\(\int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^2} \, dx\) [116]
Optimal result
Integrand size = 38, antiderivative size = 201 \[
\int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^2} \, dx=\frac {128 (5 A-11 B) c^3 \sec (e+f x) \sqrt {c-c \sin (e+f x)}}{15 a^2 f}-\frac {32 (5 A-11 B) c^2 \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{15 a^2 f}-\frac {4 (5 A-11 B) c \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{15 a^2 f}-\frac {(5 A-11 B) \sec (e+f x) (c-c \sin (e+f x))^{7/2}}{15 a^2 f}-\frac {(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{11/2}}{3 a^2 c^2 f}
\]
[Out]
-32/15*(5*A-11*B)*c^2*sec(f*x+e)*(c-c*sin(f*x+e))^(3/2)/a^2/f-4/15*(5*A-11*B)*c*sec(f*x+e)*(c-c*sin(f*x+e))^(5
/2)/a^2/f-1/15*(5*A-11*B)*sec(f*x+e)*(c-c*sin(f*x+e))^(7/2)/a^2/f-1/3*(A-B)*sec(f*x+e)^3*(c-c*sin(f*x+e))^(11/
2)/a^2/c^2/f+128/15*(5*A-11*B)*c^3*sec(f*x+e)*(c-c*sin(f*x+e))^(1/2)/a^2/f
Rubi [A] (verified)
Time = 0.40 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.00, number of
steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {3046, 2934, 2753, 2752}
\[
\int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^2} \, dx=\frac {128 c^3 (5 A-11 B) \sec (e+f x) \sqrt {c-c \sin (e+f x)}}{15 a^2 f}-\frac {(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{11/2}}{3 a^2 c^2 f}-\frac {32 c^2 (5 A-11 B) \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{15 a^2 f}-\frac {(5 A-11 B) \sec (e+f x) (c-c \sin (e+f x))^{7/2}}{15 a^2 f}-\frac {4 c (5 A-11 B) \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{15 a^2 f}
\]
[In]
Int[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(7/2))/(a + a*Sin[e + f*x])^2,x]
[Out]
(128*(5*A - 11*B)*c^3*Sec[e + f*x]*Sqrt[c - c*Sin[e + f*x]])/(15*a^2*f) - (32*(5*A - 11*B)*c^2*Sec[e + f*x]*(c
- c*Sin[e + f*x])^(3/2))/(15*a^2*f) - (4*(5*A - 11*B)*c*Sec[e + f*x]*(c - c*Sin[e + f*x])^(5/2))/(15*a^2*f) -
((5*A - 11*B)*Sec[e + f*x]*(c - c*Sin[e + f*x])^(7/2))/(15*a^2*f) - ((A - B)*Sec[e + f*x]^3*(c - c*Sin[e + f*
x])^(11/2))/(3*a^2*c^2*f)
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 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 ^4(e+f x) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{11/2} \, dx}{a^2 c^2} \\ & = -\frac {(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{11/2}}{3 a^2 c^2 f}-\frac {(5 A-11 B) \int \sec ^2(e+f x) (c-c \sin (e+f x))^{9/2} \, dx}{6 a^2 c} \\ & = -\frac {(5 A-11 B) \sec (e+f x) (c-c \sin (e+f x))^{7/2}}{15 a^2 f}-\frac {(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{11/2}}{3 a^2 c^2 f}-\frac {(2 (5 A-11 B)) \int \sec ^2(e+f x) (c-c \sin (e+f x))^{7/2} \, dx}{5 a^2} \\ & = -\frac {4 (5 A-11 B) c \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{15 a^2 f}-\frac {(5 A-11 B) \sec (e+f x) (c-c \sin (e+f x))^{7/2}}{15 a^2 f}-\frac {(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{11/2}}{3 a^2 c^2 f}-\frac {(16 (5 A-11 B) c) \int \sec ^2(e+f x) (c-c \sin (e+f x))^{5/2} \, dx}{15 a^2} \\ & = -\frac {32 (5 A-11 B) c^2 \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{15 a^2 f}-\frac {4 (5 A-11 B) c \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{15 a^2 f}-\frac {(5 A-11 B) \sec (e+f x) (c-c \sin (e+f x))^{7/2}}{15 a^2 f}-\frac {(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{11/2}}{3 a^2 c^2 f}-\frac {\left (64 (5 A-11 B) c^2\right ) \int \sec ^2(e+f x) (c-c \sin (e+f x))^{3/2} \, dx}{15 a^2} \\ & = \frac {128 (5 A-11 B) c^3 \sec (e+f x) \sqrt {c-c \sin (e+f x)}}{15 a^2 f}-\frac {32 (5 A-11 B) c^2 \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{15 a^2 f}-\frac {4 (5 A-11 B) c \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{15 a^2 f}-\frac {(5 A-11 B) \sec (e+f x) (c-c \sin (e+f x))^{7/2}}{15 a^2 f}-\frac {(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{11/2}}{3 a^2 c^2 f} \\
\end{align*}
Mathematica [A] (verified)
Time = 8.24 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.79
\[
\int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^2} \, dx=-\frac {c^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {c-c \sin (e+f x)} (-2100 A+4725 B+12 (25 A-62 B) \cos (2 (e+f x))+3 B \cos (4 (e+f x))-2730 A \sin (e+f x)+5838 B \sin (e+f x)-10 A \sin (3 (e+f x))+46 B \sin (3 (e+f x)))}{60 a^2 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (1+\sin (e+f x))^2}
\]
[In]
Integrate[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(7/2))/(a + a*Sin[e + f*x])^2,x]
[Out]
-1/60*(c^3*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c - c*Sin[e + f*x]]*(-2100*A + 4725*B + 12*(25*A - 62*B)
*Cos[2*(e + f*x)] + 3*B*Cos[4*(e + f*x)] - 2730*A*Sin[e + f*x] + 5838*B*Sin[e + f*x] - 10*A*Sin[3*(e + f*x)] +
46*B*Sin[3*(e + f*x)]))/(a^2*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(1 + Sin[e + f*x])^2)
Maple [A] (verified)
Time = 20.18 (sec) , antiderivative size = 121, normalized size of antiderivative =
0.60
| | |
| method | result | size |
| | |
| default |
\(\frac {2 c^{4} \left (\sin \left (f x +e \right )-1\right ) \left (3 B \left (\cos ^{4}\left (f x +e \right )\right )+\left (-5 A +23 B \right ) \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+\left (75 A -189 B \right ) \left (\cos ^{2}\left (f x +e \right )\right )+\left (-340 A +724 B \right ) \sin \left (f x +e \right )-300 A +684 B \right )}{15 a^{2} \left (1+\sin \left (f x +e \right )\right ) \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\) |
\(121\) |
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[In]
int((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(7/2)/(a+a*sin(f*x+e))^2,x,method=_RETURNVERBOSE)
[Out]
2/15*c^4/a^2*(sin(f*x+e)-1)/(1+sin(f*x+e))*(3*B*cos(f*x+e)^4+(-5*A+23*B)*cos(f*x+e)^2*sin(f*x+e)+(75*A-189*B)*
cos(f*x+e)^2+(-340*A+724*B)*sin(f*x+e)-300*A+684*B)/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f
Fricas [A] (verification not implemented)
none
Time = 0.26 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.66
\[
\int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^2} \, dx=-\frac {2 \, {\left (3 \, B c^{3} \cos \left (f x + e\right )^{4} + 3 \, {\left (25 \, A - 63 \, B\right )} c^{3} \cos \left (f x + e\right )^{2} - 12 \, {\left (25 \, A - 57 \, B\right )} c^{3} - {\left ({\left (5 \, A - 23 \, B\right )} c^{3} \cos \left (f x + e\right )^{2} + 4 \, {\left (85 \, A - 181 \, B\right )} c^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{15 \, {\left (a^{2} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) + a^{2} f \cos \left (f x + e\right )\right )}}
\]
[In]
integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(7/2)/(a+a*sin(f*x+e))^2,x, algorithm="fricas")
[Out]
-2/15*(3*B*c^3*cos(f*x + e)^4 + 3*(25*A - 63*B)*c^3*cos(f*x + e)^2 - 12*(25*A - 57*B)*c^3 - ((5*A - 23*B)*c^3*
cos(f*x + e)^2 + 4*(85*A - 181*B)*c^3)*sin(f*x + e))*sqrt(-c*sin(f*x + e) + c)/(a^2*f*cos(f*x + e)*sin(f*x + e
) + a^2*f*cos(f*x + e))
Sympy [F(-1)]
Timed out. \[
\int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^2} \, dx=\text {Timed out}
\]
[In]
integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))**(7/2)/(a+a*sin(f*x+e))**2,x)
[Out]
Timed out
Maxima [B] (verification not implemented)
Leaf count of result is larger than twice the leaf count of optimal. 670 vs. \(2 (181) = 362\).
Time = 0.34 (sec) , antiderivative size = 670, normalized size of antiderivative = 3.33
\[
\int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^2} \, dx=\text {Too large to display}
\]
[In]
integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(7/2)/(a+a*sin(f*x+e))^2,x, algorithm="maxima")
[Out]
-2/15*(5*(45*c^(7/2) + 138*c^(7/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 285*c^(7/2)*sin(f*x + e)^2/(cos(f*x + e)
+ 1)^2 + 544*c^(7/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 630*c^(7/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 8
12*c^(7/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 630*c^(7/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 544*c^(7/2)
*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + 285*c^(7/2)*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 + 138*c^(7/2)*sin(f*x +
e)^9/(cos(f*x + e) + 1)^9 + 45*c^(7/2)*sin(f*x + e)^10/(cos(f*x + e) + 1)^10)*A/((a^2 + 3*a^2*sin(f*x + e)/(c
os(f*x + e) + 1) + 3*a^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + a^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3)*(sin(f
*x + e)^2/(cos(f*x + e) + 1)^2 + 1)^(7/2)) - 2*(249*c^(7/2) + 747*c^(7/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 16
11*c^(7/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 2896*c^(7/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3612*c^(7/
2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 4298*c^(7/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 3612*c^(7/2)*sin(f
*x + e)^6/(cos(f*x + e) + 1)^6 + 2896*c^(7/2)*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + 1611*c^(7/2)*sin(f*x + e)^
8/(cos(f*x + e) + 1)^8 + 747*c^(7/2)*sin(f*x + e)^9/(cos(f*x + e) + 1)^9 + 249*c^(7/2)*sin(f*x + e)^10/(cos(f*
x + e) + 1)^10)*B/((a^2 + 3*a^2*sin(f*x + e)/(cos(f*x + e) + 1) + 3*a^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 +
a^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3)*(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)^(7/2)))/f
Giac [B] (verification not implemented)
Leaf count of result is larger than twice the leaf count of optimal. 774 vs. \(2 (181) = 362\).
Time = 0.75 (sec) , antiderivative size = 774, normalized size of antiderivative = 3.85
\[
\int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^2} \, dx=\text {Too large to display}
\]
[In]
integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(7/2)/(a+a*sin(f*x+e))^2,x, algorithm="giac")
[Out]
-16/15*sqrt(2)*sqrt(c)*(5*(4*A*c^3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 7*B*c^3*sgn(sin(-1/4*pi + 1/2*f*x + 1
/2*e)) + 9*A*c^3*(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) - 15*B*c^3*(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) + 3*A*c^3*(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 - 6*B*c^3*(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)/(a^2*((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)^3) - (20*A*c^3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 53*B*c^3*sgn(sin(-1/4*pi
+ 1/2*f*x + 1/2*e)) - 85*A*c^3*(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) + 235*B*c^3*(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) + 125*A*c^3*(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 - 365*B*c^3*(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 - 75*A*c^3*(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 + 165*B*c^3*(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^3*(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 - 30*B*c^3*(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^2*((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. \[
\int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^2} \, dx=\int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{7/2}}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^2} \,d x
\]
[In]
int(((A + B*sin(e + f*x))*(c - c*sin(e + f*x))^(7/2))/(a + a*sin(e + f*x))^2,x)
[Out]
int(((A + B*sin(e + f*x))*(c - c*sin(e + f*x))^(7/2))/(a + a*sin(e + f*x))^2, x)