Optimal. Leaf size=93 \[ -\frac {\tan ^{-1}\left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{4 \sqrt {b} f}-\frac {\tanh ^{-1}\left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{4 \sqrt {b} f}-\frac {\cot ^2(e+f x) \sqrt {b \sec (e+f x)}}{2 b f} \]
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Rubi [A]
time = 0.05, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps
used = 6, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2702, 294, 335,
218, 212, 209} \begin {gather*} -\frac {\text {ArcTan}\left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{4 \sqrt {b} f}-\frac {\cot ^2(e+f x) \sqrt {b \sec (e+f x)}}{2 b f}-\frac {\tanh ^{-1}\left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{4 \sqrt {b} f} \end {gather*}
Antiderivative was successfully verified.
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Rule 209
Rule 212
Rule 218
Rule 294
Rule 335
Rule 2702
Rubi steps
\begin {align*} \int \frac {\csc ^3(e+f x)}{\sqrt {b \sec (e+f x)}} \, dx &=\frac {\text {Subst}\left (\int \frac {x^{3/2}}{\left (-1+\frac {x^2}{b^2}\right )^2} \, dx,x,b \sec (e+f x)\right )}{b^3 f}\\ &=-\frac {\cot ^2(e+f x) \sqrt {b \sec (e+f x)}}{2 b f}+\frac {\text {Subst}\left (\int \frac {1}{\sqrt {x} \left (-1+\frac {x^2}{b^2}\right )} \, dx,x,b \sec (e+f x)\right )}{4 b f}\\ &=-\frac {\cot ^2(e+f x) \sqrt {b \sec (e+f x)}}{2 b f}+\frac {\text {Subst}\left (\int \frac {1}{-1+\frac {x^4}{b^2}} \, dx,x,\sqrt {b \sec (e+f x)}\right )}{2 b f}\\ &=-\frac {\cot ^2(e+f x) \sqrt {b \sec (e+f x)}}{2 b f}-\frac {\text {Subst}\left (\int \frac {1}{b-x^2} \, dx,x,\sqrt {b \sec (e+f x)}\right )}{4 f}-\frac {\text {Subst}\left (\int \frac {1}{b+x^2} \, dx,x,\sqrt {b \sec (e+f x)}\right )}{4 f}\\ &=-\frac {\tan ^{-1}\left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{4 \sqrt {b} f}-\frac {\tanh ^{-1}\left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{4 \sqrt {b} f}-\frac {\cot ^2(e+f x) \sqrt {b \sec (e+f x)}}{2 b f}\\ \end {align*}
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Mathematica [A]
time = 0.86, size = 93, normalized size = 1.00 \begin {gather*} \frac {\left (-2 \tan ^{-1}\left (\sqrt {\sec (e+f x)}\right )+\log \left (1-\sqrt {\sec (e+f x)}\right )-\log \left (1+\sqrt {\sec (e+f x)}\right )-\frac {4 \csc ^2(e+f x)}{\sec ^{\frac {3}{2}}(e+f x)}\right ) \sqrt {\sec (e+f x)}}{8 f \sqrt {b \sec (e+f x)}} \end {gather*}
Antiderivative was successfully verified.
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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(424\) vs.
\(2(73)=146\).
time = 0.25, size = 425, normalized size = 4.57
method | result | size |
default | \(-\frac {\left (-1+\cos \left (f x +e \right )\right ) \left (8 \left (\cos ^{2}\left (f x +e \right )\right ) \left (-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}\right )^{\frac {3}{2}}+16 \cos \left (f x +e \right ) \left (-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}\right )^{\frac {3}{2}}+8 \left (-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}\right )^{\frac {3}{2}}-4 \left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\left (\cos ^{2}\left (f x +e \right )\right ) \arctan \left (\frac {1}{2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\right )-\left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (-\frac {2 \left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\left (\cos ^{2}\left (f x +e \right )\right )+2 \cos \left (f x +e \right )-2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-1}{\sin \left (f x +e \right )^{2}}\right )+4 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+\arctan \left (\frac {1}{2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\right )+\ln \left (-\frac {2 \left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\left (\cos ^{2}\left (f x +e \right )\right )+2 \cos \left (f x +e \right )-2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-1}{\sin \left (f x +e \right )^{2}}\right )\right )}{8 f \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \sin \left (f x +e \right )^{4} \sqrt {\frac {b}{\cos \left (f x +e \right )}}}\) | \(425\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.52, size = 110, normalized size = 1.18 \begin {gather*} \frac {b {\left (\frac {4 \, \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{b^{2} - \frac {b^{2}}{\cos \left (f x + e\right )^{2}}} - \frac {2 \, \arctan \left (\frac {\sqrt {\frac {b}{\cos \left (f x + e\right )}}}{\sqrt {b}}\right )}{b^{\frac {3}{2}}} + \frac {\log \left (-\frac {\sqrt {b} - \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{\sqrt {b} + \sqrt {\frac {b}{\cos \left (f x + e\right )}}}\right )}{b^{\frac {3}{2}}}\right )}}{8 \, f} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 187 vs.
\(2 (77) = 154\).
time = 0.46, size = 389, normalized size = 4.18 \begin {gather*} \left [\frac {2 \, {\left (\cos \left (f x + e\right )^{2} - 1\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} {\left (\cos \left (f x + e\right ) + 1\right )}}{2 \, b}\right ) + 8 \, \sqrt {\frac {b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )^{2} - {\left (\cos \left (f x + e\right )^{2} - 1\right )} \sqrt {-b} \log \left (\frac {b \cos \left (f x + e\right )^{2} - 4 \, {\left (\cos \left (f x + e\right )^{2} - \cos \left (f x + e\right )\right )} \sqrt {-b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} - 6 \, b \cos \left (f x + e\right ) + b}{\cos \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) + 1}\right )}{16 \, {\left (b f \cos \left (f x + e\right )^{2} - b f\right )}}, \frac {2 \, {\left (\cos \left (f x + e\right )^{2} - 1\right )} \sqrt {b} \arctan \left (\frac {\sqrt {\frac {b}{\cos \left (f x + e\right )}} {\left (\cos \left (f x + e\right ) - 1\right )}}{2 \, \sqrt {b}}\right ) + 8 \, \sqrt {\frac {b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right )^{2} - 1\right )} \sqrt {b} \log \left (\frac {b \cos \left (f x + e\right )^{2} - 4 \, {\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt {b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} + 6 \, b \cos \left (f x + e\right ) + b}{\cos \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right ) + 1}\right )}{16 \, {\left (b f \cos \left (f x + e\right )^{2} - b f\right )}}\right ] \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 {\csc ^{3}{\left (e + f x \right )}}{\sqrt {b \sec {\left (e + f x \right )}}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 4.31, size = 110, normalized size = 1.18 \begin {gather*} \frac {b^{2} {\left (\frac {2 \, \sqrt {b \cos \left (f x + e\right )} \cos \left (f x + e\right )}{{\left (b^{2} \cos \left (f x + e\right )^{2} - b^{2}\right )} b} + \frac {\arctan \left (\frac {\sqrt {b \cos \left (f x + e\right )}}{\sqrt {-b}}\right )}{\sqrt {-b} b^{2}} + \frac {\arctan \left (\frac {\sqrt {b \cos \left (f x + e\right )}}{\sqrt {b}}\right )}{b^{\frac {5}{2}}}\right )}}{4 \, f \mathrm {sgn}\left (\cos \left (f x + e\right )\right )} \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 \frac {1}{{\sin \left (e+f\,x\right )}^3\,\sqrt {\frac {b}{\cos \left (e+f\,x\right )}}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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