\(\int \frac {(e+f x) \sin (c+d x)}{a+a \sin (c+d x)} \, dx\) [181]

Optimal result

Integrand size = 24, antiderivative size = 76 \[ \int \frac {(e+f x) \sin (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {e x}{a}+\frac {f x^2}{2 a}+\frac {(e+f x) \cot \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{a d}-\frac {2 f \log \left (\sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )\right )}{a d^2} \]

[Out]

e*x/a+1/2*f*x^2/a+(f*x+e)*cot(1/2*c+1/4*Pi+1/2*d*x)/a/d-2*f*ln(sin(1/2*c+1/4*Pi+1/2*d*x))/a/d^2

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {4611, 3399, 4269, 3556} \[ \int \frac {(e+f x) \sin (c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {2 f \log \left (\sin \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right )\right )}{a d^2}+\frac {(e+f x) \cot \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right )}{a d}+\frac {e x}{a}+\frac {f x^2}{2 a} \]

[In]

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

[Out]

(e*x)/a + (f*x^2)/(2*a) + ((e + f*x)*Cot[c/2 + Pi/4 + (d*x)/2])/(a*d) - (2*f*Log[Sin[c/2 + Pi/4 + (d*x)/2]])/(
a*d^2)

Rule 3399

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

Rule 3556

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 4269

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-(c + d*x)^m)*(Cot[e + f*x]/f), x
] + Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 4611

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

Rubi steps \begin{align*} \text {integral}& = \frac {\int (e+f x) \, dx}{a}-\int \frac {e+f x}{a+a \sin (c+d x)} \, dx \\ & = \frac {e x}{a}+\frac {f x^2}{2 a}-\frac {\int (e+f x) \csc ^2\left (\frac {1}{2} \left (c+\frac {\pi }{2}\right )+\frac {d x}{2}\right ) \, dx}{2 a} \\ & = \frac {e x}{a}+\frac {f x^2}{2 a}+\frac {(e+f x) \cot \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{a d}-\frac {f \int \cot \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right ) \, dx}{a d} \\ & = \frac {e x}{a}+\frac {f x^2}{2 a}+\frac {(e+f x) \cot \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{a d}-\frac {2 f \log \left (\sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )\right )}{a d^2} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(199\) vs. \(2(76)=152\).

Time = 0.57 (sec) , antiderivative size = 199, normalized size of antiderivative = 2.62 \[ \int \frac {(e+f x) \sin (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {2 d f x \cos \left (c+\frac {d x}{2}\right )+\cos \left (\frac {d x}{2}\right ) \left (d^2 x (2 e+f x)-4 f \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )-4 d e \sin \left (\frac {d x}{2}\right )-2 d f x \sin \left (\frac {d x}{2}\right )+2 d^2 e x \sin \left (c+\frac {d x}{2}\right )+d^2 f x^2 \sin \left (c+\frac {d x}{2}\right )-4 f \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin \left (c+\frac {d x}{2}\right )}{2 a d^2 \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )} \]

[In]

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

[Out]

(2*d*f*x*Cos[c + (d*x)/2] + Cos[(d*x)/2]*(d^2*x*(2*e + f*x) - 4*f*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) -
4*d*e*Sin[(d*x)/2] - 2*d*f*x*Sin[(d*x)/2] + 2*d^2*e*x*Sin[c + (d*x)/2] + d^2*f*x^2*Sin[c + (d*x)/2] - 4*f*Log[
Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sin[c + (d*x)/2])/(2*a*d^2*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[
(c + d*x)/2]))

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.25 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.16

method	result	size
risch	\(\frac {f \,x^{2}}{2 a}+\frac {e x}{a}+\frac {2 i f x}{a d}+\frac {2 i f c}{a \,d^{2}}+\frac {2 f x +2 e}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}-\frac {2 f \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a \,d^{2}}\)	\(88\)
parallelrisch	\(\frac {f \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 f \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (\left (\frac {f x}{2}+e \right ) \left (d x -2\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+x \left (\left (\frac {f x}{2}+e \right ) d +f \right )\right ) d}{d^{2} a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}\)	\(109\)
norman	\(\frac {-\frac {2 e \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a}-\frac {2 e \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {\left (d e +f \right ) x}{d a}+\frac {\left (d e -f \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a}+\frac {\left (d e -f \right ) x \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {\left (d e +f \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {f \,x^{2}}{2 a}+\frac {f \,x^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a}+\frac {f \,x^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}+\frac {f \,x^{2} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {f \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a \,d^{2}}-\frac {2 f \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a \,d^{2}}\)	\(267\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 151 vs. \(2 (62) = 124\).

Time = 0.29 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.99 \[ \int \frac {(e+f x) \sin (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {d^{2} f x^{2} + 2 \, d e + 2 \, {\left (d^{2} e + d f\right )} x + {\left (d^{2} f x^{2} + 2 \, d e + 2 \, {\left (d^{2} e + d f\right )} x\right )} \cos \left (d x + c\right ) - 2 \, {\left (f \cos \left (d x + c\right ) + f \sin \left (d x + c\right ) + f\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (d^{2} f x^{2} - 2 \, d e + 2 \, {\left (d^{2} e - d f\right )} x\right )} \sin \left (d x + c\right )}{2 \, {\left (a d^{2} \cos \left (d x + c\right ) + a d^{2} \sin \left (d x + c\right ) + a d^{2}\right )}} \]

[In]

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

[Out]

1/2*(d^2*f*x^2 + 2*d*e + 2*(d^2*e + d*f)*x + (d^2*f*x^2 + 2*d*e + 2*(d^2*e + d*f)*x)*cos(d*x + c) - 2*(f*cos(d
*x + c) + f*sin(d*x + c) + f)*log(sin(d*x + c) + 1) + (d^2*f*x^2 - 2*d*e + 2*(d^2*e - d*f)*x)*sin(d*x + c))/(a
*d^2*cos(d*x + c) + a*d^2*sin(d*x + c) + a*d^2)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 456 vs. \(2 (58) = 116\).

Time = 0.68 (sec) , antiderivative size = 456, normalized size of antiderivative = 6.00 \[ \int \frac {(e+f x) \sin (c+d x)}{a+a \sin (c+d x)} \, dx=\begin {cases} \frac {2 d^{2} e x \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2 a d^{2} \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a d^{2}} + \frac {2 d^{2} e x}{2 a d^{2} \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a d^{2}} + \frac {d^{2} f x^{2} \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2 a d^{2} \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a d^{2}} + \frac {d^{2} f x^{2}}{2 a d^{2} \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a d^{2}} + \frac {4 d e}{2 a d^{2} \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a d^{2}} - \frac {2 d f x \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2 a d^{2} \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a d^{2}} + \frac {2 d f x}{2 a d^{2} \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a d^{2}} - \frac {4 f \log {\left (\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 1 \right )} \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2 a d^{2} \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a d^{2}} - \frac {4 f \log {\left (\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 1 \right )}}{2 a d^{2} \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a d^{2}} + \frac {2 f \log {\left (\tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 1 \right )} \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2 a d^{2} \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a d^{2}} + \frac {2 f \log {\left (\tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 1 \right )}}{2 a d^{2} \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a d^{2}} & \text {for}\: d \neq 0 \\\frac {\left (e x + \frac {f x^{2}}{2}\right ) \sin {\left (c \right )}}{a \sin {\left (c \right )} + a} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((2*d**2*e*x*tan(c/2 + d*x/2)/(2*a*d**2*tan(c/2 + d*x/2) + 2*a*d**2) + 2*d**2*e*x/(2*a*d**2*tan(c/2 +
d*x/2) + 2*a*d**2) + d**2*f*x**2*tan(c/2 + d*x/2)/(2*a*d**2*tan(c/2 + d*x/2) + 2*a*d**2) + d**2*f*x**2/(2*a*d
**2*tan(c/2 + d*x/2) + 2*a*d**2) + 4*d*e/(2*a*d**2*tan(c/2 + d*x/2) + 2*a*d**2) - 2*d*f*x*tan(c/2 + d*x/2)/(2*
a*d**2*tan(c/2 + d*x/2) + 2*a*d**2) + 2*d*f*x/(2*a*d**2*tan(c/2 + d*x/2) + 2*a*d**2) - 4*f*log(tan(c/2 + d*x/2
) + 1)*tan(c/2 + d*x/2)/(2*a*d**2*tan(c/2 + d*x/2) + 2*a*d**2) - 4*f*log(tan(c/2 + d*x/2) + 1)/(2*a*d**2*tan(c
/2 + d*x/2) + 2*a*d**2) + 2*f*log(tan(c/2 + d*x/2)**2 + 1)*tan(c/2 + d*x/2)/(2*a*d**2*tan(c/2 + d*x/2) + 2*a*d
**2) + 2*f*log(tan(c/2 + d*x/2)**2 + 1)/(2*a*d**2*tan(c/2 + d*x/2) + 2*a*d**2), Ne(d, 0)), ((e*x + f*x**2/2)*s
in(c)/(a*sin(c) + a), True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 273 vs. \(2 (62) = 124\).

Time = 0.29 (sec) , antiderivative size = 273, normalized size of antiderivative = 3.59 \[ \int \frac {(e+f x) \sin (c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {4 \, c f {\left (\frac {1}{a d + \frac {a d \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}} + \frac {\arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a d}\right )} - 4 \, e {\left (\frac {\arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} + \frac {1}{a + \frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}}\right )} - \frac {{\left ({\left (d x + c\right )}^{2} \cos \left (d x + c\right )^{2} + {\left (d x + c\right )}^{2} \sin \left (d x + c\right )^{2} + 2 \, {\left (d x + c\right )}^{2} \sin \left (d x + c\right ) + {\left (d x + c\right )}^{2} + 4 \, {\left (d x + c\right )} \cos \left (d x + c\right ) - 2 \, {\left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \sin \left (d x + c\right ) + 1\right )} \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \sin \left (d x + c\right ) + 1\right )\right )} f}{a d \cos \left (d x + c\right )^{2} + a d \sin \left (d x + c\right )^{2} + 2 \, a d \sin \left (d x + c\right ) + a d}}{2 \, d} \]

[In]

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

[Out]

-1/2*(4*c*f*(1/(a*d + a*d*sin(d*x + c)/(cos(d*x + c) + 1)) + arctan(sin(d*x + c)/(cos(d*x + c) + 1))/(a*d)) -
4*e*(arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a + 1/(a + a*sin(d*x + c)/(cos(d*x + c) + 1))) - ((d*x + c)^2*cos
(d*x + c)^2 + (d*x + c)^2*sin(d*x + c)^2 + 2*(d*x + c)^2*sin(d*x + c) + (d*x + c)^2 + 4*(d*x + c)*cos(d*x + c)
- 2*(cos(d*x + c)^2 + sin(d*x + c)^2 + 2*sin(d*x + c) + 1)*log(cos(d*x + c)^2 + sin(d*x + c)^2 + 2*sin(d*x +
c) + 1))*f/(a*d*cos(d*x + c)^2 + a*d*sin(d*x + c)^2 + 2*a*d*sin(d*x + c) + a*d))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 656 vs. \(2 (62) = 124\).

Time = 0.47 (sec) , antiderivative size = 656, normalized size of antiderivative = 8.63 \[ \int \frac {(e+f x) \sin (c+d x)}{a+a \sin (c+d x)} \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/2*(d^2*f*x^2*tan(1/2*d*x)*tan(1/2*c) - d^2*f*x^2*tan(1/2*d*x) - d^2*f*x^2*tan(1/2*c) + 2*d^2*e*x*tan(1/2*d*x
)*tan(1/2*c) - d^2*f*x^2 - 2*d^2*e*x*tan(1/2*d*x) - 2*d^2*e*x*tan(1/2*c) + 2*d*f*x*tan(1/2*d*x)*tan(1/2*c) - 2
*d^2*e*x + 2*d*f*x*tan(1/2*d*x) + 2*d*f*x*tan(1/2*c) + 2*d*e*tan(1/2*d*x)*tan(1/2*c) - 2*f*log(2*(tan(1/2*d*x)
^2*tan(1/2*c)^2 - 2*tan(1/2*d*x)^2*tan(1/2*c) - 2*tan(1/2*d*x)*tan(1/2*c)^2 + tan(1/2*d*x)^2 + tan(1/2*c)^2 +
2*tan(1/2*d*x) + 2*tan(1/2*c) + 1)/(tan(1/2*d*x)^2*tan(1/2*c)^2 + tan(1/2*d*x)^2 + tan(1/2*c)^2 + 1))*tan(1/2*
d*x)*tan(1/2*c) - 2*d*f*x + 2*d*e*tan(1/2*d*x) + 2*f*log(2*(tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*tan(1/2*d*x)^2*tan
(1/2*c) - 2*tan(1/2*d*x)*tan(1/2*c)^2 + tan(1/2*d*x)^2 + tan(1/2*c)^2 + 2*tan(1/2*d*x) + 2*tan(1/2*c) + 1)/(ta
n(1/2*d*x)^2*tan(1/2*c)^2 + tan(1/2*d*x)^2 + tan(1/2*c)^2 + 1))*tan(1/2*d*x) + 2*d*e*tan(1/2*c) + 2*f*log(2*(t
an(1/2*d*x)^2*tan(1/2*c)^2 - 2*tan(1/2*d*x)^2*tan(1/2*c) - 2*tan(1/2*d*x)*tan(1/2*c)^2 + tan(1/2*d*x)^2 + tan(
1/2*c)^2 + 2*tan(1/2*d*x) + 2*tan(1/2*c) + 1)/(tan(1/2*d*x)^2*tan(1/2*c)^2 + tan(1/2*d*x)^2 + tan(1/2*c)^2 + 1
))*tan(1/2*c) - 2*d*e + 2*f*log(2*(tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*tan(1/2*d*x)^2*tan(1/2*c) - 2*tan(1/2*d*x)*
tan(1/2*c)^2 + tan(1/2*d*x)^2 + tan(1/2*c)^2 + 2*tan(1/2*d*x) + 2*tan(1/2*c) + 1)/(tan(1/2*d*x)^2*tan(1/2*c)^2
+ tan(1/2*d*x)^2 + tan(1/2*c)^2 + 1)))/(a*d^2*tan(1/2*d*x)*tan(1/2*c) - a*d^2*tan(1/2*d*x) - a*d^2*tan(1/2*c)
- a*d^2)

Mupad [B] (verification not implemented)

Time = 0.98 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.05 \[ \int \frac {(e+f x) \sin (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {f\,x^2}{2\,a}-\frac {2\,f\,\ln \left ({\mathrm {e}}^{c\,1{}\mathrm {i}}\,{\mathrm {e}}^{d\,x\,1{}\mathrm {i}}+1{}\mathrm {i}\right )}{a\,d^2}+\frac {2\,\left (e+f\,x\right )}{a\,d\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1{}\mathrm {i}\right )}+\frac {x\,\left (d\,e+f\,2{}\mathrm {i}\right )}{a\,d} \]

[In]

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

[Out]

(f*x^2)/(2*a) - (2*f*log(exp(c*1i)*exp(d*x*1i) + 1i))/(a*d^2) + (2*(e + f*x))/(a*d*(exp(c*1i + d*x*1i) + 1i))
+ (x*(f*2i + d*e))/(a*d)