3.175 \(\int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2}}{\sqrt {a+a \sin (e+f x)}} \, dx\)

Optimal. Leaf size=146 \[ \frac {2 c^2 (A-B) \cos (e+f x) \log (\sin (e+f x)+1)}{f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {c (A-B) \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}-\frac {B \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}} \]

[Out]

-1/2*B*cos(f*x+e)*(c-c*sin(f*x+e))^(3/2)/f/(a+a*sin(f*x+e))^(1/2)+2*(A-B)*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)+(A-B)*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 = 0.37, antiderivative size = 146, 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 = {2973, 2740, 2737, 2667, 31} \[ \frac {2 c^2 (A-B) \cos (e+f x) \log (\sin (e+f x)+1)}{f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {c (A-B) \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}-\frac {B \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(3/2))/Sqrt[a + a*Sin[e + f*x]],x]

[Out]

(2*(A - B)*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]]) + ((A
- B)*c*Cos[e + f*x]*Sqrt[c - c*Sin[e + f*x]])/(f*Sqrt[a + a*Sin[e + f*x]]) - (B*Cos[e + f*x]*(c - c*Sin[e + f
*x])^(3/2))/(2*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])

Rule 2973

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(B*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n)/(f*(
m + n + 1)), x] - Dist[(B*c*(m - n) - A*d*(m + n + 1))/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[
e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] &&
!LtQ[m, -2^(-1)] && NeQ[m + n + 1, 0]

Rubi steps

\begin {align*} \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2}}{\sqrt {a+a \sin (e+f x)}} \, dx &=-\frac {B \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f \sqrt {a+a \sin (e+f x)}}+(A-B) \int \frac {(c-c \sin (e+f x))^{3/2}}{\sqrt {a+a \sin (e+f x)}} \, dx\\ &=\frac {(A-B) c \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a+a \sin (e+f x)}}-\frac {B \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f \sqrt {a+a \sin (e+f x)}}+(2 (A-B) c) \int \frac {\sqrt {c-c \sin (e+f x)}}{\sqrt {a+a \sin (e+f x)}} \, dx\\ &=\frac {(A-B) c \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a+a \sin (e+f x)}}-\frac {B \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f \sqrt {a+a \sin (e+f x)}}+\frac {\left (2 a (A-B) 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 {(A-B) c \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a+a \sin (e+f x)}}-\frac {B \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f \sqrt {a+a \sin (e+f x)}}+\frac {\left (2 (A-B) 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 (A-B) 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 {(A-B) c \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a+a \sin (e+f x)}}-\frac {B \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f \sqrt {a+a \sin (e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 0.66, size = 146, normalized size = 1.00 \[ -\frac {c (\sin (e+f x)-1) \sqrt {c-c \sin (e+f x)} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right ) \left (B \cos (2 (e+f x))-4 \left ((A-2 B) \sin (e+f x)+4 (B-A) \log \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )\right )\right )}{4 f \sqrt {a (\sin (e+f x)+1)} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(3/2))/Sqrt[a + a*Sin[e + f*x]],x]

[Out]

-1/4*(c*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-1 + Sin[e + f*x])*Sqrt[c - c*Sin[e + f*x]]*(B*Cos[2*(e + f*x)]
- 4*(4*(-A + B)*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]] + (A - 2*B)*Sin[e + f*x])))/(f*(Cos[(e + f*x)/2] - S
in[(e + f*x)/2])^3*Sqrt[a*(1 + Sin[e + f*x])])

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fricas [F]  time = 2.82, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B c \cos \left (f x + e\right )^{2} - {\left (A - B\right )} c \sin \left (f x + e\right ) + {\left (A - B\right )} c\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{\sqrt {a \sin \left (f x + e\right ) + a}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

integral((B*c*cos(f*x + e)^2 - (A - B)*c*sin(f*x + e) + (A - B)*c)*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))^(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)*(A*c*sqrt(a*tan(1/2*exp(1))^2+a)*(27021597764222976*tan(1/2*exp(1))^5-81064793292668928*tan(1/
2*exp(1))^4-90071992547409920*tan(1/2*exp(1))^3+54043195528445952*tan(1/2*exp(1))^2+27021597764222976*tan(1/2*
exp(1))-9007199254740992)+B*c*sqrt(a*tan(1/2*exp(1))^2+a)*(-27021597764222976*tan(1/2*exp(1))^5+81064793292668
928*tan(1/2*exp(1))^4+90071992547409920*tan(1/2*exp(1))^3-54043195528445952*tan(1/2*exp(1))^2-2702159776422297
6*tan(1/2*exp(1))+9007199254740992)+A*c*sqrt(a*tan(1/2*exp(1))^2+a)*(54043195528445952*tan(1/2*exp(1))^6-16212
9586585337856*tan(1/2*exp(1))^5-324259173170675712*tan(1/2*exp(1))^4+540431955284459520*tan(1/2*exp(1))^3+4863
88759756013568*tan(1/2*exp(1))^2-162129586585337856*tan(1/2*exp(1)))*tan(1/4*exp(1))^5+A*c*sqrt(a*tan(1/2*exp(
1))^2+a)*(54043195528445952*tan(1/2*exp(1))^6-162129586585337856*tan(1/2*exp(1))^5-324259173170675712*tan(1/2*
exp(1))^4+540431955284459520*tan(1/2*exp(1))^3+486388759756013568*tan(1/2*exp(1))^2-162129586585337856*tan(1/2
*exp(1)))*tan(1/4*exp(1))+A*c*sqrt(a*tan(1/2*exp(1))^2+a)*(405323966463344640*tan(1/2*exp(1))^5-12159718993900
33920*tan(1/2*exp(1))^4-1351079888211148800*tan(1/2*exp(1))^3+810647932926689280*tan(1/2*exp(1))^2+40532396646
3344640*tan(1/2*exp(1))-135107988821114880)*tan(1/4*exp(1))^4+A*c*sqrt(a*tan(1/2*exp(1))^2+a)*(-27021597764222
976*tan(1/2*exp(1))^5+81064793292668928*tan(1/2*exp(1))^4+90071992547409920*tan(1/2*exp(1))^3-5404319552844595
2*tan(1/2*exp(1))^2-27021597764222976*tan(1/2*exp(1))+9007199254740992)*tan(1/4*exp(1))^6+A*c*sqrt(a*tan(1/2*e
xp(1))^2+a)*(-180143985094819840*tan(1/2*exp(1))^6+540431955284459520*tan(1/2*exp(1))^5+1080863910568919040*ta
n(1/2*exp(1))^4-1801439850948198400*tan(1/2*exp(1))^3-1621295865853378560*tan(1/2*exp(1))^2+540431955284459520
*tan(1/2*exp(1)))*tan(1/4*exp(1))^3+A*c*sqrt(a*tan(1/2*exp(1))^2+a)*(-405323966463344640*tan(1/2*exp(1))^5+121
5971899390033920*tan(1/2*exp(1))^4+1351079888211148800*tan(1/2*exp(1))^3-810647932926689280*tan(1/2*exp(1))^2-
405323966463344640*tan(1/2*exp(1))+135107988821114880)*tan(1/4*exp(1))^2+B*c*sqrt(a*tan(1/2*exp(1))^2+a)*(2702
1597764222976*tan(1/2*exp(1))^5-81064793292668928*tan(1/2*exp(1))^4-90071992547409920*tan(1/2*exp(1))^3+540431
95528445952*tan(1/2*exp(1))^2+27021597764222976*tan(1/2*exp(1))-9007199254740992)*tan(1/4*exp(1))^6+B*c*sqrt(a
*tan(1/2*exp(1))^2+a)*(180143985094819840*tan(1/2*exp(1))^6-540431955284459520*tan(1/2*exp(1))^5-1080863910568
919040*tan(1/2*exp(1))^4+1801439850948198400*tan(1/2*exp(1))^3+1621295865853378560*tan(1/2*exp(1))^2-540431955
284459520*tan(1/2*exp(1)))*tan(1/4*exp(1))^3+B*c*sqrt(a*tan(1/2*exp(1))^2+a)*(405323966463344640*tan(1/2*exp(1
))^5-1215971899390033920*tan(1/2*exp(1))^4-1351079888211148800*tan(1/2*exp(1))^3+810647932926689280*tan(1/2*ex
p(1))^2+405323966463344640*tan(1/2*exp(1))-135107988821114880)*tan(1/4*exp(1))^2+B*c*sqrt(a*tan(1/2*exp(1))^2+
a)*(-54043195528445952*tan(1/2*exp(1))^6+162129586585337856*tan(1/2*exp(1))^5+324259173170675712*tan(1/2*exp(1
))^4-540431955284459520*tan(1/2*exp(1))^3-486388759756013568*tan(1/2*exp(1))^2+162129586585337856*tan(1/2*exp(
1)))*tan(1/4*exp(1))^5+B*c*sqrt(a*tan(1/2*exp(1))^2+a)*(-54043195528445952*tan(1/2*exp(1))^6+16212958658533785
6*tan(1/2*exp(1))^5+324259173170675712*tan(1/2*exp(1))^4-540431955284459520*tan(1/2*exp(1))^3-4863887597560135
68*tan(1/2*exp(1))^2+162129586585337856*tan(1/2*exp(1)))*tan(1/4*exp(1))+B*c*sqrt(a*tan(1/2*exp(1))^2+a)*(-405
323966463344640*tan(1/2*exp(1))^5+1215971899390033920*tan(1/2*exp(1))^4+1351079888211148800*tan(1/2*exp(1))^3-
810647932926689280*tan(1/2*exp(1))^2-405323966463344640*tan(1/2*exp(1))+135107988821114880)*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))))*(t
an(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/(-9007199254740992*sqrt(2)*a
*tan(1/2*exp(1))^7-9007199254740992*sqrt(2)*a+(-9007199254740992*sqrt(2)*a*tan(1/2*exp(1))^7+27021597764222976
*sqrt(2)*a*tan(1/2*exp(1))^6+9007199254740992*sqrt(2)*a*tan(1/2*exp(1))^5+45035996273704960*sqrt(2)*a*tan(1/2*
exp(1))^4+45035996273704960*sqrt(2)*a*tan(1/2*exp(1))^3+9007199254740992*sqrt(2)*a*tan(1/2*exp(1))^2-900719925
4740992*sqrt(2)*a+27021597764222976*sqrt(2)*a*tan(1/2*exp(1)))*tan(1/4*exp(1))^6+(-27021597764222976*sqrt(2)*a
*tan(1/2*exp(1))^7+81064793292668928*sqrt(2)*a*tan(1/2*exp(1))^6+27021597764222976*sqrt(2)*a*tan(1/2*exp(1))^5
+135107988821114880*sqrt(2)*a*tan(1/2*exp(1))^4+135107988821114880*sqrt(2)*a*tan(1/2*exp(1))^3+270215977642229
76*sqrt(2)*a*tan(1/2*exp(1))^2-27021597764222976*sqrt(2)*a+81064793292668928*sqrt(2)*a*tan(1/2*exp(1)))*tan(1/
4*exp(1))^2+(-27021597764222976*sqrt(2)*a*tan(1/2*exp(1))^7+81064793292668928*sqrt(2)*a*tan(1/2*exp(1))^6+2702
1597764222976*sqrt(2)*a*tan(1/2*exp(1))^5+135107988821114880*sqrt(2)*a*tan(1/2*exp(1))^4+135107988821114880*sq
rt(2)*a*tan(1/2*exp(1))^3+27021597764222976*sqrt(2)*a*tan(1/2*exp(1))^2-27021597764222976*sqrt(2)*a+8106479329
2668928*sqrt(2)*a*tan(1/2*exp(1)))*tan(1/4*exp(1))^4+27021597764222976*sqrt(2)*a*tan(1/2*exp(1))^6+90071992547
40992*sqrt(2)*a*tan(1/2*exp(1))^5+45035996273704960*sqrt(2)*a*tan(1/2*exp(1))^4+45035996273704960*sqrt(2)*a*ta
n(1/2*exp(1))^3+9007199254740992*sqrt(2)*a*tan(1/2*exp(1))^2+27021597764222976*sqrt(2)*a*tan(1/2*exp(1)))

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maple [B]  time = 0.68, size = 501, normalized size = 3.43 \[ \frac {\left (-B \left (\cos ^{3}\left (f x +e \right )\right )+B \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+2 A \left (\cos ^{2}\left (f x +e \right )\right )+2 A \sin \left (f x +e \right ) \cos \left (f x +e \right )+4 A \cos \left (f x +e \right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )-8 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 )-4 A \sin \left (f x +e \right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )+8 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 )-3 B \left (\cos ^{2}\left (f x +e \right )\right )-4 B \sin \left (f x +e \right ) \cos \left (f x +e \right )-4 B \cos \left (f x +e \right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )+8 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 )+4 B \sin \left (f x +e \right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )-8 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 )-2 A \sin \left (f x +e \right )-4 A \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )+8 A \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 )+3 B \sin \left (f x +e \right )+4 B \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )-8 B \ln \left (\frac {1-\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-2 A +3 B \right ) \left (-c \left (\sin \left (f x +e \right )-1\right )\right )^{\frac {3}{2}}}{2 f \left (\cos ^{2}\left (f x +e \right )-\sin \left (f x +e \right ) \cos \left (f x +e \right )+\cos \left (f x +e \right )+2 \sin \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((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(1/2),x)

[Out]

1/2/f*(-B*cos(f*x+e)^3+B*cos(f*x+e)^2*sin(f*x+e)+2*A*cos(f*x+e)^2+2*A*sin(f*x+e)*cos(f*x+e)+4*A*cos(f*x+e)*ln(
2/(cos(f*x+e)+1))-8*A*cos(f*x+e)*ln((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-4*A*sin(f*x+e)*ln(2/(cos(f*x+e)+1))+
8*A*sin(f*x+e)*ln((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-3*B*cos(f*x+e)^2-4*B*sin(f*x+e)*cos(f*x+e)-4*B*cos(f*x
+e)*ln(2/(cos(f*x+e)+1))+8*B*cos(f*x+e)*ln((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+4*B*sin(f*x+e)*ln(2/(cos(f*x+
e)+1))-8*B*sin(f*x+e)*ln((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-2*A*sin(f*x+e)-4*A*ln(2/(cos(f*x+e)+1))+8*A*ln(
(1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+B*cos(f*x+e)+3*B*sin(f*x+e)+4*B*ln(2/(cos(f*x+e)+1))-8*B*ln((1-cos(f*x+e
)+sin(f*x+e))/sin(f*x+e))-2*A+3*B)*(-c*(sin(f*x+e)-1))^(3/2)/(cos(f*x+e)^2-sin(f*x+e)*cos(f*x+e)+cos(f*x+e)+2*
sin(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 \[ \int \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}}{\sqrt {a \sin \left (f x + e\right ) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*(-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 \[ \int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{3/2}}{\sqrt {a+a\,\sin \left (e+f\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*sin(e + f*x))*(c - c*sin(e + f*x))^(3/2))/(a + a*sin(e + f*x))^(1/2),x)

[Out]

int(((A + B*sin(e + f*x))*(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 \[ \int \frac {\left (- c \left (\sin {\left (e + f x \right )} - 1\right )\right )^{\frac {3}{2}} \left (A + B \sin {\left (e + f x \right )}\right )}{\sqrt {a \left (\sin {\left (e + f x \right )} + 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))**(3/2)/(a+a*sin(f*x+e))**(1/2),x)

[Out]

Integral((-c*(sin(e + f*x) - 1))**(3/2)*(A + B*sin(e + f*x))/sqrt(a*(sin(e + f*x) + 1)), x)

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