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

Optimal. Leaf size=367 \[ -\frac {2 d (c+d x) \text {Li}_2\left (\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{f^2 \sqrt {a^2-b^2}}+\frac {2 d (c+d x) \text {Li}_2\left (\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{f^2 \sqrt {a^2-b^2}}-\frac {i (c+d x)^2 \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{f \sqrt {a^2-b^2}}+\frac {i (c+d x)^2 \log \left (1-\frac {i b e^{i (e+f x)}}{\sqrt {a^2-b^2}+a}\right )}{f \sqrt {a^2-b^2}}-\frac {2 i d^2 \text {Li}_3\left (\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{f^3 \sqrt {a^2-b^2}}+\frac {2 i d^2 \text {Li}_3\left (\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{f^3 \sqrt {a^2-b^2}} \]

[Out]

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

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Rubi [A]  time = 0.82, antiderivative size = 367, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3323, 2264, 2190, 2531, 2282, 6589} \[ -\frac {2 d (c+d x) \text {PolyLog}\left (2,\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{f^2 \sqrt {a^2-b^2}}+\frac {2 d (c+d x) \text {PolyLog}\left (2,\frac {i b e^{i (e+f x)}}{\sqrt {a^2-b^2}+a}\right )}{f^2 \sqrt {a^2-b^2}}-\frac {2 i d^2 \text {PolyLog}\left (3,\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{f^3 \sqrt {a^2-b^2}}+\frac {2 i d^2 \text {PolyLog}\left (3,\frac {i b e^{i (e+f x)}}{\sqrt {a^2-b^2}+a}\right )}{f^3 \sqrt {a^2-b^2}}-\frac {i (c+d x)^2 \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{f \sqrt {a^2-b^2}}+\frac {i (c+d x)^2 \log \left (1-\frac {i b e^{i (e+f x)}}{\sqrt {a^2-b^2}+a}\right )}{f \sqrt {a^2-b^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*(c + d*x)^2*Log[1 - (I*b*E^(I*(e + f*x)))/(a - Sqrt[a^2 - b^2])])/(Sqrt[a^2 - b^2]*f) + (I*(c + d*x)^2*L
og[1 - (I*b*E^(I*(e + f*x)))/(a + Sqrt[a^2 - b^2])])/(Sqrt[a^2 - b^2]*f) - (2*d*(c + d*x)*PolyLog[2, (I*b*E^(I
*(e + f*x)))/(a - Sqrt[a^2 - b^2])])/(Sqrt[a^2 - b^2]*f^2) + (2*d*(c + d*x)*PolyLog[2, (I*b*E^(I*(e + f*x)))/(
a + Sqrt[a^2 - b^2])])/(Sqrt[a^2 - b^2]*f^2) - ((2*I)*d^2*PolyLog[3, (I*b*E^(I*(e + f*x)))/(a - Sqrt[a^2 - b^2
])])/(Sqrt[a^2 - b^2]*f^3) + ((2*I)*d^2*PolyLog[3, (I*b*E^(I*(e + f*x)))/(a + Sqrt[a^2 - b^2])])/(Sqrt[a^2 - b
^2]*f^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 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 {(c+d x)^2}{a+b \sin (e+f x)} \, dx &=2 \int \frac {e^{i (e+f x)} (c+d x)^2}{i b+2 a e^{i (e+f x)}-i b e^{2 i (e+f x)}} \, dx\\ &=-\frac {(2 i b) \int \frac {e^{i (e+f x)} (c+d x)^2}{2 a-2 \sqrt {a^2-b^2}-2 i b e^{i (e+f x)}} \, dx}{\sqrt {a^2-b^2}}+\frac {(2 i b) \int \frac {e^{i (e+f x)} (c+d x)^2}{2 a+2 \sqrt {a^2-b^2}-2 i b e^{i (e+f x)}} \, dx}{\sqrt {a^2-b^2}}\\ &=-\frac {i (c+d x)^2 \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f}+\frac {i (c+d x)^2 \log \left (1-\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f}+\frac {(2 i d) \int (c+d x) \log \left (1-\frac {2 i b e^{i (e+f x)}}{2 a-2 \sqrt {a^2-b^2}}\right ) \, dx}{\sqrt {a^2-b^2} f}-\frac {(2 i d) \int (c+d x) \log \left (1-\frac {2 i b e^{i (e+f x)}}{2 a+2 \sqrt {a^2-b^2}}\right ) \, dx}{\sqrt {a^2-b^2} f}\\ &=-\frac {i (c+d x)^2 \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f}+\frac {i (c+d x)^2 \log \left (1-\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f}-\frac {2 d (c+d x) \text {Li}_2\left (\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f^2}+\frac {2 d (c+d x) \text {Li}_2\left (\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f^2}+\frac {\left (2 d^2\right ) \int \text {Li}_2\left (\frac {2 i b e^{i (e+f x)}}{2 a-2 \sqrt {a^2-b^2}}\right ) \, dx}{\sqrt {a^2-b^2} f^2}-\frac {\left (2 d^2\right ) \int \text {Li}_2\left (\frac {2 i b e^{i (e+f x)}}{2 a+2 \sqrt {a^2-b^2}}\right ) \, dx}{\sqrt {a^2-b^2} f^2}\\ &=-\frac {i (c+d x)^2 \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f}+\frac {i (c+d x)^2 \log \left (1-\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f}-\frac {2 d (c+d x) \text {Li}_2\left (\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f^2}+\frac {2 d (c+d x) \text {Li}_2\left (\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f^2}-\frac {\left (2 i d^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 (e+f x)}\right )}{\sqrt {a^2-b^2} f^3}+\frac {\left (2 i d^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 (e+f x)}\right )}{\sqrt {a^2-b^2} f^3}\\ &=-\frac {i (c+d x)^2 \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f}+\frac {i (c+d x)^2 \log \left (1-\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f}-\frac {2 d (c+d x) \text {Li}_2\left (\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f^2}+\frac {2 d (c+d x) \text {Li}_2\left (\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f^2}-\frac {2 i d^2 \text {Li}_3\left (\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f^3}+\frac {2 i d^2 \text {Li}_3\left (\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f^3}\\ \end {align*}

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Mathematica [A]  time = 0.21, size = 296, normalized size = 0.81 \[ -\frac {i \left (\frac {2 d \left (d \text {Li}_3\left (\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )-i f (c+d x) \text {Li}_2\left (-\frac {i b e^{i (e+f x)}}{\sqrt {a^2-b^2}-a}\right )\right )}{f^2}+\frac {2 i d \left (f (c+d x) \text {Li}_2\left (\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )+i d \text {Li}_3\left (\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )\right )}{f^2}+(c+d x)^2 \log \left (1+\frac {i b e^{i (e+f x)}}{\sqrt {a^2-b^2}-a}\right )-(c+d x)^2 \log \left (1-\frac {i b e^{i (e+f x)}}{\sqrt {a^2-b^2}+a}\right )\right )}{f \sqrt {a^2-b^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [C]  time = 0.61, size = 1553, normalized size = 4.23 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.48, size = 0, normalized size = 0.00 \[ \int \frac {\left (d x +c \right )^{2}}{a +b \sin \left (f x +e \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((d*x+c)^2/(a+b*sin(f*x+e)),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((d*x+c)^2/(a+b*sin(f*x+e)),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]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (c+d\,x\right )}^2}{a+b\,\sin \left (e+f\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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