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

Optimal. Leaf size=225 \[ \frac {5 (7 A-B) \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{64 \sqrt {2} a^2 c^{5/2} f}-\frac {(A-B) \sec ^3(e+f x)}{3 a^2 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {5 (7 A-B) \sec (e+f x)}{48 a^2 c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {5 (7 A-B) \cos (e+f x)}{64 a^2 c f (c-c \sin (e+f x))^{3/2}}+\frac {(7 A-B) \sec (e+f x)}{24 a^2 c f (c-c \sin (e+f x))^{3/2}} \]

[Out]

5/64*(7*A-B)*cos(f*x+e)/a^2/c/f/(c-c*sin(f*x+e))^(3/2)+1/24*(7*A-B)*sec(f*x+e)/a^2/c/f/(c-c*sin(f*x+e))^(3/2)+
5/128*(7*A-B)*arctanh(1/2*cos(f*x+e)*c^(1/2)*2^(1/2)/(c-c*sin(f*x+e))^(1/2))/a^2/c^(5/2)/f*2^(1/2)-5/48*(7*A-B
)*sec(f*x+e)/a^2/c^2/f/(c-c*sin(f*x+e))^(1/2)-1/3*(A-B)*sec(f*x+e)^3/a^2/c^2/f/(c-c*sin(f*x+e))^(1/2)

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Rubi [A]  time = 0.48, antiderivative size = 225, 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, 2681, 2687, 2650, 2649, 206} \[ -\frac {(A-B) \sec ^3(e+f x)}{3 a^2 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {5 (7 A-B) \sec (e+f x)}{48 a^2 c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {5 (7 A-B) \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{64 \sqrt {2} a^2 c^{5/2} f}+\frac {5 (7 A-B) \cos (e+f x)}{64 a^2 c f (c-c \sin (e+f x))^{3/2}}+\frac {(7 A-B) \sec (e+f x)}{24 a^2 c f (c-c \sin (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^(5/2)),x]

[Out]

(5*(7*A - B)*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(64*Sqrt[2]*a^2*c^(5/2)*f) +
(5*(7*A - B)*Cos[e + f*x])/(64*a^2*c*f*(c - c*Sin[e + f*x])^(3/2)) + ((7*A - B)*Sec[e + f*x])/(24*a^2*c*f*(c -
c*Sin[e + f*x])^(3/2)) - (5*(7*A - B)*Sec[e + f*x])/(48*a^2*c^2*f*Sqrt[c - c*Sin[e + f*x]]) - ((A - B)*Sec[e
+ f*x]^3)/(3*a^2*c^2*f*Sqrt[c - c*Sin[e + f*x]])

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))^2 (c-c \sin (e+f x))^{5/2}} \, dx &=\frac {\int \frac {\sec ^4(e+f x) (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx}{a^2 c^2}\\ &=-\frac {(A-B) \sec ^3(e+f x)}{3 a^2 c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {(7 A-B) \int \frac {\sec ^2(e+f x)}{(c-c \sin (e+f x))^{3/2}} \, dx}{6 a^2 c}\\ &=\frac {(7 A-B) \sec (e+f x)}{24 a^2 c f (c-c \sin (e+f x))^{3/2}}-\frac {(A-B) \sec ^3(e+f x)}{3 a^2 c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {(5 (7 A-B)) \int \frac {\sec ^2(e+f x)}{\sqrt {c-c \sin (e+f x)}} \, dx}{48 a^2 c^2}\\ &=\frac {(7 A-B) \sec (e+f x)}{24 a^2 c f (c-c \sin (e+f x))^{3/2}}-\frac {5 (7 A-B) \sec (e+f x)}{48 a^2 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {(A-B) \sec ^3(e+f x)}{3 a^2 c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {(5 (7 A-B)) \int \frac {1}{(c-c \sin (e+f x))^{3/2}} \, dx}{32 a^2 c}\\ &=\frac {5 (7 A-B) \cos (e+f x)}{64 a^2 c f (c-c \sin (e+f x))^{3/2}}+\frac {(7 A-B) \sec (e+f x)}{24 a^2 c f (c-c \sin (e+f x))^{3/2}}-\frac {5 (7 A-B) \sec (e+f x)}{48 a^2 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {(A-B) \sec ^3(e+f x)}{3 a^2 c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {(5 (7 A-B)) \int \frac {1}{\sqrt {c-c \sin (e+f x)}} \, dx}{128 a^2 c^2}\\ &=\frac {5 (7 A-B) \cos (e+f x)}{64 a^2 c f (c-c \sin (e+f x))^{3/2}}+\frac {(7 A-B) \sec (e+f x)}{24 a^2 c f (c-c \sin (e+f x))^{3/2}}-\frac {5 (7 A-B) \sec (e+f x)}{48 a^2 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {(A-B) \sec ^3(e+f x)}{3 a^2 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {(5 (7 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 )}{64 a^2 c^2 f}\\ &=\frac {5 (7 A-B) \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{64 \sqrt {2} a^2 c^{5/2} f}+\frac {5 (7 A-B) \cos (e+f x)}{64 a^2 c f (c-c \sin (e+f x))^{3/2}}+\frac {(7 A-B) \sec (e+f x)}{24 a^2 c f (c-c \sin (e+f x))^{3/2}}-\frac {5 (7 A-B) \sec (e+f x)}{48 a^2 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {(A-B) \sec ^3(e+f x)}{3 a^2 c^2 f \sqrt {c-c \sin (e+f x)}}\\ \end {align*}

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Mathematica [C]  time = 1.41, size = 430, normalized size = 1.91 \[ \frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right ) \left (3 (11 A+3 B) \cos ^3(e+f x)+24 (B-3 A) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4+16 (B-A) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4+6 (11 A+3 B) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2+12 (A+B) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+24 (A+B) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^3+(-15-15 i) \sqrt [4]{-1} (7 A-B) \tan ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (\tan \left (\frac {1}{4} (e+f x)\right )+1\right )\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4\right )}{192 a^2 f (\sin (e+f x)+1)^2 (c-c \sin (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])^2*(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])*(3*(11*A + 3*B)*Cos[e + f*x]^3 +
16*(-A + B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^4 + 24*(-3*A + B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^4*(C
os[(e + f*x)/2] + Sin[(e + f*x)/2])^2 + 12*(A + B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + S
in[(e + f*x)/2])^3 - (15 + 15*I)*(-1)^(1/4)*(7*A - B)*ArcTan[(1/2 + I/2)*(-1)^(1/4)*(1 + Tan[(e + f*x)/4])]*(C
os[(e + f*x)/2] - Sin[(e + f*x)/2])^4*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3 + 24*(A + B)*Sin[(e + f*x)/2]*(C
os[(e + f*x)/2] + Sin[(e + f*x)/2])^3 + 6*(11*A + 3*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])^3))/(192*a^2*f*(1 + Sin[e + f*x])^2*(c - c*Sin[e + f*x])^(5/2))

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fricas [A]  time = 0.46, size = 279, normalized size = 1.24 \[ -\frac {15 \, \sqrt {2} {\left ({\left (7 \, A - B\right )} \cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) - {\left (7 \, A - B\right )} \cos \left (f x + e\right )^{3}\right )} \sqrt {c} \log \left (-\frac {c \cos \left (f x + e\right )^{2} - 2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} \sqrt {c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )} + 3 \, c \cos \left (f x + e\right ) + {\left (c \cos \left (f x + e\right ) - 2 \, c\right )} \sin \left (f x + e\right ) + 2 \, c}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) - 4 \, {\left (5 \, {\left (7 \, A - B\right )} \cos \left (f x + e\right )^{2} - {\left (15 \, {\left (7 \, A - B\right )} \cos \left (f x + e\right )^{2} + 56 \, A - 8 \, B\right )} \sin \left (f x + e\right ) + 8 \, A - 56 \, B\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{768 \, {\left (a^{2} c^{3} f \cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) - a^{2} c^{3} f \cos \left (f x + e\right )^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

-1/768*(15*sqrt(2)*((7*A - B)*cos(f*x + e)^3*sin(f*x + e) - (7*A - 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*(5*(7*A - B)*cos(f*x + e)^2 - (15*(7*A - B)*cos(f*x + e)^2 + 56*A - 8*B)*sin(f*x + e) + 8*A - 56*B)*sqrt
(-c*sin(f*x + e) + c))/(a^2*c^3*f*cos(f*x + e)^3*sin(f*x + e) - a^2*c^3*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))^2/(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*(-1/48*(-15*
A*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^5+9*B*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*ta
n((f*x+exp(1))/2)^2+c))^5+33*A*sqrt(c)*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^4-15*B*s
qrt(c)*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^4+22*A*c*(-sqrt(c)*tan((f*x+exp(1))/2)+s
qrt(c*tan((f*x+exp(1))/2)^2+c))^3-10*B*c*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^3-51*A
*c^2*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))-66*A*sqrt(c)*c*(-sqrt(c)*tan((f*x+exp(1))/
2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^2+21*B*c^2*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))+
30*B*sqrt(c)*c*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^2-11*A*sqrt(c)*c^2+5*B*sqrt(c)*c
^2)/a^2/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+ex
p(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))+c)^3/sign(tan((f*x+exp(1))/2)-1)+1/128*(-53*A*(-sqrt(c)*tan((f*x+exp
(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^7-29*B*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))
^7-179*A*sqrt(c)*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^6-75*B*sqrt(c)*(-sqrt(c)*tan((
f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^6-127*A*c*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1)
)/2)^2+c))^5-55*B*c*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^5+195*A*sqrt(c)*c*(-sqrt(c)
*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^4+91*B*sqrt(c)*c*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*ta
n((f*x+exp(1))/2)^2+c))^4-7*A*c^2*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^3+B*c^2*(-sqr
t(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^3+67*A*c^3*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan(
(f*x+exp(1))/2)^2+c))-121*A*sqrt(c)*c^2*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^2+27*B*
c^3*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))-65*B*sqrt(c)*c^2*(-sqrt(c)*tan((f*x+exp(1))
/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^2-15*A*sqrt(c)*c^3-7*B*sqrt(c)*c^3)/a^2/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)^4/sign(tan((f*x+exp(1))/2)-1)+1/128*(35*A-5*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^2/c^2/sqrt(-c)/sign(tan((f*x+exp(1))/2)-1))

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maple [B]  time = 1.74, size = 426, normalized size = 1.89 \[ -\frac {-86 A \,c^{\frac {7}{2}}+122 B \,c^{\frac {7}{2}}+70 A \,c^{\frac {7}{2}} \left (\sin ^{2}\left (f x +e \right )\right )-10 B \,c^{\frac {7}{2}} \left (\sin ^{2}\left (f x +e \right )\right )+322 A \,c^{\frac {7}{2}} \sin \left (f x +e \right )-46 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 )+30 B \,c^{\frac {7}{2}} \left (\sin ^{3}\left (f x +e \right )\right )+105 A \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (\sin ^{2}\left (f x +e \right )\right ) c^{2}-15 B \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (\sin ^{2}\left (f x +e \right )\right ) c^{2}-210 A \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{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^{2}+30 B \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{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^{2}+105 A \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}-15 B \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}}{384 c^{\frac {11}{2}} a^{2} \left (1+\sin \left (f x +e \right )\right ) \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))^2/(c-c*sin(f*x+e))^(5/2),x)

[Out]

-1/384/c^(11/2)/a^2*(-86*A*c^(7/2)+122*B*c^(7/2)+70*A*c^(7/2)*sin(f*x+e)^2-10*B*c^(7/2)*sin(f*x+e)^2+322*A*c^(
7/2)*sin(f*x+e)-46*B*c^(7/2)*sin(f*x+e)-210*A*c^(7/2)*sin(f*x+e)^3+30*B*c^(7/2)*sin(f*x+e)^3+105*A*(c*(1+sin(f
*x+e)))^(3/2)*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*sin(f*x+e)^2*c^2-15*B*(c*(1+sin(f*
x+e)))^(3/2)*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*sin(f*x+e)^2*c^2-210*A*(c*(1+sin(f*
x+e)))^(3/2)*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^2+30*B*(c*(1+sin(f*x+e
)))^(3/2)*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^2+105*A*(c*(1+sin(f*x+e))
)^(3/2)*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*c^2-15*B*(c*(1+sin(f*x+e)))^(3/2)*2^(1/2
)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*c^2)/(1+sin(f*x+e))/(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))^2/(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 \[ \int \frac {A+B\,\sin \left (e+f\,x\right )}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^2\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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