Optimal. Leaf size=177 \[ -\frac {a \left (a^2-2 b^2\right ) (a+b \sec (c+d x))^{1+n}}{b^4 d (1+n)}-\frac {\, _2F_1\left (1,1+n;2+n;1+\frac {b \sec (c+d x)}{a}\right ) (a+b \sec (c+d x))^{1+n}}{a d (1+n)}+\frac {\left (3 a^2-2 b^2\right ) (a+b \sec (c+d x))^{2+n}}{b^4 d (2+n)}-\frac {3 a (a+b \sec (c+d x))^{3+n}}{b^4 d (3+n)}+\frac {(a+b \sec (c+d x))^{4+n}}{b^4 d (4+n)} \]
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Rubi [A]
time = 0.14, antiderivative size = 177, normalized size of antiderivative = 1.00, number of steps
used = 5, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3970, 966,
1634, 67} \begin {gather*} -\frac {a \left (a^2-2 b^2\right ) (a+b \sec (c+d x))^{n+1}}{b^4 d (n+1)}+\frac {\left (3 a^2-2 b^2\right ) (a+b \sec (c+d x))^{n+2}}{b^4 d (n+2)}-\frac {3 a (a+b \sec (c+d x))^{n+3}}{b^4 d (n+3)}+\frac {(a+b \sec (c+d x))^{n+4}}{b^4 d (n+4)}-\frac {(a+b \sec (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {b \sec (c+d x)}{a}+1\right )}{a d (n+1)} \end {gather*}
Antiderivative was successfully verified.
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Rule 67
Rule 966
Rule 1634
Rule 3970
Rubi steps
\begin {align*} \int (a+b \sec (c+d x))^n \tan ^5(c+d x) \, dx &=\frac {\text {Subst}\left (\int \frac {(a+x)^n \left (b^2-x^2\right )^2}{x} \, dx,x,b \sec (c+d x)\right )}{b^4 d}\\ &=\frac {(a+b \sec (c+d x))^{4+n}}{b^4 d (4+n)}+\frac {\text {Subst}\left (\int \frac {(a+x)^n \left (b^4 (4+n)-a^3 (4+n) x-\left (3 a^2+2 b^2\right ) (4+n) x^2-3 a (4+n) x^3\right )}{x} \, dx,x,b \sec (c+d x)\right )}{b^4 d (4+n)}\\ &=\frac {(a+b \sec (c+d x))^{4+n}}{b^4 d (4+n)}+\frac {\text {Subst}\left (\int \left (-a \left (a^2-2 b^2\right ) (4+n) (a+x)^n+\frac {\left (4 b^4+b^4 n\right ) (a+x)^n}{x}+\left (3 a^2-2 b^2\right ) (4+n) (a+x)^{1+n}-3 a (4+n) (a+x)^{2+n}\right ) \, dx,x,b \sec (c+d x)\right )}{b^4 d (4+n)}\\ &=-\frac {a \left (a^2-2 b^2\right ) (a+b \sec (c+d x))^{1+n}}{b^4 d (1+n)}+\frac {\left (3 a^2-2 b^2\right ) (a+b \sec (c+d x))^{2+n}}{b^4 d (2+n)}-\frac {3 a (a+b \sec (c+d x))^{3+n}}{b^4 d (3+n)}+\frac {(a+b \sec (c+d x))^{4+n}}{b^4 d (4+n)}+\frac {\text {Subst}\left (\int \frac {(a+x)^n}{x} \, dx,x,b \sec (c+d x)\right )}{d}\\ &=-\frac {a \left (a^2-2 b^2\right ) (a+b \sec (c+d x))^{1+n}}{b^4 d (1+n)}-\frac {\, _2F_1\left (1,1+n;2+n;1+\frac {b \sec (c+d x)}{a}\right ) (a+b \sec (c+d x))^{1+n}}{a d (1+n)}+\frac {\left (3 a^2-2 b^2\right ) (a+b \sec (c+d x))^{2+n}}{b^4 d (2+n)}-\frac {3 a (a+b \sec (c+d x))^{3+n}}{b^4 d (3+n)}+\frac {(a+b \sec (c+d x))^{4+n}}{b^4 d (4+n)}\\ \end {align*}
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Mathematica [A]
time = 3.26, size = 298, normalized size = 1.68 \begin {gather*} -\frac {\left (n (b+a \cos (c+d x)) \left (-6 a^2 b+12 b^3-6 a^2 b n+16 b^3 n+4 b^3 n^2+3 a \left (3 a^2+b^2 \left (-8-n+n^2\right )\right ) \cos (c+d x)+2 b (1+n) \left (-3 a^2+b^2 \left (12+7 n+n^2\right )\right ) \cos (2 (c+d x))+3 a^3 \cos (3 (c+d x))-12 a b^2 \cos (3 (c+d x))-7 a b^2 n \cos (3 (c+d x))-a b^2 n^2 \cos (3 (c+d x))\right )-2 b^4 \left (24+50 n+35 n^2+10 n^3+n^4\right ) \cos ^4(c+d x) \, _2F_1\left (1,-n;1-n;\frac {a \cos (c+d x)}{b+a \cos (c+d x)}\right )\right ) \sec ^8\left (\frac {1}{2} (c+d x)\right ) (a+b \sec (c+d x))^n}{2 b^4 d n (1+n) (2+n) (3+n) (4+n) \left (-1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^4} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.12, size = 0, normalized size = 0.00 \[\int \left (a +b \sec \left (d x +c \right )\right )^{n} \left (\tan ^{5}\left (d x +c \right )\right )\, dx\]
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 (a + b \sec {\left (c + d x \right )}\right )^{n} \tan ^{5}{\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 {\mathrm {tan}\left (c+d\,x\right )}^5\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^n \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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