\(\int \cos ^2(e+f x) (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{3/2} \, dx\) [33]

Optimal result

Integrand size = 38, antiderivative size = 140 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{3/2} \, dx=\frac {4 c^2 \cos (e+f x) (a+a \sin (e+f x))^{9/2}}{105 a f \sqrt {c-c \sin (e+f x)}}+\frac {2 c \cos (e+f x) (a+a \sin (e+f x))^{9/2} \sqrt {c-c \sin (e+f x)}}{21 a f}+\frac {\cos (e+f x) (a+a \sin (e+f x))^{9/2} (c-c \sin (e+f x))^{3/2}}{7 a f} \]

[Out]

1/7*cos(f*x+e)*(a+a*sin(f*x+e))^(9/2)*(c-c*sin(f*x+e))^(3/2)/a/f+4/105*c^2*cos(f*x+e)*(a+a*sin(f*x+e))^(9/2)/a
/f/(c-c*sin(f*x+e))^(1/2)+2/21*c*cos(f*x+e)*(a+a*sin(f*x+e))^(9/2)*(c-c*sin(f*x+e))^(1/2)/a/f

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.079, Rules used = {2920, 2819, 2817} \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{3/2} \, dx=\frac {4 c^2 \cos (e+f x) (a \sin (e+f x)+a)^{9/2}}{105 a f \sqrt {c-c \sin (e+f x)}}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^{9/2} (c-c \sin (e+f x))^{3/2}}{7 a f}+\frac {2 c \cos (e+f x) (a \sin (e+f x)+a)^{9/2} \sqrt {c-c \sin (e+f x)}}{21 a f} \]

[In]

Int[Cos[e + f*x]^2*(a + a*Sin[e + f*x])^(7/2)*(c - c*Sin[e + f*x])^(3/2),x]

[Out]

(4*c^2*Cos[e + f*x]*(a + a*Sin[e + f*x])^(9/2))/(105*a*f*Sqrt[c - c*Sin[e + f*x]]) + (2*c*Cos[e + f*x]*(a + a*
Sin[e + f*x])^(9/2)*Sqrt[c - c*Sin[e + f*x]])/(21*a*f) + (Cos[e + f*x]*(a + a*Sin[e + f*x])^(9/2)*(c - c*Sin[e
+ f*x])^(3/2))/(7*a*f)

Rule 2817

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

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 (a+a \sin (e+f x))^{9/2} (c-c \sin (e+f x))^{5/2} \, dx}{a c} \\ & = \frac {\cos (e+f x) (a+a \sin (e+f x))^{9/2} (c-c \sin (e+f x))^{3/2}}{7 a f}+\frac {4 \int (a+a \sin (e+f x))^{9/2} (c-c \sin (e+f x))^{3/2} \, dx}{7 a} \\ & = \frac {2 c \cos (e+f x) (a+a \sin (e+f x))^{9/2} \sqrt {c-c \sin (e+f x)}}{21 a f}+\frac {\cos (e+f x) (a+a \sin (e+f x))^{9/2} (c-c \sin (e+f x))^{3/2}}{7 a f}+\frac {(4 c) \int (a+a \sin (e+f x))^{9/2} \sqrt {c-c \sin (e+f x)} \, dx}{21 a} \\ & = \frac {4 c^2 \cos (e+f x) (a+a \sin (e+f x))^{9/2}}{105 a f \sqrt {c-c \sin (e+f x)}}+\frac {2 c \cos (e+f x) (a+a \sin (e+f x))^{9/2} \sqrt {c-c \sin (e+f x)}}{21 a f}+\frac {\cos (e+f x) (a+a \sin (e+f x))^{9/2} (c-c \sin (e+f x))^{3/2}}{7 a f} \\ \end{align*}

Mathematica [A] (verified)

Time = 7.89 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.82 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{3/2} \, dx=\frac {a^3 c \sec (e+f x) \sqrt {a (1+\sin (e+f x))} \sqrt {c-c \sin (e+f x)} (-1050 \cos (2 (e+f x))-420 \cos (4 (e+f x))-70 \cos (6 (e+f x))+4725 \sin (e+f x)+665 \sin (3 (e+f x))+21 \sin (5 (e+f x))-15 \sin (7 (e+f x)))}{6720 f} \]

[In]

Integrate[Cos[e + f*x]^2*(a + a*Sin[e + f*x])^(7/2)*(c - c*Sin[e + f*x])^(3/2),x]

[Out]

(a^3*c*Sec[e + f*x]*Sqrt[a*(1 + Sin[e + f*x])]*Sqrt[c - c*Sin[e + f*x]]*(-1050*Cos[2*(e + f*x)] - 420*Cos[4*(e
+ f*x)] - 70*Cos[6*(e + f*x)] + 4725*Sin[e + f*x] + 665*Sin[3*(e + f*x)] + 21*Sin[5*(e + f*x)] - 15*Sin[7*(e
+ f*x)]))/(6720*f)

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.77

method	result	size
default	\(-\frac {\sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, c \,a^{3} \left (15 \left (\cos ^{5}\left (f x +e \right )\right ) \sin \left (f x +e \right )+35 \left (\cos ^{5}\left (f x +e \right )\right )-24 \left (\cos ^{3}\left (f x +e \right )\right ) \sin \left (f x +e \right )-32 \cos \left (f x +e \right ) \sin \left (f x +e \right )-64 \tan \left (f x +e \right )-35 \sec \left (f x +e \right )\right )}{105 f}\)	\(108\)

[In]

int(cos(f*x+e)^2*(a+a*sin(f*x+e))^(7/2)*(c-c*sin(f*x+e))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/105/f*(a*(1+sin(f*x+e)))^(1/2)*(-c*(sin(f*x+e)-1))^(1/2)*c*a^3*(15*cos(f*x+e)^5*sin(f*x+e)+35*cos(f*x+e)^5-
24*cos(f*x+e)^3*sin(f*x+e)-32*cos(f*x+e)*sin(f*x+e)-64*tan(f*x+e)-35*sec(f*x+e))

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.82 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{3/2} \, dx=-\frac {{\left (35 \, a^{3} c \cos \left (f x + e\right )^{6} - 35 \, a^{3} c + {\left (15 \, a^{3} c \cos \left (f x + e\right )^{6} - 24 \, a^{3} c \cos \left (f x + e\right )^{4} - 32 \, a^{3} c \cos \left (f x + e\right )^{2} - 64 \, a^{3} c\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{105 \, f \cos \left (f x + e\right )} \]

[In]

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^(7/2)*(c-c*sin(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

-1/105*(35*a^3*c*cos(f*x + e)^6 - 35*a^3*c + (15*a^3*c*cos(f*x + e)^6 - 24*a^3*c*cos(f*x + e)^4 - 32*a^3*c*cos
(f*x + e)^2 - 64*a^3*c)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(f*cos(f*x + e))

Sympy [F(-1)]

Timed out. \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{3/2} \, dx=\text {Timed out} \]

[In]

integrate(cos(f*x+e)**2*(a+a*sin(f*x+e))**(7/2)*(c-c*sin(f*x+e))**(3/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{3/2} \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {7}{2}} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{2} \,d x } \]

[In]

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^(7/2)*(c-c*sin(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^(7/2)*(-c*sin(f*x + e) + c)^(3/2)*cos(f*x + e)^2, x)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.07 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{3/2} \, dx=-\frac {128 \, {\left (15 \, a^{3} c \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{14} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 35 \, a^{3} c \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{12} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 21 \, a^{3} c \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{10} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sqrt {a} \sqrt {c}}{105 \, f} \]

[In]

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^(7/2)*(c-c*sin(f*x+e))^(3/2),x, algorithm="giac")

[Out]

-128/105*(15*a^3*c*cos(-1/4*pi + 1/2*f*x + 1/2*e)^14*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2
*f*x + 1/2*e)) - 35*a^3*c*cos(-1/4*pi + 1/2*f*x + 1/2*e)^12*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*p
i + 1/2*f*x + 1/2*e)) + 21*a^3*c*cos(-1/4*pi + 1/2*f*x + 1/2*e)^10*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin
(-1/4*pi + 1/2*f*x + 1/2*e)))*sqrt(a)*sqrt(c)/f

Mupad [B] (verification not implemented)

Time = 12.82 (sec) , antiderivative size = 319, normalized size of antiderivative = 2.28 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{3/2} \, dx=-\frac {{\mathrm {e}}^{-e\,7{}\mathrm {i}-f\,x\,7{}\mathrm {i}}\,\sqrt {c-c\,\sin \left (e+f\,x\right )}\,\left (\frac {5\,a^3\,c\,{\mathrm {e}}^{e\,7{}\mathrm {i}+f\,x\,7{}\mathrm {i}}\,\cos \left (2\,e+2\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{16\,f}+\frac {a^3\,c\,{\mathrm {e}}^{e\,7{}\mathrm {i}+f\,x\,7{}\mathrm {i}}\,\cos \left (4\,e+4\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{8\,f}+\frac {a^3\,c\,{\mathrm {e}}^{e\,7{}\mathrm {i}+f\,x\,7{}\mathrm {i}}\,\cos \left (6\,e+6\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{48\,f}-\frac {19\,a^3\,c\,{\mathrm {e}}^{e\,7{}\mathrm {i}+f\,x\,7{}\mathrm {i}}\,\sin \left (3\,e+3\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{96\,f}-\frac {a^3\,c\,{\mathrm {e}}^{e\,7{}\mathrm {i}+f\,x\,7{}\mathrm {i}}\,\sin \left (5\,e+5\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{160\,f}+\frac {a^3\,c\,{\mathrm {e}}^{e\,7{}\mathrm {i}+f\,x\,7{}\mathrm {i}}\,\sin \left (7\,e+7\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{224\,f}-\frac {45\,a^3\,c\,{\mathrm {e}}^{e\,7{}\mathrm {i}+f\,x\,7{}\mathrm {i}}\,\sin \left (e+f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{32\,f}\right )}{2\,\cos \left (e+f\,x\right )} \]

[In]

int(cos(e + f*x)^2*(a + a*sin(e + f*x))^(7/2)*(c - c*sin(e + f*x))^(3/2),x)

[Out]

-(exp(- e*7i - f*x*7i)*(c - c*sin(e + f*x))^(1/2)*((5*a^3*c*exp(e*7i + f*x*7i)*cos(2*e + 2*f*x)*(a + a*sin(e +
f*x))^(1/2))/(16*f) + (a^3*c*exp(e*7i + f*x*7i)*cos(4*e + 4*f*x)*(a + a*sin(e + f*x))^(1/2))/(8*f) + (a^3*c*e
xp(e*7i + f*x*7i)*cos(6*e + 6*f*x)*(a + a*sin(e + f*x))^(1/2))/(48*f) - (19*a^3*c*exp(e*7i + f*x*7i)*sin(3*e +
3*f*x)*(a + a*sin(e + f*x))^(1/2))/(96*f) - (a^3*c*exp(e*7i + f*x*7i)*sin(5*e + 5*f*x)*(a + a*sin(e + f*x))^(
1/2))/(160*f) + (a^3*c*exp(e*7i + f*x*7i)*sin(7*e + 7*f*x)*(a + a*sin(e + f*x))^(1/2))/(224*f) - (45*a^3*c*exp
(e*7i + f*x*7i)*sin(e + f*x)*(a + a*sin(e + f*x))^(1/2))/(32*f)))/(2*cos(e + f*x))