\(\int \frac {1}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^4} \, dx\) [268]

Optimal result

Integrand size = 26, antiderivative size = 118 \[ \int \frac {1}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^4} \, dx=\frac {\sec (e+f x)}{7 a c f (c-c \sin (e+f x))^3}+\frac {4 \sec (e+f x)}{35 a f \left (c^2-c^2 \sin (e+f x)\right )^2}+\frac {4 \sec (e+f x)}{35 a f \left (c^4-c^4 \sin (e+f x)\right )}+\frac {8 \tan (e+f x)}{35 a c^4 f} \]

[Out]

1/7*sec(f*x+e)/a/c/f/(c-c*sin(f*x+e))^3+4/35*sec(f*x+e)/a/f/(c^2-c^2*sin(f*x+e))^2+4/35*sec(f*x+e)/a/f/(c^4-c^
4*sin(f*x+e))+8/35*tan(f*x+e)/a/c^4/f

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2815, 2751, 3852, 8} \[ \int \frac {1}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^4} \, dx=\frac {8 \tan (e+f x)}{35 a c^4 f}+\frac {4 \sec (e+f x)}{35 a f \left (c^4-c^4 \sin (e+f x)\right )}+\frac {4 \sec (e+f x)}{35 a f \left (c^2-c^2 \sin (e+f x)\right )^2}+\frac {\sec (e+f x)}{7 a c f (c-c \sin (e+f x))^3} \]

[In]

Int[1/((a + a*Sin[e + f*x])*(c - c*Sin[e + f*x])^4),x]

[Out]

Sec[e + f*x]/(7*a*c*f*(c - c*Sin[e + f*x])^3) + (4*Sec[e + f*x])/(35*a*f*(c^2 - c^2*Sin[e + f*x])^2) + (4*Sec[
e + f*x])/(35*a*f*(c^4 - c^4*Sin[e + f*x])) + (8*Tan[e + f*x])/(35*a*c^4*f)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2751

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/(a*f*g*Simplify[2*m + p + 1])), x] + Dist[Simplify[m + p + 1]/(a*
Simplify[2*m + p + 1]), 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] && ILtQ[Simplify[m + p + 1], 0] && NeQ[2*m + p + 1, 0] &&  !IGtQ[m, 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]))

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {\sec ^2(e+f x)}{(c-c \sin (e+f x))^3} \, dx}{a c} \\ & = \frac {\sec (e+f x)}{7 a c f (c-c \sin (e+f x))^3}+\frac {4 \int \frac {\sec ^2(e+f x)}{(c-c \sin (e+f x))^2} \, dx}{7 a c^2} \\ & = \frac {\sec (e+f x)}{7 a c f (c-c \sin (e+f x))^3}+\frac {4 \sec (e+f x)}{35 a f \left (c^2-c^2 \sin (e+f x)\right )^2}+\frac {12 \int \frac {\sec ^2(e+f x)}{c-c \sin (e+f x)} \, dx}{35 a c^3} \\ & = \frac {\sec (e+f x)}{7 a c f (c-c \sin (e+f x))^3}+\frac {4 \sec (e+f x)}{35 a f \left (c^2-c^2 \sin (e+f x)\right )^2}+\frac {4 \sec (e+f x)}{35 a f \left (c^4-c^4 \sin (e+f x)\right )}+\frac {8 \int \sec ^2(e+f x) \, dx}{35 a c^4} \\ & = \frac {\sec (e+f x)}{7 a c f (c-c \sin (e+f x))^3}+\frac {4 \sec (e+f x)}{35 a f \left (c^2-c^2 \sin (e+f x)\right )^2}+\frac {4 \sec (e+f x)}{35 a f \left (c^4-c^4 \sin (e+f x)\right )}-\frac {8 \text {Subst}(\int 1 \, dx,x,-\tan (e+f x))}{35 a c^4 f} \\ & = \frac {\sec (e+f x)}{7 a c f (c-c \sin (e+f x))^3}+\frac {4 \sec (e+f x)}{35 a f \left (c^2-c^2 \sin (e+f x)\right )^2}+\frac {4 \sec (e+f x)}{35 a f \left (c^4-c^4 \sin (e+f x)\right )}+\frac {8 \tan (e+f x)}{35 a c^4 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.19 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.31 \[ \int \frac {1}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^4} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (9170 \cos (e+f x)+1792 \cos (2 (e+f x))-3930 \cos (3 (e+f x))-128 \cos (4 (e+f x))+1792 \sin (e+f x)-9170 \sin (2 (e+f x))-768 \sin (3 (e+f x))+655 \sin (4 (e+f x)))}{4480 a c^4 f (-1+\sin (e+f x))^4 (1+\sin (e+f x))} \]

[In]

Integrate[1/((a + a*Sin[e + f*x])*(c - c*Sin[e + f*x])^4),x]

[Out]

((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(9170*Cos[e + f*x] + 1792*Cos[2*(
e + f*x)] - 3930*Cos[3*(e + f*x)] - 128*Cos[4*(e + f*x)] + 1792*Sin[e + f*x] - 9170*Sin[2*(e + f*x)] - 768*Sin
[3*(e + f*x)] + 655*Sin[4*(e + f*x)]))/(4480*a*c^4*f*(-1 + Sin[e + f*x])^4*(1 + Sin[e + f*x]))

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 1.12 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.65

method	result	size
risch	\(-\frac {16 \left (-6 \,{\mathrm e}^{i \left (f x +e \right )}+i+14 \,{\mathrm e}^{3 i \left (f x +e \right )}-14 i {\mathrm e}^{2 i \left (f x +e \right )}\right )}{35 \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )^{7} \left ({\mathrm e}^{i \left (f x +e \right )}+i\right ) a \,c^{4} f}\)	\(77\)
parallelrisch	\(\frac {-\frac {26}{35}-10 \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+6 \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+6 \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-2 \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\frac {2 \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{5}+\frac {86 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{35}-\frac {22 \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{5}}{f a \,c^{4} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{7}}\)	\(129\)
derivativedivides	\(\frac {-\frac {1}{8 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}-\frac {8}{7 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{7}}-\frac {4}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{6}}-\frac {38}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}-\frac {9}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}-\frac {15}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {17}{4 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {15}{8 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}}{a \,c^{4} f}\)	\(133\)
default	\(\frac {-\frac {1}{8 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}-\frac {8}{7 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{7}}-\frac {4}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{6}}-\frac {38}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}-\frac {9}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}-\frac {15}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {17}{4 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {15}{8 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}}{a \,c^{4} f}\)	\(133\)
norman	\(\frac {\frac {6 \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a c f}-\frac {26}{35 a c f}-\frac {2 \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a c f}+\frac {6 \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a c f}-\frac {10 \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a c f}+\frac {2 \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{5 a c f}-\frac {22 \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{5 a c f}+\frac {86 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{35 a c f}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) c^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{7}}\)	\(195\)

[In]

int(1/(a+a*sin(f*x+e))/(c-c*sin(f*x+e))^4,x,method=_RETURNVERBOSE)

[Out]

-16/35*(-6*exp(I*(f*x+e))+I+14*exp(3*I*(f*x+e))-14*I*exp(2*I*(f*x+e)))/(exp(I*(f*x+e))-I)^7/(exp(I*(f*x+e))+I)
/a/c^4/f

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.94 \[ \int \frac {1}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^4} \, dx=\frac {8 \, \cos \left (f x + e\right )^{4} - 36 \, \cos \left (f x + e\right )^{2} + 4 \, {\left (6 \, \cos \left (f x + e\right )^{2} - 5\right )} \sin \left (f x + e\right ) + 15}{35 \, {\left (3 \, a c^{4} f \cos \left (f x + e\right )^{3} - 4 \, a c^{4} f \cos \left (f x + e\right ) - {\left (a c^{4} f \cos \left (f x + e\right )^{3} - 4 \, a c^{4} f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )}} \]

[In]

integrate(1/(a+a*sin(f*x+e))/(c-c*sin(f*x+e))^4,x, algorithm="fricas")

[Out]

1/35*(8*cos(f*x + e)^4 - 36*cos(f*x + e)^2 + 4*(6*cos(f*x + e)^2 - 5)*sin(f*x + e) + 15)/(3*a*c^4*f*cos(f*x +
e)^3 - 4*a*c^4*f*cos(f*x + e) - (a*c^4*f*cos(f*x + e)^3 - 4*a*c^4*f*cos(f*x + e))*sin(f*x + e))

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1307 vs. \(2 (97) = 194\).

Time = 4.89 (sec) , antiderivative size = 1307, normalized size of antiderivative = 11.08 \[ \int \frac {1}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^4} \, dx=\text {Too large to display} \]

[In]

integrate(1/(a+a*sin(f*x+e))/(c-c*sin(f*x+e))**4,x)

[Out]

Piecewise((-70*tan(e/2 + f*x/2)**7/(35*a*c**4*f*tan(e/2 + f*x/2)**8 - 210*a*c**4*f*tan(e/2 + f*x/2)**7 + 490*a
*c**4*f*tan(e/2 + f*x/2)**6 - 490*a*c**4*f*tan(e/2 + f*x/2)**5 + 490*a*c**4*f*tan(e/2 + f*x/2)**3 - 490*a*c**4
*f*tan(e/2 + f*x/2)**2 + 210*a*c**4*f*tan(e/2 + f*x/2) - 35*a*c**4*f) + 210*tan(e/2 + f*x/2)**6/(35*a*c**4*f*t
an(e/2 + f*x/2)**8 - 210*a*c**4*f*tan(e/2 + f*x/2)**7 + 490*a*c**4*f*tan(e/2 + f*x/2)**6 - 490*a*c**4*f*tan(e/
2 + f*x/2)**5 + 490*a*c**4*f*tan(e/2 + f*x/2)**3 - 490*a*c**4*f*tan(e/2 + f*x/2)**2 + 210*a*c**4*f*tan(e/2 + f
*x/2) - 35*a*c**4*f) - 350*tan(e/2 + f*x/2)**5/(35*a*c**4*f*tan(e/2 + f*x/2)**8 - 210*a*c**4*f*tan(e/2 + f*x/2
)**7 + 490*a*c**4*f*tan(e/2 + f*x/2)**6 - 490*a*c**4*f*tan(e/2 + f*x/2)**5 + 490*a*c**4*f*tan(e/2 + f*x/2)**3
- 490*a*c**4*f*tan(e/2 + f*x/2)**2 + 210*a*c**4*f*tan(e/2 + f*x/2) - 35*a*c**4*f) + 210*tan(e/2 + f*x/2)**4/(3
5*a*c**4*f*tan(e/2 + f*x/2)**8 - 210*a*c**4*f*tan(e/2 + f*x/2)**7 + 490*a*c**4*f*tan(e/2 + f*x/2)**6 - 490*a*c
**4*f*tan(e/2 + f*x/2)**5 + 490*a*c**4*f*tan(e/2 + f*x/2)**3 - 490*a*c**4*f*tan(e/2 + f*x/2)**2 + 210*a*c**4*f
*tan(e/2 + f*x/2) - 35*a*c**4*f) + 14*tan(e/2 + f*x/2)**3/(35*a*c**4*f*tan(e/2 + f*x/2)**8 - 210*a*c**4*f*tan(
e/2 + f*x/2)**7 + 490*a*c**4*f*tan(e/2 + f*x/2)**6 - 490*a*c**4*f*tan(e/2 + f*x/2)**5 + 490*a*c**4*f*tan(e/2 +
f*x/2)**3 - 490*a*c**4*f*tan(e/2 + f*x/2)**2 + 210*a*c**4*f*tan(e/2 + f*x/2) - 35*a*c**4*f) - 154*tan(e/2 + f
*x/2)**2/(35*a*c**4*f*tan(e/2 + f*x/2)**8 - 210*a*c**4*f*tan(e/2 + f*x/2)**7 + 490*a*c**4*f*tan(e/2 + f*x/2)**
6 - 490*a*c**4*f*tan(e/2 + f*x/2)**5 + 490*a*c**4*f*tan(e/2 + f*x/2)**3 - 490*a*c**4*f*tan(e/2 + f*x/2)**2 + 2
10*a*c**4*f*tan(e/2 + f*x/2) - 35*a*c**4*f) + 86*tan(e/2 + f*x/2)/(35*a*c**4*f*tan(e/2 + f*x/2)**8 - 210*a*c**
4*f*tan(e/2 + f*x/2)**7 + 490*a*c**4*f*tan(e/2 + f*x/2)**6 - 490*a*c**4*f*tan(e/2 + f*x/2)**5 + 490*a*c**4*f*t
an(e/2 + f*x/2)**3 - 490*a*c**4*f*tan(e/2 + f*x/2)**2 + 210*a*c**4*f*tan(e/2 + f*x/2) - 35*a*c**4*f) - 26/(35*
a*c**4*f*tan(e/2 + f*x/2)**8 - 210*a*c**4*f*tan(e/2 + f*x/2)**7 + 490*a*c**4*f*tan(e/2 + f*x/2)**6 - 490*a*c**
4*f*tan(e/2 + f*x/2)**5 + 490*a*c**4*f*tan(e/2 + f*x/2)**3 - 490*a*c**4*f*tan(e/2 + f*x/2)**2 + 210*a*c**4*f*t
an(e/2 + f*x/2) - 35*a*c**4*f), Ne(f, 0)), (x/((a*sin(e) + a)*(-c*sin(e) + c)**4), True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 319 vs. \(2 (113) = 226\).

Time = 0.22 (sec) , antiderivative size = 319, normalized size of antiderivative = 2.70 \[ \int \frac {1}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^4} \, dx=-\frac {2 \, {\left (\frac {43 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {77 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {7 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {105 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {175 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac {105 \, \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} - \frac {35 \, \sin \left (f x + e\right )^{7}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{7}} - 13\right )}}{35 \, {\left (a c^{4} - \frac {6 \, a c^{4} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {14 \, a c^{4} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {14 \, a c^{4} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {14 \, a c^{4} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} - \frac {14 \, a c^{4} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} + \frac {6 \, a c^{4} \sin \left (f x + e\right )^{7}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{7}} - \frac {a c^{4} \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}}\right )} f} \]

[In]

integrate(1/(a+a*sin(f*x+e))/(c-c*sin(f*x+e))^4,x, algorithm="maxima")

[Out]

-2/35*(43*sin(f*x + e)/(cos(f*x + e) + 1) - 77*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 7*sin(f*x + e)^3/(cos(f*x
+ e) + 1)^3 + 105*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 175*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 105*sin(f*x
+ e)^6/(cos(f*x + e) + 1)^6 - 35*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 - 13)/((a*c^4 - 6*a*c^4*sin(f*x + e)/(co
s(f*x + e) + 1) + 14*a*c^4*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 14*a*c^4*sin(f*x + e)^3/(cos(f*x + e) + 1)^3
+ 14*a*c^4*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 - 14*a*c^4*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 6*a*c^4*sin(f*
x + e)^7/(cos(f*x + e) + 1)^7 - a*c^4*sin(f*x + e)^8/(cos(f*x + e) + 1)^8)*f)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.06 \[ \int \frac {1}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^4} \, dx=-\frac {\frac {35}{a c^{4} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}} + \frac {525 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} - 1960 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 4025 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 4480 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 3143 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1176 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 243}{a c^{4} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{7}}}{280 \, f} \]

[In]

integrate(1/(a+a*sin(f*x+e))/(c-c*sin(f*x+e))^4,x, algorithm="giac")

[Out]

-1/280*(35/(a*c^4*(tan(1/2*f*x + 1/2*e) + 1)) + (525*tan(1/2*f*x + 1/2*e)^6 - 1960*tan(1/2*f*x + 1/2*e)^5 + 40
25*tan(1/2*f*x + 1/2*e)^4 - 4480*tan(1/2*f*x + 1/2*e)^3 + 3143*tan(1/2*f*x + 1/2*e)^2 - 1176*tan(1/2*f*x + 1/2
*e) + 243)/(a*c^4*(tan(1/2*f*x + 1/2*e) - 1)^7))/f

Mupad [B] (verification not implemented)

Time = 6.54 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.81 \[ \int \frac {1}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^4} \, dx=\frac {\frac {2\,\sin \left (e+f\,x\right )}{5}+\frac {2\,\cos \left (2\,e+2\,f\,x\right )}{5}-\frac {\cos \left (4\,e+4\,f\,x\right )}{35}-\frac {6\,\sin \left (3\,e+3\,f\,x\right )}{35}}{a\,c^4\,f\,\left (\frac {7\,\cos \left (e+f\,x\right )}{4}-\frac {3\,\cos \left (3\,e+3\,f\,x\right )}{4}-\frac {7\,\sin \left (2\,e+2\,f\,x\right )}{4}+\frac {\sin \left (4\,e+4\,f\,x\right )}{8}\right )} \]

[In]

int(1/((a + a*sin(e + f*x))*(c - c*sin(e + f*x))^4),x)

[Out]

((2*sin(e + f*x))/5 + (2*cos(2*e + 2*f*x))/5 - cos(4*e + 4*f*x)/35 - (6*sin(3*e + 3*f*x))/35)/(a*c^4*f*((7*cos
(e + f*x))/4 - (3*cos(3*e + 3*f*x))/4 - (7*sin(2*e + 2*f*x))/4 + sin(4*e + 4*f*x)/8))