3.3.99 \(\int \frac {(e+f x)^2 \cos ^2(c+d x)}{a+b \sin (c+d x)} \, dx\) [299]

Optimal. Leaf size=460 \[ \frac {a (e+f x)^3}{3 b^2 f}-\frac {2 f^2 \cos (c+d x)}{b d^3}+\frac {(e+f x)^2 \cos (c+d x)}{b d}+\frac {i \sqrt {a^2-b^2} (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^2 d}-\frac {i \sqrt {a^2-b^2} (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^2 d}+\frac {2 \sqrt {a^2-b^2} f (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^2 d^2}-\frac {2 \sqrt {a^2-b^2} f (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^2 d^2}+\frac {2 i \sqrt {a^2-b^2} f^2 \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^2 d^3}-\frac {2 i \sqrt {a^2-b^2} f^2 \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^2 d^3}-\frac {2 f (e+f x) \sin (c+d x)}{b d^2} \]

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

________________________________________________________________________________________

Rubi [A]
time = 0.61, antiderivative size = 460, normalized size of antiderivative = 1.00, 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 = {4621, 32, 3377, 2718, 3404, 2296, 2221, 2611, 2320, 6724} \begin {gather*} \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 {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 {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^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} \end {gather*}

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 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 4621

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 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*} \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 ) \text {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 ) \text {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 = 2.65, size = 782, normalized size = 1.70 \begin {gather*} \frac {a x \left (3 e^2+3 e f x+f^2 x^2\right )-\frac {3 i \sqrt {a^2-b^2} \left (-i \left (d^2 \left (\sqrt {a^2-b^2} f x (2 e+f x) \left (-\log \left (1+\frac {b (\cos (2 c+d x)+i \sin (2 c+d x))}{i a \cos (c)+\sqrt {\left (-a^2+b^2\right ) (\cos (c)+i \sin (c))^2}-a \sin (c)}\right )+\log \left (1-\frac {b (\cos (2 c+d x)+i \sin (2 c+d x))}{-i a \cos (c)+\sqrt {\left (-a^2+b^2\right ) (\cos (c)+i \sin (c))^2}+a \sin (c)}\right )\right ) (\cos (c)+i \sin (c))+2 e^2 \tan ^{-1}\left (\frac {b \cos (c+d x)+i (a+b \sin (c+d x))}{\sqrt {a^2-b^2}}\right ) \sqrt {-\left (\left (a^2-b^2\right ) (\cos (c)+i \sin (c))^2\right )}\right )-2 \sqrt {a^2-b^2} f^2 \text {Li}_3\left (-\frac {b (\cos (2 c+d x)+i \sin (2 c+d x))}{i a \cos (c)+\sqrt {\left (-a^2+b^2\right ) (\cos (c)+i \sin (c))^2}-a \sin (c)}\right ) (\cos (c)+i \sin (c))+2 \sqrt {a^2-b^2} f^2 \text {Li}_3\left (\frac {b (\cos (2 c+d x)+i \sin (2 c+d x))}{-i a \cos (c)+\sqrt {\left (-a^2+b^2\right ) (\cos (c)+i \sin (c))^2}+a \sin (c)}\right ) (\cos (c)+i \sin (c))\right )+2 \sqrt {a^2-b^2} d f (e+f x) \text {Li}_2\left (-\frac {b (\cos (2 c+d x)+i \sin (2 c+d x))}{i a \cos (c)+\sqrt {\left (-a^2+b^2\right ) (\cos (c)+i \sin (c))^2}-a \sin (c)}\right ) (\cos (c)+i \sin (c))-2 \sqrt {a^2-b^2} d f (e+f x) \text {Li}_2\left (\frac {b (\cos (2 c+d x)+i \sin (2 c+d x))}{-i a \cos (c)+\sqrt {\left (-a^2+b^2\right ) (\cos (c)+i \sin (c))^2}+a \sin (c)}\right ) (\cos (c)+i \sin (c))\right )}{d^3 \sqrt {-\left (\left (a^2-b^2\right ) (\cos (c)+i \sin (c))^2\right )}}+\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} \end {gather*}

Antiderivative was successfully verified.

[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)*Sqrt[a^2 - b^2]*((-I)*(d^2*(Sqrt[a^2 - b^2]*f*x*(2*e + f*x)*(-Log[1
+ (b*(Cos[2*c + d*x] + I*Sin[2*c + d*x]))/(I*a*Cos[c] + Sqrt[(-a^2 + b^2)*(Cos[c] + I*Sin[c])^2] - a*Sin[c])]
+ Log[1 - (b*(Cos[2*c + d*x] + I*Sin[2*c + d*x]))/((-I)*a*Cos[c] + Sqrt[(-a^2 + b^2)*(Cos[c] + I*Sin[c])^2] +
a*Sin[c])])*(Cos[c] + I*Sin[c]) + 2*e^2*ArcTan[(b*Cos[c + d*x] + I*(a + b*Sin[c + d*x]))/Sqrt[a^2 - b^2]]*Sqrt
[-((a^2 - b^2)*(Cos[c] + I*Sin[c])^2)]) - 2*Sqrt[a^2 - b^2]*f^2*PolyLog[3, -((b*(Cos[2*c + d*x] + I*Sin[2*c +
d*x]))/(I*a*Cos[c] + Sqrt[(-a^2 + b^2)*(Cos[c] + I*Sin[c])^2] - a*Sin[c]))]*(Cos[c] + I*Sin[c]) + 2*Sqrt[a^2 -
b^2]*f^2*PolyLog[3, (b*(Cos[2*c + d*x] + I*Sin[2*c + d*x]))/((-I)*a*Cos[c] + Sqrt[(-a^2 + b^2)*(Cos[c] + I*Si
n[c])^2] + a*Sin[c])]*(Cos[c] + I*Sin[c])) + 2*Sqrt[a^2 - b^2]*d*f*(e + f*x)*PolyLog[2, -((b*(Cos[2*c + d*x] +
I*Sin[2*c + d*x]))/(I*a*Cos[c] + Sqrt[(-a^2 + b^2)*(Cos[c] + I*Sin[c])^2] - a*Sin[c]))]*(Cos[c] + I*Sin[c]) -
2*Sqrt[a^2 - b^2]*d*f*(e + f*x)*PolyLog[2, (b*(Cos[2*c + d*x] + I*Sin[2*c + d*x]))/((-I)*a*Cos[c] + Sqrt[(-a^
2 + b^2)*(Cos[c] + I*Sin[c])^2] + a*Sin[c])]*(Cos[c] + I*Sin[c])))/(d^3*Sqrt[-((a^2 - b^2)*(Cos[c] + I*Sin[c])
^2)]) + (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 = 0.25, size = 0, normalized size = 0.00 \[\int \frac {\left (f x +e \right )^{2} \left (\cos ^{2}\left (d x +c \right )\right )}{a +b \sin \left (d x +c \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)^2*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)

________________________________________________________________________________________

Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

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 >> 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

________________________________________________________________________________________

Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1645 vs. \(2 (407) = 814\).
time = 0.55, size = 1645, normalized size = 3.58 \begin {gather*} \frac {2 \, a d^{3} f^{2} x^{3} + 6 \, a d^{3} f x^{2} e + 6 \, a d^{3} x e^{2} - 6 \, b f^{2} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} {\rm polylog}\left (3, -\frac {i \, a \cos \left (d x + c\right ) + a \sin \left (d x + c\right ) + {\left (b \cos \left (d x + c\right ) - i \, b \sin \left (d x + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}}}{b}\right ) + 6 \, b f^{2} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} {\rm polylog}\left (3, -\frac {i \, a \cos \left (d x + c\right ) + a \sin \left (d x + c\right ) - {\left (b \cos \left (d x + c\right ) - i \, b \sin \left (d x + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}}}{b}\right ) - 6 \, b f^{2} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} {\rm polylog}\left (3, -\frac {-i \, a \cos \left (d x + c\right ) + a \sin \left (d x + c\right ) + {\left (b \cos \left (d x + c\right ) + i \, b \sin \left (d x + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}}}{b}\right ) + 6 \, b f^{2} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} {\rm polylog}\left (3, -\frac {-i \, a \cos \left (d x + c\right ) + a \sin \left (d x + c\right ) - {\left (b \cos \left (d x + c\right ) + i \, b \sin \left (d x + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}}}{b}\right ) - 6 \, {\left (i \, b d f^{2} x + i \, b d f e\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} {\rm Li}_2\left (\frac {i \, a \cos \left (d x + c\right ) - a \sin \left (d x + c\right ) + {\left (b \cos \left (d x + c\right ) + i \, b \sin \left (d x + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - b}{b} + 1\right ) - 6 \, {\left (-i \, b d f^{2} x - i \, b d f e\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} {\rm Li}_2\left (\frac {i \, a \cos \left (d x + c\right ) - a \sin \left (d x + c\right ) - {\left (b \cos \left (d x + c\right ) + i \, b \sin \left (d x + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - b}{b} + 1\right ) - 6 \, {\left (-i \, b d f^{2} x - i \, b d f e\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} {\rm Li}_2\left (\frac {-i \, a \cos \left (d x + c\right ) - a \sin \left (d x + c\right ) + {\left (b \cos \left (d x + c\right ) - i \, b \sin \left (d x + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - b}{b} + 1\right ) - 6 \, {\left (i \, b d f^{2} x + i \, b d f e\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} {\rm Li}_2\left (\frac {-i \, a \cos \left (d x + c\right ) - a \sin \left (d x + c\right ) - {\left (b \cos \left (d x + c\right ) - i \, b \sin \left (d x + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - b}{b} + 1\right ) - 3 \, {\left (b c^{2} f^{2} - 2 \, b c d f e + b d^{2} e^{2}\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} \log \left (2 \, b \cos \left (d x + c\right ) + 2 i \, b \sin \left (d x + c\right ) + 2 \, b \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} + 2 i \, a\right ) - 3 \, {\left (b c^{2} f^{2} - 2 \, b c d f e + b d^{2} e^{2}\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} \log \left (2 \, b \cos \left (d x + c\right ) - 2 i \, b \sin \left (d x + c\right ) + 2 \, b \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - 2 i \, a\right ) + 3 \, {\left (b c^{2} f^{2} - 2 \, b c d f e + b d^{2} e^{2}\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} \log \left (-2 \, b \cos \left (d x + c\right ) + 2 i \, b \sin \left (d x + c\right ) + 2 \, b \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} + 2 i \, a\right ) + 3 \, {\left (b c^{2} f^{2} - 2 \, b c d f e + b d^{2} e^{2}\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} \log \left (-2 \, b \cos \left (d x + c\right ) - 2 i \, b \sin \left (d x + c\right ) + 2 \, b \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - 2 i \, a\right ) + 3 \, {\left (b d^{2} f^{2} x^{2} - b c^{2} f^{2} + 2 \, {\left (b d^{2} f x + b c d f\right )} e\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} \log \left (-\frac {i \, a \cos \left (d x + c\right ) - a \sin \left (d x + c\right ) + {\left (b \cos \left (d x + c\right ) + i \, b \sin \left (d x + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - b}{b}\right ) - 3 \, {\left (b d^{2} f^{2} x^{2} - b c^{2} f^{2} + 2 \, {\left (b d^{2} f x + b c d f\right )} e\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} \log \left (-\frac {i \, a \cos \left (d x + c\right ) - a \sin \left (d x + c\right ) - {\left (b \cos \left (d x + c\right ) + i \, b \sin \left (d x + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - b}{b}\right ) + 3 \, {\left (b d^{2} f^{2} x^{2} - b c^{2} f^{2} + 2 \, {\left (b d^{2} f x + b c d f\right )} e\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} \log \left (-\frac {-i \, a \cos \left (d x + c\right ) - a \sin \left (d x + c\right ) + {\left (b \cos \left (d x + c\right ) - i \, b \sin \left (d x + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - b}{b}\right ) - 3 \, {\left (b d^{2} f^{2} x^{2} - b c^{2} f^{2} + 2 \, {\left (b d^{2} f x + b c d f\right )} e\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} \log \left (-\frac {-i \, a \cos \left (d x + c\right ) - a \sin \left (d x + c\right ) - {\left (b \cos \left (d x + c\right ) - i \, b \sin \left (d x + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - b}{b}\right ) + 6 \, {\left (b d^{2} f^{2} x^{2} + 2 \, b d^{2} f x e + b d^{2} e^{2} - 2 \, b f^{2}\right )} \cos \left (d x + c\right ) - 12 \, {\left (b d f^{2} x + b d f e\right )} \sin \left (d x + c\right )}{6 \, b^{2} d^{3}} \end {gather*}

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

________________________________________________________________________________________

Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

________________________________________________________________________________________

Mupad [F(-1)]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \text {Hanged} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

\text{Hanged}

________________________________________________________________________________________