Optimal. Leaf size=163 \[ -\frac {\left (16 a^3-8 a^2 b+6 a b^2-5 b^3\right ) x}{16 b^4}+\frac {a^{7/2} \tan ^{-1}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{b^4 \sqrt {a+b} d}-\frac {\left (8 a^2-6 a b+5 b^2\right ) \cos (c+d x) \sin (c+d x)}{16 b^3 d}+\frac {(6 a-5 b) \cos (c+d x) \sin ^3(c+d x)}{24 b^2 d}-\frac {\cos (c+d x) \sin ^5(c+d x)}{6 b d} \]
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Rubi [A]
time = 0.23, antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps
used = 7, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3266, 481, 592,
536, 209, 211} \begin {gather*} \frac {a^{7/2} \text {ArcTan}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{b^4 d \sqrt {a+b}}-\frac {\left (8 a^2-6 a b+5 b^2\right ) \sin (c+d x) \cos (c+d x)}{16 b^3 d}-\frac {x \left (16 a^3-8 a^2 b+6 a b^2-5 b^3\right )}{16 b^4}+\frac {(6 a-5 b) \sin ^3(c+d x) \cos (c+d x)}{24 b^2 d}-\frac {\sin ^5(c+d x) \cos (c+d x)}{6 b d} \end {gather*}
Antiderivative was successfully verified.
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Rule 209
Rule 211
Rule 481
Rule 536
Rule 592
Rule 3266
Rubi steps
\begin {align*} \int \frac {\sin ^8(c+d x)}{a+b \sin ^2(c+d x)} \, dx &=\frac {\text {Subst}\left (\int \frac {x^8}{\left (1+x^2\right )^4 \left (a+(a+b) x^2\right )} \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac {\cos (c+d x) \sin ^5(c+d x)}{6 b d}+\frac {\text {Subst}\left (\int \frac {x^4 \left (5 a+(-a+5 b) x^2\right )}{\left (1+x^2\right )^3 \left (a+(a+b) x^2\right )} \, dx,x,\tan (c+d x)\right )}{6 b d}\\ &=\frac {(6 a-5 b) \cos (c+d x) \sin ^3(c+d x)}{24 b^2 d}-\frac {\cos (c+d x) \sin ^5(c+d x)}{6 b d}-\frac {\text {Subst}\left (\int \frac {x^2 \left (3 a (6 a-5 b)-3 \left (2 a^2-a b+5 b^2\right ) x^2\right )}{\left (1+x^2\right )^2 \left (a+(a+b) x^2\right )} \, dx,x,\tan (c+d x)\right )}{24 b^2 d}\\ &=-\frac {\left (8 a^2-6 a b+5 b^2\right ) \cos (c+d x) \sin (c+d x)}{16 b^3 d}+\frac {(6 a-5 b) \cos (c+d x) \sin ^3(c+d x)}{24 b^2 d}-\frac {\cos (c+d x) \sin ^5(c+d x)}{6 b d}+\frac {\text {Subst}\left (\int \frac {3 a \left (8 a^2-6 a b+5 b^2\right )-3 \left (8 a^3-2 a^2 b+a b^2-5 b^3\right ) x^2}{\left (1+x^2\right ) \left (a+(a+b) x^2\right )} \, dx,x,\tan (c+d x)\right )}{48 b^3 d}\\ &=-\frac {\left (8 a^2-6 a b+5 b^2\right ) \cos (c+d x) \sin (c+d x)}{16 b^3 d}+\frac {(6 a-5 b) \cos (c+d x) \sin ^3(c+d x)}{24 b^2 d}-\frac {\cos (c+d x) \sin ^5(c+d x)}{6 b d}+\frac {a^4 \text {Subst}\left (\int \frac {1}{a+(a+b) x^2} \, dx,x,\tan (c+d x)\right )}{b^4 d}-\frac {\left (16 a^3-8 a^2 b+6 a b^2-5 b^3\right ) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (c+d x)\right )}{16 b^4 d}\\ &=-\frac {\left (16 a^3-8 a^2 b+6 a b^2-5 b^3\right ) x}{16 b^4}+\frac {a^{7/2} \tan ^{-1}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{b^4 \sqrt {a+b} d}-\frac {\left (8 a^2-6 a b+5 b^2\right ) \cos (c+d x) \sin (c+d x)}{16 b^3 d}+\frac {(6 a-5 b) \cos (c+d x) \sin ^3(c+d x)}{24 b^2 d}-\frac {\cos (c+d x) \sin ^5(c+d x)}{6 b d}\\ \end {align*}
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Mathematica [A]
time = 0.82, size = 133, normalized size = 0.82 \begin {gather*} -\frac {12 \left (16 a^3-8 a^2 b+6 a b^2-5 b^3\right ) (c+d x)-\frac {192 a^{7/2} \tan ^{-1}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{\sqrt {a+b}}+3 b \left (16 a^2-16 a b+15 b^2\right ) \sin (2 (c+d x))+3 (2 a-3 b) b^2 \sin (4 (c+d x))+b^3 \sin (6 (c+d x))}{192 b^4 d} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.27, size = 168, normalized size = 1.03
method | result | size |
derivativedivides | \(\frac {-\frac {\frac {\left (\frac {1}{2} a^{2} b -\frac {5}{8} a \,b^{2}+\frac {11}{16} b^{3}\right ) \left (\tan ^{5}\left (d x +c \right )\right )+\left (a^{2} b -a \,b^{2}+\frac {5}{6} b^{3}\right ) \left (\tan ^{3}\left (d x +c \right )\right )+\left (\frac {1}{2} a^{2} b -\frac {3}{8} a \,b^{2}+\frac {5}{16} b^{3}\right ) \tan \left (d x +c \right )}{\left (\tan ^{2}\left (d x +c \right )+1\right )^{3}}+\frac {\left (16 a^{3}-8 a^{2} b +6 a \,b^{2}-5 b^{3}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{16}}{b^{4}}+\frac {a^{4} \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{b^{4} \sqrt {a \left (a +b \right )}}}{d}\) | \(168\) |
default | \(\frac {-\frac {\frac {\left (\frac {1}{2} a^{2} b -\frac {5}{8} a \,b^{2}+\frac {11}{16} b^{3}\right ) \left (\tan ^{5}\left (d x +c \right )\right )+\left (a^{2} b -a \,b^{2}+\frac {5}{6} b^{3}\right ) \left (\tan ^{3}\left (d x +c \right )\right )+\left (\frac {1}{2} a^{2} b -\frac {3}{8} a \,b^{2}+\frac {5}{16} b^{3}\right ) \tan \left (d x +c \right )}{\left (\tan ^{2}\left (d x +c \right )+1\right )^{3}}+\frac {\left (16 a^{3}-8 a^{2} b +6 a \,b^{2}-5 b^{3}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{16}}{b^{4}}+\frac {a^{4} \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{b^{4} \sqrt {a \left (a +b \right )}}}{d}\) | \(168\) |
risch | \(-\frac {x \,a^{3}}{b^{4}}+\frac {x \,a^{2}}{2 b^{3}}-\frac {3 a x}{8 b^{2}}+\frac {5 x}{16 b}+\frac {i {\mathrm e}^{2 i \left (d x +c \right )} a^{2}}{8 b^{3} d}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )} a}{8 b^{2} d}+\frac {15 i {\mathrm e}^{2 i \left (d x +c \right )}}{128 b d}-\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} a^{2}}{8 b^{3} d}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} a}{8 b^{2} d}-\frac {15 i {\mathrm e}^{-2 i \left (d x +c \right )}}{128 b d}+\frac {\sqrt {-a \left (a +b \right )}\, a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \sqrt {-a \left (a +b \right )}-2 a -b}{b}\right )}{2 \left (a +b \right ) d \,b^{4}}-\frac {\sqrt {-a \left (a +b \right )}\, a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \sqrt {-a \left (a +b \right )}+2 a +b}{b}\right )}{2 \left (a +b \right ) d \,b^{4}}-\frac {\sin \left (6 d x +6 c \right )}{192 b d}-\frac {\sin \left (4 d x +4 c \right ) a}{32 b^{2} d}+\frac {3 \sin \left (4 d x +4 c \right )}{64 b d}\) | \(314\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.53, size = 192, normalized size = 1.18 \begin {gather*} \frac {\frac {48 \, a^{4} \arctan \left (\frac {{\left (a + b\right )} \tan \left (d x + c\right )}{\sqrt {{\left (a + b\right )} a}}\right )}{\sqrt {{\left (a + b\right )} a} b^{4}} - \frac {3 \, {\left (8 \, a^{2} - 10 \, a b + 11 \, b^{2}\right )} \tan \left (d x + c\right )^{5} + 8 \, {\left (6 \, a^{2} - 6 \, a b + 5 \, b^{2}\right )} \tan \left (d x + c\right )^{3} + 3 \, {\left (8 \, a^{2} - 6 \, a b + 5 \, b^{2}\right )} \tan \left (d x + c\right )}{b^{3} \tan \left (d x + c\right )^{6} + 3 \, b^{3} \tan \left (d x + c\right )^{4} + 3 \, b^{3} \tan \left (d x + c\right )^{2} + b^{3}} - \frac {3 \, {\left (16 \, a^{3} - 8 \, a^{2} b + 6 \, a b^{2} - 5 \, b^{3}\right )} {\left (d x + c\right )}}{b^{4}}}{48 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.45, size = 453, normalized size = 2.78 \begin {gather*} \left [\frac {12 \, a^{3} \sqrt {-\frac {a}{a + b}} \log \left (\frac {{\left (8 \, a^{2} + 8 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (4 \, a^{2} + 5 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} - 4 \, {\left ({\left (2 \, a^{2} + 3 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{3} - {\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )\right )} \sqrt {-\frac {a}{a + b}} \sin \left (d x + c\right ) + a^{2} + 2 \, a b + b^{2}}{b^{2} \cos \left (d x + c\right )^{4} - 2 \, {\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + 2 \, a b + b^{2}}\right ) - 3 \, {\left (16 \, a^{3} - 8 \, a^{2} b + 6 \, a b^{2} - 5 \, b^{3}\right )} d x - {\left (8 \, b^{3} \cos \left (d x + c\right )^{5} + 2 \, {\left (6 \, a b^{2} - 13 \, b^{3}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (8 \, a^{2} b - 10 \, a b^{2} + 11 \, b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \, b^{4} d}, -\frac {24 \, a^{3} \sqrt {\frac {a}{a + b}} \arctan \left (\frac {{\left ({\left (2 \, a + b\right )} \cos \left (d x + c\right )^{2} - a - b\right )} \sqrt {\frac {a}{a + b}}}{2 \, a \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right ) + 3 \, {\left (16 \, a^{3} - 8 \, a^{2} b + 6 \, a b^{2} - 5 \, b^{3}\right )} d x + {\left (8 \, b^{3} \cos \left (d x + c\right )^{5} + 2 \, {\left (6 \, a b^{2} - 13 \, b^{3}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (8 \, a^{2} b - 10 \, a b^{2} + 11 \, b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \, b^{4} d}\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 = 0.45, size = 233, normalized size = 1.43 \begin {gather*} \frac {\frac {48 \, {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a + 2 \, b\right ) + \arctan \left (\frac {a \tan \left (d x + c\right ) + b \tan \left (d x + c\right )}{\sqrt {a^{2} + a b}}\right )\right )} a^{4}}{\sqrt {a^{2} + a b} b^{4}} - \frac {3 \, {\left (16 \, a^{3} - 8 \, a^{2} b + 6 \, a b^{2} - 5 \, b^{3}\right )} {\left (d x + c\right )}}{b^{4}} - \frac {24 \, a^{2} \tan \left (d x + c\right )^{5} - 30 \, a b \tan \left (d x + c\right )^{5} + 33 \, b^{2} \tan \left (d x + c\right )^{5} + 48 \, a^{2} \tan \left (d x + c\right )^{3} - 48 \, a b \tan \left (d x + c\right )^{3} + 40 \, b^{2} \tan \left (d x + c\right )^{3} + 24 \, a^{2} \tan \left (d x + c\right ) - 18 \, a b \tan \left (d x + c\right ) + 15 \, b^{2} \tan \left (d x + c\right )}{{\left (\tan \left (d x + c\right )^{2} + 1\right )}^{3} b^{3}}}{48 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 15.32, size = 2244, normalized size = 13.77 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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