\(\int \frac {\cos ^2(e+f x)}{(a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2}} \, dx\) [64]

Optimal result

Integrand size = 38, antiderivative size = 152 \[ \int \frac {\cos ^2(e+f x)}{(a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2}} \, dx=-\frac {\cos (e+f x)}{2 a c f (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{3/2}}+\frac {\cos (e+f x)}{2 a^2 c f \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}}+\frac {\text {arctanh}(\sin (e+f x)) \cos (e+f x)}{2 a^2 c^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2920, 2822, 2820, 3855} \[ \int \frac {\cos ^2(e+f x)}{(a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2}} \, dx=\frac {\cos (e+f x) \text {arctanh}(\sin (e+f x))}{2 a^2 c^2 f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {\cos (e+f x)}{2 a^2 c f \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}-\frac {\cos (e+f x)}{2 a c f (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{3/2}} \]

[In]

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

[Out]

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

Rule 2820

Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Di
st[Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]), Int[1/Cos[e + f*x], x], x] /; FreeQ[{a, b
, c, d, e, f}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0]

Rule 2822

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[b*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/(a*f*(2*m + 1))), x] + Dist[(m + n + 1)/(a*(2*m
+ 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]
&& EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && ILtQ[Simplify[m + n + 1], 0] && NeQ[m, -2^(-1)] && (SumSimplerQ[m
, 1] ||  !SumSimplerQ[n, 1])

Rule 2920

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*
(x_)])^(n_.), x_Symbol] :> Dist[1/(a^(p/2)*c^(p/2)), Int[(a + b*Sin[e + f*x])^(m + p/2)*(c + d*Sin[e + f*x])^(
n + p/2), x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[p
/2]

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {1}{(a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{3/2}} \, dx}{a c} \\ & = -\frac {\cos (e+f x)}{2 a c f (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{3/2}}+\frac {\int \frac {1}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}} \, dx}{a^2 c} \\ & = -\frac {\cos (e+f x)}{2 a c f (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{3/2}}+\frac {\cos (e+f x)}{2 a^2 c f \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}}+\frac {\int \frac {1}{\sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \, dx}{2 a^2 c^2} \\ & = -\frac {\cos (e+f x)}{2 a c f (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{3/2}}+\frac {\cos (e+f x)}{2 a^2 c f \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}}+\frac {\cos (e+f x) \int \sec (e+f x) \, dx}{2 a^2 c^2 \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \\ & = -\frac {\cos (e+f x)}{2 a c f (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{3/2}}+\frac {\cos (e+f x)}{2 a^2 c f \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}}+\frac {\text {arctanh}(\sin (e+f x)) \cos (e+f x)}{2 a^2 c^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 7.50 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.07 \[ \int \frac {\cos ^2(e+f x)}{(a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2}} \, dx=\frac {\sec (e+f x) \left (-\log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+\log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+\cos (2 (e+f x)) \left (-\log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+\log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )\right )+2 \sin (e+f x)\right )}{4 a^2 c^2 f \sqrt {a (1+\sin (e+f x))} \sqrt {c-c \sin (e+f x)}} \]

[In]

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

[Out]

(Sec[e + f*x]*(-Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] + Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]] + Cos[2*(e
+ f*x)]*(-Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] + Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]]) + 2*Sin[e + f*
x]))/(4*a^2*c^2*f*Sqrt[a*(1 + Sin[e + f*x])]*Sqrt[c - c*Sin[e + f*x]])

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.61

method	result	size
default	\(\frac {-\ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )-1\right ) \cos \left (f x +e \right )+\cos \left (f x +e \right ) \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right )+\tan \left (f x +e \right )}{2 f \,a^{2} \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, c^{2} \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}}\)	\(93\)

[In]

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

[Out]

1/2/f/a^2/(a*(1+sin(f*x+e)))^(1/2)/c^2/(-c*(sin(f*x+e)-1))^(1/2)*(-ln(csc(f*x+e)-cot(f*x+e)-1)*cos(f*x+e)+cos(
f*x+e)*ln(-cot(f*x+e)+csc(f*x+e)+1)+tan(f*x+e))

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.72 \[ \int \frac {\cos ^2(e+f x)}{(a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2}} \, dx=\left [\frac {\sqrt {a c} \cos \left (f x + e\right )^{3} \log \left (-\frac {a c \cos \left (f x + e\right )^{3} - 2 \, a c \cos \left (f x + e\right ) - 2 \, \sqrt {a c} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} \sin \left (f x + e\right )}{\cos \left (f x + e\right )^{3}}\right ) + 2 \, \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} \sin \left (f x + e\right )}{4 \, a^{3} c^{3} f \cos \left (f x + e\right )^{3}}, -\frac {\sqrt {-a c} \arctan \left (\frac {\sqrt {-a c} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{a c \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) \cos \left (f x + e\right )^{3} - \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} \sin \left (f x + e\right )}{2 \, a^{3} c^{3} f \cos \left (f x + e\right )^{3}}\right ] \]

[In]

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

[Out]

[1/4*(sqrt(a*c)*cos(f*x + e)^3*log(-(a*c*cos(f*x + e)^3 - 2*a*c*cos(f*x + e) - 2*sqrt(a*c)*sqrt(a*sin(f*x + e)
+ a)*sqrt(-c*sin(f*x + e) + c)*sin(f*x + e))/cos(f*x + e)^3) + 2*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e
) + c)*sin(f*x + e))/(a^3*c^3*f*cos(f*x + e)^3), -1/2*(sqrt(-a*c)*arctan(sqrt(-a*c)*sqrt(a*sin(f*x + e) + a)*s
qrt(-c*sin(f*x + e) + c)/(a*c*cos(f*x + e)*sin(f*x + e)))*cos(f*x + e)^3 - sqrt(a*sin(f*x + e) + a)*sqrt(-c*si
n(f*x + e) + c)*sin(f*x + e))/(a^3*c^3*f*cos(f*x + e)^3)]

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F(-2)]

Exception generated. \[ \int \frac {\cos ^2(e+f x)}{(a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2}} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:index.cc index_m i_lex_is_greater Error: Bad Argument Value

Mupad [F(-1)]

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

[In]

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

[Out]

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