Optimal. Leaf size=87 \[ -\frac {\left (8 a^2+20 a b+15 b^2\right ) x}{8 b^3}+\frac {(a+b)^{5/2} \tan ^{-1}\left (\frac {\sqrt {a+b} \tan (x)}{\sqrt {a}}\right )}{\sqrt {a} b^3}-\frac {(4 a+7 b) \cos (x) \sin (x)}{8 b^2}-\frac {\cos ^3(x) \sin (x)}{4 b} \]
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Rubi [A]
time = 0.12, antiderivative size = 87, 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 = {3270, 425, 541,
536, 209, 211} \begin {gather*} -\frac {x \left (8 a^2+20 a b+15 b^2\right )}{8 b^3}+\frac {(a+b)^{5/2} \text {ArcTan}\left (\frac {\sqrt {a+b} \tan (x)}{\sqrt {a}}\right )}{\sqrt {a} b^3}-\frac {(4 a+7 b) \sin (x) \cos (x)}{8 b^2}-\frac {\sin (x) \cos ^3(x)}{4 b} \end {gather*}
Antiderivative was successfully verified.
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Rule 209
Rule 211
Rule 425
Rule 536
Rule 541
Rule 3270
Rubi steps
\begin {align*} \int \frac {\cos ^6(x)}{a+b \sin ^2(x)} \, dx &=\text {Subst}\left (\int \frac {1}{\left (1+x^2\right )^3 \left (a+(a+b) x^2\right )} \, dx,x,\tan (x)\right )\\ &=-\frac {\cos ^3(x) \sin (x)}{4 b}+\frac {\text {Subst}\left (\int \frac {a+4 b-3 (a+b) x^2}{\left (1+x^2\right )^2 \left (a+(a+b) x^2\right )} \, dx,x,\tan (x)\right )}{4 b}\\ &=-\frac {(4 a+7 b) \cos (x) \sin (x)}{8 b^2}-\frac {\cos ^3(x) \sin (x)}{4 b}+\frac {\text {Subst}\left (\int \frac {4 a^2+9 a b+8 b^2-(a+b) (4 a+7 b) x^2}{\left (1+x^2\right ) \left (a+(a+b) x^2\right )} \, dx,x,\tan (x)\right )}{8 b^2}\\ &=-\frac {(4 a+7 b) \cos (x) \sin (x)}{8 b^2}-\frac {\cos ^3(x) \sin (x)}{4 b}+\frac {(a+b)^3 \text {Subst}\left (\int \frac {1}{a+(a+b) x^2} \, dx,x,\tan (x)\right )}{b^3}-\frac {\left (8 a^2+20 a b+15 b^2\right ) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (x)\right )}{8 b^3}\\ &=-\frac {\left (8 a^2+20 a b+15 b^2\right ) x}{8 b^3}+\frac {(a+b)^{5/2} \tan ^{-1}\left (\frac {\sqrt {a+b} \tan (x)}{\sqrt {a}}\right )}{\sqrt {a} b^3}-\frac {(4 a+7 b) \cos (x) \sin (x)}{8 b^2}-\frac {\cos ^3(x) \sin (x)}{4 b}\\ \end {align*}
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Mathematica [A]
time = 0.14, size = 79, normalized size = 0.91 \begin {gather*} \frac {(a+b)^{5/2} \tan ^{-1}\left (\frac {\sqrt {a+b} \tan (x)}{\sqrt {a}}\right )}{\sqrt {a} b^3}-\frac {4 \left (8 a^2+20 a b+15 b^2\right ) x+8 b (a+2 b) \sin (2 x)+b^2 \sin (4 x)}{32 b^3} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.29, size = 96, normalized size = 1.10
method | result | size |
default | \(\frac {\left (a +b \right )^{3} \arctan \left (\frac {\left (a +b \right ) \tan \left (x \right )}{\sqrt {a \left (a +b \right )}}\right )}{b^{3} \sqrt {a \left (a +b \right )}}-\frac {\frac {\left (\frac {1}{2} a b +\frac {7}{8} b^{2}\right ) \left (\tan ^{3}\left (x \right )\right )+\left (\frac {1}{2} a b +\frac {9}{8} b^{2}\right ) \tan \left (x \right )}{\left (\tan ^{2}\left (x \right )+1\right )^{2}}+\frac {\left (8 a^{2}+20 a b +15 b^{2}\right ) \arctan \left (\tan \left (x \right )\right )}{8}}{b^{3}}\) | \(96\) |
risch | \(-\frac {x \,a^{2}}{b^{3}}-\frac {5 a x}{2 b^{2}}-\frac {15 x}{8 b}+\frac {i {\mathrm e}^{2 i x} a}{8 b^{2}}+\frac {i {\mathrm e}^{2 i x}}{4 b}-\frac {i {\mathrm e}^{-2 i x} a}{8 b^{2}}-\frac {i {\mathrm e}^{-2 i x}}{4 b}+\frac {a \sqrt {-a \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 i x}+\frac {2 i \sqrt {-a \left (a +b \right )}-2 a -b}{b}\right )}{2 b^{3}}+\frac {\sqrt {-a \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 i x}+\frac {2 i \sqrt {-a \left (a +b \right )}-2 a -b}{b}\right )}{b^{2}}+\frac {\sqrt {-a \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 i x}+\frac {2 i \sqrt {-a \left (a +b \right )}-2 a -b}{b}\right )}{2 a b}-\frac {a \sqrt {-a \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 i x}-\frac {2 i \sqrt {-a \left (a +b \right )}+2 a +b}{b}\right )}{2 b^{3}}-\frac {\sqrt {-a \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 i x}-\frac {2 i \sqrt {-a \left (a +b \right )}+2 a +b}{b}\right )}{b^{2}}-\frac {\sqrt {-a \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 i x}-\frac {2 i \sqrt {-a \left (a +b \right )}+2 a +b}{b}\right )}{2 a b}-\frac {\sin \left (4 x \right )}{32 b}\) | \(335\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.50, size = 114, normalized size = 1.31 \begin {gather*} -\frac {{\left (4 \, a + 7 \, b\right )} \tan \left (x\right )^{3} + {\left (4 \, a + 9 \, b\right )} \tan \left (x\right )}{8 \, {\left (b^{2} \tan \left (x\right )^{4} + 2 \, b^{2} \tan \left (x\right )^{2} + b^{2}\right )}} - \frac {{\left (8 \, a^{2} + 20 \, a b + 15 \, b^{2}\right )} x}{8 \, b^{3}} + \frac {{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \arctan \left (\frac {{\left (a + b\right )} \tan \left (x\right )}{\sqrt {{\left (a + b\right )} a}}\right )}{\sqrt {{\left (a + b\right )} a} b^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.43, size = 312, normalized size = 3.59 \begin {gather*} \left [\frac {2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {-\frac {a + b}{a}} \log \left (\frac {{\left (8 \, a^{2} + 8 \, a b + b^{2}\right )} \cos \left (x\right )^{4} - 2 \, {\left (4 \, a^{2} + 5 \, a b + b^{2}\right )} \cos \left (x\right )^{2} - 4 \, {\left ({\left (2 \, a^{2} + a b\right )} \cos \left (x\right )^{3} - {\left (a^{2} + a b\right )} \cos \left (x\right )\right )} \sqrt {-\frac {a + b}{a}} \sin \left (x\right ) + a^{2} + 2 \, a b + b^{2}}{b^{2} \cos \left (x\right )^{4} - 2 \, {\left (a b + b^{2}\right )} \cos \left (x\right )^{2} + a^{2} + 2 \, a b + b^{2}}\right ) - {\left (8 \, a^{2} + 20 \, a b + 15 \, b^{2}\right )} x - {\left (2 \, b^{2} \cos \left (x\right )^{3} + {\left (4 \, a b + 7 \, b^{2}\right )} \cos \left (x\right )\right )} \sin \left (x\right )}{8 \, b^{3}}, -\frac {4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {\frac {a + b}{a}} \arctan \left (\frac {{\left ({\left (2 \, a + b\right )} \cos \left (x\right )^{2} - a - b\right )} \sqrt {\frac {a + b}{a}}}{2 \, {\left (a + b\right )} \cos \left (x\right ) \sin \left (x\right )}\right ) + {\left (8 \, a^{2} + 20 \, a b + 15 \, b^{2}\right )} x + {\left (2 \, b^{2} \cos \left (x\right )^{3} + {\left (4 \, a b + 7 \, b^{2}\right )} \cos \left (x\right )\right )} \sin \left (x\right )}{8 \, b^{3}}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.44, size = 131, normalized size = 1.51 \begin {gather*} -\frac {{\left (8 \, a^{2} + 20 \, a b + 15 \, b^{2}\right )} x}{8 \, b^{3}} + \frac {{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} {\left (\pi \left \lfloor \frac {x}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a + 2 \, b\right ) + \arctan \left (\frac {a \tan \left (x\right ) + b \tan \left (x\right )}{\sqrt {a^{2} + a b}}\right )\right )}}{\sqrt {a^{2} + a b} b^{3}} - \frac {4 \, a \tan \left (x\right )^{3} + 7 \, b \tan \left (x\right )^{3} + 4 \, a \tan \left (x\right ) + 9 \, b \tan \left (x\right )}{8 \, {\left (\tan \left (x\right )^{2} + 1\right )}^{2} b^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 15.45, size = 1804, normalized size = 20.74 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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