3.352 \(\int \frac {(a+a \sin (e+f x))^{3/2}}{\sqrt {c-c \sin (e+f x)}} \, dx\)
Optimal. Leaf size=96 \[ -\frac {2 a^2 \cos (e+f x) \log (1-\sin (e+f x))}{f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{f \sqrt {c-c \sin (e+f x)}} \]
[Out]
-2*a^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)-a*cos(f*x+e)*(a+a*sin(f*x+e
))^(1/2)/f/(c-c*sin(f*x+e))^(1/2)
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Rubi [A] time = 0.19, antiderivative size = 96, 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} \[ -\frac {2 a^2 \cos (e+f x) \log (1-\sin (e+f x))}{f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{f \sqrt {c-c \sin (e+f x)}} \]
Antiderivative was successfully verified.
[In]
Int[(a + a*Sin[e + f*x])^(3/2)/Sqrt[c - c*Sin[e + f*x]],x]
[Out]
(-2*a^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]]) - (a*Cos[e +
f*x]*Sqrt[a + a*Sin[e + f*x]])/(f*Sqrt[c - c*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 {(a+a \sin (e+f x))^{3/2}}{\sqrt {c-c \sin (e+f x)}} \, dx &=-\frac {a \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{f \sqrt {c-c \sin (e+f x)}}+(2 a) \int \frac {\sqrt {a+a \sin (e+f x)}}{\sqrt {c-c \sin (e+f x)}} \, dx\\ &=-\frac {a \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{f \sqrt {c-c \sin (e+f x)}}+\frac {\left (2 a^2 c \cos (e+f x)\right ) \int \frac {\cos (e+f x)}{c-c \sin (e+f x)} \, dx}{\sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}\\ &=-\frac {a \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{f \sqrt {c-c \sin (e+f x)}}-\frac {\left (2 a^2 \cos (e+f x)\right ) \operatorname {Subst}\left (\int \frac {1}{c+x} \, dx,x,-c \sin (e+f x)\right )}{f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}\\ &=-\frac {2 a^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 {a \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{f \sqrt {c-c \sin (e+f x)}}\\ \end {align*}
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Mathematica [A] time = 0.34, size = 113, normalized size = 1.18 \[ -\frac {(a (\sin (e+f x)+1))^{3/2} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\sin (e+f x)+4 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )\right )}{f \sqrt {c-c \sin (e+f x)} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^3} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + a*Sin[e + f*x])^(3/2)/Sqrt[c - c*Sin[e + f*x]],x]
[Out]
-(((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(a*(1 + Sin[e + f*x]))^(3/2)*(4*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/
2]] + Sin[e + f*x]))/(f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3*Sqrt[c - c*Sin[e + f*x]]))
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fricas [F] time = 0.64, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}} \sqrt {-c \sin \left (f x + e\right ) + c}}{c \sin \left (f x + e\right ) - c}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(1/2),x, algorithm="fricas")
[Out]
integral(-(a*sin(f*x + e) + a)^(3/2)*sqrt(-c*sin(f*x + e) + c)/(c*sin(f*x + e) - c), 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+a*sin(f*x+e))^(3/2)/(c-c*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*a)*(a*sqrt(c*tan(1/2*exp(1))^2+c)*(-25165824*tan(1/2*exp(1))^5-75497472*tan(1/2*exp(1))^4+8388608
0*tan(1/2*exp(1))^3+50331648*tan(1/2*exp(1))^2-25165824*tan(1/2*exp(1))-8388608)+a*sqrt(c*tan(1/2*exp(1))^2+c)
*(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)*tan(1/4*exp(1))^6+a*sqrt(c*tan(1/2*exp(1))^2+c)*(167772160*tan(1/2*exp(1))^6
+503316480*tan(1/2*exp(1))^5-1006632960*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+a*sqrt(c*tan(1/2*exp(1))^2+c)*(377487360*tan(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+377487360*tan(1/2*exp(1)
)+125829120)*tan(1/4*exp(1))^2+a*sqrt(c*tan(1/2*exp(1))^2+c)*(-50331648*tan(1/2*exp(1))^6-150994944*tan(1/2*ex
p(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+a*sqrt(c*tan(1/2*exp(1))^2+c)*(-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))+a*sqrt(c*tan(1/2*exp(1))^2+c)*(-377487360*tan(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))^2-377487360*tan(1/2*exp(1))-125829120)*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)*c*tan(1
/2*exp(1))^7+8388608*sqrt(2)*c+(-8388608*sqrt(2)*c*tan(1/2*exp(1))^7-25165824*sqrt(2)*c*tan(1/2*exp(1))^6+8388
608*sqrt(2)*c*tan(1/2*exp(1))^5-41943040*sqrt(2)*c*tan(1/2*exp(1))^4+41943040*sqrt(2)*c*tan(1/2*exp(1))^3-8388
608*sqrt(2)*c*tan(1/2*exp(1))^2+8388608*sqrt(2)*c+25165824*sqrt(2)*c*tan(1/2*exp(1)))*tan(1/4*exp(1))^6+(-2516
5824*sqrt(2)*c*tan(1/2*exp(1))^7-75497472*sqrt(2)*c*tan(1/2*exp(1))^6+25165824*sqrt(2)*c*tan(1/2*exp(1))^5-125
829120*sqrt(2)*c*tan(1/2*exp(1))^4+125829120*sqrt(2)*c*tan(1/2*exp(1))^3-25165824*sqrt(2)*c*tan(1/2*exp(1))^2+
25165824*sqrt(2)*c+75497472*sqrt(2)*c*tan(1/2*exp(1)))*tan(1/4*exp(1))^2+(-25165824*sqrt(2)*c*tan(1/2*exp(1))^
7-75497472*sqrt(2)*c*tan(1/2*exp(1))^6+25165824*sqrt(2)*c*tan(1/2*exp(1))^5-125829120*sqrt(2)*c*tan(1/2*exp(1)
)^4+125829120*sqrt(2)*c*tan(1/2*exp(1))^3-25165824*sqrt(2)*c*tan(1/2*exp(1))^2+25165824*sqrt(2)*c+75497472*sqr
t(2)*c*tan(1/2*exp(1)))*tan(1/4*exp(1))^4-25165824*sqrt(2)*c*tan(1/2*exp(1))^6+8388608*sqrt(2)*c*tan(1/2*exp(1
))^5-41943040*sqrt(2)*c*tan(1/2*exp(1))^4+41943040*sqrt(2)*c*tan(1/2*exp(1))^3-8388608*sqrt(2)*c*tan(1/2*exp(1
))^2+25165824*sqrt(2)*c*tan(1/2*exp(1)))
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maple [B] time = 0.27, size = 252, normalized size = 2.62 \[ \frac {\left (\sin \left (f x +e \right ) \cos \left (f x +e \right )-2 \sin \left (f x +e \right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )+4 \sin \left (f x +e \right ) \ln \left (-\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-\left (\cos ^{2}\left (f x +e \right )\right )-2 \cos \left (f x +e \right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )+4 \cos \left (f x +e \right ) \ln \left (-\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-\sin \left (f x +e \right )+2 \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )-4 \ln \left (-\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+1\right ) \left (a \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}}}{f \left (\sin \left (f x +e \right ) \cos \left (f x +e \right )+\cos ^{2}\left (f x +e \right )-2 \sin \left (f x +e \right )+\cos \left (f x +e \right )-2\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(1/2),x)
[Out]
1/f*(sin(f*x+e)*cos(f*x+e)-2*sin(f*x+e)*ln(2/(cos(f*x+e)+1))+4*sin(f*x+e)*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f
*x+e))-cos(f*x+e)^2-2*cos(f*x+e)*ln(2/(cos(f*x+e)+1))+4*cos(f*x+e)*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-
sin(f*x+e)+2*ln(2/(cos(f*x+e)+1))-4*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+1)*(a*(1+sin(f*x+e)))^(3/2)/(si
n(f*x+e)*cos(f*x+e)+cos(f*x+e)^2-2*sin(f*x+e)+cos(f*x+e)-2)/(-c*(sin(f*x+e)-1))^(1/2)
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}}}{\sqrt {-c \sin \left (f x + e\right ) + c}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(1/2),x, algorithm="maxima")
[Out]
integrate((a*sin(f*x + e) + a)^(3/2)/sqrt(-c*sin(f*x + e) + c), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{3/2}}{\sqrt {c-c\,\sin \left (e+f\,x\right )}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a + a*sin(e + f*x))^(3/2)/(c - c*sin(e + f*x))^(1/2),x)
[Out]
int((a + a*sin(e + f*x))^(3/2)/(c - c*sin(e + f*x))^(1/2), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}}}{\sqrt {- c \left (\sin {\left (e + f x \right )} - 1\right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))**(3/2)/(c-c*sin(f*x+e))**(1/2),x)
[Out]
Integral((a*(sin(e + f*x) + 1))**(3/2)/sqrt(-c*(sin(e + f*x) - 1)), x)
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