Optimal. Leaf size=148 \[ \frac {x}{b^3}-\frac {\sqrt {a} \left (8 a^2+20 a b+15 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{8 b^3 (a+b)^{5/2} d}+\frac {a \tan ^3(c+d x)}{4 b (a+b) d \left (a+(a+b) \tan ^2(c+d x)\right )^2}+\frac {a (4 a+7 b) \tan (c+d x)}{8 b^2 (a+b)^2 d \left (a+(a+b) \tan ^2(c+d x)\right )} \]
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Rubi [A]
time = 0.19, antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps
used = 6, 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 {\sqrt {a} \left (8 a^2+20 a b+15 b^2\right ) \text {ArcTan}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{8 b^3 d (a+b)^{5/2}}+\frac {a (4 a+7 b) \tan (c+d x)}{8 b^2 d (a+b)^2 \left ((a+b) \tan ^2(c+d x)+a\right )}+\frac {a \tan ^3(c+d x)}{4 b d (a+b) \left ((a+b) \tan ^2(c+d x)+a\right )^2}+\frac {x}{b^3} \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 ^6(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^3} \, dx &=\frac {\text {Subst}\left (\int \frac {x^6}{\left (1+x^2\right ) \left (a+(a+b) x^2\right )^3} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {a \tan ^3(c+d x)}{4 b (a+b) d \left (a+(a+b) \tan ^2(c+d x)\right )^2}-\frac {\text {Subst}\left (\int \frac {x^2 \left (3 a+(-a-4 b) x^2\right )}{\left (1+x^2\right ) \left (a+(a+b) x^2\right )^2} \, dx,x,\tan (c+d x)\right )}{4 b (a+b) d}\\ &=\frac {a \tan ^3(c+d x)}{4 b (a+b) d \left (a+(a+b) \tan ^2(c+d x)\right )^2}+\frac {a (4 a+7 b) \tan (c+d x)}{8 b^2 (a+b)^2 d \left (a+(a+b) \tan ^2(c+d x)\right )}-\frac {\text {Subst}\left (\int \frac {a (4 a+7 b)+\left (-4 a^2-9 a b-8 b^2\right ) x^2}{\left (1+x^2\right ) \left (a+(a+b) x^2\right )} \, dx,x,\tan (c+d x)\right )}{8 b^2 (a+b)^2 d}\\ &=\frac {a \tan ^3(c+d x)}{4 b (a+b) d \left (a+(a+b) \tan ^2(c+d x)\right )^2}+\frac {a (4 a+7 b) \tan (c+d x)}{8 b^2 (a+b)^2 d \left (a+(a+b) \tan ^2(c+d x)\right )}+\frac {\text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (c+d x)\right )}{b^3 d}-\frac {\left (a \left (8 a^2+20 a b+15 b^2\right )\right ) \text {Subst}\left (\int \frac {1}{a+(a+b) x^2} \, dx,x,\tan (c+d x)\right )}{8 b^3 (a+b)^2 d}\\ &=\frac {x}{b^3}-\frac {\sqrt {a} \left (8 a^2+20 a b+15 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{8 b^3 (a+b)^{5/2} d}+\frac {a \tan ^3(c+d x)}{4 b (a+b) d \left (a+(a+b) \tan ^2(c+d x)\right )^2}+\frac {a (4 a+7 b) \tan (c+d x)}{8 b^2 (a+b)^2 d \left (a+(a+b) \tan ^2(c+d x)\right )}\\ \end {align*}
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Mathematica [A]
time = 1.81, size = 134, normalized size = 0.91 \begin {gather*} \frac {8 (c+d x)-\frac {\sqrt {a} \left (8 a^2+20 a b+15 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{(a+b)^{5/2}}+\frac {a b \left (8 a^2+20 a b+9 b^2-3 b (2 a+3 b) \cos (2 (c+d x))\right ) \sin (2 (c+d x))}{(a+b)^2 (2 a+b-b \cos (2 (c+d x)))^2}}{8 b^3 d} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.34, size = 158, normalized size = 1.07
method | result | size |
derivativedivides | \(\frac {\frac {\arctan \left (\tan \left (d x +c \right )\right )}{b^{3}}-\frac {a \left (\frac {-\frac {\left (4 a +9 b \right ) b \left (\tan ^{3}\left (d x +c \right )\right )}{8 \left (a +b \right )}-\frac {a b \left (4 a +7 b \right ) \tan \left (d x +c \right )}{8 \left (a^{2}+2 a b +b^{2}\right )}}{\left (a \left (\tan ^{2}\left (d x +c \right )\right )+b \left (\tan ^{2}\left (d x +c \right )\right )+a \right )^{2}}+\frac {\left (8 a^{2}+20 a b +15 b^{2}\right ) \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{8 \left (a^{2}+2 a b +b^{2}\right ) \sqrt {a \left (a +b \right )}}\right )}{b^{3}}}{d}\) | \(158\) |
default | \(\frac {\frac {\arctan \left (\tan \left (d x +c \right )\right )}{b^{3}}-\frac {a \left (\frac {-\frac {\left (4 a +9 b \right ) b \left (\tan ^{3}\left (d x +c \right )\right )}{8 \left (a +b \right )}-\frac {a b \left (4 a +7 b \right ) \tan \left (d x +c \right )}{8 \left (a^{2}+2 a b +b^{2}\right )}}{\left (a \left (\tan ^{2}\left (d x +c \right )\right )+b \left (\tan ^{2}\left (d x +c \right )\right )+a \right )^{2}}+\frac {\left (8 a^{2}+20 a b +15 b^{2}\right ) \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{8 \left (a^{2}+2 a b +b^{2}\right ) \sqrt {a \left (a +b \right )}}\right )}{b^{3}}}{d}\) | \(158\) |
risch | \(\frac {x}{b^{3}}-\frac {i a \left (-16 b \,{\mathrm e}^{6 i \left (d x +c \right )} a^{2}-28 a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-9 b^{3} {\mathrm e}^{6 i \left (d x +c \right )}+48 a^{3} {\mathrm e}^{4 i \left (d x +c \right )}+120 b \,{\mathrm e}^{4 i \left (d x +c \right )} a^{2}+90 a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+27 b^{3} {\mathrm e}^{4 i \left (d x +c \right )}-32 b \,{\mathrm e}^{2 i \left (d x +c \right )} a^{2}-68 a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-27 b^{3} {\mathrm e}^{2 i \left (d x +c \right )}+6 a \,b^{2}+9 b^{3}\right )}{4 b^{3} \left (a +b \right )^{2} d \left (-b \,{\mathrm e}^{4 i \left (d x +c \right )}+4 a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{2 i \left (d x +c \right )}-b \right )^{2}}-\frac {\sqrt {-a \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \sqrt {-a \left (a +b \right )}-2 a -b}{b}\right ) a^{2}}{2 \left (a +b \right )^{3} d \,b^{3}}-\frac {5 \sqrt {-a \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \sqrt {-a \left (a +b \right )}-2 a -b}{b}\right ) a}{4 \left (a +b \right )^{3} d \,b^{2}}-\frac {15 \sqrt {-a \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \sqrt {-a \left (a +b \right )}-2 a -b}{b}\right )}{16 \left (a +b \right )^{3} d b}+\frac {\sqrt {-a \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \sqrt {-a \left (a +b \right )}+2 a +b}{b}\right ) a^{2}}{2 \left (a +b \right )^{3} d \,b^{3}}+\frac {5 \sqrt {-a \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \sqrt {-a \left (a +b \right )}+2 a +b}{b}\right ) a}{4 \left (a +b \right )^{3} d \,b^{2}}+\frac {15 \sqrt {-a \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \sqrt {-a \left (a +b \right )}+2 a +b}{b}\right )}{16 \left (a +b \right )^{3} d b}\) | \(551\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.55, size = 234, normalized size = 1.58 \begin {gather*} -\frac {\frac {{\left (8 \, a^{3} + 20 \, a^{2} b + 15 \, a b^{2}\right )} \arctan \left (\frac {{\left (a + b\right )} \tan \left (d x + c\right )}{\sqrt {{\left (a + b\right )} a}}\right )}{{\left (a^{2} b^{3} + 2 \, a b^{4} + b^{5}\right )} \sqrt {{\left (a + b\right )} a}} - \frac {{\left (4 \, a^{3} + 13 \, a^{2} b + 9 \, a b^{2}\right )} \tan \left (d x + c\right )^{3} + {\left (4 \, a^{3} + 7 \, a^{2} b\right )} \tan \left (d x + c\right )}{a^{4} b^{2} + 2 \, a^{3} b^{3} + a^{2} b^{4} + {\left (a^{4} b^{2} + 4 \, a^{3} b^{3} + 6 \, a^{2} b^{4} + 4 \, a b^{5} + b^{6}\right )} \tan \left (d x + c\right )^{4} + 2 \, {\left (a^{4} b^{2} + 3 \, a^{3} b^{3} + 3 \, a^{2} b^{4} + a b^{5}\right )} \tan \left (d x + c\right )^{2}} - \frac {8 \, {\left (d x + c\right )}}{b^{3}}}{8 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 420 vs.
\(2 (134) = 268\).
time = 0.47, size = 950, normalized size = 6.42 \begin {gather*} \left [\frac {32 \, {\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} d x \cos \left (d x + c\right )^{4} - 64 \, {\left (a^{3} b + 3 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4}\right )} d x \cos \left (d x + c\right )^{2} + 32 \, {\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} d x + {\left ({\left (8 \, a^{2} b^{2} + 20 \, a b^{3} + 15 \, b^{4}\right )} \cos \left (d x + c\right )^{4} + 8 \, a^{4} + 36 \, a^{3} b + 63 \, a^{2} b^{2} + 50 \, a b^{3} + 15 \, b^{4} - 2 \, {\left (8 \, a^{3} b + 28 \, a^{2} b^{2} + 35 \, a b^{3} + 15 \, b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \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 ) - 4 \, {\left (3 \, {\left (2 \, a^{2} b^{2} + 3 \, a b^{3}\right )} \cos \left (d x + c\right )^{3} - {\left (4 \, a^{3} b + 13 \, a^{2} b^{2} + 9 \, a b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{32 \, {\left ({\left (a^{2} b^{5} + 2 \, a b^{6} + b^{7}\right )} d \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{3} b^{4} + 3 \, a^{2} b^{5} + 3 \, a b^{6} + b^{7}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{4} b^{3} + 4 \, a^{3} b^{4} + 6 \, a^{2} b^{5} + 4 \, a b^{6} + b^{7}\right )} d\right )}}, \frac {16 \, {\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} d x \cos \left (d x + c\right )^{4} - 32 \, {\left (a^{3} b + 3 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4}\right )} d x \cos \left (d x + c\right )^{2} + 16 \, {\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} d x + {\left ({\left (8 \, a^{2} b^{2} + 20 \, a b^{3} + 15 \, b^{4}\right )} \cos \left (d x + c\right )^{4} + 8 \, a^{4} + 36 \, a^{3} b + 63 \, a^{2} b^{2} + 50 \, a b^{3} + 15 \, b^{4} - 2 \, {\left (8 \, a^{3} b + 28 \, a^{2} b^{2} + 35 \, a b^{3} + 15 \, b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \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 ) - 2 \, {\left (3 \, {\left (2 \, a^{2} b^{2} + 3 \, a b^{3}\right )} \cos \left (d x + c\right )^{3} - {\left (4 \, a^{3} b + 13 \, a^{2} b^{2} + 9 \, a b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{16 \, {\left ({\left (a^{2} b^{5} + 2 \, a b^{6} + b^{7}\right )} d \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{3} b^{4} + 3 \, a^{2} b^{5} + 3 \, a b^{6} + b^{7}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{4} b^{3} + 4 \, a^{3} b^{4} + 6 \, a^{2} b^{5} + 4 \, a b^{6} + b^{7}\right )} d\right )}}\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 = 0.56, size = 224, normalized size = 1.51 \begin {gather*} -\frac {\frac {{\left (8 \, a^{3} + 20 \, a^{2} b + 15 \, a b^{2}\right )} {\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 )}}{{\left (a^{2} b^{3} + 2 \, a b^{4} + b^{5}\right )} \sqrt {a^{2} + a b}} - \frac {4 \, a^{3} \tan \left (d x + c\right )^{3} + 13 \, a^{2} b \tan \left (d x + c\right )^{3} + 9 \, a b^{2} \tan \left (d x + c\right )^{3} + 4 \, a^{3} \tan \left (d x + c\right ) + 7 \, a^{2} b \tan \left (d x + c\right )}{{\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} {\left (a \tan \left (d x + c\right )^{2} + b \tan \left (d x + c\right )^{2} + a\right )}^{2}} - \frac {8 \, {\left (d x + c\right )}}{b^{3}}}{8 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 17.96, size = 2500, normalized size = 16.89 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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