\(\int \frac {3+b \sin (e+f x)}{(c+d \sin (e+f x))^2} \, dx\) [676]

Optimal result

Integrand size = 23, antiderivative size = 97 \[ \int \frac {3+b \sin (e+f x)}{(c+d \sin (e+f x))^2} \, dx=\frac {2 (3 c-b d) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{\left (c^2-d^2\right )^{3/2} f}-\frac {(b c-3 d) \cos (e+f x)}{\left (c^2-d^2\right ) f (c+d \sin (e+f x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.01, 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} \[ \int \frac {3+b \sin (e+f x)}{(c+d \sin (e+f x))^2} \, dx=\frac {2 (a c-b d) \arctan \left (\frac {c \tan \left (\frac {1}{2} (e+f x)\right )+d}{\sqrt {c^2-d^2}}\right )}{f \left (c^2-d^2\right )^{3/2}}-\frac {(b c-a d) \cos (e+f x)}{f \left (c^2-d^2\right ) (c+d \sin (e+f x))} \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.99 \[ \int \frac {3+b \sin (e+f x)}{(c+d \sin (e+f x))^2} \, dx=\frac {\frac {2 (3 c-b d) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{\left (c^2-d^2\right )^{3/2}}+\frac {(-b c+3 d) \cos (e+f x)}{(c-d) (c+d) (c+d \sin (e+f x))}}{f} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.04 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.46

method	result	size
derivativedivides	\(\frac {\frac {\frac {2 d \left (d a -c b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (c^{2}-d^{2}\right ) c}+\frac {2 \left (d a -c b \right )}{c^{2}-d^{2}}}{\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) c +2 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+c}+\frac {2 \left (a c -b d \right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (c^{2}-d^{2}\right )^{\frac {3}{2}}}}{f}\)	\(142\)
default	\(\frac {\frac {\frac {2 d \left (d a -c b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (c^{2}-d^{2}\right ) c}+\frac {2 \left (d a -c b \right )}{c^{2}-d^{2}}}{\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) c +2 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+c}+\frac {2 \left (a c -b d \right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (c^{2}-d^{2}\right )^{\frac {3}{2}}}}{f}\)	\(142\)
risch	\(\frac {2 i \left (-d a +c b \right ) \left (i d +c \,{\mathrm e}^{i \left (f x +e \right )}\right )}{d \left (c^{2}-d^{2}\right ) f \left (-i d \,{\mathrm e}^{2 i \left (f x +e \right )}+i d +2 c \,{\mathrm e}^{i \left (f x +e \right )}\right )}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}-c^{2}+d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right ) a c}{\sqrt {-c^{2}+d^{2}}\, \left (c +d \right ) \left (c -d \right ) f}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}-c^{2}+d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right ) b d}{\sqrt {-c^{2}+d^{2}}\, \left (c +d \right ) \left (c -d \right ) f}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}+c^{2}-d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right ) a c}{\sqrt {-c^{2}+d^{2}}\, \left (c +d \right ) \left (c -d \right ) f}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}+c^{2}-d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right ) b d}{\sqrt {-c^{2}+d^{2}}\, \left (c +d \right ) \left (c -d \right ) f}\)	\(396\)

[In]

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

[Out]

1/f*(2*(d*(a*d-b*c)/(c^2-d^2)/c*tan(1/2*f*x+1/2*e)+(a*d-b*c)/(c^2-d^2))/(tan(1/2*f*x+1/2*e)^2*c+2*d*tan(1/2*f*
x+1/2*e)+c)+2*(a*c-b*d)/(c^2-d^2)^(3/2)*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 = 394, normalized size of antiderivative = 4.06 \[ \int \frac {3+b \sin (e+f x)}{(c+d \sin (e+f x))^2} \, dx=\left [-\frac {{\left (a c^{2} - b c d + {\left (a c d - b d^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c^{2} + d^{2}} \log \left (\frac {{\left (2 \, c^{2} - d^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2} + 2 \, {\left (c \cos \left (f x + e\right ) \sin \left (f x + e\right ) + d \cos \left (f x + e\right )\right )} \sqrt {-c^{2} + d^{2}}}{d^{2} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2}}\right ) + 2 \, {\left (b c^{3} - a c^{2} d - b c d^{2} + a d^{3}\right )} \cos \left (f x + e\right )}{2 \, {\left ({\left (c^{4} d - 2 \, c^{2} d^{3} + d^{5}\right )} f \sin \left (f x + e\right ) + {\left (c^{5} - 2 \, c^{3} d^{2} + c d^{4}\right )} f\right )}}, -\frac {{\left (a c^{2} - b c d + {\left (a c d - b d^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {c^{2} - d^{2}} \arctan \left (-\frac {c \sin \left (f x + e\right ) + d}{\sqrt {c^{2} - d^{2}} \cos \left (f x + e\right )}\right ) + {\left (b c^{3} - a c^{2} d - b c d^{2} + a d^{3}\right )} \cos \left (f x + e\right )}{{\left (c^{4} d - 2 \, c^{2} d^{3} + d^{5}\right )} f \sin \left (f x + e\right ) + {\left (c^{5} - 2 \, c^{3} d^{2} + c d^{4}\right )} f}\right ] \]

[In]

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

[Out]

[-1/2*((a*c^2 - b*c*d + (a*c*d - b*d^2)*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*(b*c^3 - a*c^2*d - b*c*d^2 + a*d^3)*cos(f*x + e))/((c^4*d - 2*c^
2*d^3 + d^5)*f*sin(f*x + e) + (c^5 - 2*c^3*d^2 + c*d^4)*f), -((a*c^2 - b*c*d + (a*c*d - b*d^2)*sin(f*x + e))*s
qrt(c^2 - d^2)*arctan(-(c*sin(f*x + e) + d)/(sqrt(c^2 - d^2)*cos(f*x + e))) + (b*c^3 - a*c^2*d - b*c*d^2 + a*d
^3)*cos(f*x + e))/((c^4*d - 2*c^2*d^3 + d^5)*f*sin(f*x + e) + (c^5 - 2*c^3*d^2 + c*d^4)*f)]

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F(-2)]

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

[In]

integrate((a+b*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 = 0.31 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.57 \[ \int \frac {3+b \sin (e+f x)}{(c+d \sin (e+f x))^2} \, dx=\frac {2 \, {\left (\frac {{\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (c\right ) + \arctan \left (\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + d}{\sqrt {c^{2} - d^{2}}}\right )\right )} {\left (a c - b d\right )}}{{\left (c^{2} - d^{2}\right )}^{\frac {3}{2}}} - \frac {b c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - a d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + b c^{2} - a c d}{{\left (c^{3} - c d^{2}\right )} {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + c\right )}}\right )}}{f} \]

[In]

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

Mupad [B] (verification not implemented)

Time = 8.07 (sec) , antiderivative size = 214, normalized size of antiderivative = 2.21 \[ \int \frac {3+b \sin (e+f x)}{(c+d \sin (e+f x))^2} \, dx=\frac {\frac {2\,\left (a\,d-b\,c\right )}{c^2-d^2}+\frac {2\,d\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (a\,d-b\,c\right )}{c\,\left (c^2-d^2\right )}}{f\,\left (c\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+2\,d\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+c\right )}+\frac {2\,\mathrm {atan}\left (\frac {\left (\frac {2\,\left (c^2\,d-d^3\right )\,\left (a\,c-b\,d\right )}{{\left (c+d\right )}^{3/2}\,\left (c^2-d^2\right )\,{\left (c-d\right )}^{3/2}}+\frac {2\,c\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (a\,c-b\,d\right )}{{\left (c+d\right )}^{3/2}\,{\left (c-d\right )}^{3/2}}\right )\,\left (c^2-d^2\right )}{2\,\left (a\,c-b\,d\right )}\right )\,\left (a\,c-b\,d\right )}{f\,{\left (c+d\right )}^{3/2}\,{\left (c-d\right )}^{3/2}} \]

[In]

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

[Out]

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