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

Optimal. Leaf size=145 \[ -\frac {2 a^2 (A+B) \cos (e+f x) \log (1-\sin (e+f x))}{f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {a (A+B) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{f \sqrt {c-c \sin (e+f x)}}-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 f \sqrt {c-c \sin (e+f x)}} \]

[Out]

-1/2*B*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)/f/(c-c*sin(f*x+e))^(1/2)-2*a^2*(A+B)*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*(A+B)*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 = 0.38, antiderivative size = 145, 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} \[ -\frac {2 a^2 (A+B) \cos (e+f x) \log (1-\sin (e+f x))}{f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {a (A+B) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{f \sqrt {c-c \sin (e+f x)}}-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 f \sqrt {c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.69, size = 136, normalized size = 0.94 \[ -\frac {(a (\sin (e+f x)+1))^{3/2} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (4 (A+2 B) \sin (e+f x)+16 (A+B) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-B \cos (2 (e+f x))\right )}{4 f \sqrt {c-c \sin (e+f x)} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(a*(1 + Sin[e + f*x]))^(3/2)*(-(B*Cos[2*(e + f*x)]) + 16*(A + B)*L
og[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] + 4*(A + 2*B)*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 = 2.82, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B a \cos \left (f x + e\right )^{2} - {\left (A + B\right )} a \sin \left (f x + e\right ) - {\left (A + B\right )} a\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{c \sin \left (f x + e\right ) - c}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*a*cos(f*x + e)^2 - (A + B)*a*sin(f*x + e) - (A + B)*a)*sqrt(a*sin(f*x + e) + a)*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)*(A+B*sin(f*x+e))/(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*a*sqrt(c*tan(1/2*exp(1))^2+c)*(-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*a*sqrt(c*tan(1/2*exp(1))^2+c)*(-27021597764222976*tan(1/2*exp(1))^5-8106479329266
8928*tan(1/2*exp(1))^4+90071992547409920*tan(1/2*exp(1))^3+54043195528445952*tan(1/2*exp(1))^2-270215977642229
76*tan(1/2*exp(1))-9007199254740992)+A*a*sqrt(c*tan(1/2*exp(1))^2+c)*(27021597764222976*tan(1/2*exp(1))^5+8106
4793292668928*tan(1/2*exp(1))^4-90071992547409920*tan(1/2*exp(1))^3-54043195528445952*tan(1/2*exp(1))^2+270215
97764222976*tan(1/2*exp(1))+9007199254740992)*tan(1/4*exp(1))^6+A*a*sqrt(c*tan(1/2*exp(1))^2+c)*(1801439850948
19840*tan(1/2*exp(1))^6+540431955284459520*tan(1/2*exp(1))^5-1080863910568919040*tan(1/2*exp(1))^4-18014398509
48198400*tan(1/2*exp(1))^3+1621295865853378560*tan(1/2*exp(1))^2+540431955284459520*tan(1/2*exp(1)))*tan(1/4*e
xp(1))^3+A*a*sqrt(c*tan(1/2*exp(1))^2+c)*(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*exp(1))^2+405323966463344640*tan(1/2*e
xp(1))+135107988821114880)*tan(1/4*exp(1))^2+A*a*sqrt(c*tan(1/2*exp(1))^2+c)*(-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+A*a*sqrt(c*ta
n(1/2*exp(1))^2+c)*(-54043195528445952*tan(1/2*exp(1))^6-162129586585337856*tan(1/2*exp(1))^5+3242591731706757
12*tan(1/2*exp(1))^4+540431955284459520*tan(1/2*exp(1))^3-486388759756013568*tan(1/2*exp(1))^2-162129586585337
856*tan(1/2*exp(1)))*tan(1/4*exp(1))+A*a*sqrt(c*tan(1/2*exp(1))^2+c)*(-405323966463344640*tan(1/2*exp(1))^5-12
15971899390033920*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+B*a*sqrt(c*tan(1/2*exp(1))^2+c)*(270
21597764222976*tan(1/2*exp(1))^5+81064793292668928*tan(1/2*exp(1))^4-90071992547409920*tan(1/2*exp(1))^3-54043
195528445952*tan(1/2*exp(1))^2+27021597764222976*tan(1/2*exp(1))+9007199254740992)*tan(1/4*exp(1))^6+B*a*sqrt(
c*tan(1/2*exp(1))^2+c)*(180143985094819840*tan(1/2*exp(1))^6+540431955284459520*tan(1/2*exp(1))^5-108086391056
8919040*tan(1/2*exp(1))^4-1801439850948198400*tan(1/2*exp(1))^3+1621295865853378560*tan(1/2*exp(1))^2+54043195
5284459520*tan(1/2*exp(1)))*tan(1/4*exp(1))^3+B*a*sqrt(c*tan(1/2*exp(1))^2+c)*(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*e
xp(1))^2+405323966463344640*tan(1/2*exp(1))+135107988821114880)*tan(1/4*exp(1))^2+B*a*sqrt(c*tan(1/2*exp(1))^2
+c)*(-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*a*sqrt(c*tan(1/2*exp(1))^2+c)*(-54043195528445952*tan(1/2*exp(1))^6-1621295865853378
56*tan(1/2*exp(1))^5+324259173170675712*tan(1/2*exp(1))^4+540431955284459520*tan(1/2*exp(1))^3-486388759756013
568*tan(1/2*exp(1))^2-162129586585337856*tan(1/2*exp(1)))*tan(1/4*exp(1))+B*a*sqrt(c*tan(1/2*exp(1))^2+c)*(-40
5323966463344640*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))))*
(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/(-9007199254740992*sqrt(2)
*c*tan(1/2*exp(1))^7+9007199254740992*sqrt(2)*c+(-9007199254740992*sqrt(2)*c*tan(1/2*exp(1))^7-270215977642229
76*sqrt(2)*c*tan(1/2*exp(1))^6+9007199254740992*sqrt(2)*c*tan(1/2*exp(1))^5-45035996273704960*sqrt(2)*c*tan(1/
2*exp(1))^4+45035996273704960*sqrt(2)*c*tan(1/2*exp(1))^3-9007199254740992*sqrt(2)*c*tan(1/2*exp(1))^2+9007199
254740992*sqrt(2)*c+27021597764222976*sqrt(2)*c*tan(1/2*exp(1)))*tan(1/4*exp(1))^6+(-27021597764222976*sqrt(2)
*c*tan(1/2*exp(1))^7-81064793292668928*sqrt(2)*c*tan(1/2*exp(1))^6+27021597764222976*sqrt(2)*c*tan(1/2*exp(1))
^5-135107988821114880*sqrt(2)*c*tan(1/2*exp(1))^4+135107988821114880*sqrt(2)*c*tan(1/2*exp(1))^3-2702159776422
2976*sqrt(2)*c*tan(1/2*exp(1))^2+27021597764222976*sqrt(2)*c+81064793292668928*sqrt(2)*c*tan(1/2*exp(1)))*tan(
1/4*exp(1))^2+(-27021597764222976*sqrt(2)*c*tan(1/2*exp(1))^7-81064793292668928*sqrt(2)*c*tan(1/2*exp(1))^6+27
021597764222976*sqrt(2)*c*tan(1/2*exp(1))^5-135107988821114880*sqrt(2)*c*tan(1/2*exp(1))^4+135107988821114880*
sqrt(2)*c*tan(1/2*exp(1))^3-27021597764222976*sqrt(2)*c*tan(1/2*exp(1))^2+27021597764222976*sqrt(2)*c+81064793
292668928*sqrt(2)*c*tan(1/2*exp(1)))*tan(1/4*exp(1))^4-27021597764222976*sqrt(2)*c*tan(1/2*exp(1))^6+900719925
4740992*sqrt(2)*c*tan(1/2*exp(1))^5-45035996273704960*sqrt(2)*c*tan(1/2*exp(1))^4+45035996273704960*sqrt(2)*c*
tan(1/2*exp(1))^3-9007199254740992*sqrt(2)*c*tan(1/2*exp(1))^2+27021597764222976*sqrt(2)*c*tan(1/2*exp(1)))

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maple [B]  time = 0.70, size = 494, normalized size = 3.41 \[ -\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 )-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 \cos \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 )+4 A \sin \left (f x +e \right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\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 )-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 \cos \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 )+4 B \sin \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 )+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 )-4 A \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )-B \cos \left (f x +e \right )+3 B \sin \left (f x +e \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 )-4 B \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )-2 A -3 B \right ) \left (a \left (1+\sin \left (f x +e \right )\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 {-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)*(A+B*sin(f*x+e))/(c-c*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)-8*A*cos(f*x+e)*ln(
-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+4*A*cos(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))+4*A*sin(f*x+e)*ln(2/(cos(f*x+e)+1))+3*B*cos(f*x+e)^2-4*B*sin(f*x+e)*cos(f*x+e)-8*B*cos
(f*x+e)*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+4*B*cos(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))+4*B*sin(f*x+e)*ln(2/(cos(f*x+e)+1))+2*A*sin(f*x+e)+8*A*ln(-(-1+cos(f*x+e)+s
in(f*x+e))/sin(f*x+e))-4*A*ln(2/(cos(f*x+e)+1))-B*cos(f*x+e)+3*B*sin(f*x+e)+8*B*ln(-(-1+cos(f*x+e)+sin(f*x+e))
/sin(f*x+e))-4*B*ln(2/(cos(f*x+e)+1))-2*A-3*B)*(a*(1+sin(f*x+e)))^(3/2)/(cos(f*x+e)^2+sin(f*x+e)*cos(f*x+e)+co
s(f*x+e)-2*sin(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 \[ \int \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}}}{\sqrt {-c \sin \left (f x + e\right ) + c}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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