Optimal. Leaf size=164 \[ -\frac {5 \sqrt {b} (3 a-4 b) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{8 f (a+b)^{9/2}}-\frac {b (7 a-4 b) \tan (e+f x)}{8 f (a+b)^4 \left (a+b \tan ^2(e+f x)+b\right )}-\frac {a b \tan (e+f x)}{4 f (a+b)^3 \left (a+b \tan ^2(e+f x)+b\right )^2}-\frac {\cot ^3(e+f x)}{3 f (a+b)^3}-\frac {(a-2 b) \cot (e+f x)}{f (a+b)^4} \]
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Rubi [A] time = 0.25, antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {4132, 456, 1259, 1261, 205} \[ -\frac {5 \sqrt {b} (3 a-4 b) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{8 f (a+b)^{9/2}}-\frac {b (7 a-4 b) \tan (e+f x)}{8 f (a+b)^4 \left (a+b \tan ^2(e+f x)+b\right )}-\frac {a b \tan (e+f x)}{4 f (a+b)^3 \left (a+b \tan ^2(e+f x)+b\right )^2}-\frac {\cot ^3(e+f x)}{3 f (a+b)^3}-\frac {(a-2 b) \cot (e+f x)}{f (a+b)^4} \]
Antiderivative was successfully verified.
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Rule 205
Rule 456
Rule 1259
Rule 1261
Rule 4132
Rubi steps
\begin {align*} \int \frac {\csc ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1+x^2}{x^4 \left (a+b+b x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {a b \tan (e+f x)}{4 (a+b)^3 f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {b \operatorname {Subst}\left (\int \frac {-\frac {4}{b (a+b)}-\frac {4 a x^2}{b (a+b)^2}+\frac {3 a x^4}{(a+b)^3}}{x^4 \left (a+b+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{4 f}\\ &=-\frac {a b \tan (e+f x)}{4 (a+b)^3 f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {(7 a-4 b) b \tan (e+f x)}{8 (a+b)^4 f \left (a+b+b \tan ^2(e+f x)\right )}-\frac {\operatorname {Subst}\left (\int \frac {-8 b (a+b)-8 (a-b) b x^2+\frac {(7 a-4 b) b^2 x^4}{a+b}}{x^4 \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{8 b (a+b)^3 f}\\ &=-\frac {a b \tan (e+f x)}{4 (a+b)^3 f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {(7 a-4 b) b \tan (e+f x)}{8 (a+b)^4 f \left (a+b+b \tan ^2(e+f x)\right )}-\frac {\operatorname {Subst}\left (\int \left (-\frac {8 b}{x^4}+\frac {8 b (-a+2 b)}{(a+b) x^2}+\frac {5 (3 a-4 b) b^2}{(a+b) \left (a+b+b x^2\right )}\right ) \, dx,x,\tan (e+f x)\right )}{8 b (a+b)^3 f}\\ &=-\frac {(a-2 b) \cot (e+f x)}{(a+b)^4 f}-\frac {\cot ^3(e+f x)}{3 (a+b)^3 f}-\frac {a b \tan (e+f x)}{4 (a+b)^3 f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {(7 a-4 b) b \tan (e+f x)}{8 (a+b)^4 f \left (a+b+b \tan ^2(e+f x)\right )}-\frac {(5 (3 a-4 b) b) \operatorname {Subst}\left (\int \frac {1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{8 (a+b)^4 f}\\ &=-\frac {5 (3 a-4 b) \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{8 (a+b)^{9/2} f}-\frac {(a-2 b) \cot (e+f x)}{(a+b)^4 f}-\frac {\cot ^3(e+f x)}{3 (a+b)^3 f}-\frac {a b \tan (e+f x)}{4 (a+b)^3 f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {(7 a-4 b) b \tan (e+f x)}{8 (a+b)^4 f \left (a+b+b \tan ^2(e+f x)\right )}\\ \end {align*}
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Mathematica [C] time = 4.05, size = 994, normalized size = 6.06 \[ \frac {(\cos (2 (e+f x)) a+a+2 b) \sec ^6(e+f x) \left (\frac {480 (3 a-4 b) b \tan ^{-1}\left (\frac {\sec (f x) (\cos (2 e)-i \sin (2 e)) (a \sin (2 e+f x)-(a+2 b) \sin (f x))}{2 \sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}\right ) (\cos (2 (e+f x)) a+a+2 b)^2 (\cos (2 e)-i \sin (2 e))}{\sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}-\frac {\csc (e) \csc ^3(e+f x) \sec (2 e) \left (224 \sin (2 e-f x) a^4-224 \sin (2 e+f x) a^4+176 \sin (4 e+f x) a^4+48 \sin (2 e+3 f x) a^4-96 \sin (4 e+3 f x) a^4+48 \sin (6 e+3 f x) a^4+16 \sin (2 e+5 f x) a^4+16 \sin (6 e+5 f x) a^4+16 \sin (4 e+7 f x) a^4+16 \sin (8 e+7 f x) a^4+576 b \sin (2 e-f x) a^3-657 b \sin (2 e+f x) a^3+569 b \sin (4 e+f x) a^3+111 b \sin (2 e+3 f x) a^3-152 b \sin (4 e+3 f x) a^3+192 b \sin (6 e+3 f x) a^3+72 b \sin (4 e+5 f x) a^3+27 b \sin (6 e+5 f x) a^3+45 b \sin (8 e+5 f x) a^3-83 b \sin (4 e+7 f x) a^3+27 b \sin (6 e+7 f x) a^3-56 b \sin (8 e+7 f x) a^3+124 b^2 \sin (2 e-f x) a^2-538 b^2 \sin (2 e+f x) a^2+666 b^2 \sin (4 e+f x) a^2+360 b^2 \sin (2 e+3 f x) a^2+146 b^2 \sin (4 e+3 f x) a^2+558 b^2 \sin (6 e+3 f x) a^2-598 b^2 \sin (2 e+5 f x) a^2+150 b^2 \sin (4 e+5 f x) a^2-388 b^2 \sin (6 e+5 f x) a^2-60 b^2 \sin (8 e+5 f x) a^2+6 b^2 \sin (4 e+7 f x) a^2-6 b^2 \sin (6 e+7 f x) a^2-2184 b^3 \sin (2 e-f x) a+984 b^3 \sin (2 e+f x) a+1704 b^3 \sin (4 e+f x) a+312 b^3 \sin (2 e+3 f x) a-728 b^3 \sin (4 e+3 f x) a-168 b^3 \sin (6 e+3 f x) a+48 b^3 \sin (2 e+5 f x) a-48 b^3 \sin (4 e+5 f x) a+4 \left (44 a^4+122 b a^3+63 b^2 a^2+126 b^3 a+36 b^4\right ) \sin (f x)+\left (-96 a^4-71 b a^3+344 b^2 a^2-1208 b^3 a+48 b^4\right ) \sin (3 f x)+144 b^4 \sin (2 e-f x)+144 b^4 \sin (2 e+f x)-144 b^4 \sin (4 e+f x)-48 b^4 \sin (2 e+3 f x)-48 b^4 \sin (4 e+3 f x)+48 b^4 \sin (6 e+3 f x)\right )}{a}\right )}{6144 (a+b)^4 f \left (b \sec ^2(e+f x)+a\right )^3} \]
Warning: Unable to verify antiderivative.
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fricas [B] time = 0.65, size = 1009, normalized size = 6.15 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.67, size = 275, normalized size = 1.68 \[ -\frac {\frac {15 \, {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )} {\left (3 \, a b - 4 \, b^{2}\right )}}{{\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} \sqrt {a b + b^{2}}} + \frac {3 \, {\left (7 \, a b^{2} \tan \left (f x + e\right )^{3} - 4 \, b^{3} \tan \left (f x + e\right )^{3} + 9 \, a^{2} b \tan \left (f x + e\right ) + 5 \, a b^{2} \tan \left (f x + e\right ) - 4 \, b^{3} \tan \left (f x + e\right )\right )}}{{\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{2}} + \frac {8 \, {\left (3 \, a \tan \left (f x + e\right )^{2} - 6 \, b \tan \left (f x + e\right )^{2} + a + b\right )}}{{\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} \tan \left (f x + e\right )^{3}}}{24 \, f} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 1.23, size = 306, normalized size = 1.87 \[ -\frac {7 b^{2} \left (\tan ^{3}\left (f x +e \right )\right ) a}{8 f \left (a +b \right )^{4} \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2}}+\frac {b^{3} \left (\tan ^{3}\left (f x +e \right )\right )}{2 f \left (a +b \right )^{4} \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2}}-\frac {9 b \tan \left (f x +e \right ) a^{2}}{8 f \left (a +b \right )^{4} \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2}}-\frac {5 b^{2} \tan \left (f x +e \right ) a}{8 f \left (a +b \right )^{4} \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2}}+\frac {b^{3} \tan \left (f x +e \right )}{2 f \left (a +b \right )^{4} \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2}}-\frac {15 b \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right ) a}{8 f \left (a +b \right )^{4} \sqrt {\left (a +b \right ) b}}+\frac {5 b^{2} \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{2 f \left (a +b \right )^{4} \sqrt {\left (a +b \right ) b}}-\frac {1}{3 f \left (a +b \right )^{3} \tan \left (f x +e \right )^{3}}-\frac {a}{f \left (a +b \right )^{4} \tan \left (f x +e \right )}+\frac {2 b}{f \left (a +b \right )^{4} \tan \left (f x +e \right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.44, size = 323, normalized size = 1.97 \[ -\frac {\frac {15 \, {\left (3 \, a b - 4 \, b^{2}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{{\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} \sqrt {{\left (a + b\right )} b}} + \frac {15 \, {\left (3 \, a b^{2} - 4 \, b^{3}\right )} \tan \left (f x + e\right )^{6} + 25 \, {\left (3 \, a^{2} b - a b^{2} - 4 \, b^{3}\right )} \tan \left (f x + e\right )^{4} + 8 \, a^{3} + 24 \, a^{2} b + 24 \, a b^{2} + 8 \, b^{3} + 8 \, {\left (3 \, a^{3} + 2 \, a^{2} b - 5 \, a b^{2} - 4 \, b^{3}\right )} \tan \left (f x + e\right )^{2}}{{\left (a^{4} b^{2} + 4 \, a^{3} b^{3} + 6 \, a^{2} b^{4} + 4 \, a b^{5} + b^{6}\right )} \tan \left (f x + e\right )^{7} + 2 \, {\left (a^{5} b + 5 \, a^{4} b^{2} + 10 \, a^{3} b^{3} + 10 \, a^{2} b^{4} + 5 \, a b^{5} + b^{6}\right )} \tan \left (f x + e\right )^{5} + {\left (a^{6} + 6 \, a^{5} b + 15 \, a^{4} b^{2} + 20 \, a^{3} b^{3} + 15 \, a^{2} b^{4} + 6 \, a b^{5} + b^{6}\right )} \tan \left (f x + e\right )^{3}}}{24 \, f} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 6.88, size = 207, normalized size = 1.26 \[ -\frac {\frac {1}{3\,\left (a+b\right )}+\frac {25\,{\mathrm {tan}\left (e+f\,x\right )}^4\,\left (3\,a\,b-4\,b^2\right )}{24\,{\left (a+b\right )}^3}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (3\,a-4\,b\right )}{3\,{\left (a+b\right )}^2}+\frac {5\,{\mathrm {tan}\left (e+f\,x\right )}^6\,\left (3\,a\,b^2-4\,b^3\right )}{8\,{\left (a+b\right )}^4}}{f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^3\,\left (a^2+2\,a\,b+b^2\right )+{\mathrm {tan}\left (e+f\,x\right )}^5\,\left (2\,b^2+2\,a\,b\right )+b^2\,{\mathrm {tan}\left (e+f\,x\right )}^7\right )}-\frac {5\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\mathrm {tan}\left (e+f\,x\right )\,\left (a^4+4\,a^3\,b+6\,a^2\,b^2+4\,a\,b^3+b^4\right )}{{\left (a+b\right )}^{9/2}}\right )\,\left (3\,a-4\,b\right )}{8\,f\,{\left (a+b\right )}^{9/2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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