Optimal. Leaf size=205 \[ \frac{10 a^3 b^2 \sin ^3(c+d x)}{3 d}+\frac{10 a^2 b^3 \cos ^3(c+d x)}{3 d}-\frac{10 a^2 b^3 \cos (c+d x)}{d}-\frac{5 a^4 b \cos ^3(c+d x)}{3 d}-\frac{a^5 \sin ^3(c+d x)}{3 d}+\frac{a^5 \sin (c+d x)}{d}-\frac{5 a b^4 \sin ^3(c+d x)}{3 d}-\frac{5 a b^4 \sin (c+d x)}{d}+\frac{5 a b^4 \tanh ^{-1}(\sin (c+d x))}{d}-\frac{b^5 \cos ^3(c+d x)}{3 d}+\frac{2 b^5 \cos (c+d x)}{d}+\frac{b^5 \sec (c+d x)}{d} \]
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Rubi [A] time = 0.215913, antiderivative size = 205, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 10, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {3090, 2633, 2565, 30, 2564, 2592, 302, 206, 2590, 270} \[ \frac{10 a^3 b^2 \sin ^3(c+d x)}{3 d}+\frac{10 a^2 b^3 \cos ^3(c+d x)}{3 d}-\frac{10 a^2 b^3 \cos (c+d x)}{d}-\frac{5 a^4 b \cos ^3(c+d x)}{3 d}-\frac{a^5 \sin ^3(c+d x)}{3 d}+\frac{a^5 \sin (c+d x)}{d}-\frac{5 a b^4 \sin ^3(c+d x)}{3 d}-\frac{5 a b^4 \sin (c+d x)}{d}+\frac{5 a b^4 \tanh ^{-1}(\sin (c+d x))}{d}-\frac{b^5 \cos ^3(c+d x)}{3 d}+\frac{2 b^5 \cos (c+d x)}{d}+\frac{b^5 \sec (c+d x)}{d} \]
Antiderivative was successfully verified.
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Rule 3090
Rule 2633
Rule 2565
Rule 30
Rule 2564
Rule 2592
Rule 302
Rule 206
Rule 2590
Rule 270
Rubi steps
\begin{align*} \int \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx &=\int \left (a^5 \cos ^3(c+d x)+5 a^4 b \cos ^2(c+d x) \sin (c+d x)+10 a^3 b^2 \cos (c+d x) \sin ^2(c+d x)+10 a^2 b^3 \sin ^3(c+d x)+5 a b^4 \sin ^3(c+d x) \tan (c+d x)+b^5 \sin ^3(c+d x) \tan ^2(c+d x)\right ) \, dx\\ &=a^5 \int \cos ^3(c+d x) \, dx+\left (5 a^4 b\right ) \int \cos ^2(c+d x) \sin (c+d x) \, dx+\left (10 a^3 b^2\right ) \int \cos (c+d x) \sin ^2(c+d x) \, dx+\left (10 a^2 b^3\right ) \int \sin ^3(c+d x) \, dx+\left (5 a b^4\right ) \int \sin ^3(c+d x) \tan (c+d x) \, dx+b^5 \int \sin ^3(c+d x) \tan ^2(c+d x) \, dx\\ &=-\frac{a^5 \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{d}-\frac{\left (5 a^4 b\right ) \operatorname{Subst}\left (\int x^2 \, dx,x,\cos (c+d x)\right )}{d}+\frac{\left (10 a^3 b^2\right ) \operatorname{Subst}\left (\int x^2 \, dx,x,\sin (c+d x)\right )}{d}-\frac{\left (10 a^2 b^3\right ) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}+\frac{\left (5 a b^4\right ) \operatorname{Subst}\left (\int \frac{x^4}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d}-\frac{b^5 \operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^2}{x^2} \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac{10 a^2 b^3 \cos (c+d x)}{d}-\frac{5 a^4 b \cos ^3(c+d x)}{3 d}+\frac{10 a^2 b^3 \cos ^3(c+d x)}{3 d}+\frac{a^5 \sin (c+d x)}{d}-\frac{a^5 \sin ^3(c+d x)}{3 d}+\frac{10 a^3 b^2 \sin ^3(c+d x)}{3 d}+\frac{\left (5 a b^4\right ) \operatorname{Subst}\left (\int \left (-1-x^2+\frac{1}{1-x^2}\right ) \, dx,x,\sin (c+d x)\right )}{d}-\frac{b^5 \operatorname{Subst}\left (\int \left (-2+\frac{1}{x^2}+x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac{10 a^2 b^3 \cos (c+d x)}{d}+\frac{2 b^5 \cos (c+d x)}{d}-\frac{5 a^4 b \cos ^3(c+d x)}{3 d}+\frac{10 a^2 b^3 \cos ^3(c+d x)}{3 d}-\frac{b^5 \cos ^3(c+d x)}{3 d}+\frac{b^5 \sec (c+d x)}{d}+\frac{a^5 \sin (c+d x)}{d}-\frac{5 a b^4 \sin (c+d x)}{d}-\frac{a^5 \sin ^3(c+d x)}{3 d}+\frac{10 a^3 b^2 \sin ^3(c+d x)}{3 d}-\frac{5 a b^4 \sin ^3(c+d x)}{3 d}+\frac{\left (5 a b^4\right ) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac{5 a b^4 \tanh ^{-1}(\sin (c+d x))}{d}-\frac{10 a^2 b^3 \cos (c+d x)}{d}+\frac{2 b^5 \cos (c+d x)}{d}-\frac{5 a^4 b \cos ^3(c+d x)}{3 d}+\frac{10 a^2 b^3 \cos ^3(c+d x)}{3 d}-\frac{b^5 \cos ^3(c+d x)}{3 d}+\frac{b^5 \sec (c+d x)}{d}+\frac{a^5 \sin (c+d x)}{d}-\frac{5 a b^4 \sin (c+d x)}{d}-\frac{a^5 \sin ^3(c+d x)}{3 d}+\frac{10 a^3 b^2 \sin ^3(c+d x)}{3 d}-\frac{5 a b^4 \sin ^3(c+d x)}{3 d}\\ \end{align*}
Mathematica [B] time = 6.30718, size = 632, normalized size = 3.08 \[ -\frac{b \left (30 a^2 b^2+5 a^4-7 b^4\right ) \cos ^6(c+d x) (a+b \tan (c+d x))^5}{4 d (a \cos (c+d x)+b \sin (c+d x))^5}+\frac{a \left (-10 a^2 b^2+a^4+5 b^4\right ) \sin (3 (c+d x)) \cos ^5(c+d x) (a+b \tan (c+d x))^5}{12 d (a \cos (c+d x)+b \sin (c+d x))^5}-\frac{b \left (-10 a^2 b^2+5 a^4+b^4\right ) \cos (3 (c+d x)) \cos ^5(c+d x) (a+b \tan (c+d x))^5}{12 d (a \cos (c+d x)+b \sin (c+d x))^5}+\frac{a \left (10 a^2 b^2+3 a^4-25 b^4\right ) \sin (c+d x) \cos ^5(c+d x) (a+b \tan (c+d x))^5}{4 d (a \cos (c+d x)+b \sin (c+d x))^5}+\frac{b^5 \sin \left (\frac{1}{2} (c+d x)\right ) \cos ^5(c+d x) (a+b \tan (c+d x))^5}{d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b \sin (c+d x))^5}+\frac{b^5 \cos ^5(c+d x) (a+b \tan (c+d x))^5}{d (a \cos (c+d x)+b \sin (c+d x))^5}-\frac{b^5 \sin \left (\frac{1}{2} (c+d x)\right ) \cos ^5(c+d x) (a+b \tan (c+d x))^5}{d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b \sin (c+d x))^5}-\frac{5 a b^4 \cos ^5(c+d x) (a+b \tan (c+d x))^5 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{d (a \cos (c+d x)+b \sin (c+d x))^5}+\frac{5 a b^4 \cos ^5(c+d x) (a+b \tan (c+d x))^5 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{d (a \cos (c+d x)+b \sin (c+d x))^5} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.251, size = 251, normalized size = 1.2 \begin{align*}{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ){a}^{5}}{3\,d}}+{\frac{2\,{a}^{5}\sin \left ( dx+c \right ) }{3\,d}}-{\frac{5\,{a}^{4}b \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{3\,d}}+{\frac{10\,{a}^{3}{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{3\,d}}-{\frac{10\,\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{2}{a}^{2}{b}^{3}}{3\,d}}-{\frac{20\,{a}^{2}{b}^{3}\cos \left ( dx+c \right ) }{3\,d}}-{\frac{5\,a{b}^{4} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{3\,d}}-5\,{\frac{a{b}^{4}\sin \left ( dx+c \right ) }{d}}+5\,{\frac{a{b}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{{b}^{5} \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{d\cos \left ( dx+c \right ) }}+{\frac{8\,{b}^{5}\cos \left ( dx+c \right ) }{3\,d}}+{\frac{{b}^{5}\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{d}}+{\frac{4\,\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{2}{b}^{5}}{3\,d}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.22015, size = 219, normalized size = 1.07 \begin{align*} -\frac{10 \, a^{4} b \cos \left (d x + c\right )^{3} - 20 \, a^{3} b^{2} \sin \left (d x + c\right )^{3} + 2 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{5} - 20 \,{\left (\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} a^{2} b^{3} + 5 \,{\left (2 \, \sin \left (d x + c\right )^{3} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right ) + 6 \, \sin \left (d x + c\right )\right )} a b^{4} + 2 \,{\left (\cos \left (d x + c\right )^{3} - \frac{3}{\cos \left (d x + c\right )} - 6 \, \cos \left (d x + c\right )\right )} b^{5}}{6 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0.536934, size = 429, normalized size = 2.09 \begin{align*} \frac{15 \, a b^{4} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, a b^{4} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 6 \, b^{5} - 2 \,{\left (5 \, a^{4} b - 10 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{4} - 12 \,{\left (5 \, a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left ({\left (a^{5} - 10 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \cos \left (d x + c\right )^{3} + 2 \,{\left (a^{5} + 5 \, a^{3} b^{2} - 10 \, a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )} \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 = 1.2754, size = 382, normalized size = 1.86 \begin{align*} \frac{15 \, a b^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, a b^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{6 \, b^{5}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1} + \frac{2 \,{\left (3 \, a^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 15 \, a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 15 \, a^{4} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 3 \, b^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 2 \, a^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 40 \, a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 50 \, a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 60 \, a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 12 \, b^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 3 \, a^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 15 \, a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 5 \, a^{4} b - 20 \, a^{2} b^{3} + 5 \, b^{5}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{3}}}{3 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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