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

[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)

________________________________________________________________________________________

Rubi [A]  time = 0.407421, antiderivative size = 228, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, 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 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]))

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 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 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 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 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])

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*}

Mathematica [C]  time = 1.50582, 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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Maple [A]  time = 0.767, size = 246, normalized size = 1.1 \begin{align*} -{\frac{1}{1280\,{a}^{3} \left ( 1+\sin \left ( fx+e \right ) \right ) ^{2} \left ( -1+\sin \left ( fx+e \right ) \right ) \cos \left ( fx+e \right ) f} \left ( 315\, \left ( c \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{5/2}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }\sqrt{2}}{\sqrt{c}}} \right ) \left ( \sin \left ( fx+e \right ) \right ) ^{2}{c}^{2}+1176\,{c}^{9/2} \left ( \sin \left ( fx+e \right ) \right ) ^{2}-420\,{c}^{9/2} \left ( \sin \left ( fx+e \right ) \right ) ^{3}-630\,{c}^{9/2} \left ( \sin \left ( fx+e \right ) \right ) ^{4}-630\, \left ( c \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{5/2}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }\sqrt{2}}{\sqrt{c}}} \right ) \sin \left ( fx+e \right ){c}^{2}+708\,{c}^{9/2}\sin \left ( fx+e \right ) +315\, \left ( c \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{5/2}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }\sqrt{2}}{\sqrt{c}}} \right ){c}^{2}-514\,{c}^{9/2} \right ){c}^{-{\frac{13}{2}}}{\frac{1}{\sqrt{c-c\sin \left ( fx+e \right ) }}}} \end{align*}

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/(-1+sin(f*x+e))/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Fricas [A]  time = 1.20115, size = 578, normalized size = 2.54 \begin{align*} \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}} \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{2} \end{align*}

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]

sage2