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]

-2/3*a^2*c*cos(f*x+e)^3/f/(c-c*sin(f*x+e))^(3/2)+4*a^2*arctanh(1/2*cos(f*x+e)*c^(1/2)*2^(1/2)/(c-c*sin(f*x+e))
^(1/2))*2^(1/2)/f/c^(1/2)-4*a^2*cos(f*x+e)/f/(c-c*sin(f*x+e))^(1/2)

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Rubi [A]  time = 0.24, antiderivative size = 115, normalized size of antiderivative = 1.00, 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 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 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 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 {(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*}

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Mathematica [C]  time = 0.47, 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]

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

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fricas [B]  time = 0.45, size = 238, normalized size = 2.07 \[ \frac {2 \, {\left (\frac {3 \, \sqrt {2} {\left (a^{2} c \cos \left (f x + e\right ) - a^{2} c \sin \left (f x + e\right ) + a^{2} c\right )} \log \left (-\frac {\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) - 2\right )} \sin \left (f x + e\right ) + \frac {2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )}}{\sqrt {c}} + 3 \, \cos \left (f x + e\right ) + 2}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right )}{\sqrt {c}} + {\left (a^{2} \cos \left (f x + e\right )^{2} - 7 \, a^{2} \cos \left (f x + e\right ) - 8 \, a^{2} - {\left (a^{2} \cos \left (f x + e\right ) + 8 \, a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}\right )}}{3 \, {\left (c f \cos \left (f x + e\right ) - c f \sin \left (f x + e\right ) + c f\right )}} \]

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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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((a+a*sin(f*x+e))^2/(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)Unable to check si
gn: (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/sqrt(c*tan((f*x+exp(1))/2)^2+c)/(c*tan((f*x+exp(1))/2)^2+c)*(tan((f*x+exp(1))/2)*(tan((f*x+exp(1))/
2)*(3/2*a^2*c/sign(tan((f*x+exp(1))/2)-1)+7/6*a^2*c*tan((f*x+exp(1))/2)/sign(tan((f*x+exp(1))/2)-1))+3/2*a^2*c
/sign(tan((f*x+exp(1))/2)-1))+7/6*a^2*c/sign(tan((f*x+exp(1))/2)-1))+8*a^2*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)/sqrt(-c)/sign(tan((f*x+exp(1))/2)-1)+(-24*a
^2*c*atan(sqrt(c)/sqrt(-c))-16*a^2*sqrt(-c)*sqrt(c))/3/c/sqrt(-c)/sqrt(2)*sign(tan((f*x+exp(1))/2)-1))

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maple [A]  time = 0.92, size = 112, normalized size = 0.97 \[ -\frac {2 \left (\sin \left (f x +e \right )-1\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, a^{2} \left (6 c^{\frac {3}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right )-\left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}}-6 \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, c \right )}{3 c^{2} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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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