Optimal. Leaf size=117 \[ \frac {a \cos ^3(c+d x)}{3 d}-\frac {3 a \cos (c+d x)}{d}+\frac {5 a \tan ^3(c+d x)}{6 d}-\frac {5 a \tan (c+d x)}{2 d}+\frac {a \sec ^3(c+d x)}{3 d}-\frac {3 a \sec (c+d x)}{d}-\frac {a \sin ^2(c+d x) \tan ^3(c+d x)}{2 d}+\frac {5 a x}{2} \]
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Rubi [A] time = 0.13, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {2838, 2591, 288, 302, 203, 2590, 270} \[ \frac {a \cos ^3(c+d x)}{3 d}-\frac {3 a \cos (c+d x)}{d}+\frac {5 a \tan ^3(c+d x)}{6 d}-\frac {5 a \tan (c+d x)}{2 d}+\frac {a \sec ^3(c+d x)}{3 d}-\frac {3 a \sec (c+d x)}{d}-\frac {a \sin ^2(c+d x) \tan ^3(c+d x)}{2 d}+\frac {5 a x}{2} \]
Antiderivative was successfully verified.
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Rule 203
Rule 270
Rule 288
Rule 302
Rule 2590
Rule 2591
Rule 2838
Rubi steps
\begin {align*} \int \sin ^2(c+d x) (a+a \sin (c+d x)) \tan ^4(c+d x) \, dx &=a \int \sin ^2(c+d x) \tan ^4(c+d x) \, dx+a \int \sin ^3(c+d x) \tan ^4(c+d x) \, dx\\ &=-\frac {a \operatorname {Subst}\left (\int \frac {\left (1-x^2\right )^3}{x^4} \, dx,x,\cos (c+d x)\right )}{d}+\frac {a \operatorname {Subst}\left (\int \frac {x^6}{\left (1+x^2\right )^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac {a \sin ^2(c+d x) \tan ^3(c+d x)}{2 d}-\frac {a \operatorname {Subst}\left (\int \left (3+\frac {1}{x^4}-\frac {3}{x^2}-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}+\frac {(5 a) \operatorname {Subst}\left (\int \frac {x^4}{1+x^2} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=-\frac {3 a \cos (c+d x)}{d}+\frac {a \cos ^3(c+d x)}{3 d}-\frac {3 a \sec (c+d x)}{d}+\frac {a \sec ^3(c+d x)}{3 d}-\frac {a \sin ^2(c+d x) \tan ^3(c+d x)}{2 d}+\frac {(5 a) \operatorname {Subst}\left (\int \left (-1+x^2+\frac {1}{1+x^2}\right ) \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=-\frac {3 a \cos (c+d x)}{d}+\frac {a \cos ^3(c+d x)}{3 d}-\frac {3 a \sec (c+d x)}{d}+\frac {a \sec ^3(c+d x)}{3 d}-\frac {5 a \tan (c+d x)}{2 d}+\frac {5 a \tan ^3(c+d x)}{6 d}-\frac {a \sin ^2(c+d x) \tan ^3(c+d x)}{2 d}+\frac {(5 a) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=\frac {5 a x}{2}-\frac {3 a \cos (c+d x)}{d}+\frac {a \cos ^3(c+d x)}{3 d}-\frac {3 a \sec (c+d x)}{d}+\frac {a \sec ^3(c+d x)}{3 d}-\frac {5 a \tan (c+d x)}{2 d}+\frac {5 a \tan ^3(c+d x)}{6 d}-\frac {a \sin ^2(c+d x) \tan ^3(c+d x)}{2 d}\\ \end {align*}
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Mathematica [A] time = 0.41, size = 84, normalized size = 0.72 \[ \frac {a \left (-3 \sin (2 (c+d x))-33 \cos (c+d x)+\cos (3 (c+d x))-28 \tan (c+d x)+4 \sec ^3(c+d x)-36 \sec (c+d x)+4 \tan (c+d x) \sec ^2(c+d x)+30 c+30 d x\right )}{12 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.47, size = 108, normalized size = 0.92 \[ \frac {a \cos \left (d x + c\right )^{4} - 15 \, a d x \cos \left (d x + c\right ) + 29 \, a \cos \left (d x + c\right )^{2} + {\left (2 \, a \cos \left (d x + c\right )^{4} + 15 \, a d x \cos \left (d x + c\right ) - 15 \, a \cos \left (d x + c\right )^{2} - 4 \, a\right )} \sin \left (d x + c\right ) + 2 \, a}{6 \, {\left (d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - d \cos \left (d x + c\right )\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.20, size = 182, normalized size = 1.56 \[ \frac {15 \, {\left (d x + c\right )} a - \frac {3 \, a}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1} + \frac {33 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 102 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 200 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 330 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 402 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 410 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 264 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 150 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 61 \, a}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}}}{6 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.46, size = 164, normalized size = 1.40 \[ \frac {a \left (\frac {\sin ^{8}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {5 \left (\sin ^{8}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )}-\frac {5 \left (\frac {16}{5}+\sin ^{6}\left (d x +c \right )+\frac {6 \left (\sin ^{4}\left (d x +c \right )\right )}{5}+\frac {8 \left (\sin ^{2}\left (d x +c \right )\right )}{5}\right ) \cos \left (d x +c \right )}{3}\right )+a \left (\frac {\sin ^{7}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {4 \left (\sin ^{7}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )}-\frac {4 \left (\sin ^{5}\left (d x +c \right )+\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )}{3}+\frac {5 d x}{2}+\frac {5 c}{2}\right )}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.42, size = 96, normalized size = 0.82 \[ \frac {2 \, {\left (\cos \left (d x + c\right )^{3} - \frac {9 \, \cos \left (d x + c\right )^{2} - 1}{\cos \left (d x + c\right )^{3}} - 9 \, \cos \left (d x + c\right )\right )} a + {\left (2 \, \tan \left (d x + c\right )^{3} + 15 \, d x + 15 \, c - \frac {3 \, \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} - 12 \, \tan \left (d x + c\right )\right )} a}{6 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 14.96, size = 306, normalized size = 2.62 \[ \frac {5\,a\,x}{2}+\frac {\left (5\,a\,d\,x-\frac {a\,\left (30\,d\,x-30\right )}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {a\,\left (45\,d\,x-60\right )}{6}-\frac {15\,a\,d\,x}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\left (10\,a\,d\,x-\frac {a\,\left (60\,d\,x-80\right )}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {a\,\left (30\,d\,x-100\right )}{6}-5\,a\,d\,x\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (5\,a\,d\,x-\frac {a\,\left (30\,d\,x-28\right )}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (\frac {a\,\left (60\,d\,x-176\right )}{6}-10\,a\,d\,x\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {15\,a\,d\,x}{2}-\frac {a\,\left (45\,d\,x-132\right )}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\left (\frac {a\,\left (30\,d\,x-98\right )}{6}-5\,a\,d\,x\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {a\,\left (15\,d\,x-64\right )}{6}+\frac {5\,a\,d\,x}{2}}{d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}^3} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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