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

Optimal. Leaf size=305 \[ -\frac{a d \text{PolyLog}\left (2,\frac{i b e^{i (e+f x)}}{a-\sqrt{a^2-b^2}}\right )}{f^2 \left (a^2-b^2\right )^{3/2}}+\frac{a d \text{PolyLog}\left (2,\frac{i b e^{i (e+f x)}}{\sqrt{a^2-b^2}+a}\right )}{f^2 \left (a^2-b^2\right )^{3/2}}-\frac{i a (c+d x) \log \left (1-\frac{i b e^{i (e+f x)}}{a-\sqrt{a^2-b^2}}\right )}{f \left (a^2-b^2\right )^{3/2}}+\frac{i a (c+d x) \log \left (1-\frac{i b e^{i (e+f x)}}{\sqrt{a^2-b^2}+a}\right )}{f \left (a^2-b^2\right )^{3/2}}+\frac{b (c+d x) \cos (e+f x)}{f \left (a^2-b^2\right ) (a+b \sin (e+f x))}-\frac{d \log (a+b \sin (e+f x))}{f^2 \left (a^2-b^2\right )} \]

[Out]

((-I)*a*(c + d*x)*Log[1 - (I*b*E^(I*(e + f*x)))/(a - Sqrt[a^2 - b^2])])/((a^2 - b^2)^(3/2)*f) + (I*a*(c + d*x)
*Log[1 - (I*b*E^(I*(e + f*x)))/(a + Sqrt[a^2 - b^2])])/((a^2 - b^2)^(3/2)*f) - (d*Log[a + b*Sin[e + f*x]])/((a
^2 - b^2)*f^2) - (a*d*PolyLog[2, (I*b*E^(I*(e + f*x)))/(a - Sqrt[a^2 - b^2])])/((a^2 - b^2)^(3/2)*f^2) + (a*d*
PolyLog[2, (I*b*E^(I*(e + f*x)))/(a + Sqrt[a^2 - b^2])])/((a^2 - b^2)^(3/2)*f^2) + (b*(c + d*x)*Cos[e + f*x])/
((a^2 - b^2)*f*(a + b*Sin[e + f*x]))

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Rubi [A]  time = 0.550363, antiderivative size = 305, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.444, Rules used = {3324, 3323, 2264, 2190, 2279, 2391, 2668, 31} \[ -\frac{a d \text{PolyLog}\left (2,\frac{i b e^{i (e+f x)}}{a-\sqrt{a^2-b^2}}\right )}{f^2 \left (a^2-b^2\right )^{3/2}}+\frac{a d \text{PolyLog}\left (2,\frac{i b e^{i (e+f x)}}{\sqrt{a^2-b^2}+a}\right )}{f^2 \left (a^2-b^2\right )^{3/2}}-\frac{i a (c+d x) \log \left (1-\frac{i b e^{i (e+f x)}}{a-\sqrt{a^2-b^2}}\right )}{f \left (a^2-b^2\right )^{3/2}}+\frac{i a (c+d x) \log \left (1-\frac{i b e^{i (e+f x)}}{\sqrt{a^2-b^2}+a}\right )}{f \left (a^2-b^2\right )^{3/2}}+\frac{b (c+d x) \cos (e+f x)}{f \left (a^2-b^2\right ) (a+b \sin (e+f x))}-\frac{d \log (a+b \sin (e+f x))}{f^2 \left (a^2-b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*a*(c + d*x)*Log[1 - (I*b*E^(I*(e + f*x)))/(a - Sqrt[a^2 - b^2])])/((a^2 - b^2)^(3/2)*f) + (I*a*(c + d*x)
*Log[1 - (I*b*E^(I*(e + f*x)))/(a + Sqrt[a^2 - b^2])])/((a^2 - b^2)^(3/2)*f) - (d*Log[a + b*Sin[e + f*x]])/((a
^2 - b^2)*f^2) - (a*d*PolyLog[2, (I*b*E^(I*(e + f*x)))/(a - Sqrt[a^2 - b^2])])/((a^2 - b^2)^(3/2)*f^2) + (a*d*
PolyLog[2, (I*b*E^(I*(e + f*x)))/(a + Sqrt[a^2 - b^2])])/((a^2 - b^2)^(3/2)*f^2) + (b*(c + d*x)*Cos[e + f*x])/
((a^2 - b^2)*f*(a + b*Sin[e + f*x]))

Rule 3324

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

Rule 3323

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[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 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 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 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 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[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] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int \frac{c+d x}{(a+b \sin (e+f x))^2} \, dx &=\frac{b (c+d x) \cos (e+f x)}{\left (a^2-b^2\right ) f (a+b \sin (e+f x))}+\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}{\left (a^2-b^2\right ) f}\\ &=\frac{b (c+d x) \cos (e+f x)}{\left (a^2-b^2\right ) f (a+b \sin (e+f x))}+\frac{(2 a) \int \frac{e^{i (e+f x)} (c+d x)}{i b+2 a e^{i (e+f x)}-i b e^{2 i (e+f x)}} \, dx}{a^2-b^2}-\frac{d \operatorname{Subst}\left (\int \frac{1}{a+x} \, dx,x,b \sin (e+f x)\right )}{\left (a^2-b^2\right ) f^2}\\ &=-\frac{d \log (a+b \sin (e+f x))}{\left (a^2-b^2\right ) f^2}+\frac{b (c+d x) \cos (e+f x)}{\left (a^2-b^2\right ) f (a+b \sin (e+f x))}-\frac{(2 i a b) \int \frac{e^{i (e+f x)} (c+d x)}{2 a-2 \sqrt{a^2-b^2}-2 i b e^{i (e+f x)}} \, dx}{\left (a^2-b^2\right )^{3/2}}+\frac{(2 i a b) \int \frac{e^{i (e+f x)} (c+d x)}{2 a+2 \sqrt{a^2-b^2}-2 i b e^{i (e+f x)}} \, dx}{\left (a^2-b^2\right )^{3/2}}\\ &=-\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{b (c+d x) \cos (e+f x)}{\left (a^2-b^2\right ) f (a+b \sin (e+f x))}+\frac{(i a d) \int \log \left (1-\frac{2 i b e^{i (e+f x)}}{2 a-2 \sqrt{a^2-b^2}}\right ) \, dx}{\left (a^2-b^2\right )^{3/2} f}-\frac{(i a d) \int \log \left (1-\frac{2 i b e^{i (e+f x)}}{2 a+2 \sqrt{a^2-b^2}}\right ) \, dx}{\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{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{b (c+d x) \cos (e+f x)}{\left (a^2-b^2\right ) f (a+b \sin (e+f x))}+\frac{(a d) \operatorname{Subst}\left (\int \frac{\log \left (1-\frac{2 i b x}{2 a-2 \sqrt{a^2-b^2}}\right )}{x} \, dx,x,e^{i (e+f x)}\right )}{\left (a^2-b^2\right )^{3/2} f^2}-\frac{(a d) \operatorname{Subst}\left (\int \frac{\log \left (1-\frac{2 i b x}{2 a+2 \sqrt{a^2-b^2}}\right )}{x} \, dx,x,e^{i (e+f x)}\right )}{\left (a^2-b^2\right )^{3/2} f^2}\\ &=-\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 \text{Li}_2\left (\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 \text{Li}_2\left (\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))}\\ \end{align*}

Mathematica [A]  time = 0.988389, size = 236, normalized size = 0.77 \[ \frac{\frac{a \left (-d \text{PolyLog}\left (2,-\frac{i b e^{i (e+f x)}}{\sqrt{a^2-b^2}-a}\right )+d \text{PolyLog}\left (2,\frac{i b e^{i (e+f x)}}{\sqrt{a^2-b^2}+a}\right )-i f (c+d x) \left (\log \left (1+\frac{i b e^{i (e+f x)}}{\sqrt{a^2-b^2}-a}\right )-\log \left (1-\frac{i b e^{i (e+f x)}}{\sqrt{a^2-b^2}+a}\right )\right )\right )}{\sqrt{a^2-b^2}}+\frac{b f (c+d x) \cos (e+f x)}{a+b \sin (e+f x)}-d \log (a+b \sin (e+f x))}{f^2 \left (a^2-b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.789, size = 641, normalized size = 2.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(d*x+c)*(I*b+a*exp(I*(f*x+e)))/f/(a^2-b^2)/(b*exp(2*I*(f*x+e))-b+2*I*a*exp(I*(f*x+e)))-2/f^2/(-a^2+b^2)*d*ln
(exp(I*(f*x+e)))+1/f^2/(-a^2+b^2)*d*ln(I*exp(2*I*(f*x+e))*b-I*b-2*a*exp(I*(f*x+e)))+I/f^2/(-a^2+b^2)^(3/2)*d*a
*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/(-a^2+b^2)^(3/2)*d*a*dilog((I*a
+b*exp(I*(f*x+e))+(-a^2+b^2)^(1/2))/(I*a+(-a^2+b^2)^(1/2)))+1/f/(-a^2+b^2)^(3/2)*d*a*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/(-a^2+b^2)^(3/2)*d*a*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/(-a^2+b^2)^(3/2)*d*a*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/(-a^2+b^2)^(3/2)*d*a*ln((-I*a-b*exp(I*(f*x+e))+(-a^2+b^2)^(1/2))/(-I*a+(-a^2+b^2)^(1/2))
)*e-2*I/f/(-a^2+b^2)^(3/2)*a*c*arctan(1/2*(2*I*b*exp(I*(f*x+e))-2*a)/(-a^2+b^2)^(1/2))+2*I/f^2/(-a^2+b^2)^(3/2
)*a*d*e*arctan(1/2*(2*I*b*exp(I*(f*x+e))-2*a)/(-a^2+b^2)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 3.46769, size = 3549, normalized size = 11.64 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((I*a*b^2*d*sin(f*x + e) + I*a^2*b*d)*sqrt(-(a^2 - b^2)/b^2)*dilog(-1/2*(2*I*a*cos(f*x + e) + 2*a*sin(f*x
+ e) + 2*(b*cos(f*x + e) - I*b*sin(f*x + e))*sqrt(-(a^2 - b^2)/b^2) + 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(-1/2*(2*I*a*cos(f*x + e) + 2*a*sin(f*x + e) - 2*(b*cos(f*x + e) - I*b
*sin(f*x + e))*sqrt(-(a^2 - b^2)/b^2) + 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(-1/2*(-2*I*a*cos(f*x + e) + 2*a*sin(f*x + e) + 2*(b*cos(f*x + e) + I*b*sin(f*x + e))*sqrt(-(a^2 - b
^2)/b^2) + 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(-1/2*(-2*I*a*cos(f*
x + e) + 2*a*sin(f*x + e) - 2*(b*cos(f*x + e) + I*b*sin(f*x + e))*sqrt(-(a^2 - b^2)/b^2) + 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(1/2*(2*I*a*cos(f*x +
e) + 2*a*sin(f*x + e) + 2*(b*cos(f*x + e) - I*b*sin(f*x + e))*sqrt(-(a^2 - b^2)/b^2) + 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(1/2*(2*I*a*cos(f*x + e) + 2*a
*sin(f*x + e) - 2*(b*cos(f*x + e) - I*b*sin(f*x + e))*sqrt(-(a^2 - b^2)/b^2) + 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(1/2*(-2*I*a*cos(f*x + e) + 2*a*sin(f*
x + e) + 2*(b*cos(f*x + e) + I*b*sin(f*x + e))*sqrt(-(a^2 - b^2)/b^2) + 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(1/2*(-2*I*a*cos(f*x + e) + 2*a*sin(f*x + e)
- 2*(b*cos(f*x + e) + I*b*sin(f*x + e))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b) + 2*((a^2*b - b^3)*d*f*x + (a^2*b - b
^3)*c*f)*cos(f*x + e) - ((a^2*b - b^3)*d*sin(f*x + e) + (a^3 - a*b^2)*d + (a^2*b*d*e - a^2*b*c*f + (a*b^2*d*e
- a*b^2*c*f)*sin(f*x + e))*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) - ((a^2*b - b^3)*d*sin(f*x + e) + (a^3 - a*b^2)*d + (a^2*b*d*e - a^2*b*c*f + (a*b^2*d*e -
a*b^2*c*f)*sin(f*x + e))*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) - ((a^2*b - b^3)*d*sin(f*x + e) + (a^3 - a*b^2)*d - (a^2*b*d*e - a^2*b*c*f + (a*b^2*d*e -
a*b^2*c*f)*sin(f*x + e))*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) - ((a^2*b - b^3)*d*sin(f*x + e) + (a^3 - a*b^2)*d - (a^2*b*d*e - a^2*b*c*f + (a*b^2*d*e -
a*b^2*c*f)*sin(f*x + e))*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))/((a^4*b - 2*a^2*b^3 + b^5)*f^2*sin(f*x + e) + (a^5 - 2*a^3*b^2 + a*b^4)*f^2)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{d x + c}{{\left (b \sin \left (f x + e\right ) + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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