Optimal. Leaf size=299 \[ \frac {a^2 F_1\left (\frac {1}{2};\frac {1}{2} (-3+n),2;\frac {3}{2};\sin ^2(e+f x),\frac {a^2 \sin ^2(e+f x)}{a^2-b^2}\right ) \cos (e+f x) \cos ^2(e+f x)^{\frac {1}{2} (-1+n)} (d \sec (e+f x))^n \sin (e+f x)}{\left (a^2-b^2\right )^2 f}+\frac {b^2 F_1\left (\frac {1}{2};\frac {1}{2} (-1+n),2;\frac {3}{2};\sin ^2(e+f x),\frac {a^2 \sin ^2(e+f x)}{a^2-b^2}\right ) \cos (e+f x) \cos ^2(e+f x)^{\frac {1}{2} (-1+n)} (d \sec (e+f x))^n \sin (e+f x)}{\left (a^2-b^2\right )^2 f}-\frac {2 a b F_1\left (\frac {1}{2};\frac {1}{2} (-2+n),2;\frac {3}{2};\sin ^2(e+f x),\frac {a^2 \sin ^2(e+f x)}{a^2-b^2}\right ) \cos ^2(e+f x)^{n/2} (d \sec (e+f x))^n \sin (e+f x)}{\left (a^2-b^2\right )^2 f} \]
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Rubi [A]
time = 0.32, antiderivative size = 299, normalized size of antiderivative = 1.00, number of steps
used = 9, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3954, 2903,
3268, 440} \begin {gather*} \frac {a^2 \sin (e+f x) \cos (e+f x) \cos ^2(e+f x)^{\frac {n-1}{2}} (d \sec (e+f x))^n F_1\left (\frac {1}{2};\frac {n-3}{2},2;\frac {3}{2};\sin ^2(e+f x),\frac {a^2 \sin ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^2}+\frac {b^2 \sin (e+f x) \cos (e+f x) \cos ^2(e+f x)^{\frac {n-1}{2}} (d \sec (e+f x))^n F_1\left (\frac {1}{2};\frac {n-1}{2},2;\frac {3}{2};\sin ^2(e+f x),\frac {a^2 \sin ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^2}-\frac {2 a b \sin (e+f x) \cos ^2(e+f x)^{n/2} (d \sec (e+f x))^n F_1\left (\frac {1}{2};\frac {n-2}{2},2;\frac {3}{2};\sin ^2(e+f x),\frac {a^2 \sin ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^2} \end {gather*}
Antiderivative was successfully verified.
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Rule 440
Rule 2903
Rule 3268
Rule 3954
Rubi steps
\begin {align*} \int \frac {(d \sec (e+f x))^n}{(a+b \sec (e+f x))^2} \, dx &=\left (\cos ^n(e+f x) (d \sec (e+f x))^n\right ) \int \frac {\cos ^{2-n}(e+f x)}{(b+a \cos (e+f x))^2} \, dx\\ &=\left (\cos ^n(e+f x) (d \sec (e+f x))^n\right ) \int \left (\frac {b^2 \cos ^{2-n}(e+f x)}{\left (b^2-a^2 \cos ^2(e+f x)\right )^2}-\frac {2 a b \cos ^{3-n}(e+f x)}{\left (b^2-a^2 \cos ^2(e+f x)\right )^2}+\frac {a^2 \cos ^{4-n}(e+f x)}{\left (-b^2+a^2 \cos ^2(e+f x)\right )^2}\right ) \, dx\\ &=\left (a^2 \cos ^n(e+f x) (d \sec (e+f x))^n\right ) \int \frac {\cos ^{4-n}(e+f x)}{\left (-b^2+a^2 \cos ^2(e+f x)\right )^2} \, dx-\left (2 a b \cos ^n(e+f x) (d \sec (e+f x))^n\right ) \int \frac {\cos ^{3-n}(e+f x)}{\left (b^2-a^2 \cos ^2(e+f x)\right )^2} \, dx+\left (b^2 \cos ^n(e+f x) (d \sec (e+f x))^n\right ) \int \frac {\cos ^{2-n}(e+f x)}{\left (b^2-a^2 \cos ^2(e+f x)\right )^2} \, dx\\ &=\frac {\left (a^2 \cos ^{2 \left (\frac {1}{2}-\frac {n}{2}\right )+n}(e+f x) \cos ^2(e+f x)^{-\frac {1}{2}+\frac {n}{2}} (d \sec (e+f x))^n\right ) \text {Subst}\left (\int \frac {\left (1-x^2\right )^{\frac {3-n}{2}}}{\left (a^2-b^2-a^2 x^2\right )^2} \, dx,x,\sin (e+f x)\right )}{f}+\frac {\left (b^2 \cos ^{2 \left (\frac {1}{2}-\frac {n}{2}\right )+n}(e+f x) \cos ^2(e+f x)^{-\frac {1}{2}+\frac {n}{2}} (d \sec (e+f x))^n\right ) \text {Subst}\left (\int \frac {\left (1-x^2\right )^{\frac {1-n}{2}}}{\left (-a^2+b^2+a^2 x^2\right )^2} \, dx,x,\sin (e+f x)\right )}{f}-\frac {\left (2 a b \cos ^2(e+f x)^{n/2} (d \sec (e+f x))^n\right ) \text {Subst}\left (\int \frac {\left (1-x^2\right )^{\frac {2-n}{2}}}{\left (-a^2+b^2+a^2 x^2\right )^2} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac {a^2 F_1\left (\frac {1}{2};\frac {1}{2} (-3+n),2;\frac {3}{2};\sin ^2(e+f x),\frac {a^2 \sin ^2(e+f x)}{a^2-b^2}\right ) \cos (e+f x) \cos ^2(e+f x)^{\frac {1}{2} (-1+n)} (d \sec (e+f x))^n \sin (e+f x)}{\left (a^2-b^2\right )^2 f}+\frac {b^2 F_1\left (\frac {1}{2};\frac {1}{2} (-1+n),2;\frac {3}{2};\sin ^2(e+f x),\frac {a^2 \sin ^2(e+f x)}{a^2-b^2}\right ) \cos (e+f x) \cos ^2(e+f x)^{\frac {1}{2} (-1+n)} (d \sec (e+f x))^n \sin (e+f x)}{\left (a^2-b^2\right )^2 f}-\frac {2 a b F_1\left (\frac {1}{2};\frac {1}{2} (-2+n),2;\frac {3}{2};\sin ^2(e+f x),\frac {a^2 \sin ^2(e+f x)}{a^2-b^2}\right ) \cos ^2(e+f x)^{n/2} (d \sec (e+f x))^n \sin (e+f x)}{\left (a^2-b^2\right )^2 f}\\ \end {align*}
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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(13816\) vs. \(2(299)=598\).
time = 40.89, size = 13816, normalized size = 46.21 \begin {gather*} \text {Result too large to show} \end {gather*}
Warning: Unable to verify antiderivative.
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Maple [F]
time = 0.07, size = 0, normalized size = 0.00 \[\int \frac {\left (d \sec \left (f x +e \right )\right )^{n}}{\left (a +b \sec \left (f x +e \right )\right )^{2}}\, dx\]
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 \frac {\left (d \sec {\left (e + f x \right )}\right )^{n}}{\left (a + b \sec {\left (e + f x \right )}\right )^{2}}\, 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 = -0.00 \begin {gather*} \int \frac {{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^n}{{\left (a+\frac {b}{\cos \left (e+f\,x\right )}\right )}^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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