Optimal. Leaf size=146 \[ \frac {a b \left (19 a^2+16 b^2\right ) \tan (c+d x)}{6 d}+\frac {b^2 \left (26 a^2+9 b^2\right ) \tan (c+d x) \sec (c+d x)}{24 d}+\frac {\left (8 a^4+24 a^2 b^2+3 b^4\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {b \tan (c+d x) (a+b \sec (c+d x))^3}{4 d}+\frac {7 a b \tan (c+d x) (a+b \sec (c+d x))^2}{12 d} \]
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Rubi [A] time = 0.24, antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {3830, 4002, 3997, 3787, 3770, 3767, 8} \[ \frac {a b \left (19 a^2+16 b^2\right ) \tan (c+d x)}{6 d}+\frac {\left (24 a^2 b^2+8 a^4+3 b^4\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {b^2 \left (26 a^2+9 b^2\right ) \tan (c+d x) \sec (c+d x)}{24 d}+\frac {b \tan (c+d x) (a+b \sec (c+d x))^3}{4 d}+\frac {7 a b \tan (c+d x) (a+b \sec (c+d x))^2}{12 d} \]
Antiderivative was successfully verified.
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Rule 8
Rule 3767
Rule 3770
Rule 3787
Rule 3830
Rule 3997
Rule 4002
Rubi steps
\begin {align*} \int \sec (c+d x) (a+b \sec (c+d x))^4 \, dx &=\frac {b (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac {1}{4} \int \sec (c+d x) (a+b \sec (c+d x))^2 \left (4 a^2+3 b^2+7 a b \sec (c+d x)\right ) \, dx\\ &=\frac {7 a b (a+b \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac {b (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac {1}{12} \int \sec (c+d x) (a+b \sec (c+d x)) \left (a \left (12 a^2+23 b^2\right )+b \left (26 a^2+9 b^2\right ) \sec (c+d x)\right ) \, dx\\ &=\frac {b^2 \left (26 a^2+9 b^2\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {7 a b (a+b \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac {b (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac {1}{24} \int \sec (c+d x) \left (3 \left (8 a^4+24 a^2 b^2+3 b^4\right )+4 a b \left (19 a^2+16 b^2\right ) \sec (c+d x)\right ) \, dx\\ &=\frac {b^2 \left (26 a^2+9 b^2\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {7 a b (a+b \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac {b (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac {1}{6} \left (a b \left (19 a^2+16 b^2\right )\right ) \int \sec ^2(c+d x) \, dx+\frac {1}{8} \left (8 a^4+24 a^2 b^2+3 b^4\right ) \int \sec (c+d x) \, dx\\ &=\frac {\left (8 a^4+24 a^2 b^2+3 b^4\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {b^2 \left (26 a^2+9 b^2\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {7 a b (a+b \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac {b (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}-\frac {\left (a b \left (19 a^2+16 b^2\right )\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{6 d}\\ &=\frac {\left (8 a^4+24 a^2 b^2+3 b^4\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a b \left (19 a^2+16 b^2\right ) \tan (c+d x)}{6 d}+\frac {b^2 \left (26 a^2+9 b^2\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {7 a b (a+b \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac {b (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}\\ \end {align*}
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Mathematica [A] time = 0.54, size = 101, normalized size = 0.69 \[ \frac {b \tan (c+d x) \left (32 a \left (3 \left (a^2+b^2\right )+b^2 \tan ^2(c+d x)\right )+9 b \left (8 a^2+b^2\right ) \sec (c+d x)+6 b^3 \sec ^3(c+d x)\right )+3 \left (8 a^4+24 a^2 b^2+3 b^4\right ) \tanh ^{-1}(\sin (c+d x))}{24 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.51, size = 163, normalized size = 1.12 \[ \frac {3 \, {\left (8 \, a^{4} + 24 \, a^{2} b^{2} + 3 \, b^{4}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (8 \, a^{4} + 24 \, a^{2} b^{2} + 3 \, b^{4}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (32 \, a b^{3} \cos \left (d x + c\right ) + 6 \, b^{4} + 32 \, {\left (3 \, a^{3} b + 2 \, a b^{3}\right )} \cos \left (d x + c\right )^{3} + 9 \, {\left (8 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.29, size = 360, normalized size = 2.47 \[ \frac {3 \, {\left (8 \, a^{4} + 24 \, a^{2} b^{2} + 3 \, b^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (8 \, a^{4} + 24 \, a^{2} b^{2} + 3 \, b^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (96 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 72 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 96 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 15 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 288 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 72 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 160 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 288 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 72 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 160 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 9 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 96 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 72 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 96 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 1.12, size = 188, normalized size = 1.29 \[ \frac {a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {4 a^{3} b \tan \left (d x +c \right )}{d}+\frac {3 a^{2} b^{2} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{d}+\frac {3 a^{2} b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {8 a \,b^{3} \tan \left (d x +c \right )}{3 d}+\frac {4 a \,b^{3} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {b^{4} \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {3 b^{4} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {3 b^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.35, size = 180, normalized size = 1.23 \[ \frac {64 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a b^{3} - 3 \, b^{4} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 72 \, a^{2} b^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 48 \, a^{4} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 192 \, a^{3} b \tan \left (d x + c\right )}{48 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.92, size = 245, normalized size = 1.68 \[ \frac {\left (-8\,a^3\,b+6\,a^2\,b^2-8\,a\,b^3+\frac {5\,b^4}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (24\,a^3\,b-6\,a^2\,b^2+\frac {40\,a\,b^3}{3}+\frac {3\,b^4}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-24\,a^3\,b-6\,a^2\,b^2-\frac {40\,a\,b^3}{3}+\frac {3\,b^4}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (8\,a^3\,b+6\,a^2\,b^2+8\,a\,b^3+\frac {5\,b^4}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (2\,a^4+6\,a^2\,b^2+\frac {3\,b^4}{4}\right )}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \sec {\left (c + d x \right )}\right )^{4} \sec {\left (c + d x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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