Optimal. Leaf size=85 \[ -\frac {x}{a^2}+\frac {2 b \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^2 \sqrt {a-b} \sqrt {a+b} d}+\frac {\tan (c+d x)}{a d (a+b \sec (c+d x))} \]
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Rubi [A]
time = 0.11, antiderivative size = 85, 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 = {3979, 4146, 12,
3868, 2738, 214} \begin {gather*} \frac {2 b \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^2 d \sqrt {a-b} \sqrt {a+b}}-\frac {x}{a^2}+\frac {\tan (c+d x)}{a d (a+b \sec (c+d x))} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 214
Rule 2738
Rule 3868
Rule 3979
Rule 4146
Rubi steps
\begin {align*} \int \frac {\tan ^2(c+d x)}{(a+b \sec (c+d x))^2} \, dx &=\int \frac {-1+\sec ^2(c+d x)}{(a+b \sec (c+d x))^2} \, dx\\ &=\frac {\tan (c+d x)}{a d (a+b \sec (c+d x))}-\frac {\int \frac {a^2-b^2}{a+b \sec (c+d x)} \, dx}{a \left (a^2-b^2\right )}\\ &=\frac {\tan (c+d x)}{a d (a+b \sec (c+d x))}-\frac {\int \frac {1}{a+b \sec (c+d x)} \, dx}{a}\\ &=-\frac {x}{a^2}+\frac {\tan (c+d x)}{a d (a+b \sec (c+d x))}+\frac {\int \frac {1}{1+\frac {a \cos (c+d x)}{b}} \, dx}{a^2}\\ &=-\frac {x}{a^2}+\frac {\tan (c+d x)}{a d (a+b \sec (c+d x))}+\frac {2 \text {Subst}\left (\int \frac {1}{1+\frac {a}{b}+\left (1-\frac {a}{b}\right ) x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^2 d}\\ &=-\frac {x}{a^2}+\frac {2 b \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^2 \sqrt {a-b} \sqrt {a+b} d}+\frac {\tan (c+d x)}{a d (a+b \sec (c+d x))}\\ \end {align*}
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Mathematica [A]
time = 0.29, size = 80, normalized size = 0.94 \begin {gather*} -\frac {c+d x+\frac {2 b \tanh ^{-1}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}-\frac {a \sin (c+d x)}{b+a \cos (c+d x)}}{a^2 d} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.13, size = 115, normalized size = 1.35
method | result | size |
derivativedivides | \(\frac {\frac {-\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a -b}+\frac {2 b \arctanh \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}}{a^{2}}-\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2}}}{d}\) | \(115\) |
default | \(\frac {\frac {-\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a -b}+\frac {2 b \arctanh \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}}{a^{2}}-\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2}}}{d}\) | \(115\) |
risch | \(-\frac {x}{a^{2}}+\frac {2 i \left (b \,{\mathrm e}^{i \left (d x +c \right )}+a \right )}{a^{2} d \left (a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{i \left (d x +c \right )}+a \right )}-\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, d \,a^{2}}+\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, d \,a^{2}}\) | \(200\) |
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 [B] Leaf count of result is larger than twice the leaf count of optimal. 162 vs.
\(2 (76) = 152\).
time = 1.90, size = 383, normalized size = 4.51 \begin {gather*} \left [-\frac {2 \, {\left (a^{3} - a b^{2}\right )} d x \cos \left (d x + c\right ) + 2 \, {\left (a^{2} b - b^{3}\right )} d x - {\left (a b \cos \left (d x + c\right ) + b^{2}\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 ) - 2 \, {\left (a^{3} - a b^{2}\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{5} - a^{3} b^{2}\right )} d \cos \left (d x + c\right ) + {\left (a^{4} b - a^{2} b^{3}\right )} d\right )}}, -\frac {{\left (a^{3} - a b^{2}\right )} d x \cos \left (d x + c\right ) + {\left (a^{2} b - b^{3}\right )} d x - {\left (a b \cos \left (d x + c\right ) + b^{2}\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 (a^{3} - a b^{2}\right )} \sin \left (d x + c\right )}{{\left (a^{5} - a^{3} b^{2}\right )} d \cos \left (d x + c\right ) + {\left (a^{4} b - a^{2} b^{3}\right )} d}\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 {\tan ^{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 = 0.65, size = 144, normalized size = 1.69 \begin {gather*} \frac {\frac {2 \, {\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 )} b}{\sqrt {-a^{2} + b^{2}} a^{2}} - \frac {d x + c}{a^{2}} - \frac {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}}{d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 2.14, size = 551, normalized size = 6.48 \begin {gather*} -\frac {2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{a^2\,d}-\frac {b^2\,\left (a\,\sin \left (c+d\,x\right )+2\,\mathrm {atanh}\left (\frac {2\,b^3\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\left (a^2-b^2\right )}^{3/2}-a^5\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}+2\,b^5\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}-3\,a^2\,b^3\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}+a^3\,b^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}+a^4\,b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a\,b^2-a^3\right )\,\left (b\,\left (a^2-b^2\right )+a\,b^2-a^2\,b-a^3+b^3\right )}\right )\,\sqrt {a^2-b^2}\right )-a^3\,\sin \left (c+d\,x\right )+2\,a\,b\,\cos \left (c+d\,x\right )\,\mathrm {atanh}\left (\frac {2\,b^3\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\left (a^2-b^2\right )}^{3/2}-a^5\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}+2\,b^5\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}-3\,a^2\,b^3\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}+a^3\,b^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}+a^4\,b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a\,b^2-a^3\right )\,\left (b\,\left (a^2-b^2\right )+a\,b^2-a^2\,b-a^3+b^3\right )}\right )\,\sqrt {a^2-b^2}}{a^2\,d\,\left (a^2-b^2\right )\,\left (b+a\,\cos \left (c+d\,x\right )\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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