\(\int \frac {3+3 \sin (e+f x)}{(c+d \sin (e+f x))^2} \, dx\) [431]

Optimal result

Integrand size = 23, antiderivative size = 81 \[ \int \frac {3+3 \sin (e+f x)}{(c+d \sin (e+f x))^2} \, dx=\frac {6 \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{(c+d) \sqrt {c^2-d^2} f}-\frac {3 \cos (e+f x)}{(c+d) f (c+d \sin (e+f x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.02, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {2833, 12, 2739, 632, 210} \[ \int \frac {3+3 \sin (e+f x)}{(c+d \sin (e+f x))^2} \, dx=\frac {2 a \arctan \left (\frac {c \tan \left (\frac {1}{2} (e+f x)\right )+d}{\sqrt {c^2-d^2}}\right )}{f (c+d) \sqrt {c^2-d^2}}-\frac {a \cos (e+f x)}{f (c+d) (c+d \sin (e+f x))} \]

[In]

Int[(a + a*Sin[e + f*x])/(c + d*Sin[e + f*x])^2,x]

[Out]

(2*a*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/((c + d)*Sqrt[c^2 - d^2]*f) - (a*Cos[e + f*x])/((c + d)
*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 {a \cos (e+f x)}{(c+d) f (c+d \sin (e+f x))}-\frac {\int \frac {a (c-d)}{c+d \sin (e+f x)} \, dx}{-c^2+d^2} \\ & = -\frac {a \cos (e+f x)}{(c+d) f (c+d \sin (e+f x))}+\frac {a \int \frac {1}{c+d \sin (e+f x)} \, dx}{c+d} \\ & = -\frac {a \cos (e+f x)}{(c+d) f (c+d \sin (e+f x))}+\frac {(2 a) \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 )}{(c+d) f} \\ & = -\frac {a \cos (e+f x)}{(c+d) f (c+d \sin (e+f x))}-\frac {(4 a) \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 )}{(c+d) f} \\ & = \frac {2 a \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{(c+d) \sqrt {c^2-d^2} f}-\frac {a \cos (e+f x)}{(c+d) f (c+d \sin (e+f x))} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.96 (sec) , antiderivative size = 219, normalized size of antiderivative = 2.70 \[ \int \frac {3+3 \sin (e+f x)}{(c+d \sin (e+f x))^2} \, dx=\frac {3 (1+\sin (e+f x)) \left (2 \sqrt {c^2-d^2} \csc (e) \sqrt {(\cos (e)-i \sin (e))^2} (c \cos (e)+d \sin (f x))+4 d \arctan \left (\frac {\sec \left (\frac {f x}{2}\right ) (\cos (e)-i \sin (e)) \left (d \cos \left (e+\frac {f x}{2}\right )+c \sin \left (\frac {f x}{2}\right )\right )}{\sqrt {c^2-d^2} \sqrt {(\cos (e)-i \sin (e))^2}}\right ) (\cos (e)-i \sin (e)) (c+d \sin (e+f x))\right )}{2 d (c+d) \sqrt {c^2-d^2} f \sqrt {(\cos (e)-i \sin (e))^2} \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2 (c+d \sin (e+f x))} \]

[In]

Integrate[(3 + 3*Sin[e + f*x])/(c + d*Sin[e + f*x])^2,x]

[Out]

(3*(1 + Sin[e + f*x])*(2*Sqrt[c^2 - d^2]*Csc[e]*Sqrt[(Cos[e] - I*Sin[e])^2]*(c*Cos[e] + d*Sin[f*x]) + 4*d*ArcT
an[(Sec[(f*x)/2]*(Cos[e] - I*Sin[e])*(d*Cos[e + (f*x)/2] + c*Sin[(f*x)/2]))/(Sqrt[c^2 - d^2]*Sqrt[(Cos[e] - I*
Sin[e])^2])]*(Cos[e] - I*Sin[e])*(c + d*Sin[e + f*x])))/(2*d*(c + d)*Sqrt[c^2 - d^2]*f*Sqrt[(Cos[e] - I*Sin[e]
)^2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2*(c + d*Sin[e + f*x]))

Maple [A] (verified)

Time = 0.65 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.40

method	result	size
derivativedivides	\(\frac {2 a \left (\frac {-\frac {d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (c +d \right ) c}-\frac {1}{c +d}}{\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) c +2 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+c}+\frac {\arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (c +d \right ) \sqrt {c^{2}-d^{2}}}\right )}{f}\)	\(113\)
default	\(\frac {2 a \left (\frac {-\frac {d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (c +d \right ) c}-\frac {1}{c +d}}{\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) c +2 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+c}+\frac {\arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (c +d \right ) \sqrt {c^{2}-d^{2}}}\right )}{f}\)	\(113\)
risch	\(-\frac {2 a \left (i d +c \,{\mathrm e}^{i \left (f x +e \right )}\right )}{d \left (c +d \right ) f \left (d \,{\mathrm e}^{2 i \left (f x +e \right )}-d +2 i c \,{\mathrm e}^{i \left (f x +e \right )}\right )}-\frac {a \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}-c^{2}+d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right )}{\sqrt {-c^{2}+d^{2}}\, \left (c +d \right ) f}+\frac {a \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}+c^{2}-d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right )}{\sqrt {-c^{2}+d^{2}}\, \left (c +d \right ) f}\)	\(205\)

[In]

int((a+a*sin(f*x+e))/(c+d*sin(f*x+e))^2,x,method=_RETURNVERBOSE)

[Out]

2/f*a*((-d/(c+d)/c*tan(1/2*f*x+1/2*e)-1/(c+d))/(tan(1/2*f*x+1/2*e)^2*c+2*d*tan(1/2*f*x+1/2*e)+c)+1/(c+d)/(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 [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 362, normalized size of antiderivative = 4.47 \[ \int \frac {3+3 \sin (e+f x)}{(c+d \sin (e+f x))^2} \, dx=\left [-\frac {{\left (a d \sin \left (f x + e\right ) + a c\right )} \sqrt {-c^{2} + d^{2}} \log \left (\frac {{\left (2 \, c^{2} - d^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2} + 2 \, {\left (c \cos \left (f x + e\right ) \sin \left (f x + e\right ) + d \cos \left (f x + e\right )\right )} \sqrt {-c^{2} + d^{2}}}{d^{2} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2}}\right ) + 2 \, {\left (a c^{2} - a d^{2}\right )} \cos \left (f x + e\right )}{2 \, {\left ({\left (c^{3} d + c^{2} d^{2} - c d^{3} - d^{4}\right )} f \sin \left (f x + e\right ) + {\left (c^{4} + c^{3} d - c^{2} d^{2} - c d^{3}\right )} f\right )}}, -\frac {{\left (a d \sin \left (f x + e\right ) + a c\right )} \sqrt {c^{2} - d^{2}} \arctan \left (-\frac {c \sin \left (f x + e\right ) + d}{\sqrt {c^{2} - d^{2}} \cos \left (f x + e\right )}\right ) + {\left (a c^{2} - a d^{2}\right )} \cos \left (f x + e\right )}{{\left (c^{3} d + c^{2} d^{2} - c d^{3} - d^{4}\right )} f \sin \left (f x + e\right ) + {\left (c^{4} + c^{3} d - c^{2} d^{2} - c d^{3}\right )} f}\right ] \]

[In]

integrate((a+a*sin(f*x+e))/(c+d*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

[-1/2*((a*d*sin(f*x + e) + a*c)*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*(a*c^2 - a*d^2)*cos(f*x + e))/((c^3*d + c^2*d^2 - c*d^3 - d^4)*f*sin(f*x + e) + (c^4 +
c^3*d - c^2*d^2 - c*d^3)*f), -((a*d*sin(f*x + e) + a*c)*sqrt(c^2 - d^2)*arctan(-(c*sin(f*x + e) + d)/(sqrt(c^
2 - d^2)*cos(f*x + e))) + (a*c^2 - a*d^2)*cos(f*x + e))/((c^3*d + c^2*d^2 - c*d^3 - d^4)*f*sin(f*x + e) + (c^4
+ c^3*d - c^2*d^2 - c*d^3)*f)]

Sympy [F(-1)]

Timed out. \[ \int \frac {3+3 \sin (e+f x)}{(c+d \sin (e+f x))^2} \, dx=\text {Timed out} \]

[In]

integrate((a+a*sin(f*x+e))/(c+d*sin(f*x+e))**2,x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {3+3 \sin (e+f x)}{(c+d \sin (e+f x))^2} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((a+a*sin(f*x+e))/(c+d*sin(f*x+e))^2,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 [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.53 \[ \int \frac {3+3 \sin (e+f x)}{(c+d \sin (e+f x))^2} \, dx=\frac {2 \, {\left (\frac {{\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (c\right ) + \arctan \left (\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + d}{\sqrt {c^{2} - d^{2}}}\right )\right )} a}{\sqrt {c^{2} - d^{2}} {\left (c + d\right )}} - \frac {a d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + a c}{{\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + c\right )} {\left (c^{2} + c d\right )}}\right )}}{f} \]

[In]

integrate((a+a*sin(f*x+e))/(c+d*sin(f*x+e))^2,x, algorithm="giac")

[Out]

2*((pi*floor(1/2*(f*x + e)/pi + 1/2)*sgn(c) + arctan((c*tan(1/2*f*x + 1/2*e) + d)/sqrt(c^2 - d^2)))*a/(sqrt(c^
2 - d^2)*(c + d)) - (a*d*tan(1/2*f*x + 1/2*e) + a*c)/((c*tan(1/2*f*x + 1/2*e)^2 + 2*d*tan(1/2*f*x + 1/2*e) + c
)*(c^2 + c*d)))/f

Mupad [B] (verification not implemented)

Time = 7.22 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.73 \[ \int \frac {3+3 \sin (e+f x)}{(c+d \sin (e+f x))^2} \, dx=\frac {2\,a\,\mathrm {atan}\left (\frac {\left (c+d\right )\,\left (\frac {2\,a\,\left (d^2+c\,d\right )}{{\left (c+d\right )}^{5/2}\,\sqrt {c-d}}+\frac {2\,a\,c\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{{\left (c+d\right )}^{3/2}\,\sqrt {c-d}}\right )}{2\,a}\right )}{f\,{\left (c+d\right )}^{3/2}\,\sqrt {c-d}}-\frac {\frac {2\,a}{c+d}+\frac {2\,a\,d\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{c\,\left (c+d\right )}}{f\,\left (c\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+2\,d\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+c\right )} \]

[In]

int((a + a*sin(e + f*x))/(c + d*sin(e + f*x))^2,x)

[Out]

(2*a*atan(((c + d)*((2*a*(c*d + d^2))/((c + d)^(5/2)*(c - d)^(1/2)) + (2*a*c*tan(e/2 + (f*x)/2))/((c + d)^(3/2
)*(c - d)^(1/2))))/(2*a)))/(f*(c + d)^(3/2)*(c - d)^(1/2)) - ((2*a)/(c + d) + (2*a*d*tan(e/2 + (f*x)/2))/(c*(c
+ d)))/(f*(c + 2*d*tan(e/2 + (f*x)/2) + c*tan(e/2 + (f*x)/2)^2))