\(\int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{5/2}} \, dx\) [15]

Optimal result

Integrand size = 38, antiderivative size = 144 \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {\cos (e+f x) (a+a \sin (e+f x))^{3/2}}{c f (c-c \sin (e+f x))^{3/2}}+\frac {4 a^2 \cos (e+f x) \log (1-\sin (e+f x))}{c^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}+\frac {2 a \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{c^2 f \sqrt {c-c \sin (e+f x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2920, 2818, 2819, 2816, 2746, 31} \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {4 a^2 \cos (e+f x) \log (1-\sin (e+f x))}{c^2 f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {2 a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{c f (c-c \sin (e+f x))^{3/2}} \]

[In]

Int[(Cos[e + f*x]^2*(a + a*Sin[e + f*x])^(3/2))/(c - c*Sin[e + f*x])^(5/2),x]

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 2746

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rule 2816

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[a
*c*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])), Int[Cos[e + f*x]/(c + d*Sin[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 2818

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

Rule 2819

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 - 1)*((c + d*Sin[e + f*x])^n/(f*(m + n))), x] + Dist[a*((2*m - 1)/(
m + n)), 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, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[m - 1/2, 0] &&  !LtQ[n, -1] &&  !(IGtQ[n - 1/2, 0] && LtQ[n, m
]) &&  !(ILtQ[m + n, 0] && GtQ[2*m + n + 1, 0])

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]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {(a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{3/2}} \, dx}{a c} \\ & = \frac {\cos (e+f x) (a+a \sin (e+f x))^{3/2}}{c f (c-c \sin (e+f x))^{3/2}}-\frac {2 \int \frac {(a+a \sin (e+f x))^{3/2}}{\sqrt {c-c \sin (e+f x)}} \, dx}{c^2} \\ & = \frac {\cos (e+f x) (a+a \sin (e+f x))^{3/2}}{c f (c-c \sin (e+f x))^{3/2}}+\frac {2 a \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {(4 a) \int \frac {\sqrt {a+a \sin (e+f x)}}{\sqrt {c-c \sin (e+f x)}} \, dx}{c^2} \\ & = \frac {\cos (e+f x) (a+a \sin (e+f x))^{3/2}}{c f (c-c \sin (e+f x))^{3/2}}+\frac {2 a \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {\left (4 a^2 \cos (e+f x)\right ) \int \frac {\cos (e+f x)}{c-c \sin (e+f x)} \, dx}{c \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \\ & = \frac {\cos (e+f x) (a+a \sin (e+f x))^{3/2}}{c f (c-c \sin (e+f x))^{3/2}}+\frac {2 a \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {\left (4 a^2 \cos (e+f x)\right ) \text {Subst}\left (\int \frac {1}{c+x} \, dx,x,-c \sin (e+f x)\right )}{c^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \\ & = \frac {\cos (e+f x) (a+a \sin (e+f x))^{3/2}}{c f (c-c \sin (e+f x))^{3/2}}+\frac {4 a^2 \cos (e+f x) \log (1-\sin (e+f x))}{c^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}+\frac {2 a \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{c^2 f \sqrt {c-c \sin (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 7.69 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.17 \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {a \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 \sqrt {a (1+\sin (e+f x))} \left (7+\cos (2 (e+f x))+16 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+\left (2-16 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )\right ) \sin (e+f x)\right )}{2 c^2 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (-1+\sin (e+f x))^2 \sqrt {c-c \sin (e+f x)}} \]

[In]

Integrate[(Cos[e + f*x]^2*(a + a*Sin[e + f*x])^(3/2))/(c - c*Sin[e + f*x])^(5/2),x]

[Out]

(a*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3*Sqrt[a*(1 + Sin[e + f*x])]*(7 + Cos[2*(e + f*x)] + 16*Log[Cos[(e +
f*x)/2] - Sin[(e + f*x)/2]] + (2 - 16*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]])*Sin[e + f*x]))/(2*c^2*f*(Cos[(
e + f*x)/2] + Sin[(e + f*x)/2])*(-1 + Sin[e + f*x])^2*Sqrt[c - c*Sin[e + f*x]])

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.98

method	result	size
default	\(-\frac {\left (8 \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )-1\right ) \sin \left (f x +e \right )-4 \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right ) \sin \left (f x +e \right )-\left (\cos ^{2}\left (f x +e \right )\right )-8 \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )-1\right )+4 \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )-5 \sin \left (f x +e \right )+1\right ) \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, a \sec \left (f x +e \right )}{f \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, c^{2}}\)	\(141\)

[In]

int(cos(f*x+e)^2*(a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/f*(8*ln(csc(f*x+e)-cot(f*x+e)-1)*sin(f*x+e)-4*ln(2/(1+cos(f*x+e)))*sin(f*x+e)-cos(f*x+e)^2-8*ln(csc(f*x+e)-
cot(f*x+e)-1)+4*ln(2/(1+cos(f*x+e)))-5*sin(f*x+e)+1)*(a*(1+sin(f*x+e)))^(1/2)*a/(-c*(sin(f*x+e)-1))^(1/2)/c^2*
sec(f*x+e)

Fricas [F]

\[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{5/2}} \, dx=\int { \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{2}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

integral(-(a*cos(f*x + e)^2*sin(f*x + e) + a*cos(f*x + e)^2)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c
)/(3*c^3*cos(f*x + e)^2 - 4*c^3 - (c^3*cos(f*x + e)^2 - 4*c^3)*sin(f*x + e)), x)

Sympy [F]

\[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{5/2}} \, dx=\int \frac {\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}} \cos ^{2}{\left (e + f x \right )}}{\left (- c \left (\sin {\left (e + f x \right )} - 1\right )\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate(cos(f*x+e)**2*(a+a*sin(f*x+e))**(3/2)/(c-c*sin(f*x+e))**(5/2),x)

[Out]

Integral((a*(sin(e + f*x) + 1))**(3/2)*cos(e + f*x)**2/(-c*(sin(e + f*x) - 1))**(5/2), x)

Maxima [F]

\[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{5/2}} \, dx=\int { \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{2}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^(3/2)*cos(f*x + e)^2/(-c*sin(f*x + e) + c)^(5/2), x)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.94 \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {\sqrt {2} {\left (\sqrt {2} a \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 4 \, \sqrt {2} a \sqrt {c} \log \left ({\left | \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) \right |}\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - \frac {\sqrt {2} a \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}}\right )} \sqrt {a}}{c^{3} f \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} \]

[In]

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(5/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{5/2}} \, dx=\int \frac {{\cos \left (e+f\,x\right )}^2\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{3/2}}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \]

[In]

int((cos(e + f*x)^2*(a + a*sin(e + f*x))^(3/2))/(c - c*sin(e + f*x))^(5/2),x)

[Out]

int((cos(e + f*x)^2*(a + a*sin(e + f*x))^(3/2))/(c - c*sin(e + f*x))^(5/2), x)