Optimal. Leaf size=152 \[ \frac {3 b \sin (c+d x)}{a^3 d}-\frac {3 \sin (c+d x) \cos (c+d x)}{2 a^2 d}-\frac {2 b \left (2 a^2-3 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^4 d \sqrt {a-b} \sqrt {a+b}}+\frac {x \left (a^2-6 b^2\right )}{2 a^4}+\frac {\sin (c+d x) \cos ^2(c+d x)}{a d (a \cos (c+d x)+b)} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.57, antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3872, 2889, 3048, 3050, 3023, 2735, 2659, 208} \[ -\frac {2 b \left (2 a^2-3 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^4 d \sqrt {a-b} \sqrt {a+b}}+\frac {x \left (a^2-6 b^2\right )}{2 a^4}+\frac {3 b \sin (c+d x)}{a^3 d}-\frac {3 \sin (c+d x) \cos (c+d x)}{2 a^2 d}+\frac {\sin (c+d x) \cos ^2(c+d x)}{a d (a \cos (c+d x)+b)} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 208
Rule 2659
Rule 2735
Rule 2889
Rule 3023
Rule 3048
Rule 3050
Rule 3872
Rubi steps
\begin {align*} \int \frac {\sin ^2(c+d x)}{(a+b \sec (c+d x))^2} \, dx &=\int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(-b-a \cos (c+d x))^2} \, dx\\ &=\int \frac {\cos ^2(c+d x) \left (1-\cos ^2(c+d x)\right )}{(-b-a \cos (c+d x))^2} \, dx\\ &=\frac {\cos ^2(c+d x) \sin (c+d x)}{a d (b+a \cos (c+d x))}-\frac {\int \frac {\cos (c+d x) \left (2 \left (a^2-b^2\right )-3 \left (a^2-b^2\right ) \cos ^2(c+d x)\right )}{-b-a \cos (c+d x)} \, dx}{a \left (a^2-b^2\right )}\\ &=-\frac {3 \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac {\cos ^2(c+d x) \sin (c+d x)}{a d (b+a \cos (c+d x))}+\frac {\int \frac {3 b \left (a^2-b^2\right )-a \left (a^2-b^2\right ) \cos (c+d x)-6 b \left (a^2-b^2\right ) \cos ^2(c+d x)}{-b-a \cos (c+d x)} \, dx}{2 a^2 \left (a^2-b^2\right )}\\ &=\frac {3 b \sin (c+d x)}{a^3 d}-\frac {3 \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac {\cos ^2(c+d x) \sin (c+d x)}{a d (b+a \cos (c+d x))}-\frac {\int \frac {-3 a b \left (a^2-b^2\right )+\left (a^2-6 b^2\right ) \left (a^2-b^2\right ) \cos (c+d x)}{-b-a \cos (c+d x)} \, dx}{2 a^3 \left (a^2-b^2\right )}\\ &=\frac {\left (a^2-6 b^2\right ) x}{2 a^4}+\frac {3 b \sin (c+d x)}{a^3 d}-\frac {3 \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac {\cos ^2(c+d x) \sin (c+d x)}{a d (b+a \cos (c+d x))}+\frac {\left (b \left (2 a^2-3 b^2\right )\right ) \int \frac {1}{-b-a \cos (c+d x)} \, dx}{a^4}\\ &=\frac {\left (a^2-6 b^2\right ) x}{2 a^4}+\frac {3 b \sin (c+d x)}{a^3 d}-\frac {3 \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac {\cos ^2(c+d x) \sin (c+d x)}{a d (b+a \cos (c+d x))}+\frac {\left (2 b \left (2 a^2-3 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{-a-b+(a-b) x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^4 d}\\ &=\frac {\left (a^2-6 b^2\right ) x}{2 a^4}-\frac {2 b \left (2 a^2-3 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^4 \sqrt {a-b} \sqrt {a+b} d}+\frac {3 b \sin (c+d x)}{a^3 d}-\frac {3 \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac {\cos ^2(c+d x) \sin (c+d x)}{a d (b+a \cos (c+d x))}\\ \end {align*}
________________________________________________________________________________________
Mathematica [A] time = 1.33, size = 178, normalized size = 1.17 \[ \frac {\frac {16 b \left (2 a^2-3 b^2\right ) \tanh ^{-1}\left (\frac {(b-a) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+\frac {a^3 (-\sin (3 (c+d x)))-a \left (a^2-24 b^2\right ) \sin (c+d x)+4 a \left (a^2-6 b^2\right ) (c+d x) \cos (c+d x)+6 a^2 b \sin (2 (c+d x))+4 a^2 b c+4 a^2 b d x-24 b^3 c-24 b^3 d x}{a \cos (c+d x)+b}}{8 a^4 d} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
fricas [A] time = 0.55, size = 551, normalized size = 3.62 \[ \left [\frac {{\left (a^{5} - 7 \, a^{3} b^{2} + 6 \, a b^{4}\right )} d x \cos \left (d x + c\right ) + {\left (a^{4} b - 7 \, a^{2} b^{3} + 6 \, b^{5}\right )} d x - {\left (2 \, a^{2} b^{2} - 3 \, b^{4} + {\left (2 \, a^{3} b - 3 \, a b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) + {\left (6 \, a^{3} b^{2} - 6 \, a b^{4} - {\left (a^{5} - a^{3} b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (a^{4} b - a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{7} - a^{5} b^{2}\right )} d \cos \left (d x + c\right ) + {\left (a^{6} b - a^{4} b^{3}\right )} d\right )}}, \frac {{\left (a^{5} - 7 \, a^{3} b^{2} + 6 \, a b^{4}\right )} d x \cos \left (d x + c\right ) + {\left (a^{4} b - 7 \, a^{2} b^{3} + 6 \, b^{5}\right )} d x - 2 \, {\left (2 \, a^{2} b^{2} - 3 \, b^{4} + {\left (2 \, a^{3} b - 3 \, a b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) + {\left (6 \, a^{3} b^{2} - 6 \, a b^{4} - {\left (a^{5} - a^{3} b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (a^{4} b - a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{7} - a^{5} b^{2}\right )} d \cos \left (d x + c\right ) + {\left (a^{6} b - a^{4} b^{3}\right )} d\right )}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [A] time = 0.26, size = 240, normalized size = 1.58 \[ -\frac {\frac {4 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a - b\right )} a^{3}} - \frac {{\left (a^{2} - 6 \, b^{2}\right )} {\left (d x + c\right )}}{a^{4}} + \frac {4 \, {\left (2 \, a^{2} b - 3 \, b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{\sqrt {-a^{2} + b^{2}} a^{4}} - \frac {2 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a^{3}}}{2 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [B] time = 0.44, size = 325, normalized size = 2.14 \[ -\frac {2 b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{3} \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b -a -b \right )}-\frac {4 b \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{d \,a^{2} \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {6 b^{3} \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{d \,a^{4} \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {4 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b}{d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {6 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{2}}{d \,a^{4}}+\frac {\arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
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.
[In]
[Out]
________________________________________________________________________________________
mupad [B] time = 3.57, size = 1655, normalized size = 10.89 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin ^{2}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{2}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________