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

Optimal. Leaf size=224 \[ \frac {(7 A+3 B) \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{16 \sqrt {2} a^3 c^{3/2} f}-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{3/2}}{5 a^3 c^3 f}-\frac {(7 A+3 B) \sec ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{30 a^3 c^2 f}+\frac {(7 A+3 B) \cos (e+f x)}{16 a^3 f (c-c \sin (e+f x))^{3/2}}-\frac {(7 A+3 B) \sec (e+f x)}{12 a^3 c f \sqrt {c-c \sin (e+f x)}} \]

[Out]

1/16*(7*A+3*B)*cos(f*x+e)/a^3/f/(c-c*sin(f*x+e))^(3/2)-1/5*(A-B)*sec(f*x+e)^5*(c-c*sin(f*x+e))^(3/2)/a^3/c^3/f
+1/32*(7*A+3*B)*arctanh(1/2*cos(f*x+e)*c^(1/2)*2^(1/2)/(c-c*sin(f*x+e))^(1/2))/a^3/c^(3/2)/f*2^(1/2)-1/12*(7*A
+3*B)*sec(f*x+e)/a^3/c/f/(c-c*sin(f*x+e))^(1/2)-1/30*(7*A+3*B)*sec(f*x+e)^3*(c-c*sin(f*x+e))^(1/2)/a^3/c^2/f

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Rubi [A]  time = 0.48, antiderivative size = 224, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.184, Rules used = {2967, 2855, 2675, 2687, 2650, 2649, 206} \[ -\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{3/2}}{5 a^3 c^3 f}-\frac {(7 A+3 B) \sec ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{30 a^3 c^2 f}+\frac {(7 A+3 B) \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{16 \sqrt {2} a^3 c^{3/2} f}+\frac {(7 A+3 B) \cos (e+f x)}{16 a^3 f (c-c \sin (e+f x))^{3/2}}-\frac {(7 A+3 B) \sec (e+f x)}{12 a^3 c f \sqrt {c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((7*A + 3*B)*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(16*Sqrt[2]*a^3*c^(3/2)*f) +
((7*A + 3*B)*Cos[e + f*x])/(16*a^3*f*(c - c*Sin[e + f*x])^(3/2)) - ((7*A + 3*B)*Sec[e + f*x])/(12*a^3*c*f*Sqrt
[c - c*Sin[e + f*x]]) - ((7*A + 3*B)*Sec[e + f*x]^3*Sqrt[c - c*Sin[e + f*x]])/(30*a^3*c^2*f) - ((A - B)*Sec[e
+ f*x]^5*(c - c*Sin[e + f*x])^(3/2))/(5*a^3*c^3*f)

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2649

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, (b*C
os[c + d*x])/Sqrt[a + b*Sin[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2650

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^n)/(a*
d*(2*n + 1)), x] + Dist[(n + 1)/(a*(2*n + 1)), Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d},
x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

Rule 2675

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(b*(g
*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*(p + 1)), x] + Dist[(a*(m + p + 1))/(g^2*(p + 1)), Int[(
g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0]
&& GtQ[m, 0] && LeQ[p, -2*m] && IntegersQ[m + 1/2, 2*p]

Rule 2687

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> -Simp[(b*(g*
Cos[e + f*x])^(p + 1))/(a*f*g*(p + 1)*Sqrt[a + b*Sin[e + f*x]]), x] + Dist[(a*(2*p + 1))/(2*g^2*(p + 1)), Int[
(g*Cos[e + f*x])^(p + 2)/(a + b*Sin[e + f*x])^(3/2), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0]
&& LtQ[p, -1] && IntegerQ[2*p]

Rule 2855

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> -Simp[((b*c + a*d)*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*(p +
1)), x] + Dist[(b*(a*d*m + b*c*(m + p + 1)))/(a*g^2*(p + 1)), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x]
)^(m - 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, -1] && LtQ[p, -1]

Rule 2967

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a^m*c^m, Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m)*(A + B
*Sin[e + f*x]), 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] && I
ntegerQ[m] &&  !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))

Rubi steps

\begin {align*} \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^{3/2}} \, dx &=\frac {\int \sec ^6(e+f x) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx}{a^3 c^3}\\ &=-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{3/2}}{5 a^3 c^3 f}+\frac {(7 A+3 B) \int \sec ^4(e+f x) \sqrt {c-c \sin (e+f x)} \, dx}{10 a^3 c^2}\\ &=-\frac {(7 A+3 B) \sec ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{30 a^3 c^2 f}-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{3/2}}{5 a^3 c^3 f}+\frac {(7 A+3 B) \int \frac {\sec ^2(e+f x)}{\sqrt {c-c \sin (e+f x)}} \, dx}{12 a^3 c}\\ &=-\frac {(7 A+3 B) \sec (e+f x)}{12 a^3 c f \sqrt {c-c \sin (e+f x)}}-\frac {(7 A+3 B) \sec ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{30 a^3 c^2 f}-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{3/2}}{5 a^3 c^3 f}+\frac {(7 A+3 B) \int \frac {1}{(c-c \sin (e+f x))^{3/2}} \, dx}{8 a^3}\\ &=\frac {(7 A+3 B) \cos (e+f x)}{16 a^3 f (c-c \sin (e+f x))^{3/2}}-\frac {(7 A+3 B) \sec (e+f x)}{12 a^3 c f \sqrt {c-c \sin (e+f x)}}-\frac {(7 A+3 B) \sec ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{30 a^3 c^2 f}-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{3/2}}{5 a^3 c^3 f}+\frac {(7 A+3 B) \int \frac {1}{\sqrt {c-c \sin (e+f x)}} \, dx}{32 a^3 c}\\ &=\frac {(7 A+3 B) \cos (e+f x)}{16 a^3 f (c-c \sin (e+f x))^{3/2}}-\frac {(7 A+3 B) \sec (e+f x)}{12 a^3 c f \sqrt {c-c \sin (e+f x)}}-\frac {(7 A+3 B) \sec ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{30 a^3 c^2 f}-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{3/2}}{5 a^3 c^3 f}-\frac {(7 A+3 B) \operatorname {Subst}\left (\int \frac {1}{2 c-x^2} \, dx,x,-\frac {c \cos (e+f x)}{\sqrt {c-c \sin (e+f x)}}\right )}{16 a^3 c f}\\ &=\frac {(7 A+3 B) \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{16 \sqrt {2} a^3 c^{3/2} f}+\frac {(7 A+3 B) \cos (e+f x)}{16 a^3 f (c-c \sin (e+f x))^{3/2}}-\frac {(7 A+3 B) \sec (e+f x)}{12 a^3 c f \sqrt {c-c \sin (e+f x)}}-\frac {(7 A+3 B) \sec ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{30 a^3 c^2 f}-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{3/2}}{5 a^3 c^3 f}\\ \end {align*}

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Mathematica [C]  time = 1.36, size = 357, normalized size = 1.59 \[ \frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right ) \left (15 (A+B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^5+30 (A+B) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^5-30 (3 A+B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2 \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^4+24 (B-A) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2+(-15-15 i) \sqrt [4]{-1} (7 A+3 B) \tan ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (\tan \left (\frac {1}{4} (e+f x)\right )+1\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2 \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^5-40 A \cos ^2(e+f x)\right )}{240 a^3 f (\sin (e+f x)+1)^3 (c-c \sin (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-40*A*Cos[e + f*x]^2 + 24*(-A +
B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^2 - 30*(3*A + B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^2*(Cos[(e + f*
x)/2] + Sin[(e + f*x)/2])^4 + 15*(A + B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*
x)/2])^5 - (15 + 15*I)*(-1)^(1/4)*(7*A + 3*B)*ArcTan[(1/2 + I/2)*(-1)^(1/4)*(1 + Tan[(e + f*x)/4])]*(Cos[(e +
f*x)/2] - Sin[(e + f*x)/2])^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5 + 30*(A + B)*Sin[(e + f*x)/2]*(Cos[(e +
f*x)/2] + Sin[(e + f*x)/2])^5))/(240*a^3*f*(1 + Sin[e + f*x])^3*(c - c*Sin[e + f*x])^(3/2))

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fricas [A]  time = 0.46, size = 277, normalized size = 1.24 \[ \frac {15 \, \sqrt {2} {\left ({\left (7 \, A + 3 \, B\right )} \cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) + {\left (7 \, A + 3 \, B\right )} \cos \left (f x + e\right )^{3}\right )} \sqrt {c} \log \left (-\frac {c \cos \left (f x + e\right )^{2} + 2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} \sqrt {c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )} + 3 \, c \cos \left (f x + e\right ) + {\left (c \cos \left (f x + e\right ) - 2 \, c\right )} \sin \left (f x + e\right ) + 2 \, c}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) - 4 \, {\left (25 \, {\left (7 \, A + 3 \, B\right )} \cos \left (f x + e\right )^{2} + 3 \, {\left (5 \, {\left (7 \, A + 3 \, B\right )} \cos \left (f x + e\right )^{2} - 28 \, A - 12 \, B\right )} \sin \left (f x + e\right ) - 36 \, A - 84 \, B\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{960 \, {\left (a^{3} c^{2} f \cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) + a^{3} c^{2} f \cos \left (f x + e\right )^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/960*(15*sqrt(2)*((7*A + 3*B)*cos(f*x + e)^3*sin(f*x + e) + (7*A + 3*B)*cos(f*x + e)^3)*sqrt(c)*log(-(c*cos(f
*x + e)^2 + 2*sqrt(2)*sqrt(-c*sin(f*x + e) + c)*sqrt(c)*(cos(f*x + e) + sin(f*x + e) + 1) + 3*c*cos(f*x + e) +
(c*cos(f*x + e) - 2*c)*sin(f*x + e) + 2*c)/(cos(f*x + e)^2 + (cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) -
2)) - 4*(25*(7*A + 3*B)*cos(f*x + e)^2 + 3*(5*(7*A + 3*B)*cos(f*x + e)^2 - 28*A - 12*B)*sin(f*x + e) - 36*A -
84*B)*sqrt(-c*sin(f*x + e) + c))/(a^3*c^2*f*cos(f*x + e)^3*sin(f*x + e) + a^3*c^2*f*cos(f*x + e)^3)

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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))/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^(3/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)2/f*2*(1/32*(-3*A*
(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^3-3*B*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan(
(f*x+exp(1))/2)^2+c))^3+A*c*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))-A*sqrt(c)*(-sqrt(c)
*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^2+B*c*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(
1))/2)^2+c))-B*sqrt(c)*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^2-A*sqrt(c)*c-B*sqrt(c)*
c)/a^3/(-(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^2-2*sqrt(c)*(-sqrt(c)*tan((f*x+exp(1))
/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))+c)^2/c/sign(tan((f*x+exp(1))/2)-1)-1/240*(-165*A*(-sqrt(c)*tan((f*x+exp(1
))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^9+45*B*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^9
+795*A*sqrt(c)*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^8-75*B*sqrt(c)*(-sqrt(c)*tan((f*
x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^8-1100*A*c*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))
/2)^2+c))^7+60*B*c*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^7-1180*A*sqrt(c)*c*(-sqrt(c)
*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^6+60*B*sqrt(c)*c*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*ta
n((f*x+exp(1))/2)^2+c))^6+2618*A*c^2*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^5+102*B*c^
2*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^5+1130*A*sqrt(c)*c^2*(-sqrt(c)*tan((f*x+exp(1
))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^4+150*B*sqrt(c)*c^2*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1
))/2)^2+c))^4-1980*A*c^3*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^3-180*B*c^3*(-sqrt(c)*
tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^3-605*A*c^4*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x
+exp(1))/2)^2+c))-1900*A*sqrt(c)*c^3*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^2-75*B*c^4
*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))-180*B*sqrt(c)*c^3*(-sqrt(c)*tan((f*x+exp(1))/2
)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^2-77*A*sqrt(c)*c^4-3*B*sqrt(c)*c^4)/a^3/(-(-sqrt(c)*tan((f*x+exp(1))/2)+sqr
t(c*tan((f*x+exp(1))/2)^2+c))^2+2*sqrt(c)*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))+c)^5/
c/sign(tan((f*x+exp(1))/2)-1)+1/32*(7*A+3*B)*atan((-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c)+sqrt(c*tan((f*x+exp(1)
)/2)^2+c))/sqrt(2)/sqrt(-c))/sqrt(2)/a^3/sqrt(-c)/c/sign(tan((f*x+exp(1))/2)-1))

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maple [A]  time = 1.69, size = 308, normalized size = 1.38 \[ -\frac {278 A \,c^{\frac {7}{2}}-18 B \,c^{\frac {7}{2}}-350 A \,c^{\frac {7}{2}} \left (\sin ^{2}\left (f x +e \right )\right )-150 B \,c^{\frac {7}{2}} \left (\sin ^{2}\left (f x +e \right )\right )+42 A \,c^{\frac {7}{2}} \sin \left (f x +e \right )+18 B \,c^{\frac {7}{2}} \sin \left (f x +e \right )-210 A \,c^{\frac {7}{2}} \left (\sin ^{3}\left (f x +e \right )\right )-90 B \,c^{\frac {7}{2}} \left (\sin ^{3}\left (f x +e \right )\right )+45 B \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sin \left (f x +e \right ) c -105 A \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c -45 B \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c +105 A \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sin \left (f x +e \right ) c}{480 c^{\frac {9}{2}} a^{3} \left (1+\sin \left (f x +e \right )\right )^{2} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/480/c^(9/2)/a^3*(278*A*c^(7/2)-18*B*c^(7/2)-350*A*c^(7/2)*sin(f*x+e)^2-150*B*c^(7/2)*sin(f*x+e)^2+42*A*c^(7
/2)*sin(f*x+e)+18*B*c^(7/2)*sin(f*x+e)-210*A*c^(7/2)*sin(f*x+e)^3-90*B*c^(7/2)*sin(f*x+e)^3+45*B*(c*(1+sin(f*x
+e)))^(5/2)*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*sin(f*x+e)*c-105*A*(c*(1+sin(f*x+e))
)^(5/2)*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*c-45*B*(c*(1+sin(f*x+e)))^(5/2)*2^(1/2)*
arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*c+105*A*(c*(1+sin(f*x+e)))^(5/2)*2^(1/2)*arctanh(1/2*(c*
(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*sin(f*x+e)*c)/(1+sin(f*x+e))^2/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {A+B\,\sin \left (e+f\,x\right )}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^3\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \]

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*(c - c*sin(e + f*x))^(3/2)),x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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