\(\int (3+3 \sin (e+f x))^3 (c-c \sin (e+f x))^{5/2} \, dx\) [308]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [B] (verification not implemented)
   Sympy [F]
   Maxima [F]
   Giac [B] (verification not implemented)
   Mupad [F(-1)]

Optimal result

Integrand size = 28, antiderivative size = 100 \[ \int (3+3 \sin (e+f x))^3 (c-c \sin (e+f x))^{5/2} \, dx=\frac {192 c^6 \cos ^7(e+f x)}{77 f (c-c \sin (e+f x))^{7/2}}+\frac {48 c^5 \cos ^7(e+f x)}{11 f (c-c \sin (e+f x))^{5/2}}+\frac {54 c^4 \cos ^7(e+f x)}{11 f (c-c \sin (e+f x))^{3/2}} \]

[Out]

64/693*a^3*c^6*cos(f*x+e)^7/f/(c-c*sin(f*x+e))^(7/2)+16/99*a^3*c^5*cos(f*x+e)^7/f/(c-c*sin(f*x+e))^(5/2)+2/11*
a^3*c^4*cos(f*x+e)^7/f/(c-c*sin(f*x+e))^(3/2)

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.09, 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} \[ \int (3+3 \sin (e+f x))^3 (c-c \sin (e+f x))^{5/2} \, dx=\frac {64 a^3 c^6 \cos ^7(e+f x)}{693 f (c-c \sin (e+f x))^{7/2}}+\frac {16 a^3 c^5 \cos ^7(e+f x)}{99 f (c-c \sin (e+f x))^{5/2}}+\frac {2 a^3 c^4 \cos ^7(e+f x)}{11 f (c-c \sin (e+f x))^{3/2}} \]

[In]

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

[Out]

(64*a^3*c^6*Cos[e + f*x]^7)/(693*f*(c - c*Sin[e + f*x])^(7/2)) + (16*a^3*c^5*Cos[e + f*x]^7)/(99*f*(c - c*Sin[
e + f*x])^(5/2)) + (2*a^3*c^4*Cos[e + f*x]^7)/(11*f*(c - c*Sin[e + f*x])^(3/2))

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^3 c^3\right ) \int \frac {\cos ^6(e+f x)}{\sqrt {c-c \sin (e+f x)}} \, dx \\ & = \frac {2 a^3 c^4 \cos ^7(e+f x)}{11 f (c-c \sin (e+f x))^{3/2}}+\frac {1}{11} \left (8 a^3 c^4\right ) \int \frac {\cos ^6(e+f x)}{(c-c \sin (e+f x))^{3/2}} \, dx \\ & = \frac {16 a^3 c^5 \cos ^7(e+f x)}{99 f (c-c \sin (e+f x))^{5/2}}+\frac {2 a^3 c^4 \cos ^7(e+f x)}{11 f (c-c \sin (e+f x))^{3/2}}+\frac {1}{99} \left (32 a^3 c^5\right ) \int \frac {\cos ^6(e+f x)}{(c-c \sin (e+f x))^{5/2}} \, dx \\ & = \frac {64 a^3 c^6 \cos ^7(e+f x)}{693 f (c-c \sin (e+f x))^{7/2}}+\frac {16 a^3 c^5 \cos ^7(e+f x)}{99 f (c-c \sin (e+f x))^{5/2}}+\frac {2 a^3 c^4 \cos ^7(e+f x)}{11 f (c-c \sin (e+f x))^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 4.18 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.93 \[ \int (3+3 \sin (e+f x))^3 (c-c \sin (e+f x))^{5/2} \, dx=-\frac {3 c^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^7 (-365+63 \cos (2 (e+f x))+364 \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{77 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )} \]

[In]

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

[Out]

(-3*c^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^7*(-365 + 63*Cos[2*(e + f*x)] + 364*Sin[e + f*x])*Sqrt[c - c*Sin
[e + f*x]])/(77*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]))

Maple [A] (verified)

Time = 37.12 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.71

method result size
default \(-\frac {2 \left (\sin \left (f x +e \right )-1\right ) c^{3} \left (\sin \left (f x +e \right )+1\right )^{4} a^{3} \left (63 \left (\sin ^{2}\left (f x +e \right )\right )-182 \sin \left (f x +e \right )+151\right )}{693 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\) \(71\)
parts \(-\frac {2 a^{3} \left (\sin \left (f x +e \right )-1\right ) c^{3} \left (\sin \left (f x +e \right )+1\right ) \left (3 \left (\sin ^{2}\left (f x +e \right )\right )-14 \sin \left (f x +e \right )+43\right )}{15 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}-\frac {2 a^{3} \left (\sin \left (f x +e \right )-1\right ) c^{3} \left (\sin \left (f x +e \right )+1\right ) \left (63 \left (\sin ^{5}\left (f x +e \right )\right )-224 \left (\sin ^{4}\left (f x +e \right )\right )+355 \left (\sin ^{3}\left (f x +e \right )\right )-426 \left (\sin ^{2}\left (f x +e \right )\right )+568 \sin \left (f x +e \right )-1136\right )}{693 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}-\frac {2 a^{3} \left (\sin \left (f x +e \right )-1\right ) c^{3} \left (\sin \left (f x +e \right )+1\right ) \left (3 \left (\sin ^{3}\left (f x +e \right )\right )-12 \left (\sin ^{2}\left (f x +e \right )\right )+23 \sin \left (f x +e \right )-46\right )}{7 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}-\frac {2 a^{3} \left (\sin \left (f x +e \right )-1\right ) c^{3} \left (\sin \left (f x +e \right )+1\right ) \left (35 \left (\sin ^{4}\left (f x +e \right )\right )-130 \left (\sin ^{3}\left (f x +e \right )\right )+219 \left (\sin ^{2}\left (f x +e \right )\right )-292 \sin \left (f x +e \right )+584\right )}{105 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\) \(334\)

[In]

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

[Out]

-2/693*(sin(f*x+e)-1)*c^3*(sin(f*x+e)+1)^4*a^3*(63*sin(f*x+e)^2-182*sin(f*x+e)+151)/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. 234 vs. \(2 (97) = 194\).

Time = 0.26 (sec) , antiderivative size = 234, normalized size of antiderivative = 2.34 \[ \int (3+3 \sin (e+f x))^3 (c-c \sin (e+f x))^{5/2} \, dx=-\frac {2 \, {\left (63 \, a^{3} c^{2} \cos \left (f x + e\right )^{6} - 7 \, a^{3} c^{2} \cos \left (f x + e\right )^{5} + 10 \, a^{3} c^{2} \cos \left (f x + e\right )^{4} - 16 \, a^{3} c^{2} \cos \left (f x + e\right )^{3} + 32 \, a^{3} c^{2} \cos \left (f x + e\right )^{2} - 128 \, a^{3} c^{2} \cos \left (f x + e\right ) - 256 \, a^{3} c^{2} - {\left (63 \, a^{3} c^{2} \cos \left (f x + e\right )^{5} + 70 \, a^{3} c^{2} \cos \left (f x + e\right )^{4} + 80 \, a^{3} c^{2} \cos \left (f x + e\right )^{3} + 96 \, a^{3} c^{2} \cos \left (f x + e\right )^{2} + 128 \, a^{3} c^{2} \cos \left (f x + e\right ) + 256 \, a^{3} c^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{693 \, {\left (f \cos \left (f x + e\right ) - f \sin \left (f x + e\right ) + f\right )}} \]

[In]

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

[Out]

-2/693*(63*a^3*c^2*cos(f*x + e)^6 - 7*a^3*c^2*cos(f*x + e)^5 + 10*a^3*c^2*cos(f*x + e)^4 - 16*a^3*c^2*cos(f*x
+ e)^3 + 32*a^3*c^2*cos(f*x + e)^2 - 128*a^3*c^2*cos(f*x + e) - 256*a^3*c^2 - (63*a^3*c^2*cos(f*x + e)^5 + 70*
a^3*c^2*cos(f*x + e)^4 + 80*a^3*c^2*cos(f*x + e)^3 + 96*a^3*c^2*cos(f*x + e)^2 + 128*a^3*c^2*cos(f*x + e) + 25
6*a^3*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]

\[ \int (3+3 \sin (e+f x))^3 (c-c \sin (e+f x))^{5/2} \, dx=a^{3} \left (\int c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c}\, dx + \int c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c} \sin {\left (e + f x \right )}\, dx + \int \left (- 2 c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c} \sin ^{2}{\left (e + f x \right )}\right )\, dx + \int \left (- 2 c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c} \sin ^{3}{\left (e + f x \right )}\right )\, dx + \int c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c} \sin ^{4}{\left (e + f x \right )}\, dx + \int c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c} \sin ^{5}{\left (e + f x \right )}\, dx\right ) \]

[In]

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

[Out]

a**3*(Integral(c**2*sqrt(-c*sin(e + f*x) + c), x) + Integral(c**2*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x), x) +
 Integral(-2*c**2*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x)**2, x) + Integral(-2*c**2*sqrt(-c*sin(e + f*x) + c)*s
in(e + f*x)**3, x) + Integral(c**2*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x)**4, x) + Integral(c**2*sqrt(-c*sin(e
 + f*x) + c)*sin(e + f*x)**5, x))

Maxima [F]

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

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 210 vs. \(2 (97) = 194\).

Time = 0.42 (sec) , antiderivative size = 210, normalized size of antiderivative = 2.10 \[ \int (3+3 \sin (e+f x))^3 (c-c \sin (e+f x))^{5/2} \, dx=-\frac {\sqrt {2} {\left (6930 \, a^{3} c^{2} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 2310 \, a^{3} c^{2} \cos \left (-\frac {3}{4} \, \pi + \frac {3}{2} \, f x + \frac {3}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 693 \, a^{3} c^{2} \cos \left (-\frac {5}{4} \, \pi + \frac {5}{2} \, f x + \frac {5}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 495 \, a^{3} c^{2} \cos \left (-\frac {7}{4} \, \pi + \frac {7}{2} \, f x + \frac {7}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 77 \, a^{3} c^{2} \cos \left (-\frac {9}{4} \, \pi + \frac {9}{2} \, f x + \frac {9}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 63 \, a^{3} c^{2} \cos \left (-\frac {11}{4} \, \pi + \frac {11}{2} \, f x + \frac {11}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sqrt {c}}{11088 \, f} \]

[In]

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

[Out]

-1/11088*sqrt(2)*(6930*a^3*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)) + 2310*a^3*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)) - 693*a^3*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)) - 495*a^3*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)) + 77*a^3*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)) + 63*a^3*c^2*cos(-
11/4*pi + 11/2*f*x + 11/2*e)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)))*sqrt(c)/f

Mupad [F(-1)]

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

[In]

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

[Out]

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