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

Optimal. Leaf size=528 \[ \frac {2 i b f^2 \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a d^3 \sqrt {a^2-b^2}}-\frac {2 i b f^2 \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a d^3 \sqrt {a^2-b^2}}+\frac {2 b f (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a d^2 \sqrt {a^2-b^2}}-\frac {2 b f (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a d^2 \sqrt {a^2-b^2}}+\frac {i b (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a d \sqrt {a^2-b^2}}-\frac {i b (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{a d \sqrt {a^2-b^2}}-\frac {2 f^2 \text {Li}_3\left (-e^{i (c+d x)}\right )}{a d^3}+\frac {2 f^2 \text {Li}_3\left (e^{i (c+d x)}\right )}{a d^3}+\frac {2 i f (e+f x) \text {Li}_2\left (-e^{i (c+d x)}\right )}{a d^2}-\frac {2 i f (e+f x) \text {Li}_2\left (e^{i (c+d x)}\right )}{a d^2}-\frac {2 (e+f x)^2 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.95, antiderivative size = 528, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 8, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {4535, 4183, 2531, 2282, 6589, 3323, 2264, 2190} \[ \frac {2 b f (e+f x) \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a d^2 \sqrt {a^2-b^2}}-\frac {2 b f (e+f x) \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{a d^2 \sqrt {a^2-b^2}}+\frac {2 i b f^2 \text {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a d^3 \sqrt {a^2-b^2}}-\frac {2 i b f^2 \text {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{a d^3 \sqrt {a^2-b^2}}+\frac {2 i f (e+f x) \text {PolyLog}\left (2,-e^{i (c+d x)}\right )}{a d^2}-\frac {2 i f (e+f x) \text {PolyLog}\left (2,e^{i (c+d x)}\right )}{a d^2}-\frac {2 f^2 \text {PolyLog}\left (3,-e^{i (c+d x)}\right )}{a d^3}+\frac {2 f^2 \text {PolyLog}\left (3,e^{i (c+d x)}\right )}{a d^3}+\frac {i b (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a d \sqrt {a^2-b^2}}-\frac {i b (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{a d \sqrt {a^2-b^2}}-\frac {2 (e+f x)^2 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[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 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 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 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin {align*} \int \frac {(e+f x)^2 \csc (c+d x)}{a+b \sin (c+d x)} \, dx &=\frac {\int (e+f x)^2 \csc (c+d x) \, dx}{a}-\frac {b \int \frac {(e+f x)^2}{a+b \sin (c+d x)} \, dx}{a}\\ &=-\frac {2 (e+f x)^2 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}-\frac {(2 b) \int \frac {e^{i (c+d x)} (e+f x)^2}{i b+2 a e^{i (c+d x)}-i b e^{2 i (c+d x)}} \, dx}{a}-\frac {(2 f) \int (e+f x) \log \left (1-e^{i (c+d x)}\right ) \, dx}{a d}+\frac {(2 f) \int (e+f x) \log \left (1+e^{i (c+d x)}\right ) \, dx}{a d}\\ &=-\frac {2 (e+f x)^2 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}+\frac {2 i f (e+f x) \text {Li}_2\left (-e^{i (c+d x)}\right )}{a d^2}-\frac {2 i f (e+f x) \text {Li}_2\left (e^{i (c+d x)}\right )}{a d^2}+\frac {\left (2 i b^2\right ) \int \frac {e^{i (c+d x)} (e+f x)^2}{2 a-2 \sqrt {a^2-b^2}-2 i b e^{i (c+d x)}} \, dx}{a \sqrt {a^2-b^2}}-\frac {\left (2 i b^2\right ) \int \frac {e^{i (c+d x)} (e+f x)^2}{2 a+2 \sqrt {a^2-b^2}-2 i b e^{i (c+d x)}} \, dx}{a \sqrt {a^2-b^2}}-\frac {\left (2 i f^2\right ) \int \text {Li}_2\left (-e^{i (c+d x)}\right ) \, dx}{a d^2}+\frac {\left (2 i f^2\right ) \int \text {Li}_2\left (e^{i (c+d x)}\right ) \, dx}{a d^2}\\ &=-\frac {2 (e+f x)^2 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}+\frac {i b (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a \sqrt {a^2-b^2} d}-\frac {i b (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a \sqrt {a^2-b^2} d}+\frac {2 i f (e+f x) \text {Li}_2\left (-e^{i (c+d x)}\right )}{a d^2}-\frac {2 i f (e+f x) \text {Li}_2\left (e^{i (c+d x)}\right )}{a d^2}-\frac {(2 i b f) \int (e+f x) \log \left (1-\frac {2 i b e^{i (c+d x)}}{2 a-2 \sqrt {a^2-b^2}}\right ) \, dx}{a \sqrt {a^2-b^2} d}+\frac {(2 i b f) \int (e+f x) \log \left (1-\frac {2 i b e^{i (c+d x)}}{2 a+2 \sqrt {a^2-b^2}}\right ) \, dx}{a \sqrt {a^2-b^2} d}-\frac {\left (2 f^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{i (c+d x)}\right )}{a d^3}+\frac {\left (2 f^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^{i (c+d x)}\right )}{a d^3}\\ &=-\frac {2 (e+f x)^2 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}+\frac {i b (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a \sqrt {a^2-b^2} d}-\frac {i b (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a \sqrt {a^2-b^2} d}+\frac {2 i f (e+f x) \text {Li}_2\left (-e^{i (c+d x)}\right )}{a d^2}-\frac {2 i f (e+f x) \text {Li}_2\left (e^{i (c+d x)}\right )}{a d^2}+\frac {2 b f (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a \sqrt {a^2-b^2} d^2}-\frac {2 b f (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a \sqrt {a^2-b^2} d^2}-\frac {2 f^2 \text {Li}_3\left (-e^{i (c+d x)}\right )}{a d^3}+\frac {2 f^2 \text {Li}_3\left (e^{i (c+d x)}\right )}{a d^3}-\frac {\left (2 b f^2\right ) \int \text {Li}_2\left (\frac {2 i b e^{i (c+d x)}}{2 a-2 \sqrt {a^2-b^2}}\right ) \, dx}{a \sqrt {a^2-b^2} d^2}+\frac {\left (2 b f^2\right ) \int \text {Li}_2\left (\frac {2 i b e^{i (c+d x)}}{2 a+2 \sqrt {a^2-b^2}}\right ) \, dx}{a \sqrt {a^2-b^2} d^2}\\ &=-\frac {2 (e+f x)^2 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}+\frac {i b (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a \sqrt {a^2-b^2} d}-\frac {i b (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a \sqrt {a^2-b^2} d}+\frac {2 i f (e+f x) \text {Li}_2\left (-e^{i (c+d x)}\right )}{a d^2}-\frac {2 i f (e+f x) \text {Li}_2\left (e^{i (c+d x)}\right )}{a d^2}+\frac {2 b f (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a \sqrt {a^2-b^2} d^2}-\frac {2 b f (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a \sqrt {a^2-b^2} d^2}-\frac {2 f^2 \text {Li}_3\left (-e^{i (c+d x)}\right )}{a d^3}+\frac {2 f^2 \text {Li}_3\left (e^{i (c+d x)}\right )}{a d^3}+\frac {\left (2 i b f^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (\frac {i b x}{a-\sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{a \sqrt {a^2-b^2} d^3}-\frac {\left (2 i b f^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (\frac {i b x}{a+\sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{a \sqrt {a^2-b^2} d^3}\\ &=-\frac {2 (e+f x)^2 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}+\frac {i b (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a \sqrt {a^2-b^2} d}-\frac {i b (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a \sqrt {a^2-b^2} d}+\frac {2 i f (e+f x) \text {Li}_2\left (-e^{i (c+d x)}\right )}{a d^2}-\frac {2 i f (e+f x) \text {Li}_2\left (e^{i (c+d x)}\right )}{a d^2}+\frac {2 b f (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a \sqrt {a^2-b^2} d^2}-\frac {2 b f (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a \sqrt {a^2-b^2} d^2}-\frac {2 f^2 \text {Li}_3\left (-e^{i (c+d x)}\right )}{a d^3}+\frac {2 f^2 \text {Li}_3\left (e^{i (c+d x)}\right )}{a d^3}+\frac {2 i b f^2 \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a \sqrt {a^2-b^2} d^3}-\frac {2 i b f^2 \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a \sqrt {a^2-b^2} d^3}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 1.58, size = 573, normalized size = 1.09 \[ \frac {\frac {b \left (i \left (2 i d^2 e^2 \tan ^{-1}\left (\frac {i a+b e^{i (c+d x)}}{\sqrt {a^2-b^2}}\right )+2 d^2 e f x \log \left (1+\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}-a}\right )-2 d^2 e f x \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )+d^2 f^2 x^2 \log \left (1+\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}-a}\right )-d^2 f^2 x^2 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )+2 i d f (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )+2 f^2 \text {Li}_3\left (-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}-a}\right )-2 f^2 \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )\right )+2 d f (e+f x) \text {Li}_2\left (-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}-a}\right )\right )}{\sqrt {a^2-b^2}}+d^2 (e+f x)^2 \log \left (1-e^{i (c+d x)}\right )-d^2 (e+f x)^2 \log \left (1+e^{i (c+d x)}\right )+2 i d f (e+f x) \text {Li}_2\left (-e^{i (c+d x)}\right )-2 i d f (e+f x) \text {Li}_2\left (e^{i (c+d x)}\right )-2 f^2 \text {Li}_3\left (-e^{i (c+d x)}\right )+2 f^2 \text {Li}_3\left (e^{i (c+d x)}\right )}{a d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [C]  time = 0.68, size = 2432, normalized size = 4.61 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [F]  time = 0.99, size = 0, normalized size = 0.00 \[ \int \frac {\left (f x +e \right )^{2} \csc \left (d x +c \right )}{a +b \sin \left (d x +c \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

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mupad [F(-1)]  time = 0.00, size = -1, normalized size = -0.00 \[ \text {Hanged} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

\text{Hanged}

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e + f x\right )^{2} \csc {\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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