Optimal. Leaf size=90 \[ \frac {\left (3 a^2-4 a b+8 b^2\right ) \tanh ^{-1}(\sin (x))}{8 a^3}-\frac {b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sin (x)}{\sqrt {a+b}}\right )}{a^3 \sqrt {a+b}}+\frac {(3 a-4 b) \sec (x) \tan (x)}{8 a^2}+\frac {\sec ^3(x) \tan (x)}{4 a} \]
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Rubi [A]
time = 0.11, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps
used = 6, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3265, 425, 541,
536, 212, 214} \begin {gather*} -\frac {b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sin (x)}{\sqrt {a+b}}\right )}{a^3 \sqrt {a+b}}+\frac {(3 a-4 b) \tan (x) \sec (x)}{8 a^2}+\frac {\left (3 a^2-4 a b+8 b^2\right ) \tanh ^{-1}(\sin (x))}{8 a^3}+\frac {\tan (x) \sec ^3(x)}{4 a} \end {gather*}
Antiderivative was successfully verified.
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Rule 212
Rule 214
Rule 425
Rule 536
Rule 541
Rule 3265
Rubi steps
\begin {align*} \int \frac {\sec ^5(x)}{a+b \cos ^2(x)} \, dx &=\text {Subst}\left (\int \frac {1}{\left (1-x^2\right )^3 \left (a+b-b x^2\right )} \, dx,x,\sin (x)\right )\\ &=\frac {\sec ^3(x) \tan (x)}{4 a}+\frac {\text {Subst}\left (\int \frac {3 a-b-3 b x^2}{\left (1-x^2\right )^2 \left (a+b-b x^2\right )} \, dx,x,\sin (x)\right )}{4 a}\\ &=\frac {(3 a-4 b) \sec (x) \tan (x)}{8 a^2}+\frac {\sec ^3(x) \tan (x)}{4 a}+\frac {\text {Subst}\left (\int \frac {3 a^2-a b+4 b^2-(3 a-4 b) b x^2}{\left (1-x^2\right ) \left (a+b-b x^2\right )} \, dx,x,\sin (x)\right )}{8 a^2}\\ &=\frac {(3 a-4 b) \sec (x) \tan (x)}{8 a^2}+\frac {\sec ^3(x) \tan (x)}{4 a}-\frac {b^3 \text {Subst}\left (\int \frac {1}{a+b-b x^2} \, dx,x,\sin (x)\right )}{a^3}+\frac {\left (3 a^2-4 a b+8 b^2\right ) \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sin (x)\right )}{8 a^3}\\ &=\frac {\left (3 a^2-4 a b+8 b^2\right ) \tanh ^{-1}(\sin (x))}{8 a^3}-\frac {b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sin (x)}{\sqrt {a+b}}\right )}{a^3 \sqrt {a+b}}+\frac {(3 a-4 b) \sec (x) \tan (x)}{8 a^2}+\frac {\sec ^3(x) \tan (x)}{4 a}\\ \end {align*}
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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(215\) vs. \(2(90)=180\).
time = 1.34, size = 215, normalized size = 2.39 \begin {gather*} \frac {-2 \left (3 a^2-4 a b+8 b^2\right ) \log \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )+2 \left (3 a^2-4 a b+8 b^2\right ) \log \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )+\frac {8 b^{5/2} \log \left (\sqrt {a+b}-\sqrt {b} \sin (x)\right )}{\sqrt {a+b}}-\frac {8 b^{5/2} \log \left (\sqrt {a+b}+\sqrt {b} \sin (x)\right )}{\sqrt {a+b}}+\frac {a^2}{\left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )^4}-\frac {a^2}{\left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )^4}+\frac {a (-3 a+4 b)}{\left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )^2}+\frac {a (-3 a+4 b)}{-1+\sin (x)}}{16 a^3} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.25, size = 137, normalized size = 1.52
method | result | size |
default | \(-\frac {1}{16 a \left (\sin \left (x \right )+1\right )^{2}}-\frac {3 a -4 b}{16 a^{2} \left (\sin \left (x \right )+1\right )}+\frac {\left (3 a^{2}-4 a b +8 b^{2}\right ) \ln \left (\sin \left (x \right )+1\right )}{16 a^{3}}-\frac {b^{3} \arctanh \left (\frac {b \sin \left (x \right )}{\sqrt {\left (a +b \right ) b}}\right )}{a^{3} \sqrt {\left (a +b \right ) b}}+\frac {1}{16 a \left (\sin \left (x \right )-1\right )^{2}}-\frac {3 a -4 b}{16 a^{2} \left (\sin \left (x \right )-1\right )}+\frac {\left (-3 a^{2}+4 a b -8 b^{2}\right ) \ln \left (\sin \left (x \right )-1\right )}{16 a^{3}}\) | \(137\) |
risch | \(-\frac {i \left (3 a \,{\mathrm e}^{7 i x}-4 b \,{\mathrm e}^{7 i x}+11 a \,{\mathrm e}^{5 i x}-4 b \,{\mathrm e}^{5 i x}-11 a \,{\mathrm e}^{3 i x}+4 b \,{\mathrm e}^{3 i x}-3 a \,{\mathrm e}^{i x}+4 b \,{\mathrm e}^{i x}\right )}{4 \left ({\mathrm e}^{2 i x}+1\right )^{4} a^{2}}-\frac {3 \ln \left ({\mathrm e}^{i x}-i\right )}{8 a}+\frac {b \ln \left ({\mathrm e}^{i x}-i\right )}{2 a^{2}}-\frac {\ln \left ({\mathrm e}^{i x}-i\right ) b^{2}}{a^{3}}+\frac {3 \ln \left ({\mathrm e}^{i x}+i\right )}{8 a}-\frac {b \ln \left ({\mathrm e}^{i x}+i\right )}{2 a^{2}}+\frac {\ln \left ({\mathrm e}^{i x}+i\right ) b^{2}}{a^{3}}+\frac {\sqrt {\left (a +b \right ) b}\, b^{2} \ln \left ({\mathrm e}^{2 i x}-\frac {2 i \sqrt {\left (a +b \right ) b}\, {\mathrm e}^{i x}}{b}-1\right )}{2 \left (a +b \right ) a^{3}}-\frac {\sqrt {\left (a +b \right ) b}\, b^{2} \ln \left ({\mathrm e}^{2 i x}+\frac {2 i \sqrt {\left (a +b \right ) b}\, {\mathrm e}^{i x}}{b}-1\right )}{2 \left (a +b \right ) a^{3}}\) | \(265\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.48, size = 145, normalized size = 1.61 \begin {gather*} \frac {b^{3} \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^{3}} - \frac {{\left (3 \, a - 4 \, b\right )} \sin \left (x\right )^{3} - {\left (5 \, a - 4 \, b\right )} \sin \left (x\right )}{8 \, {\left (a^{2} \sin \left (x\right )^{4} - 2 \, a^{2} \sin \left (x\right )^{2} + a^{2}\right )}} + \frac {{\left (3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} \log \left (\sin \left (x\right ) + 1\right )}{16 \, a^{3}} - \frac {{\left (3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} \log \left (\sin \left (x\right ) - 1\right )}{16 \, a^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.45, size = 270, normalized size = 3.00 \begin {gather*} \left [\frac {8 \, b^{2} \sqrt {\frac {b}{a + b}} \cos \left (x\right )^{4} \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 (3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} \cos \left (x\right )^{4} \log \left (\sin \left (x\right ) + 1\right ) - {\left (3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} \cos \left (x\right )^{4} \log \left (-\sin \left (x\right ) + 1\right ) + 2 \, {\left ({\left (3 \, a^{2} - 4 \, a b\right )} \cos \left (x\right )^{2} + 2 \, a^{2}\right )} \sin \left (x\right )}{16 \, a^{3} \cos \left (x\right )^{4}}, \frac {16 \, b^{2} \sqrt {-\frac {b}{a + b}} \arctan \left (\sqrt {-\frac {b}{a + b}} \sin \left (x\right )\right ) \cos \left (x\right )^{4} + {\left (3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} \cos \left (x\right )^{4} \log \left (\sin \left (x\right ) + 1\right ) - {\left (3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} \cos \left (x\right )^{4} \log \left (-\sin \left (x\right ) + 1\right ) + 2 \, {\left ({\left (3 \, a^{2} - 4 \, a b\right )} \cos \left (x\right )^{2} + 2 \, a^{2}\right )} \sin \left (x\right )}{16 \, a^{3} \cos \left (x\right )^{4}}\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 ^{5}{\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 = 127, normalized size = 1.41 \begin {gather*} \frac {b^{3} \arctan \left (\frac {b \sin \left (x\right )}{\sqrt {-a b - b^{2}}}\right )}{\sqrt {-a b - b^{2}} a^{3}} + \frac {{\left (3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} \log \left (\sin \left (x\right ) + 1\right )}{16 \, a^{3}} - \frac {{\left (3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} \log \left (-\sin \left (x\right ) + 1\right )}{16 \, a^{3}} - \frac {3 \, a \sin \left (x\right )^{3} - 4 \, b \sin \left (x\right )^{3} - 5 \, a \sin \left (x\right ) + 4 \, b \sin \left (x\right )}{8 \, {\left (\sin \left (x\right )^{2} - 1\right )}^{2} a^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 2.64, size = 969, normalized size = 10.77 \begin {gather*} \frac {5\,a^3\,\sin \left (x\right )-3\,a^3\,{\sin \left (x\right )}^3+3\,a^3\,\mathrm {atanh}\left (\sin \left (x\right )\right )+8\,b^3\,\mathrm {atanh}\left (\sin \left (x\right )\right )-4\,a\,b^2\,\sin \left (x\right )+a^2\,b\,\sin \left (x\right )-6\,a^3\,\mathrm {atanh}\left (\sin \left (x\right )\right )\,{\sin \left (x\right )}^2+3\,a^3\,\mathrm {atanh}\left (\sin \left (x\right )\right )\,{\sin \left (x\right )}^4-16\,b^3\,\mathrm {atanh}\left (\sin \left (x\right )\right )\,{\sin \left (x\right )}^2+8\,b^3\,\mathrm {atanh}\left (\sin \left (x\right )\right )\,{\sin \left (x\right )}^4+4\,a\,b^2\,{\sin \left (x\right )}^3+a^2\,b\,{\sin \left (x\right )}^3+4\,a\,b^2\,\mathrm {atanh}\left (\sin \left (x\right )\right )-a^2\,b\,\mathrm {atanh}\left (\sin \left (x\right )\right )-8\,a\,b^2\,\mathrm {atanh}\left (\sin \left (x\right )\right )\,{\sin \left (x\right )}^2+2\,a^2\,b\,\mathrm {atanh}\left (\sin \left (x\right )\right )\,{\sin \left (x\right )}^2+4\,a\,b^2\,\mathrm {atanh}\left (\sin \left (x\right )\right )\,{\sin \left (x\right )}^4-a^2\,b\,\mathrm {atanh}\left (\sin \left (x\right )\right )\,{\sin \left (x\right )}^4+\mathrm {atan}\left (\frac {b^7\,\sin \left (x\right )\,\sqrt {b^6+a\,b^5}\,128{}\mathrm {i}-a\,\sin \left (x\right )\,{\left (b^6+a\,b^5\right )}^{3/2}\,64{}\mathrm {i}-b\,\sin \left (x\right )\,{\left (b^6+a\,b^5\right )}^{3/2}\,128{}\mathrm {i}+a\,b^6\,\sin \left (x\right )\,\sqrt {b^6+a\,b^5}\,192{}\mathrm {i}+a^6\,b\,\sin \left (x\right )\,\sqrt {b^6+a\,b^5}\,9{}\mathrm {i}+a^2\,b^5\,\sin \left (x\right )\,\sqrt {b^6+a\,b^5}\,64{}\mathrm {i}+a^3\,b^4\,\sin \left (x\right )\,\sqrt {b^6+a\,b^5}\,40{}\mathrm {i}+a^4\,b^3\,\sin \left (x\right )\,\sqrt {b^6+a\,b^5}\,25{}\mathrm {i}-a^5\,b^2\,\sin \left (x\right )\,\sqrt {b^6+a\,b^5}\,6{}\mathrm {i}}{9\,a^7\,b^3+3\,a^6\,b^4+19\,a^5\,b^5+65\,a^4\,b^6+40\,a^3\,b^7}\right )\,\sqrt {b^6+a\,b^5}\,8{}\mathrm {i}-\mathrm {atan}\left (\frac {b^7\,\sin \left (x\right )\,\sqrt {b^6+a\,b^5}\,128{}\mathrm {i}-a\,\sin \left (x\right )\,{\left (b^6+a\,b^5\right )}^{3/2}\,64{}\mathrm {i}-b\,\sin \left (x\right )\,{\left (b^6+a\,b^5\right )}^{3/2}\,128{}\mathrm {i}+a\,b^6\,\sin \left (x\right )\,\sqrt {b^6+a\,b^5}\,192{}\mathrm {i}+a^6\,b\,\sin \left (x\right )\,\sqrt {b^6+a\,b^5}\,9{}\mathrm {i}+a^2\,b^5\,\sin \left (x\right )\,\sqrt {b^6+a\,b^5}\,64{}\mathrm {i}+a^3\,b^4\,\sin \left (x\right )\,\sqrt {b^6+a\,b^5}\,40{}\mathrm {i}+a^4\,b^3\,\sin \left (x\right )\,\sqrt {b^6+a\,b^5}\,25{}\mathrm {i}-a^5\,b^2\,\sin \left (x\right )\,\sqrt {b^6+a\,b^5}\,6{}\mathrm {i}}{9\,a^7\,b^3+3\,a^6\,b^4+19\,a^5\,b^5+65\,a^4\,b^6+40\,a^3\,b^7}\right )\,{\sin \left (x\right )}^2\,\sqrt {b^6+a\,b^5}\,16{}\mathrm {i}+\mathrm {atan}\left (\frac {b^7\,\sin \left (x\right )\,\sqrt {b^6+a\,b^5}\,128{}\mathrm {i}-a\,\sin \left (x\right )\,{\left (b^6+a\,b^5\right )}^{3/2}\,64{}\mathrm {i}-b\,\sin \left (x\right )\,{\left (b^6+a\,b^5\right )}^{3/2}\,128{}\mathrm {i}+a\,b^6\,\sin \left (x\right )\,\sqrt {b^6+a\,b^5}\,192{}\mathrm {i}+a^6\,b\,\sin \left (x\right )\,\sqrt {b^6+a\,b^5}\,9{}\mathrm {i}+a^2\,b^5\,\sin \left (x\right )\,\sqrt {b^6+a\,b^5}\,64{}\mathrm {i}+a^3\,b^4\,\sin \left (x\right )\,\sqrt {b^6+a\,b^5}\,40{}\mathrm {i}+a^4\,b^3\,\sin \left (x\right )\,\sqrt {b^6+a\,b^5}\,25{}\mathrm {i}-a^5\,b^2\,\sin \left (x\right )\,\sqrt {b^6+a\,b^5}\,6{}\mathrm {i}}{9\,a^7\,b^3+3\,a^6\,b^4+19\,a^5\,b^5+65\,a^4\,b^6+40\,a^3\,b^7}\right )\,{\sin \left (x\right )}^4\,\sqrt {b^6+a\,b^5}\,8{}\mathrm {i}}{8\,a^4\,{\sin \left (x\right )}^4-16\,a^4\,{\sin \left (x\right )}^2+8\,a^4+8\,b\,a^3\,{\sin \left (x\right )}^4-16\,b\,a^3\,{\sin \left (x\right )}^2+8\,b\,a^3} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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