\(\int \frac {b+c+\cos (x)}{a+b \sin (x)} \, dx\) [4]
Optimal result
Integrand size = 14, antiderivative size = 55 \[
\int \frac {b+c+\cos (x)}{a+b \sin (x)} \, dx=\frac {2 (b+c) \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+\frac {\log (a+b \sin (x))}{b}
\]
[Out]
ln(a+b*sin(x))/b+2*(b+c)*arctan((b+a*tan(1/2*x))/(a^2-b^2)^(1/2))/(a^2-b^2)^(1/2)
Rubi [A] (verified)
Time = 0.14 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of
steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {4486, 2739, 632, 210, 2747,
31} \[
\int \frac {b+c+\cos (x)}{a+b \sin (x)} \, dx=\frac {2 (b+c) \arctan \left (\frac {a \tan \left (\frac {x}{2}\right )+b}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+\frac {\log (a+b \sin (x))}{b}
\]
[In]
Int[(b + c + Cos[x])/(a + b*Sin[x]),x]
[Out]
(2*(b + c)*ArcTan[(b + a*Tan[x/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] + Log[a + b*Sin[x]]/b
Rule 31
Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, 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 2747
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]
Rule 4486
Int[u_, x_Symbol] :> With[{v = ExpandTrig[u, x]}, Int[v, x] /; SumQ[v]] /; !InertTrigFreeQ[u]
Rubi steps \begin{align*}
\text {integral}& = \int \left (\frac {b \left (1+\frac {c}{b}\right )}{a+b \sin (x)}+\frac {\cos (x)}{a+b \sin (x)}\right ) \, dx \\ & = (b+c) \int \frac {1}{a+b \sin (x)} \, dx+\int \frac {\cos (x)}{a+b \sin (x)} \, dx \\ & = \frac {\text {Subst}\left (\int \frac {1}{a+x} \, dx,x,b \sin (x)\right )}{b}+(2 (b+c)) \text {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right ) \\ & = \frac {\log (a+b \sin (x))}{b}-(4 (b+c)) \text {Subst}\left (\int \frac {1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac {x}{2}\right )\right ) \\ & = \frac {2 (b+c) \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+\frac {\log (a+b \sin (x))}{b} \\
\end{align*}
Mathematica [A] (verified)
Time = 0.12 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00
\[
\int \frac {b+c+\cos (x)}{a+b \sin (x)} \, dx=\frac {2 (b+c) \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+\frac {\log (a+b \sin (x))}{b}
\]
[In]
Integrate[(b + c + Cos[x])/(a + b*Sin[x]),x]
[Out]
(2*(b + c)*ArcTan[(b + a*Tan[x/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] + Log[a + b*Sin[x]]/b
Maple [A] (verified)
Time = 0.76 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.98
| | |
| method | result | size |
| | |
| parts |
\(\frac {\ln \left (a +b \sin \left (x \right )\right )}{b}+\frac {2 \left (b +c \right ) \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}}\) |
\(54\) |
| default |
\(-\frac {\ln \left (1+\tan ^{2}\left (\frac {x}{2}\right )\right )}{b}+\frac {\ln \left (a \left (\tan ^{2}\left (\frac {x}{2}\right )\right )+2 b \tan \left (\frac {x}{2}\right )+a \right )+\frac {2 \left (b^{2}+b c \right ) \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}}}{b}\) |
\(86\) |
| risch |
\(\frac {i x}{b}-\frac {2 i x \,a^{2} b}{a^{2} b^{2}-b^{4}}+\frac {2 i x \,b^{3}}{a^{2} b^{2}-b^{4}}+\frac {\ln \left ({\mathrm e}^{i x}+\frac {i a \,b^{2}+i c a b +\sqrt {-a^{2} b^{4}-2 a^{2} b^{3} c -a^{2} b^{2} c^{2}+b^{6}+2 b^{5} c +b^{4} c^{2}}}{b^{2} \left (b +c \right )}\right ) a^{2}}{\left (a^{2}-b^{2}\right ) b}-\frac {b \ln \left ({\mathrm e}^{i x}+\frac {i a \,b^{2}+i c a b +\sqrt {-a^{2} b^{4}-2 a^{2} b^{3} c -a^{2} b^{2} c^{2}+b^{6}+2 b^{5} c +b^{4} c^{2}}}{b^{2} \left (b +c \right )}\right )}{a^{2}-b^{2}}+\frac {\ln \left ({\mathrm e}^{i x}+\frac {i a \,b^{2}+i c a b +\sqrt {-a^{2} b^{4}-2 a^{2} b^{3} c -a^{2} b^{2} c^{2}+b^{6}+2 b^{5} c +b^{4} c^{2}}}{b^{2} \left (b +c \right )}\right ) \sqrt {-a^{2} b^{4}-2 a^{2} b^{3} c -a^{2} b^{2} c^{2}+b^{6}+2 b^{5} c +b^{4} c^{2}}}{\left (a^{2}-b^{2}\right ) b}+\frac {\ln \left ({\mathrm e}^{i x}-\frac {-i a \,b^{2}-i c a b +\sqrt {-a^{2} b^{4}-2 a^{2} b^{3} c -a^{2} b^{2} c^{2}+b^{6}+2 b^{5} c +b^{4} c^{2}}}{b^{2} \left (b +c \right )}\right ) a^{2}}{\left (a^{2}-b^{2}\right ) b}-\frac {b \ln \left ({\mathrm e}^{i x}-\frac {-i a \,b^{2}-i c a b +\sqrt {-a^{2} b^{4}-2 a^{2} b^{3} c -a^{2} b^{2} c^{2}+b^{6}+2 b^{5} c +b^{4} c^{2}}}{b^{2} \left (b +c \right )}\right )}{a^{2}-b^{2}}-\frac {\ln \left ({\mathrm e}^{i x}-\frac {-i a \,b^{2}-i c a b +\sqrt {-a^{2} b^{4}-2 a^{2} b^{3} c -a^{2} b^{2} c^{2}+b^{6}+2 b^{5} c +b^{4} c^{2}}}{b^{2} \left (b +c \right )}\right ) \sqrt {-a^{2} b^{4}-2 a^{2} b^{3} c -a^{2} b^{2} c^{2}+b^{6}+2 b^{5} c +b^{4} c^{2}}}{\left (a^{2}-b^{2}\right ) b}\) |
\(708\) |
| | |
|
|
|
|
[In]
int((b+c+cos(x))/(a+b*sin(x)),x,method=_RETURNVERBOSE)
[Out]
ln(a+b*sin(x))/b+2*(b+c)/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*x)+2*b)/(a^2-b^2)^(1/2))
Fricas [A] (verification not implemented)
none
Time = 0.28 (sec) , antiderivative size = 250, normalized size of antiderivative = 4.55
\[
\int \frac {b+c+\cos (x)}{a+b \sin (x)} \, dx=\left [-\frac {\sqrt {-a^{2} + b^{2}} {\left (b^{2} + b c\right )} \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (x\right ) \sin \left (x\right ) + b \cos \left (x\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2}}\right ) - {\left (a^{2} - b^{2}\right )} \log \left (-b^{2} \cos \left (x\right )^{2} + 2 \, a b \sin \left (x\right ) + a^{2} + b^{2}\right )}{2 \, {\left (a^{2} b - b^{3}\right )}}, -\frac {2 \, \sqrt {a^{2} - b^{2}} {\left (b^{2} + b c\right )} \arctan \left (-\frac {a \sin \left (x\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (x\right )}\right ) - {\left (a^{2} - b^{2}\right )} \log \left (-b^{2} \cos \left (x\right )^{2} + 2 \, a b \sin \left (x\right ) + a^{2} + b^{2}\right )}{2 \, {\left (a^{2} b - b^{3}\right )}}\right ]
\]
[In]
integrate((b+c+cos(x))/(a+b*sin(x)),x, algorithm="fricas")
[Out]
[-1/2*(sqrt(-a^2 + b^2)*(b^2 + b*c)*log(((2*a^2 - b^2)*cos(x)^2 - 2*a*b*sin(x) - a^2 - b^2 + 2*(a*cos(x)*sin(x
) + b*cos(x))*sqrt(-a^2 + b^2))/(b^2*cos(x)^2 - 2*a*b*sin(x) - a^2 - b^2)) - (a^2 - b^2)*log(-b^2*cos(x)^2 + 2
*a*b*sin(x) + a^2 + b^2))/(a^2*b - b^3), -1/2*(2*sqrt(a^2 - b^2)*(b^2 + b*c)*arctan(-(a*sin(x) + b)/(sqrt(a^2
- b^2)*cos(x))) - (a^2 - b^2)*log(-b^2*cos(x)^2 + 2*a*b*sin(x) + a^2 + b^2))/(a^2*b - b^3)]
Sympy [B] (verification not implemented)
Leaf count of result is larger than twice the leaf count of optimal. 641 vs. \(2 (46) = 92\).
Time = 14.86 (sec) , antiderivative size = 641, normalized size of antiderivative = 11.65
\[
\int \frac {b+c+\cos (x)}{a+b \sin (x)} \, dx=\text {Too large to display}
\]
[In]
integrate((b+c+cos(x))/(a+b*sin(x)),x)
[Out]
Piecewise((zoo*(c*log(tan(x/2)) - log(tan(x/2)**2 + 1) + log(tan(x/2))), Eq(a, 0) & Eq(b, 0)), ((b*log(tan(x/2
)) + c*log(tan(x/2)) - log(tan(x/2)**2 + 1) + log(tan(x/2)))/b, Eq(a, 0)), (2*b/(b*tan(x/2) - b) + 2*c/(b*tan(
x/2) - b) + 2*log(tan(x/2) - 1)*tan(x/2)/(b*tan(x/2) - b) - 2*log(tan(x/2) - 1)/(b*tan(x/2) - b) - log(tan(x/2
)**2 + 1)*tan(x/2)/(b*tan(x/2) - b) + log(tan(x/2)**2 + 1)/(b*tan(x/2) - b), Eq(a, -b)), (-2*b/(b*tan(x/2) + b
) - 2*c/(b*tan(x/2) + b) + 2*log(tan(x/2) + 1)*tan(x/2)/(b*tan(x/2) + b) + 2*log(tan(x/2) + 1)/(b*tan(x/2) + b
) - log(tan(x/2)**2 + 1)*tan(x/2)/(b*tan(x/2) + b) - log(tan(x/2)**2 + 1)/(b*tan(x/2) + b), Eq(a, b)), ((c*x +
sin(x))/a, Eq(b, 0)), (-a**2*log(tan(x/2)**2 + 1)/(a**2*b - b**3) + a**2*log(tan(x/2) + b/a - sqrt(-a**2 + b*
*2)/a)/(a**2*b - b**3) + a**2*log(tan(x/2) + b/a + sqrt(-a**2 + b**2)/a)/(a**2*b - b**3) - b**2*sqrt(-a**2 + b
**2)*log(tan(x/2) + b/a - sqrt(-a**2 + b**2)/a)/(a**2*b - b**3) + b**2*sqrt(-a**2 + b**2)*log(tan(x/2) + b/a +
sqrt(-a**2 + b**2)/a)/(a**2*b - b**3) + b**2*log(tan(x/2)**2 + 1)/(a**2*b - b**3) - b**2*log(tan(x/2) + b/a -
sqrt(-a**2 + b**2)/a)/(a**2*b - b**3) - b**2*log(tan(x/2) + b/a + sqrt(-a**2 + b**2)/a)/(a**2*b - b**3) - b*c
*sqrt(-a**2 + b**2)*log(tan(x/2) + b/a - sqrt(-a**2 + b**2)/a)/(a**2*b - b**3) + b*c*sqrt(-a**2 + b**2)*log(ta
n(x/2) + b/a + sqrt(-a**2 + b**2)/a)/(a**2*b - b**3), True))
Maxima [F(-2)]
Exception generated. \[
\int \frac {b+c+\cos (x)}{a+b \sin (x)} \, dx=\text {Exception raised: ValueError}
\]
[In]
integrate((b+c+cos(x))/(a+b*sin(x)),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*b^2-4*a^2>0)', see `assume?`
for more de
Giac [A] (verification not implemented)
none
Time = 0.33 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.60
\[
\int \frac {b+c+\cos (x)}{a+b \sin (x)} \, dx=\frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, x\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} {\left (b + c\right )}}{\sqrt {a^{2} - b^{2}}} + \frac {\log \left (a \tan \left (\frac {1}{2} \, x\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, x\right ) + a\right )}{b} - \frac {\log \left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right )}{b}
\]
[In]
integrate((b+c+cos(x))/(a+b*sin(x)),x, algorithm="giac")
[Out]
2*(pi*floor(1/2*x/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*x) + b)/sqrt(a^2 - b^2)))*(b + c)/sqrt(a^2 - b^2) + log
(a*tan(1/2*x)^2 + 2*b*tan(1/2*x) + a)/b - log(tan(1/2*x)^2 + 1)/b
Mupad [B] (verification not implemented)
Time = 25.14 (sec) , antiderivative size = 1955, normalized size of antiderivative = 35.55
\[
\int \frac {b+c+\cos (x)}{a+b \sin (x)} \, dx=\text {Too large to display}
\]
[In]
int((b + c + cos(x))/(a + b*sin(x)),x)
[Out]
(2*atan((tan(x/2)*(((((b + c)^3*(96*a*b^4 - 64*a^3*b^2))/(a^2 - b^2)^(3/2) + ((2*a^2*b - 2*b^3)*(((b + c)*(32*
a*b^3 + 64*a*b^4 - ((2*a^2*b - 2*b^3)*(96*a*b^4 - 64*a^3*b^2))/(2*(b^4 - a^2*b^2)) + 64*a*b^3*c))/(a^2 - b^2)^
(1/2) - ((2*a^2*b - 2*b^3)*(b + c)*(96*a*b^4 - 64*a^3*b^2))/(2*(b^4 - a^2*b^2)*(a^2 - b^2)^(1/2))))/(2*(b^4 -
a^2*b^2)) + ((b + c)*(96*a*b^2 - 128*a*b^3 + 32*a*b^4 - 64*a^3 + 32*a*b^2*c^2 + ((2*a^2*b - 2*b^3)*(32*a*b^3 +
64*a*b^4 - ((2*a^2*b - 2*b^3)*(96*a*b^4 - 64*a^3*b^2))/(2*(b^4 - a^2*b^2)) + 64*a*b^3*c))/(2*(b^4 - a^2*b^2))
- 128*a*b^2*c + 64*a*b^3*c))/(a^2 - b^2)^(1/2))*(8*b^4*c + 4*b^5*c + 4*a^4 + 8*b^4 + 8*b^5 + 2*b^6 - 12*a^2*b
^2 - 8*a^2*b^3 - a^2*b^4 + 2*b^4*c^2 - 8*a^2*b^2*c - 2*a^2*b^3*c - a^2*b^2*c^2))/(a^3*(a^2 - b^2)^(1/2)*(2*b^3
*c + 4*a^2 - 4*b^2 + b^4 + b^2*c^2)^2) + (2*b*(b + c + 2)*(b^2*c - 2*a^2 + 2*b^2 + b^3)*(32*a*b - 64*a*b^2 + 3
2*a*b^3 - ((2*a^2*b - 2*b^3)*(96*a*b^2 - 128*a*b^3 + 32*a*b^4 - 64*a^3 + 32*a*b^2*c^2 + ((2*a^2*b - 2*b^3)*(32
*a*b^3 + 64*a*b^4 - ((2*a^2*b - 2*b^3)*(96*a*b^4 - 64*a^3*b^2))/(2*(b^4 - a^2*b^2)) + 64*a*b^3*c))/(2*(b^4 - a
^2*b^2)) - 128*a*b^2*c + 64*a*b^3*c))/(2*(b^4 - a^2*b^2)) - 64*a*b*c + ((b + c)*(((b + c)*(32*a*b^3 + 64*a*b^4
- ((2*a^2*b - 2*b^3)*(96*a*b^4 - 64*a^3*b^2))/(2*(b^4 - a^2*b^2)) + 64*a*b^3*c))/(a^2 - b^2)^(1/2) - ((2*a^2*
b - 2*b^3)*(b + c)*(96*a*b^4 - 64*a^3*b^2))/(2*(b^4 - a^2*b^2)*(a^2 - b^2)^(1/2))))/(a^2 - b^2)^(1/2) + 32*a*b
*c^2 + 64*a*b^2*c - ((2*a^2*b - 2*b^3)*(b + c)^2*(96*a*b^4 - 64*a^3*b^2))/(2*(b^4 - a^2*b^2)*(a^2 - b^2))))/(a
^3*(2*b^3*c + 4*a^2 - 4*b^2 + b^4 + b^2*c^2)^2))*(a^2 - b^2)^(3/2))/(32*a*b + 32*a*c) + ((a^2 - b^2)*(((b + c)
*(32*a^2*b - 64*a^2*b^2 + ((2*a^2*b - 2*b^3)*(32*a^2*b^2 + 32*a^2*b^3 + 32*a^2*b^2*c - (16*a^2*b^3*(2*a^2*b -
2*b^3))/(b^4 - a^2*b^2)))/(2*(b^4 - a^2*b^2)) - 64*a^2*b*c))/(a^2 - b^2)^(1/2) + ((2*a^2*b - 2*b^3)*(((b + c)*
(32*a^2*b^2 + 32*a^2*b^3 + 32*a^2*b^2*c - (16*a^2*b^3*(2*a^2*b - 2*b^3))/(b^4 - a^2*b^2)))/(a^2 - b^2)^(1/2) -
(16*a^2*b^3*(2*a^2*b - 2*b^3)*(b + c))/((b^4 - a^2*b^2)*(a^2 - b^2)^(1/2))))/(2*(b^4 - a^2*b^2)) + (32*a^2*b^
3*(b + c)^3)/(a^2 - b^2)^(3/2))*(8*b^4*c + 4*b^5*c + 4*a^4 + 8*b^4 + 8*b^5 + 2*b^6 - 12*a^2*b^2 - 8*a^2*b^3 -
a^2*b^4 + 2*b^4*c^2 - 8*a^2*b^2*c - 2*a^2*b^3*c - a^2*b^2*c^2))/(a^3*(32*a*b + 32*a*c)*(2*b^3*c + 4*a^2 - 4*b^
2 + b^4 + b^2*c^2)^2) - (2*b*(a^2 - b^2)^(3/2)*(b + c + 2)*(b^2*c - 2*a^2 + 2*b^2 + b^3)*(32*a^2*b + 32*a^2*c
- 32*a^2 - ((b + c)*(((b + c)*(32*a^2*b^2 + 32*a^2*b^3 + 32*a^2*b^2*c - (16*a^2*b^3*(2*a^2*b - 2*b^3))/(b^4 -
a^2*b^2)))/(a^2 - b^2)^(1/2) - (16*a^2*b^3*(2*a^2*b - 2*b^3)*(b + c))/((b^4 - a^2*b^2)*(a^2 - b^2)^(1/2))))/(a
^2 - b^2)^(1/2) + ((2*a^2*b - 2*b^3)*(32*a^2*b - 64*a^2*b^2 + ((2*a^2*b - 2*b^3)*(32*a^2*b^2 + 32*a^2*b^3 + 32
*a^2*b^2*c - (16*a^2*b^3*(2*a^2*b - 2*b^3))/(b^4 - a^2*b^2)))/(2*(b^4 - a^2*b^2)) - 64*a^2*b*c))/(2*(b^4 - a^2
*b^2)) + (16*a^2*b^3*(2*a^2*b - 2*b^3)*(b + c)^2)/((b^4 - a^2*b^2)*(a^2 - b^2))))/(a^3*(32*a*b + 32*a*c)*(2*b^
3*c + 4*a^2 - 4*b^2 + b^4 + b^2*c^2)^2))*(b + c))/(a^2 - b^2)^(1/2) - log(tan(x/2)^2 + 1)/b - (log((a + b*sin(
x))/(cos(x) + 1))*(2*a^2*b - 2*b^3))/(2*(b^4 - a^2*b^2))