Optimal. Leaf size=59 \[ \frac {(a-2 b) \tanh ^{-1}(\sin (x))}{2 a^2}+\frac {b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sin (x)}{\sqrt {a+b}}\right )}{a^2 \sqrt {a+b}}+\frac {\sec (x) \tan (x)}{2 a} \]
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Rubi [A]
time = 0.06, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps
used = 5, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3265, 425, 536,
212, 214} \begin {gather*} \frac {b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sin (x)}{\sqrt {a+b}}\right )}{a^2 \sqrt {a+b}}+\frac {(a-2 b) \tanh ^{-1}(\sin (x))}{2 a^2}+\frac {\tan (x) \sec (x)}{2 a} \end {gather*}
Antiderivative was successfully verified.
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Rule 212
Rule 214
Rule 425
Rule 536
Rule 3265
Rubi steps
\begin {align*} \int \frac {\sec ^3(x)}{a+b \cos ^2(x)} \, dx &=\text {Subst}\left (\int \frac {1}{\left (1-x^2\right )^2 \left (a+b-b x^2\right )} \, dx,x,\sin (x)\right )\\ &=\frac {\sec (x) \tan (x)}{2 a}+\frac {\text {Subst}\left (\int \frac {a-b-b x^2}{\left (1-x^2\right ) \left (a+b-b x^2\right )} \, dx,x,\sin (x)\right )}{2 a}\\ &=\frac {\sec (x) \tan (x)}{2 a}+\frac {(a-2 b) \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sin (x)\right )}{2 a^2}+\frac {b^2 \text {Subst}\left (\int \frac {1}{a+b-b x^2} \, dx,x,\sin (x)\right )}{a^2}\\ &=\frac {(a-2 b) \tanh ^{-1}(\sin (x))}{2 a^2}+\frac {b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sin (x)}{\sqrt {a+b}}\right )}{a^2 \sqrt {a+b}}+\frac {\sec (x) \tan (x)}{2 a}\\ \end {align*}
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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(152\) vs. \(2(59)=118\).
time = 0.43, size = 152, normalized size = 2.58 \begin {gather*} \frac {-2 (a-2 b) \log \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )+2 (a-2 b) \log \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )-\frac {2 b^{3/2} \log \left (\sqrt {a+b}-\sqrt {b} \sin (x)\right )}{\sqrt {a+b}}+\frac {2 b^{3/2} \log \left (\sqrt {a+b}+\sqrt {b} \sin (x)\right )}{\sqrt {a+b}}+\frac {a}{\left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )^2}-\frac {a}{\left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )^2}}{4 a^2} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.20, size = 82, normalized size = 1.39
method | result | size |
default | \(-\frac {1}{4 a \left (\sin \left (x \right )+1\right )}+\frac {\left (a -2 b \right ) \ln \left (\sin \left (x \right )+1\right )}{4 a^{2}}+\frac {b^{2} \arctanh \left (\frac {b \sin \left (x \right )}{\sqrt {\left (a +b \right ) b}}\right )}{a^{2} \sqrt {\left (a +b \right ) b}}-\frac {1}{4 a \left (\sin \left (x \right )-1\right )}+\frac {\left (-a +2 b \right ) \ln \left (\sin \left (x \right )-1\right )}{4 a^{2}}\) | \(82\) |
risch | \(-\frac {i \left ({\mathrm e}^{3 i x}-{\mathrm e}^{i x}\right )}{\left ({\mathrm e}^{2 i x}+1\right )^{2} a}-\frac {\ln \left ({\mathrm e}^{i x}-i\right )}{2 a}+\frac {b \ln \left ({\mathrm e}^{i x}-i\right )}{a^{2}}+\frac {\ln \left ({\mathrm e}^{i x}+i\right )}{2 a}-\frac {b \ln \left ({\mathrm e}^{i x}+i\right )}{a^{2}}+\frac {\sqrt {\left (a +b \right ) b}\, b \ln \left ({\mathrm e}^{2 i x}+\frac {2 i \sqrt {\left (a +b \right ) b}\, {\mathrm e}^{i x}}{b}-1\right )}{2 \left (a +b \right ) a^{2}}-\frac {\sqrt {\left (a +b \right ) b}\, b \ln \left ({\mathrm e}^{2 i x}-\frac {2 i \sqrt {\left (a +b \right ) b}\, {\mathrm e}^{i x}}{b}-1\right )}{2 \left (a +b \right ) a^{2}}\) | \(175\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.49, size = 92, normalized size = 1.56 \begin {gather*} -\frac {b^{2} \log \left (\frac {b \sin \left (x\right ) - \sqrt {{\left (a + b\right )} b}}{b \sin \left (x\right ) + \sqrt {{\left (a + b\right )} b}}\right )}{2 \, \sqrt {{\left (a + b\right )} b} a^{2}} + \frac {{\left (a - 2 \, b\right )} \log \left (\sin \left (x\right ) + 1\right )}{4 \, a^{2}} - \frac {{\left (a - 2 \, b\right )} \log \left (\sin \left (x\right ) - 1\right )}{4 \, a^{2}} - \frac {\sin \left (x\right )}{2 \, {\left (a \sin \left (x\right )^{2} - a\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.48, size = 186, normalized size = 3.15 \begin {gather*} \left [\frac {2 \, b \sqrt {\frac {b}{a + b}} \cos \left (x\right )^{2} \log \left (-\frac {b \cos \left (x\right )^{2} - 2 \, {\left (a + b\right )} \sqrt {\frac {b}{a + b}} \sin \left (x\right ) - a - 2 \, b}{b \cos \left (x\right )^{2} + a}\right ) + {\left (a - 2 \, b\right )} \cos \left (x\right )^{2} \log \left (\sin \left (x\right ) + 1\right ) - {\left (a - 2 \, b\right )} \cos \left (x\right )^{2} \log \left (-\sin \left (x\right ) + 1\right ) + 2 \, a \sin \left (x\right )}{4 \, a^{2} \cos \left (x\right )^{2}}, -\frac {4 \, b \sqrt {-\frac {b}{a + b}} \arctan \left (\sqrt {-\frac {b}{a + b}} \sin \left (x\right )\right ) \cos \left (x\right )^{2} - {\left (a - 2 \, b\right )} \cos \left (x\right )^{2} \log \left (\sin \left (x\right ) + 1\right ) + {\left (a - 2 \, b\right )} \cos \left (x\right )^{2} \log \left (-\sin \left (x\right ) + 1\right ) - 2 \, a \sin \left (x\right )}{4 \, a^{2} \cos \left (x\right )^{2}}\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 {\sec ^{3}{\left (x \right )}}{a + b \cos ^{2}{\left (x \right )}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.41, size = 85, normalized size = 1.44 \begin {gather*} -\frac {b^{2} \arctan \left (\frac {b \sin \left (x\right )}{\sqrt {-a b - b^{2}}}\right )}{\sqrt {-a b - b^{2}} a^{2}} + \frac {{\left (a - 2 \, b\right )} \log \left (\sin \left (x\right ) + 1\right )}{4 \, a^{2}} - \frac {{\left (a - 2 \, b\right )} \log \left (-\sin \left (x\right ) + 1\right )}{4 \, a^{2}} - \frac {\sin \left (x\right )}{2 \, {\left (\sin \left (x\right )^{2} - 1\right )} a} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 2.53, size = 483, normalized size = 8.19 \begin {gather*} -\frac {a^2\,\sin \left (x\right )+a^2\,\mathrm {atanh}\left (\sin \left (x\right )\right )-2\,b^2\,\mathrm {atanh}\left (\sin \left (x\right )\right )+a\,b\,\sin \left (x\right )-a\,b\,\mathrm {atanh}\left (\sin \left (x\right )\right )-a^2\,\mathrm {atanh}\left (\sin \left (x\right )\right )\,{\sin \left (x\right )}^2+2\,b^2\,\mathrm {atanh}\left (\sin \left (x\right )\right )\,{\sin \left (x\right )}^2+a\,b\,\mathrm {atanh}\left (\sin \left (x\right )\right )\,{\sin \left (x\right )}^2+\mathrm {atan}\left (\frac {b^5\,\sin \left (x\right )\,\sqrt {b^4+a\,b^3}\,8{}\mathrm {i}-a\,\sin \left (x\right )\,{\left (b^4+a\,b^3\right )}^{3/2}\,4{}\mathrm {i}-b\,\sin \left (x\right )\,{\left (b^4+a\,b^3\right )}^{3/2}\,8{}\mathrm {i}+a\,b^4\,\sin \left (x\right )\,\sqrt {b^4+a\,b^3}\,12{}\mathrm {i}+a^4\,b\,\sin \left (x\right )\,\sqrt {b^4+a\,b^3}\,1{}\mathrm {i}+a^2\,b^3\,\sin \left (x\right )\,\sqrt {b^4+a\,b^3}\,1{}\mathrm {i}-a^3\,b^2\,\sin \left (x\right )\,\sqrt {b^4+a\,b^3}\,2{}\mathrm {i}}{-a^5\,b^2+a^4\,b^3+5\,a^3\,b^4+3\,a^2\,b^5}\right )\,\sqrt {b^4+a\,b^3}\,2{}\mathrm {i}-\mathrm {atan}\left (\frac {b^5\,\sin \left (x\right )\,\sqrt {b^4+a\,b^3}\,8{}\mathrm {i}-a\,\sin \left (x\right )\,{\left (b^4+a\,b^3\right )}^{3/2}\,4{}\mathrm {i}-b\,\sin \left (x\right )\,{\left (b^4+a\,b^3\right )}^{3/2}\,8{}\mathrm {i}+a\,b^4\,\sin \left (x\right )\,\sqrt {b^4+a\,b^3}\,12{}\mathrm {i}+a^4\,b\,\sin \left (x\right )\,\sqrt {b^4+a\,b^3}\,1{}\mathrm {i}+a^2\,b^3\,\sin \left (x\right )\,\sqrt {b^4+a\,b^3}\,1{}\mathrm {i}-a^3\,b^2\,\sin \left (x\right )\,\sqrt {b^4+a\,b^3}\,2{}\mathrm {i}}{-a^5\,b^2+a^4\,b^3+5\,a^3\,b^4+3\,a^2\,b^5}\right )\,{\sin \left (x\right )}^2\,\sqrt {b^4+a\,b^3}\,2{}\mathrm {i}}{2\,a^3\,{\sin \left (x\right )}^2-2\,a^3+2\,b\,a^2\,{\sin \left (x\right )}^2-2\,b\,a^2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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