3.130 \(\int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=258 \[ \frac {7 (9 A+B) \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{128 \sqrt {2} a^3 c^{5/2} f}-\frac {(A-B) \sec ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{5 a^3 c^3 f}-\frac {(9 A+B) \sec ^3(e+f x)}{30 a^3 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {7 (9 A+B) \sec (e+f x)}{96 a^3 c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {7 (9 A+B) \cos (e+f x)}{128 a^3 c f (c-c \sin (e+f x))^{3/2}}+\frac {7 (9 A+B) \sec (e+f x)}{240 a^3 c f (c-c \sin (e+f x))^{3/2}} \]

[Out]

7/128*(9*A+B)*cos(f*x+e)/a^3/c/f/(c-c*sin(f*x+e))^(3/2)+7/240*(9*A+B)*sec(f*x+e)/a^3/c/f/(c-c*sin(f*x+e))^(3/2
)+7/256*(9*A+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^(5/2)/f*2^(1/2)-7/96*(9*A
+B)*sec(f*x+e)/a^3/c^2/f/(c-c*sin(f*x+e))^(1/2)-1/30*(9*A+B)*sec(f*x+e)^3/a^3/c^2/f/(c-c*sin(f*x+e))^(1/2)-1/5
*(A-B)*sec(f*x+e)^5*(c-c*sin(f*x+e))^(1/2)/a^3/c^3/f

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Rubi [A]  time = 0.56, antiderivative size = 258, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.184, Rules used = {2967, 2855, 2687, 2681, 2650, 2649, 206} \[ -\frac {(A-B) \sec ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{5 a^3 c^3 f}-\frac {(9 A+B) \sec ^3(e+f x)}{30 a^3 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {7 (9 A+B) \sec (e+f x)}{96 a^3 c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {7 (9 A+B) \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{128 \sqrt {2} a^3 c^{5/2} f}+\frac {7 (9 A+B) \cos (e+f x)}{128 a^3 c f (c-c \sin (e+f x))^{3/2}}+\frac {7 (9 A+B) \sec (e+f x)}{240 a^3 c f (c-c \sin (e+f x))^{3/2}} \]

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])^(5/2)),x]

[Out]

(7*(9*A + B)*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(128*Sqrt[2]*a^3*c^(5/2)*f) +
(7*(9*A + B)*Cos[e + f*x])/(128*a^3*c*f*(c - c*Sin[e + f*x])^(3/2)) + (7*(9*A + B)*Sec[e + f*x])/(240*a^3*c*f
*(c - c*Sin[e + f*x])^(3/2)) - (7*(9*A + B)*Sec[e + f*x])/(96*a^3*c^2*f*Sqrt[c - c*Sin[e + f*x]]) - ((9*A + B)
*Sec[e + f*x]^3)/(30*a^3*c^2*f*Sqrt[c - c*Sin[e + f*x]]) - ((A - B)*Sec[e + f*x]^5*Sqrt[c - c*Sin[e + f*x]])/(
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 2681

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*(2*m + p + 1)), x] + Dist[(m + p + 1)/(a*(2*m + p + 1)),
Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^
2, 0] && LtQ[m, -1] && NeQ[2*m + p + 1, 0] && IntegersQ[2*m, 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))^{5/2}} \, dx &=\frac {\int \sec ^6(e+f x) (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx}{a^3 c^3}\\ &=-\frac {(A-B) \sec ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{5 a^3 c^3 f}+\frac {(9 A+B) \int \frac {\sec ^4(e+f x)}{\sqrt {c-c \sin (e+f x)}} \, dx}{10 a^3 c^2}\\ &=-\frac {(9 A+B) \sec ^3(e+f x)}{30 a^3 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {(A-B) \sec ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{5 a^3 c^3 f}+\frac {(7 (9 A+B)) \int \frac {\sec ^2(e+f x)}{(c-c \sin (e+f x))^{3/2}} \, dx}{60 a^3 c}\\ &=\frac {7 (9 A+B) \sec (e+f x)}{240 a^3 c f (c-c \sin (e+f x))^{3/2}}-\frac {(9 A+B) \sec ^3(e+f x)}{30 a^3 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {(A-B) \sec ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{5 a^3 c^3 f}+\frac {(7 (9 A+B)) \int \frac {\sec ^2(e+f x)}{\sqrt {c-c \sin (e+f x)}} \, dx}{96 a^3 c^2}\\ &=\frac {7 (9 A+B) \sec (e+f x)}{240 a^3 c f (c-c \sin (e+f x))^{3/2}}-\frac {7 (9 A+B) \sec (e+f x)}{96 a^3 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {(9 A+B) \sec ^3(e+f x)}{30 a^3 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {(A-B) \sec ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{5 a^3 c^3 f}+\frac {(7 (9 A+B)) \int \frac {1}{(c-c \sin (e+f x))^{3/2}} \, dx}{64 a^3 c}\\ &=\frac {7 (9 A+B) \cos (e+f x)}{128 a^3 c f (c-c \sin (e+f x))^{3/2}}+\frac {7 (9 A+B) \sec (e+f x)}{240 a^3 c f (c-c \sin (e+f x))^{3/2}}-\frac {7 (9 A+B) \sec (e+f x)}{96 a^3 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {(9 A+B) \sec ^3(e+f x)}{30 a^3 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {(A-B) \sec ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{5 a^3 c^3 f}+\frac {(7 (9 A+B)) \int \frac {1}{\sqrt {c-c \sin (e+f x)}} \, dx}{256 a^3 c^2}\\ &=\frac {7 (9 A+B) \cos (e+f x)}{128 a^3 c f (c-c \sin (e+f x))^{3/2}}+\frac {7 (9 A+B) \sec (e+f x)}{240 a^3 c f (c-c \sin (e+f x))^{3/2}}-\frac {7 (9 A+B) \sec (e+f x)}{96 a^3 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {(9 A+B) \sec ^3(e+f x)}{30 a^3 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {(A-B) \sec ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{5 a^3 c^3 f}-\frac {(7 (9 A+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 )}{128 a^3 c^2 f}\\ &=\frac {7 (9 A+B) \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{128 \sqrt {2} a^3 c^{5/2} f}+\frac {7 (9 A+B) \cos (e+f x)}{128 a^3 c f (c-c \sin (e+f x))^{3/2}}+\frac {7 (9 A+B) \sec (e+f x)}{240 a^3 c f (c-c \sin (e+f x))^{3/2}}-\frac {7 (9 A+B) \sec (e+f x)}{96 a^3 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {(9 A+B) \sec ^3(e+f x)}{30 a^3 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {(A-B) \sec ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{5 a^3 c^3 f}\\ \end {align*}

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Mathematica [C]  time = 2.29, size = 479, normalized size = 1.86 \[ \frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right ) \left (15 (15 A+7 B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^5+60 (A+B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^5+30 (15 A+7 B) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2 \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^5+120 (A+B) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^5+80 (B-3 A) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^2+96 (B-A) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4+(-105-105 i) \sqrt [4]{-1} (9 A+B) \tan ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (\tan \left (\frac {1}{4} (e+f x)\right )+1\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^5-720 A \cos ^4(e+f x)\right )}{1920 a^3 f (\sin (e+f x)+1)^3 (c-c \sin (e+f x))^{5/2}} \]

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])^(5/2)),x]

[Out]

((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-720*A*Cos[e + f*x]^4 + 96*(-A +
B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^4 + 80*(-3*A + B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^4*(Cos[(e +
f*x)/2] + Sin[(e + f*x)/2])^2 + 60*(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*A + 7*B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5 -
(105 + 105*I)*(-1)^(1/4)*(9*A + 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])^4*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5 + 120*(A + B)*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] +
Sin[(e + f*x)/2])^5 + 30*(15*A + 7*B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^2*Sin[(e + f*x)/2]*(Cos[(e + f*x)
/2] + Sin[(e + f*x)/2])^5))/(1920*a^3*f*(1 + Sin[e + f*x])^3*(c - c*Sin[e + f*x])^(5/2))

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fricas [A]  time = 0.49, size = 238, normalized size = 0.92 \[ \frac {105 \, \sqrt {2} {\left (9 \, A + B\right )} \sqrt {c} \cos \left (f x + e\right )^{5} \log \left (-\frac {c \cos \left (f x + e\right )^{2} + 2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} \sqrt {c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )} + 3 \, c \cos \left (f x + e\right ) + {\left (c \cos \left (f x + e\right ) - 2 \, c\right )} \sin \left (f x + e\right ) + 2 \, c}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) - 4 \, {\left (105 \, {\left (9 \, A + B\right )} \cos \left (f x + e\right )^{4} - 14 \, {\left (9 \, A + B\right )} \cos \left (f x + e\right )^{2} - 2 \, {\left (35 \, {\left (9 \, A + B\right )} \cos \left (f x + e\right )^{2} + 216 \, A + 24 \, B\right )} \sin \left (f x + e\right ) - 48 \, A - 432 \, B\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{7680 \, a^{3} c^{3} f \cos \left (f x + e\right )^{5}} \]

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))^(5/2),x, algorithm="fricas")

[Out]

1/7680*(105*sqrt(2)*(9*A + B)*sqrt(c)*cos(f*x + e)^5*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*(105*(9*A + B)*cos(f*x + e)^4 - 14*
(9*A + B)*cos(f*x + e)^2 - 2*(35*(9*A + B)*cos(f*x + e)^2 + 216*A + 24*B)*sin(f*x + e) - 48*A - 432*B)*sqrt(-c
*sin(f*x + e) + c))/(a^3*c^3*f*cos(f*x + e)^5)

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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))^(5/2),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*(1/256*(-65*
A*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^7-41*B*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*t
an((f*x+exp(1))/2)^2+c))^7-231*A*sqrt(c)*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^6-127*
B*sqrt(c)*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^6-163*A*c*(-sqrt(c)*tan((f*x+exp(1))/
2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^5-91*B*c*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^5+
247*A*sqrt(c)*c*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^4+143*B*sqrt(c)*c*(-sqrt(c)*tan
((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^4-11*A*c^2*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp
(1))/2)^2+c))^3-3*B*c^2*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^3+87*A*c^3*(-sqrt(c)*ta
n((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))-149*A*sqrt(c)*c^2*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan(
(f*x+exp(1))/2)^2+c))^2+47*B*c^3*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))-93*B*sqrt(c)*c
^2*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^2-19*A*sqrt(c)*c^3-11*B*sqrt(c)*c^3)/a^3/c^2
/(-(-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)+sq
rt(c*tan((f*x+exp(1))/2)^2+c))+c)^4/sign(tan((f*x+exp(1))/2)-1)-1/240*(-120*A*(-sqrt(c)*tan((f*x+exp(1))/2)+sq
rt(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+630*A*s
qrt(c)*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^8-165*B*sqrt(c)*(-sqrt(c)*tan((f*x+exp(1
))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^8-900*A*c*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c)
)^7+200*B*c*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^7-960*A*sqrt(c)*c*(-sqrt(c)*tan((f*
x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^6+220*B*sqrt(c)*c*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+
exp(1))/2)^2+c))^6+2244*A*c^2*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^5-374*B*c^2*(-sqr
t(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^5+1020*A*sqrt(c)*c^2*(-sqrt(c)*tan((f*x+exp(1))/2)+s
qrt(c*tan((f*x+exp(1))/2)^2+c))^4-110*B*sqrt(c)*c^2*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2
+c))^4-1740*A*c^3*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^3+240*B*c^3*(-sqrt(c)*tan((f*
x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^3-540*A*c^4*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1)
)/2)^2+c))-1680*A*sqrt(c)*c^3*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^2+65*B*c^4*(-sqrt
(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))+220*B*sqrt(c)*c^3*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(
c*tan((f*x+exp(1))/2)^2+c))^2-66*A*sqrt(c)*c^4+11*B*sqrt(c)*c^4)/a^3/c^2/(-(-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)^5/si
gn(tan((f*x+exp(1))/2)-1)+1/256*(63*A+7*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/c^2/sqrt(-c)/sign(tan((f*x+exp(1))/2)-1))

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maple [A]  time = 2.06, size = 410, normalized size = 1.59 \[ -\frac {\left (1260 c^{\frac {9}{2}} A +140 c^{\frac {9}{2}} B \right ) \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right )+\left (-1890 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (c +c \sin \left (f x +e \right )\right )^{\frac {5}{2}} c^{2} A +864 c^{\frac {9}{2}} A -210 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (c +c \sin \left (f x +e \right )\right )^{\frac {5}{2}} c^{2} B +96 c^{\frac {9}{2}} B \right ) \sin \left (f x +e \right )+\left (-1890 c^{\frac {9}{2}} A -210 c^{\frac {9}{2}} B \right ) \left (\cos ^{4}\left (f x +e \right )\right )+\left (-945 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (c +c \sin \left (f x +e \right )\right )^{\frac {5}{2}} c^{2} A +252 c^{\frac {9}{2}} A -105 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (c +c \sin \left (f x +e \right )\right )^{\frac {5}{2}} c^{2} B +28 c^{\frac {9}{2}} B \right ) \left (\cos ^{2}\left (f x +e \right )\right )+1890 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (c +c \sin \left (f x +e \right )\right )^{\frac {5}{2}} c^{2} A +96 c^{\frac {9}{2}} A +210 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (c +c \sin \left (f x +e \right )\right )^{\frac {5}{2}} c^{2} B +864 c^{\frac {9}{2}} B}{3840 c^{\frac {13}{2}} a^{3} \left (1+\sin \left (f x +e \right )\right )^{2} \left (\sin \left (f x +e \right )-1\right ) \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f} \]

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))^(5/2),x)

[Out]

-1/3840/c^(13/2)/a^3*((1260*c^(9/2)*A+140*c^(9/2)*B)*sin(f*x+e)*cos(f*x+e)^2+(-1890*2^(1/2)*arctanh(1/2*(c+c*s
in(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*(c+c*sin(f*x+e))^(5/2)*c^2*A+864*c^(9/2)*A-210*2^(1/2)*arctanh(1/2*(c+c*sin(
f*x+e))^(1/2)*2^(1/2)/c^(1/2))*(c+c*sin(f*x+e))^(5/2)*c^2*B+96*c^(9/2)*B)*sin(f*x+e)+(-1890*c^(9/2)*A-210*c^(9
/2)*B)*cos(f*x+e)^4+(-945*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*(c+c*sin(f*x+e))^(5/2)*c
^2*A+252*c^(9/2)*A-105*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*(c+c*sin(f*x+e))^(5/2)*c^2*
B+28*c^(9/2)*B)*cos(f*x+e)^2+1890*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*(c+c*sin(f*x+e))
^(5/2)*c^2*A+96*c^(9/2)*A+210*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*(c+c*sin(f*x+e))^(5/
2)*c^2*B+864*c^(9/2)*B)/(1+sin(f*x+e))^2/(sin(f*x+e)-1)/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))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {A+B\,\sin \left (e+f\,x\right )}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^3\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \]

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))^(5/2)),x)

[Out]

int((A + B*sin(e + f*x))/((a + a*sin(e + f*x))^3*(c - c*sin(e + f*x))^(5/2)), 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))**(5/2),x)

[Out]

Timed out

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