Optimal. Leaf size=119 \[ \frac {a^3 \log (\cos (c+d x))}{d}+\frac {b \left (3 a^2-b^2\right ) \sec ^3(c+d x)}{3 d}+\frac {a \left (a^2-3 b^2\right ) \sec ^2(c+d x)}{2 d}-\frac {3 a^2 b \sec (c+d x)}{d}+\frac {3 a b^2 \sec ^4(c+d x)}{4 d}+\frac {b^3 \sec ^5(c+d x)}{5 d} \]
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Rubi [A] time = 0.22, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4397, 2837, 12, 894} \[ \frac {b \left (3 a^2-b^2\right ) \sec ^3(c+d x)}{3 d}+\frac {a \left (a^2-3 b^2\right ) \sec ^2(c+d x)}{2 d}-\frac {3 a^2 b \sec (c+d x)}{d}+\frac {a^3 \log (\cos (c+d x))}{d}+\frac {3 a b^2 \sec ^4(c+d x)}{4 d}+\frac {b^3 \sec ^5(c+d x)}{5 d} \]
Antiderivative was successfully verified.
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Rule 12
Rule 894
Rule 2837
Rule 4397
Rubi steps
\begin {align*} \int \sec ^3(c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx &=\int (b+a \cos (c+d x))^3 \sec ^3(c+d x) \tan ^3(c+d x) \, dx\\ &=-\frac {\operatorname {Subst}\left (\int \frac {a^6 (b+x)^3 \left (a^2-x^2\right )}{x^6} \, dx,x,a \cos (c+d x)\right )}{a^3 d}\\ &=-\frac {a^3 \operatorname {Subst}\left (\int \frac {(b+x)^3 \left (a^2-x^2\right )}{x^6} \, dx,x,a \cos (c+d x)\right )}{d}\\ &=-\frac {a^3 \operatorname {Subst}\left (\int \left (\frac {a^2 b^3}{x^6}+\frac {3 a^2 b^2}{x^5}+\frac {3 a^2 b-b^3}{x^4}+\frac {a^2-3 b^2}{x^3}-\frac {3 b}{x^2}-\frac {1}{x}\right ) \, dx,x,a \cos (c+d x)\right )}{d}\\ &=\frac {a^3 \log (\cos (c+d x))}{d}-\frac {3 a^2 b \sec (c+d x)}{d}+\frac {a \left (a^2-3 b^2\right ) \sec ^2(c+d x)}{2 d}+\frac {b \left (3 a^2-b^2\right ) \sec ^3(c+d x)}{3 d}+\frac {3 a b^2 \sec ^4(c+d x)}{4 d}+\frac {b^3 \sec ^5(c+d x)}{5 d}\\ \end {align*}
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Mathematica [A] time = 0.32, size = 99, normalized size = 0.83 \[ \frac {60 a^3 \log (\cos (c+d x))-20 b \left (b^2-3 a^2\right ) \sec ^3(c+d x)+30 a \left (a^2-3 b^2\right ) \sec ^2(c+d x)-180 a^2 b \sec (c+d x)+45 a b^2 \sec ^4(c+d x)+12 b^3 \sec ^5(c+d x)}{60 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.65, size = 109, normalized size = 0.92 \[ \frac {60 \, a^{3} \cos \left (d x + c\right )^{5} \log \left (-\cos \left (d x + c\right )\right ) - 180 \, a^{2} b \cos \left (d x + c\right )^{4} + 45 \, a b^{2} \cos \left (d x + c\right ) + 30 \, {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 12 \, b^{3} + 20 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2}}{60 \, d \cos \left (d x + c\right )^{5}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.14, size = 252, normalized size = 2.12 \[ \frac {a^{3} \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}+\frac {a^{3} \ln \left (\cos \left (d x +c \right )\right )}{d}+\frac {a^{2} b \left (\sin ^{4}\left (d x +c \right )\right )}{d \cos \left (d x +c \right )^{3}}-\frac {a^{2} b \left (\sin ^{4}\left (d x +c \right )\right )}{d \cos \left (d x +c \right )}-\frac {\cos \left (d x +c \right ) \left (\sin ^{2}\left (d x +c \right )\right ) a^{2} b}{d}-\frac {2 a^{2} b \cos \left (d x +c \right )}{d}+\frac {3 a \,b^{2} \left (\sin ^{4}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}+\frac {b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{5 d \cos \left (d x +c \right )^{5}}+\frac {b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{15 d \cos \left (d x +c \right )^{3}}-\frac {b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{15 d \cos \left (d x +c \right )}-\frac {b^{3} \cos \left (d x +c \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{15 d}-\frac {2 b^{3} \cos \left (d x +c \right )}{15 d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.36, size = 128, normalized size = 1.08 \[ -\frac {30 \, a^{3} {\left (\frac {1}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right )^{2} - 1\right )\right )} - \frac {45 \, {\left (2 \, \sin \left (d x + c\right )^{2} - 1\right )} a b^{2}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} + \frac {60 \, {\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} a^{2} b}{\cos \left (d x + c\right )^{3}} + \frac {4 \, {\left (5 \, \cos \left (d x + c\right )^{2} - 3\right )} b^{3}}{\cos \left (d x + c\right )^{5}}}{60 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.15, size = 220, normalized size = 1.85 \[ -\frac {2\,a^3\,\mathrm {atanh}\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (6\,a^3+12\,a^2\,b-12\,a\,b^2+4\,b^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (-6\,a^3-28\,a^2\,b+12\,a\,b^2+\frac {4\,b^3}{3}\right )-2\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,a^2\,b+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a^3+20\,a^2\,b+\frac {4\,b^3}{3}\right )-\frac {4\,b^3}{15}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]
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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