3.384 \(\int \frac {(c-c \sin (e+f x))^{3/2}}{\sqrt {a+a \sin (e+f x)}} \, dx\)
Optimal. Leaf size=93 \[ \frac {2 c^2 \cos (e+f x) \log (\sin (e+f x)+1)}{f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {c \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}} \]
[Out]
2*c^2*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)+c*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 = 0.19, antiderivative size = 93, normalized size of antiderivative = 1.00,
number of steps used = 4, number of rules used = 4, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used =
{2740, 2737, 2667, 31} \[ \frac {2 c^2 \cos (e+f x) \log (\sin (e+f x)+1)}{f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {c \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}} \]
Antiderivative was successfully verified.
[In]
Int[(c - c*Sin[e + f*x])^(3/2)/Sqrt[a + a*Sin[e + f*x]],x]
[Out]
(2*c^2*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]]) + (c*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 2740
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Sim
p[(b*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n)/(f*(m + n)), x] + Dist[(a*(2*m - 1))/(m
+ n), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && E
qQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[m - 1/2, 0] && !LtQ[n, -1] && !(IGtQ[n - 1/2, 0] && LtQ[n, m])
&& !(ILtQ[m + n, 0] && GtQ[2*m + n + 1, 0])
Rubi steps
\begin {align*} \int \frac {(c-c \sin (e+f x))^{3/2}}{\sqrt {a+a \sin (e+f x)}} \, dx &=\frac {c \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a+a \sin (e+f x)}}+(2 c) \int \frac {\sqrt {c-c \sin (e+f x)}}{\sqrt {a+a \sin (e+f x)}} \, dx\\ &=\frac {c \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a+a \sin (e+f x)}}+\frac {\left (2 a c^2 \cos (e+f x)\right ) \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 {c \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a+a \sin (e+f x)}}+\frac {\left (2 c^2 \cos (e+f x)\right ) \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 {2 c^2 \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 {c \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 [A] time = 0.31, size = 119, normalized size = 1.28 \[ \frac {c (\sin (e+f x)-1) \sqrt {c-c \sin (e+f x)} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right ) \left (\sin (e+f x)-4 \log \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )\right )}{f \sqrt {a (\sin (e+f x)+1)} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3} \]
Antiderivative was successfully verified.
[In]
Integrate[(c - c*Sin[e + f*x])^(3/2)/Sqrt[a + a*Sin[e + f*x]],x]
[Out]
(c*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-1 + Sin[e + f*x])*(-4*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]] + Si
n[e + f*x])*Sqrt[c - c*Sin[e + f*x]])/(f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3*Sqrt[a*(1 + Sin[e + f*x])])
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fricas [F] time = 0.63, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}}{\sqrt {a \sin \left (f x + e\right ) + a}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c-c*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(1/2),x, algorithm="fricas")
[Out]
integral((-c*sin(f*x + e) + c)^(3/2)/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((c-c*sin(f*x+e))^(3/2)/(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)*(c*sqrt(a*tan(1/2*exp(1))^2+a)*(25165824*tan(1/2*exp(1))^5-75497472*tan(1/2*exp(1))^4-83886080
*tan(1/2*exp(1))^3+50331648*tan(1/2*exp(1))^2+25165824*tan(1/2*exp(1))-8388608)+c*sqrt(a*tan(1/2*exp(1))^2+a)*
(50331648*tan(1/2*exp(1))^6-150994944*tan(1/2*exp(1))^5-301989888*tan(1/2*exp(1))^4+503316480*tan(1/2*exp(1))^
3+452984832*tan(1/2*exp(1))^2-150994944*tan(1/2*exp(1)))*tan(1/4*exp(1))^5+c*sqrt(a*tan(1/2*exp(1))^2+a)*(5033
1648*tan(1/2*exp(1))^6-150994944*tan(1/2*exp(1))^5-301989888*tan(1/2*exp(1))^4+503316480*tan(1/2*exp(1))^3+452
984832*tan(1/2*exp(1))^2-150994944*tan(1/2*exp(1)))*tan(1/4*exp(1))+c*sqrt(a*tan(1/2*exp(1))^2+a)*(377487360*t
an(1/2*exp(1))^5-1132462080*tan(1/2*exp(1))^4-1258291200*tan(1/2*exp(1))^3+754974720*tan(1/2*exp(1))^2+3774873
60*tan(1/2*exp(1))-125829120)*tan(1/4*exp(1))^4+c*sqrt(a*tan(1/2*exp(1))^2+a)*(-25165824*tan(1/2*exp(1))^5+754
97472*tan(1/2*exp(1))^4+83886080*tan(1/2*exp(1))^3-50331648*tan(1/2*exp(1))^2-25165824*tan(1/2*exp(1))+8388608
)*tan(1/4*exp(1))^6+c*sqrt(a*tan(1/2*exp(1))^2+a)*(-167772160*tan(1/2*exp(1))^6+503316480*tan(1/2*exp(1))^5+10
06632960*tan(1/2*exp(1))^4-1677721600*tan(1/2*exp(1))^3-1509949440*tan(1/2*exp(1))^2+503316480*tan(1/2*exp(1))
)*tan(1/4*exp(1))^3+c*sqrt(a*tan(1/2*exp(1))^2+a)*(-377487360*tan(1/2*exp(1))^5+1132462080*tan(1/2*exp(1))^4+1
258291200*tan(1/2*exp(1))^3-754974720*tan(1/2*exp(1))^2-377487360*tan(1/2*exp(1))+125829120)*tan(1/4*exp(1))^2
)*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+838860
8*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+838860
8*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+(-251658
24*sqrt(2)*a*tan(1/2*exp(1))^7+75497472*sqrt(2)*a*tan(1/2*exp(1))^6+25165824*sqrt(2)*a*tan(1/2*exp(1))^5+12582
9120*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))^2-25
165824*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+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))^4+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+25165824*sqrt(2)*a*tan(1/2*exp(1)))
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maple [B] time = 0.27, size = 261, normalized size = 2.81 \[ -\frac {\left (\sin \left (f x +e \right ) \cos \left (f x +e \right )+4 \sin \left (f x +e \right ) \ln \left (-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-2 \sin \left (f x +e \right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )+\cos ^{2}\left (f x +e \right )-4 \cos \left (f x +e \right ) \ln \left (-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+2 \cos \left (f x +e \right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )-\sin \left (f x +e \right )+4 \ln \left (-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-2 \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )-1\right ) \left (-c \left (\sin \left (f x +e \right )-1\right )\right )^{\frac {3}{2}}}{f \left (\sin \left (f x +e \right ) \cos \left (f x +e \right )-\left (\cos ^{2}\left (f x +e \right )\right )-2 \sin \left (f x +e \right )-\cos \left (f x +e \right )+2\right ) \sqrt {a \left (1+\sin \left (f x +e \right )\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((c-c*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(1/2),x)
[Out]
-1/f*(sin(f*x+e)*cos(f*x+e)+4*sin(f*x+e)*ln(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))-2*sin(f*x+e)*ln(2/(cos(f*x
+e)+1))+cos(f*x+e)^2-4*cos(f*x+e)*ln(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))+2*cos(f*x+e)*ln(2/(cos(f*x+e)+1))
-sin(f*x+e)+4*ln(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))-2*ln(2/(cos(f*x+e)+1))-1)*(-c*(sin(f*x+e)-1))^(3/2)/(
sin(f*x+e)*cos(f*x+e)-cos(f*x+e)^2-2*sin(f*x+e)-cos(f*x+e)+2)/(a*(1+sin(f*x+e)))^(1/2)
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}}{\sqrt {a \sin \left (f x + e\right ) + a}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c-c*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(1/2),x, algorithm="maxima")
[Out]
integrate((-c*sin(f*x + e) + c)^(3/2)/sqrt(a*sin(f*x + e) + a), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{3/2}}{\sqrt {a+a\,\sin \left (e+f\,x\right )}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((c - c*sin(e + f*x))^(3/2)/(a + a*sin(e + f*x))^(1/2),x)
[Out]
int((c - c*sin(e + f*x))^(3/2)/(a + a*sin(e + f*x))^(1/2), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (- c \left (\sin {\left (e + f x \right )} - 1\right )\right )^{\frac {3}{2}}}{\sqrt {a \left (\sin {\left (e + f x \right )} + 1\right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c-c*sin(f*x+e))**(3/2)/(a+a*sin(f*x+e))**(1/2),x)
[Out]
Integral((-c*(sin(e + f*x) - 1))**(3/2)/sqrt(a*(sin(e + f*x) + 1)), x)
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