\(\int \frac {(e+f x)^2 \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx\) [225]

Optimal result

Integrand size = 28, antiderivative size = 479 \[ \int \frac {(e+f x)^2 \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {a (e+f x)^3}{3 b^2 f}+\frac {2 f^2 \cos (c+d x)}{b d^3}-\frac {(e+f x)^2 \cos (c+d x)}{b d}-\frac {i a^2 (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^2 \sqrt {a^2-b^2} d}+\frac {i a^2 (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^2 \sqrt {a^2-b^2} d}-\frac {2 a^2 f (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^2 \sqrt {a^2-b^2} d^2}+\frac {2 a^2 f (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^2 \sqrt {a^2-b^2} d^2}-\frac {2 i a^2 f^2 \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^2 \sqrt {a^2-b^2} d^3}+\frac {2 i a^2 f^2 \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^2 \sqrt {a^2-b^2} d^3}+\frac {2 f (e+f x) \sin (c+d x)}{b d^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.67 (sec) , antiderivative size = 479, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {4611, 3377, 2718, 32, 3404, 2296, 2221, 2611, 2320, 6724} \[ \int \frac {(e+f x)^2 \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {2 i a^2 f^2 \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^2 d^3 \sqrt {a^2-b^2}}+\frac {2 i a^2 f^2 \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^2 d^3 \sqrt {a^2-b^2}}-\frac {2 a^2 f (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^2 d^2 \sqrt {a^2-b^2}}+\frac {2 a^2 f (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^2 d^2 \sqrt {a^2-b^2}}-\frac {i a^2 (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^2 d \sqrt {a^2-b^2}}+\frac {i a^2 (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b^2 d \sqrt {a^2-b^2}}-\frac {a (e+f x)^3}{3 b^2 f}+\frac {2 f^2 \cos (c+d x)}{b d^3}+\frac {2 f (e+f x) \sin (c+d x)}{b d^2}-\frac {(e+f x)^2 \cos (c+d x)}{b d} \]

[In]

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

[Out]

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

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 2221

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]
- 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 2296

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 2320

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 2611

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 2718

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

Rule 3377

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

Rule 3404

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 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]

Rule 6724

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

Mathematica [A] (verified)

Time = 2.36 (sec) , antiderivative size = 531, normalized size of antiderivative = 1.11 \[ \int \frac {(e+f x)^2 \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {-a x \left (3 e^2+3 e f x+f^2 x^2\right )+\frac {3 i a^2 \left (-2 \sqrt {a^2-b^2} d f (e+f x) \operatorname {PolyLog}\left (2,\frac {b e^{i (c+d x)}}{-i a+\sqrt {-a^2+b^2}}\right )+2 \sqrt {a^2-b^2} d f (e+f x) \operatorname {PolyLog}\left (2,-\frac {b e^{i (c+d x)}}{i a+\sqrt {-a^2+b^2}}\right )-i \left (d^2 \left (2 \sqrt {-a^2+b^2} e^2 \arctan \left (\frac {i a+b e^{i (c+d x)}}{\sqrt {a^2-b^2}}\right )+\sqrt {a^2-b^2} f x (2 e+f x) \left (\log \left (1-\frac {b e^{i (c+d x)}}{-i a+\sqrt {-a^2+b^2}}\right )-\log \left (1+\frac {b e^{i (c+d x)}}{i a+\sqrt {-a^2+b^2}}\right )\right )\right )+2 \sqrt {a^2-b^2} f^2 \operatorname {PolyLog}\left (3,\frac {b e^{i (c+d x)}}{-i a+\sqrt {-a^2+b^2}}\right )-2 \sqrt {a^2-b^2} f^2 \operatorname {PolyLog}\left (3,-\frac {b e^{i (c+d x)}}{i a+\sqrt {-a^2+b^2}}\right )\right )\right )}{\sqrt {-\left (a^2-b^2\right )^2} d^3}-\frac {3 b \cos (d x) \left (\left (-2 f^2+d^2 (e+f x)^2\right ) \cos (c)-2 d f (e+f x) \sin (c)\right )}{d^3}+\frac {3 b \left (2 d f (e+f x) \cos (c)+\left (-2 f^2+d^2 (e+f x)^2\right ) \sin (c)\right ) \sin (d x)}{d^3}}{3 b^2} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

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. 1855 vs. \(2 (419) = 838\).

Time = 0.51 (sec) , antiderivative size = 1855, normalized size of antiderivative = 3.87 \[ \int \frac {(e+f x)^2 \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Too large to display} \]

[In]

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

[Out]

-1/6*(2*(a^3 - a*b^2)*d^3*f^2*x^3 + 6*(a^3 - a*b^2)*d^3*e*f*x^2 + 6*(a^3 - a*b^2)*d^3*e^2*x - 6*a^2*b*f^2*sqrt
(-(a^2 - b^2)/b^2)*polylog(3, -(I*a*cos(d*x + c) + a*sin(d*x + c) + (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-
(a^2 - b^2)/b^2))/b) + 6*a^2*b*f^2*sqrt(-(a^2 - b^2)/b^2)*polylog(3, -(I*a*cos(d*x + c) + a*sin(d*x + c) - (b*
cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2))/b) - 6*a^2*b*f^2*sqrt(-(a^2 - b^2)/b^2)*polylog(3, -(
-I*a*cos(d*x + c) + a*sin(d*x + c) + (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2))/b) + 6*a^2*b*
f^2*sqrt(-(a^2 - b^2)/b^2)*polylog(3, -(-I*a*cos(d*x + c) + a*sin(d*x + c) - (b*cos(d*x + c) + I*b*sin(d*x + c
))*sqrt(-(a^2 - b^2)/b^2))/b) - 6*(I*a^2*b*d*f^2*x + I*a^2*b*d*e*f)*sqrt(-(a^2 - b^2)/b^2)*dilog((I*a*cos(d*x
+ c) - a*sin(d*x + c) + (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 1) - 6*(-I*a^2*b*d
*f^2*x - I*a^2*b*d*e*f)*sqrt(-(a^2 - b^2)/b^2)*dilog((I*a*cos(d*x + c) - a*sin(d*x + c) - (b*cos(d*x + c) + I*
b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 1) - 6*(-I*a^2*b*d*f^2*x - I*a^2*b*d*e*f)*sqrt(-(a^2 - b^2)/b^
2)*dilog((-I*a*cos(d*x + c) - a*sin(d*x + c) + (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)
/b + 1) - 6*(I*a^2*b*d*f^2*x + I*a^2*b*d*e*f)*sqrt(-(a^2 - b^2)/b^2)*dilog((-I*a*cos(d*x + c) - a*sin(d*x + c)
- (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 1) - 3*(a^2*b*d^2*e^2 - 2*a^2*b*c*d*e*f
+ a^2*b*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) - 3*(a^2*b*d^2*e^2 - 2*a^2*b*c*d*e*f + a^2*b*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) + 3*(a^2*b*d^2*e^2 - 2*a^2*b*c*d*e*f + a^2*b*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) + 3*
(a^2*b*d^2*e^2 - 2*a^2*b*c*d*e*f + a^2*b*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) + 3*(a^2*b*d^2*f^2*x^2 + 2*a^2*b*d^2*e*f*x + 2*a^2*b*c*d*e*f - a^2
*b*c^2*f^2)*sqrt(-(a^2 - b^2)/b^2)*log(-(I*a*cos(d*x + c) - a*sin(d*x + c) + (b*cos(d*x + c) + I*b*sin(d*x + c
))*sqrt(-(a^2 - b^2)/b^2) - b)/b) - 3*(a^2*b*d^2*f^2*x^2 + 2*a^2*b*d^2*e*f*x + 2*a^2*b*c*d*e*f - a^2*b*c^2*f^2
)*sqrt(-(a^2 - b^2)/b^2)*log(-(I*a*cos(d*x + c) - a*sin(d*x + c) - (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(
a^2 - b^2)/b^2) - b)/b) + 3*(a^2*b*d^2*f^2*x^2 + 2*a^2*b*d^2*e*f*x + 2*a^2*b*c*d*e*f - a^2*b*c^2*f^2)*sqrt(-(a
^2 - b^2)/b^2)*log(-(-I*a*cos(d*x + c) - a*sin(d*x + c) + (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2
)/b^2) - b)/b) - 3*(a^2*b*d^2*f^2*x^2 + 2*a^2*b*d^2*e*f*x + 2*a^2*b*c*d*e*f - a^2*b*c^2*f^2)*sqrt(-(a^2 - b^2)
/b^2)*log(-(-I*a*cos(d*x + c) - a*sin(d*x + c) - (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) -
b)/b) + 6*((a^2*b - b^3)*d^2*f^2*x^2 + 2*(a^2*b - b^3)*d^2*e*f*x + (a^2*b - b^3)*d^2*e^2 - 2*(a^2*b - b^3)*f^2
)*cos(d*x + c) - 12*((a^2*b - b^3)*d*f^2*x + (a^2*b - b^3)*d*e*f)*sin(d*x + c))/((a^2*b^2 - b^4)*d^3)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F(-2)]

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

[In]

integrate((f*x+e)^2*sin(d*x+c)^2/(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 de

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

\text{Hanged}