\(\int \frac {3+b \sin (e+f x)}{(c+d \sin (e+f x))^3} \, dx\) [677]

Optimal result

Integrand size = 23, antiderivative size = 161 \[ \int \frac {3+b \sin (e+f x)}{(c+d \sin (e+f x))^3} \, dx=-\frac {\left (3 b c d-3 \left (2 c^2+d^2\right )\right ) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{\left (c^2-d^2\right )^{5/2} f}-\frac {(b c-3 d) \cos (e+f x)}{2 \left (c^2-d^2\right ) f (c+d \sin (e+f x))^2}+\frac {\left (9 c d-b \left (c^2+2 d^2\right )\right ) \cos (e+f x)}{2 \left (c^2-d^2\right )^2 f (c+d \sin (e+f x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.02, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {2833, 12, 2739, 632, 210} \[ \int \frac {3+b \sin (e+f x)}{(c+d \sin (e+f x))^3} \, dx=-\frac {\left (3 b c d-a \left (2 c^2+d^2\right )\right ) \arctan \left (\frac {c \tan \left (\frac {1}{2} (e+f x)\right )+d}{\sqrt {c^2-d^2}}\right )}{f \left (c^2-d^2\right )^{5/2}}+\frac {\left (3 a c d-b \left (c^2+2 d^2\right )\right ) \cos (e+f x)}{2 f \left (c^2-d^2\right )^2 (c+d \sin (e+f x))}-\frac {(b c-a d) \cos (e+f x)}{2 f \left (c^2-d^2\right ) (c+d \sin (e+f x))^2} \]

[In]

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

[Out]

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

Rubi steps \begin{align*} \text {integral}& = -\frac {(b c-a d) \cos (e+f x)}{2 \left (c^2-d^2\right ) f (c+d \sin (e+f x))^2}-\frac {\int \frac {-2 (a c-b d)-(b c-a d) \sin (e+f x)}{(c+d \sin (e+f x))^2} \, dx}{2 \left (c^2-d^2\right )} \\ & = -\frac {(b c-a d) \cos (e+f x)}{2 \left (c^2-d^2\right ) f (c+d \sin (e+f x))^2}+\frac {\left (3 a c d-b \left (c^2+2 d^2\right )\right ) \cos (e+f x)}{2 \left (c^2-d^2\right )^2 f (c+d \sin (e+f x))}+\frac {\int \frac {-3 b c d+a \left (2 c^2+d^2\right )}{c+d \sin (e+f x)} \, dx}{2 \left (c^2-d^2\right )^2} \\ & = -\frac {(b c-a d) \cos (e+f x)}{2 \left (c^2-d^2\right ) f (c+d \sin (e+f x))^2}+\frac {\left (3 a c d-b \left (c^2+2 d^2\right )\right ) \cos (e+f x)}{2 \left (c^2-d^2\right )^2 f (c+d \sin (e+f x))}-\frac {\left (3 b c d-a \left (2 c^2+d^2\right )\right ) \int \frac {1}{c+d \sin (e+f x)} \, dx}{2 \left (c^2-d^2\right )^2} \\ & = -\frac {(b c-a d) \cos (e+f x)}{2 \left (c^2-d^2\right ) f (c+d \sin (e+f x))^2}+\frac {\left (3 a c d-b \left (c^2+2 d^2\right )\right ) \cos (e+f x)}{2 \left (c^2-d^2\right )^2 f (c+d \sin (e+f x))}-\frac {\left (3 b c d-a \left (2 c^2+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 )}{\left (c^2-d^2\right )^2 f} \\ & = -\frac {(b c-a d) \cos (e+f x)}{2 \left (c^2-d^2\right ) f (c+d \sin (e+f x))^2}+\frac {\left (3 a c d-b \left (c^2+2 d^2\right )\right ) \cos (e+f x)}{2 \left (c^2-d^2\right )^2 f (c+d \sin (e+f x))}+\frac {\left (2 \left (3 b c d-a \left (2 c^2+d^2\right )\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 )}{\left (c^2-d^2\right )^2 f} \\ & = -\frac {\left (3 b c d-a \left (2 c^2+d^2\right )\right ) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{\left (c^2-d^2\right )^{5/2} f}-\frac {(b c-a d) \cos (e+f x)}{2 \left (c^2-d^2\right ) f (c+d \sin (e+f x))^2}+\frac {\left (3 a c d-b \left (c^2+2 d^2\right )\right ) \cos (e+f x)}{2 \left (c^2-d^2\right )^2 f (c+d \sin (e+f x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.56 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.95 \[ \int \frac {3+b \sin (e+f x)}{(c+d \sin (e+f x))^3} \, dx=\frac {\frac {6 \left (2 c^2-b c d+d^2\right ) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{\left (c^2-d^2\right )^{5/2}}+\frac {(-b c+3 d) \cos (e+f x)}{(c-d) (c+d) (c+d \sin (e+f x))^2}-\frac {\left (-9 c d+b \left (c^2+2 d^2\right )\right ) \cos (e+f x)}{(c-d)^2 (c+d)^2 (c+d \sin (e+f x))}}{2 f} \]

[In]

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

[Out]

((6*(2*c^2 - b*c*d + d^2)*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/(c^2 - d^2)^(5/2) + ((-(b*c) + 3*d
)*Cos[e + f*x])/((c - d)*(c + d)*(c + d*Sin[e + f*x])^2) - ((-9*c*d + b*(c^2 + 2*d^2))*Cos[e + f*x])/((c - d)^
2*(c + d)^2*(c + d*Sin[e + f*x])))/(2*f)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(350\) vs. \(2(155)=310\).

Time = 1.46 (sec) , antiderivative size = 351, normalized size of antiderivative = 2.18

method	result	size
derivativedivides	\(\frac {\frac {\frac {d \left (5 a \,c^{2} d -2 a \,d^{3}-3 c^{3} b \right ) \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c \left (c^{4}-2 c^{2} d^{2}+d^{4}\right )}+\frac {\left (4 a \,c^{4} d +7 a \,c^{2} d^{3}-2 a \,d^{5}-2 c^{5} b -5 b \,c^{3} d^{2}-2 b c \,d^{4}\right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{\left (c^{4}-2 c^{2} d^{2}+d^{4}\right ) c^{2}}+\frac {d \left (11 a \,c^{2} d -2 a \,d^{3}-5 c^{3} b -4 c \,d^{2} b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{c \left (c^{4}-2 c^{2} d^{2}+d^{4}\right )}+\frac {2 \left (4 a \,c^{2} d -a \,d^{3}-2 c^{3} b -c \,d^{2} b \right )}{2 c^{4}-4 c^{2} d^{2}+2 d^{4}}}{{\left (\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) c +2 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+c \right )}^{2}}+\frac {\left (2 c^{2} a +a \,d^{2}-3 c d b \right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (c^{4}-2 c^{2} d^{2}+d^{4}\right ) \sqrt {c^{2}-d^{2}}}}{f}\)	\(351\)
default	\(\frac {\frac {\frac {d \left (5 a \,c^{2} d -2 a \,d^{3}-3 c^{3} b \right ) \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c \left (c^{4}-2 c^{2} d^{2}+d^{4}\right )}+\frac {\left (4 a \,c^{4} d +7 a \,c^{2} d^{3}-2 a \,d^{5}-2 c^{5} b -5 b \,c^{3} d^{2}-2 b c \,d^{4}\right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{\left (c^{4}-2 c^{2} d^{2}+d^{4}\right ) c^{2}}+\frac {d \left (11 a \,c^{2} d -2 a \,d^{3}-5 c^{3} b -4 c \,d^{2} b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{c \left (c^{4}-2 c^{2} d^{2}+d^{4}\right )}+\frac {2 \left (4 a \,c^{2} d -a \,d^{3}-2 c^{3} b -c \,d^{2} b \right )}{2 c^{4}-4 c^{2} d^{2}+2 d^{4}}}{{\left (\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) c +2 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+c \right )}^{2}}+\frac {\left (2 c^{2} a +a \,d^{2}-3 c d b \right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (c^{4}-2 c^{2} d^{2}+d^{4}\right ) \sqrt {c^{2}-d^{2}}}}{f}\)	\(351\)
risch	\(\frac {i \left (2 i d^{2} a \,c^{2} {\mathrm e}^{3 i \left (f x +e \right )}+i d^{4} a \,{\mathrm e}^{3 i \left (f x +e \right )}-3 i d^{3} b c \,{\mathrm e}^{3 i \left (f x +e \right )}-10 i a \,c^{2} d^{2} {\mathrm e}^{i \left (f x +e \right )}+i a \,d^{4} {\mathrm e}^{i \left (f x +e \right )}+4 i b \,c^{3} d \,{\mathrm e}^{i \left (f x +e \right )}+5 i b c \,d^{3} {\mathrm e}^{i \left (f x +e \right )}-6 d a \,c^{3} {\mathrm e}^{2 i \left (f x +e \right )}-3 d^{3} a c \,{\mathrm e}^{2 i \left (f x +e \right )}+2 b \,c^{4} {\mathrm e}^{2 i \left (f x +e \right )}+5 d^{2} b \,c^{2} {\mathrm e}^{2 i \left (f x +e \right )}+2 d^{4} b \,{\mathrm e}^{2 i \left (f x +e \right )}+3 a c \,d^{3}-b \,c^{2} d^{2}-2 b \,d^{4}\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^{2}-d^{2}\right )^{2} f d}-\frac {\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} a}{\sqrt {-c^{2}+d^{2}}\, \left (c +d \right )^{2} \left (c -d \right )^{2} f}-\frac {\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 ) a \,d^{2}}{2 \sqrt {-c^{2}+d^{2}}\, \left (c +d \right )^{2} \left (c -d \right )^{2} f}+\frac {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 ) c d b}{2 \sqrt {-c^{2}+d^{2}}\, \left (c +d \right )^{2} \left (c -d \right )^{2} f}+\frac {\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} a}{\sqrt {-c^{2}+d^{2}}\, \left (c +d \right )^{2} \left (c -d \right )^{2} f}+\frac {\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 ) a \,d^{2}}{2 \sqrt {-c^{2}+d^{2}}\, \left (c +d \right )^{2} \left (c -d \right )^{2} f}-\frac {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 ) c d b}{2 \sqrt {-c^{2}+d^{2}}\, \left (c +d \right )^{2} \left (c -d \right )^{2} f}\)	\(767\)

[In]

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

[Out]

1/f*(2*(1/2*d*(5*a*c^2*d-2*a*d^3-3*b*c^3)/c/(c^4-2*c^2*d^2+d^4)*tan(1/2*f*x+1/2*e)^3+1/2*(4*a*c^4*d+7*a*c^2*d^
3-2*a*d^5-2*b*c^5-5*b*c^3*d^2-2*b*c*d^4)/(c^4-2*c^2*d^2+d^4)/c^2*tan(1/2*f*x+1/2*e)^2+1/2*d*(11*a*c^2*d-2*a*d^
3-5*b*c^3-4*b*c*d^2)/c/(c^4-2*c^2*d^2+d^4)*tan(1/2*f*x+1/2*e)+1/2*(4*a*c^2*d-a*d^3-2*b*c^3-b*c*d^2)/(c^4-2*c^2
*d^2+d^4))/(tan(1/2*f*x+1/2*e)^2*c+2*d*tan(1/2*f*x+1/2*e)+c)^2+(2*a*c^2+a*d^2-3*b*c*d)/(c^4-2*c^2*d^2+d^4)/(c^
2-d^2)^(1/2)*arctan(1/2*(2*c*tan(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. 354 vs. \(2 (155) = 310\).

Time = 0.30 (sec) , antiderivative size = 793, normalized size of antiderivative = 4.93 \[ \int \frac {3+b \sin (e+f x)}{(c+d \sin (e+f x))^3} \, dx=\left [\frac {2 \, {\left (b c^{4} d - 3 \, a c^{3} d^{2} + b c^{2} d^{3} + 3 \, a c d^{4} - 2 \, b d^{5}\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + {\left (2 \, a c^{4} - 3 \, b c^{3} d + 3 \, a c^{2} d^{2} - 3 \, b c d^{3} + a d^{4} - {\left (2 \, a c^{2} d^{2} - 3 \, b c d^{3} + a d^{4}\right )} \cos \left (f x + e\right )^{2} + 2 \, {\left (2 \, a c^{3} d - 3 \, b c^{2} d^{2} + a c d^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c^{2} + d^{2}} \log \left (\frac {{\left (2 \, c^{2} - d^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2} + 2 \, {\left (c \cos \left (f x + e\right ) \sin \left (f x + e\right ) + d \cos \left (f x + e\right )\right )} \sqrt {-c^{2} + d^{2}}}{d^{2} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2}}\right ) + 2 \, {\left (2 \, b c^{5} - 4 \, a c^{4} d - b c^{3} d^{2} + 5 \, a c^{2} d^{3} - b c d^{4} - a d^{5}\right )} \cos \left (f x + e\right )}{4 \, {\left ({\left (c^{6} d^{2} - 3 \, c^{4} d^{4} + 3 \, c^{2} d^{6} - d^{8}\right )} f \cos \left (f x + e\right )^{2} - 2 \, {\left (c^{7} d - 3 \, c^{5} d^{3} + 3 \, c^{3} d^{5} - c d^{7}\right )} f \sin \left (f x + e\right ) - {\left (c^{8} - 2 \, c^{6} d^{2} + 2 \, c^{2} d^{6} - d^{8}\right )} f\right )}}, \frac {{\left (b c^{4} d - 3 \, a c^{3} d^{2} + b c^{2} d^{3} + 3 \, a c d^{4} - 2 \, b d^{5}\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + {\left (2 \, a c^{4} - 3 \, b c^{3} d + 3 \, a c^{2} d^{2} - 3 \, b c d^{3} + a d^{4} - {\left (2 \, a c^{2} d^{2} - 3 \, b c d^{3} + a d^{4}\right )} \cos \left (f x + e\right )^{2} + 2 \, {\left (2 \, a c^{3} d - 3 \, b c^{2} d^{2} + a c d^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {c^{2} - d^{2}} \arctan \left (-\frac {c \sin \left (f x + e\right ) + d}{\sqrt {c^{2} - d^{2}} \cos \left (f x + e\right )}\right ) + {\left (2 \, b c^{5} - 4 \, a c^{4} d - b c^{3} d^{2} + 5 \, a c^{2} d^{3} - b c d^{4} - a d^{5}\right )} \cos \left (f x + e\right )}{2 \, {\left ({\left (c^{6} d^{2} - 3 \, c^{4} d^{4} + 3 \, c^{2} d^{6} - d^{8}\right )} f \cos \left (f x + e\right )^{2} - 2 \, {\left (c^{7} d - 3 \, c^{5} d^{3} + 3 \, c^{3} d^{5} - c d^{7}\right )} f \sin \left (f x + e\right ) - {\left (c^{8} - 2 \, c^{6} d^{2} + 2 \, c^{2} d^{6} - d^{8}\right )} f\right )}}\right ] \]

[In]

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

[Out]

[1/4*(2*(b*c^4*d - 3*a*c^3*d^2 + b*c^2*d^3 + 3*a*c*d^4 - 2*b*d^5)*cos(f*x + e)*sin(f*x + e) + (2*a*c^4 - 3*b*c
^3*d + 3*a*c^2*d^2 - 3*b*c*d^3 + a*d^4 - (2*a*c^2*d^2 - 3*b*c*d^3 + a*d^4)*cos(f*x + e)^2 + 2*(2*a*c^3*d - 3*b
*c^2*d^2 + a*c*d^3)*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*b*c^5 - 4*a*c^4*d - b*c^3*d^2 + 5*a*c^2*d^3 - b*c*d^4 - a*d^5)*cos(f*x + e))/((c^
6*d^2 - 3*c^4*d^4 + 3*c^2*d^6 - d^8)*f*cos(f*x + e)^2 - 2*(c^7*d - 3*c^5*d^3 + 3*c^3*d^5 - c*d^7)*f*sin(f*x +
e) - (c^8 - 2*c^6*d^2 + 2*c^2*d^6 - d^8)*f), 1/2*((b*c^4*d - 3*a*c^3*d^2 + b*c^2*d^3 + 3*a*c*d^4 - 2*b*d^5)*co
s(f*x + e)*sin(f*x + e) + (2*a*c^4 - 3*b*c^3*d + 3*a*c^2*d^2 - 3*b*c*d^3 + a*d^4 - (2*a*c^2*d^2 - 3*b*c*d^3 +
a*d^4)*cos(f*x + e)^2 + 2*(2*a*c^3*d - 3*b*c^2*d^2 + a*c*d^3)*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*b*c^5 - 4*a*c^4*d - b*c^3*d^2 + 5*a*c^2*d^3 - b*c*d^4 - a*d^5)
*cos(f*x + e))/((c^6*d^2 - 3*c^4*d^4 + 3*c^2*d^6 - d^8)*f*cos(f*x + e)^2 - 2*(c^7*d - 3*c^5*d^3 + 3*c^3*d^5 -
c*d^7)*f*sin(f*x + e) - (c^8 - 2*c^6*d^2 + 2*c^2*d^6 - d^8)*f)]

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F(-2)]

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

[In]

integrate((a+b*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. 411 vs. \(2 (155) = 310\).

Time = 0.32 (sec) , antiderivative size = 411, normalized size of antiderivative = 2.55 \[ \int \frac {3+b \sin (e+f x)}{(c+d \sin (e+f x))^3} \, dx=\frac {\frac {{\left (2 \, a c^{2} - 3 \, b c d + a d^{2}\right )} {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (c\right ) + \arctan \left (\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + d}{\sqrt {c^{2} - d^{2}}}\right )\right )}}{{\left (c^{4} - 2 \, c^{2} d^{2} + d^{4}\right )} \sqrt {c^{2} - d^{2}}} - \frac {3 \, b c^{4} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 5 \, a c^{3} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 2 \, a c d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 2 \, b c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 4 \, a c^{4} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 5 \, b c^{3} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 7 \, a c^{2} d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, b c d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, a d^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 5 \, b c^{4} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 11 \, a c^{3} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 4 \, b c^{2} d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 2 \, a c d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 2 \, b c^{5} - 4 \, a c^{4} d + b c^{3} d^{2} + a c^{2} d^{3}}{{\left (c^{6} - 2 \, c^{4} d^{2} + c^{2} d^{4}\right )} {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + c\right )}^{2}}}{f} \]

[In]

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

[Out]

((2*a*c^2 - 3*b*c*d + a*d^2)*(pi*floor(1/2*(f*x + e)/pi + 1/2)*sgn(c) + arctan((c*tan(1/2*f*x + 1/2*e) + d)/sq
rt(c^2 - d^2)))/((c^4 - 2*c^2*d^2 + d^4)*sqrt(c^2 - d^2)) - (3*b*c^4*d*tan(1/2*f*x + 1/2*e)^3 - 5*a*c^3*d^2*ta
n(1/2*f*x + 1/2*e)^3 + 2*a*c*d^4*tan(1/2*f*x + 1/2*e)^3 + 2*b*c^5*tan(1/2*f*x + 1/2*e)^2 - 4*a*c^4*d*tan(1/2*f
*x + 1/2*e)^2 + 5*b*c^3*d^2*tan(1/2*f*x + 1/2*e)^2 - 7*a*c^2*d^3*tan(1/2*f*x + 1/2*e)^2 + 2*b*c*d^4*tan(1/2*f*
x + 1/2*e)^2 + 2*a*d^5*tan(1/2*f*x + 1/2*e)^2 + 5*b*c^4*d*tan(1/2*f*x + 1/2*e) - 11*a*c^3*d^2*tan(1/2*f*x + 1/
2*e) + 4*b*c^2*d^3*tan(1/2*f*x + 1/2*e) + 2*a*c*d^4*tan(1/2*f*x + 1/2*e) + 2*b*c^5 - 4*a*c^4*d + b*c^3*d^2 + a
*c^2*d^3)/((c^6 - 2*c^4*d^2 + c^2*d^4)*(c*tan(1/2*f*x + 1/2*e)^2 + 2*d*tan(1/2*f*x + 1/2*e) + c)^2))/f

Mupad [B] (verification not implemented)

Time = 9.98 (sec) , antiderivative size = 477, normalized size of antiderivative = 2.96 \[ \int \frac {3+b \sin (e+f x)}{(c+d \sin (e+f x))^3} \, dx=\frac {\mathrm {atan}\left (\frac {\left (\frac {\left (2\,c^4\,d-4\,c^2\,d^3+2\,d^5\right )\,\left (2\,a\,c^2-3\,b\,c\,d+a\,d^2\right )}{2\,{\left (c+d\right )}^{5/2}\,{\left (c-d\right )}^{5/2}\,\left (c^4-2\,c^2\,d^2+d^4\right )}+\frac {c\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (2\,a\,c^2-3\,b\,c\,d+a\,d^2\right )}{{\left (c+d\right )}^{5/2}\,{\left (c-d\right )}^{5/2}}\right )\,\left (c^4-2\,c^2\,d^2+d^4\right )}{2\,a\,c^2-3\,b\,c\,d+a\,d^2}\right )\,\left (2\,a\,c^2-3\,b\,c\,d+a\,d^2\right )}{f\,{\left (c+d\right )}^{5/2}\,{\left (c-d\right )}^{5/2}}-\frac {\frac {2\,b\,c^3-4\,a\,c^2\,d+b\,c\,d^2+a\,d^3}{c^4-2\,c^2\,d^2+d^4}+\frac {d\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (3\,b\,c^3-5\,a\,c^2\,d+2\,a\,d^3\right )}{c\,\left (c^4-2\,c^2\,d^2+d^4\right )}+\frac {d\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (5\,b\,c^3-11\,a\,c^2\,d+4\,b\,c\,d^2+2\,a\,d^3\right )}{c\,\left (c^4-2\,c^2\,d^2+d^4\right )}+\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (c^2+2\,d^2\right )\,\left (2\,b\,c^3-4\,a\,c^2\,d+b\,c\,d^2+a\,d^3\right )}{c^2\,\left (c^4-2\,c^2\,d^2+d^4\right )}}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (2\,c^2+4\,d^2\right )+c^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+c^2+4\,c\,d\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+4\,c\,d\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )} \]

[In]

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

[Out]

(atan(((((2*c^4*d + 2*d^5 - 4*c^2*d^3)*(2*a*c^2 + a*d^2 - 3*b*c*d))/(2*(c + d)^(5/2)*(c - d)^(5/2)*(c^4 + d^4
- 2*c^2*d^2)) + (c*tan(e/2 + (f*x)/2)*(2*a*c^2 + a*d^2 - 3*b*c*d))/((c + d)^(5/2)*(c - d)^(5/2)))*(c^4 + d^4 -
2*c^2*d^2))/(2*a*c^2 + a*d^2 - 3*b*c*d))*(2*a*c^2 + a*d^2 - 3*b*c*d))/(f*(c + d)^(5/2)*(c - d)^(5/2)) - ((a*d
^3 + 2*b*c^3 - 4*a*c^2*d + b*c*d^2)/(c^4 + d^4 - 2*c^2*d^2) + (d*tan(e/2 + (f*x)/2)^3*(2*a*d^3 + 3*b*c^3 - 5*a
*c^2*d))/(c*(c^4 + d^4 - 2*c^2*d^2)) + (d*tan(e/2 + (f*x)/2)*(2*a*d^3 + 5*b*c^3 - 11*a*c^2*d + 4*b*c*d^2))/(c*
(c^4 + d^4 - 2*c^2*d^2)) + (tan(e/2 + (f*x)/2)^2*(c^2 + 2*d^2)*(a*d^3 + 2*b*c^3 - 4*a*c^2*d + b*c*d^2))/(c^2*(
c^4 + d^4 - 2*c^2*d^2)))/(f*(tan(e/2 + (f*x)/2)^2*(2*c^2 + 4*d^2) + c^2*tan(e/2 + (f*x)/2)^4 + c^2 + 4*c*d*tan
(e/2 + (f*x)/2)^3 + 4*c*d*tan(e/2 + (f*x)/2)))