Optimal. Leaf size=147 \[ \frac{b \left (3 a^2-b^2\right ) (a+b \tan (c+d x))^2}{2 d}+\frac{4 a b^2 \left (a^2-b^2\right ) \tan (c+d x)}{d}-\frac{b \left (-10 a^2 b^2+5 a^4+b^4\right ) \log (\cos (c+d x))}{d}+a x \left (-10 a^2 b^2+a^4+5 b^4\right )+\frac{b (a+b \tan (c+d x))^4}{4 d}+\frac{2 a b (a+b \tan (c+d x))^3}{3 d} \]
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Rubi [A] time = 0.230165, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {3086, 3482, 3528, 3525, 3475} \[ \frac{b \left (3 a^2-b^2\right ) (a+b \tan (c+d x))^2}{2 d}+\frac{4 a b^2 \left (a^2-b^2\right ) \tan (c+d x)}{d}-\frac{b \left (-10 a^2 b^2+5 a^4+b^4\right ) \log (\cos (c+d x))}{d}+a x \left (-10 a^2 b^2+a^4+5 b^4\right )+\frac{b (a+b \tan (c+d x))^4}{4 d}+\frac{2 a b (a+b \tan (c+d x))^3}{3 d} \]
Antiderivative was successfully verified.
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Rule 3086
Rule 3482
Rule 3528
Rule 3525
Rule 3475
Rubi steps
\begin{align*} \int \sec ^5(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx &=\int (a+b \tan (c+d x))^5 \, dx\\ &=\frac{b (a+b \tan (c+d x))^4}{4 d}+\int (a+b \tan (c+d x))^3 \left (a^2-b^2+2 a b \tan (c+d x)\right ) \, dx\\ &=\frac{2 a b (a+b \tan (c+d x))^3}{3 d}+\frac{b (a+b \tan (c+d x))^4}{4 d}+\int (a+b \tan (c+d x))^2 \left (a \left (a^2-3 b^2\right )+b \left (3 a^2-b^2\right ) \tan (c+d x)\right ) \, dx\\ &=\frac{b \left (3 a^2-b^2\right ) (a+b \tan (c+d x))^2}{2 d}+\frac{2 a b (a+b \tan (c+d x))^3}{3 d}+\frac{b (a+b \tan (c+d x))^4}{4 d}+\int (a+b \tan (c+d x)) \left (a^4-6 a^2 b^2+b^4+4 a b \left (a^2-b^2\right ) \tan (c+d x)\right ) \, dx\\ &=a \left (a^4-10 a^2 b^2+5 b^4\right ) x+\frac{4 a b^2 \left (a^2-b^2\right ) \tan (c+d x)}{d}+\frac{b \left (3 a^2-b^2\right ) (a+b \tan (c+d x))^2}{2 d}+\frac{2 a b (a+b \tan (c+d x))^3}{3 d}+\frac{b (a+b \tan (c+d x))^4}{4 d}+\left (b \left (5 a^4-10 a^2 b^2+b^4\right )\right ) \int \tan (c+d x) \, dx\\ &=a \left (a^4-10 a^2 b^2+5 b^4\right ) x-\frac{b \left (5 a^4-10 a^2 b^2+b^4\right ) \log (\cos (c+d x))}{d}+\frac{4 a b^2 \left (a^2-b^2\right ) \tan (c+d x)}{d}+\frac{b \left (3 a^2-b^2\right ) (a+b \tan (c+d x))^2}{2 d}+\frac{2 a b (a+b \tan (c+d x))^3}{3 d}+\frac{b (a+b \tan (c+d x))^4}{4 d}\\ \end{align*}
Mathematica [C] time = 0.754068, size = 126, normalized size = 0.86 \[ \frac{-6 b^3 \left (b^2-10 a^2\right ) \tan ^2(c+d x)+60 a b^2 \left (2 a^2-b^2\right ) \tan (c+d x)+20 a b^4 \tan ^3(c+d x)+6 (b-i a)^5 \log (-\tan (c+d x)+i)+6 (b+i a)^5 \log (\tan (c+d x)+i)+3 b^5 \tan ^4(c+d x)}{12 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.264, size = 202, normalized size = 1.4 \begin{align*}{a}^{5}x+{\frac{{a}^{5}c}{d}}-5\,{\frac{{a}^{4}b\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}-10\,{a}^{3}{b}^{2}x+10\,{\frac{\tan \left ( dx+c \right ){a}^{3}{b}^{2}}{d}}-10\,{\frac{{a}^{3}{b}^{2}c}{d}}+5\,{\frac{{a}^{2}{b}^{3} \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{d}}+10\,{\frac{{a}^{2}{b}^{3}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}+{\frac{5\,a{b}^{4} \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{3\,d}}-5\,{\frac{a{b}^{4}\tan \left ( dx+c \right ) }{d}}+5\,a{b}^{4}x+5\,{\frac{a{b}^{4}c}{d}}+{\frac{{b}^{5} \left ( \tan \left ( dx+c \right ) \right ) ^{4}}{4\,d}}-{\frac{{b}^{5} \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{2\,d}}-{\frac{{b}^{5}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.74333, size = 235, normalized size = 1.6 \begin{align*} \frac{12 \,{\left (d x + c\right )} a^{5} - 120 \,{\left (d x + c - \tan \left (d x + c\right )\right )} a^{3} b^{2} + 20 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, d x + 3 \, c - 3 \, \tan \left (d x + c\right )\right )} a b^{4} + 3 \, b^{5}{\left (\frac{4 \, \sin \left (d x + c\right )^{2} - 3}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 2 \, \log \left (\sin \left (d x + c\right )^{2} - 1\right )\right )} - 60 \, a^{2} b^{3}{\left (\frac{1}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right )^{2} - 1\right )\right )} - 30 \, a^{4} b \log \left (-\sin \left (d x + c\right )^{2} + 1\right )}{12 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0.533899, size = 367, normalized size = 2.5 \begin{align*} \frac{12 \,{\left (a^{5} - 10 \, a^{3} b^{2} + 5 \, a b^{4}\right )} d x \cos \left (d x + c\right )^{4} - 12 \,{\left (5 \, a^{4} b - 10 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{4} \log \left (-\cos \left (d x + c\right )\right ) + 3 \, b^{5} + 12 \,{\left (5 \, a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2} + 20 \,{\left (a b^{4} \cos \left (d x + c\right ) + 2 \,{\left (3 \, a^{3} b^{2} - 2 \, a b^{4}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{4}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.27717, size = 194, normalized size = 1.32 \begin{align*} \frac{3 \, b^{5} \tan \left (d x + c\right )^{4} + 20 \, a b^{4} \tan \left (d x + c\right )^{3} + 60 \, a^{2} b^{3} \tan \left (d x + c\right )^{2} - 6 \, b^{5} \tan \left (d x + c\right )^{2} + 120 \, a^{3} b^{2} \tan \left (d x + c\right ) - 60 \, a b^{4} \tan \left (d x + c\right ) + 12 \,{\left (a^{5} - 10 \, a^{3} b^{2} + 5 \, a b^{4}\right )}{\left (d x + c\right )} + 6 \,{\left (5 \, a^{4} b - 10 \, a^{2} b^{3} + b^{5}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{12 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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