\left (\cos (c+d x) \sqrt{1-\frac{a+b \tan (c+d x)}{a-\frac{b^2}{\sqrt{-b^2}}}} \sqrt{1-\frac{a+b \tan (c+d x)}{a+\frac{b^2}{\sqrt{-b^2}}}}\right ) \operatorname{Subst}\left (\int \frac{x^n}{\sqrt{1-\frac{x}{a-\sqrt{-b^2}}} \sqrt{1-\frac{x}{a+\sqrt{-b^2}}}} \, dx,x,a+b \tan (c+d x)\right )}{b d}\\ &=\frac{F_1\left (1+n;\frac{1}{2},\frac{1}{2};2+n;\frac{a+b \tan (c+d x)}{a-\sqrt{-b^2}},\frac{a+b \tan (c+d x)}{a+\sqrt{-b^2}}\right ) \cos (c+d x) (a+b \tan (c+d x))^{1+n} \sqrt{1-\frac{a+b \tan (c+d x)}{a-\sqrt{-b^2}}} \sqrt{1-\frac{a+b \tan (c+d x)}{a+\sqrt{-b^2}}}}{b d (1+n)}\\ \end{align*}
Mathematica [C] time = 3.75919, size = 340, normalized size = 2.14 \[ \frac{2 (n+2) \left (a^2+b^2\right )^2 \cos ^3(c+d x) (\tan (c+d x)-i) (\tan (c+d x)+i) (a+b \tan (c+d x))^{n+1} F_1\left (n+1;\frac{1}{2},\frac{1}{2};n+2;\frac{a+b \tan (c+d x)}{a-i b},\frac{a+b \tan (c+d x)}{a+i b}\right )}{b d (n+1) (a-i b) (a+i b) \left (2 (n+2) \left (a^2+b^2\right ) F_1\left (n+1;\frac{1}{2},\frac{1}{2};n+2;\frac{a+b \tan (c+d x)}{a-i b},\frac{a+b \tan (c+d x)}{a+i b}\right )+(a+b \tan (c+d x)) \left ((a-i b) F_1\left (n+2;\frac{1}{2},\frac{3}{2};n+3;\frac{a+b \tan (c+d x)}{a-i b},\frac{a+b \tan (c+d x)}{a+i b}\right )+(a+i b) F_1\left (n+2;\frac{3}{2},\frac{1}{2};n+3;\frac{a+b \tan (c+d x)}{a-i b},\frac{a+b \tan (c+d x)}{a+i b}\right )\right )\right )} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.346, size = 0, normalized size = 0. \begin{align*} \int \sec \left ( dx+c \right ) \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} \sec \left (d x + c\right )\,{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} \sec \left (d x + c\right ), x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \tan{\left (c + d x \right )}\right )^{n} \sec{\left (c + d x \right )}\, dx \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} \sec \left (d x + c\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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