\(\int \frac {1}{(3+3 \sin (e+f x)) (c+d \sin (e+f x))^3} \, dx\) [460]

Optimal result

Integrand size = 25, antiderivative size = 204 \[ \int \frac {1}{(3+3 \sin (e+f x)) (c+d \sin (e+f x))^3} \, dx=-\frac {d \left (2 c^2+2 c d+d^2\right ) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{(c-d) \left (c^2-d^2\right )^{5/2} f}-\frac {d (2 c+3 d) \cos (e+f x)}{6 (c-d)^2 (c+d) f (c+d \sin (e+f x))^2}-\frac {\cos (e+f x)}{(c-d) f (3+3 \sin (e+f x)) (c+d \sin (e+f x))^2}-\frac {d (2 c+d) (c+4 d) \cos (e+f x)}{6 (c-d)^3 (c+d)^2 f (c+d \sin (e+f x))} \]

[Out]

-3*d*(2*c^2+2*c*d+d^2)*arctan((d+c*tan(1/2*f*x+1/2*e))/(c^2-d^2)^(1/2))/a/(c-d)/(c^2-d^2)^(5/2)/f-1/2*d*(2*c+3
*d)*cos(f*x+e)/a/(c-d)^2/(c+d)/f/(c+d*sin(f*x+e))^2-cos(f*x+e)/(c-d)/f/(a+a*sin(f*x+e))/(c+d*sin(f*x+e))^2-1/2
*d*(2*c+d)*(c+4*d)*cos(f*x+e)/a/(c-d)^3/(c+d)^2/f/(c+d*sin(f*x+e))

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {2847, 2833, 12, 2739, 632, 210} \[ \int \frac {1}{(3+3 \sin (e+f x)) (c+d \sin (e+f x))^3} \, dx=-\frac {3 d \left (2 c^2+2 c d+d^2\right ) \arctan \left (\frac {c \tan \left (\frac {1}{2} (e+f x)\right )+d}{\sqrt {c^2-d^2}}\right )}{a f (c-d) \left (c^2-d^2\right )^{5/2}}-\frac {d (2 c+d) (c+4 d) \cos (e+f x)}{2 a f (c-d)^3 (c+d)^2 (c+d \sin (e+f x))}-\frac {d (2 c+3 d) \cos (e+f x)}{2 a f (c-d)^2 (c+d) (c+d \sin (e+f x))^2}-\frac {\cos (e+f x)}{f (c-d) (a \sin (e+f x)+a) (c+d \sin (e+f x))^2} \]

[In]

Int[1/((a + a*Sin[e + f*x])*(c + d*Sin[e + f*x])^3),x]

[Out]

(-3*d*(2*c^2 + 2*c*d + d^2)*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/(a*(c - d)*(c^2 - d^2)^(5/2)*f)
- (d*(2*c + 3*d)*Cos[e + f*x])/(2*a*(c - d)^2*(c + d)*f*(c + d*Sin[e + f*x])^2) - Cos[e + f*x]/((c - d)*f*(a +
a*Sin[e + f*x])*(c + d*Sin[e + f*x])^2) - (d*(2*c + d)*(c + 4*d)*Cos[e + f*x])/(2*a*(c - d)^3*(c + d)^2*f*(c
+ d*Sin[e + f*x]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2739

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[2*(e/d), Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 2833

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

Rule 2847

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-b
^2)*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(a*f*(b*c - a*d)*(a + b*Sin[e + f*x]))), x] + Dist[d/(a*(b*c -
a*d)), Int[(c + d*Sin[e + f*x])^n*(a*n - b*(n + 1)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && Ne
Q[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, 0] && (IntegerQ[2*n] || EqQ[c, 0])

Rubi steps \begin{align*} \text {integral}& = -\frac {\cos (e+f x)}{(c-d) f (a+a \sin (e+f x)) (c+d \sin (e+f x))^2}+\frac {d \int \frac {-3 a+2 a \sin (e+f x)}{(c+d \sin (e+f x))^3} \, dx}{a^2 (c-d)} \\ & = -\frac {d (2 c+3 d) \cos (e+f x)}{2 a (c-d)^2 (c+d) f (c+d \sin (e+f x))^2}-\frac {\cos (e+f x)}{(c-d) f (a+a \sin (e+f x)) (c+d \sin (e+f x))^2}-\frac {d \int \frac {2 a (3 c+2 d)-a (2 c+3 d) \sin (e+f x)}{(c+d \sin (e+f x))^2} \, dx}{2 a^2 (c-d)^2 (c+d)} \\ & = -\frac {d (2 c+3 d) \cos (e+f x)}{2 a (c-d)^2 (c+d) f (c+d \sin (e+f x))^2}-\frac {\cos (e+f x)}{(c-d) f (a+a \sin (e+f x)) (c+d \sin (e+f x))^2}-\frac {d (2 c+d) (c+4 d) \cos (e+f x)}{2 a (c-d)^3 (c+d)^2 f (c+d \sin (e+f x))}+\frac {d \int -\frac {3 a \left (2 c^2+2 c d+d^2\right )}{c+d \sin (e+f x)} \, dx}{2 a^2 (c-d)^3 (c+d)^2} \\ & = -\frac {d (2 c+3 d) \cos (e+f x)}{2 a (c-d)^2 (c+d) f (c+d \sin (e+f x))^2}-\frac {\cos (e+f x)}{(c-d) f (a+a \sin (e+f x)) (c+d \sin (e+f x))^2}-\frac {d (2 c+d) (c+4 d) \cos (e+f x)}{2 a (c-d)^3 (c+d)^2 f (c+d \sin (e+f x))}-\frac {\left (3 d \left (2 c^2+2 c d+d^2\right )\right ) \int \frac {1}{c+d \sin (e+f x)} \, dx}{2 a (c-d)^3 (c+d)^2} \\ & = -\frac {d (2 c+3 d) \cos (e+f x)}{2 a (c-d)^2 (c+d) f (c+d \sin (e+f x))^2}-\frac {\cos (e+f x)}{(c-d) f (a+a \sin (e+f x)) (c+d \sin (e+f x))^2}-\frac {d (2 c+d) (c+4 d) \cos (e+f x)}{2 a (c-d)^3 (c+d)^2 f (c+d \sin (e+f x))}-\frac {\left (3 d \left (2 c^2+2 c d+d^2\right )\right ) \text {Subst}\left (\int \frac {1}{c+2 d x+c x^2} \, dx,x,\tan \left (\frac {1}{2} (e+f x)\right )\right )}{a (c-d)^3 (c+d)^2 f} \\ & = -\frac {d (2 c+3 d) \cos (e+f x)}{2 a (c-d)^2 (c+d) f (c+d \sin (e+f x))^2}-\frac {\cos (e+f x)}{(c-d) f (a+a \sin (e+f x)) (c+d \sin (e+f x))^2}-\frac {d (2 c+d) (c+4 d) \cos (e+f x)}{2 a (c-d)^3 (c+d)^2 f (c+d \sin (e+f x))}+\frac {\left (6 d \left (2 c^2+2 c d+d^2\right )\right ) \text {Subst}\left (\int \frac {1}{-4 \left (c^2-d^2\right )-x^2} \, dx,x,2 d+2 c \tan \left (\frac {1}{2} (e+f x)\right )\right )}{a (c-d)^3 (c+d)^2 f} \\ & = -\frac {3 d \left (2 c^2+2 c d+d^2\right ) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{a (c-d)^3 (c+d)^2 \sqrt {c^2-d^2} f}-\frac {d (2 c+3 d) \cos (e+f x)}{2 a (c-d)^2 (c+d) f (c+d \sin (e+f x))^2}-\frac {\cos (e+f x)}{(c-d) f (a+a \sin (e+f x)) (c+d \sin (e+f x))^2}-\frac {d (2 c+d) (c+4 d) \cos (e+f x)}{2 a (c-d)^3 (c+d)^2 f (c+d \sin (e+f x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.32 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.11 \[ \int \frac {1}{(3+3 \sin (e+f x)) (c+d \sin (e+f x))^3} \, dx=\frac {\cos (e+f x) \left (-\frac {6 d \left (2 c^2+2 c d+d^2\right ) \text {arctanh}\left (\frac {\sqrt {c-d} \sqrt {1-\sin (e+f x)}}{\sqrt {-c-d} \sqrt {1+\sin (e+f x)}}\right )}{(-c-d)^{3/2} (c-d)^{5/2} \sqrt {\cos ^2(e+f x)}}+\frac {2 c^2+9 c d+4 d^2}{(c-d)^2 (c+d) (1+\sin (e+f x))}-\frac {d}{(1+\sin (e+f x)) (c+d \sin (e+f x))^2}-\frac {d (4 c+d)}{(c-d) (c+d) (1+\sin (e+f x)) (c+d \sin (e+f x))}\right )}{6 (-c+d) (c+d) f} \]

[In]

Integrate[1/((3 + 3*Sin[e + f*x])*(c + d*Sin[e + f*x])^3),x]

[Out]

(Cos[e + f*x]*((-6*d*(2*c^2 + 2*c*d + d^2)*ArcTanh[(Sqrt[c - d]*Sqrt[1 - Sin[e + f*x]])/(Sqrt[-c - d]*Sqrt[1 +
Sin[e + f*x]])])/((-c - d)^(3/2)*(c - d)^(5/2)*Sqrt[Cos[e + f*x]^2]) + (2*c^2 + 9*c*d + 4*d^2)/((c - d)^2*(c
+ d)*(1 + Sin[e + f*x])) - d/((1 + Sin[e + f*x])*(c + d*Sin[e + f*x])^2) - (d*(4*c + d))/((c - d)*(c + d)*(1 +
Sin[e + f*x])*(c + d*Sin[e + f*x]))))/(6*(-c + d)*(c + d)*f)

Maple [A] (verified)

Time = 2.27 (sec) , antiderivative size = 327, normalized size of antiderivative = 1.60

method	result	size
derivativedivides	\(\frac {-\frac {2}{\left (c -d \right )^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}-\frac {2 d \left (\frac {\frac {d^{2} \left (7 c^{2}+2 c d -2 d^{2}\right ) \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 c \left (c^{2}+2 c d +d^{2}\right )}+\frac {d \left (6 c^{4}+2 c^{3} d +11 c^{2} d^{2}+4 d^{3} c -2 d^{4}\right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 c^{2} \left (c^{2}+2 c d +d^{2}\right )}+\frac {d^{2} \left (17 c^{2}+6 c d -2 d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2 \left (c^{2}+2 c d +d^{2}\right ) c}+\frac {d \left (6 c^{2}+2 c d -d^{2}\right )}{2 c^{2}+4 c d +2 d^{2}}}{{\left (\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) c +2 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+c \right )}^{2}}+\frac {3 \left (2 c^{2}+2 c d +d^{2}\right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{2 \left (c^{2}+2 c d +d^{2}\right ) \sqrt {c^{2}-d^{2}}}\right )}{\left (c -d \right )^{3}}}{f a}\)	\(327\)
default	\(\frac {-\frac {2}{\left (c -d \right )^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}-\frac {2 d \left (\frac {\frac {d^{2} \left (7 c^{2}+2 c d -2 d^{2}\right ) \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 c \left (c^{2}+2 c d +d^{2}\right )}+\frac {d \left (6 c^{4}+2 c^{3} d +11 c^{2} d^{2}+4 d^{3} c -2 d^{4}\right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 c^{2} \left (c^{2}+2 c d +d^{2}\right )}+\frac {d^{2} \left (17 c^{2}+6 c d -2 d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2 \left (c^{2}+2 c d +d^{2}\right ) c}+\frac {d \left (6 c^{2}+2 c d -d^{2}\right )}{2 c^{2}+4 c d +2 d^{2}}}{{\left (\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) c +2 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+c \right )}^{2}}+\frac {3 \left (2 c^{2}+2 c d +d^{2}\right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{2 \left (c^{2}+2 c d +d^{2}\right ) \sqrt {c^{2}-d^{2}}}\right )}{\left (c -d \right )^{3}}}{f a}\)	\(327\)
risch	\(-\frac {8 i c^{3} d \,{\mathrm e}^{i \left (f x +e \right )}-3 i d^{4} {\mathrm e}^{3 i \left (f x +e \right )}-18 i c^{3} d \,{\mathrm e}^{3 i \left (f x +e \right )}-15 i c \,d^{3} {\mathrm e}^{3 i \left (f x +e \right )}-6 c^{2} d^{2} {\mathrm e}^{4 i \left (f x +e \right )}-6 c \,d^{3} {\mathrm e}^{4 i \left (f x +e \right )}-3 d^{4} {\mathrm e}^{4 i \left (f x +e \right )}+i d^{4} {\mathrm e}^{i \left (f x +e \right )}+19 i c \,d^{3} {\mathrm e}^{i \left (f x +e \right )}+32 i c^{2} d^{2} {\mathrm e}^{i \left (f x +e \right )}-24 i c^{2} d^{2} {\mathrm e}^{3 i \left (f x +e \right )}+8 c^{4} {\mathrm e}^{2 i \left (f x +e \right )}+26 d \,c^{3} {\mathrm e}^{2 i \left (f x +e \right )}+32 d^{2} c^{2} {\mathrm e}^{2 i \left (f x +e \right )}+19 d^{3} c \,{\mathrm e}^{2 i \left (f x +e \right )}+5 d^{4} {\mathrm e}^{2 i \left (f x +e \right )}-2 c^{2} d^{2}-9 d^{3} c -4 d^{4}}{\left ({\mathrm e}^{i \left (f x +e \right )}+i\right ) \left (-i d \,{\mathrm e}^{2 i \left (f x +e \right )}+i d +2 c \,{\mathrm e}^{i \left (f x +e \right )}\right )^{2} \left (c +d \right )^{2} \left (c -d \right )^{3} f a}-\frac {3 d \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}+c^{2}-d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right ) c^{2}}{\sqrt {-c^{2}+d^{2}}\, \left (c +d \right )^{2} \left (c -d \right )^{3} f a}-\frac {3 d^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}+c^{2}-d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right ) c}{\sqrt {-c^{2}+d^{2}}\, \left (c +d \right )^{2} \left (c -d \right )^{3} f a}-\frac {3 d^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}+c^{2}-d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right )}{2 \sqrt {-c^{2}+d^{2}}\, \left (c +d \right )^{2} \left (c -d \right )^{3} f a}+\frac {3 d \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}-c^{2}+d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right ) c^{2}}{\sqrt {-c^{2}+d^{2}}\, \left (c +d \right )^{2} \left (c -d \right )^{3} f a}+\frac {3 d^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}-c^{2}+d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right ) c}{\sqrt {-c^{2}+d^{2}}\, \left (c +d \right )^{2} \left (c -d \right )^{3} f a}+\frac {3 d^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}-c^{2}+d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right )}{2 \sqrt {-c^{2}+d^{2}}\, \left (c +d \right )^{2} \left (c -d \right )^{3} f a}\)	\(847\)

[In]

int(1/(a+a*sin(f*x+e))/(c+d*sin(f*x+e))^3,x,method=_RETURNVERBOSE)

[Out]

2/f/a*(-1/(c-d)^3/(tan(1/2*f*x+1/2*e)+1)-1/(c-d)^3*d*((1/2*d^2*(7*c^2+2*c*d-2*d^2)/c/(c^2+2*c*d+d^2)*tan(1/2*f
*x+1/2*e)^3+1/2*d*(6*c^4+2*c^3*d+11*c^2*d^2+4*c*d^3-2*d^4)/c^2/(c^2+2*c*d+d^2)*tan(1/2*f*x+1/2*e)^2+1/2*d^2*(1
7*c^2+6*c*d-2*d^2)/(c^2+2*c*d+d^2)/c*tan(1/2*f*x+1/2*e)+1/2*d*(6*c^2+2*c*d-d^2)/(c^2+2*c*d+d^2))/(tan(1/2*f*x+
1/2*e)^2*c+2*d*tan(1/2*f*x+1/2*e)+c)^2+3/2*(2*c^2+2*c*d+d^2)/(c^2+2*c*d+d^2)/(c^2-d^2)^(1/2)*arctan(1/2*(2*c*t
an(1/2*f*x+1/2*e)+2*d)/(c^2-d^2)^(1/2))))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1140 vs. \(2 (204) = 408\).

Time = 0.35 (sec) , antiderivative size = 2365, normalized size of antiderivative = 11.59 \[ \int \frac {1}{(3+3 \sin (e+f x)) (c+d \sin (e+f x))^3} \, dx=\text {Too large to display} \]

[In]

integrate(1/(a+a*sin(f*x+e))/(c+d*sin(f*x+e))^3,x, algorithm="fricas")

[Out]

[1/4*(4*c^6 - 12*c^4*d^2 + 12*c^2*d^4 - 4*d^6 - 2*(2*c^4*d^2 + 9*c^3*d^3 + 2*c^2*d^4 - 9*c*d^5 - 4*d^6)*cos(f*
x + e)^3 + 2*(4*c^5*d + 12*c^4*d^2 - 2*c^3*d^3 - 15*c^2*d^4 - 2*c*d^5 + 3*d^6)*cos(f*x + e)^2 - 3*(2*c^4*d + 6
*c^3*d^2 + 7*c^2*d^3 + 4*c*d^4 + d^5 - (2*c^2*d^3 + 2*c*d^4 + d^5)*cos(f*x + e)^3 - (4*c^3*d^2 + 6*c^2*d^3 + 4
*c*d^4 + d^5)*cos(f*x + e)^2 + (2*c^4*d + 2*c^3*d^2 + 3*c^2*d^3 + 2*c*d^4 + d^5)*cos(f*x + e) + (2*c^4*d + 6*c
^3*d^2 + 7*c^2*d^3 + 4*c*d^4 + d^5 - (2*c^2*d^3 + 2*c*d^4 + d^5)*cos(f*x + e)^2 + 2*(2*c^3*d^2 + 2*c^2*d^3 + c
*d^4)*cos(f*x + e))*sin(f*x + e))*sqrt(-c^2 + d^2)*log(((2*c^2 - d^2)*cos(f*x + e)^2 - 2*c*d*sin(f*x + e) - c^
2 - d^2 + 2*(c*cos(f*x + e)*sin(f*x + e) + d*cos(f*x + e))*sqrt(-c^2 + d^2))/(d^2*cos(f*x + e)^2 - 2*c*d*sin(f
*x + e) - c^2 - d^2)) + 2*(2*c^6 + 4*c^5*d + 8*c^4*d^2 + 7*c^3*d^3 - 7*c^2*d^4 - 11*c*d^5 - 3*d^6)*cos(f*x + e
) - 2*(2*c^6 - 6*c^4*d^2 + 6*c^2*d^4 - 2*d^6 - (2*c^4*d^2 + 9*c^3*d^3 + 2*c^2*d^4 - 9*c*d^5 - 4*d^6)*cos(f*x +
e)^2 - (4*c^5*d + 14*c^4*d^2 + 7*c^3*d^3 - 13*c^2*d^4 - 11*c*d^5 - d^6)*cos(f*x + e))*sin(f*x + e))/((a*c^7*d
^2 - a*c^6*d^3 - 3*a*c^5*d^4 + 3*a*c^4*d^5 + 3*a*c^3*d^6 - 3*a*c^2*d^7 - a*c*d^8 + a*d^9)*f*cos(f*x + e)^3 + (
2*a*c^8*d - a*c^7*d^2 - 7*a*c^6*d^3 + 3*a*c^5*d^4 + 9*a*c^4*d^5 - 3*a*c^3*d^6 - 5*a*c^2*d^7 + a*c*d^8 + a*d^9)
*f*cos(f*x + e)^2 - (a*c^9 - a*c^8*d - 2*a*c^7*d^2 + 2*a*c^6*d^3 + 2*a*c^3*d^6 - 2*a*c^2*d^7 - a*c*d^8 + a*d^9
)*f*cos(f*x + e) - (a*c^9 + a*c^8*d - 4*a*c^7*d^2 - 4*a*c^6*d^3 + 6*a*c^5*d^4 + 6*a*c^4*d^5 - 4*a*c^3*d^6 - 4*
a*c^2*d^7 + a*c*d^8 + a*d^9)*f + ((a*c^7*d^2 - a*c^6*d^3 - 3*a*c^5*d^4 + 3*a*c^4*d^5 + 3*a*c^3*d^6 - 3*a*c^2*d
^7 - a*c*d^8 + a*d^9)*f*cos(f*x + e)^2 - 2*(a*c^8*d - a*c^7*d^2 - 3*a*c^6*d^3 + 3*a*c^5*d^4 + 3*a*c^4*d^5 - 3*
a*c^3*d^6 - a*c^2*d^7 + a*c*d^8)*f*cos(f*x + e) - (a*c^9 + a*c^8*d - 4*a*c^7*d^2 - 4*a*c^6*d^3 + 6*a*c^5*d^4 +
6*a*c^4*d^5 - 4*a*c^3*d^6 - 4*a*c^2*d^7 + a*c*d^8 + a*d^9)*f)*sin(f*x + e)), 1/2*(2*c^6 - 6*c^4*d^2 + 6*c^2*d
^4 - 2*d^6 - (2*c^4*d^2 + 9*c^3*d^3 + 2*c^2*d^4 - 9*c*d^5 - 4*d^6)*cos(f*x + e)^3 + (4*c^5*d + 12*c^4*d^2 - 2*
c^3*d^3 - 15*c^2*d^4 - 2*c*d^5 + 3*d^6)*cos(f*x + e)^2 - 3*(2*c^4*d + 6*c^3*d^2 + 7*c^2*d^3 + 4*c*d^4 + d^5 -
(2*c^2*d^3 + 2*c*d^4 + d^5)*cos(f*x + e)^3 - (4*c^3*d^2 + 6*c^2*d^3 + 4*c*d^4 + d^5)*cos(f*x + e)^2 + (2*c^4*d
+ 2*c^3*d^2 + 3*c^2*d^3 + 2*c*d^4 + d^5)*cos(f*x + e) + (2*c^4*d + 6*c^3*d^2 + 7*c^2*d^3 + 4*c*d^4 + d^5 - (2
*c^2*d^3 + 2*c*d^4 + d^5)*cos(f*x + e)^2 + 2*(2*c^3*d^2 + 2*c^2*d^3 + c*d^4)*cos(f*x + e))*sin(f*x + e))*sqrt(
c^2 - d^2)*arctan(-(c*sin(f*x + e) + d)/(sqrt(c^2 - d^2)*cos(f*x + e))) + (2*c^6 + 4*c^5*d + 8*c^4*d^2 + 7*c^3
*d^3 - 7*c^2*d^4 - 11*c*d^5 - 3*d^6)*cos(f*x + e) - (2*c^6 - 6*c^4*d^2 + 6*c^2*d^4 - 2*d^6 - (2*c^4*d^2 + 9*c^
3*d^3 + 2*c^2*d^4 - 9*c*d^5 - 4*d^6)*cos(f*x + e)^2 - (4*c^5*d + 14*c^4*d^2 + 7*c^3*d^3 - 13*c^2*d^4 - 11*c*d^
5 - d^6)*cos(f*x + e))*sin(f*x + e))/((a*c^7*d^2 - a*c^6*d^3 - 3*a*c^5*d^4 + 3*a*c^4*d^5 + 3*a*c^3*d^6 - 3*a*c
^2*d^7 - a*c*d^8 + a*d^9)*f*cos(f*x + e)^3 + (2*a*c^8*d - a*c^7*d^2 - 7*a*c^6*d^3 + 3*a*c^5*d^4 + 9*a*c^4*d^5
- 3*a*c^3*d^6 - 5*a*c^2*d^7 + a*c*d^8 + a*d^9)*f*cos(f*x + e)^2 - (a*c^9 - a*c^8*d - 2*a*c^7*d^2 + 2*a*c^6*d^3
+ 2*a*c^3*d^6 - 2*a*c^2*d^7 - a*c*d^8 + a*d^9)*f*cos(f*x + e) - (a*c^9 + a*c^8*d - 4*a*c^7*d^2 - 4*a*c^6*d^3
+ 6*a*c^5*d^4 + 6*a*c^4*d^5 - 4*a*c^3*d^6 - 4*a*c^2*d^7 + a*c*d^8 + a*d^9)*f + ((a*c^7*d^2 - a*c^6*d^3 - 3*a*c
^5*d^4 + 3*a*c^4*d^5 + 3*a*c^3*d^6 - 3*a*c^2*d^7 - a*c*d^8 + a*d^9)*f*cos(f*x + e)^2 - 2*(a*c^8*d - a*c^7*d^2
- 3*a*c^6*d^3 + 3*a*c^5*d^4 + 3*a*c^4*d^5 - 3*a*c^3*d^6 - a*c^2*d^7 + a*c*d^8)*f*cos(f*x + e) - (a*c^9 + a*c^8
*d - 4*a*c^7*d^2 - 4*a*c^6*d^3 + 6*a*c^5*d^4 + 6*a*c^4*d^5 - 4*a*c^3*d^6 - 4*a*c^2*d^7 + a*c*d^8 + a*d^9)*f)*s
in(f*x + e))]

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(3+3 \sin (e+f x)) (c+d \sin (e+f x))^3} \, dx=\text {Timed out} \]

[In]

integrate(1/(a+a*sin(f*x+e))/(c+d*sin(f*x+e))**3,x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(3+3 \sin (e+f x)) (c+d \sin (e+f x))^3} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(1/(a+a*sin(f*x+e))/(c+d*sin(f*x+e))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*d^2-4*c^2>0)', see `assume?`
for more de

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 466 vs. \(2 (204) = 408\).

Time = 0.36 (sec) , antiderivative size = 466, normalized size of antiderivative = 2.28 \[ \int \frac {1}{(3+3 \sin (e+f x)) (c+d \sin (e+f x))^3} \, dx=-\frac {\frac {3 \, {\left (2 \, c^{2} d + 2 \, c d^{2} + d^{3}\right )} {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (c\right ) + \arctan \left (\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + d}{\sqrt {c^{2} - d^{2}}}\right )\right )}}{{\left (a c^{5} - a c^{4} d - 2 \, a c^{3} d^{2} + 2 \, a c^{2} d^{3} + a c d^{4} - a d^{5}\right )} \sqrt {c^{2} - d^{2}}} + \frac {7 \, c^{3} d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 2 \, c^{2} d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 2 \, c d^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 6 \, c^{4} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, c^{3} d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 11 \, c^{2} d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 4 \, c d^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 2 \, d^{6} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 17 \, c^{3} d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 6 \, c^{2} d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 2 \, c d^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 6 \, c^{4} d^{2} + 2 \, c^{3} d^{3} - c^{2} d^{4}}{{\left (a c^{7} - a c^{6} d - 2 \, a c^{5} d^{2} + 2 \, a c^{4} d^{3} + a c^{3} d^{4} - a c^{2} d^{5}\right )} {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + c\right )}^{2}} + \frac {2}{{\left (a c^{3} - 3 \, a c^{2} d + 3 \, a c d^{2} - a d^{3}\right )} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}}}{f} \]

[In]

integrate(1/(a+a*sin(f*x+e))/(c+d*sin(f*x+e))^3,x, algorithm="giac")

[Out]

-(3*(2*c^2*d + 2*c*d^2 + d^3)*(pi*floor(1/2*(f*x + e)/pi + 1/2)*sgn(c) + arctan((c*tan(1/2*f*x + 1/2*e) + d)/s
qrt(c^2 - d^2)))/((a*c^5 - a*c^4*d - 2*a*c^3*d^2 + 2*a*c^2*d^3 + a*c*d^4 - a*d^5)*sqrt(c^2 - d^2)) + (7*c^3*d^
3*tan(1/2*f*x + 1/2*e)^3 + 2*c^2*d^4*tan(1/2*f*x + 1/2*e)^3 - 2*c*d^5*tan(1/2*f*x + 1/2*e)^3 + 6*c^4*d^2*tan(1
/2*f*x + 1/2*e)^2 + 2*c^3*d^3*tan(1/2*f*x + 1/2*e)^2 + 11*c^2*d^4*tan(1/2*f*x + 1/2*e)^2 + 4*c*d^5*tan(1/2*f*x
+ 1/2*e)^2 - 2*d^6*tan(1/2*f*x + 1/2*e)^2 + 17*c^3*d^3*tan(1/2*f*x + 1/2*e) + 6*c^2*d^4*tan(1/2*f*x + 1/2*e)
- 2*c*d^5*tan(1/2*f*x + 1/2*e) + 6*c^4*d^2 + 2*c^3*d^3 - c^2*d^4)/((a*c^7 - a*c^6*d - 2*a*c^5*d^2 + 2*a*c^4*d^
3 + a*c^3*d^4 - a*c^2*d^5)*(c*tan(1/2*f*x + 1/2*e)^2 + 2*d*tan(1/2*f*x + 1/2*e) + c)^2) + 2/((a*c^3 - 3*a*c^2*
d + 3*a*c*d^2 - a*d^3)*(tan(1/2*f*x + 1/2*e) + 1)))/f

Mupad [B] (verification not implemented)

Time = 10.32 (sec) , antiderivative size = 753, normalized size of antiderivative = 3.69 \[ \int \frac {1}{(3+3 \sin (e+f x)) (c+d \sin (e+f x))^3} \, dx=\frac {3\,d\,\mathrm {atan}\left (\frac {\frac {3\,d\,\left (2\,c^2+2\,c\,d+d^2\right )\,\left (-2\,a\,c^5\,d+2\,a\,c^4\,d^2+4\,a\,c^3\,d^3-4\,a\,c^2\,d^4-2\,a\,c\,d^5+2\,a\,d^6\right )}{2\,a\,{\left (c+d\right )}^{5/2}\,{\left (c-d\right )}^{7/2}}-\frac {3\,c\,d\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (2\,c^2+2\,c\,d+d^2\right )\,\left (a\,c^5-a\,c^4\,d-2\,a\,c^3\,d^2+2\,a\,c^2\,d^3+a\,c\,d^4-a\,d^5\right )}{a\,{\left (c+d\right )}^{5/2}\,{\left (c-d\right )}^{7/2}}}{6\,c^2\,d+6\,c\,d^2+3\,d^3}\right )\,\left (2\,c^2+2\,c\,d+d^2\right )}{a\,f\,{\left (c+d\right )}^{5/2}\,{\left (c-d\right )}^{7/2}}-\frac {\frac {2\,c^4+4\,c^3\,d+8\,c^2\,d^2+2\,c\,d^3-d^4}{\left (c+d\right )\,\left (c^2-d^2\right )\,\left (c^2-2\,c\,d+d^2\right )}-\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (8\,c^5\,d+22\,c^4\,d^2+17\,c^3\,d^3+13\,c^2\,d^4+2\,c\,d^5-2\,d^6\right )}{c^2\,\left (c^2-2\,c\,d+d^2\right )\,\left (-c^3-c^2\,d+c\,d^2+d^3\right )}+\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (4\,c^5+4\,c^4\,d+14\,c^3\,d^2+21\,c^2\,d^3+4\,c\,d^4-2\,d^5\right )}{c^2\,\left (c^2-d^2\right )\,\left (c^2-2\,c\,d+d^2\right )}-\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (2\,c^5+4\,c^4\,d+2\,c^3\,d^2+7\,c^2\,d^3+2\,c\,d^4-2\,d^5\right )}{c\,\left (c^2-2\,c\,d+d^2\right )\,\left (-c^3-c^2\,d+c\,d^2+d^3\right )}+\frac {\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (8\,c^4\,d+22\,c^3\,d^2+27\,c^2\,d^3+5\,c\,d^4-2\,d^5\right )}{c\,\left (c+d\right )\,\left (c^2-d^2\right )\,\left (c^2-2\,c\,d+d^2\right )}}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (2\,a\,c^2+4\,a\,c\,d+4\,a\,d^2\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (2\,a\,c^2+4\,a\,c\,d+4\,a\,d^2\right )+a\,c^2+\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (a\,c^2+4\,a\,d\,c\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (a\,c^2+4\,a\,d\,c\right )+a\,c^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5\right )} \]

[In]

int(1/((a + a*sin(e + f*x))*(c + d*sin(e + f*x))^3),x)

[Out]

(3*d*atan(((3*d*(2*c*d + 2*c^2 + d^2)*(2*a*d^6 - 4*a*c^2*d^4 + 4*a*c^3*d^3 + 2*a*c^4*d^2 - 2*a*c*d^5 - 2*a*c^5
*d))/(2*a*(c + d)^(5/2)*(c - d)^(7/2)) - (3*c*d*tan(e/2 + (f*x)/2)*(2*c*d + 2*c^2 + d^2)*(a*c^5 - a*d^5 + 2*a*
c^2*d^3 - 2*a*c^3*d^2 + a*c*d^4 - a*c^4*d))/(a*(c + d)^(5/2)*(c - d)^(7/2)))/(6*c*d^2 + 6*c^2*d + 3*d^3))*(2*c
*d + 2*c^2 + d^2))/(a*f*(c + d)^(5/2)*(c - d)^(7/2)) - ((2*c*d^3 + 4*c^3*d + 2*c^4 - d^4 + 8*c^2*d^2)/((c + d)
*(c^2 - d^2)*(c^2 - 2*c*d + d^2)) - (tan(e/2 + (f*x)/2)^3*(2*c*d^5 + 8*c^5*d - 2*d^6 + 13*c^2*d^4 + 17*c^3*d^3
+ 22*c^4*d^2))/(c^2*(c^2 - 2*c*d + d^2)*(c*d^2 - c^2*d - c^3 + d^3)) + (tan(e/2 + (f*x)/2)^2*(4*c*d^4 + 4*c^4
*d + 4*c^5 - 2*d^5 + 21*c^2*d^3 + 14*c^3*d^2))/(c^2*(c^2 - d^2)*(c^2 - 2*c*d + d^2)) - (tan(e/2 + (f*x)/2)^4*(
2*c*d^4 + 4*c^4*d + 2*c^5 - 2*d^5 + 7*c^2*d^3 + 2*c^3*d^2))/(c*(c^2 - 2*c*d + d^2)*(c*d^2 - c^2*d - c^3 + d^3)
) + (tan(e/2 + (f*x)/2)*(5*c*d^4 + 8*c^4*d - 2*d^5 + 27*c^2*d^3 + 22*c^3*d^2))/(c*(c + d)*(c^2 - d^2)*(c^2 - 2
*c*d + d^2)))/(f*(tan(e/2 + (f*x)/2)^2*(2*a*c^2 + 4*a*d^2 + 4*a*c*d) + tan(e/2 + (f*x)/2)^3*(2*a*c^2 + 4*a*d^2
+ 4*a*c*d) + a*c^2 + tan(e/2 + (f*x)/2)*(a*c^2 + 4*a*c*d) + tan(e/2 + (f*x)/2)^4*(a*c^2 + 4*a*c*d) + a*c^2*ta
n(e/2 + (f*x)/2)^5))