\(\int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{5/2}} \, dx\) [37]

Optimal result

Integrand size = 38, antiderivative size = 238 \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {\cos (e+f x) (a+a \sin (e+f x))^{7/2}}{c f (c-c \sin (e+f x))^{3/2}}+\frac {32 a^4 \cos (e+f x) \log (1-\sin (e+f x))}{c^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}+\frac {16 a^3 \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {4 a^2 \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {4 a \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{3 c^2 f \sqrt {c-c \sin (e+f x)}} \]

[Out]

cos(f*x+e)*(a+a*sin(f*x+e))^(7/2)/c/f/(c-c*sin(f*x+e))^(3/2)+4*a^2*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)/c^2/f/(c-
c*sin(f*x+e))^(1/2)+4/3*a*cos(f*x+e)*(a+a*sin(f*x+e))^(5/2)/c^2/f/(c-c*sin(f*x+e))^(1/2)+32*a^4*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)+16*a^3*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 = 0.54 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2920, 2818, 2819, 2816, 2746, 31} \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {32 a^4 \cos (e+f x) \log (1-\sin (e+f x))}{c^2 f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {16 a^3 \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {4 a^2 \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {4 a \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{3 c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{c f (c-c \sin (e+f x))^{3/2}} \]

[In]

Int[(Cos[e + f*x]^2*(a + a*Sin[e + f*x])^(7/2))/(c - c*Sin[e + f*x])^(5/2),x]

[Out]

(Cos[e + f*x]*(a + a*Sin[e + f*x])^(7/2))/(c*f*(c - c*Sin[e + f*x])^(3/2)) + (32*a^4*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]]) + (16*a^3*Cos[e + f*x]*Sqrt[a + a*Sin[e +
f*x]])/(c^2*f*Sqrt[c - c*Sin[e + f*x]]) + (4*a^2*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(c^2*f*Sqrt[c - c*S
in[e + f*x]]) + (4*a*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(3*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))^{9/2}}{(c-c \sin (e+f x))^{3/2}} \, dx}{a c} \\ & = \frac {\cos (e+f x) (a+a \sin (e+f x))^{7/2}}{c f (c-c \sin (e+f x))^{3/2}}-\frac {4 \int \frac {(a+a \sin (e+f x))^{7/2}}{\sqrt {c-c \sin (e+f x)}} \, dx}{c^2} \\ & = \frac {\cos (e+f x) (a+a \sin (e+f x))^{7/2}}{c f (c-c \sin (e+f x))^{3/2}}+\frac {4 a \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{3 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {(8 a) \int \frac {(a+a \sin (e+f x))^{5/2}}{\sqrt {c-c \sin (e+f x)}} \, dx}{c^2} \\ & = \frac {\cos (e+f x) (a+a \sin (e+f x))^{7/2}}{c f (c-c \sin (e+f x))^{3/2}}+\frac {4 a^2 \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {4 a \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{3 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {\left (16 a^2\right ) \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))^{7/2}}{c f (c-c \sin (e+f x))^{3/2}}+\frac {16 a^3 \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {4 a^2 \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {4 a \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{3 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {\left (32 a^3\right ) \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))^{7/2}}{c f (c-c \sin (e+f x))^{3/2}}+\frac {16 a^3 \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {4 a^2 \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {4 a \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{3 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {\left (32 a^4 \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))^{7/2}}{c f (c-c \sin (e+f x))^{3/2}}+\frac {16 a^3 \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {4 a^2 \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {4 a \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{3 c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {\left (32 a^4 \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))^{7/2}}{c f (c-c \sin (e+f x))^{3/2}}+\frac {32 a^4 \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 {16 a^3 \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {4 a^2 \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {4 a \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{3 c^2 f \sqrt {c-c \sin (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 10.24 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.82 \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{5/2}} \, dx=-\frac {a^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 \sqrt {a (1+\sin (e+f x))} \left (-177-172 \cos (2 (e+f x))+\cos (4 (e+f x))-1536 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-396 \sin (e+f x)+1536 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sin (e+f x)-16 \sin (3 (e+f x))\right )}{24 c^2 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (-1+\sin (e+f x))^2 \sqrt {c-c \sin (e+f x)}} \]

[In]

Integrate[(Cos[e + f*x]^2*(a + a*Sin[e + f*x])^(7/2))/(c - c*Sin[e + f*x])^(5/2),x]

[Out]

-1/24*(a^3*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3*Sqrt[a*(1 + Sin[e + f*x])]*(-177 - 172*Cos[2*(e + f*x)] + C
os[4*(e + f*x)] - 1536*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] - 396*Sin[e + f*x] + 1536*Log[Cos[(e + f*x)/2]
- Sin[(e + f*x)/2]]*Sin[e + f*x] - 16*Sin[3*(e + f*x)]))/(c^2*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-1 + S
in[e + f*x])^2*Sqrt[c - c*Sin[e + f*x]])

Maple [A] (verified)

Time = 0.29 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.71

method	result	size
default	\(\frac {\left (-\left (\cos ^{4}\left (f x +e \right )\right )+8 \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-192 \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )-1\right ) \sin \left (f x +e \right )+96 \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right ) \sin \left (f x +e \right )+44 \left (\cos ^{2}\left (f x +e \right )\right )+91 \sin \left (f x +e \right )+192 \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )-1\right )-96 \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )-43\right ) \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, a^{3} \sec \left (f x +e \right )}{3 f \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, c^{2}}\)	\(169\)

[In]

int(cos(f*x+e)^2*(a+a*sin(f*x+e))^(7/2)/(c-c*sin(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/3/f*(-cos(f*x+e)^4+8*cos(f*x+e)^2*sin(f*x+e)-192*ln(csc(f*x+e)-cot(f*x+e)-1)*sin(f*x+e)+96*ln(2/(1+cos(f*x+e
)))*sin(f*x+e)+44*cos(f*x+e)^2+91*sin(f*x+e)+192*ln(csc(f*x+e)-cot(f*x+e)-1)-96*ln(2/(1+cos(f*x+e)))-43)*(a*(1
+sin(f*x+e)))^(1/2)*a^3/(-c*(sin(f*x+e)-1))^(1/2)/c^2*sec(f*x+e)

Fricas [F]

\[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{5/2}} \, dx=\int { \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {7}{2}} \cos \left (f x + e\right )^{2}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^(7/2)/(c-c*sin(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

integral((3*a^3*cos(f*x + e)^4 - 4*a^3*cos(f*x + e)^2 + (a^3*cos(f*x + e)^4 - 4*a^3*cos(f*x + e)^2)*sin(f*x +
e))*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(-1)]

Timed out. \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate(cos(f*x+e)**2*(a+a*sin(f*x+e))**(7/2)/(c-c*sin(f*x+e))**(5/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{5/2}} \, dx=\int { \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {7}{2}} \cos \left (f x + e\right )^{2}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^(7/2)/(c-c*sin(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^(7/2)*cos(f*x + e)^2/(-c*sin(f*x + e) + c)^(5/2), x)

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.75 \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{5/2}} \, dx=-\frac {8 \, a^{\frac {7}{2}} \sqrt {c} {\left (\frac {12 \, \log \left (-\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )}{c^{3} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {3}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )} c^{3} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {c^{6} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} + 3 \, c^{6} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 9 \, c^{6} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}}{c^{9} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}\right )} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{3 \, f} \]

[In]

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^(7/2)/(c-c*sin(f*x+e))^(5/2),x, algorithm="giac")

[Out]

-8/3*a^(7/2)*sqrt(c)*(12*log(-cos(-1/4*pi + 1/2*f*x + 1/2*e)^2 + 1)/(c^3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)))
- 3/((cos(-1/4*pi + 1/2*f*x + 1/2*e)^2 - 1)*c^3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))) + (c^6*cos(-1/4*pi + 1/2*
f*x + 1/2*e)^6 + 3*c^6*cos(-1/4*pi + 1/2*f*x + 1/2*e)^4 + 9*c^6*cos(-1/4*pi + 1/2*f*x + 1/2*e)^2)/(c^9*sgn(sin
(-1/4*pi + 1/2*f*x + 1/2*e))))*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))/f

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{5/2}} \, dx=\int \frac {{\cos \left (e+f\,x\right )}^2\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{7/2}}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \]

[In]

int((cos(e + f*x)^2*(a + a*sin(e + f*x))^(7/2))/(c - c*sin(e + f*x))^(5/2),x)

[Out]

int((cos(e + f*x)^2*(a + a*sin(e + f*x))^(7/2))/(c - c*sin(e + f*x))^(5/2), x)