Optimal. Leaf size=84 \[ -\frac {2 b^2 \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{3/2}}-\frac {\sec (c+d x) (b-a \sin (c+d x))}{d \left (a^2-b^2\right )} \]
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Rubi [A] time = 0.09, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2696, 12, 2660, 618, 204} \[ -\frac {2 b^2 \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{3/2}}-\frac {\sec (c+d x) (b-a \sin (c+d x))}{d \left (a^2-b^2\right )} \]
Antiderivative was successfully verified.
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Rule 12
Rule 204
Rule 618
Rule 2660
Rule 2696
Rubi steps
\begin {align*} \int \frac {\sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx &=-\frac {\sec (c+d x) (b-a \sin (c+d x))}{\left (a^2-b^2\right ) d}+\frac {\int \frac {b^2}{a+b \sin (c+d x)} \, dx}{-a^2+b^2}\\ &=-\frac {\sec (c+d x) (b-a \sin (c+d x))}{\left (a^2-b^2\right ) d}-\frac {b^2 \int \frac {1}{a+b \sin (c+d x)} \, dx}{a^2-b^2}\\ &=-\frac {\sec (c+d x) (b-a \sin (c+d x))}{\left (a^2-b^2\right ) d}-\frac {\left (2 b^2\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 ) d}\\ &=-\frac {\sec (c+d x) (b-a \sin (c+d x))}{\left (a^2-b^2\right ) d}+\frac {\left (4 b^2\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 ) d}\\ &=-\frac {2 b^2 \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 )^{3/2} d}-\frac {\sec (c+d x) (b-a \sin (c+d x))}{\left (a^2-b^2\right ) d}\\ \end {align*}
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Mathematica [A] time = 0.32, size = 152, normalized size = 1.81 \[ \frac {\sqrt {a^2-b^2} (-a \sin (c+d x)+b (-\cos (c+d x))+b)+2 b^2 \cos (c+d x) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{d (b-a) (a+b) \sqrt {a^2-b^2} \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.50, size = 305, normalized size = 3.63 \[ \left [\frac {\sqrt {-a^{2} + b^{2}} b^{2} \cos \left (d x + c\right ) \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) - 2 \, a^{2} b + 2 \, b^{3} + 2 \, {\left (a^{3} - a b^{2}\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} d \cos \left (d x + c\right )}, \frac {\sqrt {a^{2} - b^{2}} b^{2} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) \cos \left (d x + c\right ) - a^{2} b + b^{3} + {\left (a^{3} - a b^{2}\right )} \sin \left (d x + c\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} d \cos \left (d x + c\right )}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 1.17, size = 107, normalized size = 1.27 \[ -\frac {2 \, {\left (\frac {{\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} b^{2}}{{\left (a^{2} - b^{2}\right )}^{\frac {3}{2}}} + \frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b}{{\left (a^{2} - b^{2}\right )} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}}\right )}}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.16, size = 117, normalized size = 1.39 \[ -\frac {2}{d \left (2 a +2 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {2}{d \left (2 a -2 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {2 b^{2} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d \left (a -b \right ) \left (a +b \right ) \sqrt {a^{2}-b^{2}}} \]
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 = 5.26, size = 149, normalized size = 1.77 \[ \frac {\frac {2\,b}{a^2-b^2}-\frac {2\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^2-b^2}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}-\frac {2\,b^2\,\mathrm {atan}\left (\frac {\frac {b^2\,\left (2\,a^2\,b-2\,b^3\right )}{{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}+\frac {2\,a\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2-b^2\right )}{{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}}{2\,b^2}\right )}{d\,{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}} \]
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 \[ \int \frac {\sec ^{2}{\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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