\(\int (3+3 \sin (e+f x))^m (c-c \sin (e+f x))^{5/2} \, dx\) [411]

Optimal result

Integrand size = 28, antiderivative size = 160 \[ \int (3+3 \sin (e+f x))^m (c-c \sin (e+f x))^{5/2} \, dx=\frac {64 c^3 \cos (e+f x) (3+3 \sin (e+f x))^m}{f (5+2 m) \left (3+8 m+4 m^2\right ) \sqrt {c-c \sin (e+f x)}}+\frac {16 c^2 \cos (e+f x) (3+3 \sin (e+f x))^m \sqrt {c-c \sin (e+f x)}}{f \left (15+16 m+4 m^2\right )}+\frac {2 c \cos (e+f x) (3+3 \sin (e+f x))^m (c-c \sin (e+f x))^{3/2}}{f (5+2 m)} \]

[Out]

2*c*cos(f*x+e)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(3/2)/f/(5+2*m)+64*c^3*cos(f*x+e)*(a+a*sin(f*x+e))^m/f/(5+2
*m)/(4*m^2+8*m+3)/(c-c*sin(f*x+e))^(1/2)+16*c^2*cos(f*x+e)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(1/2)/f/(4*m^2+
16*m+15)

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {2819, 2817} \[ \int (3+3 \sin (e+f x))^m (c-c \sin (e+f x))^{5/2} \, dx=\frac {64 c^3 \cos (e+f x) (a \sin (e+f x)+a)^m}{f (2 m+5) \left (4 m^2+8 m+3\right ) \sqrt {c-c \sin (e+f x)}}+\frac {16 c^2 \cos (e+f x) \sqrt {c-c \sin (e+f x)} (a \sin (e+f x)+a)^m}{f \left (4 m^2+16 m+15\right )}+\frac {2 c \cos (e+f x) (c-c \sin (e+f x))^{3/2} (a \sin (e+f x)+a)^m}{f (2 m+5)} \]

[In]

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

[Out]

(64*c^3*Cos[e + f*x]*(a + a*Sin[e + f*x])^m)/(f*(5 + 2*m)*(3 + 8*m + 4*m^2)*Sqrt[c - c*Sin[e + f*x]]) + (16*c^
2*Cos[e + f*x]*(a + a*Sin[e + f*x])^m*Sqrt[c - c*Sin[e + f*x]])/(f*(15 + 16*m + 4*m^2)) + (2*c*Cos[e + f*x]*(a
+ a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(3/2))/(f*(5 + 2*m))

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])

Rubi steps \begin{align*} \text {integral}& = \frac {2 c \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{3/2}}{f (5+2 m)}+\frac {(8 c) \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{3/2} \, dx}{5+2 m} \\ & = \frac {16 c^2 \cos (e+f x) (a+a \sin (e+f x))^m \sqrt {c-c \sin (e+f x)}}{f \left (15+16 m+4 m^2\right )}+\frac {2 c \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{3/2}}{f (5+2 m)}+\frac {\left (32 c^2\right ) \int (a+a \sin (e+f x))^m \sqrt {c-c \sin (e+f x)} \, dx}{15+16 m+4 m^2} \\ & = \frac {64 c^3 \cos (e+f x) (a+a \sin (e+f x))^m}{f \left (15+46 m+36 m^2+8 m^3\right ) \sqrt {c-c \sin (e+f x)}}+\frac {16 c^2 \cos (e+f x) (a+a \sin (e+f x))^m \sqrt {c-c \sin (e+f x)}}{f \left (15+16 m+4 m^2\right )}+\frac {2 c \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{3/2}}{f (5+2 m)} \\ \end{align*}

Mathematica [A] (verified)

Time = 6.52 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.94 \[ \int (3+3 \sin (e+f x))^m (c-c \sin (e+f x))^{5/2} \, dx=-\frac {3^m c^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (1+\sin (e+f x))^m \sqrt {c-c \sin (e+f x)} \left (-89-56 m-12 m^2+\left (3+8 m+4 m^2\right ) \cos (2 (e+f x))+4 \left (7+16 m+4 m^2\right ) \sin (e+f x)\right )}{f (1+2 m) (3+2 m) (5+2 m) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )} \]

[In]

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

[Out]

-((3^m*c^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(1 + Sin[e + f*x])^m*Sqrt[c - c*Sin[e + f*x]]*(-89 - 56*m - 1
2*m^2 + (3 + 8*m + 4*m^2)*Cos[2*(e + f*x)] + 4*(7 + 16*m + 4*m^2)*Sin[e + f*x]))/(f*(1 + 2*m)*(3 + 2*m)*(5 + 2
*m)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])))

Maple [F]

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.67 \[ \int (3+3 \sin (e+f x))^m (c-c \sin (e+f x))^{5/2} \, dx=-\frac {2 \, {\left ({\left (4 \, c^{2} m^{2} + 8 \, c^{2} m + 3 \, c^{2}\right )} \cos \left (f x + e\right )^{3} - {\left (4 \, c^{2} m^{2} + 24 \, c^{2} m + 11 \, c^{2}\right )} \cos \left (f x + e\right )^{2} - 32 \, c^{2} - 2 \, {\left (4 \, c^{2} m^{2} + 16 \, c^{2} m + 23 \, c^{2}\right )} \cos \left (f x + e\right ) + {\left ({\left (4 \, c^{2} m^{2} + 8 \, c^{2} m + 3 \, c^{2}\right )} \cos \left (f x + e\right )^{2} - 32 \, c^{2} + 2 \, {\left (4 \, c^{2} m^{2} + 16 \, c^{2} m + 7 \, c^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{8 \, f m^{3} + 36 \, f m^{2} + 46 \, f m + {\left (8 \, f m^{3} + 36 \, f m^{2} + 46 \, f m + 15 \, f\right )} \cos \left (f x + e\right ) - {\left (8 \, f m^{3} + 36 \, f m^{2} + 46 \, f m + 15 \, f\right )} \sin \left (f x + e\right ) + 15 \, f} \]

[In]

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

[Out]

-2*((4*c^2*m^2 + 8*c^2*m + 3*c^2)*cos(f*x + e)^3 - (4*c^2*m^2 + 24*c^2*m + 11*c^2)*cos(f*x + e)^2 - 32*c^2 - 2
*(4*c^2*m^2 + 16*c^2*m + 23*c^2)*cos(f*x + e) + ((4*c^2*m^2 + 8*c^2*m + 3*c^2)*cos(f*x + e)^2 - 32*c^2 + 2*(4*
c^2*m^2 + 16*c^2*m + 7*c^2)*cos(f*x + e))*sin(f*x + e))*sqrt(-c*sin(f*x + e) + c)*(a*sin(f*x + e) + a)^m/(8*f*
m^3 + 36*f*m^2 + 46*f*m + (8*f*m^3 + 36*f*m^2 + 46*f*m + 15*f)*cos(f*x + e) - (8*f*m^3 + 36*f*m^2 + 46*f*m + 1
5*f)*sin(f*x + e) + 15*f)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 290, normalized size of antiderivative = 1.81 \[ \int (3+3 \sin (e+f x))^m (c-c \sin (e+f x))^{5/2} \, dx=-\frac {2 \, {\left ({\left (4 \, m^{2} + 24 \, m + 43\right )} a^{m} c^{\frac {5}{2}} - \frac {{\left (12 \, m^{2} + 40 \, m - 15\right )} a^{m} c^{\frac {5}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {2 \, {\left (4 \, m^{2} + 8 \, m + 35\right )} a^{m} c^{\frac {5}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {2 \, {\left (4 \, m^{2} + 8 \, m + 35\right )} a^{m} c^{\frac {5}{2}} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac {{\left (12 \, m^{2} + 40 \, m - 15\right )} a^{m} c^{\frac {5}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {{\left (4 \, m^{2} + 24 \, m + 43\right )} a^{m} c^{\frac {5}{2}} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )} e^{\left (2 \, m \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right ) - m \log \left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )\right )}}{{\left (8 \, m^{3} + 36 \, m^{2} + 46 \, m + 15\right )} f {\left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}^{\frac {5}{2}}} \]

[In]

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

[Out]

-2*((4*m^2 + 24*m + 43)*a^m*c^(5/2) - (12*m^2 + 40*m - 15)*a^m*c^(5/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 2*(4*
m^2 + 8*m + 35)*a^m*c^(5/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 2*(4*m^2 + 8*m + 35)*a^m*c^(5/2)*sin(f*x + e
)^3/(cos(f*x + e) + 1)^3 - (12*m^2 + 40*m - 15)*a^m*c^(5/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + (4*m^2 + 24*
m + 43)*a^m*c^(5/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5)*e^(2*m*log(sin(f*x + e)/(cos(f*x + e) + 1) + 1) - m*l
og(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1))/((8*m^3 + 36*m^2 + 46*m + 15)*f*(sin(f*x + e)^2/(cos(f*x + e) + 1
)^2 + 1)^(5/2))

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 10.05 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.02 \[ \int (3+3 \sin (e+f x))^m (c-c \sin (e+f x))^{5/2} \, dx=\frac {c^2\,{\left (a\,\left (\sin \left (e+f\,x\right )+1\right )\right )}^m\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (3\,\cos \left (3\,e+3\,f\,x\right )-175\,\cos \left (e+f\,x\right )+28\,\sin \left (2\,e+2\,f\,x\right )+16\,m^2\,\sin \left (2\,e+2\,f\,x\right )-104\,m\,\cos \left (e+f\,x\right )+8\,m\,\cos \left (3\,e+3\,f\,x\right )-20\,m^2\,\cos \left (e+f\,x\right )+64\,m\,\sin \left (2\,e+2\,f\,x\right )+4\,m^2\,\cos \left (3\,e+3\,f\,x\right )\right )}{2\,f\,\left (\sin \left (e+f\,x\right )-1\right )\,\left (8\,m^3+36\,m^2+46\,m+15\right )} \]

[In]

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

[Out]

(c^2*(a*(sin(e + f*x) + 1))^m*(-c*(sin(e + f*x) - 1))^(1/2)*(3*cos(3*e + 3*f*x) - 175*cos(e + f*x) + 28*sin(2*
e + 2*f*x) + 16*m^2*sin(2*e + 2*f*x) - 104*m*cos(e + f*x) + 8*m*cos(3*e + 3*f*x) - 20*m^2*cos(e + f*x) + 64*m*
sin(2*e + 2*f*x) + 4*m^2*cos(3*e + 3*f*x)))/(2*f*(sin(e + f*x) - 1)*(46*m + 36*m^2 + 8*m^3 + 15))