Integrand size = 19, antiderivative size = 116 \[ \int (a+b \sin (c+d x)) \tan ^5(c+d x) \, dx=-\frac {(8 a+15 b) \log (1-\sin (c+d x))}{16 d}-\frac {(8 a-15 b) \log (1+\sin (c+d x))}{16 d}-\frac {15 b \sin (c+d x)}{8 d}-\frac {(4 a+5 b \sin (c+d x)) \tan ^2(c+d x)}{8 d}+\frac {(a+b \sin (c+d x)) \tan ^4(c+d x)}{4 d} \]
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Time = 0.07 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {2800, 833, 788, 647, 31} \[ \int (a+b \sin (c+d x)) \tan ^5(c+d x) \, dx=-\frac {(8 a+15 b) \log (1-\sin (c+d x))}{16 d}-\frac {(8 a-15 b) \log (\sin (c+d x)+1)}{16 d}+\frac {\tan ^4(c+d x) (a+b \sin (c+d x))}{4 d}-\frac {\tan ^2(c+d x) (4 a+5 b \sin (c+d x))}{8 d}-\frac {15 b \sin (c+d x)}{8 d} \]
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Rule 31
Rule 647
Rule 788
Rule 833
Rule 2800
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x^5 (a+x)}{\left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d} \\ & = \frac {(a+b \sin (c+d x)) \tan ^4(c+d x)}{4 d}-\frac {\text {Subst}\left (\int \frac {x^3 \left (4 a b^2+5 b^2 x\right )}{\left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 b^2 d} \\ & = -\frac {(4 a+5 b \sin (c+d x)) \tan ^2(c+d x)}{8 d}+\frac {(a+b \sin (c+d x)) \tan ^4(c+d x)}{4 d}+\frac {\text {Subst}\left (\int \frac {x \left (8 a b^4+15 b^4 x\right )}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{8 b^4 d} \\ & = -\frac {15 b \sin (c+d x)}{8 d}-\frac {(4 a+5 b \sin (c+d x)) \tan ^2(c+d x)}{8 d}+\frac {(a+b \sin (c+d x)) \tan ^4(c+d x)}{4 d}-\frac {\text {Subst}\left (\int \frac {-15 b^6-8 a b^4 x}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{8 b^4 d} \\ & = -\frac {15 b \sin (c+d x)}{8 d}-\frac {(4 a+5 b \sin (c+d x)) \tan ^2(c+d x)}{8 d}+\frac {(a+b \sin (c+d x)) \tan ^4(c+d x)}{4 d}+\frac {(8 a-15 b) \text {Subst}\left (\int \frac {1}{-b-x} \, dx,x,b \sin (c+d x)\right )}{16 d}+\frac {(8 a+15 b) \text {Subst}\left (\int \frac {1}{b-x} \, dx,x,b \sin (c+d x)\right )}{16 d} \\ & = -\frac {(8 a+15 b) \log (1-\sin (c+d x))}{16 d}-\frac {(8 a-15 b) \log (1+\sin (c+d x))}{16 d}-\frac {15 b \sin (c+d x)}{8 d}-\frac {(4 a+5 b \sin (c+d x)) \tan ^2(c+d x)}{8 d}+\frac {(a+b \sin (c+d x)) \tan ^4(c+d x)}{4 d} \\ \end{align*}
Time = 0.23 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.17 \[ \int (a+b \sin (c+d x)) \tan ^5(c+d x) \, dx=\frac {15 b \text {arctanh}(\sin (c+d x))}{8 d}+\frac {15 b \sec (c+d x) \tan (c+d x)}{8 d}-\frac {15 b \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {5 b \sec (c+d x) \tan ^3(c+d x)}{d}-\frac {b \sin (c+d x) \tan ^4(c+d x)}{d}-\frac {a \left (4 \log (\cos (c+d x))+2 \tan ^2(c+d x)-\tan ^4(c+d x)\right )}{4 d} \]
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Time = 0.74 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.04
method | result | size |
derivativedivides | \(\frac {a \left (\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+b \left (\frac {\sin ^{7}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {3 \left (\sin ^{7}\left (d x +c \right )\right )}{8 \cos \left (d x +c \right )^{2}}-\frac {3 \left (\sin ^{5}\left (d x +c \right )\right )}{8}-\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {15 \sin \left (d x +c \right )}{8}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(121\) |
default | \(\frac {a \left (\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+b \left (\frac {\sin ^{7}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {3 \left (\sin ^{7}\left (d x +c \right )\right )}{8 \cos \left (d x +c \right )^{2}}-\frac {3 \left (\sin ^{5}\left (d x +c \right )\right )}{8}-\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {15 \sin \left (d x +c \right )}{8}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(121\) |
parallelrisch | \(\frac {16 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) a \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-16 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a +\frac {15 b}{8}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-16 \left (a -\frac {15 b}{8}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-4 a \cos \left (2 d x +2 c \right )+3 \cos \left (4 d x +4 c \right ) a -2 b \sin \left (5 d x +5 c \right )-5 b \sin \left (d x +c \right )-15 b \sin \left (3 d x +3 c \right )+a}{4 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\) | \(208\) |
risch | \(i a x +\frac {i b \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}-\frac {i b \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {2 i a c}{d}+\frac {i \left (16 i a \,{\mathrm e}^{6 i \left (d x +c \right )}+9 b \,{\mathrm e}^{7 i \left (d x +c \right )}+16 i a \,{\mathrm e}^{4 i \left (d x +c \right )}+b \,{\mathrm e}^{5 i \left (d x +c \right )}+16 i a \,{\mathrm e}^{2 i \left (d x +c \right )}-b \,{\mathrm e}^{3 i \left (d x +c \right )}-9 b \,{\mathrm e}^{i \left (d x +c \right )}\right )}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}+\frac {15 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b}{8 d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}-\frac {15 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b}{8 d}\) | \(229\) |
norman | \(\frac {-\frac {15 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {10 b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {9 b \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {10 b \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {15 b \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {6 a \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {6 a \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 a \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {a \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {\left (8 a -15 b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}-\frac {\left (8 a +15 b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}\) | \(252\) |
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Time = 0.33 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.98 \[ \int (a+b \sin (c+d x)) \tan ^5(c+d x) \, dx=-\frac {{\left (8 \, a - 15 \, b\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (8 \, a + 15 \, b\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 16 \, a \cos \left (d x + c\right )^{2} + 2 \, {\left (8 \, b \cos \left (d x + c\right )^{4} + 9 \, b \cos \left (d x + c\right )^{2} - 2 \, b\right )} \sin \left (d x + c\right ) - 4 \, a}{16 \, d \cos \left (d x + c\right )^{4}} \]
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Timed out. \[ \int (a+b \sin (c+d x)) \tan ^5(c+d x) \, dx=\text {Timed out} \]
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Time = 0.19 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.93 \[ \int (a+b \sin (c+d x)) \tan ^5(c+d x) \, dx=-\frac {{\left (8 \, a - 15 \, b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (8 \, a + 15 \, b\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) + 16 \, b \sin \left (d x + c\right ) - \frac {2 \, {\left (9 \, b \sin \left (d x + c\right )^{3} + 8 \, a \sin \left (d x + c\right )^{2} - 7 \, b \sin \left (d x + c\right ) - 6 \, a\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{16 \, d} \]
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Time = 0.35 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.93 \[ \int (a+b \sin (c+d x)) \tan ^5(c+d x) \, dx=-\frac {{\left (8 \, a - 15 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) + {\left (8 \, a + 15 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) + 16 \, b \sin \left (d x + c\right ) - \frac {2 \, {\left (6 \, a \sin \left (d x + c\right )^{4} + 9 \, b \sin \left (d x + c\right )^{3} - 4 \, a \sin \left (d x + c\right )^{2} - 7 \, b \sin \left (d x + c\right )\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]
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Time = 12.39 (sec) , antiderivative size = 261, normalized size of antiderivative = 2.25 \[ \int (a+b \sin (c+d x)) \tan ^5(c+d x) \, dx=\frac {a\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left (a-\frac {15\,b}{8}\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left (a+\frac {15\,b}{8}\right )}{d}-\frac {\frac {15\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{4}+2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-10\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-6\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\frac {9\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{2}-6\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-10\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {15\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]
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