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

Optimal. Leaf size=156 \[ \frac {15 \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{32 \sqrt {2} a c^{5/2} f}-\frac {5 \sec (e+f x)}{8 a c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {15 \cos (e+f x)}{32 a c f (c-c \sin (e+f x))^{3/2}}+\frac {\sec (e+f x)}{4 a c f (c-c \sin (e+f x))^{3/2}} \]

[Out]

15/32*cos(f*x+e)/a/c/f/(c-c*sin(f*x+e))^(3/2)+1/4*sec(f*x+e)/a/c/f/(c-c*sin(f*x+e))^(3/2)+15/64*arctanh(1/2*co
s(f*x+e)*c^(1/2)*2^(1/2)/(c-c*sin(f*x+e))^(1/2))/a/c^(5/2)/f*2^(1/2)-5/8*sec(f*x+e)/a/c^2/f/(c-c*sin(f*x+e))^(
1/2)

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Rubi [A]  time = 0.27, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {2736, 2681, 2687, 2650, 2649, 206} \[ -\frac {5 \sec (e+f x)}{8 a c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {15 \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{32 \sqrt {2} a c^{5/2} f}+\frac {15 \cos (e+f x)}{32 a c f (c-c \sin (e+f x))^{3/2}}+\frac {\sec (e+f x)}{4 a c f (c-c \sin (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(15*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(32*Sqrt[2]*a*c^(5/2)*f) + (15*Cos[e +
f*x])/(32*a*c*f*(c - c*Sin[e + f*x])^(3/2)) + Sec[e + f*x]/(4*a*c*f*(c - c*Sin[e + f*x])^(3/2)) - (5*Sec[e +
f*x])/(8*a*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 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)) (c-c \sin (e+f x))^{5/2}} \, dx &=\frac {\int \frac {\sec ^2(e+f x)}{(c-c \sin (e+f x))^{3/2}} \, dx}{a c}\\ &=\frac {\sec (e+f x)}{4 a c f (c-c \sin (e+f x))^{3/2}}+\frac {5 \int \frac {\sec ^2(e+f x)}{\sqrt {c-c \sin (e+f x)}} \, dx}{8 a c^2}\\ &=\frac {\sec (e+f x)}{4 a c f (c-c \sin (e+f x))^{3/2}}-\frac {5 \sec (e+f x)}{8 a c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {15 \int \frac {1}{(c-c \sin (e+f x))^{3/2}} \, dx}{16 a c}\\ &=\frac {15 \cos (e+f x)}{32 a c f (c-c \sin (e+f x))^{3/2}}+\frac {\sec (e+f x)}{4 a c f (c-c \sin (e+f x))^{3/2}}-\frac {5 \sec (e+f x)}{8 a c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {15 \int \frac {1}{\sqrt {c-c \sin (e+f x)}} \, dx}{64 a c^2}\\ &=\frac {15 \cos (e+f x)}{32 a c f (c-c \sin (e+f x))^{3/2}}+\frac {\sec (e+f x)}{4 a c f (c-c \sin (e+f x))^{3/2}}-\frac {5 \sec (e+f x)}{8 a c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {15 \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 )}{32 a c^2 f}\\ &=\frac {15 \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{32 \sqrt {2} a c^{5/2} f}+\frac {15 \cos (e+f x)}{32 a c f (c-c \sin (e+f x))^{3/2}}+\frac {\sec (e+f x)}{4 a c f (c-c \sin (e+f x))^{3/2}}-\frac {5 \sec (e+f x)}{8 a c^2 f \sqrt {c-c \sin (e+f x)}}\\ \end {align*}

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Mathematica [C]  time = 0.84, size = 162, normalized size = 1.04 \[ \frac {\left (\frac {1}{128}+\frac {i}{128}\right ) \cos (e+f x) \left ((1-i) (40 \sin (e+f x)+15 \cos (2 (e+f x))-9)-60 \sqrt [4]{-1} \tan ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (\tan \left (\frac {1}{4} (e+f x)\right )+1\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )\right )}{a c^2 f (\sin (e+f x)-1)^2 (\sin (e+f x)+1) \sqrt {c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1/128 + I/128)*Cos[e + f*x]*(-60*(-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]) + (1 - I)*(-9 + 15*Cos[2*(e + f*x)] + 40*S
in[e + f*x])))/(a*c^2*f*(-1 + Sin[e + f*x])^2*(1 + Sin[e + f*x])*Sqrt[c - c*Sin[e + f*x]])

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fricas [A]  time = 0.47, size = 241, normalized size = 1.54 \[ \frac {15 \, \sqrt {2} {\left (\cos \left (f x + e\right )^{3} + 2 \, \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, \cos \left (f x + e\right )\right )} \sqrt {c} \log \left (-\frac {c \cos \left (f x + e\right )^{2} + 2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} \sqrt {c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )} + 3 \, c \cos \left (f x + e\right ) + {\left (c \cos \left (f x + e\right ) - 2 \, c\right )} \sin \left (f x + e\right ) + 2 \, c}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) - 4 \, {\left (15 \, \cos \left (f x + e\right )^{2} + 20 \, \sin \left (f x + e\right ) - 12\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{128 \, {\left (a c^{3} f \cos \left (f x + e\right )^{3} + 2 \, a c^{3} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a c^{3} f \cos \left (f x + e\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/128*(15*sqrt(2)*(cos(f*x + e)^3 + 2*cos(f*x + e)*sin(f*x + e) - 2*cos(f*x + e))*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*(15*cos(f*x + e)^2 + 20*sin(f*x + e) - 12)*sqrt(-c*sin(f*x + e) + c))/(a*c^3*f*cos(f*x + e)^3 + 2*a*c^3*f*co
s(f*x + e)*sin(f*x + e) - 2*a*c^3*f*cos(f*x + e))

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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(1/(a+a*sin(f*x+e))/(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*(-1/8*(sqrt(
c)*tan((f*x+exp(1))/2)-sqrt(c)-sqrt(c*tan((f*x+exp(1))/2)^2+c))/c^2/a/(-(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*t
an((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)/sign(ta
n((f*x+exp(1))/2)-1)+1/64*(-41*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^7-127*sqrt(c)*(-
sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^6-91*c*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((
f*x+exp(1))/2)^2+c))^5+143*sqrt(c)*c*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^4-3*c^2*(-
sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^3+47*c^3*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan
((f*x+exp(1))/2)^2+c))-93*sqrt(c)*c^2*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^2-11*sqrt
(c)*c^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+e
xp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))+c)^4/a/sign(tan((f*x+exp(1))/2)-1)+15/64*atan((-sqrt(c)*tan((f*x+ex
p(1))/2)+sqrt(c)+sqrt(c*tan((f*x+exp(1))/2)^2+c))/sqrt(2)/sqrt(-c))/sqrt(2)/c^2/sqrt(-c)/a/sign(tan((f*x+exp(1
))/2)-1))

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maple [A]  time = 1.52, size = 210, normalized size = 1.35 \[ -\frac {15 \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (\sin ^{2}\left (f x +e \right )\right ) c^{2}-30 c^{\frac {5}{2}} \left (\sin ^{2}\left (f x +e \right )\right )-30 \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sin \left (f x +e \right ) c^{2}+40 c^{\frac {5}{2}} \sin \left (f x +e \right )+15 \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}+6 c^{\frac {5}{2}}}{64 c^{\frac {9}{2}} a \left (\sin \left (f x +e \right )-1\right ) \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/64/c^(9/2)/a*(15*(c*(1+sin(f*x+e)))^(1/2)*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^2-30*c^(5/2)*sin(f*x+e)^2-30*(c*(1+sin(f*x+e)))^(1/2)*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+40*c^(5/2)*sin(f*x+e)+15*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)*arctanh(1/2*(c*(1+s
in(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*c^2+6*c^(5/2))/(sin(f*x+e)-1)/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (a \sin \left (f x + e\right ) + a\right )} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{\left (a+a\,\sin \left (e+f\,x\right )\right )\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {1}{c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c} \sin ^{3}{\left (e + f x \right )} - c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c} \sin ^{2}{\left (e + f x \right )} - c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c} \sin {\left (e + f x \right )} + c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c}}\, dx}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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