\(\int \frac {c+d x}{a+b \sin (e+f x)} \, dx\) [165]

Optimal result

Integrand size = 18, antiderivative size = 234 \[ \int \frac {c+d x}{a+b \sin (e+f x)} \, dx=-\frac {i (c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f}+\frac {i (c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f}-\frac {d \operatorname {PolyLog}\left (2,\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f^2}+\frac {d \operatorname {PolyLog}\left (2,\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.78 \[ \int \frac {c+d x}{a+b \sin (e+f x)} \, dx=\frac {-i f (c+d x) \left (\log \left (1+\frac {i b e^{i (e+f x)}}{-a+\sqrt {a^2-b^2}}\right )-\log \left (1-\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )\right )-d \operatorname {PolyLog}\left (2,-\frac {i b e^{i (e+f x)}}{-a+\sqrt {a^2-b^2}}\right )+d \operatorname {PolyLog}\left (2,\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.87 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {3042, 3804, 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.

\(\displaystyle \int \frac {c+d x}{a+b \sin (e+f x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {c+d x}{a+b \sin (e+f x)}dx\)

\(\Big \downarrow \) 3804

\(\displaystyle 2 \int \frac {e^{i (e+f x)} (c+d x)}{2 e^{i (e+f x)} a-i b e^{2 i (e+f x)}+i b}dx\)

\(\Big \downarrow \) 2694

\(\displaystyle 2 \left (\frac {i b \int \frac {e^{i (e+f x)} (c+d x)}{2 \left (a-i b e^{i (e+f x)}+\sqrt {a^2-b^2}\right )}dx}{\sqrt {a^2-b^2}}-\frac {i b \int \frac {e^{i (e+f x)} (c+d x)}{2 \left (a-i b e^{i (e+f x)}-\sqrt {a^2-b^2}\right )}dx}{\sqrt {a^2-b^2}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle 2 \left (\frac {i b \int \frac {e^{i (e+f x)} (c+d x)}{a-i b e^{i (e+f x)}+\sqrt {a^2-b^2}}dx}{2 \sqrt {a^2-b^2}}-\frac {i b \int \frac {e^{i (e+f x)} (c+d x)}{a-i b e^{i (e+f x)}-\sqrt {a^2-b^2}}dx}{2 \sqrt {a^2-b^2}}\right )\)

\(\Big \downarrow \) 2620

\(\displaystyle 2 \left (\frac {i b \left (\frac {(c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{\sqrt {a^2-b^2}+a}\right )}{b f}-\frac {d \int \log \left (1-\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )dx}{b f}\right )}{2 \sqrt {a^2-b^2}}-\frac {i b \left (\frac {(c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{b f}-\frac {d \int \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )dx}{b f}\right )}{2 \sqrt {a^2-b^2}}\right )\)

\(\Big \downarrow \) 2715

\(\displaystyle 2 \left (\frac {i b \left (\frac {i d \int e^{-i (e+f x)} \log \left (1-\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )de^{i (e+f x)}}{b f^2}+\frac {(c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{\sqrt {a^2-b^2}+a}\right )}{b f}\right )}{2 \sqrt {a^2-b^2}}-\frac {i b \left (\frac {i d \int e^{-i (e+f x)} \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )de^{i (e+f x)}}{b f^2}+\frac {(c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{b f}\right )}{2 \sqrt {a^2-b^2}}\right )\)

\(\Big \downarrow \) 2838

\(\displaystyle 2 \left (\frac {i b \left (\frac {(c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{\sqrt {a^2-b^2}+a}\right )}{b f}-\frac {i d \operatorname {PolyLog}\left (2,\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{b f^2}\right )}{2 \sqrt {a^2-b^2}}-\frac {i b \left (\frac {(c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{b f}-\frac {i d \operatorname {PolyLog}\left (2,\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{b f^2}\right )}{2 \sqrt {a^2-b^2}}\right )\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2620

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]
 

rule 2694

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]
 

rule 2715

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]
 

rule 2838

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 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3804

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

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 491 vs. \(2 (204 ) = 408\).

Time = 0.69 (sec) , antiderivative size = 492, normalized size of antiderivative = 2.10

method	result	size
risch	\(\frac {2 i c \arctan \left (\frac {2 i b \,{\mathrm e}^{i \left (f x +e \right )}-2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{f \sqrt {-a^{2}+b^{2}}}+\frac {d \ln \left (\frac {-i a -b \,{\mathrm e}^{i \left (f x +e \right )}+\sqrt {-a^{2}+b^{2}}}{-i a +\sqrt {-a^{2}+b^{2}}}\right ) x}{f \sqrt {-a^{2}+b^{2}}}-\frac {d \ln \left (\frac {i a +b \,{\mathrm e}^{i \left (f x +e \right )}+\sqrt {-a^{2}+b^{2}}}{i a +\sqrt {-a^{2}+b^{2}}}\right ) x}{f \sqrt {-a^{2}+b^{2}}}+\frac {d \ln \left (\frac {-i a -b \,{\mathrm e}^{i \left (f x +e \right )}+\sqrt {-a^{2}+b^{2}}}{-i a +\sqrt {-a^{2}+b^{2}}}\right ) e}{f^{2} \sqrt {-a^{2}+b^{2}}}-\frac {d \ln \left (\frac {i a +b \,{\mathrm e}^{i \left (f x +e \right )}+\sqrt {-a^{2}+b^{2}}}{i a +\sqrt {-a^{2}+b^{2}}}\right ) e}{f^{2} \sqrt {-a^{2}+b^{2}}}-\frac {i d \operatorname {dilog}\left (\frac {-i a -b \,{\mathrm e}^{i \left (f x +e \right )}+\sqrt {-a^{2}+b^{2}}}{-i a +\sqrt {-a^{2}+b^{2}}}\right )}{f^{2} \sqrt {-a^{2}+b^{2}}}+\frac {i d \operatorname {dilog}\left (\frac {i a +b \,{\mathrm e}^{i \left (f x +e \right )}+\sqrt {-a^{2}+b^{2}}}{i a +\sqrt {-a^{2}+b^{2}}}\right )}{f^{2} \sqrt {-a^{2}+b^{2}}}-\frac {2 i d e \arctan \left (\frac {2 i b \,{\mathrm e}^{i \left (f x +e \right )}-2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{f^{2} \sqrt {-a^{2}+b^{2}}}\)	\(492\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 997 vs. \(2 (200) = 400\).

Time = 0.21 (sec) , antiderivative size = 997, normalized size of antiderivative = 4.26 \[ \int \frac {c+d x}{a+b \sin (e+f x)} \, dx=\text {Too large to display} \] Input:

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

Output:

1/2*(I*b*d*sqrt(-(a^2 - b^2)/b^2)*dilog((I*a*cos(f*x + e) - a*sin(f*x + e) 
 + (b*cos(f*x + e) + I*b*sin(f*x + e))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 1) 
- I*b*d*sqrt(-(a^2 - b^2)/b^2)*dilog((I*a*cos(f*x + e) - a*sin(f*x + e) - 
(b*cos(f*x + e) + I*b*sin(f*x + e))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 1) - I 
*b*d*sqrt(-(a^2 - b^2)/b^2)*dilog((-I*a*cos(f*x + e) - a*sin(f*x + e) + (b 
*cos(f*x + e) - I*b*sin(f*x + e))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 1) + I*b 
*d*sqrt(-(a^2 - b^2)/b^2)*dilog((-I*a*cos(f*x + e) - a*sin(f*x + e) - (b*c 
os(f*x + e) - I*b*sin(f*x + e))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 1) - (b*d* 
e - b*c*f)*sqrt(-(a^2 - b^2)/b^2)*log(2*b*cos(f*x + e) + 2*I*b*sin(f*x + e 
) + 2*b*sqrt(-(a^2 - b^2)/b^2) + 2*I*a) - (b*d*e - b*c*f)*sqrt(-(a^2 - b^2 
)/b^2)*log(2*b*cos(f*x + e) - 2*I*b*sin(f*x + e) + 2*b*sqrt(-(a^2 - b^2)/b 
^2) - 2*I*a) + (b*d*e - b*c*f)*sqrt(-(a^2 - b^2)/b^2)*log(-2*b*cos(f*x + e 
) + 2*I*b*sin(f*x + e) + 2*b*sqrt(-(a^2 - b^2)/b^2) + 2*I*a) + (b*d*e - b* 
c*f)*sqrt(-(a^2 - b^2)/b^2)*log(-2*b*cos(f*x + e) - 2*I*b*sin(f*x + e) + 2 
*b*sqrt(-(a^2 - b^2)/b^2) - 2*I*a) - (b*d*f*x + b*d*e)*sqrt(-(a^2 - b^2)/b 
^2)*log(-(I*a*cos(f*x + e) - a*sin(f*x + e) + (b*cos(f*x + e) + I*b*sin(f* 
x + e))*sqrt(-(a^2 - b^2)/b^2) - b)/b) + (b*d*f*x + b*d*e)*sqrt(-(a^2 - b^ 
2)/b^2)*log(-(I*a*cos(f*x + e) - a*sin(f*x + e) - (b*cos(f*x + e) + I*b*si 
n(f*x + e))*sqrt(-(a^2 - b^2)/b^2) - b)/b) - (b*d*f*x + b*d*e)*sqrt(-(a^2 
- b^2)/b^2)*log(-(-I*a*cos(f*x + e) - a*sin(f*x + e) + (b*cos(f*x + e) ...
 

Sympy [F]

\[ \int \frac {c+d x}{a+b \sin (e+f x)} \, dx=\int \frac {c + d x}{a + b \sin {\left (e + f x \right )}}\, dx \] Input:

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

Output:

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

Maxima [F(-2)]

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

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

Giac [F]

\[ \int \frac {c+d x}{a+b \sin (e+f x)} \, dx=\int { \frac {d x + c}{b \sin \left (f x + e\right ) + a} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {c+d x}{a+b \sin (e+f x)} \, dx=\int \frac {c+d\,x}{a+b\,\sin \left (e+f\,x\right )} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {c+d x}{a+b \sin (e+f x)} \, dx=\frac {2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) c +\left (\int \frac {x}{\sin \left (f x +e \right ) b +a}d x \right ) a^{2} d f -\left (\int \frac {x}{\sin \left (f x +e \right ) b +a}d x \right ) b^{2} d f}{f \left (a^{2}-b^{2}\right )} \] Input:

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

Output:

(2*sqrt(a**2 - b**2)*atan((tan((e + f*x)/2)*a + b)/sqrt(a**2 - b**2))*c + 
int(x/(sin(e + f*x)*b + a),x)*a**2*d*f - int(x/(sin(e + f*x)*b + a),x)*b** 
2*d*f)/(f*(a**2 - b**2))