\(\int \frac {a+a \sin (e+f x)}{(c-c \sin (e+f x))^{3/2}} \, dx\) [295]

Optimal result

Integrand size = 26, antiderivative size = 76 \[ \int \frac {a+a \sin (e+f x)}{(c-c \sin (e+f x))^{3/2}} \, dx=-\frac {a \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{\sqrt {2} c^{3/2} f}+\frac {a \cos (e+f x)}{f (c-c \sin (e+f x))^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2815, 2759, 2728, 212} \[ \int \frac {a+a \sin (e+f x)}{(c-c \sin (e+f x))^{3/2}} \, dx=\frac {a \cos (e+f x)}{f (c-c \sin (e+f x))^{3/2}}-\frac {a \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{\sqrt {2} c^{3/2} f} \]

[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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2728

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 2759

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 2815

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*} \text {integral}& = (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 \text {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 \text {arctanh}\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] (verified)

Time = 1.62 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.41 \[ \int \frac {a+a \sin (e+f x)}{(c-c \sin (e+f x))^{3/2}} \, dx=\frac {a \sec (e+f x) \left (2 \sqrt {c} (1+\sin (e+f x))-\sqrt {2} \arctan \left (\frac {\sqrt {-c (1+\sin (e+f x))}}{\sqrt {2} \sqrt {c}}\right ) (-1+\sin (e+f x)) \sqrt {-c (1+\sin (e+f x))}\right )}{2 c^{3/2} f \sqrt {c-c \sin (e+f x)}} \]

[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]])

Maple [A] (verified)

Time = 1.89 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.58

method	result	size
default	\(\frac {a \left (\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c \sin \left (f x +e \right )-\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c +2 \sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {c}\right ) \sqrt {c \left (\sin \left (f x +e \right )+1\right )}}{2 c^{\frac {5}{2}} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(120\)
parts	\(-\frac {a \left (\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2} \sin \left (f x +e \right )-\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}-2 \sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, c^{\frac {3}{2}}\right ) \sqrt {c \left (\sin \left (f x +e \right )+1\right )}}{4 c^{\frac {7}{2}} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}+\frac {a \left (3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c \sin \left (f x +e \right )-3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c +2 \sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {c}\right ) \sqrt {c \left (\sin \left (f x +e \right )+1\right )}}{4 c^{\frac {5}{2}} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(245\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 255 vs. \(2 (65) = 130\).

Time = 0.36 (sec) , antiderivative size = 255, normalized size of antiderivative = 3.36 \[ \int \frac {a+a \sin (e+f x)}{(c-c \sin (e+f x))^{3/2}} \, dx=\frac {\frac {\sqrt {2} {\left (a c \cos \left (f x + e\right )^{2} - a c \cos \left (f x + e\right ) - 2 \, a c + {\left (a c \cos \left (f x + e\right ) + 2 \, a c\right )} \sin \left (f x + e\right )\right )} \log \left (-\frac {\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) - 2\right )} \sin \left (f x + e\right ) - \frac {2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )}}{\sqrt {c}} + 3 \, \cos \left (f x + e\right ) + 2}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right )}{\sqrt {c}} - 4 \, {\left (a \cos \left (f x + e\right ) + a \sin \left (f x + e\right ) + a\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{4 \, {\left (c^{2} f \cos \left (f x + e\right )^{2} - c^{2} f \cos \left (f x + e\right ) - 2 \, c^{2} f + {\left (c^{2} f \cos \left (f x + e\right ) + 2 \, c^{2} f\right )} \sin \left (f x + e\right )\right )}} \]

[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))

Sympy [F]

\[ \int \frac {a+a \sin (e+f x)}{(c-c \sin (e+f x))^{3/2}} \, dx=a \left (\int \frac {\sin {\left (e + f x \right )}}{- c \sqrt {- c \sin {\left (e + f x \right )} + c} \sin {\left (e + f x \right )} + c \sqrt {- c \sin {\left (e + f x \right )} + c}}\, dx + \int \frac {1}{- c \sqrt {- c \sin {\left (e + f x \right )} + c} \sin {\left (e + f x \right )} + c \sqrt {- c \sin {\left (e + f x \right )} + c}}\, dx\right ) \]

[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))

Maxima [F]

\[ \int \frac {a+a \sin (e+f x)}{(c-c \sin (e+f x))^{3/2}} \, dx=\int { \frac {a \sin \left (f x + e\right ) + a}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \]

[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)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 211 vs. \(2 (65) = 130\).

Time = 0.34 (sec) , antiderivative size = 211, normalized size of antiderivative = 2.78 \[ \int \frac {a+a \sin (e+f x)}{(c-c \sin (e+f x))^{3/2}} \, dx=-\frac {\frac {2 \, \sqrt {2} a \log \left (-\frac {\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1}\right )}{c^{\frac {3}{2}} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {\sqrt {2} a {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}}{c^{\frac {3}{2}} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {\sqrt {2} {\left (a \sqrt {c} + \frac {2 \, a \sqrt {c} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1}\right )} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}}{c^{2} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{8 \, f} \]

[In]

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

[Out]

-1/8*(2*sqrt(2)*a*log(-(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1))/(c^(3/2)*sgn
(sin(-1/4*pi + 1/2*f*x + 1/2*e))) + sqrt(2)*a*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(c^(3/2)*(cos(-1/4*pi + 1/2
*f*x + 1/2*e) + 1)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))) - sqrt(2)*(a*sqrt(c) + 2*a*sqrt(c)*(cos(-1/4*pi + 1/2*
f*x + 1/2*e) - 1)/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1))*(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)/(c^2*(cos(-1/4*pi
+ 1/2*f*x + 1/2*e) - 1)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))))/f

Mupad [F(-1)]

Timed out. \[ \int \frac {a+a \sin (e+f x)}{(c-c \sin (e+f x))^{3/2}} \, dx=\int \frac {a+a\,\sin \left (e+f\,x\right )}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \]

[In]

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

[Out]

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