3.1.4 \(\int \frac {b+c+\cos (x)}{a+b \sin (x)} \, dx\) [4]

Optimal. Leaf size=55 \[ \frac {2 (b+c) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+\frac {\log (a+b \sin (x))}{b} \]

[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)

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Rubi [A]
time = 0.08, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {4486, 2739, 632, 210, 2747, 31} \begin {gather*} \frac {2 (b+c) \text {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} \end {gather*}

Antiderivative was successfully verified.

[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*} \int \frac {b+c+\cos (x)}{a+b \sin (x)} \, dx &=\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) \tan ^{-1}\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*}

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Mathematica [A]
time = 0.07, size = 55, normalized size = 1.00 \begin {gather*} \frac {2 (b+c) \tan ^{-1}\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 {gather*}

Antiderivative was successfully verified.

[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

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Maple [A]
time = 0.33, size = 86, normalized size = 1.56

method	result	size
default	\(\frac {\ln \left (a \left (\tan ^{2}\left (\frac {x}{2}\right )\right )+2 b \tan \left (\frac {x}{2}\right )+a \right )+\frac {2 \left (b^{2}+b c \right ) \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}}}{b}-\frac {\ln \left (1+\tan ^{2}\left (\frac {x}{2}\right )\right )}{b}\)	\(86\)
risch	\(\frac {i x}{b}-\frac {2 i x \,a^{2} b}{a^{2} b^{2}-b^{4}}+\frac {2 i x \,b^{3}}{a^{2} b^{2}-b^{4}}+\frac {\ln \left ({\mathrm e}^{i x}+\frac {i a \,b^{2}+i c a b +\sqrt {-a^{2} b^{4}-2 a^{2} b^{3} c -a^{2} b^{2} c^{2}+b^{6}+2 b^{5} c +b^{4} c^{2}}}{b^{2} \left (b +c \right )}\right ) a^{2}}{\left (a^{2}-b^{2}\right ) b}-\frac {b \ln \left ({\mathrm e}^{i x}+\frac {i a \,b^{2}+i c a b +\sqrt {-a^{2} b^{4}-2 a^{2} b^{3} c -a^{2} b^{2} c^{2}+b^{6}+2 b^{5} c +b^{4} c^{2}}}{b^{2} \left (b +c \right )}\right )}{a^{2}-b^{2}}+\frac {\ln \left ({\mathrm e}^{i x}+\frac {i a \,b^{2}+i c a b +\sqrt {-a^{2} b^{4}-2 a^{2} b^{3} c -a^{2} b^{2} c^{2}+b^{6}+2 b^{5} c +b^{4} c^{2}}}{b^{2} \left (b +c \right )}\right ) \sqrt {-a^{2} b^{4}-2 a^{2} b^{3} c -a^{2} b^{2} c^{2}+b^{6}+2 b^{5} c +b^{4} c^{2}}}{\left (a^{2}-b^{2}\right ) b}+\frac {\ln \left ({\mathrm e}^{i x}-\frac {-i a \,b^{2}-i c a b +\sqrt {-a^{2} b^{4}-2 a^{2} b^{3} c -a^{2} b^{2} c^{2}+b^{6}+2 b^{5} c +b^{4} c^{2}}}{b^{2} \left (b +c \right )}\right ) a^{2}}{\left (a^{2}-b^{2}\right ) b}-\frac {b \ln \left ({\mathrm e}^{i x}-\frac {-i a \,b^{2}-i c a b +\sqrt {-a^{2} b^{4}-2 a^{2} b^{3} c -a^{2} b^{2} c^{2}+b^{6}+2 b^{5} c +b^{4} c^{2}}}{b^{2} \left (b +c \right )}\right )}{a^{2}-b^{2}}-\frac {\ln \left ({\mathrm e}^{i x}-\frac {-i a \,b^{2}-i c a b +\sqrt {-a^{2} b^{4}-2 a^{2} b^{3} c -a^{2} b^{2} c^{2}+b^{6}+2 b^{5} c +b^{4} c^{2}}}{b^{2} \left (b +c \right )}\right ) \sqrt {-a^{2} b^{4}-2 a^{2} b^{3} c -a^{2} b^{2} c^{2}+b^{6}+2 b^{5} c +b^{4} c^{2}}}{\left (a^{2}-b^{2}\right ) b}\)	\(708\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b+c+cos(x))/(a+b*sin(x)),x,method=_RETURNVERBOSE)

[Out]

2/b*(1/2*ln(a*tan(1/2*x)^2+2*b*tan(1/2*x)+a)+(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)))-1/b*ln(1+tan(1/2*x)^2)

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]
time = 0.44, size = 250, normalized size = 4.55 \begin {gather*} \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 ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)]

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 641 vs. \(2 (46) = 92\).
time = 12.00, size = 641, normalized size = 11.65 \begin {gather*} \begin {cases} \tilde {\infty } \left (c \log {\left (\tan {\left (\frac {x}{2} \right )} \right )} - \log {\left (\tan ^{2}{\left (\frac {x}{2} \right )} + 1 \right )} + \log {\left (\tan {\left (\frac {x}{2} \right )} \right )}\right ) & \text {for}\: a = 0 \wedge b = 0 \\\frac {b \log {\left (\tan {\left (\frac {x}{2} \right )} \right )} + c \log {\left (\tan {\left (\frac {x}{2} \right )} \right )} - \log {\left (\tan ^{2}{\left (\frac {x}{2} \right )} + 1 \right )} + \log {\left (\tan {\left (\frac {x}{2} \right )} \right )}}{b} & \text {for}\: a = 0 \\\frac {2 b}{b \tan {\left (\frac {x}{2} \right )} - b} + \frac {2 c}{b \tan {\left (\frac {x}{2} \right )} - b} + \frac {2 \log {\left (\tan {\left (\frac {x}{2} \right )} - 1 \right )} \tan {\left (\frac {x}{2} \right )}}{b \tan {\left (\frac {x}{2} \right )} - b} - \frac {2 \log {\left (\tan {\left (\frac {x}{2} \right )} - 1 \right )}}{b \tan {\left (\frac {x}{2} \right )} - b} - \frac {\log {\left (\tan ^{2}{\left (\frac {x}{2} \right )} + 1 \right )} \tan {\left (\frac {x}{2} \right )}}{b \tan {\left (\frac {x}{2} \right )} - b} + \frac {\log {\left (\tan ^{2}{\left (\frac {x}{2} \right )} + 1 \right )}}{b \tan {\left (\frac {x}{2} \right )} - b} & \text {for}\: a = - b \\- \frac {2 b}{b \tan {\left (\frac {x}{2} \right )} + b} - \frac {2 c}{b \tan {\left (\frac {x}{2} \right )} + b} + \frac {2 \log {\left (\tan {\left (\frac {x}{2} \right )} + 1 \right )} \tan {\left (\frac {x}{2} \right )}}{b \tan {\left (\frac {x}{2} \right )} + b} + \frac {2 \log {\left (\tan {\left (\frac {x}{2} \right )} + 1 \right )}}{b \tan {\left (\frac {x}{2} \right )} + b} - \frac {\log {\left (\tan ^{2}{\left (\frac {x}{2} \right )} + 1 \right )} \tan {\left (\frac {x}{2} \right )}}{b \tan {\left (\frac {x}{2} \right )} + b} - \frac {\log {\left (\tan ^{2}{\left (\frac {x}{2} \right )} + 1 \right )}}{b \tan {\left (\frac {x}{2} \right )} + b} & \text {for}\: a = b \\\frac {c x + \sin {\left (x \right )}}{a} & \text {for}\: b = 0 \\- \frac {a^{2} \log {\left (\tan ^{2}{\left (\frac {x}{2} \right )} + 1 \right )}}{a^{2} b - b^{3}} + \frac {a^{2} \log {\left (\tan {\left (\frac {x}{2} \right )} + \frac {b}{a} - \frac {\sqrt {- a^{2} + b^{2}}}{a} \right )}}{a^{2} b - b^{3}} + \frac {a^{2} \log {\left (\tan {\left (\frac {x}{2} \right )} + \frac {b}{a} + \frac {\sqrt {- a^{2} + b^{2}}}{a} \right )}}{a^{2} b - b^{3}} - \frac {b^{2} \sqrt {- a^{2} + b^{2}} \log {\left (\tan {\left (\frac {x}{2} \right )} + \frac {b}{a} - \frac {\sqrt {- a^{2} + b^{2}}}{a} \right )}}{a^{2} b - b^{3}} + \frac {b^{2} \sqrt {- a^{2} + b^{2}} \log {\left (\tan {\left (\frac {x}{2} \right )} + \frac {b}{a} + \frac {\sqrt {- a^{2} + b^{2}}}{a} \right )}}{a^{2} b - b^{3}} + \frac {b^{2} \log {\left (\tan ^{2}{\left (\frac {x}{2} \right )} + 1 \right )}}{a^{2} b - b^{3}} - \frac {b^{2} \log {\left (\tan {\left (\frac {x}{2} \right )} + \frac {b}{a} - \frac {\sqrt {- a^{2} + b^{2}}}{a} \right )}}{a^{2} b - b^{3}} - \frac {b^{2} \log {\left (\tan {\left (\frac {x}{2} \right )} + \frac {b}{a} + \frac {\sqrt {- a^{2} + b^{2}}}{a} \right )}}{a^{2} b - b^{3}} - \frac {b c \sqrt {- a^{2} + b^{2}} \log {\left (\tan {\left (\frac {x}{2} \right )} + \frac {b}{a} - \frac {\sqrt {- a^{2} + b^{2}}}{a} \right )}}{a^{2} b - b^{3}} + \frac {b c \sqrt {- a^{2} + b^{2}} \log {\left (\tan {\left (\frac {x}{2} \right )} + \frac {b}{a} + \frac {\sqrt {- a^{2} + b^{2}}}{a} \right )}}{a^{2} b - b^{3}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Giac [A]
time = 0.42, size = 88, normalized size = 1.60 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Mupad [B]
time = 24.22, size = 1955, normalized size = 35.55 \begin {gather*} \frac {2\,\mathrm {atan}\left (\frac {\mathrm {tan}\left (\frac {x}{2}\right )\,\left (\frac {\left (\frac {{\left (b+c\right )}^3\,\left (96\,a\,b^4-64\,a^3\,b^2\right )}{{\left (a^2-b^2\right )}^{3/2}}+\frac {\left (2\,a^2\,b-2\,b^3\right )\,\left (\frac {\left (b+c\right )\,\left (32\,a\,b^3+64\,a\,b^4-\frac {\left (2\,a^2\,b-2\,b^3\right )\,\left (96\,a\,b^4-64\,a^3\,b^2\right )}{2\,\left (b^4-a^2\,b^2\right )}+64\,a\,b^3\,c\right )}{\sqrt {a^2-b^2}}-\frac {\left (2\,a^2\,b-2\,b^3\right )\,\left (b+c\right )\,\left (96\,a\,b^4-64\,a^3\,b^2\right )}{2\,\left (b^4-a^2\,b^2\right )\,\sqrt {a^2-b^2}}\right )}{2\,\left (b^4-a^2\,b^2\right )}+\frac {\left (b+c\right )\,\left (96\,a\,b^2-128\,a\,b^3+32\,a\,b^4-64\,a^3+32\,a\,b^2\,c^2+\frac {\left (2\,a^2\,b-2\,b^3\right )\,\left (32\,a\,b^3+64\,a\,b^4-\frac {\left (2\,a^2\,b-2\,b^3\right )\,\left (96\,a\,b^4-64\,a^3\,b^2\right )}{2\,\left (b^4-a^2\,b^2\right )}+64\,a\,b^3\,c\right )}{2\,\left (b^4-a^2\,b^2\right )}-128\,a\,b^2\,c+64\,a\,b^3\,c\right )}{\sqrt {a^2-b^2}}\right )\,\left (4\,a^4-a^2\,b^4-2\,a^2\,b^3\,c-8\,a^2\,b^3-a^2\,b^2\,c^2-8\,a^2\,b^2\,c-12\,a^2\,b^2+2\,b^6+4\,b^5\,c+8\,b^5+2\,b^4\,c^2+8\,b^4\,c+8\,b^4\right )}{a^3\,\sqrt {a^2-b^2}\,{\left (4\,a^2+b^4+2\,b^3\,c+b^2\,c^2-4\,b^2\right )}^2}+\frac {2\,b\,\left (b+c+2\right )\,\left (b^2\,c-2\,a^2+2\,b^2+b^3\right )\,\left (32\,a\,b-64\,a\,b^2+32\,a\,b^3-\frac {\left (2\,a^2\,b-2\,b^3\right )\,\left (96\,a\,b^2-128\,a\,b^3+32\,a\,b^4-64\,a^3+32\,a\,b^2\,c^2+\frac {\left (2\,a^2\,b-2\,b^3\right )\,\left (32\,a\,b^3+64\,a\,b^4-\frac {\left (2\,a^2\,b-2\,b^3\right )\,\left (96\,a\,b^4-64\,a^3\,b^2\right )}{2\,\left (b^4-a^2\,b^2\right )}+64\,a\,b^3\,c\right )}{2\,\left (b^4-a^2\,b^2\right )}-128\,a\,b^2\,c+64\,a\,b^3\,c\right )}{2\,\left (b^4-a^2\,b^2\right )}-64\,a\,b\,c+\frac {\left (b+c\right )\,\left (\frac {\left (b+c\right )\,\left (32\,a\,b^3+64\,a\,b^4-\frac {\left (2\,a^2\,b-2\,b^3\right )\,\left (96\,a\,b^4-64\,a^3\,b^2\right )}{2\,\left (b^4-a^2\,b^2\right )}+64\,a\,b^3\,c\right )}{\sqrt {a^2-b^2}}-\frac {\left (2\,a^2\,b-2\,b^3\right )\,\left (b+c\right )\,\left (96\,a\,b^4-64\,a^3\,b^2\right )}{2\,\left (b^4-a^2\,b^2\right )\,\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+32\,a\,b\,c^2+64\,a\,b^2\,c-\frac {\left (2\,a^2\,b-2\,b^3\right )\,{\left (b+c\right )}^2\,\left (96\,a\,b^4-64\,a^3\,b^2\right )}{2\,\left (b^4-a^2\,b^2\right )\,\left (a^2-b^2\right )}\right )}{a^3\,{\left (4\,a^2+b^4+2\,b^3\,c+b^2\,c^2-4\,b^2\right )}^2}\right )\,{\left (a^2-b^2\right )}^{3/2}}{32\,a\,b+32\,a\,c}+\frac {\left (a^2-b^2\right )\,\left (\frac {\left (b+c\right )\,\left (32\,a^2\,b-64\,a^2\,b^2+\frac {\left (2\,a^2\,b-2\,b^3\right )\,\left (32\,a^2\,b^2+32\,a^2\,b^3+32\,a^2\,b^2\,c-\frac {16\,a^2\,b^3\,\left (2\,a^2\,b-2\,b^3\right )}{b^4-a^2\,b^2}\right )}{2\,\left (b^4-a^2\,b^2\right )}-64\,a^2\,b\,c\right )}{\sqrt {a^2-b^2}}+\frac {\left (2\,a^2\,b-2\,b^3\right )\,\left (\frac {\left (b+c\right )\,\left (32\,a^2\,b^2+32\,a^2\,b^3+32\,a^2\,b^2\,c-\frac {16\,a^2\,b^3\,\left (2\,a^2\,b-2\,b^3\right )}{b^4-a^2\,b^2}\right )}{\sqrt {a^2-b^2}}-\frac {16\,a^2\,b^3\,\left (2\,a^2\,b-2\,b^3\right )\,\left (b+c\right )}{\left (b^4-a^2\,b^2\right )\,\sqrt {a^2-b^2}}\right )}{2\,\left (b^4-a^2\,b^2\right )}+\frac {32\,a^2\,b^3\,{\left (b+c\right )}^3}{{\left (a^2-b^2\right )}^{3/2}}\right )\,\left (4\,a^4-a^2\,b^4-2\,a^2\,b^3\,c-8\,a^2\,b^3-a^2\,b^2\,c^2-8\,a^2\,b^2\,c-12\,a^2\,b^2+2\,b^6+4\,b^5\,c+8\,b^5+2\,b^4\,c^2+8\,b^4\,c+8\,b^4\right )}{a^3\,\left (32\,a\,b+32\,a\,c\right )\,{\left (4\,a^2+b^4+2\,b^3\,c+b^2\,c^2-4\,b^2\right )}^2}-\frac {2\,b\,{\left (a^2-b^2\right )}^{3/2}\,\left (b+c+2\right )\,\left (b^2\,c-2\,a^2+2\,b^2+b^3\right )\,\left (32\,a^2\,b+32\,a^2\,c-32\,a^2-\frac {\left (b+c\right )\,\left (\frac {\left (b+c\right )\,\left (32\,a^2\,b^2+32\,a^2\,b^3+32\,a^2\,b^2\,c-\frac {16\,a^2\,b^3\,\left (2\,a^2\,b-2\,b^3\right )}{b^4-a^2\,b^2}\right )}{\sqrt {a^2-b^2}}-\frac {16\,a^2\,b^3\,\left (2\,a^2\,b-2\,b^3\right )\,\left (b+c\right )}{\left (b^4-a^2\,b^2\right )\,\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+\frac {\left (2\,a^2\,b-2\,b^3\right )\,\left (32\,a^2\,b-64\,a^2\,b^2+\frac {\left (2\,a^2\,b-2\,b^3\right )\,\left (32\,a^2\,b^2+32\,a^2\,b^3+32\,a^2\,b^2\,c-\frac {16\,a^2\,b^3\,\left (2\,a^2\,b-2\,b^3\right )}{b^4-a^2\,b^2}\right )}{2\,\left (b^4-a^2\,b^2\right )}-64\,a^2\,b\,c\right )}{2\,\left (b^4-a^2\,b^2\right )}+\frac {16\,a^2\,b^3\,\left (2\,a^2\,b-2\,b^3\right )\,{\left (b+c\right )}^2}{\left (b^4-a^2\,b^2\right )\,\left (a^2-b^2\right )}\right )}{a^3\,\left (32\,a\,b+32\,a\,c\right )\,{\left (4\,a^2+b^4+2\,b^3\,c+b^2\,c^2-4\,b^2\right )}^2}\right )\,\left (b+c\right )}{\sqrt {a^2-b^2}}-\frac {\ln \left ({\mathrm {tan}\left (\frac {x}{2}\right )}^2+1\right )}{b}-\frac {\ln \left (\frac {a+b\,\sin \left (x\right )}{\cos \left (x\right )+1}\right )\,\left (2\,a^2\,b-2\,b^3\right )}{2\,\left (b^4-a^2\,b^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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