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

Optimal. Leaf size=408 \[ \frac {2 i a f^2 \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d^3 \sqrt {a^2-b^2}}-\frac {2 i a f^2 \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b d^3 \sqrt {a^2-b^2}}+\frac {2 a f (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d^2 \sqrt {a^2-b^2}}-\frac {2 a f (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b d^2 \sqrt {a^2-b^2}}+\frac {i a (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d \sqrt {a^2-b^2}}-\frac {i a (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d \sqrt {a^2-b^2}}+\frac {(e+f x)^3}{3 b f} \]

[Out]

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

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Rubi [A]  time = 0.86, antiderivative size = 408, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {4515, 32, 3323, 2264, 2190, 2531, 2282, 6589} \[ \frac {2 a f (e+f x) \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d^2 \sqrt {a^2-b^2}}-\frac {2 a f (e+f x) \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d^2 \sqrt {a^2-b^2}}+\frac {2 i a f^2 \text {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d^3 \sqrt {a^2-b^2}}-\frac {2 i a f^2 \text {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d^3 \sqrt {a^2-b^2}}+\frac {i a (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d \sqrt {a^2-b^2}}-\frac {i a (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d \sqrt {a^2-b^2}}+\frac {(e+f x)^3}{3 b f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 4515

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

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Mathematica [A]  time = 2.29, size = 445, normalized size = 1.09 \[ \frac {x \left (3 e^2+3 e f x+f^2 x^2\right )}{3 b}-\frac {i a \left (-i \left (d^2 \left (2 e^2 \sqrt {b^2-a^2} \tan ^{-1}\left (\frac {i a+b e^{i (c+d x)}}{\sqrt {a^2-b^2}}\right )+f x \sqrt {a^2-b^2} (2 e+f x) \left (\log \left (1-\frac {b e^{i (c+d x)}}{\sqrt {b^2-a^2}-i a}\right )-\log \left (1+\frac {b e^{i (c+d x)}}{\sqrt {b^2-a^2}+i a}\right )\right )\right )+2 f^2 \sqrt {a^2-b^2} \text {Li}_3\left (\frac {b e^{i (c+d x)}}{\sqrt {b^2-a^2}-i a}\right )-2 f^2 \sqrt {a^2-b^2} \text {Li}_3\left (-\frac {b e^{i (c+d x)}}{i a+\sqrt {b^2-a^2}}\right )\right )-2 d f \sqrt {a^2-b^2} (e+f x) \text {Li}_2\left (\frac {b e^{i (c+d x)}}{\sqrt {b^2-a^2}-i a}\right )+2 d f \sqrt {a^2-b^2} (e+f x) \text {Li}_2\left (-\frac {b e^{i (c+d x)}}{i a+\sqrt {b^2-a^2}}\right )\right )}{b d^3 \sqrt {-\left (a^2-b^2\right )^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x*(3*e^2 + 3*e*f*x + f^2*x^2))/(3*b) - (I*a*(-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]] + 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]))])))/(b*Sqrt[-(a^2 - b^2)^2]
*d^3)

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fricas [C]  time = 0.70, size = 1660, normalized size = 4.07 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*(a^2 - b^2)*d^3*f^2*x^3 + 6*(a^2 - b^2)*d^3*e*f*x^2 + 6*(a^2 - b^2)*d^3*e^2*x + 6*a*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*a*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*a*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*a*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*I*a*b*d*f^2*x + 6*I*a*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*a*b*d*f^2*x - 6*I*a*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*a*b*d*f^
2*x - 6*I*a*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*a*b*d*f^2*x + 6*I*a*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*(a*b*d^2*e^2 - 2*a*b*c*d*e*f + a*b*c^2*f^2)*sqrt(-(a^2 - b^2)/b^2)*log(2*b*co
s(d*x + c) + 2*I*b*sin(d*x + c) + 2*b*sqrt(-(a^2 - b^2)/b^2) + 2*I*a) - 3*(a*b*d^2*e^2 - 2*a*b*c*d*e*f + a*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*b*d^2*e^2 - 2*a*b*c*d*e*f + a*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*b*d^2*e^2 - 2*a*b*c*d*e*f + a*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*b*d^2*f^2*x^2 + 2*
a*b*d^2*e*f*x + 2*a*b*c*d*e*f - a*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*(a*b*d^2*f^2*x^2 + 2*a*b*d^2
*e*f*x + 2*a*b*c*d*e*f - a*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*(a*b*d^2*f^2*x^2 + 2*a*b*d^2*e*f*x
+ 2*a*b*c*d*e*f - a*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*c
os(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b) + 3*(a*b*d^2*f^2*x^2 + 2*a*b*d^2*e*f*x + 2*a*
b*c*d*e*f - a*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))/((a^2*b - b^3)*d^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.10, size = 0, normalized size = 0.00 \[ \int \frac {\left (f x +e \right )^{2} \sin \left (d x +c \right )}{a +b \sin \left (d x +c \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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