3.176 \(\int \frac {(A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{\sqrt {a+a \sin (e+f x)}} \, dx\)
Optimal. Leaf size=96 \[ \frac {c (A-B) \cos (e+f x) \log (\sin (e+f x)+1)}{f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {B \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}} \]
[Out]
(A-B)*c*cos(f*x+e)*ln(1+sin(f*x+e))/f/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(1/2)-B*cos(f*x+e)*(c-c*sin(f*x+
e))^(1/2)/f/(a+a*sin(f*x+e))^(1/2)
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Rubi [A] time = 0.32, antiderivative size = 96, normalized size of antiderivative = 1.00,
number of steps used = 5, number of rules used = 5, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used =
{2971, 2738, 2737, 2667, 31} \[ \frac {c (A-B) \cos (e+f x) \log (\sin (e+f x)+1)}{f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {B \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}} \]
Antiderivative was successfully verified.
[In]
Int[((A + B*Sin[e + f*x])*Sqrt[c - c*Sin[e + f*x]])/Sqrt[a + a*Sin[e + f*x]],x]
[Out]
((A - B)*c*Cos[e + f*x]*Log[1 + Sin[e + f*x]])/(f*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c - c*Sin[e + f*x]]) - (B*Cos[
e + f*x]*Sqrt[c - c*Sin[e + f*x]])/(f*Sqrt[a + a*Sin[e + f*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 2737
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(
a*c*Cos[e + f*x])/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]), Int[Cos[e + f*x]/(c + d*Sin[e + f*x]),
x], x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0]
Rule 2738
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[
(-2*b*Cos[e + f*x]*(c + d*Sin[e + f*x])^n)/(f*(2*n + 1)*Sqrt[a + b*Sin[e + f*x]]), x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[n, -2^(-1)]
Rule 2971
Int[Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[B/d, Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n + 1), x], x
] - Dist[(B*c - A*d)/d, Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f
, A, B, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0]
Rubi steps
\begin {align*} \int \frac {(A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{\sqrt {a+a \sin (e+f x)}} \, dx &=\frac {B \int \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)} \, dx}{a}-(-A+B) \int \frac {\sqrt {c-c \sin (e+f x)}}{\sqrt {a+a \sin (e+f x)}} \, dx\\ &=-\frac {B \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a+a \sin (e+f x)}}-\frac {(a (-A+B) c \cos (e+f x)) \int \frac {\cos (e+f x)}{a+a \sin (e+f x)} \, dx}{\sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}\\ &=-\frac {B \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a+a \sin (e+f x)}}-\frac {((-A+B) c \cos (e+f x)) \operatorname {Subst}\left (\int \frac {1}{a+x} \, dx,x,a \sin (e+f x)\right )}{f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}\\ &=\frac {(A-B) c \cos (e+f x) \log (1+\sin (e+f x))}{f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}-\frac {B \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a+a \sin (e+f x)}}\\ \end {align*}
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Mathematica [C] time = 1.13, size = 119, normalized size = 1.24 \[ \frac {\sqrt {c-c \sin (e+f x)} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right ) \left (B \sin (e+f x)+(A-B) \left (2 \log \left (e^{i (e+f x)}+i\right )-i f x\right )\right )}{f \sqrt {a (\sin (e+f x)+1)} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )} \]
Antiderivative was successfully verified.
[In]
Integrate[((A + B*Sin[e + f*x])*Sqrt[c - c*Sin[e + f*x]])/Sqrt[a + a*Sin[e + f*x]],x]
[Out]
((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*((A - B)*((-I)*f*x + 2*Log[I + E^(I*(e + f*x))]) + B*Sin[e + f*x])*Sqrt
[c - c*Sin[e + f*x]])/(f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x])])
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fricas [F] time = 1.36, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B \sin \left (f x + e\right ) + A\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{\sqrt {a \sin \left (f x + e\right ) + a}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e))^(1/2),x, algorithm="fricas")
[Out]
integral((B*sin(f*x + e) + A)*sqrt(-c*sin(f*x + e) + c)/sqrt(a*sin(f*x + e) + a), x)
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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((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e))^(1/2),x, algorithm="giac")
[Out]
Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (8*pi/x/2)>(-8*pi/
x/2)4*sqrt(2*c)*(A*sqrt(a*tan(1/2*exp(1))^2+a)*(12582912*tan(1/2*exp(1))^5-37748736*tan(1/2*exp(1))^4-41943040
*tan(1/2*exp(1))^3+25165824*tan(1/2*exp(1))^2+12582912*tan(1/2*exp(1))-4194304)+B*sqrt(a*tan(1/2*exp(1))^2+a)*
(-12582912*tan(1/2*exp(1))^5+37748736*tan(1/2*exp(1))^4+41943040*tan(1/2*exp(1))^3-25165824*tan(1/2*exp(1))^2-
12582912*tan(1/2*exp(1))+4194304)+A*sqrt(a*tan(1/2*exp(1))^2+a)*(25165824*tan(1/2*exp(1))^6-75497472*tan(1/2*e
xp(1))^5-150994944*tan(1/2*exp(1))^4+251658240*tan(1/2*exp(1))^3+226492416*tan(1/2*exp(1))^2-75497472*tan(1/2*
exp(1)))*tan(1/4*exp(1))^5+A*sqrt(a*tan(1/2*exp(1))^2+a)*(25165824*tan(1/2*exp(1))^6-75497472*tan(1/2*exp(1))^
5-150994944*tan(1/2*exp(1))^4+251658240*tan(1/2*exp(1))^3+226492416*tan(1/2*exp(1))^2-75497472*tan(1/2*exp(1))
)*tan(1/4*exp(1))+A*sqrt(a*tan(1/2*exp(1))^2+a)*(188743680*tan(1/2*exp(1))^5-566231040*tan(1/2*exp(1))^4-62914
5600*tan(1/2*exp(1))^3+377487360*tan(1/2*exp(1))^2+188743680*tan(1/2*exp(1))-62914560)*tan(1/4*exp(1))^4+A*sqr
t(a*tan(1/2*exp(1))^2+a)*(-12582912*tan(1/2*exp(1))^5+37748736*tan(1/2*exp(1))^4+41943040*tan(1/2*exp(1))^3-25
165824*tan(1/2*exp(1))^2-12582912*tan(1/2*exp(1))+4194304)*tan(1/4*exp(1))^6+A*sqrt(a*tan(1/2*exp(1))^2+a)*(-8
3886080*tan(1/2*exp(1))^6+251658240*tan(1/2*exp(1))^5+503316480*tan(1/2*exp(1))^4-838860800*tan(1/2*exp(1))^3-
754974720*tan(1/2*exp(1))^2+251658240*tan(1/2*exp(1)))*tan(1/4*exp(1))^3+A*sqrt(a*tan(1/2*exp(1))^2+a)*(-18874
3680*tan(1/2*exp(1))^5+566231040*tan(1/2*exp(1))^4+629145600*tan(1/2*exp(1))^3-377487360*tan(1/2*exp(1))^2-188
743680*tan(1/2*exp(1))+62914560)*tan(1/4*exp(1))^2+B*sqrt(a*tan(1/2*exp(1))^2+a)*(12582912*tan(1/2*exp(1))^5-3
7748736*tan(1/2*exp(1))^4-41943040*tan(1/2*exp(1))^3+25165824*tan(1/2*exp(1))^2+12582912*tan(1/2*exp(1))-41943
04)*tan(1/4*exp(1))^6+B*sqrt(a*tan(1/2*exp(1))^2+a)*(83886080*tan(1/2*exp(1))^6-251658240*tan(1/2*exp(1))^5-50
3316480*tan(1/2*exp(1))^4+838860800*tan(1/2*exp(1))^3+754974720*tan(1/2*exp(1))^2-251658240*tan(1/2*exp(1)))*t
an(1/4*exp(1))^3+B*sqrt(a*tan(1/2*exp(1))^2+a)*(188743680*tan(1/2*exp(1))^5-566231040*tan(1/2*exp(1))^4-629145
600*tan(1/2*exp(1))^3+377487360*tan(1/2*exp(1))^2+188743680*tan(1/2*exp(1))-62914560)*tan(1/4*exp(1))^2+B*sqrt
(a*tan(1/2*exp(1))^2+a)*(-25165824*tan(1/2*exp(1))^6+75497472*tan(1/2*exp(1))^5+150994944*tan(1/2*exp(1))^4-25
1658240*tan(1/2*exp(1))^3-226492416*tan(1/2*exp(1))^2+75497472*tan(1/2*exp(1)))*tan(1/4*exp(1))^5+B*sqrt(a*tan
(1/2*exp(1))^2+a)*(-25165824*tan(1/2*exp(1))^6+75497472*tan(1/2*exp(1))^5+150994944*tan(1/2*exp(1))^4-25165824
0*tan(1/2*exp(1))^3-226492416*tan(1/2*exp(1))^2+75497472*tan(1/2*exp(1)))*tan(1/4*exp(1))+B*sqrt(a*tan(1/2*exp
(1))^2+a)*(-188743680*tan(1/2*exp(1))^5+566231040*tan(1/2*exp(1))^4+629145600*tan(1/2*exp(1))^3-377487360*tan(
1/2*exp(1))^2-188743680*tan(1/2*exp(1))+62914560)*tan(1/4*exp(1))^4)*ln(abs(2*tan(1/2*exp(1))^3+6*tan(1/2*exp(
1))^2+(tan(1/2*(1/2*f*x+2*exp(1)))-1/tan(1/2*(1/2*f*x+2*exp(1))))*(tan(1/2*exp(1))^3-3*tan(1/2*exp(1))^2-3*tan
(1/2*exp(1))+1)-6*tan(1/2*exp(1))-2))/f/(-8388608*sqrt(2)*a*tan(1/2*exp(1))^7-8388608*sqrt(2)*a+(-8388608*sqrt
(2)*a*tan(1/2*exp(1))^7+25165824*sqrt(2)*a*tan(1/2*exp(1))^6+8388608*sqrt(2)*a*tan(1/2*exp(1))^5+41943040*sqrt
(2)*a*tan(1/2*exp(1))^4+41943040*sqrt(2)*a*tan(1/2*exp(1))^3+8388608*sqrt(2)*a*tan(1/2*exp(1))^2-8388608*sqrt(
2)*a+25165824*sqrt(2)*a*tan(1/2*exp(1)))*tan(1/4*exp(1))^6+(-25165824*sqrt(2)*a*tan(1/2*exp(1))^7+75497472*sqr
t(2)*a*tan(1/2*exp(1))^6+25165824*sqrt(2)*a*tan(1/2*exp(1))^5+125829120*sqrt(2)*a*tan(1/2*exp(1))^4+125829120*
sqrt(2)*a*tan(1/2*exp(1))^3+25165824*sqrt(2)*a*tan(1/2*exp(1))^2-25165824*sqrt(2)*a+75497472*sqrt(2)*a*tan(1/2
*exp(1)))*tan(1/4*exp(1))^2+(-25165824*sqrt(2)*a*tan(1/2*exp(1))^7+75497472*sqrt(2)*a*tan(1/2*exp(1))^6+251658
24*sqrt(2)*a*tan(1/2*exp(1))^5+125829120*sqrt(2)*a*tan(1/2*exp(1))^4+125829120*sqrt(2)*a*tan(1/2*exp(1))^3+251
65824*sqrt(2)*a*tan(1/2*exp(1))^2-25165824*sqrt(2)*a+75497472*sqrt(2)*a*tan(1/2*exp(1)))*tan(1/4*exp(1))^4+251
65824*sqrt(2)*a*tan(1/2*exp(1))^6+8388608*sqrt(2)*a*tan(1/2*exp(1))^5+41943040*sqrt(2)*a*tan(1/2*exp(1))^4+419
43040*sqrt(2)*a*tan(1/2*exp(1))^3+8388608*sqrt(2)*a*tan(1/2*exp(1))^2+25165824*sqrt(2)*a*tan(1/2*exp(1)))
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maple [B] time = 0.68, size = 399, normalized size = 4.16 \[ -\frac {\left (2 A \cos \left (f x +e \right ) \ln \left (\frac {1-\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-A \cos \left (f x +e \right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )-2 A \sin \left (f x +e \right ) \ln \left (\frac {1-\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+A \sin \left (f x +e \right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )+B \left (\cos ^{2}\left (f x +e \right )\right )+B \sin \left (f x +e \right ) \cos \left (f x +e \right )-2 B \cos \left (f x +e \right ) \ln \left (\frac {1-\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+B \cos \left (f x +e \right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )+2 B \sin \left (f x +e \right ) \ln \left (\frac {1-\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-B \sin \left (f x +e \right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )-2 A \ln \left (\frac {1-\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+A \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )-B \sin \left (f x +e \right )+2 B \ln \left (\frac {1-\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-B \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )-B \right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}}{f \left (-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )\right ) \sqrt {a \left (1+\sin \left (f x +e \right )\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e))^(1/2),x)
[Out]
-1/f*(2*A*cos(f*x+e)*ln((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-A*cos(f*x+e)*ln(2/(cos(f*x+e)+1))-2*A*sin(f*x+e)
*ln((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+A*sin(f*x+e)*ln(2/(cos(f*x+e)+1))+B*cos(f*x+e)^2+B*sin(f*x+e)*cos(f*
x+e)-2*B*cos(f*x+e)*ln((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+B*cos(f*x+e)*ln(2/(cos(f*x+e)+1))+2*B*sin(f*x+e)*
ln((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-B*sin(f*x+e)*ln(2/(cos(f*x+e)+1))-2*A*ln((1-cos(f*x+e)+sin(f*x+e))/si
n(f*x+e))+A*ln(2/(cos(f*x+e)+1))-B*sin(f*x+e)+2*B*ln((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-B*ln(2/(cos(f*x+e)+
1))-B)*(-c*(sin(f*x+e)-1))^(1/2)/(-1+cos(f*x+e)+sin(f*x+e))/(a*(1+sin(f*x+e)))^(1/2)
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maxima [A] time = 0.79, size = 176, normalized size = 1.83 \[ \frac {B {\left (\frac {2 \, \sqrt {c} \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{\sqrt {a}} - \frac {\sqrt {c} \log \left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}{\sqrt {a}} - \frac {2 \, \sqrt {a} \sqrt {c} \sin \left (f x + e\right )}{{\left (a + \frac {a \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (f x + e\right ) + 1\right )}}\right )} - A {\left (\frac {2 \, \sqrt {c} \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{\sqrt {a}} - \frac {\sqrt {c} \log \left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}{\sqrt {a}}\right )}}{f} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e))^(1/2),x, algorithm="maxima")
[Out]
(B*(2*sqrt(c)*log(sin(f*x + e)/(cos(f*x + e) + 1) + 1)/sqrt(a) - sqrt(c)*log(sin(f*x + e)^2/(cos(f*x + e) + 1)
^2 + 1)/sqrt(a) - 2*sqrt(a)*sqrt(c)*sin(f*x + e)/((a + a*sin(f*x + e)^2/(cos(f*x + e) + 1)^2)*(cos(f*x + e) +
1))) - A*(2*sqrt(c)*log(sin(f*x + e)/(cos(f*x + e) + 1) + 1)/sqrt(a) - sqrt(c)*log(sin(f*x + e)^2/(cos(f*x + e
) + 1)^2 + 1)/sqrt(a)))/f
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,\sqrt {c-c\,\sin \left (e+f\,x\right )}}{\sqrt {a+a\,\sin \left (e+f\,x\right )}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(((A + B*sin(e + f*x))*(c - c*sin(e + f*x))^(1/2))/(a + a*sin(e + f*x))^(1/2),x)
[Out]
int(((A + B*sin(e + f*x))*(c - c*sin(e + f*x))^(1/2))/(a + a*sin(e + f*x))^(1/2), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {- c \left (\sin {\left (e + f x \right )} - 1\right )} \left (A + B \sin {\left (e + f x \right )}\right )}{\sqrt {a \left (\sin {\left (e + f x \right )} + 1\right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))**(1/2)/(a+a*sin(f*x+e))**(1/2),x)
[Out]
Integral(sqrt(-c*(sin(e + f*x) - 1))*(A + B*sin(e + f*x))/sqrt(a*(sin(e + f*x) + 1)), x)
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