3.124 \(\int (c+d x)^2 (a+a \cos (e+f x))^2 \, dx\)

Optimal. Leaf size=168 \[ \frac {a^2 d (c+d x) \cos ^2(e+f x)}{2 f^2}+\frac {4 a^2 d (c+d x) \cos (e+f x)}{f^2}+\frac {2 a^2 (c+d x)^2 \sin (e+f x)}{f}+\frac {a^2 (c+d x)^2 \sin (e+f x) \cos (e+f x)}{2 f}+\frac {a^2 (c+d x)^3}{2 d}-\frac {4 a^2 d^2 \sin (e+f x)}{f^3}-\frac {a^2 d^2 \sin (e+f x) \cos (e+f x)}{4 f^3}-\frac {a^2 d^2 x}{4 f^2} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 168, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {3317, 3296, 2637, 3311, 32, 2635, 8} \[ \frac {a^2 d (c+d x) \cos ^2(e+f x)}{2 f^2}+\frac {4 a^2 d (c+d x) \cos (e+f x)}{f^2}+\frac {2 a^2 (c+d x)^2 \sin (e+f x)}{f}+\frac {a^2 (c+d x)^2 \sin (e+f x) \cos (e+f x)}{2 f}+\frac {a^2 (c+d x)^3}{2 d}-\frac {4 a^2 d^2 \sin (e+f x)}{f^3}-\frac {a^2 d^2 \sin (e+f x) \cos (e+f x)}{4 f^3}-\frac {a^2 d^2 x}{4 f^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

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

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 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

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 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 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])

Rubi steps

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

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Mathematica [A]  time = 0.65, size = 193, normalized size = 1.15 \[ \frac {a^2 \left (16 c^2 f^2 \sin (e+f x)+2 c^2 f^2 \sin (2 (e+f x))+12 c^2 f^3 x+32 c d f^2 x \sin (e+f x)+4 c d f^2 x \sin (2 (e+f x))+32 d f (c+d x) \cos (e+f x)+2 d f (c+d x) \cos (2 (e+f x))+12 c d f^3 x^2+16 d^2 f^2 x^2 \sin (e+f x)+2 d^2 f^2 x^2 \sin (2 (e+f x))-32 d^2 \sin (e+f x)-d^2 \sin (2 (e+f x))+4 d^2 f^3 x^3\right )}{8 f^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(12*c^2*f^3*x + 12*c*d*f^3*x^2 + 4*d^2*f^3*x^3 + 32*d*f*(c + d*x)*Cos[e + f*x] + 2*d*f*(c + d*x)*Cos[2*(e
+ f*x)] - 32*d^2*Sin[e + f*x] + 16*c^2*f^2*Sin[e + f*x] + 32*c*d*f^2*x*Sin[e + f*x] + 16*d^2*f^2*x^2*Sin[e +
f*x] - d^2*Sin[2*(e + f*x)] + 2*c^2*f^2*Sin[2*(e + f*x)] + 4*c*d*f^2*x*Sin[2*(e + f*x)] + 2*d^2*f^2*x^2*Sin[2*
(e + f*x)]))/(8*f^3)

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fricas [A]  time = 0.58, size = 212, normalized size = 1.26 \[ \frac {2 \, a^{2} d^{2} f^{3} x^{3} + 6 \, a^{2} c d f^{3} x^{2} + 2 \, {\left (a^{2} d^{2} f x + a^{2} c d f\right )} \cos \left (f x + e\right )^{2} + {\left (6 \, a^{2} c^{2} f^{3} - a^{2} d^{2} f\right )} x + 16 \, {\left (a^{2} d^{2} f x + a^{2} c d f\right )} \cos \left (f x + e\right ) + {\left (8 \, a^{2} d^{2} f^{2} x^{2} + 16 \, a^{2} c d f^{2} x + 8 \, a^{2} c^{2} f^{2} - 16 \, a^{2} d^{2} + {\left (2 \, a^{2} d^{2} f^{2} x^{2} + 4 \, a^{2} c d f^{2} x + 2 \, a^{2} c^{2} f^{2} - a^{2} d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{4 \, f^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*a^2*d^2*f^3*x^3 + 6*a^2*c*d*f^3*x^2 + 2*(a^2*d^2*f*x + a^2*c*d*f)*cos(f*x + e)^2 + (6*a^2*c^2*f^3 - a^2
*d^2*f)*x + 16*(a^2*d^2*f*x + a^2*c*d*f)*cos(f*x + e) + (8*a^2*d^2*f^2*x^2 + 16*a^2*c*d*f^2*x + 8*a^2*c^2*f^2
- 16*a^2*d^2 + (2*a^2*d^2*f^2*x^2 + 4*a^2*c*d*f^2*x + 2*a^2*c^2*f^2 - a^2*d^2)*cos(f*x + e))*sin(f*x + e))/f^3

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giac [A]  time = 0.81, size = 207, normalized size = 1.23 \[ \frac {1}{2} \, a^{2} d^{2} x^{3} + \frac {3}{2} \, a^{2} c d x^{2} + \frac {3}{2} \, a^{2} c^{2} x + \frac {{\left (a^{2} d^{2} f x + a^{2} c d f\right )} \cos \left (2 \, f x + 2 \, e\right )}{4 \, f^{3}} + \frac {4 \, {\left (a^{2} d^{2} f x + a^{2} c d f\right )} \cos \left (f x + e\right )}{f^{3}} + \frac {{\left (2 \, a^{2} d^{2} f^{2} x^{2} + 4 \, a^{2} c d f^{2} x + 2 \, a^{2} c^{2} f^{2} - a^{2} d^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )}{8 \, f^{3}} + \frac {2 \, {\left (a^{2} d^{2} f^{2} x^{2} + 2 \, a^{2} c d f^{2} x + a^{2} c^{2} f^{2} - 2 \, a^{2} d^{2}\right )} \sin \left (f x + e\right )}{f^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.05, size = 564, normalized size = 3.36 \[ \frac {\frac {a^{2} d^{2} \left (\left (f x +e \right )^{2} \left (\frac {\cos \left (f x +e \right ) \sin \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )+\frac {\left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right )}{2}-\frac {\cos \left (f x +e \right ) \sin \left (f x +e \right )}{4}-\frac {f x}{4}-\frac {e}{4}-\frac {\left (f x +e \right )^{3}}{3}\right )}{f^{2}}+\frac {2 a^{2} c d \left (\left (f x +e \right ) \left (\frac {\cos \left (f x +e \right ) \sin \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )-\frac {\left (f x +e \right )^{2}}{4}-\frac {\left (\sin ^{2}\left (f x +e \right )\right )}{4}\right )}{f}-\frac {2 a^{2} d^{2} e \left (\left (f x +e \right ) \left (\frac {\cos \left (f x +e \right ) \sin \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )-\frac {\left (f x +e \right )^{2}}{4}-\frac {\left (\sin ^{2}\left (f x +e \right )\right )}{4}\right )}{f^{2}}+a^{2} c^{2} \left (\frac {\cos \left (f x +e \right ) \sin \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )-\frac {2 a^{2} c d e \left (\frac {\cos \left (f x +e \right ) \sin \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )}{f}+\frac {a^{2} d^{2} e^{2} \left (\frac {\cos \left (f x +e \right ) \sin \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )}{f^{2}}+\frac {2 a^{2} d^{2} \left (\left (f x +e \right )^{2} \sin \left (f x +e \right )-2 \sin \left (f x +e \right )+2 \left (f x +e \right ) \cos \left (f x +e \right )\right )}{f^{2}}+\frac {4 a^{2} c d \left (\cos \left (f x +e \right )+\left (f x +e \right ) \sin \left (f x +e \right )\right )}{f}-\frac {4 a^{2} d^{2} e \left (\cos \left (f x +e \right )+\left (f x +e \right ) \sin \left (f x +e \right )\right )}{f^{2}}+2 a^{2} c^{2} \sin \left (f x +e \right )-\frac {4 a^{2} c d e \sin \left (f x +e \right )}{f}+\frac {2 a^{2} d^{2} e^{2} \sin \left (f x +e \right )}{f^{2}}+\frac {a^{2} d^{2} \left (f x +e \right )^{3}}{3 f^{2}}+\frac {a^{2} c d \left (f x +e \right )^{2}}{f}-\frac {a^{2} d^{2} e \left (f x +e \right )^{2}}{f^{2}}+a^{2} c^{2} \left (f x +e \right )-\frac {2 a^{2} c d e \left (f x +e \right )}{f}+\frac {a^{2} d^{2} e^{2} \left (f x +e \right )}{f^{2}}}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 1.00, size = 494, normalized size = 2.94 \[ \frac {6 \, {\left (2 \, f x + 2 \, e + \sin \left (2 \, f x + 2 \, e\right )\right )} a^{2} c^{2} + 24 \, {\left (f x + e\right )} a^{2} c^{2} + \frac {8 \, {\left (f x + e\right )}^{3} a^{2} d^{2}}{f^{2}} - \frac {24 \, {\left (f x + e\right )}^{2} a^{2} d^{2} e}{f^{2}} + \frac {6 \, {\left (2 \, f x + 2 \, e + \sin \left (2 \, f x + 2 \, e\right )\right )} a^{2} d^{2} e^{2}}{f^{2}} + \frac {24 \, {\left (f x + e\right )} a^{2} d^{2} e^{2}}{f^{2}} + \frac {24 \, {\left (f x + e\right )}^{2} a^{2} c d}{f} - \frac {12 \, {\left (2 \, f x + 2 \, e + \sin \left (2 \, f x + 2 \, e\right )\right )} a^{2} c d e}{f} - \frac {48 \, {\left (f x + e\right )} a^{2} c d e}{f} + 48 \, a^{2} c^{2} \sin \left (f x + e\right ) + \frac {48 \, a^{2} d^{2} e^{2} \sin \left (f x + e\right )}{f^{2}} - \frac {96 \, a^{2} c d e \sin \left (f x + e\right )}{f} - \frac {6 \, {\left (2 \, {\left (f x + e\right )}^{2} + 2 \, {\left (f x + e\right )} \sin \left (2 \, f x + 2 \, e\right ) + \cos \left (2 \, f x + 2 \, e\right )\right )} a^{2} d^{2} e}{f^{2}} - \frac {96 \, {\left ({\left (f x + e\right )} \sin \left (f x + e\right ) + \cos \left (f x + e\right )\right )} a^{2} d^{2} e}{f^{2}} + \frac {6 \, {\left (2 \, {\left (f x + e\right )}^{2} + 2 \, {\left (f x + e\right )} \sin \left (2 \, f x + 2 \, e\right ) + \cos \left (2 \, f x + 2 \, e\right )\right )} a^{2} c d}{f} + \frac {96 \, {\left ({\left (f x + e\right )} \sin \left (f x + e\right ) + \cos \left (f x + e\right )\right )} a^{2} c d}{f} + \frac {{\left (4 \, {\left (f x + e\right )}^{3} + 6 \, {\left (f x + e\right )} \cos \left (2 \, f x + 2 \, e\right ) + 3 \, {\left (2 \, {\left (f x + e\right )}^{2} - 1\right )} \sin \left (2 \, f x + 2 \, e\right )\right )} a^{2} d^{2}}{f^{2}} + \frac {48 \, {\left (2 \, {\left (f x + e\right )} \cos \left (f x + e\right ) + {\left ({\left (f x + e\right )}^{2} - 2\right )} \sin \left (f x + e\right )\right )} a^{2} d^{2}}{f^{2}}}{24 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.59, size = 255, normalized size = 1.52 \[ \frac {8\,a^2\,c^2\,f^2\,\sin \left (e+f\,x\right )-\frac {a^2\,d^2\,\sin \left (2\,e+2\,f\,x\right )}{2}-16\,a^2\,d^2\,\sin \left (e+f\,x\right )+6\,a^2\,c^2\,f^3\,x+a^2\,c^2\,f^2\,\sin \left (2\,e+2\,f\,x\right )+2\,a^2\,d^2\,f^3\,x^3+a^2\,c\,d\,f\,\cos \left (2\,e+2\,f\,x\right )+16\,a^2\,d^2\,f\,x\,\cos \left (e+f\,x\right )+a^2\,d^2\,f^2\,x^2\,\sin \left (2\,e+2\,f\,x\right )+6\,a^2\,c\,d\,f^3\,x^2+a^2\,d^2\,f\,x\,\cos \left (2\,e+2\,f\,x\right )+16\,a^2\,c\,d\,f\,\cos \left (e+f\,x\right )+8\,a^2\,d^2\,f^2\,x^2\,\sin \left (e+f\,x\right )+16\,a^2\,c\,d\,f^2\,x\,\sin \left (e+f\,x\right )+2\,a^2\,c\,d\,f^2\,x\,\sin \left (2\,e+2\,f\,x\right )}{4\,f^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(8*a^2*c^2*f^2*sin(e + f*x) - (a^2*d^2*sin(2*e + 2*f*x))/2 - 16*a^2*d^2*sin(e + f*x) + 6*a^2*c^2*f^3*x + a^2*c
^2*f^2*sin(2*e + 2*f*x) + 2*a^2*d^2*f^3*x^3 + a^2*c*d*f*cos(2*e + 2*f*x) + 16*a^2*d^2*f*x*cos(e + f*x) + a^2*d
^2*f^2*x^2*sin(2*e + 2*f*x) + 6*a^2*c*d*f^3*x^2 + a^2*d^2*f*x*cos(2*e + 2*f*x) + 16*a^2*c*d*f*cos(e + f*x) + 8
*a^2*d^2*f^2*x^2*sin(e + f*x) + 16*a^2*c*d*f^2*x*sin(e + f*x) + 2*a^2*c*d*f^2*x*sin(2*e + 2*f*x))/(4*f^3)

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sympy [A]  time = 1.59, size = 456, normalized size = 2.71 \[ \begin {cases} \frac {a^{2} c^{2} x \sin ^{2}{\left (e + f x \right )}}{2} + \frac {a^{2} c^{2} x \cos ^{2}{\left (e + f x \right )}}{2} + a^{2} c^{2} x + \frac {a^{2} c^{2} \sin {\left (e + f x \right )} \cos {\left (e + f x \right )}}{2 f} + \frac {2 a^{2} c^{2} \sin {\left (e + f x \right )}}{f} + \frac {a^{2} c d x^{2} \sin ^{2}{\left (e + f x \right )}}{2} + \frac {a^{2} c d x^{2} \cos ^{2}{\left (e + f x \right )}}{2} + a^{2} c d x^{2} + \frac {a^{2} c d x \sin {\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} + \frac {4 a^{2} c d x \sin {\left (e + f x \right )}}{f} - \frac {a^{2} c d \sin ^{2}{\left (e + f x \right )}}{2 f^{2}} + \frac {4 a^{2} c d \cos {\left (e + f x \right )}}{f^{2}} + \frac {a^{2} d^{2} x^{3} \sin ^{2}{\left (e + f x \right )}}{6} + \frac {a^{2} d^{2} x^{3} \cos ^{2}{\left (e + f x \right )}}{6} + \frac {a^{2} d^{2} x^{3}}{3} + \frac {a^{2} d^{2} x^{2} \sin {\left (e + f x \right )} \cos {\left (e + f x \right )}}{2 f} + \frac {2 a^{2} d^{2} x^{2} \sin {\left (e + f x \right )}}{f} - \frac {a^{2} d^{2} x \sin ^{2}{\left (e + f x \right )}}{4 f^{2}} + \frac {a^{2} d^{2} x \cos ^{2}{\left (e + f x \right )}}{4 f^{2}} + \frac {4 a^{2} d^{2} x \cos {\left (e + f x \right )}}{f^{2}} - \frac {a^{2} d^{2} \sin {\left (e + f x \right )} \cos {\left (e + f x \right )}}{4 f^{3}} - \frac {4 a^{2} d^{2} \sin {\left (e + f x \right )}}{f^{3}} & \text {for}\: f \neq 0 \\\left (a \cos {\relax (e )} + a\right )^{2} \left (c^{2} x + c d x^{2} + \frac {d^{2} x^{3}}{3}\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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