3.158 \(\int (c+d x)^2 (a+b \sin (e+f x))^2 \, dx\)

Optimal. Leaf size=182 \[ \frac{a^2 (c+d x)^3}{3 d}+\frac{4 a b d (c+d x) \sin (e+f x)}{f^2}-\frac{2 a b (c+d x)^2 \cos (e+f x)}{f}+\frac{4 a b d^2 \cos (e+f x)}{f^3}+\frac{b^2 d (c+d x) \sin ^2(e+f x)}{2 f^2}-\frac{b^2 (c+d x)^2 \sin (e+f x) \cos (e+f x)}{2 f}+\frac{b^2 (c+d x)^3}{6 d}+\frac{b^2 d^2 \sin (e+f x) \cos (e+f x)}{4 f^3}-\frac{b^2 d^2 x}{4 f^2} \]

[Out]

-(b^2*d^2*x)/(4*f^2) + (a^2*(c + d*x)^3)/(3*d) + (b^2*(c + d*x)^3)/(6*d) + (4*a*b*d^2*Cos[e + f*x])/f^3 - (2*a
*b*(c + d*x)^2*Cos[e + f*x])/f + (4*a*b*d*(c + d*x)*Sin[e + f*x])/f^2 + (b^2*d^2*Cos[e + f*x]*Sin[e + f*x])/(4
*f^3) - (b^2*(c + d*x)^2*Cos[e + f*x]*Sin[e + f*x])/(2*f) + (b^2*d*(c + d*x)*Sin[e + f*x]^2)/(2*f^2)

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Rubi [A]  time = 0.191988, antiderivative size = 182, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, Rules used = {3317, 3296, 2638, 3311, 32, 2635, 8} \[ \frac{a^2 (c+d x)^3}{3 d}+\frac{4 a b d (c+d x) \sin (e+f x)}{f^2}-\frac{2 a b (c+d x)^2 \cos (e+f x)}{f}+\frac{4 a b d^2 \cos (e+f x)}{f^3}+\frac{b^2 d (c+d x) \sin ^2(e+f x)}{2 f^2}-\frac{b^2 (c+d x)^2 \sin (e+f x) \cos (e+f x)}{2 f}+\frac{b^2 (c+d x)^3}{6 d}+\frac{b^2 d^2 \sin (e+f x) \cos (e+f x)}{4 f^3}-\frac{b^2 d^2 x}{4 f^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b^2*d^2*x)/(4*f^2) + (a^2*(c + d*x)^3)/(3*d) + (b^2*(c + d*x)^3)/(6*d) + (4*a*b*d^2*Cos[e + f*x])/f^3 - (2*a
*b*(c + d*x)^2*Cos[e + f*x])/f + (4*a*b*d*(c + d*x)*Sin[e + f*x])/f^2 + (b^2*d^2*Cos[e + f*x]*Sin[e + f*x])/(4
*f^3) - (b^2*(c + d*x)^2*Cos[e + f*x]*Sin[e + f*x])/(2*f) + (b^2*d*(c + d*x)*Sin[e + f*x]^2)/(2*f^2)

Rule 3317

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

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 3311

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

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 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int (c+d x)^2 (a+b \sin (e+f x))^2 \, dx &=\int \left (a^2 (c+d x)^2+2 a b (c+d x)^2 \sin (e+f x)+b^2 (c+d x)^2 \sin ^2(e+f x)\right ) \, dx\\ &=\frac{a^2 (c+d x)^3}{3 d}+(2 a b) \int (c+d x)^2 \sin (e+f x) \, dx+b^2 \int (c+d x)^2 \sin ^2(e+f x) \, dx\\ &=\frac{a^2 (c+d x)^3}{3 d}-\frac{2 a b (c+d x)^2 \cos (e+f x)}{f}-\frac{b^2 (c+d x)^2 \cos (e+f x) \sin (e+f x)}{2 f}+\frac{b^2 d (c+d x) \sin ^2(e+f x)}{2 f^2}+\frac{1}{2} b^2 \int (c+d x)^2 \, dx-\frac{\left (b^2 d^2\right ) \int \sin ^2(e+f x) \, dx}{2 f^2}+\frac{(4 a b d) \int (c+d x) \cos (e+f x) \, dx}{f}\\ &=\frac{a^2 (c+d x)^3}{3 d}+\frac{b^2 (c+d x)^3}{6 d}-\frac{2 a b (c+d x)^2 \cos (e+f x)}{f}+\frac{4 a b d (c+d x) \sin (e+f x)}{f^2}+\frac{b^2 d^2 \cos (e+f x) \sin (e+f x)}{4 f^3}-\frac{b^2 (c+d x)^2 \cos (e+f x) \sin (e+f x)}{2 f}+\frac{b^2 d (c+d x) \sin ^2(e+f x)}{2 f^2}-\frac{\left (4 a b d^2\right ) \int \sin (e+f x) \, dx}{f^2}-\frac{\left (b^2 d^2\right ) \int 1 \, dx}{4 f^2}\\ &=-\frac{b^2 d^2 x}{4 f^2}+\frac{a^2 (c+d x)^3}{3 d}+\frac{b^2 (c+d x)^3}{6 d}+\frac{4 a b d^2 \cos (e+f x)}{f^3}-\frac{2 a b (c+d x)^2 \cos (e+f x)}{f}+\frac{4 a b d (c+d x) \sin (e+f x)}{f^2}+\frac{b^2 d^2 \cos (e+f x) \sin (e+f x)}{4 f^3}-\frac{b^2 (c+d x)^2 \cos (e+f x) \sin (e+f x)}{2 f}+\frac{b^2 d (c+d x) \sin ^2(e+f x)}{2 f^2}\\ \end{align*}

Mathematica [A]  time = 0.726044, size = 249, normalized size = 1.37 \[ \frac{24 a^2 c^2 f^3 x+24 a^2 c d f^3 x^2+8 a^2 d^2 f^3 x^3-48 a b \left (c^2 f^2+2 c d f^2 x+d^2 \left (f^2 x^2-2\right )\right ) \cos (e+f x)+96 a b c d f \sin (e+f x)+96 a b d^2 f x \sin (e+f x)-6 b^2 c^2 f^2 \sin (2 (e+f x))+12 b^2 c^2 f^3 x-12 b^2 c d f^2 x \sin (2 (e+f x))-6 b^2 d f (c+d x) \cos (2 (e+f x))+12 b^2 c d f^3 x^2-6 b^2 d^2 f^2 x^2 \sin (2 (e+f x))+3 b^2 d^2 \sin (2 (e+f x))+4 b^2 d^2 f^3 x^3}{24 f^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(24*a^2*c^2*f^3*x + 12*b^2*c^2*f^3*x + 24*a^2*c*d*f^3*x^2 + 12*b^2*c*d*f^3*x^2 + 8*a^2*d^2*f^3*x^3 + 4*b^2*d^2
*f^3*x^3 - 48*a*b*(c^2*f^2 + 2*c*d*f^2*x + d^2*(-2 + f^2*x^2))*Cos[e + f*x] - 6*b^2*d*f*(c + d*x)*Cos[2*(e + f
*x)] + 96*a*b*c*d*f*Sin[e + f*x] + 96*a*b*d^2*f*x*Sin[e + f*x] + 3*b^2*d^2*Sin[2*(e + f*x)] - 6*b^2*c^2*f^2*Si
n[2*(e + f*x)] - 12*b^2*c*d*f^2*x*Sin[2*(e + f*x)] - 6*b^2*d^2*f^2*x^2*Sin[2*(e + f*x)])/(24*f^3)

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Maple [B]  time = 0.013, size = 561, normalized size = 3.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*(1/3*a^2/f^2*d^2*(f*x+e)^3+a^2/f*c*d*(f*x+e)^2-a^2/f^2*d^2*e*(f*x+e)^2+a^2*c^2*(f*x+e)-2*a^2/f*c*d*e*(f*x+
e)+a^2/f^2*d^2*e^2*(f*x+e)+2/f^2*a*b*d^2*(-(f*x+e)^2*cos(f*x+e)+2*cos(f*x+e)+2*(f*x+e)*sin(f*x+e))+4/f*a*b*c*d
*(sin(f*x+e)-(f*x+e)*cos(f*x+e))-4/f^2*a*b*d^2*e*(sin(f*x+e)-(f*x+e)*cos(f*x+e))-2*a*b*c^2*cos(f*x+e)+4/f*a*b*
c*d*e*cos(f*x+e)-2/f^2*a*b*d^2*e^2*cos(f*x+e)+1/f^2*b^2*d^2*((f*x+e)^2*(-1/2*sin(f*x+e)*cos(f*x+e)+1/2*f*x+1/2
*e)-1/2*(f*x+e)*cos(f*x+e)^2+1/4*sin(f*x+e)*cos(f*x+e)+1/4*f*x+1/4*e-1/3*(f*x+e)^3)+2/f*b^2*c*d*((f*x+e)*(-1/2
*sin(f*x+e)*cos(f*x+e)+1/2*f*x+1/2*e)-1/4*(f*x+e)^2+1/4*sin(f*x+e)^2)-2/f^2*b^2*d^2*e*((f*x+e)*(-1/2*sin(f*x+e
)*cos(f*x+e)+1/2*f*x+1/2*e)-1/4*(f*x+e)^2+1/4*sin(f*x+e)^2)+b^2*c^2*(-1/2*sin(f*x+e)*cos(f*x+e)+1/2*f*x+1/2*e)
-2/f*b^2*c*d*e*(-1/2*sin(f*x+e)*cos(f*x+e)+1/2*f*x+1/2*e)+1/f^2*b^2*d^2*e^2*(-1/2*sin(f*x+e)*cos(f*x+e)+1/2*f*
x+1/2*e))

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Maxima [B]  time = 1.04999, size = 678, normalized size = 3.73 \begin{align*} \frac{24 \,{\left (f x + e\right )} a^{2} c^{2} + 6 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} b^{2} c^{2} + \frac{8 \,{\left (f x + e\right )}^{3} a^{2} d^{2}}{f^{2}} - \frac{24 \,{\left (f x + e\right )}^{2} a^{2} d^{2} e}{f^{2}} + \frac{24 \,{\left (f x + e\right )} a^{2} d^{2} e^{2}}{f^{2}} + \frac{6 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} b^{2} d^{2} e^{2}}{f^{2}} + \frac{24 \,{\left (f x + e\right )}^{2} a^{2} c d}{f} - \frac{48 \,{\left (f x + e\right )} a^{2} c d e}{f} - \frac{12 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} b^{2} c d e}{f} - 48 \, a b c^{2} \cos \left (f x + e\right ) - \frac{48 \, a b d^{2} e^{2} \cos \left (f x + e\right )}{f^{2}} + \frac{96 \, a b c d e \cos \left (f x + e\right )}{f} + \frac{96 \,{\left ({\left (f x + e\right )} \cos \left (f x + e\right ) - \sin \left (f x + e\right )\right )} a b d^{2} e}{f^{2}} - \frac{6 \,{\left (2 \,{\left (f x + e\right )}^{2} - 2 \,{\left (f x + e\right )} \sin \left (2 \, f x + 2 \, e\right ) - \cos \left (2 \, f x + 2 \, e\right )\right )} b^{2} d^{2} e}{f^{2}} - \frac{96 \,{\left ({\left (f x + e\right )} \cos \left (f x + e\right ) - \sin \left (f x + e\right )\right )} a b c d}{f} + \frac{6 \,{\left (2 \,{\left (f x + e\right )}^{2} - 2 \,{\left (f x + e\right )} \sin \left (2 \, f x + 2 \, e\right ) - \cos \left (2 \, f x + 2 \, e\right )\right )} b^{2} c d}{f} - \frac{48 \,{\left ({\left ({\left (f x + e\right )}^{2} - 2\right )} \cos \left (f x + e\right ) - 2 \,{\left (f x + e\right )} \sin \left (f x + e\right )\right )} a b d^{2}}{f^{2}} + \frac{{\left (4 \,{\left (f x + e\right )}^{3} - 6 \,{\left (f x + e\right )} \cos \left (2 \, f x + 2 \, e\right ) - 3 \,{\left (2 \,{\left (f x + e\right )}^{2} - 1\right )} \sin \left (2 \, f x + 2 \, e\right )\right )} b^{2} d^{2}}{f^{2}}}{24 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(24*(f*x + e)*a^2*c^2 + 6*(2*f*x + 2*e - sin(2*f*x + 2*e))*b^2*c^2 + 8*(f*x + e)^3*a^2*d^2/f^2 - 24*(f*x
+ e)^2*a^2*d^2*e/f^2 + 24*(f*x + e)*a^2*d^2*e^2/f^2 + 6*(2*f*x + 2*e - sin(2*f*x + 2*e))*b^2*d^2*e^2/f^2 + 24*
(f*x + e)^2*a^2*c*d/f - 48*(f*x + e)*a^2*c*d*e/f - 12*(2*f*x + 2*e - sin(2*f*x + 2*e))*b^2*c*d*e/f - 48*a*b*c^
2*cos(f*x + e) - 48*a*b*d^2*e^2*cos(f*x + e)/f^2 + 96*a*b*c*d*e*cos(f*x + e)/f + 96*((f*x + e)*cos(f*x + e) -
sin(f*x + e))*a*b*d^2*e/f^2 - 6*(2*(f*x + e)^2 - 2*(f*x + e)*sin(2*f*x + 2*e) - cos(2*f*x + 2*e))*b^2*d^2*e/f^
2 - 96*((f*x + e)*cos(f*x + e) - sin(f*x + e))*a*b*c*d/f + 6*(2*(f*x + e)^2 - 2*(f*x + e)*sin(2*f*x + 2*e) - c
os(2*f*x + 2*e))*b^2*c*d/f - 48*(((f*x + e)^2 - 2)*cos(f*x + e) - 2*(f*x + e)*sin(f*x + e))*a*b*d^2/f^2 + (4*(
f*x + e)^3 - 6*(f*x + e)*cos(2*f*x + 2*e) - 3*(2*(f*x + e)^2 - 1)*sin(2*f*x + 2*e))*b^2*d^2/f^2)/f

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Fricas [A]  time = 2.08423, size = 495, normalized size = 2.72 \begin{align*} \frac{2 \,{\left (2 \, a^{2} + b^{2}\right )} d^{2} f^{3} x^{3} + 6 \,{\left (2 \, a^{2} + b^{2}\right )} c d f^{3} x^{2} - 6 \,{\left (b^{2} d^{2} f x + b^{2} c d f\right )} \cos \left (f x + e\right )^{2} + 3 \,{\left (2 \,{\left (2 \, a^{2} + b^{2}\right )} c^{2} f^{3} + b^{2} d^{2} f\right )} x - 24 \,{\left (a b d^{2} f^{2} x^{2} + 2 \, a b c d f^{2} x + a b c^{2} f^{2} - 2 \, a b d^{2}\right )} \cos \left (f x + e\right ) + 3 \,{\left (16 \, a b d^{2} f x + 16 \, a b c d f -{\left (2 \, b^{2} d^{2} f^{2} x^{2} + 4 \, b^{2} c d f^{2} x + 2 \, b^{2} c^{2} f^{2} - b^{2} d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{12 \, f^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(2*(2*a^2 + b^2)*d^2*f^3*x^3 + 6*(2*a^2 + b^2)*c*d*f^3*x^2 - 6*(b^2*d^2*f*x + b^2*c*d*f)*cos(f*x + e)^2 +
3*(2*(2*a^2 + b^2)*c^2*f^3 + b^2*d^2*f)*x - 24*(a*b*d^2*f^2*x^2 + 2*a*b*c*d*f^2*x + a*b*c^2*f^2 - 2*a*b*d^2)*
cos(f*x + e) + 3*(16*a*b*d^2*f*x + 16*a*b*c*d*f - (2*b^2*d^2*f^2*x^2 + 4*b^2*c*d*f^2*x + 2*b^2*c^2*f^2 - b^2*d
^2)*cos(f*x + e))*sin(f*x + e))/f^3

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Sympy [A]  time = 2.06795, size = 456, normalized size = 2.51 \begin{align*} \begin{cases} a^{2} c^{2} x + a^{2} c d x^{2} + \frac{a^{2} d^{2} x^{3}}{3} - \frac{2 a b c^{2} \cos{\left (e + f x \right )}}{f} - \frac{4 a b c d x \cos{\left (e + f x \right )}}{f} + \frac{4 a b c d \sin{\left (e + f x \right )}}{f^{2}} - \frac{2 a b d^{2} x^{2} \cos{\left (e + f x \right )}}{f} + \frac{4 a b d^{2} x \sin{\left (e + f x \right )}}{f^{2}} + \frac{4 a b d^{2} \cos{\left (e + f x \right )}}{f^{3}} + \frac{b^{2} c^{2} x \sin ^{2}{\left (e + f x \right )}}{2} + \frac{b^{2} c^{2} x \cos ^{2}{\left (e + f x \right )}}{2} - \frac{b^{2} c^{2} \sin{\left (e + f x \right )} \cos{\left (e + f x \right )}}{2 f} + \frac{b^{2} c d x^{2} \sin ^{2}{\left (e + f x \right )}}{2} + \frac{b^{2} c d x^{2} \cos ^{2}{\left (e + f x \right )}}{2} - \frac{b^{2} c d x \sin{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} - \frac{b^{2} c d \cos ^{2}{\left (e + f x \right )}}{2 f^{2}} + \frac{b^{2} d^{2} x^{3} \sin ^{2}{\left (e + f x \right )}}{6} + \frac{b^{2} d^{2} x^{3} \cos ^{2}{\left (e + f x \right )}}{6} - \frac{b^{2} d^{2} x^{2} \sin{\left (e + f x \right )} \cos{\left (e + f x \right )}}{2 f} + \frac{b^{2} d^{2} x \sin ^{2}{\left (e + f x \right )}}{4 f^{2}} - \frac{b^{2} d^{2} x \cos ^{2}{\left (e + f x \right )}}{4 f^{2}} + \frac{b^{2} d^{2} \sin{\left (e + f x \right )} \cos{\left (e + f x \right )}}{4 f^{3}} & \text{for}\: f \neq 0 \\\left (a + b \sin{\left (e \right )}\right )^{2} \left (c^{2} x + c d x^{2} + \frac{d^{2} x^{3}}{3}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**2*c**2*x + a**2*c*d*x**2 + a**2*d**2*x**3/3 - 2*a*b*c**2*cos(e + f*x)/f - 4*a*b*c*d*x*cos(e + f*
x)/f + 4*a*b*c*d*sin(e + f*x)/f**2 - 2*a*b*d**2*x**2*cos(e + f*x)/f + 4*a*b*d**2*x*sin(e + f*x)/f**2 + 4*a*b*d
**2*cos(e + f*x)/f**3 + b**2*c**2*x*sin(e + f*x)**2/2 + b**2*c**2*x*cos(e + f*x)**2/2 - b**2*c**2*sin(e + f*x)
*cos(e + f*x)/(2*f) + b**2*c*d*x**2*sin(e + f*x)**2/2 + b**2*c*d*x**2*cos(e + f*x)**2/2 - b**2*c*d*x*sin(e + f
*x)*cos(e + f*x)/f - b**2*c*d*cos(e + f*x)**2/(2*f**2) + b**2*d**2*x**3*sin(e + f*x)**2/6 + b**2*d**2*x**3*cos
(e + f*x)**2/6 - b**2*d**2*x**2*sin(e + f*x)*cos(e + f*x)/(2*f) + b**2*d**2*x*sin(e + f*x)**2/(4*f**2) - b**2*
d**2*x*cos(e + f*x)**2/(4*f**2) + b**2*d**2*sin(e + f*x)*cos(e + f*x)/(4*f**3), Ne(f, 0)), ((a + b*sin(e))**2*
(c**2*x + c*d*x**2 + d**2*x**3/3), True))

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Giac [A]  time = 1.12622, size = 309, normalized size = 1.7 \begin{align*} \frac{1}{3} \, a^{2} d^{2} x^{3} + \frac{1}{6} \, b^{2} d^{2} x^{3} + a^{2} c d x^{2} + \frac{1}{2} \, b^{2} c d x^{2} + a^{2} c^{2} x + \frac{1}{2} \, b^{2} c^{2} x - \frac{{\left (b^{2} d^{2} f x + b^{2} c d f\right )} \cos \left (2 \, f x + 2 \, e\right )}{4 \, f^{3}} - \frac{2 \,{\left (a b d^{2} f^{2} x^{2} + 2 \, a b c d f^{2} x + a b c^{2} f^{2} - 2 \, a b d^{2}\right )} \cos \left (f x + e\right )}{f^{3}} - \frac{{\left (2 \, b^{2} d^{2} f^{2} x^{2} + 4 \, b^{2} c d f^{2} x + 2 \, b^{2} c^{2} f^{2} - b^{2} d^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )}{8 \, f^{3}} + \frac{4 \,{\left (a b d^{2} f x + a b c d f\right )} \sin \left (f x + e\right )}{f^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*a^2*d^2*x^3 + 1/6*b^2*d^2*x^3 + a^2*c*d*x^2 + 1/2*b^2*c*d*x^2 + a^2*c^2*x + 1/2*b^2*c^2*x - 1/4*(b^2*d^2*f
*x + b^2*c*d*f)*cos(2*f*x + 2*e)/f^3 - 2*(a*b*d^2*f^2*x^2 + 2*a*b*c*d*f^2*x + a*b*c^2*f^2 - 2*a*b*d^2)*cos(f*x
+ e)/f^3 - 1/8*(2*b^2*d^2*f^2*x^2 + 4*b^2*c*d*f^2*x + 2*b^2*c^2*f^2 - b^2*d^2)*sin(2*f*x + 2*e)/f^3 + 4*(a*b*
d^2*f*x + a*b*c*d*f)*sin(f*x + e)/f^3