3.2.80 \(\int \frac {(e+f x)^2 \sin (c+d x)}{a+a \sin (c+d x)} \, dx\) [180]

Optimal. Leaf size=129 \[ \frac {i (e+f x)^2}{a d}+\frac {(e+f x)^3}{3 a f}+\frac {(e+f x)^2 \cot \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{a d}-\frac {4 f (e+f x) \log \left (1-i e^{i (c+d x)}\right )}{a d^2}+\frac {4 i f^2 \text {Li}_2\left (i e^{i (c+d x)}\right )}{a d^3} \]

[Out]

I*(f*x+e)^2/a/d+1/3*(f*x+e)^3/a/f+(f*x+e)^2*cot(1/2*c+1/4*Pi+1/2*d*x)/a/d-4*f*(f*x+e)*ln(1-I*exp(I*(d*x+c)))/a
/d^2+4*I*f^2*polylog(2,I*exp(I*(d*x+c)))/a/d^3

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Rubi [A]
time = 0.16, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {4611, 32, 3399, 4269, 3798, 2221, 2317, 2438} \begin {gather*} \frac {4 i f^2 \text {PolyLog}\left (2,i e^{i (c+d x)}\right )}{a d^3}-\frac {4 f (e+f x) \log \left (1-i e^{i (c+d x)}\right )}{a d^2}+\frac {(e+f x)^2 \cot \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right )}{a d}+\frac {i (e+f x)^2}{a d}+\frac {(e+f x)^3}{3 a f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(I*(e + f*x)^2)/(a*d) + (e + f*x)^3/(3*a*f) + ((e + f*x)^2*Cot[c/2 + Pi/4 + (d*x)/2])/(a*d) - (4*f*(e + f*x)*L
og[1 - I*E^(I*(c + d*x))])/(a*d^2) + ((4*I)*f^2*PolyLog[2, I*E^(I*(c + d*x))])/(a*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 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 2317

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 3399

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

Rule 3798

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol] :> Simp[I*((c + d*x)^(m + 1)/(d*(
m + 1))), x] - Dist[2*I, Int[(c + d*x)^m*E^(2*I*k*Pi)*(E^(2*I*(e + f*x))/(1 + E^(2*I*k*Pi)*E^(2*I*(e + f*x))))
, x], x] /; FreeQ[{c, d, e, f}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 4269

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-(c + d*x)^m)*(Cot[e + f*x]/f), x
] + Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[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]

Rubi steps

\begin {align*} \int \frac {(e+f x)^2 \sin (c+d x)}{a+a \sin (c+d x)} \, dx &=\frac {\int (e+f x)^2 \, dx}{a}-\int \frac {(e+f x)^2}{a+a \sin (c+d x)} \, dx\\ &=\frac {(e+f x)^3}{3 a f}-\frac {\int (e+f x)^2 \csc ^2\left (\frac {1}{2} \left (c+\frac {\pi }{2}\right )+\frac {d x}{2}\right ) \, dx}{2 a}\\ &=\frac {(e+f x)^3}{3 a f}+\frac {(e+f x)^2 \cot \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{a d}-\frac {(2 f) \int (e+f x) \cot \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right ) \, dx}{a d}\\ &=\frac {i (e+f x)^2}{a d}+\frac {(e+f x)^3}{3 a f}+\frac {(e+f x)^2 \cot \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{a d}-\frac {(4 f) \int \frac {e^{2 i \left (\frac {c}{2}+\frac {d x}{2}\right )} (e+f x)}{1-i e^{2 i \left (\frac {c}{2}+\frac {d x}{2}\right )}} \, dx}{a d}\\ &=\frac {i (e+f x)^2}{a d}+\frac {(e+f x)^3}{3 a f}+\frac {(e+f x)^2 \cot \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{a d}-\frac {4 f (e+f x) \log \left (1-i e^{i (c+d x)}\right )}{a d^2}+\frac {\left (4 f^2\right ) \int \log \left (1-i e^{2 i \left (\frac {c}{2}+\frac {d x}{2}\right )}\right ) \, dx}{a d^2}\\ &=\frac {i (e+f x)^2}{a d}+\frac {(e+f x)^3}{3 a f}+\frac {(e+f x)^2 \cot \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{a d}-\frac {4 f (e+f x) \log \left (1-i e^{i (c+d x)}\right )}{a d^2}-\frac {\left (4 i f^2\right ) \text {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{2 i \left (\frac {c}{2}+\frac {d x}{2}\right )}\right )}{a d^3}\\ &=\frac {i (e+f x)^2}{a d}+\frac {(e+f x)^3}{3 a f}+\frac {(e+f x)^2 \cot \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{a d}-\frac {4 f (e+f x) \log \left (1-i e^{i (c+d x)}\right )}{a d^2}+\frac {4 i f^2 \text {Li}_2\left (i e^{i (c+d x)}\right )}{a d^3}\\ \end {align*}

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Mathematica [A]
time = 0.99, size = 181, normalized size = 1.40 \begin {gather*} \frac {x \left (3 e^2+3 e f x+f^2 x^2\right )+\frac {6 i f \left (2 i d (e+f x) \log (1-i \cos (c+d x)+\sin (c+d x))+2 f \text {Li}_2(i \cos (c+d x)-\sin (c+d x))+\frac {d^2 x (2 e+f x) (\cos (c)+i \sin (c))}{\cos (c)+i (1+\sin (c))}\right )}{d^3}-\frac {6 (e+f x)^2 \sin \left (\frac {d x}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}}{3 a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(3*e^2 + 3*e*f*x + f^2*x^2) + ((6*I)*f*((2*I)*d*(e + f*x)*Log[1 - I*Cos[c + d*x] + Sin[c + d*x]] + 2*f*Poly
Log[2, I*Cos[c + d*x] - Sin[c + d*x]] + (d^2*x*(2*e + f*x)*(Cos[c] + I*Sin[c]))/(Cos[c] + I*(1 + Sin[c]))))/d^
3 - (6*(e + f*x)^2*Sin[(d*x)/2])/(d*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])))/(3*a)

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 292 vs. \(2 (113 ) = 226\).
time = 0.11, size = 293, normalized size = 2.27

method	result	size
risch	\(\frac {f^{2} x^{3}}{3 a}+\frac {f e \,x^{2}}{a}+\frac {e^{2} x}{a}+\frac {e^{3}}{3 a f}+\frac {2 x^{2} f^{2}+4 e f x +2 e^{2}}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}-\frac {4 f \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) e}{a \,d^{2}}+\frac {4 f \ln \left ({\mathrm e}^{i \left (d x +c \right )}\right ) e}{a \,d^{2}}+\frac {2 i f^{2} x^{2}}{a d}+\frac {4 i f^{2} c x}{a \,d^{2}}+\frac {2 i f^{2} c^{2}}{a \,d^{3}}-\frac {4 f^{2} \ln \left (1-i {\mathrm e}^{i \left (d x +c \right )}\right ) x}{a \,d^{2}}-\frac {4 f^{2} \ln \left (1-i {\mathrm e}^{i \left (d x +c \right )}\right ) c}{a \,d^{3}}+\frac {4 i f^{2} \polylog \left (2, i {\mathrm e}^{i \left (d x +c \right )}\right )}{a \,d^{3}}+\frac {4 f^{2} c \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a \,d^{3}}-\frac {4 f^{2} c \ln \left ({\mathrm e}^{i \left (d x +c \right )}\right )}{a \,d^{3}}\)	\(293\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3/a*f^2*x^3+1/a*f*e*x^2+1/a*e^2*x+1/3/a/f*e^3+2*(f^2*x^2+2*e*f*x+e^2)/d/a/(exp(I*(d*x+c))+I)-4/a/d^2*f*ln(ex
p(I*(d*x+c))+I)*e+4/a/d^2*f*ln(exp(I*(d*x+c)))*e+2*I/a/d*f^2*x^2+4*I/a/d^2*f^2*c*x+2*I/a/d^3*f^2*c^2-4/a/d^2*f
^2*ln(1-I*exp(I*(d*x+c)))*x-4/a/d^3*f^2*ln(1-I*exp(I*(d*x+c)))*c+4*I*f^2*polylog(2,I*exp(I*(d*x+c)))/a/d^3+4/a
/d^3*f^2*c*ln(exp(I*(d*x+c))+I)-4/a/d^3*f^2*c*ln(exp(I*(d*x+c)))

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Maxima [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 409 vs. \(2 (112) = 224\).
time = 0.60, size = 409, normalized size = 3.17 \begin {gather*} \frac {d^{3} f^{2} x^{3} + 3 \, d^{3} f x^{2} e + 3 \, d^{3} x e^{2} - 6 i \, d^{2} e^{2} - 12 \, {\left (d f \cos \left (d x + c\right ) e + i \, d f e \sin \left (d x + c\right ) + i \, d f e\right )} \arctan \left (\sin \left (d x + c\right ) + 1, \cos \left (d x + c\right )\right ) + 12 \, {\left (d f^{2} x \cos \left (d x + c\right ) + i \, d f^{2} x \sin \left (d x + c\right ) + i \, d f^{2} x\right )} \arctan \left (\cos \left (d x + c\right ), \sin \left (d x + c\right ) + 1\right ) - {\left (i \, d^{3} f^{2} x^{3} - 3 \, {\left (-i \, d^{3} f e + 2 \, d^{2} f^{2}\right )} x^{2} - 3 \, {\left (-i \, d^{3} e^{2} + 4 \, d^{2} f e\right )} x\right )} \cos \left (d x + c\right ) + 12 \, {\left (f^{2} \cos \left (d x + c\right ) + i \, f^{2} \sin \left (d x + c\right ) + i \, f^{2}\right )} {\rm Li}_2\left (i \, e^{\left (i \, d x + i \, c\right )}\right ) - 6 \, {\left (d f^{2} x + d f e - {\left (i \, d f^{2} x + i \, d f e\right )} \cos \left (d x + c\right ) + {\left (d f^{2} x + d f e\right )} \sin \left (d x + c\right )\right )} \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \sin \left (d x + c\right ) + 1\right ) + {\left (d^{3} f^{2} x^{3} + 3 \, {\left (d^{3} f e + 2 i \, d^{2} f^{2}\right )} x^{2} + 3 \, {\left (d^{3} e^{2} + 4 i \, d^{2} f e\right )} x\right )} \sin \left (d x + c\right )}{-3 i \, a d^{3} \cos \left (d x + c\right ) + 3 \, a d^{3} \sin \left (d x + c\right ) + 3 \, a d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(d^3*f^2*x^3 + 3*d^3*f*x^2*e + 3*d^3*x*e^2 - 6*I*d^2*e^2 - 12*(d*f*cos(d*x + c)*e + I*d*f*e*sin(d*x + c) + I*d
*f*e)*arctan2(sin(d*x + c) + 1, cos(d*x + c)) + 12*(d*f^2*x*cos(d*x + c) + I*d*f^2*x*sin(d*x + c) + I*d*f^2*x)
*arctan2(cos(d*x + c), sin(d*x + c) + 1) - (I*d^3*f^2*x^3 - 3*(-I*d^3*f*e + 2*d^2*f^2)*x^2 - 3*(-I*d^3*e^2 + 4
*d^2*f*e)*x)*cos(d*x + c) + 12*(f^2*cos(d*x + c) + I*f^2*sin(d*x + c) + I*f^2)*dilog(I*e^(I*d*x + I*c)) - 6*(d
*f^2*x + d*f*e - (I*d*f^2*x + I*d*f*e)*cos(d*x + c) + (d*f^2*x + d*f*e)*sin(d*x + c))*log(cos(d*x + c)^2 + sin
(d*x + c)^2 + 2*sin(d*x + c) + 1) + (d^3*f^2*x^3 + 3*(d^3*f*e + 2*I*d^2*f^2)*x^2 + 3*(d^3*e^2 + 4*I*d^2*f*e)*x
)*sin(d*x + c))/(-3*I*a*d^3*cos(d*x + c) + 3*a*d^3*sin(d*x + c) + 3*a*d^3)

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 587 vs. \(2 (112) = 224\).
time = 0.39, size = 587, normalized size = 4.55 \begin {gather*} \frac {d^{3} f^{2} x^{3} + 3 \, d^{2} f^{2} x^{2} + {\left (d^{3} f^{2} x^{3} + 3 \, d^{2} f^{2} x^{2} + 3 \, {\left (d^{3} x + d^{2}\right )} e^{2} + 3 \, {\left (d^{3} f x^{2} + 2 \, d^{2} f x\right )} e\right )} \cos \left (d x + c\right ) - 6 \, {\left (-i \, f^{2} \cos \left (d x + c\right ) - i \, f^{2} \sin \left (d x + c\right ) - i \, f^{2}\right )} {\rm Li}_2\left (i \, \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right ) - 6 \, {\left (i \, f^{2} \cos \left (d x + c\right ) + i \, f^{2} \sin \left (d x + c\right ) + i \, f^{2}\right )} {\rm Li}_2\left (-i \, \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right ) + 3 \, {\left (d^{3} x + d^{2}\right )} e^{2} + 3 \, {\left (d^{3} f x^{2} + 2 \, d^{2} f x\right )} e + 6 \, {\left (c f^{2} - d f e + {\left (c f^{2} - d f e\right )} \cos \left (d x + c\right ) + {\left (c f^{2} - d f e\right )} \sin \left (d x + c\right )\right )} \log \left (\cos \left (d x + c\right ) + i \, \sin \left (d x + c\right ) + i\right ) - 6 \, {\left (d f^{2} x + c f^{2} + {\left (d f^{2} x + c f^{2}\right )} \cos \left (d x + c\right ) + {\left (d f^{2} x + c f^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (i \, \cos \left (d x + c\right ) + \sin \left (d x + c\right ) + 1\right ) - 6 \, {\left (d f^{2} x + c f^{2} + {\left (d f^{2} x + c f^{2}\right )} \cos \left (d x + c\right ) + {\left (d f^{2} x + c f^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (-i \, \cos \left (d x + c\right ) + \sin \left (d x + c\right ) + 1\right ) + 6 \, {\left (c f^{2} - d f e + {\left (c f^{2} - d f e\right )} \cos \left (d x + c\right ) + {\left (c f^{2} - d f e\right )} \sin \left (d x + c\right )\right )} \log \left (-\cos \left (d x + c\right ) + i \, \sin \left (d x + c\right ) + i\right ) + {\left (d^{3} f^{2} x^{3} - 3 \, d^{2} f^{2} x^{2} + 3 \, {\left (d^{3} x - d^{2}\right )} e^{2} + 3 \, {\left (d^{3} f x^{2} - 2 \, d^{2} f x\right )} e\right )} \sin \left (d x + c\right )}{3 \, {\left (a d^{3} \cos \left (d x + c\right ) + a d^{3} \sin \left (d x + c\right ) + a d^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(d^3*f^2*x^3 + 3*d^2*f^2*x^2 + (d^3*f^2*x^3 + 3*d^2*f^2*x^2 + 3*(d^3*x + d^2)*e^2 + 3*(d^3*f*x^2 + 2*d^2*f
*x)*e)*cos(d*x + c) - 6*(-I*f^2*cos(d*x + c) - I*f^2*sin(d*x + c) - I*f^2)*dilog(I*cos(d*x + c) - sin(d*x + c)
) - 6*(I*f^2*cos(d*x + c) + I*f^2*sin(d*x + c) + I*f^2)*dilog(-I*cos(d*x + c) - sin(d*x + c)) + 3*(d^3*x + d^2
)*e^2 + 3*(d^3*f*x^2 + 2*d^2*f*x)*e + 6*(c*f^2 - d*f*e + (c*f^2 - d*f*e)*cos(d*x + c) + (c*f^2 - d*f*e)*sin(d*
x + c))*log(cos(d*x + c) + I*sin(d*x + c) + I) - 6*(d*f^2*x + c*f^2 + (d*f^2*x + c*f^2)*cos(d*x + c) + (d*f^2*
x + c*f^2)*sin(d*x + c))*log(I*cos(d*x + c) + sin(d*x + c) + 1) - 6*(d*f^2*x + c*f^2 + (d*f^2*x + c*f^2)*cos(d
*x + c) + (d*f^2*x + c*f^2)*sin(d*x + c))*log(-I*cos(d*x + c) + sin(d*x + c) + 1) + 6*(c*f^2 - d*f*e + (c*f^2
- d*f*e)*cos(d*x + c) + (c*f^2 - d*f*e)*sin(d*x + c))*log(-cos(d*x + c) + I*sin(d*x + c) + I) + (d^3*f^2*x^3 -
3*d^2*f^2*x^2 + 3*(d^3*x - d^2)*e^2 + 3*(d^3*f*x^2 - 2*d^2*f*x)*e)*sin(d*x + c))/(a*d^3*cos(d*x + c) + a*d^3*
sin(d*x + c) + a*d^3)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {e^{2} \sin {\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx + \int \frac {f^{2} x^{2} \sin {\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx + \int \frac {2 e f x \sin {\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(e**2*sin(c + d*x)/(sin(c + d*x) + 1), x) + Integral(f**2*x**2*sin(c + d*x)/(sin(c + d*x) + 1), x) +
Integral(2*e*f*x*sin(c + d*x)/(sin(c + d*x) + 1), x))/a

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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*sin(d*x+c)/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sin \left (c+d\,x\right )\,{\left (e+f\,x\right )}^2}{a+a\,\sin \left (c+d\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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