\(\int \frac {(c-c \sin (e+f x))^4}{a+a \sin (e+f x)} \, dx\) [261]

Optimal result

Integrand size = 26, antiderivative size = 118 \[ \int \frac {(c-c \sin (e+f x))^4}{a+a \sin (e+f x)} \, dx=-\frac {35 c^4 x}{2 a}-\frac {35 c^4 \cos ^3(e+f x)}{3 a f}-\frac {35 c^4 \cos (e+f x) \sin (e+f x)}{2 a f}-\frac {2 a^3 c^4 \cos ^7(e+f x)}{f (a+a \sin (e+f x))^4}-\frac {14 a c^4 \cos ^5(e+f x)}{f (a+a \sin (e+f x))^2} \]

[Out]

-35/2*c^4*x/a-35/3*c^4*cos(f*x+e)^3/a/f-35/2*c^4*cos(f*x+e)*sin(f*x+e)/a/f-2*a^3*c^4*cos(f*x+e)^7/f/(a+a*sin(f
*x+e))^4-14*a*c^4*cos(f*x+e)^5/f/(a+a*sin(f*x+e))^2

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {2815, 2759, 2761, 2715, 8} \[ \int \frac {(c-c \sin (e+f x))^4}{a+a \sin (e+f x)} \, dx=-\frac {2 a^3 c^4 \cos ^7(e+f x)}{f (a \sin (e+f x)+a)^4}-\frac {35 c^4 \cos ^3(e+f x)}{3 a f}-\frac {14 a c^4 \cos ^5(e+f x)}{f (a \sin (e+f x)+a)^2}-\frac {35 c^4 \sin (e+f x) \cos (e+f x)}{2 a f}-\frac {35 c^4 x}{2 a} \]

[In]

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

[Out]

(-35*c^4*x)/(2*a) - (35*c^4*Cos[e + f*x]^3)/(3*a*f) - (35*c^4*Cos[e + f*x]*Sin[e + f*x])/(2*a*f) - (2*a^3*c^4*
Cos[e + f*x]^7)/(f*(a + a*Sin[e + f*x])^4) - (14*a*c^4*Cos[e + f*x]^5)/(f*(a + a*Sin[e + f*x])^2)

Rule 8

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

Rule 2715

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

Rule 2759

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[2*g*(g
*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(2*m + p + 1))), x] + Dist[g^2*((p - 1)/(b^2*(2*m +
p + 1))), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, g}, x] && Eq
Q[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] && NeQ[2*m + p + 1, 0] &&  !ILtQ[m + p + 1, 0] && IntegersQ[2*m, 2*
p]

Rule 2761

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

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^4 c^4\right ) \int \frac {\cos ^8(e+f x)}{(a+a \sin (e+f x))^5} \, dx \\ & = -\frac {2 a^3 c^4 \cos ^7(e+f x)}{f (a+a \sin (e+f x))^4}-\left (7 a^2 c^4\right ) \int \frac {\cos ^6(e+f x)}{(a+a \sin (e+f x))^3} \, dx \\ & = -\frac {2 a^3 c^4 \cos ^7(e+f x)}{f (a+a \sin (e+f x))^4}-\frac {14 a c^4 \cos ^5(e+f x)}{f (a+a \sin (e+f x))^2}-\left (35 c^4\right ) \int \frac {\cos ^4(e+f x)}{a+a \sin (e+f x)} \, dx \\ & = -\frac {35 c^4 \cos ^3(e+f x)}{3 a f}-\frac {2 a^3 c^4 \cos ^7(e+f x)}{f (a+a \sin (e+f x))^4}-\frac {14 a c^4 \cos ^5(e+f x)}{f (a+a \sin (e+f x))^2}-\frac {\left (35 c^4\right ) \int \cos ^2(e+f x) \, dx}{a} \\ & = -\frac {35 c^4 \cos ^3(e+f x)}{3 a f}-\frac {35 c^4 \cos (e+f x) \sin (e+f x)}{2 a f}-\frac {2 a^3 c^4 \cos ^7(e+f x)}{f (a+a \sin (e+f x))^4}-\frac {14 a c^4 \cos ^5(e+f x)}{f (a+a \sin (e+f x))^2}-\frac {\left (35 c^4\right ) \int 1 \, dx}{2 a} \\ & = -\frac {35 c^4 x}{2 a}-\frac {35 c^4 \cos ^3(e+f x)}{3 a f}-\frac {35 c^4 \cos (e+f x) \sin (e+f x)}{2 a f}-\frac {2 a^3 c^4 \cos ^7(e+f x)}{f (a+a \sin (e+f x))^4}-\frac {14 a c^4 \cos ^5(e+f x)}{f (a+a \sin (e+f x))^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 12.83 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.48 \[ \int \frac {(c-c \sin (e+f x))^4}{a+a \sin (e+f x)} \, dx=-\frac {c^4 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (-1+\sin (e+f x))^4 \left (\sin \left (\frac {1}{2} (e+f x)\right ) (-384+210 e+210 f x+141 \cos (e+f x)-\cos (3 (e+f x))-15 \sin (2 (e+f x)))+\cos \left (\frac {1}{2} (e+f x)\right ) (210 e+210 f x+141 \cos (e+f x)-\cos (3 (e+f x))-15 \sin (2 (e+f x)))\right )}{12 a f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^8 (1+\sin (e+f x))} \]

[In]

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

[Out]

-1/12*(c^4*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-1 + Sin[e + f*x])^4*(Sin[(e + f*x)/2]*(-384 + 210*e + 210*f
*x + 141*Cos[e + f*x] - Cos[3*(e + f*x)] - 15*Sin[2*(e + f*x)]) + Cos[(e + f*x)/2]*(210*e + 210*f*x + 141*Cos[
e + f*x] - Cos[3*(e + f*x)] - 15*Sin[2*(e + f*x)])))/(a*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^8*(1 + Sin[e +
f*x]))

Maple [A] (verified)

Time = 0.86 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.67

method	result	size
parallelrisch	\(\frac {c^{4} \left (-420 \cos \left (f x +e \right ) f x +15 \sin \left (3 f x +3 e \right )+\cos \left (4 f x +4 e \right )-140 \cos \left (2 f x +2 e \right )+399 \sin \left (f x +e \right )-664 \cos \left (f x +e \right )-525\right )}{24 a f \cos \left (f x +e \right )}\)	\(79\)
derivativedivides	\(\frac {2 c^{4} \left (-\frac {\frac {5 \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}+11 \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+24 \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-\frac {5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2}+\frac {35}{3}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{3}}-\frac {35 \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}-\frac {16}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}\right )}{f a}\)	\(109\)
default	\(\frac {2 c^{4} \left (-\frac {\frac {5 \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}+11 \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+24 \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-\frac {5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2}+\frac {35}{3}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{3}}-\frac {35 \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}-\frac {16}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}\right )}{f a}\)	\(109\)
risch	\(-\frac {35 c^{4} x}{2 a}-\frac {47 c^{4} {\mathrm e}^{i \left (f x +e \right )}}{8 f a}-\frac {47 c^{4} {\mathrm e}^{-i \left (f x +e \right )}}{8 f a}-\frac {32 c^{4}}{f a \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}+\frac {c^{4} \cos \left (3 f x +3 e \right )}{12 f a}+\frac {5 c^{4} \sin \left (2 f x +2 e \right )}{4 f a}\)	\(116\)
norman	\(\frac {-\frac {257 c^{4} \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f a}-\frac {155 c^{4} \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f a}-\frac {37 c^{4} \left (\tan ^{8}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f a}-\frac {27 c^{4} \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f a}-\frac {35 c^{4} x}{2 a}-\frac {166 c^{4}}{3 f a}-\frac {35 c^{4} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2 a}-\frac {70 c^{4} x \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a}-\frac {70 c^{4} x \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a}-\frac {105 c^{4} x \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a}-\frac {105 c^{4} x \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a}-\frac {70 c^{4} x \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a}-\frac {70 c^{4} x \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a}-\frac {35 c^{4} x \left (\tan ^{8}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 a}-\frac {35 c^{4} x \left (\tan ^{9}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 a}-\frac {583 c^{4} \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 f a}-\frac {199 c^{4} \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 f a}-\frac {55 c^{4} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{3 f a}-\frac {75 c^{4} \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f a}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{4} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}\)	\(403\)

[In]

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

[Out]

1/24*c^4/a/f*(-420*cos(f*x+e)*f*x+15*sin(3*f*x+3*e)+cos(4*f*x+4*e)-140*cos(2*f*x+2*e)+399*sin(f*x+e)-664*cos(f
*x+e)-525)/cos(f*x+e)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.32 \[ \int \frac {(c-c \sin (e+f x))^4}{a+a \sin (e+f x)} \, dx=\frac {2 \, c^{4} \cos \left (f x + e\right )^{4} - 13 \, c^{4} \cos \left (f x + e\right )^{3} - 105 \, c^{4} f x - 72 \, c^{4} \cos \left (f x + e\right )^{2} - 96 \, c^{4} - 3 \, {\left (35 \, c^{4} f x + 51 \, c^{4}\right )} \cos \left (f x + e\right ) + {\left (2 \, c^{4} \cos \left (f x + e\right )^{3} - 105 \, c^{4} f x + 15 \, c^{4} \cos \left (f x + e\right )^{2} - 57 \, c^{4} \cos \left (f x + e\right ) + 96 \, c^{4}\right )} \sin \left (f x + e\right )}{6 \, {\left (a f \cos \left (f x + e\right ) + a f \sin \left (f x + e\right ) + a f\right )}} \]

[In]

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

[Out]

1/6*(2*c^4*cos(f*x + e)^4 - 13*c^4*cos(f*x + e)^3 - 105*c^4*f*x - 72*c^4*cos(f*x + e)^2 - 96*c^4 - 3*(35*c^4*f
*x + 51*c^4)*cos(f*x + e) + (2*c^4*cos(f*x + e)^3 - 105*c^4*f*x + 15*c^4*cos(f*x + e)^2 - 57*c^4*cos(f*x + e)
+ 96*c^4)*sin(f*x + e))/(a*f*cos(f*x + e) + a*f*sin(f*x + e) + a*f)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2108 vs. \(2 (112) = 224\).

Time = 3.66 (sec) , antiderivative size = 2108, normalized size of antiderivative = 17.86 \[ \int \frac {(c-c \sin (e+f x))^4}{a+a \sin (e+f x)} \, dx=\text {Too large to display} \]

[In]

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

[Out]

Piecewise((-105*c**4*f*x*tan(e/2 + f*x/2)**7/(6*a*f*tan(e/2 + f*x/2)**7 + 6*a*f*tan(e/2 + f*x/2)**6 + 18*a*f*t
an(e/2 + f*x/2)**5 + 18*a*f*tan(e/2 + f*x/2)**4 + 18*a*f*tan(e/2 + f*x/2)**3 + 18*a*f*tan(e/2 + f*x/2)**2 + 6*
a*f*tan(e/2 + f*x/2) + 6*a*f) - 105*c**4*f*x*tan(e/2 + f*x/2)**6/(6*a*f*tan(e/2 + f*x/2)**7 + 6*a*f*tan(e/2 +
f*x/2)**6 + 18*a*f*tan(e/2 + f*x/2)**5 + 18*a*f*tan(e/2 + f*x/2)**4 + 18*a*f*tan(e/2 + f*x/2)**3 + 18*a*f*tan(
e/2 + f*x/2)**2 + 6*a*f*tan(e/2 + f*x/2) + 6*a*f) - 315*c**4*f*x*tan(e/2 + f*x/2)**5/(6*a*f*tan(e/2 + f*x/2)**
7 + 6*a*f*tan(e/2 + f*x/2)**6 + 18*a*f*tan(e/2 + f*x/2)**5 + 18*a*f*tan(e/2 + f*x/2)**4 + 18*a*f*tan(e/2 + f*x
/2)**3 + 18*a*f*tan(e/2 + f*x/2)**2 + 6*a*f*tan(e/2 + f*x/2) + 6*a*f) - 315*c**4*f*x*tan(e/2 + f*x/2)**4/(6*a*
f*tan(e/2 + f*x/2)**7 + 6*a*f*tan(e/2 + f*x/2)**6 + 18*a*f*tan(e/2 + f*x/2)**5 + 18*a*f*tan(e/2 + f*x/2)**4 +
18*a*f*tan(e/2 + f*x/2)**3 + 18*a*f*tan(e/2 + f*x/2)**2 + 6*a*f*tan(e/2 + f*x/2) + 6*a*f) - 315*c**4*f*x*tan(e
/2 + f*x/2)**3/(6*a*f*tan(e/2 + f*x/2)**7 + 6*a*f*tan(e/2 + f*x/2)**6 + 18*a*f*tan(e/2 + f*x/2)**5 + 18*a*f*ta
n(e/2 + f*x/2)**4 + 18*a*f*tan(e/2 + f*x/2)**3 + 18*a*f*tan(e/2 + f*x/2)**2 + 6*a*f*tan(e/2 + f*x/2) + 6*a*f)
- 315*c**4*f*x*tan(e/2 + f*x/2)**2/(6*a*f*tan(e/2 + f*x/2)**7 + 6*a*f*tan(e/2 + f*x/2)**6 + 18*a*f*tan(e/2 + f
*x/2)**5 + 18*a*f*tan(e/2 + f*x/2)**4 + 18*a*f*tan(e/2 + f*x/2)**3 + 18*a*f*tan(e/2 + f*x/2)**2 + 6*a*f*tan(e/
2 + f*x/2) + 6*a*f) - 105*c**4*f*x*tan(e/2 + f*x/2)/(6*a*f*tan(e/2 + f*x/2)**7 + 6*a*f*tan(e/2 + f*x/2)**6 + 1
8*a*f*tan(e/2 + f*x/2)**5 + 18*a*f*tan(e/2 + f*x/2)**4 + 18*a*f*tan(e/2 + f*x/2)**3 + 18*a*f*tan(e/2 + f*x/2)*
*2 + 6*a*f*tan(e/2 + f*x/2) + 6*a*f) - 105*c**4*f*x/(6*a*f*tan(e/2 + f*x/2)**7 + 6*a*f*tan(e/2 + f*x/2)**6 + 1
8*a*f*tan(e/2 + f*x/2)**5 + 18*a*f*tan(e/2 + f*x/2)**4 + 18*a*f*tan(e/2 + f*x/2)**3 + 18*a*f*tan(e/2 + f*x/2)*
*2 + 6*a*f*tan(e/2 + f*x/2) + 6*a*f) - 222*c**4*tan(e/2 + f*x/2)**6/(6*a*f*tan(e/2 + f*x/2)**7 + 6*a*f*tan(e/2
+ f*x/2)**6 + 18*a*f*tan(e/2 + f*x/2)**5 + 18*a*f*tan(e/2 + f*x/2)**4 + 18*a*f*tan(e/2 + f*x/2)**3 + 18*a*f*t
an(e/2 + f*x/2)**2 + 6*a*f*tan(e/2 + f*x/2) + 6*a*f) - 162*c**4*tan(e/2 + f*x/2)**5/(6*a*f*tan(e/2 + f*x/2)**7
+ 6*a*f*tan(e/2 + f*x/2)**6 + 18*a*f*tan(e/2 + f*x/2)**5 + 18*a*f*tan(e/2 + f*x/2)**4 + 18*a*f*tan(e/2 + f*x/
2)**3 + 18*a*f*tan(e/2 + f*x/2)**2 + 6*a*f*tan(e/2 + f*x/2) + 6*a*f) - 708*c**4*tan(e/2 + f*x/2)**4/(6*a*f*tan
(e/2 + f*x/2)**7 + 6*a*f*tan(e/2 + f*x/2)**6 + 18*a*f*tan(e/2 + f*x/2)**5 + 18*a*f*tan(e/2 + f*x/2)**4 + 18*a*
f*tan(e/2 + f*x/2)**3 + 18*a*f*tan(e/2 + f*x/2)**2 + 6*a*f*tan(e/2 + f*x/2) + 6*a*f) - 288*c**4*tan(e/2 + f*x/
2)**3/(6*a*f*tan(e/2 + f*x/2)**7 + 6*a*f*tan(e/2 + f*x/2)**6 + 18*a*f*tan(e/2 + f*x/2)**5 + 18*a*f*tan(e/2 + f
*x/2)**4 + 18*a*f*tan(e/2 + f*x/2)**3 + 18*a*f*tan(e/2 + f*x/2)**2 + 6*a*f*tan(e/2 + f*x/2) + 6*a*f) - 834*c**
4*tan(e/2 + f*x/2)**2/(6*a*f*tan(e/2 + f*x/2)**7 + 6*a*f*tan(e/2 + f*x/2)**6 + 18*a*f*tan(e/2 + f*x/2)**5 + 18
*a*f*tan(e/2 + f*x/2)**4 + 18*a*f*tan(e/2 + f*x/2)**3 + 18*a*f*tan(e/2 + f*x/2)**2 + 6*a*f*tan(e/2 + f*x/2) +
6*a*f) - 110*c**4*tan(e/2 + f*x/2)/(6*a*f*tan(e/2 + f*x/2)**7 + 6*a*f*tan(e/2 + f*x/2)**6 + 18*a*f*tan(e/2 + f
*x/2)**5 + 18*a*f*tan(e/2 + f*x/2)**4 + 18*a*f*tan(e/2 + f*x/2)**3 + 18*a*f*tan(e/2 + f*x/2)**2 + 6*a*f*tan(e/
2 + f*x/2) + 6*a*f) - 332*c**4/(6*a*f*tan(e/2 + f*x/2)**7 + 6*a*f*tan(e/2 + f*x/2)**6 + 18*a*f*tan(e/2 + f*x/2
)**5 + 18*a*f*tan(e/2 + f*x/2)**4 + 18*a*f*tan(e/2 + f*x/2)**3 + 18*a*f*tan(e/2 + f*x/2)**2 + 6*a*f*tan(e/2 +
f*x/2) + 6*a*f), Ne(f, 0)), (x*(-c*sin(e) + c)**4/(a*sin(e) + a), True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 720 vs. \(2 (112) = 224\).

Time = 0.30 (sec) , antiderivative size = 720, normalized size of antiderivative = 6.10 \[ \int \frac {(c-c \sin (e+f x))^4}{a+a \sin (e+f x)} \, dx=\text {Too large to display} \]

[In]

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

[Out]

-1/3*(c^4*((7*sin(f*x + e)/(cos(f*x + e) + 1) + 39*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 24*sin(f*x + e)^3/(co
s(f*x + e) + 1)^3 + 24*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 9*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 9*sin(f*x
+ e)^6/(cos(f*x + e) + 1)^6 + 16)/(a + a*sin(f*x + e)/(cos(f*x + e) + 1) + 3*a*sin(f*x + e)^2/(cos(f*x + e) +
1)^2 + 3*a*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*a*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 3*a*sin(f*x + e)^5
/(cos(f*x + e) + 1)^5 + a*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + a*sin(f*x + e)^7/(cos(f*x + e) + 1)^7) + 9*arc
tan(sin(f*x + e)/(cos(f*x + e) + 1))/a) + 12*c^4*((sin(f*x + e)/(cos(f*x + e) + 1) + 5*sin(f*x + e)^2/(cos(f*x
+ e) + 1)^2 + 3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 4)/(a + a*sin(f
*x + e)/(cos(f*x + e) + 1) + 2*a*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 2*a*sin(f*x + e)^3/(cos(f*x + e) + 1)^3
+ a*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + a*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) + 3*arctan(sin(f*x + e)/(cos
(f*x + e) + 1))/a) + 36*c^4*((sin(f*x + e)/(cos(f*x + e) + 1) + sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 2)/(a +
a*sin(f*x + e)/(cos(f*x + e) + 1) + a*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + a*sin(f*x + e)^3/(cos(f*x + e) + 1
)^3) + arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a) + 24*c^4*(arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a + 1/(a +
a*sin(f*x + e)/(cos(f*x + e) + 1))) + 6*c^4/(a + a*sin(f*x + e)/(cos(f*x + e) + 1)))/f

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.08 \[ \int \frac {(c-c \sin (e+f x))^4}{a+a \sin (e+f x)} \, dx=-\frac {\frac {105 \, {\left (f x + e\right )} c^{4}}{a} + \frac {192 \, c^{4}}{a {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}} + \frac {2 \, {\left (15 \, c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 66 \, c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 144 \, c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 15 \, c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 70 \, c^{4}\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )}^{3} a}}{6 \, f} \]

[In]

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

[Out]

-1/6*(105*(f*x + e)*c^4/a + 192*c^4/(a*(tan(1/2*f*x + 1/2*e) + 1)) + 2*(15*c^4*tan(1/2*f*x + 1/2*e)^5 + 66*c^4
*tan(1/2*f*x + 1/2*e)^4 + 144*c^4*tan(1/2*f*x + 1/2*e)^2 - 15*c^4*tan(1/2*f*x + 1/2*e) + 70*c^4)/((tan(1/2*f*x
+ 1/2*e)^2 + 1)^3*a))/f

Mupad [B] (verification not implemented)

Time = 10.56 (sec) , antiderivative size = 290, normalized size of antiderivative = 2.46 \[ \int \frac {(c-c \sin (e+f x))^4}{a+a \sin (e+f x)} \, dx=\frac {\frac {35\,c^4\,\left (e+f\,x\right )}{2}+\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (\frac {35\,c^4\,\left (e+f\,x\right )}{2}-\frac {c^4\,\left (105\,e+105\,f\,x+110\right )}{6}\right )-\frac {c^4\,\left (105\,e+105\,f\,x+332\right )}{6}+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6\,\left (\frac {35\,c^4\,\left (e+f\,x\right )}{2}-\frac {c^4\,\left (105\,e+105\,f\,x+222\right )}{6}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5\,\left (\frac {105\,c^4\,\left (e+f\,x\right )}{2}-\frac {c^4\,\left (315\,e+315\,f\,x+162\right )}{6}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (\frac {105\,c^4\,\left (e+f\,x\right )}{2}-\frac {c^4\,\left (315\,e+315\,f\,x+288\right )}{6}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (\frac {105\,c^4\,\left (e+f\,x\right )}{2}-\frac {c^4\,\left (315\,e+315\,f\,x+708\right )}{6}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (\frac {105\,c^4\,\left (e+f\,x\right )}{2}-\frac {c^4\,\left (315\,e+315\,f\,x+834\right )}{6}\right )}{a\,f\,\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+1\right )\,{\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )}^3}-\frac {35\,c^4\,x}{2\,a} \]

[In]

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

[Out]

((35*c^4*(e + f*x))/2 + tan(e/2 + (f*x)/2)*((35*c^4*(e + f*x))/2 - (c^4*(105*e + 105*f*x + 110))/6) - (c^4*(10
5*e + 105*f*x + 332))/6 + tan(e/2 + (f*x)/2)^6*((35*c^4*(e + f*x))/2 - (c^4*(105*e + 105*f*x + 222))/6) + tan(
e/2 + (f*x)/2)^5*((105*c^4*(e + f*x))/2 - (c^4*(315*e + 315*f*x + 162))/6) + tan(e/2 + (f*x)/2)^3*((105*c^4*(e
+ f*x))/2 - (c^4*(315*e + 315*f*x + 288))/6) + tan(e/2 + (f*x)/2)^4*((105*c^4*(e + f*x))/2 - (c^4*(315*e + 31
5*f*x + 708))/6) + tan(e/2 + (f*x)/2)^2*((105*c^4*(e + f*x))/2 - (c^4*(315*e + 315*f*x + 834))/6))/(a*f*(tan(e
/2 + (f*x)/2) + 1)*(tan(e/2 + (f*x)/2)^2 + 1)^3) - (35*c^4*x)/(2*a)