\(\int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}}{a+a \sin (e+f x)} \, dx\) [109]

Optimal result

Integrand size = 38, antiderivative size = 159 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}}{a+a \sin (e+f x)} \, dx=-\frac {32 (5 A-7 B) c^3 \cos (e+f x)}{15 a f \sqrt {c-c \sin (e+f x)}}-\frac {8 (5 A-7 B) c^2 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{15 a f}-\frac {(5 A-7 B) c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{5 a f}-\frac {(A-B) \sec (e+f x) (c-c \sin (e+f x))^{7/2}}{a c f} \]

[Out]

-1/5*(5*A-7*B)*c*cos(f*x+e)*(c-c*sin(f*x+e))^(3/2)/a/f-(A-B)*sec(f*x+e)*(c-c*sin(f*x+e))^(7/2)/a/c/f-32/15*(5*
A-7*B)*c^3*cos(f*x+e)/a/f/(c-c*sin(f*x+e))^(1/2)-8/15*(5*A-7*B)*c^2*cos(f*x+e)*(c-c*sin(f*x+e))^(1/2)/a/f

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {3046, 2934, 2726, 2725} \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}}{a+a \sin (e+f x)} \, dx=-\frac {32 c^3 (5 A-7 B) \cos (e+f x)}{15 a f \sqrt {c-c \sin (e+f x)}}-\frac {8 c^2 (5 A-7 B) \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{15 a f}-\frac {c (5 A-7 B) \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{5 a f}-\frac {(A-B) \sec (e+f x) (c-c \sin (e+f x))^{7/2}}{a c f} \]

[In]

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

[Out]

(-32*(5*A - 7*B)*c^3*Cos[e + f*x])/(15*a*f*Sqrt[c - c*Sin[e + f*x]]) - (8*(5*A - 7*B)*c^2*Cos[e + f*x]*Sqrt[c
- c*Sin[e + f*x]])/(15*a*f) - ((5*A - 7*B)*c*Cos[e + f*x]*(c - c*Sin[e + f*x])^(3/2))/(5*a*f) - ((A - B)*Sec[e
+ f*x]*(c - c*Sin[e + f*x])^(7/2))/(a*c*f)

Rule 2725

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

Rule 2726

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

Rule 2934

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

Rule 3046

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a^m*c^m, Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m)*(A + B
*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && I
ntegerQ[m] &&  !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))

Rubi steps \begin{align*} \text {integral}& = \frac {\int \sec ^2(e+f x) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx}{a c} \\ & = -\frac {(A-B) \sec (e+f x) (c-c \sin (e+f x))^{7/2}}{a c f}-\frac {(5 A-7 B) \int (c-c \sin (e+f x))^{5/2} \, dx}{2 a} \\ & = -\frac {(5 A-7 B) c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{5 a f}-\frac {(A-B) \sec (e+f x) (c-c \sin (e+f x))^{7/2}}{a c f}-\frac {(4 (5 A-7 B) c) \int (c-c \sin (e+f x))^{3/2} \, dx}{5 a} \\ & = -\frac {8 (5 A-7 B) c^2 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{15 a f}-\frac {(5 A-7 B) c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{5 a f}-\frac {(A-B) \sec (e+f x) (c-c \sin (e+f x))^{7/2}}{a c f}-\frac {\left (16 (5 A-7 B) c^2\right ) \int \sqrt {c-c \sin (e+f x)} \, dx}{15 a} \\ & = -\frac {32 (5 A-7 B) c^3 \cos (e+f x)}{15 a f \sqrt {c-c \sin (e+f x)}}-\frac {8 (5 A-7 B) c^2 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{15 a f}-\frac {(5 A-7 B) c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{5 a f}-\frac {(A-B) \sec (e+f x) (c-c \sin (e+f x))^{7/2}}{a c f} \\ \end{align*}

Mathematica [A] (verified)

Time = 11.90 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.84 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}}{a+a \sin (e+f x)} \, dx=-\frac {c^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {c-c \sin (e+f x)} (450 A-600 B+2 (5 A-16 B) \cos (2 (e+f x))+25 (8 A-13 B) \sin (e+f x)+3 B \sin (3 (e+f x)))}{30 a f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (1+\sin (e+f x))} \]

[In]

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

[Out]

-1/30*(c^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c - c*Sin[e + f*x]]*(450*A - 600*B + 2*(5*A - 16*B)*Cos[
2*(e + f*x)] + 25*(8*A - 13*B)*Sin[e + f*x] + 3*B*Sin[3*(e + f*x)]))/(a*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]
)*(1 + Sin[e + f*x]))

Maple [A] (verified)

Time = 3.15 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.60

method	result	size
default	\(-\frac {2 c^{3} \left (\sin \left (f x +e \right )-1\right ) \left (-3 B \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+\left (-5 A +16 B \right ) \left (\cos ^{2}\left (f x +e \right )\right )+\left (-50 A +82 B \right ) \sin \left (f x +e \right )-110 A +142 B \right )}{15 a \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(95\)

[In]

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

[Out]

-2/15*c^3/a*(sin(f*x+e)-1)*(-3*B*cos(f*x+e)^2*sin(f*x+e)+(-5*A+16*B)*cos(f*x+e)^2+(-50*A+82*B)*sin(f*x+e)-110*
A+142*B)/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.60 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}}{a+a \sin (e+f x)} \, dx=-\frac {2 \, {\left ({\left (5 \, A - 16 \, B\right )} c^{2} \cos \left (f x + e\right )^{2} + 2 \, {\left (55 \, A - 71 \, B\right )} c^{2} + {\left (3 \, B c^{2} \cos \left (f x + e\right )^{2} + 2 \, {\left (25 \, A - 41 \, B\right )} c^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{15 \, a f \cos \left (f x + e\right )} \]

[In]

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

[Out]

-2/15*((5*A - 16*B)*c^2*cos(f*x + e)^2 + 2*(55*A - 71*B)*c^2 + (3*B*c^2*cos(f*x + e)^2 + 2*(25*A - 41*B)*c^2)*
sin(f*x + e))*sqrt(-c*sin(f*x + e) + c)/(a*f*cos(f*x + e))

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 386 vs. \(2 (145) = 290\).

Time = 0.32 (sec) , antiderivative size = 386, normalized size of antiderivative = 2.43 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}}{a+a \sin (e+f x)} \, dx=\frac {2 \, {\left (\frac {5 \, {\left (23 \, c^{\frac {5}{2}} + \frac {20 \, c^{\frac {5}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {65 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {40 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {65 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {20 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac {23 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}}\right )} A}{{\left (a + \frac {a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )} {\left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}^{\frac {5}{2}}} - \frac {2 \, {\left (79 \, c^{\frac {5}{2}} + \frac {79 \, c^{\frac {5}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {205 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {170 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {205 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {79 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac {79 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}}\right )} B}{{\left (a + \frac {a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )} {\left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}^{\frac {5}{2}}}\right )}}{15 \, f} \]

[In]

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

[Out]

2/15*(5*(23*c^(5/2) + 20*c^(5/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 65*c^(5/2)*sin(f*x + e)^2/(cos(f*x + e) + 1
)^2 + 40*c^(5/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 65*c^(5/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 20*c^(
5/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 23*c^(5/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6)*A/((a + a*sin(f*x +
e)/(cos(f*x + e) + 1))*(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)^(5/2)) - 2*(79*c^(5/2) + 79*c^(5/2)*sin(f*x
+ e)/(cos(f*x + e) + 1) + 205*c^(5/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 170*c^(5/2)*sin(f*x + e)^3/(cos(f*
x + e) + 1)^3 + 205*c^(5/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 79*c^(5/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)
^5 + 79*c^(5/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6)*B/((a + a*sin(f*x + e)/(cos(f*x + e) + 1))*(sin(f*x + e)^
2/(cos(f*x + e) + 1)^2 + 1)^(5/2)))/f

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 573 vs. \(2 (145) = 290\).

Time = 0.44 (sec) , antiderivative size = 573, normalized size of antiderivative = 3.60 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}}{a+a \sin (e+f x)} \, dx=\text {Too large to display} \]

[In]

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

[Out]

8/15*sqrt(2)*sqrt(c)*(15*(A*c^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - B*c^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)
))/(a*((cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) + 1)) - (25*A*c^2*sgn(sin(-1/
4*pi + 1/2*f*x + 1/2*e)) - 41*B*c^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 110*A*c^2*(cos(-1/4*pi + 1/2*f*x + 1
/2*e) - 1)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) + 190*B*c^2*(cos(-1/4*pi +
1/2*f*x + 1/2*e) - 1)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) + 160*A*c^2*(c
os(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^
2 - 320*B*c^2*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*
x + 1/2*e) + 1)^2 - 90*A*c^2*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(cos(-
1/4*pi + 1/2*f*x + 1/2*e) + 1)^3 + 90*B*c^2*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^3*sgn(sin(-1/4*pi + 1/2*f*x +
1/2*e))/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^3 + 15*A*c^2*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^4*sgn(sin(-1/4
*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^4 - 15*B*c^2*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)
^4*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^4)/(a*((cos(-1/4*pi + 1/2*f*x + 1/
2*e) - 1)/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) - 1)^5))/f

Mupad [F(-1)]

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

[In]

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

[Out]

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