Optimal. Leaf size=150 \[ -\frac {8 a^3 (e \cos (c+d x))^{6-2 m} (a+a \sin (c+d x))^{-3+m}}{d e (5-m) \left (12-7 m+m^2\right )}-\frac {4 a^2 (e \cos (c+d x))^{6-2 m} (a+a \sin (c+d x))^{-2+m}}{d e \left (20-9 m+m^2\right )}-\frac {a (e \cos (c+d x))^{6-2 m} (a+a \sin (c+d x))^{-1+m}}{d e (5-m)} \]
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Rubi [A]
time = 0.17, antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps
used = 3, number of rules used = 2, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {2753, 2752}
\begin {gather*} -\frac {8 a^3 (a \sin (c+d x)+a)^{m-3} (e \cos (c+d x))^{6-2 m}}{d e (5-m) \left (m^2-7 m+12\right )}-\frac {4 a^2 (a \sin (c+d x)+a)^{m-2} (e \cos (c+d x))^{6-2 m}}{d e \left (m^2-9 m+20\right )}-\frac {a (a \sin (c+d x)+a)^{m-1} (e \cos (c+d x))^{6-2 m}}{d e (5-m)} \end {gather*}
Antiderivative was successfully verified.
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Rule 2752
Rule 2753
Rubi steps
\begin {align*} \int (e \cos (c+d x))^{5-2 m} (a+a \sin (c+d x))^m \, dx &=-\frac {a (e \cos (c+d x))^{6-2 m} (a+a \sin (c+d x))^{-1+m}}{d e (5-m)}+\frac {(4 a) \int (e \cos (c+d x))^{5-2 m} (a+a \sin (c+d x))^{-1+m} \, dx}{5-m}\\ &=-\frac {4 a^2 (e \cos (c+d x))^{6-2 m} (a+a \sin (c+d x))^{-2+m}}{d e \left (20-9 m+m^2\right )}-\frac {a (e \cos (c+d x))^{6-2 m} (a+a \sin (c+d x))^{-1+m}}{d e (5-m)}+\frac {\left (8 a^2\right ) \int (e \cos (c+d x))^{5-2 m} (a+a \sin (c+d x))^{-2+m} \, dx}{20-9 m+m^2}\\ &=-\frac {8 a^3 (e \cos (c+d x))^{6-2 m} (a+a \sin (c+d x))^{-3+m}}{d e (3-m) \left (20-9 m+m^2\right )}-\frac {4 a^2 (e \cos (c+d x))^{6-2 m} (a+a \sin (c+d x))^{-2+m}}{d e \left (20-9 m+m^2\right )}-\frac {a (e \cos (c+d x))^{6-2 m} (a+a \sin (c+d x))^{-1+m}}{d e (5-m)}\\ \end {align*}
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Mathematica [A]
time = 0.40, size = 105, normalized size = 0.70 \begin {gather*} \frac {e^5 \cos ^6(c+d x) (e \cos (c+d x))^{-2 m} (a (1+\sin (c+d x)))^m \left (32-11 m+m^2+2 \left (18-9 m+m^2\right ) \sin (c+d x)+\left (12-7 m+m^2\right ) \sin ^2(c+d x)\right )}{d (-5+m) (-4+m) (-3+m) (1+\sin (c+d x))^3} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.21, size = 0, normalized size = 0.00 \[\int \left (e \cos \left (d x +c \right )\right )^{5-2 m} \left (a +a \sin \left (d x +c \right )\right )^{m}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 689 vs.
\(2 (145) = 290\).
time = 0.55, size = 689, normalized size = 4.59 \begin {gather*} \frac {{\left ({\left (m^{2} - 11 \, m + 32\right )} a^{m} e^{5} - \frac {2 \, {\left (m^{2} - 15 \, m + 60\right )} a^{m} e^{5} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {{\left (3 \, m^{2} - m - 160\right )} a^{m} e^{5} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {8 \, {\left (m^{2} - 7 \, m - 20\right )} a^{m} e^{5} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {2 \, {\left (m^{2} + 5 \, m + 160\right )} a^{m} e^{5} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {4 \, {\left (3 \, m^{2} - 13 \, m + 116\right )} a^{m} e^{5} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {2 \, {\left (m^{2} + 5 \, m + 160\right )} a^{m} e^{5} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {8 \, {\left (m^{2} - 7 \, m - 20\right )} a^{m} e^{5} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {{\left (3 \, m^{2} - m - 160\right )} a^{m} e^{5} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {2 \, {\left (m^{2} - 15 \, m + 60\right )} a^{m} e^{5} \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} + \frac {{\left (m^{2} - 11 \, m + 32\right )} a^{m} e^{5} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}}\right )} e^{\left (-2 \, m \log \left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right ) + m \log \left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )\right )}}{{\left (m^{3} e^{\left (2 \, m\right )} - 12 \, m^{2} e^{\left (2 \, m\right )} + 47 \, m e^{\left (2 \, m\right )} + \frac {5 \, {\left (m^{3} e^{\left (2 \, m\right )} - 12 \, m^{2} e^{\left (2 \, m\right )} + 47 \, m e^{\left (2 \, m\right )} - 60 \, e^{\left (2 \, m\right )}\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {10 \, {\left (m^{3} e^{\left (2 \, m\right )} - 12 \, m^{2} e^{\left (2 \, m\right )} + 47 \, m e^{\left (2 \, m\right )} - 60 \, e^{\left (2 \, m\right )}\right )} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {10 \, {\left (m^{3} e^{\left (2 \, m\right )} - 12 \, m^{2} e^{\left (2 \, m\right )} + 47 \, m e^{\left (2 \, m\right )} - 60 \, e^{\left (2 \, m\right )}\right )} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {5 \, {\left (m^{3} e^{\left (2 \, m\right )} - 12 \, m^{2} e^{\left (2 \, m\right )} + 47 \, m e^{\left (2 \, m\right )} - 60 \, e^{\left (2 \, m\right )}\right )} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac {{\left (m^{3} e^{\left (2 \, m\right )} - 12 \, m^{2} e^{\left (2 \, m\right )} + 47 \, m e^{\left (2 \, m\right )} - 60 \, e^{\left (2 \, m\right )}\right )} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} - 60 \, e^{\left (2 \, m\right )}\right )} d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 315 vs.
\(2 (145) = 290\).
time = 0.42, size = 315, normalized size = 2.10 \begin {gather*} -\frac {{\left ({\left (m^{2} - 7 \, m + 12\right )} \cos \left (d x + c\right )^{3} - {\left (m^{2} - 11 \, m + 24\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (m^{2} - 9 \, m + 22\right )} \cos \left (d x + c\right ) - {\left ({\left (m^{2} - 7 \, m + 12\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (m^{2} - 9 \, m + 18\right )} \cos \left (d x + c\right ) - 8\right )} \sin \left (d x + c\right ) - 8\right )} \left (\cos \left (d x + c\right ) e\right )^{-2 \, m + 5} {\left (a \sin \left (d x + c\right ) + a\right )}^{m}}{4 \, d m^{3} - {\left (d m^{3} - 12 \, d m^{2} + 47 \, d m - 60 \, d\right )} \cos \left (d x + c\right )^{3} - 48 \, d m^{2} - 3 \, {\left (d m^{3} - 12 \, d m^{2} + 47 \, d m - 60 \, d\right )} \cos \left (d x + c\right )^{2} + 188 \, d m + 2 \, {\left (d m^{3} - 12 \, d m^{2} + 47 \, d m - 60 \, d\right )} \cos \left (d x + c\right ) + {\left (4 \, d m^{3} - 48 \, d m^{2} - {\left (d m^{3} - 12 \, d m^{2} + 47 \, d m - 60 \, d\right )} \cos \left (d x + c\right )^{2} + 188 \, d m + 2 \, {\left (d m^{3} - 12 \, d m^{2} + 47 \, d m - 60 \, d\right )} \cos \left (d x + c\right ) - 240 \, d\right )} \sin \left (d x + c\right ) - 240 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 13.48, size = 601, normalized size = 4.01 \begin {gather*} -\frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^m\,\left (-\frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^{5-2\,m}\,\left (m^2-7\,m+12\right )}{d\,\left (m^3-12\,m^2+47\,m-60\right )}+\frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^{5-2\,m}\,\left (\cos \left (c+d\,x\right )+\sin \left (c+d\,x\right )\,1{}\mathrm {i}\right )\,\left (m^2\,3{}\mathrm {i}-m\,29{}\mathrm {i}+60{}\mathrm {i}\right )}{d\,\left (m^3-12\,m^2+47\,m-60\right )}-\frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^{5-2\,m}\,\left (\cos \left (5\,c+5\,d\,x\right )+\sin \left (5\,c+5\,d\,x\right )\,1{}\mathrm {i}\right )\,\left (m^2\,1{}\mathrm {i}-m\,7{}\mathrm {i}+12{}\mathrm {i}\right )}{d\,\left (m^3-12\,m^2+47\,m-60\right )}+\frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^{5-2\,m}\,\left (\cos \left (4\,c+4\,d\,x\right )+\sin \left (4\,c+4\,d\,x\right )\,1{}\mathrm {i}\right )\,\left (3\,m^2-29\,m+60\right )}{d\,\left (m^3-12\,m^2+47\,m-60\right )}+\frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^{5-2\,m}\,\left (\cos \left (2\,c+2\,d\,x\right )+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )\,\left (2\,m^2-22\,m+80\right )}{d\,\left (m^3-12\,m^2+47\,m-60\right )}+\frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^{5-2\,m}\,\left (\cos \left (3\,c+3\,d\,x\right )+\sin \left (3\,c+3\,d\,x\right )\,1{}\mathrm {i}\right )\,\left (m^2\,2{}\mathrm {i}-m\,22{}\mathrm {i}+80{}\mathrm {i}\right )}{d\,\left (m^3-12\,m^2+47\,m-60\right )}\right )}{5\,\cos \left (c+d\,x\right )+\sin \left (c+d\,x\right )\,5{}\mathrm {i}-10\,\cos \left (3\,c+3\,d\,x\right )+\cos \left (5\,c+5\,d\,x\right )-\sin \left (3\,c+3\,d\,x\right )\,10{}\mathrm {i}+\sin \left (5\,c+5\,d\,x\right )\,1{}\mathrm {i}+\frac {m^3\,1{}\mathrm {i}-m^2\,12{}\mathrm {i}+m\,47{}\mathrm {i}-60{}\mathrm {i}}{m^3-12\,m^2+47\,m-60}-\frac {10\,\left (\cos \left (2\,c+2\,d\,x\right )+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )\,\left (m^3\,1{}\mathrm {i}-m^2\,12{}\mathrm {i}+m\,47{}\mathrm {i}-60{}\mathrm {i}\right )}{m^3-12\,m^2+47\,m-60}+\frac {5\,\left (\cos \left (4\,c+4\,d\,x\right )+\sin \left (4\,c+4\,d\,x\right )\,1{}\mathrm {i}\right )\,\left (m^3\,1{}\mathrm {i}-m^2\,12{}\mathrm {i}+m\,47{}\mathrm {i}-60{}\mathrm {i}\right )}{m^3-12\,m^2+47\,m-60}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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