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

Optimal. Leaf size=124 \[ -\frac{\sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{3 a^2 c^2 f}-\frac{\sec (e+f x) \sqrt{c-c \sin (e+f x)}}{2 a^2 c f}+\frac{\tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{2 \sqrt{2} a^2 \sqrt{c} f} \]

[Out]

ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])]/(2*Sqrt[2]*a^2*Sqrt[c]*f) - (Sec[e + f*x]*S
qrt[c - c*Sin[e + f*x]])/(2*a^2*c*f) - (Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(3/2))/(3*a^2*c^2*f)

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Rubi [A]  time = 0.224182, antiderivative size = 124, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2736, 2675, 2649, 206} \[ -\frac{\sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{3 a^2 c^2 f}-\frac{\sec (e+f x) \sqrt{c-c \sin (e+f x)}}{2 a^2 c f}+\frac{\tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{2 \sqrt{2} a^2 \sqrt{c} f} \]

Antiderivative was successfully verified.

[In]

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

Mathematica [C]  time = 0.481597, size = 109, normalized size = 0.88 \[ \frac{\cos (e+f x) \left (-3 \sin (e+f x)+(-3-3 i) \sqrt [4]{-1} \tan ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt [4]{-1} \left (\tan \left (\frac{1}{4} (e+f x)\right )+1\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3-5\right )}{6 a^2 f (\sin (e+f x)+1)^2 \sqrt{c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[e + f*x]*(-5 - (3 + 3*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])^3 - 3*Sin[e + f*x]))/(6*a^2*f*(1 + Sin[e + f*x])^2*Sqrt[c - c*Sin[e + f*x]])

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Maple [A]  time = 0.681, size = 109, normalized size = 0.9 \begin{align*} -{\frac{-1+\sin \left ( fx+e \right ) }{12\,{a}^{2} \left ( 1+\sin \left ( fx+e \right ) \right ) \cos \left ( fx+e \right ) f} \left ( -6\,{c}^{7/2}\sin \left ( fx+e \right ) +3\,\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} \left ( c \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{3/2}-10\,{c}^{7/2} \right ){c}^{-{\frac{7}{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))^2/(c-c*sin(f*x+e))^(1/2),x)

[Out]

-1/12*(-1+sin(f*x+e))*(-6*c^(7/2)*sin(f*x+e)+3*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*c
^2*(c*(1+sin(f*x+e)))^(3/2)-10*c^(7/2))/a^2/c^(7/2)/(1+sin(f*x+e))/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (a \sin \left (f x + e\right ) + a\right )}^{2} \sqrt{-c \sin \left (f x + e\right ) + c}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.15062, size = 563, normalized size = 4.54 \begin{align*} \frac{3 \, \sqrt{2}{\left (\cos \left (f x + e\right ) \sin \left (f x + e\right ) + \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 \, \sqrt{-c \sin \left (f x + e\right ) + c}{\left (3 \, \sin \left (f x + e\right ) + 5\right )}}{24 \,{\left (a^{2} c f \cos \left (f x + e\right ) \sin \left (f x + e\right ) + a^{2} c f \cos \left (f x + e\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*sqrt(2)*(cos(f*x + e)*sin(f*x + e) + 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*sqrt(-c*sin(f*x + e)
+ c)*(3*sin(f*x + e) + 5))/(a^2*c*f*cos(f*x + e)*sin(f*x + e) + a^2*c*f*cos(f*x + e))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{1}{\sqrt{- c \sin{\left (e + f x \right )} + c} \sin ^{2}{\left (e + f x \right )} + 2 \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^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.70229, size = 664, normalized size = 5.35 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*((30*sqrt(2)*c*arctan(sqrt(c)/sqrt(-c)) - 42*c*arctan(sqrt(c)/sqrt(-c)) - 15*sqrt(2)*sqrt(-c)*sqrt(c) + 22
*sqrt(-c)*sqrt(c))*sgn(tan(1/2*f*x + 1/2*e) - 1)/(7*sqrt(2)*a^2*sqrt(-c)*c - 10*a^2*sqrt(-c)*c) + 3*sqrt(2)*ar
ctan(-1/2*sqrt(2)*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c) - sqrt(c))/sqrt(-c))/(a^2
*sqrt(-c)*sgn(tan(1/2*f*x + 1/2*e) - 1)) + 2*(9*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2
+ c))^5 + 15*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^4*sqrt(c) - 10*(sqrt(c)*tan(1
/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^3*c - 30*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*
f*x + 1/2*e)^2 + c))^2*c^(3/2) + 21*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))*c^2 -
5*c^(5/2))/(((sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^2 + 2*(sqrt(c)*tan(1/2*f*x +
1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))*sqrt(c) - c)^3*a^2*sgn(tan(1/2*f*x + 1/2*e) - 1)))/f