3.2.2 \(\int (c+d x)^2 (a+a \sin (e+f x))^2 \, dx\) [102]

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

[Out]

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

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Rubi [A]
time = 0.12, 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 = {3398, 3377, 2718, 3392, 32, 2715, 8} \begin {gather*} \frac {a^2 d (c+d x) \sin ^2(e+f x)}{2 f^2}+\frac {4 a^2 d (c+d x) \sin (e+f x)}{f^2}-\frac {2 a^2 (c+d x)^2 \cos (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 \cos (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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(a^2*d^2*x)/f^2 + (a^2*(c + d*x)^3)/(2*d) + (4*a^2*d^2*Cos[e + f*x])/f^3 - (2*a^2*(c + d*x)^2*Cos[e + f*x
])/f + (4*a^2*d*(c + d*x)*Sin[e + f*x])/f^2 + (a^2*d^2*Cos[e + f*x]*Sin[e + f*x])/(4*f^3) - (a^2*(c + d*x)^2*C
os[e + f*x]*Sin[e + f*x])/(2*f) + (a^2*d*(c + d*x)*Sin[e + f*x]^2)/(2*f^2)

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 2715

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] && Integ
erQ[2*n]

Rule 2718

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

Rule 3377

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 3392

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 3398

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

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Mathematica [A]
time = 0.39, size = 182, normalized size = 1.08 \begin {gather*} \frac {a^2 \left (12 c^2 f^3 x+12 c d f^3 x^2+4 d^2 f^3 x^3-16 \left (c^2 f^2+2 c d f^2 x+d^2 \left (-2+f^2 x^2\right )\right ) \cos (e+f x)-2 d f (c+d x) \cos (2 (e+f x))+32 c d f \sin (e+f x)+32 d^2 f x \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))\right )}{8 f^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^2*(a + a*Sin[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 - 16*(c^2*f^2 + 2*c*d*f^2*x + d^2*(-2 + f^2*x^2))*Cos[e +
f*x] - 2*d*f*(c + d*x)*Cos[2*(e + f*x)] + 32*c*d*f*Sin[e + f*x] + 32*d^2*f*x*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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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(566\) vs. \(2(158)=316\).
time = 0.10, size = 567, normalized size = 3.38 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 549 vs. \(2 (166) = 332\).
time = 0.31, size = 549, normalized size = 3.27 \begin {gather*} \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} c d}{f} - 48 \, a^{2} c^{2} \cos \left (f x + e\right ) - \frac {24 \, {\left (f x + e\right )}^{2} a^{2} d^{2} e}{f^{2}} - \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} + \frac {96 \, a^{2} c d \cos \left (f x + e\right ) e}{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} c d}{f} - \frac {96 \, {\left ({\left (f x + e\right )} \cos \left (f x + e\right ) - \sin \left (f x + e\right )\right )} a^{2} c d}{f} + \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 {48 \, a^{2} d^{2} \cos \left (f x + e\right ) e^{2}}{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} 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^{2} d^{2} e}{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 )} a^{2} d^{2}}{f^{2}} - \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^{2} d^{2}}{f^{2}}}{24 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^2*(a+a*sin(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*c*d/f - 48*a^2*c^2*cos(f*x + e) - 24*(f*x + e)^2*a^2*d^2*e/f^2 - 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 + 96*a^2*c*d*cos(f*x + e)*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*c*d/f - 96*((f*x + e)*cos(f*x + e) - sin(f*x + e))*a^2*c*d/f + 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 - 48*a^2*d^2*cos(f*x + e)*e^2/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*d^2*e/f^2 + 96*((f*x + e)*cos(f*x + e) - s
in(f*x + e))*a^2*d^2*e/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))*a^2*d^2/f^2 - 48*(((f*x + e)^2 - 2)*cos(f*x + e) - 2*(f*x + e)*sin(f*x + e))*a^2*d^2/f^2)/f

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Fricas [A]
time = 0.39, size = 216, normalized size = 1.29 \begin {gather*} \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 - 8 \, {\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 )} \cos \left (f x + e\right ) + {\left (16 \, a^{2} d^{2} f x + 16 \, a^{2} c d f - {\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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^2*(a+a*sin(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 - 8*(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)*cos(f*x + e) + (16*a^2*d^2*f*x + 1
6*a^2*c*d*f - (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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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 456 vs. \(2 (163) = 326\).
time = 0.29, size = 456, normalized size = 2.71 \begin {gather*} \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} \cos {\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 \cos {\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 \sin {\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} \cos {\left (e + f x \right )}}{f} + \frac {a^{2} d^{2} x \sin ^{2}{\left (e + f x \right )}}{4 f^{2}} + \frac {4 a^{2} d^{2} x \sin {\left (e + f x \right )}}{f^{2}} - \frac {a^{2} d^{2} x \cos ^{2}{\left (e + f x \right )}}{4 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} \cos {\left (e + f x \right )}}{f^{3}} & \text {for}\: f \neq 0 \\\left (a \sin {\left (e \right )} + a\right )^{2} \left (c^{2} x + c d x^{2} + \frac {d^{2} x^{3}}{3}\right ) & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Giac [A]
time = 1.75, size = 207, normalized size = 1.23 \begin {gather*} \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 {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 )} \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 {4 \, {\left (a^{2} d^{2} f x + a^{2} c d f\right )} \sin \left (f x + e\right )}{f^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Mupad [B]
time = 0.97, size = 255, normalized size = 1.52 \begin {gather*} -\frac {8\,a^2\,c^2\,f^2\,\cos \left (e+f\,x\right )-\frac {a^2\,d^2\,\sin \left (2\,e+2\,f\,x\right )}{2}-16\,a^2\,d^2\,\cos \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\,\sin \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\,\sin \left (e+f\,x\right )+8\,a^2\,d^2\,f^2\,x^2\,\cos \left (e+f\,x\right )+16\,a^2\,c\,d\,f^2\,x\,\cos \left (e+f\,x\right )+2\,a^2\,c\,d\,f^2\,x\,\sin \left (2\,e+2\,f\,x\right )}{4\,f^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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