3.187 \(\int \frac {x}{a+b \cos (c+d x)} \, dx\)

Optimal. Leaf size=214 \[ -\frac {\text {Li}_2\left (-\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d^2 \sqrt {a^2-b^2}}+\frac {\text {Li}_2\left (-\frac {b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d^2 \sqrt {a^2-b^2}}-\frac {i x \log \left (1+\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d \sqrt {a^2-b^2}}+\frac {i x \log \left (1+\frac {b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{d \sqrt {a^2-b^2}} \]

[Out]

-I*x*ln(1+b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/d/(a^2-b^2)^(1/2)+I*x*ln(1+b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)
))/d/(a^2-b^2)^(1/2)-polylog(2,-b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/d^2/(a^2-b^2)^(1/2)+polylog(2,-b*exp(I*(
d*x+c))/(a+(a^2-b^2)^(1/2)))/d^2/(a^2-b^2)^(1/2)

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Rubi [A]  time = 0.40, antiderivative size = 214, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {3321, 2264, 2190, 2279, 2391} \[ -\frac {\text {Li}_2\left (-\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d^2 \sqrt {a^2-b^2}}+\frac {\text {Li}_2\left (-\frac {b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d^2 \sqrt {a^2-b^2}}-\frac {i x \log \left (1+\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d \sqrt {a^2-b^2}}+\frac {i x \log \left (1+\frac {b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{d \sqrt {a^2-b^2}} \]

Antiderivative was successfully verified.

[In]

Int[x/(a + b*Cos[c + d*x]),x]

[Out]

((-I)*x*Log[1 + (b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(Sqrt[a^2 - b^2]*d) + (I*x*Log[1 + (b*E^(I*(c + d*
x)))/(a + Sqrt[a^2 - b^2])])/(Sqrt[a^2 - b^2]*d) - PolyLog[2, -((b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2]))]/(S
qrt[a^2 - b^2]*d^2) + PolyLog[2, -((b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2]))]/(Sqrt[a^2 - b^2]*d^2)

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2264

Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.)*(F_)^(v_)), x_Symbol] :> With[{q =
Rt[b^2 - 4*a*c, 2]}, Dist[(2*c)/q, Int[((f + g*x)^m*F^u)/(b - q + 2*c*F^u), x], x] - Dist[(2*c)/q, Int[((f +
g*x)^m*F^u)/(b + q + 2*c*F^u), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[v, 2*u] && LinearQ[u, x] && NeQ[
b^2 - 4*a*c, 0] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 3321

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + Pi*(k_.) + (f_.)*(x_)]), x_Symbol] :> Dist[2, Int[((c
+ d*x)^m*E^(I*Pi*(k - 1/2))*E^(I*(e + f*x)))/(b + 2*a*E^(I*Pi*(k - 1/2))*E^(I*(e + f*x)) - b*E^(2*I*k*Pi)*E^(
2*I*(e + f*x))), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[2*k] && NeQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {x}{a+b \cos (c+d x)} \, dx &=2 \int \frac {e^{i (c+d x)} x}{b+2 a e^{i (c+d x)}+b e^{2 i (c+d x)}} \, dx\\ &=\frac {(2 b) \int \frac {e^{i (c+d x)} x}{2 a-2 \sqrt {a^2-b^2}+2 b e^{i (c+d x)}} \, dx}{\sqrt {a^2-b^2}}-\frac {(2 b) \int \frac {e^{i (c+d x)} x}{2 a+2 \sqrt {a^2-b^2}+2 b e^{i (c+d x)}} \, dx}{\sqrt {a^2-b^2}}\\ &=-\frac {i x \log \left (1+\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} d}+\frac {i x \log \left (1+\frac {b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} d}+\frac {i \int \log \left (1+\frac {2 b e^{i (c+d x)}}{2 a-2 \sqrt {a^2-b^2}}\right ) \, dx}{\sqrt {a^2-b^2} d}-\frac {i \int \log \left (1+\frac {2 b e^{i (c+d x)}}{2 a+2 \sqrt {a^2-b^2}}\right ) \, dx}{\sqrt {a^2-b^2} d}\\ &=-\frac {i x \log \left (1+\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} d}+\frac {i x \log \left (1+\frac {b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} d}+\frac {\operatorname {Subst}\left (\int \frac {\log \left (1+\frac {2 b x}{2 a-2 \sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{\sqrt {a^2-b^2} d^2}-\frac {\operatorname {Subst}\left (\int \frac {\log \left (1+\frac {2 b x}{2 a+2 \sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{\sqrt {a^2-b^2} d^2}\\ &=-\frac {i x \log \left (1+\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} d}+\frac {i x \log \left (1+\frac {b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} d}-\frac {\text {Li}_2\left (-\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} d^2}+\frac {\text {Li}_2\left (-\frac {b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} d^2}\\ \end {align*}

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Mathematica [B]  time = 0.83, size = 756, normalized size = 3.53 \[ \frac {i \left (\text {Li}_2\left (\frac {\left (a-i \sqrt {b^2-a^2}\right ) \left (a+b-\sqrt {b^2-a^2} \tan \left (\frac {1}{2} (c+d x)\right )\right )}{b \left (a+b+\sqrt {b^2-a^2} \tan \left (\frac {1}{2} (c+d x)\right )\right )}\right )-\text {Li}_2\left (\frac {\left (a+i \sqrt {b^2-a^2}\right ) \left (a+b-\sqrt {b^2-a^2} \tan \left (\frac {1}{2} (c+d x)\right )\right )}{b \left (a+b+\sqrt {b^2-a^2} \tan \left (\frac {1}{2} (c+d x)\right )\right )}\right )\right )+2 (c+d x) \tanh ^{-1}\left (\frac {(a+b) \cot \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^2-a^2}}\right )-2 \left (\cos ^{-1}\left (-\frac {a}{b}\right )+c\right ) \tanh ^{-1}\left (\frac {(b-a) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^2-a^2}}\right )-\log \left (\frac {(a+b) \left (-i \sqrt {b^2-a^2}-a+b\right ) \left (1+i \tan \left (\frac {1}{2} (c+d x)\right )\right )}{b \left (\sqrt {b^2-a^2} \tan \left (\frac {1}{2} (c+d x)\right )+a+b\right )}\right ) \left (\cos ^{-1}\left (-\frac {a}{b}\right )-2 i \tanh ^{-1}\left (\frac {(b-a) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^2-a^2}}\right )\right )-\log \left (\frac {(a+b) \left (\sqrt {b^2-a^2}+i a-i b\right ) \left (\tan \left (\frac {1}{2} (c+d x)\right )+i\right )}{b \left (\sqrt {b^2-a^2} \tan \left (\frac {1}{2} (c+d x)\right )+a+b\right )}\right ) \left (\cos ^{-1}\left (-\frac {a}{b}\right )+2 i \tanh ^{-1}\left (\frac {(b-a) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^2-a^2}}\right )\right )+\log \left (\frac {\sqrt {b^2-a^2} e^{-\frac {1}{2} i (c+d x)}}{\sqrt {2} \sqrt {b} \sqrt {a+b \cos (c+d x)}}\right ) \left (2 i \tanh ^{-1}\left (\frac {(b-a) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^2-a^2}}\right )-2 i \tanh ^{-1}\left (\frac {(a+b) \cot \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^2-a^2}}\right )+\cos ^{-1}\left (-\frac {a}{b}\right )\right )+\log \left (\frac {\sqrt {b^2-a^2} e^{\frac {1}{2} i (c+d x)}}{\sqrt {2} \sqrt {b} \sqrt {a+b \cos (c+d x)}}\right ) \left (\cos ^{-1}\left (-\frac {a}{b}\right )+2 i \left (\tanh ^{-1}\left (\frac {(a+b) \cot \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^2-a^2}}\right )-\tanh ^{-1}\left (\frac {(b-a) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^2-a^2}}\right )\right )\right )}{d^2 \sqrt {b^2-a^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x/(a + b*Cos[c + d*x]),x]

[Out]

(2*(c + d*x)*ArcTanh[((a + b)*Cot[(c + d*x)/2])/Sqrt[-a^2 + b^2]] - 2*(c + ArcCos[-(a/b)])*ArcTanh[((-a + b)*T
an[(c + d*x)/2])/Sqrt[-a^2 + b^2]] + (ArcCos[-(a/b)] - (2*I)*ArcTanh[((a + b)*Cot[(c + d*x)/2])/Sqrt[-a^2 + b^
2]] + (2*I)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])*Log[Sqrt[-a^2 + b^2]/(Sqrt[2]*Sqrt[b]*E^((I
/2)*(c + d*x))*Sqrt[a + b*Cos[c + d*x]])] + (ArcCos[-(a/b)] + (2*I)*(ArcTanh[((a + b)*Cot[(c + d*x)/2])/Sqrt[-
a^2 + b^2]] - ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]]))*Log[(Sqrt[-a^2 + b^2]*E^((I/2)*(c + d*x)
))/(Sqrt[2]*Sqrt[b]*Sqrt[a + b*Cos[c + d*x]])] - (ArcCos[-(a/b)] - (2*I)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/S
qrt[-a^2 + b^2]])*Log[((a + b)*(-a + b - I*Sqrt[-a^2 + b^2])*(1 + I*Tan[(c + d*x)/2]))/(b*(a + b + Sqrt[-a^2 +
b^2]*Tan[(c + d*x)/2]))] - (ArcCos[-(a/b)] + (2*I)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])*Log
[((a + b)*(I*a - I*b + Sqrt[-a^2 + b^2])*(I + Tan[(c + d*x)/2]))/(b*(a + b + Sqrt[-a^2 + b^2]*Tan[(c + d*x)/2]
))] + I*(PolyLog[2, ((a - I*Sqrt[-a^2 + b^2])*(a + b - Sqrt[-a^2 + b^2]*Tan[(c + d*x)/2]))/(b*(a + b + Sqrt[-a
^2 + b^2]*Tan[(c + d*x)/2]))] - PolyLog[2, ((a + I*Sqrt[-a^2 + b^2])*(a + b - Sqrt[-a^2 + b^2]*Tan[(c + d*x)/2
]))/(b*(a + b + Sqrt[-a^2 + b^2]*Tan[(c + d*x)/2]))]))/(Sqrt[-a^2 + b^2]*d^2)

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fricas [B]  time = 0.96, size = 917, normalized size = 4.29 \[ -\frac {2 i \, b c \sqrt {\frac {a^{2} - b^{2}}{b^{2}}} \log \left (2 \, b \cos \left (d x + c\right ) + 2 i \, b \sin \left (d x + c\right ) + 2 \, b \sqrt {\frac {a^{2} - b^{2}}{b^{2}}} + 2 \, a\right ) - 2 i \, b c \sqrt {\frac {a^{2} - b^{2}}{b^{2}}} \log \left (2 \, b \cos \left (d x + c\right ) - 2 i \, b \sin \left (d x + c\right ) + 2 \, b \sqrt {\frac {a^{2} - b^{2}}{b^{2}}} + 2 \, a\right ) + 2 i \, b c \sqrt {\frac {a^{2} - b^{2}}{b^{2}}} \log \left (-2 \, b \cos \left (d x + c\right ) + 2 i \, b \sin \left (d x + c\right ) + 2 \, b \sqrt {\frac {a^{2} - b^{2}}{b^{2}}} - 2 \, a\right ) - 2 i \, b c \sqrt {\frac {a^{2} - b^{2}}{b^{2}}} \log \left (-2 \, b \cos \left (d x + c\right ) - 2 i \, b \sin \left (d x + c\right ) + 2 \, b \sqrt {\frac {a^{2} - b^{2}}{b^{2}}} - 2 \, a\right ) + 2 \, b \sqrt {\frac {a^{2} - b^{2}}{b^{2}}} {\rm Li}_2\left (-\frac {a \cos \left (d x + c\right ) + i \, a \sin \left (d x + c\right ) + {\left (b \cos \left (d x + c\right ) + i \, b \sin \left (d x + c\right )\right )} \sqrt {\frac {a^{2} - b^{2}}{b^{2}}} + b}{b} + 1\right ) - 2 \, b \sqrt {\frac {a^{2} - b^{2}}{b^{2}}} {\rm Li}_2\left (-\frac {a \cos \left (d x + c\right ) + i \, a \sin \left (d x + c\right ) - {\left (b \cos \left (d x + c\right ) + i \, b \sin \left (d x + c\right )\right )} \sqrt {\frac {a^{2} - b^{2}}{b^{2}}} + b}{b} + 1\right ) + 2 \, b \sqrt {\frac {a^{2} - b^{2}}{b^{2}}} {\rm Li}_2\left (-\frac {a \cos \left (d x + c\right ) - i \, a \sin \left (d x + c\right ) + {\left (b \cos \left (d x + c\right ) - i \, b \sin \left (d x + c\right )\right )} \sqrt {\frac {a^{2} - b^{2}}{b^{2}}} + b}{b} + 1\right ) - 2 \, b \sqrt {\frac {a^{2} - b^{2}}{b^{2}}} {\rm Li}_2\left (-\frac {a \cos \left (d x + c\right ) - i \, a \sin \left (d x + c\right ) - {\left (b \cos \left (d x + c\right ) - i \, b \sin \left (d x + c\right )\right )} \sqrt {\frac {a^{2} - b^{2}}{b^{2}}} + b}{b} + 1\right ) + 2 \, {\left (i \, b d x + i \, b c\right )} \sqrt {\frac {a^{2} - b^{2}}{b^{2}}} \log \left (\frac {a \cos \left (d x + c\right ) + i \, a \sin \left (d x + c\right ) + {\left (b \cos \left (d x + c\right ) + i \, b \sin \left (d x + c\right )\right )} \sqrt {\frac {a^{2} - b^{2}}{b^{2}}} + b}{b}\right ) + 2 \, {\left (-i \, b d x - i \, b c\right )} \sqrt {\frac {a^{2} - b^{2}}{b^{2}}} \log \left (\frac {a \cos \left (d x + c\right ) + i \, a \sin \left (d x + c\right ) - {\left (b \cos \left (d x + c\right ) + i \, b \sin \left (d x + c\right )\right )} \sqrt {\frac {a^{2} - b^{2}}{b^{2}}} + b}{b}\right ) + 2 \, {\left (-i \, b d x - i \, b c\right )} \sqrt {\frac {a^{2} - b^{2}}{b^{2}}} \log \left (\frac {a \cos \left (d x + c\right ) - i \, a \sin \left (d x + c\right ) + {\left (b \cos \left (d x + c\right ) - i \, b \sin \left (d x + c\right )\right )} \sqrt {\frac {a^{2} - b^{2}}{b^{2}}} + b}{b}\right ) + 2 \, {\left (i \, b d x + i \, b c\right )} \sqrt {\frac {a^{2} - b^{2}}{b^{2}}} \log \left (\frac {a \cos \left (d x + c\right ) - i \, a \sin \left (d x + c\right ) - {\left (b \cos \left (d x + c\right ) - i \, b \sin \left (d x + c\right )\right )} \sqrt {\frac {a^{2} - b^{2}}{b^{2}}} + b}{b}\right )}{4 \, {\left (a^{2} - b^{2}\right )} d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a+b*cos(d*x+c)),x, algorithm="fricas")

[Out]

-1/4*(2*I*b*c*sqrt((a^2 - b^2)/b^2)*log(2*b*cos(d*x + c) + 2*I*b*sin(d*x + c) + 2*b*sqrt((a^2 - b^2)/b^2) + 2*
a) - 2*I*b*c*sqrt((a^2 - b^2)/b^2)*log(2*b*cos(d*x + c) - 2*I*b*sin(d*x + c) + 2*b*sqrt((a^2 - b^2)/b^2) + 2*a
) + 2*I*b*c*sqrt((a^2 - b^2)/b^2)*log(-2*b*cos(d*x + c) + 2*I*b*sin(d*x + c) + 2*b*sqrt((a^2 - b^2)/b^2) - 2*a
) - 2*I*b*c*sqrt((a^2 - b^2)/b^2)*log(-2*b*cos(d*x + c) - 2*I*b*sin(d*x + c) + 2*b*sqrt((a^2 - b^2)/b^2) - 2*a
) + 2*b*sqrt((a^2 - b^2)/b^2)*dilog(-(a*cos(d*x + c) + I*a*sin(d*x + c) + (b*cos(d*x + c) + I*b*sin(d*x + c))*
sqrt((a^2 - b^2)/b^2) + b)/b + 1) - 2*b*sqrt((a^2 - b^2)/b^2)*dilog(-(a*cos(d*x + c) + I*a*sin(d*x + c) - (b*c
os(d*x + c) + I*b*sin(d*x + c))*sqrt((a^2 - b^2)/b^2) + b)/b + 1) + 2*b*sqrt((a^2 - b^2)/b^2)*dilog(-(a*cos(d*
x + c) - I*a*sin(d*x + c) + (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt((a^2 - b^2)/b^2) + b)/b + 1) - 2*b*sqrt((
a^2 - b^2)/b^2)*dilog(-(a*cos(d*x + c) - I*a*sin(d*x + c) - (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt((a^2 - b^
2)/b^2) + b)/b + 1) + 2*(I*b*d*x + I*b*c)*sqrt((a^2 - b^2)/b^2)*log((a*cos(d*x + c) + I*a*sin(d*x + c) + (b*co
s(d*x + c) + I*b*sin(d*x + c))*sqrt((a^2 - b^2)/b^2) + b)/b) + 2*(-I*b*d*x - I*b*c)*sqrt((a^2 - b^2)/b^2)*log(
(a*cos(d*x + c) + I*a*sin(d*x + c) - (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt((a^2 - b^2)/b^2) + b)/b) + 2*(-I
*b*d*x - I*b*c)*sqrt((a^2 - b^2)/b^2)*log((a*cos(d*x + c) - I*a*sin(d*x + c) + (b*cos(d*x + c) - I*b*sin(d*x +
c))*sqrt((a^2 - b^2)/b^2) + b)/b) + 2*(I*b*d*x + I*b*c)*sqrt((a^2 - b^2)/b^2)*log((a*cos(d*x + c) - I*a*sin(d
*x + c) - (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt((a^2 - b^2)/b^2) + b)/b))/((a^2 - b^2)*d^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{b \cos \left (d x + c\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a+b*cos(d*x+c)),x, algorithm="giac")

[Out]

integrate(x/(b*cos(d*x + c) + a), x)

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maple [B]  time = 0.09, size = 414, normalized size = 1.93 \[ -\frac {i \ln \left (\frac {-b \,{\mathrm e}^{i \left (d x +c \right )}+\sqrt {a^{2}-b^{2}}-a}{-a +\sqrt {a^{2}-b^{2}}}\right ) x}{d \sqrt {a^{2}-b^{2}}}+\frac {i \ln \left (\frac {b \,{\mathrm e}^{i \left (d x +c \right )}+\sqrt {a^{2}-b^{2}}+a}{a +\sqrt {a^{2}-b^{2}}}\right ) x}{d \sqrt {a^{2}-b^{2}}}-\frac {i \ln \left (\frac {-b \,{\mathrm e}^{i \left (d x +c \right )}+\sqrt {a^{2}-b^{2}}-a}{-a +\sqrt {a^{2}-b^{2}}}\right ) c}{d^{2} \sqrt {a^{2}-b^{2}}}+\frac {i \ln \left (\frac {b \,{\mathrm e}^{i \left (d x +c \right )}+\sqrt {a^{2}-b^{2}}+a}{a +\sqrt {a^{2}-b^{2}}}\right ) c}{d^{2} \sqrt {a^{2}-b^{2}}}-\frac {\dilog \left (\frac {-b \,{\mathrm e}^{i \left (d x +c \right )}+\sqrt {a^{2}-b^{2}}-a}{-a +\sqrt {a^{2}-b^{2}}}\right )}{d^{2} \sqrt {a^{2}-b^{2}}}+\frac {\dilog \left (\frac {b \,{\mathrm e}^{i \left (d x +c \right )}+\sqrt {a^{2}-b^{2}}+a}{a +\sqrt {a^{2}-b^{2}}}\right )}{d^{2} \sqrt {a^{2}-b^{2}}}+\frac {2 i c \arctan \left (\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}+2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{d^{2} \sqrt {-a^{2}+b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a+b*cos(d*x+c)),x)

[Out]

-I/d/(a^2-b^2)^(1/2)*ln((-b*exp(I*(d*x+c))+(a^2-b^2)^(1/2)-a)/(-a+(a^2-b^2)^(1/2)))*x+I/d/(a^2-b^2)^(1/2)*ln((
b*exp(I*(d*x+c))+(a^2-b^2)^(1/2)+a)/(a+(a^2-b^2)^(1/2)))*x-I/d^2/(a^2-b^2)^(1/2)*ln((-b*exp(I*(d*x+c))+(a^2-b^
2)^(1/2)-a)/(-a+(a^2-b^2)^(1/2)))*c+I/d^2/(a^2-b^2)^(1/2)*ln((b*exp(I*(d*x+c))+(a^2-b^2)^(1/2)+a)/(a+(a^2-b^2)
^(1/2)))*c-1/d^2/(a^2-b^2)^(1/2)*dilog((-b*exp(I*(d*x+c))+(a^2-b^2)^(1/2)-a)/(-a+(a^2-b^2)^(1/2)))+1/d^2/(a^2-
b^2)^(1/2)*dilog((b*exp(I*(d*x+c))+(a^2-b^2)^(1/2)+a)/(a+(a^2-b^2)^(1/2)))+2*I/d^2*c/(-a^2+b^2)^(1/2)*arctan(1
/2*(2*b*exp(I*(d*x+c))+2*a)/(-a^2+b^2)^(1/2))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a+b*cos(d*x+c)),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 details)Is 4*b^2-4*a^2 positive or negative?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x}{a+b\,\cos \left (c+d\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a + b*cos(c + d*x)),x)

[Out]

int(x/(a + b*cos(c + d*x)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{a + b \cos {\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a+b*cos(d*x+c)),x)

[Out]

Integral(x/(a + b*cos(c + d*x)), x)

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