Optimal. Leaf size=203 \[ -\frac {4 a^2 b \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2} d}-\frac {2 b^3 \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2} d}-\frac {\sin (c+d x)}{2 (a+b)^2 d (1-\cos (c+d x))}+\frac {\sin (c+d x)}{2 (a-b)^2 d (1+\cos (c+d x))}+\frac {a b^2 \sin (c+d x)}{\left (a^2-b^2\right )^2 d (b+a \cos (c+d x))} \]
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Rubi [A]
time = 0.34, antiderivative size = 203, normalized size of antiderivative = 1.00, number of steps
used = 11, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3957, 2810,
2727, 2743, 12, 2738, 214} \begin {gather*} \frac {a b^2 \sin (c+d x)}{d \left (a^2-b^2\right )^2 (a \cos (c+d x)+b)}-\frac {4 a^2 b \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{5/2} (a+b)^{5/2}}-\frac {2 b^3 \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{5/2} (a+b)^{5/2}}-\frac {\sin (c+d x)}{2 d (a+b)^2 (1-\cos (c+d x))}+\frac {\sin (c+d x)}{2 d (a-b)^2 (\cos (c+d x)+1)} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 214
Rule 2727
Rule 2738
Rule 2743
Rule 2810
Rule 3957
Rubi steps
\begin {align*} \int \frac {\csc ^2(c+d x)}{(a+b \sec (c+d x))^2} \, dx &=\int \frac {\cot ^2(c+d x)}{(-b-a \cos (c+d x))^2} \, dx\\ &=\int \left (-\frac {1}{2 (a+b)^2 (-1+\cos (c+d x))}+\frac {1}{2 (a-b)^2 (1+\cos (c+d x))}-\frac {b^2}{\left (-a^2+b^2\right ) (b+a \cos (c+d x))^2}-\frac {2 a^2 b}{\left (a^2-b^2\right )^2 (b+a \cos (c+d x))}\right ) \, dx\\ &=\frac {\int \frac {1}{1+\cos (c+d x)} \, dx}{2 (a-b)^2}-\frac {\int \frac {1}{-1+\cos (c+d x)} \, dx}{2 (a+b)^2}-\frac {\left (2 a^2 b\right ) \int \frac {1}{b+a \cos (c+d x)} \, dx}{\left (a^2-b^2\right )^2}+\frac {b^2 \int \frac {1}{(b+a \cos (c+d x))^2} \, dx}{a^2-b^2}\\ &=-\frac {\sin (c+d x)}{2 (a+b)^2 d (1-\cos (c+d x))}+\frac {\sin (c+d x)}{2 (a-b)^2 d (1+\cos (c+d x))}+\frac {a b^2 \sin (c+d x)}{\left (a^2-b^2\right )^2 d (b+a \cos (c+d x))}-\frac {b^2 \int \frac {b}{b+a \cos (c+d x)} \, dx}{\left (a^2-b^2\right )^2}-\frac {\left (4 a^2 b\right ) \text {Subst}\left (\int \frac {1}{a+b+(-a+b) x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{\left (a^2-b^2\right )^2 d}\\ &=-\frac {4 a^2 b \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2} d}-\frac {\sin (c+d x)}{2 (a+b)^2 d (1-\cos (c+d x))}+\frac {\sin (c+d x)}{2 (a-b)^2 d (1+\cos (c+d x))}+\frac {a b^2 \sin (c+d x)}{\left (a^2-b^2\right )^2 d (b+a \cos (c+d x))}-\frac {b^3 \int \frac {1}{b+a \cos (c+d x)} \, dx}{\left (a^2-b^2\right )^2}\\ &=-\frac {4 a^2 b \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2} d}-\frac {\sin (c+d x)}{2 (a+b)^2 d (1-\cos (c+d x))}+\frac {\sin (c+d x)}{2 (a-b)^2 d (1+\cos (c+d x))}+\frac {a b^2 \sin (c+d x)}{\left (a^2-b^2\right )^2 d (b+a \cos (c+d x))}-\frac {\left (2 b^3\right ) \text {Subst}\left (\int \frac {1}{a+b+(-a+b) x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{\left (a^2-b^2\right )^2 d}\\ &=-\frac {4 a^2 b \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2} d}-\frac {2 b^3 \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2} d}-\frac {\sin (c+d x)}{2 (a+b)^2 d (1-\cos (c+d x))}+\frac {\sin (c+d x)}{2 (a-b)^2 d (1+\cos (c+d x))}+\frac {a b^2 \sin (c+d x)}{\left (a^2-b^2\right )^2 d (b+a \cos (c+d x))}\\ \end {align*}
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Mathematica [A]
time = 0.59, size = 128, normalized size = 0.63 \begin {gather*} \frac {\frac {4 b \left (2 a^2+b^2\right ) \tanh ^{-1}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}-\frac {\cot \left (\frac {1}{2} (c+d x)\right )}{(a+b)^2}+\frac {\frac {2 a b^2 \sin (c+d x)}{(a+b)^2 (b+a \cos (c+d x))}+\tan \left (\frac {1}{2} (c+d x)\right )}{(a-b)^2}}{2 d} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.18, size = 162, normalized size = 0.80
method | result | size |
derivativedivides | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a^{2}-4 b a +2 b^{2}}-\frac {1}{2 \left (a +b \right )^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {2 b \left (-\frac {a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a -b}-\frac {\left (2 a^{2}+b^{2}\right ) \arctanh \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2}}}{d}\) | \(162\) |
default | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a^{2}-4 b a +2 b^{2}}-\frac {1}{2 \left (a +b \right )^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {2 b \left (-\frac {a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a -b}-\frac {\left (2 a^{2}+b^{2}\right ) \arctanh \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2}}}{d}\) | \(162\) |
risch | \(-\frac {2 i \left (-2 b \,a^{2} {\mathrm e}^{3 i \left (d x +c \right )}-b^{3} {\mathrm e}^{3 i \left (d x +c \right )}+a^{3} {\mathrm e}^{2 i \left (d x +c \right )}-4 a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+3 b^{3} {\mathrm e}^{i \left (d x +c \right )}+a^{3}+2 b^{2} a \right )}{\left (a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{i \left (d x +c \right )}+a \right ) \left (a^{2}-b^{2}\right )^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) d}-\frac {2 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right ) a^{2}}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}-\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}+\frac {2 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right ) a^{2}}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}+\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}\) | \(473\) |
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 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 3.53, size = 526, normalized size = 2.59 \begin {gather*} \left [\frac {6 \, a^{3} b^{2} - 6 \, a b^{4} + {\left (2 \, a^{2} b^{2} + b^{4} + {\left (2 \, a^{3} b + 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 ) \sin \left (d x + c\right ) - 2 \, {\left (a^{5} + a^{3} b^{2} - 2 \, a b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )}{2 \, {\left ({\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} d \cos \left (d x + c\right ) + {\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} d\right )} \sin \left (d x + c\right )}, \frac {3 \, a^{3} b^{2} - 3 \, a b^{4} - {\left (2 \, a^{2} b^{2} + b^{4} + {\left (2 \, a^{3} b + 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 ) \sin \left (d x + c\right ) - {\left (a^{5} + a^{3} b^{2} - 2 \, a b^{4}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )}{{\left ({\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} d \cos \left (d x + c\right ) + {\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} d\right )} \sin \left (d x + c\right )}\right ] \end {gather*}
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 \begin {gather*} \int \frac {\csc ^{2}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{2}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.49, size = 289, normalized size = 1.42 \begin {gather*} \frac {\frac {4 \, {\left (2 \, a^{2} b + 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 )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {-a^{2} + b^{2}}} + \frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{2} - 2 \, a b + b^{2}} - \frac {a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 7 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a^{3} + a^{2} b + a b^{2} - b^{3}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 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 ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}}{2 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 1.79, size = 245, normalized size = 1.21 \begin {gather*} \frac {\frac {a^2-2\,a\,b+b^2}{a+b}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (a^3-3\,a^2\,b+7\,a\,b^2-b^3\right )}{{\left (a+b\right )}^2}}{d\,\left (\left (2\,a^3-6\,a^2\,b+6\,a\,b^2-2\,b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (-2\,a^3+2\,a^2\,b+2\,a\,b^2-2\,b^3\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d\,{\left (a-b\right )}^2}+\frac {b\,\mathrm {atan}\left (\frac {1{}\mathrm {i}\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^4-2{}\mathrm {i}\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b^2+1{}\mathrm {i}\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^4}{{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{3/2}}\right )\,\left (2\,a^2+b^2\right )\,2{}\mathrm {i}}{d\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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