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

Optimal. Leaf size=76 \[ \frac{a \cos (e+f x)}{f (c-c \sin (e+f x))^{3/2}}-\frac{a \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{\sqrt{2} c^{3/2} f} \]

[Out]

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

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Rubi [A]  time = 0.140707, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {2736, 2680, 2649, 206} \[ \frac{a \cos (e+f x)}{f (c-c \sin (e+f x))^{3/2}}-\frac{a \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{\sqrt{2} c^{3/2} f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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)}{(c-c \sin (e+f x))^{3/2}} \, dx &=(a c) \int \frac{\cos ^2(e+f x)}{(c-c \sin (e+f x))^{5/2}} \, dx\\ &=\frac{a \cos (e+f x)}{f (c-c \sin (e+f x))^{3/2}}-\frac{a \int \frac{1}{\sqrt{c-c \sin (e+f x)}} \, dx}{2 c}\\ &=\frac{a \cos (e+f x)}{f (c-c \sin (e+f x))^{3/2}}+\frac{a \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{a \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{\sqrt{2} c^{3/2} f}+\frac{a \cos (e+f x)}{f (c-c \sin (e+f x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.622697, size = 107, normalized size = 1.41 \[ \frac{a \sec (e+f x) \left (2 \sqrt{c} (\sin (e+f x)+1)-\sqrt{2} (\sin (e+f x)-1) \sqrt{-c (\sin (e+f x)+1)} \tan ^{-1}\left (\frac{\sqrt{-c (\sin (e+f x)+1)}}{\sqrt{2} \sqrt{c}}\right )\right )}{2 c^{3/2} f \sqrt{c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Sec[e + f*x]*(2*Sqrt[c]*(1 + Sin[e + f*x]) - Sqrt[2]*ArcTan[Sqrt[-(c*(1 + Sin[e + f*x]))]/(Sqrt[2]*Sqrt[c])
]*(-1 + Sin[e + f*x])*Sqrt[-(c*(1 + Sin[e + f*x]))]))/(2*c^(3/2)*f*Sqrt[c - c*Sin[e + f*x]])

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Maple [A]  time = 0.425, size = 120, normalized size = 1.6 \begin{align*}{\frac{a}{2\,f\cos \left ( fx+e \right ) } \left ( \sqrt{2}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }{\frac{1}{\sqrt{c}}}} \right ) c\sin \left ( fx+e \right ) -\sqrt{2}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }{\frac{1}{\sqrt{c}}}} \right ) c+2\,\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }\sqrt{c} \right ) \sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }{c}^{-{\frac{5}{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((a+a*sin(f*x+e))/(c-c*sin(f*x+e))^(3/2),x)

[Out]

1/2/c^(5/2)*a*(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)-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))*(c*(1+sin(f*x+e)))^(1/2)/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{a \sin \left (f x + e\right ) + a}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.06673, size = 687, normalized size = 9.04 \begin{align*} \frac{\frac{\sqrt{2}{\left (a c \cos \left (f x + e\right )^{2} - a c \cos \left (f x + e\right ) - 2 \, a c +{\left (a c \cos \left (f x + e\right ) + 2 \, a 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 \cos \left (f x + e\right ) + a \sin \left (f x + e\right ) + a\right )} \sqrt{-c \sin \left (f x + e\right ) + c}}{4 \,{\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(sqrt(2)*(a*c*cos(f*x + e)^2 - a*c*cos(f*x + e) - 2*a*c + (a*c*cos(f*x + e) + 2*a*c)*sin(f*x + e))*log(-(c
os(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*cos(f*x + e) + a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c))/(c^2*f*cos(f*x + e)^2 - c^2*f*co
s(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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a \left (\int \frac{\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{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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*(Integral(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) + 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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Giac [B]  time = 2.38424, size = 428, normalized size = 5.63 \begin{align*} -\frac{\frac{\sqrt{2} a \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} c \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )} - \frac{2 \,{\left (3 \,{\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}\right )}^{3} a -{\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}\right )}^{2} a \sqrt{c} -{\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}\right )} a c - a c^{\frac{3}{2}}\right )}}{{\left ({\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}\right )}^{2} - 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}\right )} \sqrt{c} - c\right )}^{2} c \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(sqrt(2)*a*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) - sqrt(c))/
sqrt(-c))/(sqrt(-c)*c*sgn(tan(1/2*f*x + 1/2*e) - 1)) - 2*(3*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x
+ 1/2*e)^2 + c))^3*a - (sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^2*a*sqrt(c) - (sqr
t(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))*a*c - a*c^(3/2))/(((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)^2*c*sgn(tan(1/2*f*x + 1/2*e) - 1)))/f