Optimal. Leaf size=38 \[ \frac {A \sec ^{-n}(c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d (1+n)} \]
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Rubi [A]
time = 0.66, antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps
used = 16, number of rules used = 6, integrand size = 102, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {4108, 3913,
3910, 134, 138, 4171} \begin {gather*} \frac {A \sin (c+d x) \sec ^{-n}(c+d x) (a \sec (c+d x)+a)^n}{d (n+1)} \end {gather*}
Antiderivative was successfully verified.
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Rule 134
Rule 138
Rule 3910
Rule 3913
Rule 4108
Rule 4171
Rubi steps
\begin {align*} \int \left (\frac {\sec ^{-n}(c+d x) (a+a \sec (c+d x))^n (-a (B+A n+B n)-a C (1+n) \sec (c+d x))}{a (1+n)}+\sec ^{-1-n}(c+d x) (a+a \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )\right ) \, dx &=\frac {\int \sec ^{-n}(c+d x) (a+a \sec (c+d x))^n (-a (B+A n+B n)-a C (1+n) \sec (c+d x)) \, dx}{a (1+n)}+\int \sec ^{-1-n}(c+d x) (a+a \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\\ &=\frac {A \sec ^{-n}(c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d (1+n)}-\frac {C \int \sec ^{-n}(c+d x) (a+a \sec (c+d x))^{1+n} \, dx}{a}+\frac {\int \sec ^{-n}(c+d x) (a+a \sec (c+d x))^n (a (B+A n+B n)+a C (1+n) \sec (c+d x)) \, dx}{a (1+n)}-\left (B-C+\frac {A n}{1+n}\right ) \int \sec ^{-n}(c+d x) (a+a \sec (c+d x))^n \, dx\\ &=\frac {A \sec ^{-n}(c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d (1+n)}+\frac {C \int \sec ^{-n}(c+d x) (a+a \sec (c+d x))^{1+n} \, dx}{a}+\frac {(B+A n+B n-C (1+n)) \int \sec ^{-n}(c+d x) (a+a \sec (c+d x))^n \, dx}{1+n}-\left (C (1+\sec (c+d x))^{-n} (a+a \sec (c+d x))^n\right ) \int \sec ^{-n}(c+d x) (1+\sec (c+d x))^{1+n} \, dx-\left (\left (B-C+\frac {A n}{1+n}\right ) (1+\sec (c+d x))^{-n} (a+a \sec (c+d x))^n\right ) \int \sec ^{-n}(c+d x) (1+\sec (c+d x))^n \, dx\\ &=\frac {A \sec ^{-n}(c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d (1+n)}+\left (C (1+\sec (c+d x))^{-n} (a+a \sec (c+d x))^n\right ) \int \sec ^{-n}(c+d x) (1+\sec (c+d x))^{1+n} \, dx+\frac {\left ((B+A n+B n-C (1+n)) (1+\sec (c+d x))^{-n} (a+a \sec (c+d x))^n\right ) \int \sec ^{-n}(c+d x) (1+\sec (c+d x))^n \, dx}{1+n}-\frac {\left (C (1+\sec (c+d x))^{-\frac {1}{2}-n} (a+a \sec (c+d x))^n \tan (c+d x)\right ) \text {Subst}\left (\int \frac {(1-x)^{-1-n} (2-x)^{\frac {1}{2}+n}}{\sqrt {x}} \, dx,x,1-\sec (c+d x)\right )}{d \sqrt {1-\sec (c+d x)}}-\frac {\left (\left (B-C+\frac {A n}{1+n}\right ) (1+\sec (c+d x))^{-\frac {1}{2}-n} (a+a \sec (c+d x))^n \tan (c+d x)\right ) \text {Subst}\left (\int \frac {(1-x)^{-1-n} (2-x)^{-\frac {1}{2}+n}}{\sqrt {x}} \, dx,x,1-\sec (c+d x)\right )}{d \sqrt {1-\sec (c+d x)}}\\ &=\frac {A \sec ^{-n}(c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d (1+n)}-\frac {\left (B-C+\frac {A n}{1+n}\right ) \, _2F_1\left (\frac {1}{2}-n,-n;1-n;-\frac {2 \sec (c+d x)}{1-\sec (c+d x)}\right ) \sec ^{1-n}(c+d x) \left (\frac {1+\sec (c+d x)}{1-\sec (c+d x)}\right )^{\frac {1}{2}-n} (a+a \sec (c+d x))^n \sin (c+d x)}{d n (1+\sec (c+d x))}-\frac {2^{\frac {3}{2}+n} C F_1\left (\frac {1}{2};1+n,-\frac {1}{2}-n;\frac {3}{2};1-\sec (c+d x),\frac {1}{2} (1-\sec (c+d x))\right ) (1+\sec (c+d x))^{-\frac {1}{2}-n} (a+a \sec (c+d x))^n \tan (c+d x)}{d}+\frac {\left (C (1+\sec (c+d x))^{-\frac {1}{2}-n} (a+a \sec (c+d x))^n \tan (c+d x)\right ) \text {Subst}\left (\int \frac {(1-x)^{-1-n} (2-x)^{\frac {1}{2}+n}}{\sqrt {x}} \, dx,x,1-\sec (c+d x)\right )}{d \sqrt {1-\sec (c+d x)}}+\frac {\left ((B+A n+B n-C (1+n)) (1+\sec (c+d x))^{-\frac {1}{2}-n} (a+a \sec (c+d x))^n \tan (c+d x)\right ) \text {Subst}\left (\int \frac {(1-x)^{-1-n} (2-x)^{-\frac {1}{2}+n}}{\sqrt {x}} \, dx,x,1-\sec (c+d x)\right )}{d (1+n) \sqrt {1-\sec (c+d x)}}\\ &=\frac {A \sec ^{-n}(c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d (1+n)}-\frac {\left (B-C+\frac {A n}{1+n}\right ) \, _2F_1\left (\frac {1}{2}-n,-n;1-n;-\frac {2 \sec (c+d x)}{1-\sec (c+d x)}\right ) \sec ^{1-n}(c+d x) \left (\frac {1+\sec (c+d x)}{1-\sec (c+d x)}\right )^{\frac {1}{2}-n} (a+a \sec (c+d x))^n \sin (c+d x)}{d n (1+\sec (c+d x))}+\frac {(B+A n+B n-C (1+n)) \, _2F_1\left (\frac {1}{2}-n,-n;1-n;-\frac {2 \sec (c+d x)}{1-\sec (c+d x)}\right ) \sec ^{1-n}(c+d x) \left (\frac {1+\sec (c+d x)}{1-\sec (c+d x)}\right )^{\frac {1}{2}-n} (a+a \sec (c+d x))^n \sin (c+d x)}{d n (1+n) (1+\sec (c+d x))}\\ \end {align*}
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Mathematica [A]
time = 0.17, size = 38, normalized size = 1.00 \begin {gather*} \frac {A \sec ^{-n}(c+d x) (a (1+\sec (c+d x)))^n \sin (c+d x)}{d (1+n)} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.41, size = 0, normalized size = 0.00 \[\int \frac {\left (a +a \sec \left (d x +c \right )\right )^{n} \left (-a \left (A n +B n +B \right )-a C \left (1+n \right ) \sec \left (d x +c \right )\right ) \left (\sec ^{-n}\left (d x +c \right )\right )}{a \left (1+n \right )}+\left (\sec ^{-1-n}\left (d x +c \right )\right ) \left (a +a \sec \left (d x +c \right )\right )^{n} \left (A +B \sec \left (d x +c \right )+C \left (\sec ^{2}\left (d x +c \right )\right )\right )\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 310 vs.
\(2 (38) = 76\).
time = 4.30, size = 310, normalized size = 8.16 \begin {gather*} \frac {{\left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )}^{n} A a^{n} \cos \left (-{\left (d n + d\right )} x + 2 \, n \arctan \left (\sin \left (d x + c\right ), \cos \left (d x + c\right ) + 1\right ) - c\right ) \sin \left (c n\right ) - {\left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )}^{n} A a^{n} \cos \left (-{\left (d n - d\right )} x + 2 \, n \arctan \left (\sin \left (d x + c\right ), \cos \left (d x + c\right ) + 1\right ) + c\right ) \sin \left (c n\right ) - {\left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )}^{n} A a^{n} \cos \left (c n\right ) \sin \left (-{\left (d n + d\right )} x + 2 \, n \arctan \left (\sin \left (d x + c\right ), \cos \left (d x + c\right ) + 1\right ) - c\right ) + {\left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )}^{n} A a^{n} \cos \left (c n\right ) \sin \left (-{\left (d n - d\right )} x + 2 \, n \arctan \left (\sin \left (d x + c\right ), \cos \left (d x + c\right ) + 1\right ) + c\right )}{2 \, {\left ({\left (d n + d\right )} 2^{n} \cos \left (c n\right )^{2} + {\left (d n + d\right )} 2^{n} \sin \left (c n\right )^{2}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 3.04, size = 58, normalized size = 1.53 \begin {gather*} \frac {A \left (\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}\right )^{n} \frac {1}{\cos \left (d x + c\right )}^{-n - 1} \sin \left (d x + c\right )}{{\left (d n + d\right )} \cos \left (d x + c\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \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 = -0.03 \begin {gather*} \int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^n\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{n+1}}-\frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^n\,\left (a\,\left (B+A\,n+B\,n\right )+\frac {C\,a\,\left (n+1\right )}{\cos \left (c+d\,x\right )}\right )}{a\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^n\,\left (n+1\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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