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

Optimal. Leaf size=115 \[ -\frac{2 a^2 c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{3/2}}-\frac{4 a^2 \cos (e+f x)}{f \sqrt{c-c \sin (e+f x)}}+\frac{4 \sqrt{2} a^2 \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{\sqrt{c} f} \]

[Out]

(4*Sqrt[2]*a^2*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(Sqrt[c]*f) - (2*a^2*c*Cos[
e + f*x]^3)/(3*f*(c - c*Sin[e + f*x])^(3/2)) - (4*a^2*Cos[e + f*x])/(f*Sqrt[c - c*Sin[e + f*x]])

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Rubi [A]  time = 0.244165, antiderivative size = 115, 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, 2679, 2649, 206} \[ -\frac{2 a^2 c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{3/2}}-\frac{4 a^2 \cos (e+f x)}{f \sqrt{c-c \sin (e+f x)}}+\frac{4 \sqrt{2} a^2 \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{\sqrt{c} f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Sqrt[2]*a^2*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(Sqrt[c]*f) - (2*a^2*c*Cos[
e + f*x]^3)/(3*f*(c - c*Sin[e + f*x])^(3/2)) - (4*a^2*Cos[e + f*x])/(f*Sqrt[c - c*Sin[e + f*x]])

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 2679

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(g*(g*
Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + p)), x] + Dist[(g^2*(p - 1))/(a*(m + p)), 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]
&& LtQ[m, -1] && GtQ[p, 1] && (GtQ[m, -2] || EqQ[2*m + p + 1, 0] || (EqQ[m, -2] && IntegerQ[p])) && NeQ[m + p,
0] && IntegersQ[2*m, 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{(a+a \sin (e+f x))^2}{\sqrt{c-c \sin (e+f x)}} \, dx &=\left (a^2 c^2\right ) \int \frac{\cos ^4(e+f x)}{(c-c \sin (e+f x))^{5/2}} \, dx\\ &=-\frac{2 a^2 c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{3/2}}+\left (2 a^2 c\right ) \int \frac{\cos ^2(e+f x)}{(c-c \sin (e+f x))^{3/2}} \, dx\\ &=-\frac{2 a^2 c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{3/2}}-\frac{4 a^2 \cos (e+f x)}{f \sqrt{c-c \sin (e+f x)}}+\left (4 a^2\right ) \int \frac{1}{\sqrt{c-c \sin (e+f x)}} \, dx\\ &=-\frac{2 a^2 c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{3/2}}-\frac{4 a^2 \cos (e+f x)}{f \sqrt{c-c \sin (e+f x)}}-\frac{\left (8 a^2\right ) \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 )}{f}\\ &=\frac{4 \sqrt{2} a^2 \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{\sqrt{c} f}-\frac{2 a^2 c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{3/2}}-\frac{4 a^2 \cos (e+f x)}{f \sqrt{c-c \sin (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 0.443477, size = 130, normalized size = 1.13 \[ -\frac{a^2 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (15 \sin \left (\frac{1}{2} (e+f x)\right )+\sin \left (\frac{3}{2} (e+f x)\right )+15 \cos \left (\frac{1}{2} (e+f x)\right )-\cos \left (\frac{3}{2} (e+f x)\right )+(24+24 i) \sqrt [4]{-1} \tan ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt [4]{-1} \left (\tan \left (\frac{1}{4} (e+f x)\right )+1\right )\right )\right )}{3 f \sqrt{c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*((24 + 24*I)*(-1)^(1/4)*ArcTan[(1/2 + I/2)*(-1)^(1/4)*(1 + Tan[(e
+ f*x)/4])] + 15*Cos[(e + f*x)/2] - Cos[(3*(e + f*x))/2] + 15*Sin[(e + f*x)/2] + Sin[(3*(e + f*x))/2]))/(3*f*S
qrt[c - c*Sin[e + f*x]])

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Maple [A]  time = 0.674, size = 112, normalized size = 1. \begin{align*} -{\frac{ \left ( -2+2\,\sin \left ( fx+e \right ) \right ){a}^{2}}{3\,{c}^{2}\cos \left ( fx+e \right ) f}\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) } \left ( 6\,{c}^{3/2}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }\sqrt{2}}{\sqrt{c}}} \right ) - \left ( c \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{{\frac{3}{2}}}-6\,\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }c \right ){\frac{1}{\sqrt{c-c\sin \left ( fx+e \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(-1+sin(f*x+e))*(c*(1+sin(f*x+e)))^(1/2)*a^2*(6*c^(3/2)*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1
/2)/c^(1/2))-(c*(1+sin(f*x+e)))^(3/2)-6*(c*(1+sin(f*x+e)))^(1/2)*c)/c^2/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{{\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((a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.09188, size = 630, normalized size = 5.48 \begin{align*} \frac{2 \,{\left (\frac{3 \, \sqrt{2}{\left (a^{2} c \cos \left (f x + e\right ) - a^{2} c \sin \left (f x + e\right ) + a^{2} c\right )} \log \left (-\frac{\cos \left (f x + e\right )^{2} +{\left (\cos \left (f x + e\right ) - 2\right )} \sin \left (f x + e\right ) + \frac{2 \, \sqrt{2} \sqrt{-c \sin \left (f x + e\right ) + c}{\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )}}{\sqrt{c}} + 3 \, \cos \left (f x + e\right ) + 2}{\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 )}{\sqrt{c}} +{\left (a^{2} \cos \left (f x + e\right )^{2} - 7 \, a^{2} \cos \left (f x + e\right ) - 8 \, a^{2} -{\left (a^{2} \cos \left (f x + e\right ) + 8 \, a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt{-c \sin \left (f x + e\right ) + c}\right )}}{3 \,{\left (c f \cos \left (f x + e\right ) - c f \sin \left (f x + e\right ) + c f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(3*sqrt(2)*(a^2*c*cos(f*x + e) - a^2*c*sin(f*x + e) + a^2*c)*log(-(cos(f*x + e)^2 + (cos(f*x + e) - 2)*sin
(f*x + e) + 2*sqrt(2)*sqrt(-c*sin(f*x + e) + c)*(cos(f*x + e) + sin(f*x + e) + 1)/sqrt(c) + 3*cos(f*x + e) + 2
)/(cos(f*x + e)^2 + (cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2))/sqrt(c) + (a^2*cos(f*x + e)^2 - 7*a^2
*cos(f*x + e) - 8*a^2 - (a^2*cos(f*x + e) + 8*a^2)*sin(f*x + e))*sqrt(-c*sin(f*x + e) + c))/(c*f*cos(f*x + e)
- c*f*sin(f*x + e) + c*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 3.50761, size = 375, normalized size = 3.26 \begin{align*} \frac{\frac{24 \, \sqrt{2} a^{2} \arctan \left (-\frac{\sqrt{2}{\left (\sqrt{c} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - \sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + c} - \sqrt{c}\right )}}{2 \, \sqrt{-c}}\right )}{\sqrt{-c} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )} + \frac{{\left ({\left (\frac{7 \, a^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right ) \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{c^{5}} + \frac{9 \, a^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )}{c^{5}}\right )} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + \frac{9 \, a^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )}{c^{5}}\right )} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + \frac{7 \, a^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )}{c^{5}}}{{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + c\right )}^{\frac{3}{2}}} - \frac{8 \, \sqrt{2}{\left (3 \, a^{2} c^{\frac{15}{2}} \arctan \left (\frac{\sqrt{c}}{\sqrt{-c}}\right ) + a^{2} \sqrt{-c} c\right )} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )}{\sqrt{-c} c^{\frac{15}{2}}}}{3 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(24*sqrt(2)*a^2*arctan(-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) - s
qrt(c))/sqrt(-c))/(sqrt(-c)*sgn(tan(1/2*f*x + 1/2*e) - 1)) + (((7*a^2*sgn(tan(1/2*f*x + 1/2*e) - 1)*tan(1/2*f*
x + 1/2*e)/c^5 + 9*a^2*sgn(tan(1/2*f*x + 1/2*e) - 1)/c^5)*tan(1/2*f*x + 1/2*e) + 9*a^2*sgn(tan(1/2*f*x + 1/2*e
) - 1)/c^5)*tan(1/2*f*x + 1/2*e) + 7*a^2*sgn(tan(1/2*f*x + 1/2*e) - 1)/c^5)/(c*tan(1/2*f*x + 1/2*e)^2 + c)^(3/
2) - 8*sqrt(2)*(3*a^2*c^(15/2)*arctan(sqrt(c)/sqrt(-c)) + a^2*sqrt(-c)*c)*sgn(tan(1/2*f*x + 1/2*e) - 1)/(sqrt(
-c)*c^(15/2)))/f