Optimal. Leaf size=394 \[ -\frac {\tan (e+f x)}{2 a (c-d) f (1+\sec (e+f x)) \sqrt {a+a \sec (e+f x)}}+\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right ) \tan (e+f x)}{\sqrt {a} c f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {\sqrt {2} (c-2 d) \tanh ^{-1}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {2} \sqrt {a}}\right ) \tan (e+f x)}{\sqrt {a} (c-d)^2 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {2} \sqrt {a}}\right ) \tan (e+f x)}{2 \sqrt {2} \sqrt {a} (c-d) f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {2 d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right ) \tan (e+f x)}{\sqrt {a} c (c-d)^2 \sqrt {c+d} f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}} \]
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Rubi [A]
time = 0.23, antiderivative size = 394, normalized size of antiderivative = 1.00, number of steps
used = 12, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {4025, 186, 65,
212, 44, 214} \begin {gather*} -\frac {2 d^{5/2} \tan (e+f x) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right )}{\sqrt {a} c f (c-d)^2 \sqrt {c+d} \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}-\frac {\tan (e+f x)}{2 a f (c-d) (\sec (e+f x)+1) \sqrt {a \sec (e+f x)+a}}-\frac {\tan (e+f x) \tanh ^{-1}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} \sqrt {a} f (c-d) \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}-\frac {\sqrt {2} (c-2 d) \tan (e+f x) \tanh ^{-1}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {a} f (c-d)^2 \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}+\frac {2 \tan (e+f x) \tanh ^{-1}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right )}{\sqrt {a} c f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}} \end {gather*}
Antiderivative was successfully verified.
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Rule 44
Rule 65
Rule 186
Rule 212
Rule 214
Rule 4025
Rubi steps
\begin {align*} \int \frac {1}{(a+a \sec (e+f x))^{3/2} (c+d \sec (e+f x))} \, dx &=-\frac {\left (a^2 \tan (e+f x)\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {a-a x} (a+a x)^2 (c+d x)} \, dx,x,\sec (e+f x)\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\\ &=-\frac {\left (a^2 \tan (e+f x)\right ) \text {Subst}\left (\int \left (\frac {1}{a^2 c x \sqrt {a-a x}}-\frac {1}{a^2 (c-d) (1+x)^2 \sqrt {a-a x}}+\frac {-c+2 d}{a^2 (c-d)^2 (1+x) \sqrt {a-a x}}-\frac {d^3}{a^2 c (c-d)^2 \sqrt {a-a x} (c+d x)}\right ) \, dx,x,\sec (e+f x)\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\\ &=-\frac {\tan (e+f x) \text {Subst}\left (\int \frac {1}{x \sqrt {a-a x}} \, dx,x,\sec (e+f x)\right )}{c f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}+\frac {((c-2 d) \tan (e+f x)) \text {Subst}\left (\int \frac {1}{(1+x) \sqrt {a-a x}} \, dx,x,\sec (e+f x)\right )}{(c-d)^2 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}+\frac {\tan (e+f x) \text {Subst}\left (\int \frac {1}{(1+x)^2 \sqrt {a-a x}} \, dx,x,\sec (e+f x)\right )}{(c-d) f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}+\frac {\left (d^3 \tan (e+f x)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a-a x} (c+d x)} \, dx,x,\sec (e+f x)\right )}{c (c-d)^2 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\\ &=-\frac {\tan (e+f x)}{2 a (c-d) f (1+\sec (e+f x)) \sqrt {a+a \sec (e+f x)}}+\frac {(2 \tan (e+f x)) \text {Subst}\left (\int \frac {1}{1-\frac {x^2}{a}} \, dx,x,\sqrt {a-a \sec (e+f x)}\right )}{a c f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {(2 (c-2 d) \tan (e+f x)) \text {Subst}\left (\int \frac {1}{2-\frac {x^2}{a}} \, dx,x,\sqrt {a-a \sec (e+f x)}\right )}{a (c-d)^2 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}+\frac {\tan (e+f x) \text {Subst}\left (\int \frac {1}{(1+x) \sqrt {a-a x}} \, dx,x,\sec (e+f x)\right )}{4 (c-d) f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {\left (2 d^3 \tan (e+f x)\right ) \text {Subst}\left (\int \frac {1}{c+d-\frac {d x^2}{a}} \, dx,x,\sqrt {a-a \sec (e+f x)}\right )}{a c (c-d)^2 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\\ &=-\frac {\tan (e+f x)}{2 a (c-d) f (1+\sec (e+f x)) \sqrt {a+a \sec (e+f x)}}+\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right ) \tan (e+f x)}{\sqrt {a} c f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {\sqrt {2} (c-2 d) \tanh ^{-1}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {2} \sqrt {a}}\right ) \tan (e+f x)}{\sqrt {a} (c-d)^2 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {2 d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right ) \tan (e+f x)}{\sqrt {a} c (c-d)^2 \sqrt {c+d} f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {\tan (e+f x) \text {Subst}\left (\int \frac {1}{2-\frac {x^2}{a}} \, dx,x,\sqrt {a-a \sec (e+f x)}\right )}{2 a (c-d) f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\\ &=-\frac {\tan (e+f x)}{2 a (c-d) f (1+\sec (e+f x)) \sqrt {a+a \sec (e+f x)}}+\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right ) \tan (e+f x)}{\sqrt {a} c f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {\sqrt {2} (c-2 d) \tanh ^{-1}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {2} \sqrt {a}}\right ) \tan (e+f x)}{\sqrt {a} (c-d)^2 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {2} \sqrt {a}}\right ) \tan (e+f x)}{2 \sqrt {2} \sqrt {a} (c-d) f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {2 d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right ) \tan (e+f x)}{\sqrt {a} c (c-d)^2 \sqrt {c+d} f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\\ \end {align*}
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Mathematica [C] Result contains higher order function than in optimal. Order 4 vs. order 3 in
optimal.
time = 35.32, size = 378865, normalized size = 961.59 \begin {gather*} \text {Result too large to show} \end {gather*}
Warning: Unable to verify antiderivative.
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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(2079\) vs.
\(2(334)=668\).
time = 5.36, size = 2080, normalized size = 5.28
method | result | size |
default | \(\text {Expression too large to display}\) | \(2080\) |
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(-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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a \left (\sec {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}} \left (c + d \sec {\left (e + f x \right )}\right )}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \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 {1}{{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{3/2}\,\left (c+\frac {d}{\cos \left (e+f\,x\right )}\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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