Optimal. Leaf size=88 \[ \frac {i 2^{\frac {1}{2}+n} a \, _2F_1\left (\frac {1}{2},\frac {1}{2}-n;\frac {3}{2};\frac {1}{2} (1-i \tan (c+d x))\right ) \sec (c+d x) (1+i \tan (c+d x))^{\frac {1}{2}-n} (a+i a \tan (c+d x))^{-1+n}}{d} \]
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Rubi [A]
time = 0.11, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3586, 3604, 72,
71} \begin {gather*} \frac {i a 2^{n+\frac {1}{2}} \sec (c+d x) (1+i \tan (c+d x))^{\frac {1}{2}-n} (a+i a \tan (c+d x))^{n-1} \, _2F_1\left (\frac {1}{2},\frac {1}{2}-n;\frac {3}{2};\frac {1}{2} (1-i \tan (c+d x))\right )}{d} \end {gather*}
Antiderivative was successfully verified.
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Rule 71
Rule 72
Rule 3586
Rule 3604
Rubi steps
\begin {align*} \int \sec (c+d x) (a+i a \tan (c+d x))^n \, dx &=\frac {\sec (c+d x) \int \sqrt {a-i a \tan (c+d x)} (a+i a \tan (c+d x))^{\frac {1}{2}+n} \, dx}{\sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}\\ &=\frac {\left (a^2 \sec (c+d x)\right ) \text {Subst}\left (\int \frac {(a+i a x)^{-\frac {1}{2}+n}}{\sqrt {a-i a x}} \, dx,x,\tan (c+d x)\right )}{d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}\\ &=\frac {\left (2^{-\frac {1}{2}+n} a^2 \sec (c+d x) (a+i a \tan (c+d x))^{-1+n} \left (\frac {a+i a \tan (c+d x)}{a}\right )^{\frac {1}{2}-n}\right ) \text {Subst}\left (\int \frac {\left (\frac {1}{2}+\frac {i x}{2}\right )^{-\frac {1}{2}+n}}{\sqrt {a-i a x}} \, dx,x,\tan (c+d x)\right )}{d \sqrt {a-i a \tan (c+d x)}}\\ &=\frac {i 2^{\frac {1}{2}+n} a \, _2F_1\left (\frac {1}{2},\frac {1}{2}-n;\frac {3}{2};\frac {1}{2} (1-i \tan (c+d x))\right ) \sec (c+d x) (1+i \tan (c+d x))^{\frac {1}{2}-n} (a+i a \tan (c+d x))^{-1+n}}{d}\\ \end {align*}
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Mathematica [A]
time = 8.82, size = 146, normalized size = 1.66 \begin {gather*} -\frac {i 2^{1+n} \left (e^{i d x}\right )^n \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{1+n} \left (1+e^{2 i (c+d x)}\right )^{1+n} \, _2F_1\left (\frac {1}{2}+n,1+n;\frac {3}{2}+n;-e^{2 i (c+d x)}\right ) \sec ^{-n}(c+d x) (\cos (d x)+i \sin (d x))^{-n} (a+i a \tan (c+d x))^n}{d (1+2 n)} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.26, size = 0, normalized size = 0.00 \[\int \sec \left (d x +c \right ) \left (a +i a \tan \left (d x +c \right )\right )^{n}\, dx\]
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 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{n} \sec {\left (c + d x \right )}\, dx \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^n}{\cos \left (c+d\,x\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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