Optimal. Leaf size=299 \[ \frac{a^2 \sin (e+f x) \cos (e+f x) \cos ^2(e+f x)^{\frac{n-1}{2}} (d \sec (e+f x))^n F_1\left (\frac{1}{2};\frac{n-3}{2},2;\frac{3}{2};\sin ^2(e+f x),\frac{a^2 \sin ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^2}+\frac{b^2 \sin (e+f x) \cos (e+f x) \cos ^2(e+f x)^{\frac{n-1}{2}} (d \sec (e+f x))^n F_1\left (\frac{1}{2};\frac{n-1}{2},2;\frac{3}{2};\sin ^2(e+f x),\frac{a^2 \sin ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^2}-\frac{2 a b \sin (e+f x) \cos ^2(e+f x)^{n/2} (d \sec (e+f x))^n F_1\left (\frac{1}{2};\frac{n-2}{2},2;\frac{3}{2};\sin ^2(e+f x),\frac{a^2 \sin ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^2} \]
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Rubi [A] time = 0.443625, antiderivative size = 299, normalized size of antiderivative = 1., 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 = {3869, 2824, 3189, 429} \[ \frac{a^2 \sin (e+f x) \cos (e+f x) \cos ^2(e+f x)^{\frac{n-1}{2}} (d \sec (e+f x))^n F_1\left (\frac{1}{2};\frac{n-3}{2},2;\frac{3}{2};\sin ^2(e+f x),\frac{a^2 \sin ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^2}+\frac{b^2 \sin (e+f x) \cos (e+f x) \cos ^2(e+f x)^{\frac{n-1}{2}} (d \sec (e+f x))^n F_1\left (\frac{1}{2};\frac{n-1}{2},2;\frac{3}{2};\sin ^2(e+f x),\frac{a^2 \sin ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^2}-\frac{2 a b \sin (e+f x) \cos ^2(e+f x)^{n/2} (d \sec (e+f x))^n F_1\left (\frac{1}{2};\frac{n-2}{2},2;\frac{3}{2};\sin ^2(e+f x),\frac{a^2 \sin ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^2} \]
Antiderivative was successfully verified.
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Rule 3869
Rule 2824
Rule 3189
Rule 429
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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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*}
Mathematica [B] time = 32.7371, size = 10428, normalized size = 34.88 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.244, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( d\sec \left ( fx+e \right ) \right ) ^{n}}{ \left ( a+b\sec \left ( fx+e \right ) \right ) ^{2}}}\, 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 \frac{\left (d \sec \left (f x + e\right )\right )^{n}}{{\left (b \sec \left (f x + e\right ) + a\right )}^{2}}\,{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 (\frac{\left (d \sec \left (f x + e\right )\right )^{n}}{b^{2} \sec \left (f x + e\right )^{2} + 2 \, a b \sec \left (f x + e\right ) + a^{2}}, 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 \frac{\left (d \sec{\left (e + f x \right )}\right )^{n}}{\left (a + b \sec{\left (e + f x \right )}\right )^{2}}\, 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 \frac{\left (d \sec \left (f x + e\right )\right )^{n}}{{\left (b \sec \left (f x + e\right ) + a\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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