Optimal. Leaf size=116 \[ -\frac {(3 a-5 b) \sqrt {b} \text {ArcTan}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{2 a^{7/2} f}-\frac {(a-2 b) \cot (e+f x)}{a^3 f}-\frac {\cot ^3(e+f x)}{3 a^2 f}-\frac {(a-b) b \tan (e+f x)}{2 a^3 f \left (a+b \tan ^2(e+f x)\right )} \]
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Rubi [A]
time = 0.10, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps
used = 5, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3744, 467,
1275, 211} \begin {gather*} -\frac {\sqrt {b} (3 a-5 b) \text {ArcTan}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{2 a^{7/2} f}-\frac {b (a-b) \tan (e+f x)}{2 a^3 f \left (a+b \tan ^2(e+f x)\right )}-\frac {(a-2 b) \cot (e+f x)}{a^3 f}-\frac {\cot ^3(e+f x)}{3 a^2 f} \end {gather*}
Antiderivative was successfully verified.
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Rule 211
Rule 467
Rule 1275
Rule 3744
Rubi steps
\begin {align*} \int \frac {\csc ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx &=\frac {\text {Subst}\left (\int \frac {1+x^2}{x^4 \left (a+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {(a-b) b \tan (e+f x)}{2 a^3 f \left (a+b \tan ^2(e+f x)\right )}-\frac {b \text {Subst}\left (\int \frac {-\frac {2}{a b}-\frac {2 (a-b) x^2}{a^2 b}+\frac {(a-b) x^4}{a^3}}{x^4 \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{2 f}\\ &=-\frac {(a-b) b \tan (e+f x)}{2 a^3 f \left (a+b \tan ^2(e+f x)\right )}-\frac {b \text {Subst}\left (\int \left (-\frac {2}{a^2 b x^4}-\frac {2 (a-2 b)}{a^3 b x^2}+\frac {3 a-5 b}{a^3 \left (a+b x^2\right )}\right ) \, dx,x,\tan (e+f x)\right )}{2 f}\\ &=-\frac {(a-2 b) \cot (e+f x)}{a^3 f}-\frac {\cot ^3(e+f x)}{3 a^2 f}-\frac {(a-b) b \tan (e+f x)}{2 a^3 f \left (a+b \tan ^2(e+f x)\right )}-\frac {((3 a-5 b) b) \text {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\tan (e+f x)\right )}{2 a^3 f}\\ &=-\frac {(3 a-5 b) \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{2 a^{7/2} f}-\frac {(a-2 b) \cot (e+f x)}{a^3 f}-\frac {\cot ^3(e+f x)}{3 a^2 f}-\frac {(a-b) b \tan (e+f x)}{2 a^3 f \left (a+b \tan ^2(e+f x)\right )}\\ \end {align*}
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Mathematica [A]
time = 0.66, size = 112, normalized size = 0.97 \begin {gather*} \frac {3 \sqrt {b} (-3 a+5 b) \text {ArcTan}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )+\sqrt {a} \left (-2 \cot (e+f x) \left (2 a-6 b+a \csc ^2(e+f x)\right )+\frac {3 b (-a+b) \sin (2 (e+f x))}{a+b+(a-b) \cos (2 (e+f x))}\right )}{6 a^{7/2} f} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.31, size = 100, normalized size = 0.86
method | result | size |
derivativedivides | \(\frac {-\frac {1}{3 a^{2} \tan \left (f x +e \right )^{3}}-\frac {a -2 b}{a^{3} \tan \left (f x +e \right )}-\frac {b \left (\frac {\left (\frac {a}{2}-\frac {b}{2}\right ) \tan \left (f x +e \right )}{a +b \left (\tan ^{2}\left (f x +e \right )\right )}+\frac {\left (3 a -5 b \right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{a^{3}}}{f}\) | \(100\) |
default | \(\frac {-\frac {1}{3 a^{2} \tan \left (f x +e \right )^{3}}-\frac {a -2 b}{a^{3} \tan \left (f x +e \right )}-\frac {b \left (\frac {\left (\frac {a}{2}-\frac {b}{2}\right ) \tan \left (f x +e \right )}{a +b \left (\tan ^{2}\left (f x +e \right )\right )}+\frac {\left (3 a -5 b \right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{a^{3}}}{f}\) | \(100\) |
risch | \(\frac {i \left (9 a b \,{\mathrm e}^{8 i \left (f x +e \right )}-15 b^{2} {\mathrm e}^{8 i \left (f x +e \right )}+12 a^{2} {\mathrm e}^{6 i \left (f x +e \right )}-6 a b \,{\mathrm e}^{6 i \left (f x +e \right )}+60 b^{2} {\mathrm e}^{6 i \left (f x +e \right )}+20 a^{2} {\mathrm e}^{4 i \left (f x +e \right )}+4 a b \,{\mathrm e}^{4 i \left (f x +e \right )}-90 b^{2} {\mathrm e}^{4 i \left (f x +e \right )}+4 a^{2} {\mathrm e}^{2 i \left (f x +e \right )}-26 a b \,{\mathrm e}^{2 i \left (f x +e \right )}+60 b^{2} {\mathrm e}^{2 i \left (f x +e \right )}-4 a^{2}+19 a b -15 b^{2}\right )}{3 f \,a^{3} \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{3} \left (a \,{\mathrm e}^{4 i \left (f x +e \right )}-b \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+2 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a -b \right )}+\frac {3 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-a b}+a +b}{a -b}\right )}{4 a^{3} f}-\frac {5 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-a b}+a +b}{a -b}\right ) b}{4 a^{4} f}-\frac {3 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-a b}-a -b}{a -b}\right )}{4 a^{3} f}+\frac {5 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-a b}-a -b}{a -b}\right ) b}{4 a^{4} f}\) | \(435\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.52, size = 120, normalized size = 1.03 \begin {gather*} -\frac {\frac {3 \, {\left (3 \, a b - 5 \, b^{2}\right )} \tan \left (f x + e\right )^{4} + 2 \, {\left (3 \, a^{2} - 5 \, a b\right )} \tan \left (f x + e\right )^{2} + 2 \, a^{2}}{a^{3} b \tan \left (f x + e\right )^{5} + a^{4} \tan \left (f x + e\right )^{3}} + \frac {3 \, {\left (3 \, a b - 5 \, b^{2}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3}}}{6 \, 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. 261 vs.
\(2 (107) = 214\).
time = 3.07, size = 615, normalized size = 5.30 \begin {gather*} \left [-\frac {4 \, {\left (4 \, a^{2} - 19 \, a b + 15 \, b^{2}\right )} \cos \left (f x + e\right )^{5} - 8 \, {\left (3 \, a^{2} - 14 \, a b + 15 \, b^{2}\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left ({\left (3 \, a^{2} - 8 \, a b + 5 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - {\left (3 \, a^{2} - 11 \, a b + 10 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 3 \, a b + 5 \, b^{2}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {{\left (a^{2} + 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \, {\left ({\left (a^{2} + a b\right )} \cos \left (f x + e\right )^{3} - a b \cos \left (f x + e\right )\right )} \sqrt {-\frac {b}{a}} \sin \left (f x + e\right ) + b^{2}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 2 \, {\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + b^{2}}\right ) \sin \left (f x + e\right ) - 12 \, {\left (3 \, a b - 5 \, b^{2}\right )} \cos \left (f x + e\right )}{24 \, {\left ({\left (a^{4} - a^{3} b\right )} f \cos \left (f x + e\right )^{4} - a^{3} b f - {\left (a^{4} - 2 \, a^{3} b\right )} f \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}, -\frac {2 \, {\left (4 \, a^{2} - 19 \, a b + 15 \, b^{2}\right )} \cos \left (f x + e\right )^{5} - 4 \, {\left (3 \, a^{2} - 14 \, a b + 15 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - 3 \, {\left ({\left (3 \, a^{2} - 8 \, a b + 5 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - {\left (3 \, a^{2} - 11 \, a b + 10 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 3 \, a b + 5 \, b^{2}\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {{\left ({\left (a + b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {b}{a}}}{2 \, b \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) - 6 \, {\left (3 \, a b - 5 \, b^{2}\right )} \cos \left (f x + e\right )}{12 \, {\left ({\left (a^{4} - a^{3} b\right )} f \cos \left (f x + e\right )^{4} - a^{3} b f - {\left (a^{4} - 2 \, a^{3} b\right )} f \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\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 ^{4}{\left (e + f x \right )}}{\left (a + b \tan ^{2}{\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 [A]
time = 0.73, size = 142, normalized size = 1.22 \begin {gather*} -\frac {\frac {3 \, {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right )\right )} {\left (3 \, a b - 5 \, b^{2}\right )}}{\sqrt {a b} a^{3}} + \frac {3 \, {\left (a b \tan \left (f x + e\right ) - b^{2} \tan \left (f x + e\right )\right )}}{{\left (b \tan \left (f x + e\right )^{2} + a\right )} a^{3}} + \frac {2 \, {\left (3 \, a \tan \left (f x + e\right )^{2} - 6 \, b \tan \left (f x + e\right )^{2} + a\right )}}{a^{3} \tan \left (f x + e\right )^{3}}}{6 \, f} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 11.37, size = 108, normalized size = 0.93 \begin {gather*} -\frac {\frac {1}{3\,a}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (3\,a-5\,b\right )}{3\,a^2}+\frac {b\,{\mathrm {tan}\left (e+f\,x\right )}^4\,\left (3\,a-5\,b\right )}{2\,a^3}}{f\,\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^5+a\,{\mathrm {tan}\left (e+f\,x\right )}^3\right )}-\frac {\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\mathrm {tan}\left (e+f\,x\right )}{\sqrt {a}}\right )\,\left (3\,a-5\,b\right )}{2\,a^{7/2}\,f} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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