Optimal. Leaf size=205 \[ -\frac {a^5 \sin ^3(c+d x)}{3 d}+\frac {a^5 \sin (c+d x)}{d}-\frac {5 a^4 b \cos ^3(c+d x)}{3 d}+\frac {10 a^3 b^2 \sin ^3(c+d x)}{3 d}+\frac {10 a^2 b^3 \cos ^3(c+d x)}{3 d}-\frac {10 a^2 b^3 \cos (c+d x)}{d}-\frac {5 a b^4 \sin ^3(c+d x)}{3 d}-\frac {5 a b^4 \sin (c+d x)}{d}+\frac {5 a b^4 \tanh ^{-1}(\sin (c+d x))}{d}-\frac {b^5 \cos ^3(c+d x)}{3 d}+\frac {2 b^5 \cos (c+d x)}{d}+\frac {b^5 \sec (c+d x)}{d} \]
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Rubi [A] time = 0.22, antiderivative size = 205, normalized size of antiderivative = 1.00, 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 30
Rule 206
Rule 270
Rule 302
Rule 2564
Rule 2565
Rule 2590
Rule 2592
Rule 2633
Rule 3090
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*}
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Mathematica [B] time = 6.30, size = 632, normalized size = 3.08 \[ -\frac {b \left (5 a^4+30 a^2 b^2-7 b^4\right ) \cos ^6(c+d x) (a+b \tan (c+d x))^5}{4 d (a \cos (c+d x)+b \sin (c+d x))^5}+\frac {a \left (a^4-10 a^2 b^2+5 b^4\right ) \sin (3 (c+d x)) \cos ^5(c+d x) (a+b \tan (c+d x))^5}{12 d (a \cos (c+d x)+b \sin (c+d x))^5}-\frac {b \left (5 a^4-10 a^2 b^2+b^4\right ) \cos (3 (c+d x)) \cos ^5(c+d x) (a+b \tan (c+d x))^5}{12 d (a \cos (c+d x)+b \sin (c+d x))^5}+\frac {a \left (3 a^4+10 a^2 b^2-25 b^4\right ) \sin (c+d x) \cos ^5(c+d x) (a+b \tan (c+d x))^5}{4 d (a \cos (c+d x)+b \sin (c+d x))^5}+\frac {b^5 \sin \left (\frac {1}{2} (c+d x)\right ) \cos ^5(c+d x) (a+b \tan (c+d x))^5}{d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b \sin (c+d x))^5}+\frac {b^5 \cos ^5(c+d x) (a+b \tan (c+d x))^5}{d (a \cos (c+d x)+b \sin (c+d x))^5}-\frac {b^5 \sin \left (\frac {1}{2} (c+d x)\right ) \cos ^5(c+d x) (a+b \tan (c+d x))^5}{d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b \sin (c+d x))^5}-\frac {5 a b^4 \cos ^5(c+d x) (a+b \tan (c+d x))^5 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d (a \cos (c+d x)+b \sin (c+d x))^5}+\frac {5 a b^4 \cos ^5(c+d x) (a+b \tan (c+d x))^5 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{d (a \cos (c+d x)+b \sin (c+d x))^5} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.61, size = 177, normalized size = 0.86 \[ \frac {15 \, a b^{4} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, a b^{4} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 6 \, b^{5} - 2 \, {\left (5 \, a^{4} b - 10 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{4} - 12 \, {\left (5 \, a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left ({\left (a^{5} - 10 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (a^{5} + 5 \, a^{3} b^{2} - 10 \, a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.56, size = 283, normalized size = 1.38 \[ \frac {15 \, a b^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, a b^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {6 \, b^{5}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} + \frac {2 \, {\left (3 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 15 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 15 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 3 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 2 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 40 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 50 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 60 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 12 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 5 \, a^{4} b - 20 \, a^{2} b^{3} + 5 \, b^{5}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{3 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.24, size = 251, normalized size = 1.22 \[ \frac {\left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) a^{5}}{3 d}+\frac {2 a^{5} \sin \left (d x +c \right )}{3 d}-\frac {5 a^{4} b \left (\cos ^{3}\left (d x +c \right )\right )}{3 d}+\frac {10 a^{3} b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{3 d}-\frac {10 \cos \left (d x +c \right ) \left (\sin ^{2}\left (d x +c \right )\right ) a^{2} b^{3}}{3 d}-\frac {20 a^{2} b^{3} \cos \left (d x +c \right )}{3 d}-\frac {5 a \,b^{4} \left (\sin ^{3}\left (d x +c \right )\right )}{3 d}-\frac {5 a \,b^{4} \sin \left (d x +c \right )}{d}+\frac {5 a \,b^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {b^{5} \left (\sin ^{6}\left (d x +c \right )\right )}{d \cos \left (d x +c \right )}+\frac {8 b^{5} \cos \left (d x +c \right )}{3 d}+\frac {b^{5} \cos \left (d x +c \right ) \left (\sin ^{4}\left (d x +c \right )\right )}{d}+\frac {4 \cos \left (d x +c \right ) \left (\sin ^{2}\left (d x +c \right )\right ) b^{5}}{3 d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.33, size = 162, normalized size = 0.79 \[ -\frac {10 \, a^{4} b \cos \left (d x + c\right )^{3} - 20 \, a^{3} b^{2} \sin \left (d x + c\right )^{3} + 2 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{5} - 20 \, {\left (\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} a^{2} b^{3} + 5 \, {\left (2 \, \sin \left (d x + c\right )^{3} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right ) + 6 \, \sin \left (d x + c\right )\right )} a b^{4} + 2 \, {\left (\cos \left (d x + c\right )^{3} - \frac {3}{\cos \left (d x + c\right )} - 6 \, \cos \left (d x + c\right )\right )} b^{5}}{6 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.98, size = 277, normalized size = 1.35 \[ \frac {10\,a\,b^4\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (10\,a\,b^4-2\,a^5\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (10\,a^4\,b-40\,a^2\,b^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {2\,a^5}{3}-\frac {80\,a^3\,b^2}{3}+\frac {70\,a\,b^4}{3}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {2\,a^5}{3}-\frac {80\,a^3\,b^2}{3}+\frac {70\,a\,b^4}{3}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {10\,a^4\,b}{3}-\frac {80\,a^2\,b^3}{3}+\frac {32\,b^5}{3}\right )+\frac {10\,a^4\,b}{3}-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (10\,a\,b^4-2\,a^5\right )-\frac {16\,b^5}{3}+\frac {40\,a^2\,b^3}{3}-10\,a^4\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{d\,\left (-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]
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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