\(\int \frac {a+a \sin (e+f x)}{(c-c \sin (e+f x))^{5/2}} \, dx\) [296]

Optimal result

Integrand size = 26, antiderivative size = 113 \[ \int \frac {a+a \sin (e+f x)}{(c-c \sin (e+f x))^{5/2}} \, dx=-\frac {a \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{8 \sqrt {2} c^{5/2} f}+\frac {a \cos (e+f x)}{2 f (c-c \sin (e+f x))^{5/2}}-\frac {a \cos (e+f x)}{8 c f (c-c \sin (e+f x))^{3/2}} \]

[Out]

1/2*a*cos(f*x+e)/f/(c-c*sin(f*x+e))^(5/2)-1/8*a*cos(f*x+e)/c/f/(c-c*sin(f*x+e))^(3/2)-1/16*a*arctanh(1/2*cos(f
*x+e)*c^(1/2)*2^(1/2)/(c-c*sin(f*x+e))^(1/2))/c^(5/2)/f*2^(1/2)

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {2815, 2759, 2729, 2728, 212} \[ \int \frac {a+a \sin (e+f x)}{(c-c \sin (e+f x))^{5/2}} \, dx=-\frac {a \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{8 \sqrt {2} c^{5/2} f}-\frac {a \cos (e+f x)}{8 c f (c-c \sin (e+f x))^{3/2}}+\frac {a \cos (e+f x)}{2 f (c-c \sin (e+f x))^{5/2}} \]

[In]

Int[(a + a*Sin[e + f*x])/(c - c*Sin[e + f*x])^(5/2),x]

[Out]

-1/8*(a*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(Sqrt[2]*c^(5/2)*f) + (a*Cos[e + f
*x])/(2*f*(c - c*Sin[e + f*x])^(5/2)) - (a*Cos[e + f*x])/(8*c*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 2729

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*Cos[c + d*x]*((a + b*Sin[c + d*x])^n/(a*d
*(2*n + 1))), x] + Dist[(n + 1)/(a*(2*n + 1)), Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d},
x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

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))^{7/2}} \, dx \\ & = \frac {a \cos (e+f x)}{2 f (c-c \sin (e+f x))^{5/2}}-\frac {a \int \frac {1}{(c-c \sin (e+f x))^{3/2}} \, dx}{4 c} \\ & = \frac {a \cos (e+f x)}{2 f (c-c \sin (e+f x))^{5/2}}-\frac {a \cos (e+f x)}{8 c f (c-c \sin (e+f x))^{3/2}}-\frac {a \int \frac {1}{\sqrt {c-c \sin (e+f x)}} \, dx}{16 c^2} \\ & = \frac {a \cos (e+f x)}{2 f (c-c \sin (e+f x))^{5/2}}-\frac {a \cos (e+f x)}{8 c 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 )}{8 c^2 f} \\ & = -\frac {a \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{8 \sqrt {2} c^{5/2} f}+\frac {a \cos (e+f x)}{2 f (c-c \sin (e+f x))^{5/2}}-\frac {a \cos (e+f x)}{8 c f (c-c \sin (e+f x))^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.12 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.56 \[ \int \frac {a+a \sin (e+f x)}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {a \left (-2 \sqrt {c} (-7+\cos (2 (e+f x))-8 \sin (e+f x))+2 \sqrt {2} \arctan \left (\frac {\sqrt {-c (1+\sin (e+f x))}}{\sqrt {2} \sqrt {c}}\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \sqrt {-c (1+\sin (e+f x))}\right )}{32 c^{5/2} f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {c-c \sin (e+f x)}} \]

[In]

Integrate[(a + a*Sin[e + f*x])/(c - c*Sin[e + f*x])^(5/2),x]

[Out]

(a*(-2*Sqrt[c]*(-7 + Cos[2*(e + f*x)] - 8*Sin[e + f*x]) + 2*Sqrt[2]*ArcTan[Sqrt[-(c*(1 + Sin[e + f*x]))]/(Sqrt
[2]*Sqrt[c])]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^4*Sqrt[-(c*(1 + Sin[e + f*x]))]))/(32*c^(5/2)*f*(Cos[(e +
f*x)/2] - Sin[(e + f*x)/2])^3*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c - c*Sin[e + f*x]])

Maple [A] (verified)

Time = 2.09 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.65

method	result	size
default	\(\frac {a \left (\operatorname {arctanh}\left (\frac {\sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (\sin ^{2}\left (f x +e \right )\right ) \sqrt {2}\, c^{3}-2 \,\operatorname {arctanh}\left (\frac {\sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sin \left (f x +e \right ) \sqrt {2}\, c^{3}-2 \left (c \left (\sin \left (f x +e \right )+1\right )\right )^{\frac {3}{2}} c^{\frac {3}{2}}-4 \sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, c^{\frac {5}{2}}+\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{3}\right ) \sqrt {c \left (\sin \left (f x +e \right )+1\right )}}{16 c^{\frac {11}{2}} \left (\sin \left (f x +e \right )-1\right ) \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(187\)
parts	\(\frac {a \left (-3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (\sin ^{2}\left (f x +e \right )\right ) c^{2}+6 \sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, c^{\frac {3}{2}} \sin \left (f x +e \right )+6 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2} \sin \left (f x +e \right )-14 \sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, c^{\frac {3}{2}}-3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}\right ) \sqrt {c \left (\sin \left (f x +e \right )+1\right )}}{32 c^{\frac {9}{2}} \left (\sin \left (f x +e \right )-1\right ) \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}+\frac {a \left (5 \,\operatorname {arctanh}\left (\frac {\sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (\sin ^{2}\left (f x +e \right )\right ) \sqrt {2}\, c^{3}+12 \sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, c^{\frac {5}{2}}-10 \left (c \left (\sin \left (f x +e \right )+1\right )\right )^{\frac {3}{2}} c^{\frac {3}{2}}-10 \,\operatorname {arctanh}\left (\frac {\sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sin \left (f x +e \right ) \sqrt {2}\, c^{3}+5 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{3}\right ) \sqrt {c \left (\sin \left (f x +e \right )+1\right )}}{32 c^{\frac {11}{2}} \left (\sin \left (f x +e \right )-1\right ) \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(384\)

[In]

int((a+a*sin(f*x+e))/(c-c*sin(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/16*a*(arctanh(1/2*(c*(sin(f*x+e)+1))^(1/2)*2^(1/2)/c^(1/2))*sin(f*x+e)^2*2^(1/2)*c^3-2*arctanh(1/2*(c*(sin(f
*x+e)+1))^(1/2)*2^(1/2)/c^(1/2))*sin(f*x+e)*2^(1/2)*c^3-2*(c*(sin(f*x+e)+1))^(3/2)*c^(3/2)-4*(c*(sin(f*x+e)+1)
)^(1/2)*c^(5/2)+2^(1/2)*arctanh(1/2*(c*(sin(f*x+e)+1))^(1/2)*2^(1/2)/c^(1/2))*c^3)*(c*(sin(f*x+e)+1))^(1/2)/c^
(11/2)/(sin(f*x+e)-1)/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. 336 vs. \(2 (94) = 188\).

Time = 0.28 (sec) , antiderivative size = 336, normalized size of antiderivative = 2.97 \[ \int \frac {a+a \sin (e+f x)}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {\sqrt {2} {\left (a \cos \left (f x + e\right )^{3} + 3 \, a \cos \left (f x + e\right )^{2} - 2 \, a \cos \left (f x + e\right ) - {\left (a \cos \left (f x + e\right )^{2} - 2 \, a \cos \left (f x + e\right ) - 4 \, a\right )} \sin \left (f x + e\right ) - 4 \, a\right )} \sqrt {c} \log \left (-\frac {c \cos \left (f x + e\right )^{2} - 2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} \sqrt {c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )} + 3 \, c \cos \left (f x + e\right ) + {\left (c \cos \left (f x + e\right ) - 2 \, c\right )} \sin \left (f x + e\right ) + 2 \, c}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) + 4 \, {\left (a \cos \left (f x + e\right )^{2} - 3 \, a \cos \left (f x + e\right ) - {\left (a \cos \left (f x + e\right ) + 4 \, a\right )} \sin \left (f x + e\right ) - 4 \, a\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{32 \, {\left (c^{3} f \cos \left (f x + e\right )^{3} + 3 \, c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{3} f - {\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{3} f\right )} \sin \left (f x + e\right )\right )}} \]

[In]

integrate((a+a*sin(f*x+e))/(c-c*sin(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

1/32*(sqrt(2)*(a*cos(f*x + e)^3 + 3*a*cos(f*x + e)^2 - 2*a*cos(f*x + e) - (a*cos(f*x + e)^2 - 2*a*cos(f*x + e)
- 4*a)*sin(f*x + e) - 4*a)*sqrt(c)*log(-(c*cos(f*x + e)^2 - 2*sqrt(2)*sqrt(-c*sin(f*x + e) + c)*sqrt(c)*(cos(
f*x + e) + sin(f*x + e) + 1) + 3*c*cos(f*x + e) + (c*cos(f*x + e) - 2*c)*sin(f*x + e) + 2*c)/(cos(f*x + e)^2 +
(cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2)) + 4*(a*cos(f*x + e)^2 - 3*a*cos(f*x + e) - (a*cos(f*x +
e) + 4*a)*sin(f*x + e) - 4*a)*sqrt(-c*sin(f*x + e) + c))/(c^3*f*cos(f*x + e)^3 + 3*c^3*f*cos(f*x + e)^2 - 2*c^
3*f*cos(f*x + e) - 4*c^3*f - (c^3*f*cos(f*x + e)^2 - 2*c^3*f*cos(f*x + e) - 4*c^3*f)*sin(f*x + e))

Sympy [F]

\[ \int \frac {a+a \sin (e+f x)}{(c-c \sin (e+f x))^{5/2}} \, dx=a \left (\int \frac {\sin {\left (e + f x \right )}}{c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c} \sin ^{2}{\left (e + f x \right )} - 2 c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c} \sin {\left (e + f x \right )} + c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c}}\, dx + \int \frac {1}{c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c} \sin ^{2}{\left (e + f x \right )} - 2 c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c} \sin {\left (e + f x \right )} + c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c}}\, dx\right ) \]

[In]

integrate((a+a*sin(f*x+e))/(c-c*sin(f*x+e))**(5/2),x)

[Out]

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

Maxima [F]

\[ \int \frac {a+a \sin (e+f x)}{(c-c \sin (e+f x))^{5/2}} \, dx=\int { \frac {a \sin \left (f x + e\right ) + a}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate((a+a*sin(f*x+e))/(c-c*sin(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)/(-c*sin(f*x + e) + c)^(5/2), x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 218 vs. \(2 (94) = 188\).

Time = 0.40 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.93 \[ \int \frac {a+a \sin (e+f x)}{(c-c \sin (e+f x))^{5/2}} \, dx=-\frac {\frac {2 \, \sqrt {2} a \log \left (\frac {{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{2}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2}}\right )}{c^{\frac {5}{2}} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {\sqrt {2} a {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{2}}{c^{\frac {5}{2}} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {\sqrt {2} {\left (a \sqrt {c} - \frac {2 \, a \sqrt {c} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{2}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2}}{c^{3} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{128 \, f} \]

[In]

integrate((a+a*sin(f*x+e))/(c-c*sin(f*x+e))^(5/2),x, algorithm="giac")

[Out]

-1/128*(2*sqrt(2)*a*log((cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^2/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2)/(c^(5/2
)*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)^2/(c^(5/2)*(cos(-1/4*p
i + 1/2*f*x + 1/2*e) + 1)^2*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)^2/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2/(c
^3*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))))/f

Mupad [F(-1)]

Timed out. \[ \int \frac {a+a \sin (e+f x)}{(c-c \sin (e+f x))^{5/2}} \, dx=\int \frac {a+a\,\sin \left (e+f\,x\right )}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \]

[In]

int((a + a*sin(e + f*x))/(c - c*sin(e + f*x))^(5/2),x)

[Out]

int((a + a*sin(e + f*x))/(c - c*sin(e + f*x))^(5/2), x)