\(\int \frac {c+d x}{(a+b \sin (e+f x))^2} \, dx\) [170]

Optimal result

Integrand size = 18, antiderivative size = 305 \[ \int \frac {c+d x}{(a+b \sin (e+f x))^2} \, dx=-\frac {i a (c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f}+\frac {i a (c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f}-\frac {d \log (a+b \sin (e+f x))}{\left (a^2-b^2\right ) f^2}-\frac {a d \operatorname {PolyLog}\left (2,\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f^2}+\frac {a d \operatorname {PolyLog}\left (2,\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f^2}+\frac {b (c+d x) \cos (e+f x)}{\left (a^2-b^2\right ) f (a+b \sin (e+f x))} \] Output:

-I*a*(d*x+c)*ln(1-I*b*exp(I*(f*x+e))/(a-(a^2-b^2)^(1/2)))/(a^2-b^2)^(3/2)/ 
f+I*a*(d*x+c)*ln(1-I*b*exp(I*(f*x+e))/(a+(a^2-b^2)^(1/2)))/(a^2-b^2)^(3/2) 
/f-d*ln(a+b*sin(f*x+e))/(a^2-b^2)/f^2-a*d*polylog(2,I*b*exp(I*(f*x+e))/(a- 
(a^2-b^2)^(1/2)))/(a^2-b^2)^(3/2)/f^2+a*d*polylog(2,I*b*exp(I*(f*x+e))/(a+ 
(a^2-b^2)^(1/2)))/(a^2-b^2)^(3/2)/f^2+b*(d*x+c)*cos(f*x+e)/(a^2-b^2)/f/(a+ 
b*sin(f*x+e))
 

Mathematica [A] (verified)

Time = 1.27 (sec) , antiderivative size = 236, normalized size of antiderivative = 0.77 \[ \int \frac {c+d x}{(a+b \sin (e+f x))^2} \, dx=\frac {-d \log (a+b \sin (e+f x))+\frac {a \left (-i f (c+d x) \left (\log \left (1+\frac {i b e^{i (e+f x)}}{-a+\sqrt {a^2-b^2}}\right )-\log \left (1-\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )\right )-d \operatorname {PolyLog}\left (2,-\frac {i b e^{i (e+f x)}}{-a+\sqrt {a^2-b^2}}\right )+d \operatorname {PolyLog}\left (2,\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )\right )}{\sqrt {a^2-b^2}}+\frac {b f (c+d x) \cos (e+f x)}{a+b \sin (e+f x)}}{\left (a^2-b^2\right ) f^2} \] Input:

Integrate[(c + d*x)/(a + b*Sin[e + f*x])^2,x]
 

Output:

(-(d*Log[a + b*Sin[e + f*x]]) + (a*((-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 + Sqr 
t[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] + (b*f*(c + d*x)*Cos[e + f*x])/(a + b*Sin[e + f*x]))/((a^2 - b^2 
)*f^2)
 

Rubi [A] (verified)

Time = 1.17 (sec) , antiderivative size = 317, normalized size of antiderivative = 1.04, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.611, Rules used = {3042, 3805, 3042, 3147, 16, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3805

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3147

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

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 3804

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

\(\Big \downarrow \) 2694

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

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

Input:

Int[(c + d*x)/(a + b*Sin[e + f*x])^2,x]
 

Output:

-((d*Log[a + b*Sin[e + f*x]])/((a^2 - b^2)*f^2)) + (2*a*(((-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]))/(a^2 - b^2) + (b*(c + d*x)*Cos[e + f*x 
])/((a^2 - b^2)*f*(a + b*Sin[e + f*x]))
 

Defintions of rubi rules used

rule 16

Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + 
b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
 

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 3147

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m 
_.), x_Symbol] :> Simp[1/(b^p*f)   Subst[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] && IntegerQ[(p 
 - 1)/2] && NeQ[a^2 - b^2, 0]
 

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]
 

rule 3805

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^2, x_ 
Symbol] :> Simp[b*(c + d*x)^m*(Cos[e + f*x]/(f*(a^2 - b^2)*(a + b*Sin[e + f 
*x]))), x] + (Simp[a/(a^2 - b^2)   Int[(c + d*x)^m/(a + b*Sin[e + f*x]), x] 
, x] - Simp[b*d*(m/(f*(a^2 - b^2)))   Int[(c + d*x)^(m - 1)*(Cos[e + f*x]/( 
a + b*Sin[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. 640 vs. \(2 (275 ) = 550\).

Time = 3.91 (sec) , antiderivative size = 641, normalized size of antiderivative = 2.10

method	result	size
risch	\(\frac {2 \left (d x +c \right ) \left (i b +a \,{\mathrm e}^{i \left (f x +e \right )}\right )}{f \left (a^{2}-b^{2}\right ) \left ({\mathrm e}^{2 i \left (f x +e \right )} b -b +2 i a \,{\mathrm e}^{i \left (f x +e \right )}\right )}-\frac {2 d \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{\left (-a^{2}+b^{2}\right ) f^{2}}+\frac {d \ln \left (i {\mathrm e}^{2 i \left (f x +e \right )} b -i b -2 a \,{\mathrm e}^{i \left (f x +e \right )}\right )}{\left (-a^{2}+b^{2}\right ) f^{2}}+\frac {2 i a d e \arctan \left (\frac {2 i b \,{\mathrm e}^{i \left (f x +e \right )}-2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{\left (-a^{2}+b^{2}\right )^{\frac {3}{2}} f^{2}}-\frac {2 i a c \arctan \left (\frac {2 i b \,{\mathrm e}^{i \left (f x +e \right )}-2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{\left (-a^{2}+b^{2}\right )^{\frac {3}{2}} f}-\frac {a d \ln \left (\frac {-i a -b \,{\mathrm e}^{i \left (f x +e \right )}+\sqrt {-a^{2}+b^{2}}}{-i a +\sqrt {-a^{2}+b^{2}}}\right ) x}{\left (-a^{2}+b^{2}\right )^{\frac {3}{2}} f}+\frac {a d \ln \left (\frac {i a +b \,{\mathrm e}^{i \left (f x +e \right )}+\sqrt {-a^{2}+b^{2}}}{i a +\sqrt {-a^{2}+b^{2}}}\right ) x}{\left (-a^{2}+b^{2}\right )^{\frac {3}{2}} f}-\frac {a d \ln \left (\frac {-i a -b \,{\mathrm e}^{i \left (f x +e \right )}+\sqrt {-a^{2}+b^{2}}}{-i a +\sqrt {-a^{2}+b^{2}}}\right ) e}{\left (-a^{2}+b^{2}\right )^{\frac {3}{2}} f^{2}}+\frac {a d \ln \left (\frac {i a +b \,{\mathrm e}^{i \left (f x +e \right )}+\sqrt {-a^{2}+b^{2}}}{i a +\sqrt {-a^{2}+b^{2}}}\right ) e}{\left (-a^{2}+b^{2}\right )^{\frac {3}{2}} f^{2}}+\frac {i a d \operatorname {dilog}\left (\frac {-i a -b \,{\mathrm e}^{i \left (f x +e \right )}+\sqrt {-a^{2}+b^{2}}}{-i a +\sqrt {-a^{2}+b^{2}}}\right )}{\left (-a^{2}+b^{2}\right )^{\frac {3}{2}} f^{2}}-\frac {i a d \operatorname {dilog}\left (\frac {i a +b \,{\mathrm e}^{i \left (f x +e \right )}+\sqrt {-a^{2}+b^{2}}}{i a +\sqrt {-a^{2}+b^{2}}}\right )}{\left (-a^{2}+b^{2}\right )^{\frac {3}{2}} f^{2}}\)	\(641\)

Input:

int((d*x+c)/(a+b*sin(f*x+e))^2,x,method=_RETURNVERBOSE)
 

Output:

2*(d*x+c)*(I*b+a*exp(I*(f*x+e)))/f/(a^2-b^2)/(exp(2*I*(f*x+e))*b-b+2*I*a*e 
xp(I*(f*x+e)))-2/(-a^2+b^2)/f^2*d*ln(exp(I*(f*x+e)))+1/(-a^2+b^2)/f^2*d*ln 
(I*exp(2*I*(f*x+e))*b-I*b-2*a*exp(I*(f*x+e)))+2*I/(-a^2+b^2)^(3/2)/f^2*a*d 
*e*arctan(1/2*(2*I*b*exp(I*(f*x+e))-2*a)/(-a^2+b^2)^(1/2))-2*I/(-a^2+b^2)^ 
(3/2)/f*a*c*arctan(1/2*(2*I*b*exp(I*(f*x+e))-2*a)/(-a^2+b^2)^(1/2))-1/(-a^ 
2+b^2)^(3/2)/f*a*d*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/(-a^2+b^2)^(3/2)/f*a*d*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/(-a^2+b^2)^(3/2)/f^2*a*d*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/(-a^2+b^2)^( 
3/2)/f^2*a*d*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/(-a^2+b^2)^(3/2)/f^2*a*d*dilog((-I*a-b*exp(I*(f*x+e))+(-a^2+b^2) 
^(1/2))/(-I*a+(-a^2+b^2)^(1/2)))-I/(-a^2+b^2)^(3/2)/f^2*a*d*dilog((I*a+b*e 
xp(I*(f*x+e))+(-a^2+b^2)^(1/2))/(I*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. 1512 vs. \(2 (267) = 534\).

Time = 0.29 (sec) , antiderivative size = 1512, normalized size of antiderivative = 4.96 \[ \int \frac {c+d x}{(a+b \sin (e+f x))^2} \, dx=\text {Too large to display} \] Input:

integrate((d*x+c)/(a+b*sin(f*x+e))^2,x, algorithm="fricas")
 

Output:

1/2*((I*a*b^2*d*sin(f*x + e) + I*a^2*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*a*b^2*d*sin(f*x + e) - I*a^2*b*d)*sqr 
t(-(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*a*b^2*d* 
sin(f*x + e) - I*a^2*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*a*b^2*d*sin(f*x + e) + I*a^2*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) - (a^2*b*d*f*x + a^2*b*d*e + 
(a*b^2*d*f*x + a*b^2*d*e)*sin(f*x + e))*sqrt(-(a^2 - b^2)/b^2)*log(-(I*a*c 
os(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) + (a^2*b*d*f*x + a^2*b*d*e + (a*b^2*d*f*x + a*b^2* 
d*e)*sin(f*x + 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) - (a^2*b*d*f*x + a^2*b*d*e + (a*b^2*d*f*x + a*b^2*d*e)*sin(f*x + e))*sq 
rt(-(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) + (a^2*b*d*f*x + 
a^2*b*d*e + (a*b^2*d*f*x + a*b^2*d*e)*sin(f*x + 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...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {c+d x}{(a+b \sin (e+f x))^2} \, dx=\text {Timed out} \] Input:

integrate((d*x+c)/(a+b*sin(f*x+e))**2,x)
                                                                                    
                                                                                    
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {c+d x}{(a+b \sin (e+f x))^2} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((d*x+c)/(a+b*sin(f*x+e))^2,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]

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

integrate((d*x+c)/(a+b*sin(f*x+e))^2,x, algorithm="giac")
 

Output:

integrate((d*x + c)/(b*sin(f*x + e) + a)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {c+d x}{(a+b \sin (e+f x))^2} \, dx=\text {Hanged} \] Input:

int((c + d*x)/(a + b*sin(e + f*x))^2,x)
 

Output:

\text{Hanged}
 

Reduce [F]

\[ \int \frac {c+d x}{(a+b \sin (e+f x))^2} \, dx=\frac {2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) \sin \left (f x +e \right ) a b c +2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) a^{2} c +\cos \left (f x +e \right ) a^{2} b c -\cos \left (f x +e \right ) b^{3} c +\left (\int \frac {x}{\sin \left (f x +e \right )^{2} b^{2}+2 \sin \left (f x +e \right ) a b +a^{2}}d x \right ) \sin \left (f x +e \right ) a^{4} b d f -2 \left (\int \frac {x}{\sin \left (f x +e \right )^{2} b^{2}+2 \sin \left (f x +e \right ) a b +a^{2}}d x \right ) \sin \left (f x +e \right ) a^{2} b^{3} d f +\left (\int \frac {x}{\sin \left (f x +e \right )^{2} b^{2}+2 \sin \left (f x +e \right ) a b +a^{2}}d x \right ) \sin \left (f x +e \right ) b^{5} d f +\left (\int \frac {x}{\sin \left (f x +e \right )^{2} b^{2}+2 \sin \left (f x +e \right ) a b +a^{2}}d x \right ) a^{5} d f -2 \left (\int \frac {x}{\sin \left (f x +e \right )^{2} b^{2}+2 \sin \left (f x +e \right ) a b +a^{2}}d x \right ) a^{3} b^{2} d f +\left (\int \frac {x}{\sin \left (f x +e \right )^{2} b^{2}+2 \sin \left (f x +e \right ) a b +a^{2}}d x \right ) a \,b^{4} d f}{f \left (\sin \left (f x +e \right ) a^{4} b -2 \sin \left (f x +e \right ) a^{2} b^{3}+\sin \left (f x +e \right ) b^{5}+a^{5}-2 a^{3} b^{2}+a \,b^{4}\right )} \] Input:

int((d*x+c)/(a+b*sin(f*x+e))^2,x)
 

Output:

(2*sqrt(a**2 - b**2)*atan((tan((e + f*x)/2)*a + b)/sqrt(a**2 - b**2))*sin( 
e + f*x)*a*b*c + 2*sqrt(a**2 - b**2)*atan((tan((e + f*x)/2)*a + b)/sqrt(a* 
*2 - b**2))*a**2*c + cos(e + f*x)*a**2*b*c - cos(e + f*x)*b**3*c + int(x/( 
sin(e + f*x)**2*b**2 + 2*sin(e + f*x)*a*b + a**2),x)*sin(e + f*x)*a**4*b*d 
*f - 2*int(x/(sin(e + f*x)**2*b**2 + 2*sin(e + f*x)*a*b + a**2),x)*sin(e + 
 f*x)*a**2*b**3*d*f + int(x/(sin(e + f*x)**2*b**2 + 2*sin(e + f*x)*a*b + a 
**2),x)*sin(e + f*x)*b**5*d*f + int(x/(sin(e + f*x)**2*b**2 + 2*sin(e + f* 
x)*a*b + a**2),x)*a**5*d*f - 2*int(x/(sin(e + f*x)**2*b**2 + 2*sin(e + f*x 
)*a*b + a**2),x)*a**3*b**2*d*f + int(x/(sin(e + f*x)**2*b**2 + 2*sin(e + f 
*x)*a*b + a**2),x)*a*b**4*d*f)/(f*(sin(e + f*x)*a**4*b - 2*sin(e + f*x)*a* 
*2*b**3 + sin(e + f*x)*b**5 + a**5 - 2*a**3*b**2 + a*b**4))