Optimal. Leaf size=177 \[ \frac {\left (2 a^2-b^2\right ) \cot (c+d x)}{d}+\frac {\left (4 a^2-7 b^2\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {5}{8} x \left (4 a^2-3 b^2\right )-\frac {a^2 \cot ^3(c+d x)}{3 d}-\frac {5 a b \cos ^3(c+d x)}{3 d}-\frac {5 a b \cos (c+d x)}{d}-\frac {a b \cos ^3(c+d x) \cot ^2(c+d x)}{d}+\frac {5 a b \tanh ^{-1}(\cos (c+d x))}{d}-\frac {b^2 \sin (c+d x) \cos ^3(c+d x)}{4 d} \]
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Rubi [A] time = 0.44, antiderivative size = 177, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {2911, 2592, 288, 302, 206, 456, 1259, 1261, 203} \[ \frac {\left (2 a^2-b^2\right ) \cot (c+d x)}{d}+\frac {\left (4 a^2-7 b^2\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {5}{8} x \left (4 a^2-3 b^2\right )-\frac {a^2 \cot ^3(c+d x)}{3 d}-\frac {5 a b \cos ^3(c+d x)}{3 d}-\frac {5 a b \cos (c+d x)}{d}-\frac {a b \cos ^3(c+d x) \cot ^2(c+d x)}{d}+\frac {5 a b \tanh ^{-1}(\cos (c+d x))}{d}-\frac {b^2 \sin (c+d x) \cos ^3(c+d x)}{4 d} \]
Antiderivative was successfully verified.
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Rule 203
Rule 206
Rule 288
Rule 302
Rule 456
Rule 1259
Rule 1261
Rule 2592
Rule 2911
Rubi steps
\begin {align*} \int \cos ^2(c+d x) \cot ^4(c+d x) (a+b \sin (c+d x))^2 \, dx &=(2 a b) \int \cos ^3(c+d x) \cot ^3(c+d x) \, dx+\int \cos ^2(c+d x) \cot ^4(c+d x) \left (a^2+b^2 \sin ^2(c+d x)\right ) \, dx\\ &=\frac {\operatorname {Subst}\left (\int \frac {a^2+\left (a^2+b^2\right ) x^2}{x^4 \left (1+x^2\right )^3} \, dx,x,\tan (c+d x)\right )}{d}-\frac {(2 a b) \operatorname {Subst}\left (\int \frac {x^6}{\left (1-x^2\right )^2} \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac {a b \cos ^3(c+d x) \cot ^2(c+d x)}{d}-\frac {b^2 \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {\operatorname {Subst}\left (\int \frac {-4 a^2-4 b^2 x^2+3 b^2 x^4}{x^4 \left (1+x^2\right )^2} \, dx,x,\tan (c+d x)\right )}{4 d}+\frac {(5 a b) \operatorname {Subst}\left (\int \frac {x^4}{1-x^2} \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac {a b \cos ^3(c+d x) \cot ^2(c+d x)}{d}+\frac {\left (4 a^2-7 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 d}-\frac {b^2 \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {\operatorname {Subst}\left (\int \frac {-8 a^2+8 \left (a^2-b^2\right ) x^2+\left (-4 a^2+7 b^2\right ) x^4}{x^4 \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{8 d}+\frac {(5 a b) \operatorname {Subst}\left (\int \left (-1-x^2+\frac {1}{1-x^2}\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac {5 a b \cos (c+d x)}{d}-\frac {5 a b \cos ^3(c+d x)}{3 d}-\frac {a b \cos ^3(c+d x) \cot ^2(c+d x)}{d}+\frac {\left (4 a^2-7 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 d}-\frac {b^2 \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {\operatorname {Subst}\left (\int \left (-\frac {8 a^2}{x^4}+\frac {8 \left (2 a^2-b^2\right )}{x^2}-\frac {5 \left (4 a^2-3 b^2\right )}{1+x^2}\right ) \, dx,x,\tan (c+d x)\right )}{8 d}+\frac {(5 a b) \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\cos (c+d x)\right )}{d}\\ &=\frac {5 a b \tanh ^{-1}(\cos (c+d x))}{d}-\frac {5 a b \cos (c+d x)}{d}-\frac {5 a b \cos ^3(c+d x)}{3 d}+\frac {\left (2 a^2-b^2\right ) \cot (c+d x)}{d}-\frac {a b \cos ^3(c+d x) \cot ^2(c+d x)}{d}-\frac {a^2 \cot ^3(c+d x)}{3 d}+\frac {\left (4 a^2-7 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 d}-\frac {b^2 \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {\left (5 \left (4 a^2-3 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (c+d x)\right )}{8 d}\\ &=\frac {5}{8} \left (4 a^2-3 b^2\right ) x+\frac {5 a b \tanh ^{-1}(\cos (c+d x))}{d}-\frac {5 a b \cos (c+d x)}{d}-\frac {5 a b \cos ^3(c+d x)}{3 d}+\frac {\left (2 a^2-b^2\right ) \cot (c+d x)}{d}-\frac {a b \cos ^3(c+d x) \cot ^2(c+d x)}{d}-\frac {a^2 \cot ^3(c+d x)}{3 d}+\frac {\left (4 a^2-7 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 d}-\frac {b^2 \cos ^3(c+d x) \sin (c+d x)}{4 d}\\ \end {align*}
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Mathematica [A] time = 6.28, size = 336, normalized size = 1.90 \[ \frac {5 \left (4 a^2-3 b^2\right ) (c+d x)}{8 d}+\frac {\left (a^2-2 b^2\right ) \sin (2 (c+d x))}{4 d}+\frac {\csc \left (\frac {1}{2} (c+d x)\right ) \left (7 a^2 \cos \left (\frac {1}{2} (c+d x)\right )-3 b^2 \cos \left (\frac {1}{2} (c+d x)\right )\right )}{6 d}+\frac {\sec \left (\frac {1}{2} (c+d x)\right ) \left (3 b^2 \sin \left (\frac {1}{2} (c+d x)\right )-7 a^2 \sin \left (\frac {1}{2} (c+d x)\right )\right )}{6 d}-\frac {a^2 \cot \left (\frac {1}{2} (c+d x)\right ) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{24 d}+\frac {a^2 \tan \left (\frac {1}{2} (c+d x)\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right )}{24 d}-\frac {9 a b \cos (c+d x)}{2 d}-\frac {a b \cos (3 (c+d x))}{6 d}-\frac {a b \csc ^2\left (\frac {1}{2} (c+d x)\right )}{4 d}+\frac {a b \sec ^2\left (\frac {1}{2} (c+d x)\right )}{4 d}-\frac {5 a b \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {5 a b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{d}-\frac {b^2 \sin (4 (c+d x))}{32 d} \]
Warning: Unable to verify antiderivative.
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fricas [A] time = 0.68, size = 252, normalized size = 1.42 \[ \frac {6 \, b^{2} \cos \left (d x + c\right )^{7} - 3 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )} \cos \left (d x + c\right )^{5} + 20 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )} \cos \left (d x + c\right )^{3} + 60 \, {\left (a b \cos \left (d x + c\right )^{2} - a b\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 60 \, {\left (a b \cos \left (d x + c\right )^{2} - a b\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 15 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )} \cos \left (d x + c\right ) - {\left (16 \, a b \cos \left (d x + c\right )^{5} - 15 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )} d x \cos \left (d x + c\right )^{2} + 80 \, a b \cos \left (d x + c\right )^{3} + 15 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )} d x - 120 \, a b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )} \sin \left (d x + c\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.31, size = 366, normalized size = 2.07 \[ \frac {a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 120 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - 27 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 15 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )} {\left (d x + c\right )} + \frac {220 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 27 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 6 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}} - \frac {2 \, {\left (12 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 27 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 144 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 12 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 336 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 12 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 304 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 27 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 112 \, a b\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{24 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.52, size = 321, normalized size = 1.81 \[ -\frac {a^{2} \left (\cos ^{7}\left (d x +c \right )\right )}{3 d \sin \left (d x +c \right )^{3}}+\frac {4 a^{2} \left (\cos ^{7}\left (d x +c \right )\right )}{3 d \sin \left (d x +c \right )}+\frac {4 a^{2} \left (\cos ^{5}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 d}+\frac {5 a^{2} \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 d}+\frac {5 a^{2} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}+\frac {5 a^{2} x}{2}+\frac {5 a^{2} c}{2 d}-\frac {a b \left (\cos ^{7}\left (d x +c \right )\right )}{d \sin \left (d x +c \right )^{2}}-\frac {a b \left (\cos ^{5}\left (d x +c \right )\right )}{d}-\frac {5 a b \left (\cos ^{3}\left (d x +c \right )\right )}{3 d}-\frac {5 a b \cos \left (d x +c \right )}{d}-\frac {5 a b \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{d}-\frac {b^{2} \left (\cos ^{7}\left (d x +c \right )\right )}{d \sin \left (d x +c \right )}-\frac {b^{2} \left (\cos ^{5}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{d}-\frac {5 b^{2} \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{4 d}-\frac {15 b^{2} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{8 d}-\frac {15 b^{2} x}{8}-\frac {15 b^{2} c}{8 d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.58, size = 189, normalized size = 1.07 \[ \frac {4 \, {\left (15 \, d x + 15 \, c + \frac {15 \, \tan \left (d x + c\right )^{4} + 10 \, \tan \left (d x + c\right )^{2} - 2}{\tan \left (d x + c\right )^{5} + \tan \left (d x + c\right )^{3}}\right )} a^{2} - 4 \, {\left (4 \, \cos \left (d x + c\right )^{3} - \frac {6 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} + 24 \, \cos \left (d x + c\right ) - 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} a b - 3 \, {\left (15 \, d x + 15 \, c + \frac {15 \, \tan \left (d x + c\right )^{4} + 25 \, \tan \left (d x + c\right )^{2} + 8}{\tan \left (d x + c\right )^{5} + 2 \, \tan \left (d x + c\right )^{3} + \tan \left (d x + c\right )}\right )} b^{2}}{24 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 12.10, size = 665, normalized size = 3.76 \[ \frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,d}-\frac {\frac {a^2}{3}-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (42\,a^2-34\,b^2\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (\frac {83\,a^2}{3}-14\,b^2\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {182\,a^2}{3}-26\,b^2\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {23\,a^2}{3}-4\,b^2\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,\left (a^2+14\,b^2\right )+\frac {248\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+\frac {644\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{3}+232\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+98\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+2\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+48\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {9\,a^2}{8}-\frac {b^2}{2}\right )}{d}+\frac {\mathrm {atan}\left (\frac {\left (\frac {a^2\,5{}\mathrm {i}}{2}-\frac {b^2\,15{}\mathrm {i}}{8}\right )\,\left (\frac {15\,b^2}{4}-5\,a^2+6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {a^2\,5{}\mathrm {i}}{2}-\frac {b^2\,15{}\mathrm {i}}{8}\right )+10\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,1{}\mathrm {i}-\left (\frac {a^2\,5{}\mathrm {i}}{2}-\frac {b^2\,15{}\mathrm {i}}{8}\right )\,\left (5\,a^2-\frac {15\,b^2}{4}+6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {a^2\,5{}\mathrm {i}}{2}-\frac {b^2\,15{}\mathrm {i}}{8}\right )-10\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,1{}\mathrm {i}}{\left (\frac {a^2\,5{}\mathrm {i}}{2}-\frac {b^2\,15{}\mathrm {i}}{8}\right )\,\left (\frac {15\,b^2}{4}-5\,a^2+6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {a^2\,5{}\mathrm {i}}{2}-\frac {b^2\,15{}\mathrm {i}}{8}\right )+10\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )+\left (\frac {a^2\,5{}\mathrm {i}}{2}-\frac {b^2\,15{}\mathrm {i}}{8}\right )\,\left (5\,a^2-\frac {15\,b^2}{4}+6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {a^2\,5{}\mathrm {i}}{2}-\frac {b^2\,15{}\mathrm {i}}{8}\right )-10\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )+\frac {75\,a\,b^3}{2}-50\,a^3\,b+2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (25\,a^4-\frac {75\,a^2\,b^2}{2}+\frac {225\,b^4}{16}\right )}\right )\,\left (5\,a^2-\frac {15\,b^2}{4}\right )}{d}+\frac {a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4\,d}-\frac {5\,a\,b\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d} \]
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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