Optimal. Leaf size=87 \[ -\frac {2^{-m-\frac {1}{2}} (1-\sin (c+d x))^{m+\frac {1}{2}} (a \sin (c+d x)+a)^m (e \cos (c+d x))^{-2 m-1} \, _2F_1\left (-\frac {1}{2},\frac {1}{2} (2 m+3);\frac {1}{2};\frac {1}{2} (\sin (c+d x)+1)\right )}{d e} \]
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Rubi [A] time = 0.10, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2689, 7, 70, 69} \[ -\frac {2^{-m-\frac {1}{2}} (1-\sin (c+d x))^{m+\frac {1}{2}} (a \sin (c+d x)+a)^m (e \cos (c+d x))^{-2 m-1} \, _2F_1\left (-\frac {1}{2},\frac {1}{2} (2 m+3);\frac {1}{2};\frac {1}{2} (\sin (c+d x)+1)\right )}{d e} \]
Antiderivative was successfully verified.
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Rule 7
Rule 69
Rule 70
Rule 2689
Rubi steps
\begin {align*} \int (e \cos (c+d x))^{-2-2 m} (a+a \sin (c+d x))^m \, dx &=\frac {\left (a^2 (e \cos (c+d x))^{-1-2 m} (a-a \sin (c+d x))^{\frac {1}{2} (1+2 m)} (a+a \sin (c+d x))^{\frac {1}{2} (1+2 m)}\right ) \operatorname {Subst}\left (\int (a-a x)^{\frac {1}{2} (-3-2 m)} (a+a x)^{\frac {1}{2} (-3-2 m)+m} \, dx,x,\sin (c+d x)\right )}{d e}\\ &=\frac {\left (a^2 (e \cos (c+d x))^{-1-2 m} (a-a \sin (c+d x))^{\frac {1}{2} (1+2 m)} (a+a \sin (c+d x))^{\frac {1}{2} (1+2 m)}\right ) \operatorname {Subst}\left (\int \frac {(a-a x)^{\frac {1}{2} (-3-2 m)}}{(a+a x)^{3/2}} \, dx,x,\sin (c+d x)\right )}{d e}\\ &=\frac {\left (2^{-\frac {3}{2}-m} a (e \cos (c+d x))^{-1-2 m} (a-a \sin (c+d x))^{-\frac {1}{2}-m+\frac {1}{2} (1+2 m)} \left (\frac {a-a \sin (c+d x)}{a}\right )^{\frac {1}{2}+m} (a+a \sin (c+d x))^{\frac {1}{2} (1+2 m)}\right ) \operatorname {Subst}\left (\int \frac {\left (\frac {1}{2}-\frac {x}{2}\right )^{\frac {1}{2} (-3-2 m)}}{(a+a x)^{3/2}} \, dx,x,\sin (c+d x)\right )}{d e}\\ &=-\frac {2^{-\frac {1}{2}-m} (e \cos (c+d x))^{-1-2 m} \, _2F_1\left (-\frac {1}{2},\frac {1}{2} (3+2 m);\frac {1}{2};\frac {1}{2} (1+\sin (c+d x))\right ) (1-\sin (c+d x))^{\frac {1}{2}+m} (a+a \sin (c+d x))^m}{d e}\\ \end {align*}
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Mathematica [A] time = 0.22, size = 87, normalized size = 1.00 \[ \frac {\sqrt {\sin (c+d x)+1} (a (\sin (c+d x)+1))^m (e \cos (c+d x))^{-2 m-1} \, _2F_1\left (\frac {3}{2},-m-\frac {1}{2};\frac {1}{2}-m;\frac {1}{2} (1-\sin (c+d x))\right )}{\sqrt {2} e (2 d m+d)} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (e \cos \left (d x + c\right )\right )^{-2 \, m - 2} {\left (a \sin \left (d x + c\right ) + a\right )}^{m}, x\right ) \]
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 \[ \int \left (e \cos \left (d x + c\right )\right )^{-2 \, m - 2} {\left (a \sin \left (d x + c\right ) + a\right )}^{m}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 1.29, size = 0, normalized size = 0.00 \[ \int \left (e \cos \left (d x +c \right )\right )^{-2-2 m} \left (a +a \sin \left (d x +c \right )\right )^{m}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (e \cos \left (d x + c\right )\right )^{-2 \, m - 2} {\left (a \sin \left (d x + c\right ) + a\right )}^{m}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^m}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{2\,m+2}} \,d x \]
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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