3.337 \(\int \frac {1}{(a+a \sin (e+f x))^3 \sqrt {c-c \sin (e+f x)}} \, dx\)

Optimal. Leaf size=160 \[ -\frac {\sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 c^3 f}-\frac {\sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{6 a^3 c^2 f}-\frac {\sec (e+f x) \sqrt {c-c \sin (e+f x)}}{4 a^3 c f}+\frac {\tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{4 \sqrt {2} a^3 \sqrt {c} f} \]

[Out]

-1/6*sec(f*x+e)^3*(c-c*sin(f*x+e))^(3/2)/a^3/c^2/f-1/5*sec(f*x+e)^5*(c-c*sin(f*x+e))^(5/2)/a^3/c^3/f+1/8*arcta
nh(1/2*cos(f*x+e)*c^(1/2)*2^(1/2)/(c-c*sin(f*x+e))^(1/2))/a^3/f*2^(1/2)/c^(1/2)-1/4*sec(f*x+e)*(c-c*sin(f*x+e)
)^(1/2)/a^3/c/f

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Rubi [A]  time = 0.29, antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2736, 2675, 2649, 206} \[ -\frac {\sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 c^3 f}-\frac {\sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{6 a^3 c^2 f}-\frac {\sec (e+f x) \sqrt {c-c \sin (e+f x)}}{4 a^3 c f}+\frac {\tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{4 \sqrt {2} a^3 \sqrt {c} f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])]/(4*Sqrt[2]*a^3*Sqrt[c]*f) - (Sec[e + f*x]*S
qrt[c - c*Sin[e + f*x]])/(4*a^3*c*f) - (Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(3/2))/(6*a^3*c^2*f) - (Sec[e + f*
x]^5*(c - c*Sin[e + f*x])^(5/2))/(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 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 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 \sqrt {c-c \sin (e+f x)}} \, dx &=\frac {\int \sec ^6(e+f x) (c-c \sin (e+f x))^{5/2} \, dx}{a^3 c^3}\\ &=-\frac {\sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 c^3 f}+\frac {\int \sec ^4(e+f x) (c-c \sin (e+f x))^{3/2} \, dx}{2 a^3 c^2}\\ &=-\frac {\sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{6 a^3 c^2 f}-\frac {\sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 c^3 f}+\frac {\int \sec ^2(e+f x) \sqrt {c-c \sin (e+f x)} \, dx}{4 a^3 c}\\ &=-\frac {\sec (e+f x) \sqrt {c-c \sin (e+f x)}}{4 a^3 c f}-\frac {\sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{6 a^3 c^2 f}-\frac {\sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 c^3 f}+\frac {\int \frac {1}{\sqrt {c-c \sin (e+f x)}} \, dx}{8 a^3}\\ &=-\frac {\sec (e+f x) \sqrt {c-c \sin (e+f x)}}{4 a^3 c f}-\frac {\sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{6 a^3 c^2 f}-\frac {\sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 c^3 f}-\frac {\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 )}{4 a^3 f}\\ &=\frac {\tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{4 \sqrt {2} a^3 \sqrt {c} f}-\frac {\sec (e+f x) \sqrt {c-c \sin (e+f x)}}{4 a^3 c f}-\frac {\sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{6 a^3 c^2 f}-\frac {\sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 c^3 f}\\ \end {align*}

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Mathematica [C]  time = 0.61, size = 189, normalized size = 1.18 \[ \frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right ) \left (-15 \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^4-10 \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^2+(-15-15 i) \sqrt [4]{-1} \tan ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (\tan \left (\frac {1}{4} (e+f x)\right )+1\right )\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^5-12\right )}{60 a^3 f (\sin (e+f x)+1)^3 \sqrt {c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + a*Sin[e + f*x])^3*Sqrt[c - c*Sin[e + f*x]]),x]

[Out]

((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-12 - 10*(Cos[(e + f*x)/2] + Sin
[(e + f*x)/2])^2 - 15*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4 - (15 + 15*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])^5))/(60*a^3*f*(1 + Sin[e + f*x])^3*Sqrt[c
- c*Sin[e + f*x]])

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fricas [A]  time = 0.48, size = 241, normalized size = 1.51 \[ \frac {15 \, \sqrt {2} {\left (\cos \left (f x + e\right )^{3} - 2 \, \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, \cos \left (f x + e\right )\right )} \sqrt {c} \log \left (-\frac {c \cos \left (f x + e\right )^{2} + 2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} \sqrt {c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )} + 3 \, c \cos \left (f x + e\right ) + {\left (c \cos \left (f x + e\right ) - 2 \, c\right )} \sin \left (f x + e\right ) + 2 \, c}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) - 4 \, {\left (15 \, \cos \left (f x + e\right )^{2} - 40 \, \sin \left (f x + e\right ) - 52\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{240 \, {\left (a^{3} c f \cos \left (f x + e\right )^{3} - 2 \, a^{3} c f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a^{3} c f \cos \left (f x + e\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(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 - 40*sin(f*x + e) - 52)*sqrt(-c*sin(f*x + e) + c))/(a^3*c*f*cos(f*x + e)^3 - 2*a^3*c*f*co
s(f*x + e)*sin(f*x + e) - 2*a^3*c*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))^3/(c-c*sin(f*x+e))^(1/2),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*((870*c*sqrt(2
)*atan(sqrt(c)/sqrt(-c))-1230*c*atan(sqrt(c)/sqrt(-c))-850*sqrt(-c)*sqrt(2)*sqrt(c)+1203*sqrt(-c)*sqrt(c))/(49
20*a^3*c*sqrt(-c)*sqrt(2)-6960*a^3*c*sqrt(-c))*sign(tan((f*x+exp(1))/2)-1)+2*(1/120*(105*(-sqrt(c)*tan((f*x+ex
p(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^9-435*sqrt(c)*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/
2)^2+c))^8+580*c*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^7+620*sqrt(c)*c*(-sqrt(c)*tan(
(f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^6-1258*c^2*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(
1))/2)^2+c))^5-490*sqrt(c)*c^2*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^4+900*c^3*(-sqrt
(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^3+265*c^4*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f
*x+exp(1))/2)^2+c))+860*sqrt(c)*c^3*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^2+37*sqrt(c
)*c^4)/a^3/(-(-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/sign(tan((f*x+exp(1))/2)-1)+1/8*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/sqrt(-c)/sign(tan((f*x+exp(1))/2)-1)
))

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maple [A]  time = 1.18, size = 122, normalized size = 0.76 \[ -\frac {\left (\sin \left (f x +e \right )-1\right ) \left (-30 c^{\frac {11}{2}} \left (\sin ^{2}\left (f x +e \right )\right )-80 c^{\frac {11}{2}} \sin \left (f x +e \right )-74 c^{\frac {11}{2}}+15 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{3} \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}}\right )}{120 a^{3} c^{\frac {11}{2}} \left (1+\sin \left (f x +e \right )\right )^{2} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f} \]

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))^(1/2),x)

[Out]

-1/120*(sin(f*x+e)-1)*(-30*c^(11/2)*sin(f*x+e)^2-80*c^(11/2)*sin(f*x+e)-74*c^(11/2)+15*2^(1/2)*arctanh(1/2*(c*
(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*c^3*(c*(1+sin(f*x+e)))^(5/2))/a^3/c^(11/2)/(1+sin(f*x+e))^2/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))^(1/2),x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^3\,\sqrt {c-c\,\sin \left (e+f\,x\right )}} \,d x \]

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))^(1/2)),x)

[Out]

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

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

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))**(1/2),x)

[Out]

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

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