Optimal. Leaf size=135 \[ \frac {\left (4 a^2+5 b^2\right ) \tan ^3(c+d x)}{15 d}+\frac {\left (4 a^2+5 b^2\right ) \tan (c+d x)}{5 d}+\frac {a^2 \tan (c+d x) \sec ^4(c+d x)}{5 d}+\frac {3 a b \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac {a b \tan (c+d x) \sec ^3(c+d x)}{2 d}+\frac {3 a b \tan (c+d x) \sec (c+d x)}{4 d} \]
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Rubi [A] time = 0.11, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2789, 3768, 3770, 3012, 3767} \[ \frac {\left (4 a^2+5 b^2\right ) \tan ^3(c+d x)}{15 d}+\frac {\left (4 a^2+5 b^2\right ) \tan (c+d x)}{5 d}+\frac {a^2 \tan (c+d x) \sec ^4(c+d x)}{5 d}+\frac {3 a b \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac {a b \tan (c+d x) \sec ^3(c+d x)}{2 d}+\frac {3 a b \tan (c+d x) \sec (c+d x)}{4 d} \]
Antiderivative was successfully verified.
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Rule 2789
Rule 3012
Rule 3767
Rule 3768
Rule 3770
Rubi steps
\begin {align*} \int (a+b \cos (c+d x))^2 \sec ^6(c+d x) \, dx &=(2 a b) \int \sec ^5(c+d x) \, dx+\int \left (a^2+b^2 \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx\\ &=\frac {a b \sec ^3(c+d x) \tan (c+d x)}{2 d}+\frac {a^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{2} (3 a b) \int \sec ^3(c+d x) \, dx+\frac {1}{5} \left (4 a^2+5 b^2\right ) \int \sec ^4(c+d x) \, dx\\ &=\frac {3 a b \sec (c+d x) \tan (c+d x)}{4 d}+\frac {a b \sec ^3(c+d x) \tan (c+d x)}{2 d}+\frac {a^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{4} (3 a b) \int \sec (c+d x) \, dx-\frac {\left (4 a^2+5 b^2\right ) \operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{5 d}\\ &=\frac {3 a b \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac {\left (4 a^2+5 b^2\right ) \tan (c+d x)}{5 d}+\frac {3 a b \sec (c+d x) \tan (c+d x)}{4 d}+\frac {a b \sec ^3(c+d x) \tan (c+d x)}{2 d}+\frac {a^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {\left (4 a^2+5 b^2\right ) \tan ^3(c+d x)}{15 d}\\ \end {align*}
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Mathematica [A] time = 0.56, size = 118, normalized size = 0.87 \[ \frac {a^2 \left (\frac {1}{5} \tan ^5(c+d x)+\frac {2}{3} \tan ^3(c+d x)+\tan (c+d x)\right )}{d}+\frac {a b \tan (c+d x) \sec ^3(c+d x)}{2 d}+\frac {3 a b \left (\tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \sec (c+d x)\right )}{4 d}+\frac {b^2 \left (\frac {1}{3} \tan ^3(c+d x)+\tan (c+d x)\right )}{d} \]
Antiderivative was successfully verified.
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fricas [A] time = 1.43, size = 136, normalized size = 1.01 \[ \frac {45 \, a b \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 45 \, a b \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (45 \, a b \cos \left (d x + c\right )^{3} + 8 \, {\left (4 \, a^{2} + 5 \, b^{2}\right )} \cos \left (d x + c\right )^{4} + 30 \, a b \cos \left (d x + c\right ) + 4 \, {\left (4 \, a^{2} + 5 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 12 \, a^{2}\right )} \sin \left (d x + c\right )}{120 \, d \cos \left (d x + c\right )^{5}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.61, size = 272, normalized size = 2.01 \[ \frac {45 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 45 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (60 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 75 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 60 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 80 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 30 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 160 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 232 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 200 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 80 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 30 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 160 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 60 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 75 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 60 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{60 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.10, size = 157, normalized size = 1.16 \[ \frac {8 a^{2} \tan \left (d x +c \right )}{15 d}+\frac {a^{2} \left (\sec ^{4}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{5 d}+\frac {4 a^{2} \left (\sec ^{2}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{15 d}+\frac {a b \left (\sec ^{3}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{2 d}+\frac {3 a b \sec \left (d x +c \right ) \tan \left (d x +c \right )}{4 d}+\frac {3 a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{4 d}+\frac {2 b^{2} \tan \left (d x +c \right )}{3 d}+\frac {b^{2} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.55, size = 132, normalized size = 0.98 \[ \frac {8 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} a^{2} + 40 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} b^{2} - 15 \, a b {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{120 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.23, size = 221, normalized size = 1.64 \[ \frac {3\,a\,b\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,d}-\frac {\left (2\,a^2-\frac {5\,a\,b}{2}+2\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (-\frac {8\,a^2}{3}+a\,b-\frac {16\,b^2}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {116\,a^2}{15}+\frac {20\,b^2}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {8\,a^2}{3}-a\,b-\frac {16\,b^2}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,a^2+\frac {5\,a\,b}{2}+2\,b^2\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]
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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