Optimal. Leaf size=142 \[ -\frac{2 a c \tanh ^{-1}\left (\frac{b-(a-c) \tan \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2-c^2}}\right )}{\left (b^2-c^2\right ) \sqrt{a^2+b^2-c^2}}+\frac{b \log \left (-(a-c) \tan ^2\left (\frac{x}{2}\right )+a+2 b \tan \left (\frac{x}{2}\right )+c\right )}{b^2-c^2}-\frac{\log \left (1-\tan \left (\frac{x}{2}\right )\right )}{b+c}-\frac{\log \left (\tan \left (\frac{x}{2}\right )+1\right )}{b-c} \]
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Rubi [A] time = 0.514689, antiderivative size = 142, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.471, Rules used = {4397, 1075, 634, 618, 206, 628, 633, 31} \[ -\frac{2 a c \tanh ^{-1}\left (\frac{b-(a-c) \tan \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2-c^2}}\right )}{\left (b^2-c^2\right ) \sqrt{a^2+b^2-c^2}}+\frac{b \log \left (-(a-c) \tan ^2\left (\frac{x}{2}\right )+a+2 b \tan \left (\frac{x}{2}\right )+c\right )}{b^2-c^2}-\frac{\log \left (1-\tan \left (\frac{x}{2}\right )\right )}{b+c}-\frac{\log \left (\tan \left (\frac{x}{2}\right )+1\right )}{b-c} \]
Antiderivative was successfully verified.
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Rule 4397
Rule 1075
Rule 634
Rule 618
Rule 206
Rule 628
Rule 633
Rule 31
Rubi steps
\begin{align*} \int \frac{\sec ^2(x)}{a+c \sec (x)+b \tan (x)} \, dx &=\int \frac{\sec (x)}{c+a \cos (x)+b \sin (x)} \, dx\\ &=2 \operatorname{Subst}\left (\int \frac{1+x^2}{\left (1-x^2\right ) \left (a+c+2 b x-(a-c) x^2\right )} \, dx,x,\tan \left (\frac{x}{2}\right )\right )\\ &=-\frac{\operatorname{Subst}\left (\int \frac{4 c-4 b x}{1-x^2} \, dx,x,\tan \left (\frac{x}{2}\right )\right )}{2 \left (b^2-c^2\right )}-\frac{\operatorname{Subst}\left (\int \frac{-4 b^2+(-a+c)^2-(a+c)^2-4 b (-a+c) x}{a+c+2 b x+(-a+c) x^2} \, dx,x,\tan \left (\frac{x}{2}\right )\right )}{2 \left (b^2-c^2\right )}\\ &=\frac{\operatorname{Subst}\left (\int \frac{1}{-1-x} \, dx,x,\tan \left (\frac{x}{2}\right )\right )}{b-c}+\frac{\operatorname{Subst}\left (\int \frac{1}{1-x} \, dx,x,\tan \left (\frac{x}{2}\right )\right )}{b+c}+\frac{b \operatorname{Subst}\left (\int \frac{2 b+2 (-a+c) x}{a+c+2 b x+(-a+c) x^2} \, dx,x,\tan \left (\frac{x}{2}\right )\right )}{b^2-c^2}+\frac{(2 a c) \operatorname{Subst}\left (\int \frac{1}{a+c+2 b x+(-a+c) x^2} \, dx,x,\tan \left (\frac{x}{2}\right )\right )}{b^2-c^2}\\ &=-\frac{\log \left (1-\tan \left (\frac{x}{2}\right )\right )}{b+c}-\frac{\log \left (1+\tan \left (\frac{x}{2}\right )\right )}{b-c}+\frac{b \log \left (a+c+2 b \tan \left (\frac{x}{2}\right )-(a-c) \tan ^2\left (\frac{x}{2}\right )\right )}{b^2-c^2}-\frac{(4 a c) \operatorname{Subst}\left (\int \frac{1}{4 \left (a^2+b^2-c^2\right )-x^2} \, dx,x,2 b+2 (-a+c) \tan \left (\frac{x}{2}\right )\right )}{b^2-c^2}\\ &=-\frac{2 a c \tanh ^{-1}\left (\frac{b-(a-c) \tan \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2-c^2}}\right )}{\left (b^2-c^2\right ) \sqrt{a^2+b^2-c^2}}-\frac{\log \left (1-\tan \left (\frac{x}{2}\right )\right )}{b+c}-\frac{\log \left (1+\tan \left (\frac{x}{2}\right )\right )}{b-c}+\frac{b \log \left (a+c+2 b \tan \left (\frac{x}{2}\right )-(a-c) \tan ^2\left (\frac{x}{2}\right )\right )}{b^2-c^2}\\ \end{align*}
Mathematica [A] time = 0.289371, size = 120, normalized size = 0.85 \[ \frac{\frac{2 a c \tanh ^{-1}\left (\frac{(c-a) \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2+b^2-c^2}}\right )}{\sqrt{a^2+b^2-c^2}}-b \log (a \cos (x)+b \sin (x)+c)+(b-c) \log \left (\cos \left (\frac{x}{2}\right )-\sin \left (\frac{x}{2}\right )\right )+(b+c) \log \left (\sin \left (\frac{x}{2}\right )+\cos \left (\frac{x}{2}\right )\right )}{(c-b) (b+c)} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.059, size = 430, normalized size = 3. \begin{align*}{\frac{ab}{ \left ( b-c \right ) \left ( b+c \right ) \left ( a-c \right ) }\ln \left ( a \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}- \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}c-2\,b\tan \left ( x/2 \right ) -a-c \right ) }-{\frac{cb}{ \left ( b-c \right ) \left ( b+c \right ) \left ( a-c \right ) }\ln \left ( a \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}- \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}c-2\,b\tan \left ( x/2 \right ) -a-c \right ) }-2\,{\frac{ac}{ \left ( b-c \right ) \left ( b+c \right ) \sqrt{-{a}^{2}-{b}^{2}+{c}^{2}}}\arctan \left ( 1/2\,{\frac{2\, \left ( a-c \right ) \tan \left ( x/2 \right ) -2\,b}{\sqrt{-{a}^{2}-{b}^{2}+{c}^{2}}}} \right ) }-2\,{\frac{{b}^{2}}{ \left ( b-c \right ) \left ( b+c \right ) \sqrt{-{a}^{2}-{b}^{2}+{c}^{2}}}\arctan \left ( 1/2\,{\frac{2\, \left ( a-c \right ) \tan \left ( x/2 \right ) -2\,b}{\sqrt{-{a}^{2}-{b}^{2}+{c}^{2}}}} \right ) }+2\,{\frac{a{b}^{2}}{ \left ( b-c \right ) \left ( b+c \right ) \sqrt{-{a}^{2}-{b}^{2}+{c}^{2}} \left ( a-c \right ) }\arctan \left ( 1/2\,{\frac{2\, \left ( a-c \right ) \tan \left ( x/2 \right ) -2\,b}{\sqrt{-{a}^{2}-{b}^{2}+{c}^{2}}}} \right ) }-2\,{\frac{c{b}^{2}}{ \left ( b-c \right ) \left ( b+c \right ) \sqrt{-{a}^{2}-{b}^{2}+{c}^{2}} \left ( a-c \right ) }\arctan \left ( 1/2\,{\frac{2\, \left ( a-c \right ) \tan \left ( x/2 \right ) -2\,b}{\sqrt{-{a}^{2}-{b}^{2}+{c}^{2}}}} \right ) }-2\,{\frac{\ln \left ( 1+\tan \left ( x/2 \right ) \right ) }{-2\,c+2\,b}}-2\,{\frac{\ln \left ( \tan \left ( x/2 \right ) -1 \right ) }{2\,b+2\,c}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 21.9246, size = 1539, normalized size = 10.84 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{2}{\left (x \right )}}{a + b \tan{\left (x \right )} + c \sec{\left (x \right )}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.23609, size = 217, normalized size = 1.53 \begin{align*} \frac{2 \,{\left (\pi \left \lfloor \frac{x}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (-2 \, a + 2 \, c\right ) + \arctan \left (-\frac{a \tan \left (\frac{1}{2} \, x\right ) - c \tan \left (\frac{1}{2} \, x\right ) - b}{\sqrt{-a^{2} - b^{2} + c^{2}}}\right )\right )} a c}{\sqrt{-a^{2} - b^{2} + c^{2}}{\left (b^{2} - c^{2}\right )}} + \frac{b \log \left (-a \tan \left (\frac{1}{2} \, x\right )^{2} + c \tan \left (\frac{1}{2} \, x\right )^{2} + 2 \, b \tan \left (\frac{1}{2} \, x\right ) + a + c\right )}{b^{2} - c^{2}} - \frac{\log \left ({\left | \tan \left (\frac{1}{2} \, x\right ) + 1 \right |}\right )}{b - c} - \frac{\log \left ({\left | \tan \left (\frac{1}{2} \, x\right ) - 1 \right |}\right )}{b + c} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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