Integrand size = 14, antiderivative size = 214 \[ \int \frac {x}{a+b \cos (c+d x)} \, dx=-\frac {i x \log \left (1+\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} d}+\frac {i x \log \left (1+\frac {b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} d}-\frac {\operatorname {PolyLog}\left (2,-\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} d^2}+\frac {\operatorname {PolyLog}\left (2,-\frac {b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} d^2} \] Output:
-I*x*ln(1+b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/(a^2-b^2)^(1/2)/d+I*x*ln(1 +b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/(a^2-b^2)^(1/2)/d-polylog(2,-b*exp( I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/(a^2-b^2)^(1/2)/d^2+polylog(2,-b*exp(I*(d* x+c))/(a+(a^2-b^2)^(1/2)))/(a^2-b^2)^(1/2)/d^2
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(756\) vs. \(2(214)=428\).
Time = 0.96 (sec) , antiderivative size = 756, normalized size of antiderivative = 3.53 \[ \int \frac {x}{a+b \cos (c+d x)} \, dx =\text {Too large to display} \] Input:
Integrate[x/(a + b*Cos[c + d*x]),x]
Output:
(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)*Tan[(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[(Sqr t[-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])/Sqrt[-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)
Time = 0.79 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 3802, 2694, 27, 2620, 2715, 2838}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {x}{a+b \cos (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {x}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 3802 |
\(\displaystyle 2 \int \frac {e^{i (c+d x)} x}{2 e^{i (c+d x)} a+b e^{2 i (c+d x)}+b}dx\) |
\(\Big \downarrow \) 2694 |
\(\displaystyle 2 \left (\frac {b \int \frac {e^{i (c+d x)} x}{2 \left (a+b e^{i (c+d x)}-\sqrt {a^2-b^2}\right )}dx}{\sqrt {a^2-b^2}}-\frac {b \int \frac {e^{i (c+d x)} x}{2 \left (a+b e^{i (c+d x)}+\sqrt {a^2-b^2}\right )}dx}{\sqrt {a^2-b^2}}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 \left (\frac {b \int \frac {e^{i (c+d x)} x}{a+b e^{i (c+d x)}-\sqrt {a^2-b^2}}dx}{2 \sqrt {a^2-b^2}}-\frac {b \int \frac {e^{i (c+d x)} x}{a+b e^{i (c+d x)}+\sqrt {a^2-b^2}}dx}{2 \sqrt {a^2-b^2}}\right )\) |
\(\Big \downarrow \) 2620 |
\(\displaystyle 2 \left (\frac {b \left (\frac {i \int \log \left (\frac {e^{i (c+d x)} b}{a-\sqrt {a^2-b^2}}+1\right )dx}{b d}-\frac {i x \log \left (1+\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}-\frac {b \left (\frac {i \int \log \left (\frac {e^{i (c+d x)} b}{a+\sqrt {a^2-b^2}}+1\right )dx}{b d}-\frac {i x \log \left (1+\frac {b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}\right )\) |
\(\Big \downarrow \) 2715 |
\(\displaystyle 2 \left (\frac {b \left (\frac {\int e^{-i (c+d x)} \log \left (\frac {e^{i (c+d x)} b}{a-\sqrt {a^2-b^2}}+1\right )de^{i (c+d x)}}{b d^2}-\frac {i x \log \left (1+\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}-\frac {b \left (\frac {\int e^{-i (c+d x)} \log \left (\frac {e^{i (c+d x)} b}{a+\sqrt {a^2-b^2}}+1\right )de^{i (c+d x)}}{b d^2}-\frac {i x \log \left (1+\frac {b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}\right )\) |
\(\Big \downarrow \) 2838 |
\(\displaystyle 2 \left (\frac {b \left (-\frac {\operatorname {PolyLog}\left (2,-\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d^2}-\frac {i x \log \left (1+\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}-\frac {b \left (-\frac {\operatorname {PolyLog}\left (2,-\frac {b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b d^2}-\frac {i x \log \left (1+\frac {b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}\right )\) |
Input:
Int[x/(a + b*Cos[c + d*x]),x]
Output:
2*((b*(((-I)*x*Log[1 + (b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b*d) - PolyLog[2, -((b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2]))]/(b*d^2)))/(2*Sqr t[a^2 - b^2]) - (b*(((-I)*x*Log[1 + (b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^ 2])])/(b*d) - PolyLog[2, -((b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2]))]/(b* d^2)))/(2*Sqrt[a^2 - b^2]))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si mp[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]
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]}, Simp[2*(c/q) Int [(f + g*x)^m*(F^u/(b - q + 2*c*F^u)), x], x] - Simp[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]
Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Simp[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]
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]
Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + Pi*(k_.) + (f_.)*( x_)]), x_Symbol] :> Simp[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]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 413 vs. \(2 (188 ) = 376\).
Time = 0.78 (sec) , antiderivative size = 414, normalized size of antiderivative = 1.93
method | result | size |
risch | \(-\frac {i \ln \left (\frac {-{\mathrm e}^{i \left (d x +c \right )} b +\sqrt {a^{2}-b^{2}}-a}{-a +\sqrt {a^{2}-b^{2}}}\right ) x}{d \sqrt {a^{2}-b^{2}}}+\frac {i \ln \left (\frac {{\mathrm e}^{i \left (d x +c \right )} b +\sqrt {a^{2}-b^{2}}+a}{a +\sqrt {a^{2}-b^{2}}}\right ) x}{d \sqrt {a^{2}-b^{2}}}-\frac {i \ln \left (\frac {-{\mathrm e}^{i \left (d x +c \right )} b +\sqrt {a^{2}-b^{2}}-a}{-a +\sqrt {a^{2}-b^{2}}}\right ) c}{d^{2} \sqrt {a^{2}-b^{2}}}+\frac {i \ln \left (\frac {{\mathrm e}^{i \left (d x +c \right )} b +\sqrt {a^{2}-b^{2}}+a}{a +\sqrt {a^{2}-b^{2}}}\right ) c}{d^{2} \sqrt {a^{2}-b^{2}}}-\frac {\operatorname {dilog}\left (\frac {-{\mathrm e}^{i \left (d x +c \right )} b +\sqrt {a^{2}-b^{2}}-a}{-a +\sqrt {a^{2}-b^{2}}}\right )}{d^{2} \sqrt {a^{2}-b^{2}}}+\frac {\operatorname {dilog}\left (\frac {{\mathrm e}^{i \left (d x +c \right )} b +\sqrt {a^{2}-b^{2}}+a}{a +\sqrt {a^{2}-b^{2}}}\right )}{d^{2} \sqrt {a^{2}-b^{2}}}+\frac {2 i c \arctan \left (\frac {2 \,{\mathrm e}^{i \left (d x +c \right )} b +2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{d^{2} \sqrt {-a^{2}+b^{2}}}\) | \(414\) |
Input:
int(x/(a+cos(d*x+c)*b),x,method=_RETURNVERBOSE)
Output:
-I/d/(a^2-b^2)^(1/2)*ln((-exp(I*(d*x+c))*b+(a^2-b^2)^(1/2)-a)/(-a+(a^2-b^2 )^(1/2)))*x+I/d/(a^2-b^2)^(1/2)*ln((exp(I*(d*x+c))*b+(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((-exp(I*(d*x+c))*b+(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((exp(I*(d*x+c)) *b+(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(( -exp(I*(d*x+c))*b+(a^2-b^2)^(1/2)-a)/(-a+(a^2-b^2)^(1/2)))+1/d^2/(a^2-b^2) ^(1/2)*dilog((exp(I*(d*x+c))*b+(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*exp(I*(d*x+c))*b+2*a)/(-a^2+b^2)^(1/ 2))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 915 vs. \(2 (184) = 368\).
Time = 0.21 (sec) , antiderivative size = 915, normalized size of antiderivative = 4.28 \[ \int \frac {x}{a+b \cos (c+d x)} \, dx =\text {Too large to display} \] Input:
integrate(x/(a+b*cos(d*x+c)),x, algorithm="fricas")
Output:
1/2*(-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) + 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) - 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) + 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) - 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) + 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) - 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) + 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))*sq rt((a^2 - b^2)/b^2) + b)/b + 1) - (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) - (-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) - (-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) - (I*b*d*x + I*b*c)*sqrt((a^2 - b^...
\[ \int \frac {x}{a+b \cos (c+d x)} \, dx=\int \frac {x}{a + b \cos {\left (c + d x \right )}}\, dx \] Input:
integrate(x/(a+b*cos(d*x+c)),x)
Output:
Integral(x/(a + b*cos(c + d*x)), x)
Exception generated. \[ \int \frac {x}{a+b \cos (c+d x)} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(x/(a+b*cos(d*x+c)),x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f or more de
\[ \int \frac {x}{a+b \cos (c+d x)} \, dx=\int { \frac {x}{b \cos \left (d x + c\right ) + a} \,d x } \] Input:
integrate(x/(a+b*cos(d*x+c)),x, algorithm="giac")
Output:
integrate(x/(b*cos(d*x + c) + a), x)
Timed out. \[ \int \frac {x}{a+b \cos (c+d x)} \, dx=\int \frac {x}{a+b\,\cos \left (c+d\,x\right )} \,d x \] Input:
int(x/(a + b*cos(c + d*x)),x)
Output:
int(x/(a + b*cos(c + d*x)), x)
\[ \int \frac {x}{a+b \cos (c+d x)} \, dx=\frac {4 \left (\int \frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} x}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a^{2}-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b^{2}+a^{2}+2 a b +b^{2}}d x \right ) a b +4 \left (\int \frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} x}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a^{2}-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b^{2}+a^{2}+2 a b +b^{2}}d x \right ) b^{2}+x^{2}}{2 a +2 b} \] Input:
int(x/(a+b*cos(d*x+c)),x)
Output:
(4*int((tan((c + d*x)/2)**2*x)/(tan((c + d*x)/2)**2*a**2 - tan((c + d*x)/2 )**2*b**2 + a**2 + 2*a*b + b**2),x)*a*b + 4*int((tan((c + d*x)/2)**2*x)/(t an((c + d*x)/2)**2*a**2 - tan((c + d*x)/2)**2*b**2 + a**2 + 2*a*b + b**2), x)*b**2 + x**2)/(2*(a + b))