3.174 \(\int (g \cos (e+f x))^{1-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1+m} \, dx\)
Optimal. Leaf size=57 \[ -\frac {g \log (1-\sin (e+f x)) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^m (g \cos (e+f x))^{-2 m}}{c f} \]
[Out]
-g*ln(1-sin(f*x+e))*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^m/c/f/((g*cos(f*x+e))^(2*m))
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Rubi [A] time = 0.23, antiderivative size = 57, normalized size of antiderivative = 1.00,
number of steps used = 4, number of rules used = 4, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used =
{2843, 12, 2667, 31} \[ -\frac {g \log (1-\sin (e+f x)) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^m (g \cos (e+f x))^{-2 m}}{c f} \]
Antiderivative was successfully verified.
[In]
Int[(g*Cos[e + f*x])^(1 - 2*m)*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(-1 + m),x]
[Out]
-((g*Log[1 - Sin[e + f*x]]*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^m)/(c*f*(g*Cos[e + f*x])^(2*m)))
Rule 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]
Rule 31
Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]
Rule 2667
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] || !IntegerQ[m + 1/2])
Rule 2843
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) +
(f_.)*(x_)])^(n_), x_Symbol] :> Dist[(a^IntPart[m]*c^IntPart[m]*(a + b*Sin[e + f*x])^FracPart[m]*(c + d*Sin[e
+ f*x])^FracPart[m])/(g^(2*IntPart[m])*(g*Cos[e + f*x])^(2*FracPart[m])), Int[(g*Cos[e + f*x])^(2*m + p)/(c +
d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0]
&& EqQ[2*m + p - 1, 0] && EqQ[m - n - 1, 0]
Rubi steps
\begin {align*} \int (g \cos (e+f x))^{1-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1+m} \, dx &=\left ((g \cos (e+f x))^{-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^m\right ) \int \frac {g \cos (e+f x)}{c-c \sin (e+f x)} \, dx\\ &=\left (g (g \cos (e+f x))^{-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^m\right ) \int \frac {\cos (e+f x)}{c-c \sin (e+f x)} \, dx\\ &=-\frac {\left (g (g \cos (e+f x))^{-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^m\right ) \operatorname {Subst}\left (\int \frac {1}{c+x} \, dx,x,-c \sin (e+f x)\right )}{c f}\\ &=-\frac {g (g \cos (e+f x))^{-2 m} \log (1-\sin (e+f x)) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^m}{c f}\\ \end {align*}
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Mathematica [B] time = 84.63, size = 132, normalized size = 2.32 \[ \frac {g 2^{m+1} \cos ^{2 m}\left (\frac {1}{4} (2 e+2 f x+\pi )\right ) (a (\sin (e+f x)+1))^m (c-c \sin (e+f x))^m (g \cos (e+f x))^{-2 m} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^{-2 m} \left (\log \left (\csc ^2\left (\frac {1}{8} (2 e+2 f x+3 \pi )\right )\right )-\log \left (\tan \left (\frac {1}{8} (-2 e-2 f x+\pi )\right )\right )\right )}{c f} \]
Antiderivative was successfully verified.
[In]
Integrate[(g*Cos[e + f*x])^(1 - 2*m)*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(-1 + m),x]
[Out]
(2^(1 + m)*g*Cos[(2*e + Pi + 2*f*x)/4]^(2*m)*(Log[Csc[(2*e + 3*Pi + 2*f*x)/8]^2] - Log[Tan[(-2*e + Pi - 2*f*x)
/8]])*(a*(1 + Sin[e + f*x]))^m*(c - c*Sin[e + f*x])^m)/(c*f*(g*Cos[e + f*x])^(2*m)*(Cos[(e + f*x)/2] - Sin[(e
+ f*x)/2])^(2*m))
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fricas [A] time = 0.50, size = 30, normalized size = 0.53 \[ -\frac {a \left (\frac {a c}{g^{2}}\right )^{m - 1} \log \left (-\sin \left (f x + e\right ) + 1\right )}{f g} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((g*cos(f*x+e))^(1-2*m)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-1+m),x, algorithm="fricas")
[Out]
-a*(a*c/g^2)^(m - 1)*log(-sin(f*x + e) + 1)/(f*g)
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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((g*cos(f*x+e))^(1-2*m)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-1+m),x, algorithm="giac")
[Out]
Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)(4*pi*exp(m*ln(abs(a))+m*ln(abs(c))-2*m*ln(abs(g))-ln(abs(c))+
ln(abs(g)))*floor((2*f*x-4*pi*floor((f*x+pi+exp(1))*1/2/pi)+pi+2*exp(1))*1/4/pi)*tan((4*m*pi*floor(-(sign(a)-2
)/4)+4*m*pi*floor(-(sign(c)-4)/4)+m*pi*sign(a)+m*pi*sign(c)-2*m*pi*sign(g)-4*pi*floor(-(sign(c)-4)/4)-pi*sign(
c)+pi*sign(g)+pi)/4)^2-4*pi*exp(m*ln(abs(a))+m*ln(abs(c))-2*m*ln(abs(g))-ln(abs(c))+ln(abs(g)))*floor((2*f*x-4
*pi*floor((f*x+pi+exp(1))*1/2/pi)+pi+2*exp(1))*1/4/pi)+4*pi*exp(m*ln(abs(a))+m*ln(abs(c))-2*m*ln(abs(g))-ln(ab
s(c))+ln(abs(g)))*floor((f*x+pi+exp(1))*1/2/pi)*tan((4*m*pi*floor(-(sign(a)-2)/4)+4*m*pi*floor(-(sign(c)-4)/4)
+m*pi*sign(a)+m*pi*sign(c)-2*m*pi*sign(g)-4*pi*floor(-(sign(c)-4)/4)-pi*sign(c)+pi*sign(g)+pi)/4)^2-4*pi*exp(m
*ln(abs(a))+m*ln(abs(c))-2*m*ln(abs(g))-ln(abs(c))+ln(abs(g)))*floor((f*x+pi+exp(1))*1/2/pi)+2*pi*exp(m*ln(abs
(a))+m*ln(abs(c))-2*m*ln(abs(g))-ln(abs(c))+ln(abs(g)))*sign(tan((f*x+exp(1))/2)^2-1)*tan((4*m*pi*floor(-(sign
(a)-2)/4)+4*m*pi*floor(-(sign(c)-4)/4)+m*pi*sign(a)+m*pi*sign(c)-2*m*pi*sign(g)-4*pi*floor(-(sign(c)-4)/4)-pi*
sign(c)+pi*sign(g)+pi)/4)^2-2*pi*exp(m*ln(abs(a))+m*ln(abs(c))-2*m*ln(abs(g))-ln(abs(c))+ln(abs(g)))*sign(tan(
(f*x+exp(1))/2)^2-1)+3*pi*exp(m*ln(abs(a))+m*ln(abs(c))-2*m*ln(abs(g))-ln(abs(c))+ln(abs(g)))*tan((4*m*pi*floo
r(-(sign(a)-2)/4)+4*m*pi*floor(-(sign(c)-4)/4)+m*pi*sign(a)+m*pi*sign(c)-2*m*pi*sign(g)-4*pi*floor(-(sign(c)-4
)/4)-pi*sign(c)+pi*sign(g)+pi)/4)^2-3*pi*exp(m*ln(abs(a))+m*ln(abs(c))-2*m*ln(abs(g))-ln(abs(c))+ln(abs(g)))-2
*exp(1)*exp(m*ln(abs(a))+m*ln(abs(c))-2*m*ln(abs(g))-ln(abs(c))+ln(abs(g)))*tan((4*m*pi*floor(-(sign(a)-2)/4)+
4*m*pi*floor(-(sign(c)-4)/4)+m*pi*sign(a)+m*pi*sign(c)-2*m*pi*sign(g)-4*pi*floor(-(sign(c)-4)/4)-pi*sign(c)+pi
*sign(g)+pi)/4)^2+2*exp(1)*exp(m*ln(abs(a))+m*ln(abs(c))-2*m*ln(abs(g))-ln(abs(c))+ln(abs(g)))-4*exp(m*ln(abs(
a))+m*ln(abs(c))-2*m*ln(abs(g))-ln(abs(c))+ln(abs(g)))*ln((2*tan((f*x+exp(1))/2)^2-4*tan((f*x+exp(1))/2)+2)/(t
an((f*x+exp(1))/2)^2+1))*tan((4*m*pi*floor(-(sign(a)-2)/4)+4*m*pi*floor(-(sign(c)-4)/4)+m*pi*sign(a)+m*pi*sign
(c)-2*m*pi*sign(g)-4*pi*floor(-(sign(c)-4)/4)-pi*sign(c)+pi*sign(g)+pi)/4))/(2*f*tan((4*m*pi*floor(-(sign(a)-2
)/4)+4*m*pi*floor(-(sign(c)-4)/4)+m*pi*sign(a)+m*pi*sign(c)-2*m*pi*sign(g)-4*pi*floor(-(sign(c)-4)/4)-pi*sign(
c)+pi*sign(g)+pi)/4)^2+2*f)
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maple [C] time = 13.42, size = 8507, normalized size = 149.25 \[ \text {output too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((g*cos(f*x+e))^(1-2*m)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-1+m),x)
[Out]
result too large to display
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (g \cos \left (f x + e\right )\right )^{-2 \, m + 1} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{m - 1}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((g*cos(f*x+e))^(1-2*m)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-1+m),x, algorithm="maxima")
[Out]
integrate((g*cos(f*x + e))^(-2*m + 1)*(a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c)^(m - 1), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.02 \[ \int {\left (g\,\cos \left (e+f\,x\right )\right )}^{1-2\,m}\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{m-1} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((g*cos(e + f*x))^(1 - 2*m)*(a + a*sin(e + f*x))^m*(c - c*sin(e + f*x))^(m - 1),x)
[Out]
int((g*cos(e + f*x))^(1 - 2*m)*(a + a*sin(e + f*x))^m*(c - c*sin(e + f*x))^(m - 1), x)
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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.
[In]
integrate((g*cos(f*x+e))**(1-2*m)*(a+a*sin(f*x+e))**m*(c-c*sin(f*x+e))**(-1+m),x)
[Out]
Timed out
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