3.339 \(\int \frac {1}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=228 \[ \frac {63 \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{128 \sqrt {2} a^3 c^{5/2} f}-\frac {\sec ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{5 a^3 c^3 f}-\frac {3 \sec ^3(e+f x)}{10 a^3 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {21 \sec (e+f x)}{32 a^3 c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {63 \cos (e+f x)}{128 a^3 c f (c-c \sin (e+f x))^{3/2}}+\frac {21 \sec (e+f x)}{80 a^3 c f (c-c \sin (e+f x))^{3/2}} \]

[Out]

63/128*cos(f*x+e)/a^3/c/f/(c-c*sin(f*x+e))^(3/2)+21/80*sec(f*x+e)/a^3/c/f/(c-c*sin(f*x+e))^(3/2)+63/256*arctan
h(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)-21/32*sec(f*x+e)/a^3/c^2/f/(c-c
*sin(f*x+e))^(1/2)-3/10*sec(f*x+e)^3/a^3/c^2/f/(c-c*sin(f*x+e))^(1/2)-1/5*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 = 0.41, antiderivative size = 228, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2736, 2675, 2687, 2681, 2650, 2649, 206} \[ -\frac {\sec ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{5 a^3 c^3 f}-\frac {3 \sec ^3(e+f x)}{10 a^3 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {21 \sec (e+f x)}{32 a^3 c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {63 \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{128 \sqrt {2} a^3 c^{5/2} f}+\frac {63 \cos (e+f x)}{128 a^3 c f (c-c \sin (e+f x))^{3/2}}+\frac {21 \sec (e+f x)}{80 a^3 c f (c-c \sin (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x])^(5/2)),x]

[Out]

(63*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) + (63*Cos[
e + f*x])/(128*a^3*c*f*(c - c*Sin[e + f*x])^(3/2)) + (21*Sec[e + f*x])/(80*a^3*c*f*(c - c*Sin[e + f*x])^(3/2))
- (21*Sec[e + f*x])/(32*a^3*c^2*f*Sqrt[c - c*Sin[e + f*x]]) - (3*Sec[e + f*x]^3)/(10*a^3*c^2*f*Sqrt[c - c*Sin
[e + f*x]]) - (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 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 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 2736

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[a^m*c^m, Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] &&  !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0,
n, m] || LtQ[m, n, 0]))

Rubi steps

\begin {align*} \int \frac {1}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^{5/2}} \, dx &=\frac {\int \sec ^6(e+f x) \sqrt {c-c \sin (e+f x)} \, dx}{a^3 c^3}\\ &=-\frac {\sec ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{5 a^3 c^3 f}+\frac {9 \int \frac {\sec ^4(e+f x)}{\sqrt {c-c \sin (e+f x)}} \, dx}{10 a^3 c^2}\\ &=-\frac {3 \sec ^3(e+f x)}{10 a^3 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {\sec ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{5 a^3 c^3 f}+\frac {21 \int \frac {\sec ^2(e+f x)}{(c-c \sin (e+f x))^{3/2}} \, dx}{20 a^3 c}\\ &=\frac {21 \sec (e+f x)}{80 a^3 c f (c-c \sin (e+f x))^{3/2}}-\frac {3 \sec ^3(e+f x)}{10 a^3 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {\sec ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{5 a^3 c^3 f}+\frac {21 \int \frac {\sec ^2(e+f x)}{\sqrt {c-c \sin (e+f x)}} \, dx}{32 a^3 c^2}\\ &=\frac {21 \sec (e+f x)}{80 a^3 c f (c-c \sin (e+f x))^{3/2}}-\frac {21 \sec (e+f x)}{32 a^3 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {3 \sec ^3(e+f x)}{10 a^3 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {\sec ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{5 a^3 c^3 f}+\frac {63 \int \frac {1}{(c-c \sin (e+f x))^{3/2}} \, dx}{64 a^3 c}\\ &=\frac {63 \cos (e+f x)}{128 a^3 c f (c-c \sin (e+f x))^{3/2}}+\frac {21 \sec (e+f x)}{80 a^3 c f (c-c \sin (e+f x))^{3/2}}-\frac {21 \sec (e+f x)}{32 a^3 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {3 \sec ^3(e+f x)}{10 a^3 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {\sec ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{5 a^3 c^3 f}+\frac {63 \int \frac {1}{\sqrt {c-c \sin (e+f x)}} \, dx}{256 a^3 c^2}\\ &=\frac {63 \cos (e+f x)}{128 a^3 c f (c-c \sin (e+f x))^{3/2}}+\frac {21 \sec (e+f x)}{80 a^3 c f (c-c \sin (e+f x))^{3/2}}-\frac {21 \sec (e+f x)}{32 a^3 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {3 \sec ^3(e+f x)}{10 a^3 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {\sec ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{5 a^3 c^3 f}-\frac {63 \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 {63 \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 {63 \cos (e+f x)}{128 a^3 c f (c-c \sin (e+f x))^{3/2}}+\frac {21 \sec (e+f x)}{80 a^3 c f (c-c \sin (e+f x))^{3/2}}-\frac {21 \sec (e+f x)}{32 a^3 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {3 \sec ^3(e+f x)}{10 a^3 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {\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 = 1.54, size = 443, normalized size = 1.94 \[ \frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right ) \left (-240 \cos ^4(e+f x)+75 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^5+20 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^5+150 \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2 \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^5+40 \sin \left (\frac {1}{2} (e+f x)\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^5-80 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^2-32 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4+(-315-315 i) \sqrt [4]{-1} \tan ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (\tan \left (\frac {1}{4} (e+f x)\right )+1\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^5\right )}{640 a^3 f (\sin (e+f x)+1)^3 (c-c \sin (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((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])*(-240*Cos[e + f*x]^4 - 32*(Cos[(e
+ f*x)/2] - Sin[(e + f*x)/2])^4 - 80*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^4*(Cos[(e + f*x)/2] + Sin[(e + f*x
)/2])^2 + 20*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5 + 75*(Cos[(e + f*x)
/2] - Sin[(e + f*x)/2])^3*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5 - (315 + 315*I)*(-1)^(1/4)*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 + 40*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5 + 150*(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))/(640*a^3*f*(1 + Sin[e + f*x])^3*(c - c*Sin[e +
f*x])^(5/2))

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fricas [A]  time = 0.47, size = 208, normalized size = 0.91 \[ \frac {315 \, \sqrt {2} \sqrt {c} \cos \left (f x + e\right )^{5} \log \left (-\frac {c \cos \left (f x + e\right )^{2} + 2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} \sqrt {c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )} + 3 \, c \cos \left (f x + e\right ) + {\left (c \cos \left (f x + e\right ) - 2 \, c\right )} \sin \left (f x + e\right ) + 2 \, c}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) - 4 \, {\left (315 \, \cos \left (f x + e\right )^{4} - 42 \, \cos \left (f x + e\right )^{2} - 6 \, {\left (35 \, \cos \left (f x + e\right )^{2} + 24\right )} \sin \left (f x + e\right ) - 16\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{2560 \, a^{3} c^{3} f \cos \left (f x + e\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

1/2560*(315*sqrt(2)*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*(315*cos(f*x + e)^4 - 42*cos(f*x + e)^2 - 6*(
35*cos(f*x + e)^2 + 24)*sin(f*x + e) - 16)*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 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab
le to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (4*p
i/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)Unab
le 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*p
i/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)Unab
le 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*p
i/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)Unab
le to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to ch
eck sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable
to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)
Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check
[abs(sin((f*t_nostep+exp(1))/2-pi/4))]Unable to check sign: (8*pi/t_nostep/2)>(-8*pi/t_nostep/2)Discontinuiti
es at zeroes of sin((f*t_nostep+exp(1))/2-pi/4) were not checkedUnable to check sign: (4*pi/t_nostep/2)>(-4*pi
/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>
(-4*pi/t_nostep/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)Unab
le to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to ch
eck sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Warning, integration of abs or sig
n assumes constant sign by intervals (correct if the argument is real):Check [abs(t_nostep-1)]Evaluation time:
0.44Not invertible Error: Bad Argument Value

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maple [A]  time = 1.30, size = 246, normalized size = 1.08 \[ -\frac {315 \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (\sin ^{2}\left (f x +e \right )\right ) c^{2}+1176 c^{\frac {9}{2}} \left (\sin ^{2}\left (f x +e \right )\right )-420 c^{\frac {9}{2}} \left (\sin ^{3}\left (f x +e \right )\right )-630 c^{\frac {9}{2}} \left (\sin ^{4}\left (f x +e \right )\right )-630 \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sin \left (f x +e \right ) c^{2}+708 c^{\frac {9}{2}} \sin \left (f x +e \right )+315 \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}-514 c^{\frac {9}{2}}}{1280 c^{\frac {13}{2}} a^{3} \left (1+\sin \left (f x +e \right )\right )^{2} \left (\sin \left (f x +e \right )-1\right ) \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^(5/2),x)

[Out]

-1/1280/c^(13/2)/a^3*(315*(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)^2*c^2+1176*c^(9/2)*sin(f*x+e)^2-420*c^(9/2)*sin(f*x+e)^3-630*c^(9/2)*sin(f*x+e)^4-630*(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^2+708*c^(9/2)*sin(f*
x+e)+315*(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^2-514*c^(9/2
))/(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(1/(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 \[ \int \frac {1}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^3\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + a*sin(e + f*x))^3*(c - c*sin(e + f*x))^(5/2)),x)

[Out]

int(1/((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(1/(a+a*sin(f*x+e))**3/(c-c*sin(f*x+e))**(5/2),x)

[Out]

Timed out

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