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

Optimal. Leaf size=460 \[ \frac{2 f \sqrt{a^2-b^2} (e+f x) \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b^2 d^2}-\frac{2 f \sqrt{a^2-b^2} (e+f x) \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{b^2 d^2}+\frac{2 i f^2 \sqrt{a^2-b^2} \text{PolyLog}\left (3,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b^2 d^3}-\frac{2 i f^2 \sqrt{a^2-b^2} \text{PolyLog}\left (3,\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{b^2 d^3}+\frac{i \sqrt{a^2-b^2} (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b^2 d}-\frac{i \sqrt{a^2-b^2} (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{b^2 d}+\frac{a (e+f x)^3}{3 b^2 f}-\frac{2 f (e+f x) \sin (c+d x)}{b d^2}-\frac{2 f^2 \cos (c+d x)}{b d^3}+\frac{(e+f x)^2 \cos (c+d x)}{b d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.928734, antiderivative size = 460, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 10, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {4525, 32, 3296, 2638, 3323, 2264, 2190, 2531, 2282, 6589} \[ \frac{2 f \sqrt{a^2-b^2} (e+f x) \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b^2 d^2}-\frac{2 f \sqrt{a^2-b^2} (e+f x) \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{b^2 d^2}+\frac{2 i f^2 \sqrt{a^2-b^2} \text{PolyLog}\left (3,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b^2 d^3}-\frac{2 i f^2 \sqrt{a^2-b^2} \text{PolyLog}\left (3,\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{b^2 d^3}+\frac{i \sqrt{a^2-b^2} (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b^2 d}-\frac{i \sqrt{a^2-b^2} (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{b^2 d}+\frac{a (e+f x)^3}{3 b^2 f}-\frac{2 f (e+f x) \sin (c+d x)}{b d^2}-\frac{2 f^2 \cos (c+d x)}{b d^3}+\frac{(e+f x)^2 \cos (c+d x)}{b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4525

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

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 3296

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 2638

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

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

Mathematica [A]  time = 2.62308, size = 536, normalized size = 1.17 \[ \frac{\frac{3 i \left (b^2-a^2\right ) \left (-i \left (2 f^2 \sqrt{a^2-b^2} \text{PolyLog}\left (3,\frac{b e^{i (c+d x)}}{\sqrt{b^2-a^2}-i a}\right )-2 f^2 \sqrt{a^2-b^2} \text{PolyLog}\left (3,-\frac{b e^{i (c+d x)}}{\sqrt{b^2-a^2}+i a}\right )+d^2 \left (2 e^2 \sqrt{b^2-a^2} \tan ^{-1}\left (\frac{i a+b e^{i (c+d x)}}{\sqrt{a^2-b^2}}\right )+f x \sqrt{a^2-b^2} (2 e+f x) \left (\log \left (1-\frac{b e^{i (c+d x)}}{\sqrt{b^2-a^2}-i a}\right )-\log \left (1+\frac{b e^{i (c+d x)}}{\sqrt{b^2-a^2}+i a}\right )\right )\right )\right )-2 d f \sqrt{a^2-b^2} (e+f x) \text{PolyLog}\left (2,\frac{b e^{i (c+d x)}}{\sqrt{b^2-a^2}-i a}\right )+2 d f \sqrt{a^2-b^2} (e+f x) \text{PolyLog}\left (2,-\frac{b e^{i (c+d x)}}{\sqrt{b^2-a^2}+i a}\right )\right )}{d^3 \sqrt{-\left (a^2-b^2\right )^2}}+a x \left (3 e^2+3 e f x+f^2 x^2\right )+\frac{3 b \cos (d x) \left (\cos (c) \left (d^2 (e+f x)^2-2 f^2\right )-2 d f \sin (c) (e+f x)\right )}{d^3}-\frac{3 b \sin (d x) \left (\sin (c) \left (d^2 (e+f x)^2-2 f^2\right )+2 d f \cos (c) (e+f x)\right )}{d^3}}{3 b^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((e + f*x)^2*Cos[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 + b^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
+ Sqrt[-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]] + Sq
rt[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]  time = 0.956, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( fx+e \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{a+b\sin \left ( dx+c \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [C]  time = 3.1454, size = 3931, normalized size = 8.55 \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)^2*cos(d*x+c)^2/(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/6*(2*a*d^3*f^2*x^3 + 6*a*d^3*e*f*x^2 + 6*a*d^3*e^2*x + 6*b*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) - 6*b*f^2*s
qrt(-(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) - 6*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*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
*b*d*f^2*x - 6*I*b*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) + (6*I*b*d*f^2*x + 6*I*b*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) + (6*I*b*d*f^2*x + 6*I*b*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) + (-6*I*b
*d*f^2*x - 6*I*b*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) - 3*(b*d^2*e^2 - 2*b*c*d*e*f + b*c^2*f^2)*sqr
t(-(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*(b*d^2
*e^2 - 2*b*c*d*e*f + 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*(b*d^2*e^2 - 2*b*c*d*e*f + 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*(b*d^2*e^2 - 2*b*c*d*e*f + 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*(b*d^2*f^
2*x^2 + 2*b*d^2*e*f*x + 2*b*c*d*e*f - b*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) + 3*(b*d^2*f^2*x^2 + 2*b*d^2
*e*f*x + 2*b*c*d*e*f - b*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) - 3*(b*d^2*f^2*x^2 + 2*b*d^2*e*f*x + 2*b*c*
d*e*f - b*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) + 3*(b*d^2*f^2*x^2 + 2*b*d^2*e*f*x + 2*b*c*d*e*f - b*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) + 6*(b*d^2*f^2*x^2 + 2*b*d^2*e*f*x + b*d^2*e^2 - 2*b*f^2)*cos(d*x + c)
- 12*(b*d*f^2*x + b*d*e*f)*sin(d*x + c))/(b^2*d^3)

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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