\(\int (a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{5/2} \, dx\) [299]

Optimal result

Integrand size = 28, antiderivative size = 109 \[ \int (a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{5/2} \, dx=\frac {64 a^2 c^5 \cos ^5(e+f x)}{315 f (c-c \sin (e+f x))^{5/2}}+\frac {16 a^2 c^4 \cos ^5(e+f x)}{63 f (c-c \sin (e+f x))^{3/2}}+\frac {2 a^2 c^3 \cos ^5(e+f x)}{9 f \sqrt {c-c \sin (e+f x)}} \]

[Out]

64/315*a^2*c^5*cos(f*x+e)^5/f/(c-c*sin(f*x+e))^(5/2)+16/63*a^2*c^4*cos(f*x+e)^5/f/(c-c*sin(f*x+e))^(3/2)+2/9*a
^2*c^3*cos(f*x+e)^5/f/(c-c*sin(f*x+e))^(1/2)

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {2815, 2753, 2752} \[ \int (a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{5/2} \, dx=\frac {64 a^2 c^5 \cos ^5(e+f x)}{315 f (c-c \sin (e+f x))^{5/2}}+\frac {16 a^2 c^4 \cos ^5(e+f x)}{63 f (c-c \sin (e+f x))^{3/2}}+\frac {2 a^2 c^3 \cos ^5(e+f x)}{9 f \sqrt {c-c \sin (e+f x)}} \]

[In]

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

[Out]

(64*a^2*c^5*Cos[e + f*x]^5)/(315*f*(c - c*Sin[e + f*x])^(5/2)) + (16*a^2*c^4*Cos[e + f*x]^5)/(63*f*(c - c*Sin[
e + f*x])^(3/2)) + (2*a^2*c^3*Cos[e + f*x]^5)/(9*f*Sqrt[c - c*Sin[e + f*x]])

Rule 2752

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b*(g*C
os[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^(m - 1)/(f*g*(m - 1))), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && Eq
Q[a^2 - b^2, 0] && EqQ[2*m + p - 1, 0] && NeQ[m, 1]

Rule 2753

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(-b)*(
g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^(m - 1)/(f*g*(m + p))), x] + Dist[a*((2*m + p - 1)/(m + p)), Int
[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2,
0] && IGtQ[Simplify[(2*m + p - 1)/2], 0] && NeQ[m + p, 0]

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}& = \left (a^2 c^2\right ) \int \cos ^4(e+f x) \sqrt {c-c \sin (e+f x)} \, dx \\ & = \frac {2 a^2 c^3 \cos ^5(e+f x)}{9 f \sqrt {c-c \sin (e+f x)}}+\frac {1}{9} \left (8 a^2 c^3\right ) \int \frac {\cos ^4(e+f x)}{\sqrt {c-c \sin (e+f x)}} \, dx \\ & = \frac {16 a^2 c^4 \cos ^5(e+f x)}{63 f (c-c \sin (e+f x))^{3/2}}+\frac {2 a^2 c^3 \cos ^5(e+f x)}{9 f \sqrt {c-c \sin (e+f x)}}+\frac {1}{63} \left (32 a^2 c^4\right ) \int \frac {\cos ^4(e+f x)}{(c-c \sin (e+f x))^{3/2}} \, dx \\ & = \frac {64 a^2 c^5 \cos ^5(e+f x)}{315 f (c-c \sin (e+f x))^{5/2}}+\frac {16 a^2 c^4 \cos ^5(e+f x)}{63 f (c-c \sin (e+f x))^{3/2}}+\frac {2 a^2 c^3 \cos ^5(e+f x)}{9 f \sqrt {c-c \sin (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.85 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.61 \[ \int (a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{5/2} \, dx=\frac {a^2 c^2 \cos ^5(e+f x) (-249+35 \cos (2 (e+f x))+220 \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{315 f (-1+\sin (e+f x))^3} \]

[In]

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

[Out]

(a^2*c^2*Cos[e + f*x]^5*(-249 + 35*Cos[2*(e + f*x)] + 220*Sin[e + f*x])*Sqrt[c - c*Sin[e + f*x]])/(315*f*(-1 +
Sin[e + f*x])^3)

Maple [A] (verified)

Time = 7.72 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.65

method	result	size
default	\(-\frac {2 \left (\sin \left (f x +e \right )-1\right ) c^{3} \left (\sin \left (f x +e \right )+1\right )^{3} a^{2} \left (35 \left (\sin ^{2}\left (f x +e \right )\right )-110 \sin \left (f x +e \right )+107\right )}{315 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(71\)
parts	\(-\frac {2 a^{2} \left (\sin \left (f x +e \right )-1\right ) c^{3} \left (\sin \left (f x +e \right )+1\right ) \left (3 \left (\sin ^{2}\left (f x +e \right )\right )-14 \sin \left (f x +e \right )+43\right )}{15 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}-\frac {2 a^{2} \left (\sin \left (f x +e \right )-1\right ) c^{3} \left (\sin \left (f x +e \right )+1\right ) \left (35 \left (\sin ^{4}\left (f x +e \right )\right )-130 \left (\sin ^{3}\left (f x +e \right )\right )+219 \left (\sin ^{2}\left (f x +e \right )\right )-292 \sin \left (f x +e \right )+584\right )}{315 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}-\frac {4 a^{2} \left (\sin \left (f x +e \right )-1\right ) c^{3} \left (\sin \left (f x +e \right )+1\right ) \left (3 \left (\sin ^{3}\left (f x +e \right )\right )-12 \left (\sin ^{2}\left (f x +e \right )\right )+23 \sin \left (f x +e \right )-46\right )}{21 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(236\)

[In]

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

[Out]

-2/315*(sin(f*x+e)-1)*c^3*(sin(f*x+e)+1)^3*a^2*(35*sin(f*x+e)^2-110*sin(f*x+e)+107)/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. 201 vs. \(2 (97) = 194\).

Time = 0.27 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.84 \[ \int (a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{5/2} \, dx=\frac {2 \, {\left (35 \, a^{2} c^{2} \cos \left (f x + e\right )^{5} - 5 \, a^{2} c^{2} \cos \left (f x + e\right )^{4} + 8 \, a^{2} c^{2} \cos \left (f x + e\right )^{3} - 16 \, a^{2} c^{2} \cos \left (f x + e\right )^{2} + 64 \, a^{2} c^{2} \cos \left (f x + e\right ) + 128 \, a^{2} c^{2} + {\left (35 \, a^{2} c^{2} \cos \left (f x + e\right )^{4} + 40 \, a^{2} c^{2} \cos \left (f x + e\right )^{3} + 48 \, a^{2} c^{2} \cos \left (f x + e\right )^{2} + 64 \, a^{2} c^{2} \cos \left (f x + e\right ) + 128 \, a^{2} c^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{315 \, {\left (f \cos \left (f x + e\right ) - f \sin \left (f x + e\right ) + f\right )}} \]

[In]

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

[Out]

2/315*(35*a^2*c^2*cos(f*x + e)^5 - 5*a^2*c^2*cos(f*x + e)^4 + 8*a^2*c^2*cos(f*x + e)^3 - 16*a^2*c^2*cos(f*x +
e)^2 + 64*a^2*c^2*cos(f*x + e) + 128*a^2*c^2 + (35*a^2*c^2*cos(f*x + e)^4 + 40*a^2*c^2*cos(f*x + e)^3 + 48*a^2
*c^2*cos(f*x + e)^2 + 64*a^2*c^2*cos(f*x + e) + 128*a^2*c^2)*sin(f*x + e))*sqrt(-c*sin(f*x + e) + c)/(f*cos(f*
x + e) - f*sin(f*x + e) + f)

Sympy [F]

\[ \int (a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{5/2} \, dx=a^{2} \left (\int c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c}\, dx + \int \left (- 2 c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c} \sin ^{2}{\left (e + f x \right )}\right )\, dx + \int c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c} \sin ^{4}{\left (e + f x \right )}\, dx\right ) \]

[In]

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

[Out]

a**2*(Integral(c**2*sqrt(-c*sin(e + f*x) + c), x) + Integral(-2*c**2*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x)**2
, x) + Integral(c**2*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x)**4, x))

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.62 \[ \int (a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{5/2} \, dx=-\frac {\sqrt {2} {\left (1890 \, a^{2} c^{2} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 420 \, a^{2} c^{2} \cos \left (-\frac {3}{4} \, \pi + \frac {3}{2} \, f x + \frac {3}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 252 \, a^{2} c^{2} \cos \left (-\frac {5}{4} \, \pi + \frac {5}{2} \, f x + \frac {5}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 45 \, a^{2} c^{2} \cos \left (-\frac {7}{4} \, \pi + \frac {7}{2} \, f x + \frac {7}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 35 \, a^{2} c^{2} \cos \left (-\frac {9}{4} \, \pi + \frac {9}{2} \, f x + \frac {9}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sqrt {c}}{2520 \, f} \]

[In]

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

[Out]

-1/2520*sqrt(2)*(1890*a^2*c^2*cos(-1/4*pi + 1/2*f*x + 1/2*e)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) + 420*a^2*c^2
*cos(-3/4*pi + 3/2*f*x + 3/2*e)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 252*a^2*c^2*cos(-5/4*pi + 5/2*f*x + 5/2*
e)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 45*a^2*c^2*cos(-7/4*pi + 7/2*f*x + 7/2*e)*sgn(sin(-1/4*pi + 1/2*f*x +
1/2*e)) + 35*a^2*c^2*cos(-9/4*pi + 9/2*f*x + 9/2*e)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)))*sqrt(c)/f

Mupad [F(-1)]

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

[In]

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

[Out]

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