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

Optimal. Leaf size=370 \[ -\frac{b^2 f \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{a^2 d^2 \sqrt{a^2-b^2}}+\frac{b^2 f \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{a^2 d^2 \sqrt{a^2-b^2}}-\frac{i b f \text{PolyLog}\left (2,-e^{i (c+d x)}\right )}{a^2 d^2}+\frac{i b f \text{PolyLog}\left (2,e^{i (c+d x)}\right )}{a^2 d^2}-\frac{i b^2 (e+f x) \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{a^2 d \sqrt{a^2-b^2}}+\frac{i b^2 (e+f x) \log \left (1-\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{a^2 d \sqrt{a^2-b^2}}+\frac{2 b (e+f x) \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a^2 d}+\frac{f \log (\sin (c+d x))}{a d^2}-\frac{(e+f x) \cot (c+d x)}{a d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.616411, antiderivative size = 370, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 9, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.346, Rules used = {4535, 4184, 3475, 4183, 2279, 2391, 3323, 2264, 2190} \[ -\frac{b^2 f \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{a^2 d^2 \sqrt{a^2-b^2}}+\frac{b^2 f \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{a^2 d^2 \sqrt{a^2-b^2}}-\frac{i b f \text{PolyLog}\left (2,-e^{i (c+d x)}\right )}{a^2 d^2}+\frac{i b f \text{PolyLog}\left (2,e^{i (c+d x)}\right )}{a^2 d^2}-\frac{i b^2 (e+f x) \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{a^2 d \sqrt{a^2-b^2}}+\frac{i b^2 (e+f x) \log \left (1-\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{a^2 d \sqrt{a^2-b^2}}+\frac{2 b (e+f x) \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a^2 d}+\frac{f \log (\sin (c+d x))}{a d^2}-\frac{(e+f x) \cot (c+d x)}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4535

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

Rule 4184

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 3475

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

Rule 4183

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

Rubi steps

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

Mathematica [B]  time = 11.2584, size = 933, normalized size = 2.52 \[ \frac{(d e+d f x) \left (\frac{2 (d e-c f) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}-\frac{i f \left (\log \left (1-i \tan \left (\frac{1}{2} (c+d x)\right )\right ) \log \left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )+\sqrt{b^2-a^2}}{-i a+b+\sqrt{b^2-a^2}}\right )+\text{PolyLog}\left (2,\frac{a \left (1-i \tan \left (\frac{1}{2} (c+d x)\right )\right )}{a+i \left (b+\sqrt{b^2-a^2}\right )}\right )\right )}{\sqrt{b^2-a^2}}+\frac{i f \left (\log \left (i \tan \left (\frac{1}{2} (c+d x)\right )+1\right ) \log \left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )+\sqrt{b^2-a^2}}{i a+b+\sqrt{b^2-a^2}}\right )+\text{PolyLog}\left (2,\frac{a \left (i \tan \left (\frac{1}{2} (c+d x)\right )+1\right )}{a-i \left (b+\sqrt{b^2-a^2}\right )}\right )\right )}{\sqrt{b^2-a^2}}+\frac{i f \left (\log \left (1-i \tan \left (\frac{1}{2} (c+d x)\right )\right ) \log \left (-\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )-\sqrt{b^2-a^2}}{i a-b+\sqrt{b^2-a^2}}\right )+\text{PolyLog}\left (2,\frac{a \left (\tan \left (\frac{1}{2} (c+d x)\right )+i\right )}{i a-b+\sqrt{b^2-a^2}}\right )\right )}{\sqrt{b^2-a^2}}-\frac{i f \left (\log \left (i \tan \left (\frac{1}{2} (c+d x)\right )+1\right ) \log \left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )-\sqrt{b^2-a^2}}{i a+b-\sqrt{b^2-a^2}}\right )+\text{PolyLog}\left (2,\frac{i \tan \left (\frac{1}{2} (c+d x)\right ) a+a}{a+i \left (\sqrt{b^2-a^2}-b\right )}\right )\right )}{\sqrt{b^2-a^2}}\right ) b^2}{a^2 d^2 \left (d e-c f+i f \log \left (1-i \tan \left (\frac{1}{2} (c+d x)\right )\right )-i f \log \left (i \tan \left (\frac{1}{2} (c+d x)\right )+1\right )\right )}-\frac{e \log \left (\tan \left (\frac{1}{2} (c+d x)\right )\right ) b}{a^2 d}+\frac{c f \log \left (\tan \left (\frac{1}{2} (c+d x)\right )\right ) b}{a^2 d^2}-\frac{f \left ((c+d x) \left (\log \left (1-e^{i (c+d x)}\right )-\log \left (1+e^{i (c+d x)}\right )\right )+i \left (\text{PolyLog}\left (2,-e^{i (c+d x)}\right )-\text{PolyLog}\left (2,e^{i (c+d x)}\right )\right )\right ) b}{a^2 d^2}+\frac{\left (-d e \cos \left (\frac{1}{2} (c+d x)\right )+c f \cos \left (\frac{1}{2} (c+d x)\right )-f (c+d x) \cos \left (\frac{1}{2} (c+d x)\right )\right ) \csc \left (\frac{1}{2} (c+d x)\right )}{2 a d^2}+\frac{f \log (\sin (c+d x))}{a d^2}+\frac{\sec \left (\frac{1}{2} (c+d x)\right ) \left (d e \sin \left (\frac{1}{2} (c+d x)\right )-c f \sin \left (\frac{1}{2} (c+d x)\right )+f (c+d x) \sin \left (\frac{1}{2} (c+d x)\right )\right )}{2 a d^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-(d*e*Cos[(c + d*x)/2]) + c*f*Cos[(c + d*x)/2] - f*(c + d*x)*Cos[(c + d*x)/2])*Csc[(c + d*x)/2])/(2*a*d^2) +
(f*Log[Sin[c + d*x]])/(a*d^2) - (b*e*Log[Tan[(c + d*x)/2]])/(a^2*d) + (b*c*f*Log[Tan[(c + d*x)/2]])/(a^2*d^2)
- (b*f*((c + d*x)*(Log[1 - E^(I*(c + d*x))] - Log[1 + E^(I*(c + d*x))]) + I*(PolyLog[2, -E^(I*(c + d*x))] - P
olyLog[2, E^(I*(c + d*x))])))/(a^2*d^2) + (b^2*(d*e + d*f*x)*((2*(d*e - c*f)*ArcTan[(b + a*Tan[(c + d*x)/2])/S
qrt[a^2 - b^2]])/Sqrt[a^2 - b^2] - (I*f*(Log[1 - I*Tan[(c + d*x)/2]]*Log[(b + Sqrt[-a^2 + b^2] + a*Tan[(c + d*
x)/2])/((-I)*a + b + Sqrt[-a^2 + b^2])] + PolyLog[2, (a*(1 - I*Tan[(c + d*x)/2]))/(a + I*(b + Sqrt[-a^2 + b^2]
))]))/Sqrt[-a^2 + b^2] + (I*f*(Log[1 + I*Tan[(c + d*x)/2]]*Log[(b + Sqrt[-a^2 + b^2] + a*Tan[(c + d*x)/2])/(I*
a + b + Sqrt[-a^2 + b^2])] + PolyLog[2, (a*(1 + I*Tan[(c + d*x)/2]))/(a - I*(b + Sqrt[-a^2 + b^2]))]))/Sqrt[-a
^2 + b^2] + (I*f*(Log[1 - I*Tan[(c + d*x)/2]]*Log[-((b - Sqrt[-a^2 + b^2] + a*Tan[(c + d*x)/2])/(I*a - b + Sqr
t[-a^2 + b^2]))] + PolyLog[2, (a*(I + Tan[(c + d*x)/2]))/(I*a - b + Sqrt[-a^2 + b^2])]))/Sqrt[-a^2 + b^2] - (I
*f*(Log[1 + I*Tan[(c + d*x)/2]]*Log[(b - Sqrt[-a^2 + b^2] + a*Tan[(c + d*x)/2])/(I*a + b - Sqrt[-a^2 + b^2])]
+ PolyLog[2, (a + I*a*Tan[(c + d*x)/2])/(a + I*(-b + Sqrt[-a^2 + b^2]))]))/Sqrt[-a^2 + b^2]))/(a^2*d^2*(d*e -
c*f + I*f*Log[1 - I*Tan[(c + d*x)/2]] - I*f*Log[1 + I*Tan[(c + d*x)/2]])) + (Sec[(c + d*x)/2]*(d*e*Sin[(c + d*
x)/2] - c*f*Sin[(c + d*x)/2] + f*(c + d*x)*Sin[(c + d*x)/2]))/(2*a*d^2)

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Maple [B]  time = 0.194, size = 766, 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((f*x+e)*csc(d*x+c)^2/(a+b*sin(d*x+c)),x)

[Out]

2*I/d/a^2*b^2*e/(-a^2+b^2)^(1/2)*arctan(1/2*(2*I*b*exp(I*(d*x+c))-2*a)/(-a^2+b^2)^(1/2))-I/d^2/a^2*b*f*dilog(e
xp(I*(d*x+c))+1)-2*I*(f*x+e)/d/a/(exp(2*I*(d*x+c))-1)-2/d^2/a*f*ln(exp(I*(d*x+c)))+I/d^2/a^2*b^2*f/(-a^2+b^2)^
(1/2)*dilog((I*a+b*exp(I*(d*x+c))+(-a^2+b^2)^(1/2))/(I*a+(-a^2+b^2)^(1/2)))-I/d^2/a^2*b*f*dilog(exp(I*(d*x+c))
)+1/d/a^2*b^2*f/(-a^2+b^2)^(1/2)*ln((I*a+b*exp(I*(d*x+c))-(-a^2+b^2)^(1/2))/(I*a-(-a^2+b^2)^(1/2)))*x+1/d^2/a^
2*b^2*f/(-a^2+b^2)^(1/2)*ln((I*a+b*exp(I*(d*x+c))-(-a^2+b^2)^(1/2))/(I*a-(-a^2+b^2)^(1/2)))*c+1/d^2/a*f*ln(exp
(I*(d*x+c))-1)+1/d^2/a*f*ln(exp(I*(d*x+c))+1)-1/d/a^2*b*e*ln(exp(I*(d*x+c))-1)+1/d/a^2*b*e*ln(exp(I*(d*x+c))+1
)+1/d^2/a^2*b*f*c*ln(exp(I*(d*x+c))-1)+1/d/a^2*b*f*ln(exp(I*(d*x+c))+1)*x-2*I/d^2/a^2*b^2*f*c/(-a^2+b^2)^(1/2)
*arctan(1/2*(2*I*b*exp(I*(d*x+c))-2*a)/(-a^2+b^2)^(1/2))-1/d/a^2*b^2*f/(-a^2+b^2)^(1/2)*ln((I*a+b*exp(I*(d*x+c
))+(-a^2+b^2)^(1/2))/(I*a+(-a^2+b^2)^(1/2)))*x-1/d^2/a^2*b^2*f/(-a^2+b^2)^(1/2)*ln((I*a+b*exp(I*(d*x+c))+(-a^2
+b^2)^(1/2))/(I*a+(-a^2+b^2)^(1/2)))*c-I/d^2/a^2*b^2*f/(-a^2+b^2)^(1/2)*dilog((I*a+b*exp(I*(d*x+c))-(-a^2+b^2)
^(1/2))/(I*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((f*x+e)*csc(d*x+c)^2/(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 4.48821, size = 4132, normalized size = 11.17 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out