Optimal. Leaf size=180 \[ \frac {\cos ^3(e+f x)}{3 f (a-b)^3}-\frac {(a+2 b) \cos (e+f x)}{f (a-b)^4}-\frac {b (7 a+4 b) \sec (e+f x)}{8 f (a-b)^4 \left (a+b \sec ^2(e+f x)-b\right )}-\frac {a b \sec (e+f x)}{4 f (a-b)^3 \left (a+b \sec ^2(e+f x)-b\right )^2}-\frac {5 \sqrt {b} (3 a+4 b) \tan ^{-1}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{8 f (a-b)^{9/2}} \]
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Rubi [A] time = 0.25, antiderivative size = 180, 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 = {3664, 456, 1259, 1261, 205} \[ \frac {\cos ^3(e+f x)}{3 f (a-b)^3}-\frac {(a+2 b) \cos (e+f x)}{f (a-b)^4}-\frac {b (7 a+4 b) \sec (e+f x)}{8 f (a-b)^4 \left (a+b \sec ^2(e+f x)-b\right )}-\frac {a b \sec (e+f x)}{4 f (a-b)^3 \left (a+b \sec ^2(e+f x)-b\right )^2}-\frac {5 \sqrt {b} (3 a+4 b) \tan ^{-1}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{8 f (a-b)^{9/2}} \]
Antiderivative was successfully verified.
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Rule 205
Rule 456
Rule 1259
Rule 1261
Rule 3664
Rubi steps
\begin {align*} \int \frac {\sin ^3(e+f x)}{\left (a+b \tan ^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,\sec (e+f x)\right )}{f}\\ &=-\frac {a b \sec (e+f x)}{4 (a-b)^3 f \left (a-b+b \sec ^2(e+f x)\right )^2}-\frac {b \operatorname {Subst}\left (\int \frac {\frac {4}{(a-b) b}-\frac {4 a x^2}{(a-b)^2 b}+\frac {3 a x^4}{(a-b)^3}}{x^4 \left (a-b+b x^2\right )^2} \, dx,x,\sec (e+f x)\right )}{4 f}\\ &=-\frac {a b \sec (e+f x)}{4 (a-b)^3 f \left (a-b+b \sec ^2(e+f x)\right )^2}-\frac {b (7 a+4 b) \sec (e+f x)}{8 (a-b)^4 f \left (a-b+b \sec ^2(e+f x)\right )}-\frac {\operatorname {Subst}\left (\int \frac {8 (a-b) b-8 b (a+b) x^2+\frac {b^2 (7 a+4 b) x^4}{a-b}}{x^4 \left (a-b+b x^2\right )} \, dx,x,\sec (e+f x)\right )}{8 (a-b)^3 b f}\\ &=-\frac {a b \sec (e+f x)}{4 (a-b)^3 f \left (a-b+b \sec ^2(e+f x)\right )^2}-\frac {b (7 a+4 b) \sec (e+f x)}{8 (a-b)^4 f \left (a-b+b \sec ^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 b^2 (3 a+4 b)}{(a-b) \left (a-b+b x^2\right )}\right ) \, dx,x,\sec (e+f x)\right )}{8 (a-b)^3 b f}\\ &=-\frac {(a+2 b) \cos (e+f x)}{(a-b)^4 f}+\frac {\cos ^3(e+f x)}{3 (a-b)^3 f}-\frac {a b \sec (e+f x)}{4 (a-b)^3 f \left (a-b+b \sec ^2(e+f x)\right )^2}-\frac {b (7 a+4 b) \sec (e+f x)}{8 (a-b)^4 f \left (a-b+b \sec ^2(e+f x)\right )}-\frac {(5 b (3 a+4 b)) \operatorname {Subst}\left (\int \frac {1}{a-b+b x^2} \, dx,x,\sec (e+f x)\right )}{8 (a-b)^4 f}\\ &=-\frac {5 \sqrt {b} (3 a+4 b) \tan ^{-1}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{8 (a-b)^{9/2} f}-\frac {(a+2 b) \cos (e+f x)}{(a-b)^4 f}+\frac {\cos ^3(e+f x)}{3 (a-b)^3 f}-\frac {a b \sec (e+f x)}{4 (a-b)^3 f \left (a-b+b \sec ^2(e+f x)\right )^2}-\frac {b (7 a+4 b) \sec (e+f x)}{8 (a-b)^4 f \left (a-b+b \sec ^2(e+f x)\right )}\\ \end {align*}
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Mathematica [A] time = 5.95, size = 230, normalized size = 1.28 \[ \frac {\frac {2 \left (3 \cos (e+f x) \left (a \left (\frac {4 b^2}{((a-b) \cos (2 (e+f x))+a+b)^2}-\frac {9 b}{(a-b) \cos (2 (e+f x))+a+b}-3\right )+b \left (-\frac {4 b}{(a-b) \cos (2 (e+f x))+a+b}-9\right )\right )+(a-b) \cos (3 (e+f x))\right )}{(a-b)^4}+\frac {15 \sqrt {b} (3 a+4 b) \tan ^{-1}\left (\frac {\sqrt {a-b}-\sqrt {a} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {b}}\right )}{(a-b)^{9/2}}+\frac {15 \sqrt {b} (3 a+4 b) \tan ^{-1}\left (\frac {\sqrt {a-b}+\sqrt {a} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {b}}\right )}{(a-b)^{9/2}}}{24 f} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.67, size = 775, normalized size = 4.31 \[ \left [\frac {16 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )^{7} - 16 \, {\left (3 \, a^{3} - 2 \, a^{2} b - 5 \, a b^{2} + 4 \, b^{3}\right )} \cos \left (f x + e\right )^{5} - 50 \, {\left (3 \, a^{2} b + a b^{2} - 4 \, b^{3}\right )} \cos \left (f x + e\right )^{3} + 15 \, {\left ({\left (3 \, a^{3} - 2 \, a^{2} b - 5 \, a b^{2} + 4 \, b^{3}\right )} \cos \left (f x + e\right )^{4} + 3 \, a b^{2} + 4 \, b^{3} + 2 \, {\left (3 \, a^{2} b + a b^{2} - 4 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-\frac {b}{a - b}} \log \left (\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + 2 \, {\left (a - b\right )} \sqrt {-\frac {b}{a - b}} \cos \left (f x + e\right ) - b}{{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}\right ) - 30 \, {\left (3 \, a b^{2} + 4 \, b^{3}\right )} \cos \left (f x + e\right )}{48 \, {\left ({\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 )} f \cos \left (f x + e\right )^{4} + 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 )} f \cos \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 )} f\right )}}, \frac {8 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )^{7} - 8 \, {\left (3 \, a^{3} - 2 \, a^{2} b - 5 \, a b^{2} + 4 \, b^{3}\right )} \cos \left (f x + e\right )^{5} - 25 \, {\left (3 \, a^{2} b + a b^{2} - 4 \, b^{3}\right )} \cos \left (f x + e\right )^{3} - 15 \, {\left ({\left (3 \, a^{3} - 2 \, a^{2} b - 5 \, a b^{2} + 4 \, b^{3}\right )} \cos \left (f x + e\right )^{4} + 3 \, a b^{2} + 4 \, b^{3} + 2 \, {\left (3 \, a^{2} b + a b^{2} - 4 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {b}{a - b}} \arctan \left (-\frac {{\left (a - b\right )} \sqrt {\frac {b}{a - b}} \cos \left (f x + e\right )}{b}\right ) - 15 \, {\left (3 \, a b^{2} + 4 \, b^{3}\right )} \cos \left (f x + e\right )}{24 \, {\left ({\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 )} f \cos \left (f x + e\right )^{4} + 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 )} f \cos \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 )} f\right )}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 7.40, size = 563, normalized size = 3.13 \[ \frac {a^{6} f^{17} \cos \left (f x + e\right )^{3} - 6 \, a^{5} b f^{17} \cos \left (f x + e\right )^{3} + 15 \, a^{4} b^{2} f^{17} \cos \left (f x + e\right )^{3} - 20 \, a^{3} b^{3} f^{17} \cos \left (f x + e\right )^{3} + 15 \, a^{2} b^{4} f^{17} \cos \left (f x + e\right )^{3} - 6 \, a b^{5} f^{17} \cos \left (f x + e\right )^{3} + b^{6} f^{17} \cos \left (f x + e\right )^{3} - 3 \, a^{6} f^{17} \cos \left (f x + e\right ) + 9 \, a^{5} b f^{17} \cos \left (f x + e\right ) - 30 \, a^{3} b^{3} f^{17} \cos \left (f x + e\right ) + 45 \, a^{2} b^{4} f^{17} \cos \left (f x + e\right ) - 27 \, a b^{5} f^{17} \cos \left (f x + e\right ) + 6 \, b^{6} f^{17} \cos \left (f x + e\right )}{3 \, {\left (a^{9} f^{18} - 9 \, a^{8} b f^{18} + 36 \, a^{7} b^{2} f^{18} - 84 \, a^{6} b^{3} f^{18} + 126 \, a^{5} b^{4} f^{18} - 126 \, a^{4} b^{5} f^{18} + 84 \, a^{3} b^{6} f^{18} - 36 \, a^{2} b^{7} f^{18} + 9 \, a b^{8} f^{18} - b^{9} f^{18}\right )}} + \frac {5 \, {\left (3 \, a b + 4 \, b^{2}\right )} \arctan \left (\frac {a \cos \left (f x + e\right ) - b \cos \left (f x + e\right )}{\sqrt {a b - b^{2}}}\right )}{8 \, {\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} \sqrt {a b - b^{2}} f} - \frac {\frac {9 \, a^{2} b \cos \left (f x + e\right )^{3}}{f} - \frac {5 \, a b^{2} \cos \left (f x + e\right )^{3}}{f} - \frac {4 \, b^{3} \cos \left (f x + e\right )^{3}}{f} + \frac {7 \, a b^{2} \cos \left (f x + e\right )}{f} + \frac {4 \, b^{3} \cos \left (f x + e\right )}{f}}{8 \, {\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} {\left (a \cos \left (f x + e\right )^{2} - b \cos \left (f x + e\right )^{2} + b\right )}^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.59, size = 504, normalized size = 2.80 \[ \frac {a \left (\cos ^{3}\left (f x +e \right )\right )}{3 f \left (a^{3}-3 a^{2} b +3 b^{2} a -b^{3}\right ) \left (a -b \right )}-\frac {b \left (\cos ^{3}\left (f x +e \right )\right )}{3 f \left (a^{3}-3 a^{2} b +3 b^{2} a -b^{3}\right ) \left (a -b \right )}-\frac {a \cos \left (f x +e \right )}{f \left (a^{3}-3 a^{2} b +3 b^{2} a -b^{3}\right ) \left (a -b \right )}-\frac {2 \cos \left (f x +e \right ) b}{f \left (a^{3}-3 a^{2} b +3 b^{2} a -b^{3}\right ) \left (a -b \right )}-\frac {9 b \,a^{2} \left (\cos ^{3}\left (f x +e \right )\right )}{8 f \left (a -b \right )^{4} \left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )^{2}}+\frac {5 b^{2} \left (\cos ^{3}\left (f x +e \right )\right ) a}{8 f \left (a -b \right )^{4} \left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )^{2}}+\frac {b^{3} \left (\cos ^{3}\left (f x +e \right )\right )}{2 f \left (a -b \right )^{4} \left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )^{2}}-\frac {7 b^{2} a \cos \left (f x +e \right )}{8 f \left (a -b \right )^{4} \left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )^{2}}-\frac {b^{3} \cos \left (f x +e \right )}{2 f \left (a -b \right )^{4} \left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )^{2}}+\frac {15 b \arctan \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )}{\sqrt {\left (a -b \right ) b}}\right ) a}{8 f \left (a -b \right )^{4} \sqrt {\left (a -b \right ) b}}+\frac {5 b^{2} \arctan \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )}{\sqrt {\left (a -b \right ) b}}\right )}{2 f \left (a -b \right )^{4} \sqrt {\left (a -b \right ) b}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 15.45, size = 1154, normalized size = 6.41 \[ \text {result too large to display} \]
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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