Optimal. Leaf size=137 \[ -\frac {\sec ^3(c+d x) (b-a \sin (c+d x))}{3 d \left (a^2-b^2\right )}+\frac {2 b^4 \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{5/2}}+\frac {\sec (c+d x) \left (a \left (2 a^2-5 b^2\right ) \sin (c+d x)+3 b^3\right )}{3 d \left (a^2-b^2\right )^2} \]
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Rubi [A] time = 0.25, antiderivative size = 137, 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 = {2696, 2866, 12, 2660, 618, 204} \[ \frac {2 b^4 \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{5/2}}-\frac {\sec ^3(c+d x) (b-a \sin (c+d x))}{3 d \left (a^2-b^2\right )}+\frac {\sec (c+d x) \left (a \left (2 a^2-5 b^2\right ) \sin (c+d x)+3 b^3\right )}{3 d \left (a^2-b^2\right )^2} \]
Antiderivative was successfully verified.
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Rule 12
Rule 204
Rule 618
Rule 2660
Rule 2696
Rule 2866
Rubi steps
\begin {align*} \int \frac {\sec ^4(c+d x)}{a+b \sin (c+d x)} \, dx &=-\frac {\sec ^3(c+d x) (b-a \sin (c+d x))}{3 \left (a^2-b^2\right ) d}-\frac {\int \frac {\sec ^2(c+d x) \left (-2 a^2+3 b^2-2 a b \sin (c+d x)\right )}{a+b \sin (c+d x)} \, dx}{3 \left (a^2-b^2\right )}\\ &=-\frac {\sec ^3(c+d x) (b-a \sin (c+d x))}{3 \left (a^2-b^2\right ) d}+\frac {\sec (c+d x) \left (3 b^3+a \left (2 a^2-5 b^2\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^2 d}+\frac {\int \frac {3 b^4}{a+b \sin (c+d x)} \, dx}{3 \left (a^2-b^2\right )^2}\\ &=-\frac {\sec ^3(c+d x) (b-a \sin (c+d x))}{3 \left (a^2-b^2\right ) d}+\frac {\sec (c+d x) \left (3 b^3+a \left (2 a^2-5 b^2\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^2 d}+\frac {b^4 \int \frac {1}{a+b \sin (c+d x)} \, dx}{\left (a^2-b^2\right )^2}\\ &=-\frac {\sec ^3(c+d x) (b-a \sin (c+d x))}{3 \left (a^2-b^2\right ) d}+\frac {\sec (c+d x) \left (3 b^3+a \left (2 a^2-5 b^2\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^2 d}+\frac {\left (2 b^4\right ) \operatorname {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{\left (a^2-b^2\right )^2 d}\\ &=-\frac {\sec ^3(c+d x) (b-a \sin (c+d x))}{3 \left (a^2-b^2\right ) d}+\frac {\sec (c+d x) \left (3 b^3+a \left (2 a^2-5 b^2\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^2 d}-\frac {\left (4 b^4\right ) \operatorname {Subst}\left (\int \frac {1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{\left (a^2-b^2\right )^2 d}\\ &=\frac {2 b^4 \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2} d}-\frac {\sec ^3(c+d x) (b-a \sin (c+d x))}{3 \left (a^2-b^2\right ) d}+\frac {\sec (c+d x) \left (3 b^3+a \left (2 a^2-5 b^2\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^2 d}\\ \end {align*}
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Mathematica [A] time = 1.34, size = 202, normalized size = 1.47 \[ \frac {\frac {24 b^4 \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac {\sec ^3(c+d x) \left (6 a^3 \sin (c+d x)+2 a^3 \sin (3 (c+d x))+\frac {3}{2} b \left (a^2-7 b^2\right ) \cos (c+d x)+\frac {1}{2} a^2 b \cos (3 (c+d x))-4 a^2 b-9 a b^2 \sin (c+d x)-5 a b^2 \sin (3 (c+d x))+6 b^3 \cos (2 (c+d x))-\frac {7}{2} b^3 \cos (3 (c+d x))+10 b^3\right )}{(a-b)^2 (a+b)^2}}{12 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.51, size = 466, normalized size = 3.40 \[ \left [-\frac {3 \, \sqrt {-a^{2} + b^{2}} b^{4} \cos \left (d x + c\right )^{3} \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) + 2 \, a^{4} b - 4 \, a^{2} b^{3} + 2 \, b^{5} - 6 \, {\left (a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4} + {\left (2 \, a^{5} - 7 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{3}}, -\frac {3 \, \sqrt {a^{2} - b^{2}} b^{4} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) \cos \left (d x + c\right )^{3} + a^{4} b - 2 \, a^{2} b^{3} + b^{5} - 3 \, {\left (a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2} - {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4} + {\left (2 \, a^{5} - 7 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{3 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{3}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.53, size = 273, normalized size = 1.99 \[ \frac {2 \, {\left (\frac {3 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} b^{4}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} - b^{2}}} - \frac {3 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 6 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 2 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 8 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{2} b + 4 \, b^{3}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}\right )}}{3 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.19, size = 270, normalized size = 1.97 \[ -\frac {2}{3 d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (2 a +2 b \right )}-\frac {1}{d \left (2 a +2 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {a}{d \left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {3 b}{2 d \left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {2 b^{4} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d \left (a -b \right )^{2} \left (a +b \right )^{2} \sqrt {a^{2}-b^{2}}}-\frac {2}{3 d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3} \left (2 a -2 b \right )}+\frac {1}{d \left (2 a -2 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {a}{d \left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {3 b}{2 d \left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \]
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 = 7.95, size = 387, normalized size = 2.82 \[ \frac {\frac {2\,\left (a^2\,b-4\,b^3\right )}{3\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a\,b^2-a^3\right )}{a^4-2\,a^2\,b^2+b^4}+\frac {4\,b^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{a^4-2\,a^2\,b^2+b^4}-\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (4\,a\,b^2-a^3\right )}{3\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (2\,a\,b^2-a^3\right )}{a^4-2\,a^2\,b^2+b^4}+\frac {2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (a^2-2\,b^2\right )}{a^4-2\,a^2\,b^2+b^4}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}+\frac {2\,b^4\,\mathrm {atan}\left (\frac {\frac {b^4\,\left (2\,a^4\,b-4\,a^2\,b^3+2\,b^5\right )}{{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}}+\frac {2\,a\,b^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^4-2\,a^2\,b^2+b^4\right )}{{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}}}{2\,b^4}\right )}{d\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}} \]
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 \[ \int \frac {\sec ^{4}{\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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