3.303 \(\int \frac {(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=115 \[ -\frac {3 \sqrt {2} a^2 \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{c^{3/2} f}+\frac {a^2 c \cos ^3(e+f x)}{f (c-c \sin (e+f x))^{5/2}}+\frac {3 a^2 \cos (e+f x)}{c f \sqrt {c-c \sin (e+f x)}} \]

[Out]

a^2*c*cos(f*x+e)^3/f/(c-c*sin(f*x+e))^(5/2)-3*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)/c^(3/2)/f+3*a^2*cos(f*x+e)/c/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 = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {2736, 2680, 2679, 2649, 206} \[ -\frac {3 \sqrt {2} a^2 \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{c^{3/2} f}+\frac {a^2 c \cos ^3(e+f x)}{f (c-c \sin (e+f x))^{5/2}}+\frac {3 a^2 \cos (e+f x)}{c f \sqrt {c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^2/(c - c*Sin[e + f*x])^(3/2),x]

[Out]

(-3*Sqrt[2]*a^2*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(c^(3/2)*f) + (a^2*c*Cos[e
+ f*x]^3)/(f*(c - c*Sin[e + f*x])^(5/2)) + (3*a^2*Cos[e + f*x])/(c*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 2680

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(2*g*(
g*Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(2*m + p + 1)), x] + Dist[(g^2*(p - 1))/(b^2*(2*m +
p + 1)), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, g}, x] && Eq
Q[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] && NeQ[2*m + p + 1, 0] &&  !ILtQ[m + p + 1, 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}{(c-c \sin (e+f x))^{3/2}} \, dx &=\left (a^2 c^2\right ) \int \frac {\cos ^4(e+f x)}{(c-c \sin (e+f x))^{7/2}} \, dx\\ &=\frac {a^2 c \cos ^3(e+f x)}{f (c-c \sin (e+f x))^{5/2}}-\frac {1}{2} \left (3 a^2\right ) \int \frac {\cos ^2(e+f x)}{(c-c \sin (e+f x))^{3/2}} \, dx\\ &=\frac {a^2 c \cos ^3(e+f x)}{f (c-c \sin (e+f x))^{5/2}}+\frac {3 a^2 \cos (e+f x)}{c f \sqrt {c-c \sin (e+f x)}}-\frac {\left (3 a^2\right ) \int \frac {1}{\sqrt {c-c \sin (e+f x)}} \, dx}{c}\\ &=\frac {a^2 c \cos ^3(e+f x)}{f (c-c \sin (e+f x))^{5/2}}+\frac {3 a^2 \cos (e+f x)}{c f \sqrt {c-c \sin (e+f x)}}+\frac {\left (6 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 )}{c f}\\ &=-\frac {3 \sqrt {2} a^2 \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{c^{3/2} f}+\frac {a^2 c \cos ^3(e+f x)}{f (c-c \sin (e+f x))^{5/2}}+\frac {3 a^2 \cos (e+f x)}{c f \sqrt {c-c \sin (e+f x)}}\\ \end {align*}

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Mathematica [C]  time = 0.69, size = 149, normalized size = 1.30 \[ -\frac {a^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (3 \sin \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {3}{2} (e+f x)\right )+3 \cos \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {3}{2} (e+f x)\right )-(6+6 i) \sqrt [4]{-1} (\sin (e+f x)-1) \tan ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (\tan \left (\frac {1}{4} (e+f x)\right )+1\right )\right )\right )}{c f (\sin (e+f x)-1) \sqrt {c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])^2/(c - c*Sin[e + f*x])^(3/2),x]

[Out]

-((a^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(3*Cos[(e + f*x)/2] + Cos[(3*(e + f*x))/2] + 3*Sin[(e + f*x)/2] -
(6 + 6*I)*(-1)^(1/4)*ArcTan[(1/2 + I/2)*(-1)^(1/4)*(1 + Tan[(e + f*x)/4])]*(-1 + Sin[e + f*x]) - Sin[(3*(e +
f*x))/2]))/(c*f*(-1 + Sin[e + f*x])*Sqrt[c - c*Sin[e + f*x]]))

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fricas [B]  time = 0.46, size = 299, normalized size = 2.60 \[ \frac {\frac {3 \, \sqrt {2} {\left (a^{2} c \cos \left (f x + e\right )^{2} - a^{2} c \cos \left (f x + e\right ) - 2 \, a^{2} c + {\left (a^{2} c \cos \left (f x + e\right ) + 2 \, a^{2} c\right )} \sin \left (f x + e\right )\right )} \log \left (-\frac {\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) - 2\right )} \sin \left (f x + e\right ) - \frac {2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )}}{\sqrt {c}} + 3 \, \cos \left (f x + e\right ) + 2}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right )}{\sqrt {c}} - 4 \, {\left (a^{2} \cos \left (f x + e\right )^{2} + 2 \, a^{2} \cos \left (f x + e\right ) + a^{2} - {\left (a^{2} \cos \left (f x + e\right ) - a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{2 \, {\left (c^{2} f \cos \left (f x + e\right )^{2} - c^{2} f \cos \left (f x + e\right ) - 2 \, c^{2} f + {\left (c^{2} f \cos \left (f x + e\right ) + 2 \, c^{2} f\right )} \sin \left (f x + e\right )\right )}} \]

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))^(3/2),x, algorithm="fricas")

[Out]

1/2*(3*sqrt(2)*(a^2*c*cos(f*x + e)^2 - a^2*c*cos(f*x + e) - 2*a^2*c + (a^2*c*cos(f*x + e) + 2*a^2*c)*sin(f*x +
e))*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) - 4*(a^2*cos(f*x + e)^2 + 2*a^2*cos(f*x + e) + a^2 - (a^2*cos(f*x + e) - a^2)*sin(f*x + e
))*sqrt(-c*sin(f*x + e) + c))/(c^2*f*cos(f*x + e)^2 - c^2*f*cos(f*x + e) - 2*c^2*f + (c^2*f*cos(f*x + e) + 2*c
^2*f)*sin(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((a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^(3/2),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)2/f*(2*sqrt(c*tan((f*x+exp(1))/2)^2+c)*(-1/2*a^2/c/sign(tan((f*x+exp(1))/2)-1)-1/2*a
^2*tan((f*x+exp(1))/2)/c/sign(tan((f*x+exp(1))/2)-1))/(c*tan((f*x+exp(1))/2)^2+c)+2*(-(3*a^2*(-sqrt(c)*tan((f*
x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^3-a^2*c*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)
^2+c))+a^2*sqrt(c)*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^2+a^2*sqrt(c)*c)/(-(-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)^2/c/sign(tan((f*x+exp(1))/2)-1)-3*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)/c/sign(tan((f*x+exp(1))/2)-1)))

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maple [A]  time = 0.78, size = 145, normalized size = 1.26 \[ \frac {a^{2} \left (3 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c \sin \left (f x +e \right )-3 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c -2 \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {c}\, \sin \left (f x +e \right )+4 \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {c}\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}}{c^{\frac {5}{2}} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f} \]

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))^(3/2),x)

[Out]

a^2*(3*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*c*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)))^(1/2)*c^(1/2)*sin(f*x+e)+4*(c*(1+sin(f*x+e)))^(1/2)
*c^(1/2))*(c*(1+sin(f*x+e)))^(1/2)/c^(5/2)/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 \[ \int \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{2}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}}\,{d x} \]

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))^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^2}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \]

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))^(3/2),x)

[Out]

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

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

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))**(3/2),x)

[Out]

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

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