3.1.17 \(\int (c+d x)^3 \cos ^3(a+b x) \, dx\) [17]

Optimal. Leaf size=175 \[ -\frac {40 d^3 \cos (a+b x)}{9 b^4}+\frac {2 d (c+d x)^2 \cos (a+b x)}{b^2}-\frac {2 d^3 \cos ^3(a+b x)}{27 b^4}+\frac {d (c+d x)^2 \cos ^3(a+b x)}{3 b^2}-\frac {40 d^2 (c+d x) \sin (a+b x)}{9 b^3}+\frac {2 (c+d x)^3 \sin (a+b x)}{3 b}-\frac {2 d^2 (c+d x) \cos ^2(a+b x) \sin (a+b x)}{9 b^3}+\frac {(c+d x)^3 \cos ^2(a+b x) \sin (a+b x)}{3 b} \]

[Out]

-40/9*d^3*cos(b*x+a)/b^4+2*d*(d*x+c)^2*cos(b*x+a)/b^2-2/27*d^3*cos(b*x+a)^3/b^4+1/3*d*(d*x+c)^2*cos(b*x+a)^3/b
^2-40/9*d^2*(d*x+c)*sin(b*x+a)/b^3+2/3*(d*x+c)^3*sin(b*x+a)/b-2/9*d^2*(d*x+c)*cos(b*x+a)^2*sin(b*x+a)/b^3+1/3*
(d*x+c)^3*cos(b*x+a)^2*sin(b*x+a)/b

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Rubi [A]
time = 0.11, antiderivative size = 175, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3392, 3377, 2718, 3391} \begin {gather*} -\frac {2 d^3 \cos ^3(a+b x)}{27 b^4}-\frac {40 d^3 \cos (a+b x)}{9 b^4}-\frac {40 d^2 (c+d x) \sin (a+b x)}{9 b^3}-\frac {2 d^2 (c+d x) \sin (a+b x) \cos ^2(a+b x)}{9 b^3}+\frac {d (c+d x)^2 \cos ^3(a+b x)}{3 b^2}+\frac {2 d (c+d x)^2 \cos (a+b x)}{b^2}+\frac {2 (c+d x)^3 \sin (a+b x)}{3 b}+\frac {(c+d x)^3 \sin (a+b x) \cos ^2(a+b x)}{3 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^3*Cos[a + b*x]^3,x]

[Out]

(-40*d^3*Cos[a + b*x])/(9*b^4) + (2*d*(c + d*x)^2*Cos[a + b*x])/b^2 - (2*d^3*Cos[a + b*x]^3)/(27*b^4) + (d*(c
+ d*x)^2*Cos[a + b*x]^3)/(3*b^2) - (40*d^2*(c + d*x)*Sin[a + b*x])/(9*b^3) + (2*(c + d*x)^3*Sin[a + b*x])/(3*b
) - (2*d^2*(c + d*x)*Cos[a + b*x]^2*Sin[a + b*x])/(9*b^3) + ((c + d*x)^3*Cos[a + b*x]^2*Sin[a + b*x])/(3*b)

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 3391

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

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]

Rubi steps

\begin {align*} \int (c+d x)^3 \cos ^3(a+b x) \, dx &=\frac {d (c+d x)^2 \cos ^3(a+b x)}{3 b^2}+\frac {(c+d x)^3 \cos ^2(a+b x) \sin (a+b x)}{3 b}+\frac {2}{3} \int (c+d x)^3 \cos (a+b x) \, dx-\frac {\left (2 d^2\right ) \int (c+d x) \cos ^3(a+b x) \, dx}{3 b^2}\\ &=-\frac {2 d^3 \cos ^3(a+b x)}{27 b^4}+\frac {d (c+d x)^2 \cos ^3(a+b x)}{3 b^2}+\frac {2 (c+d x)^3 \sin (a+b x)}{3 b}-\frac {2 d^2 (c+d x) \cos ^2(a+b x) \sin (a+b x)}{9 b^3}+\frac {(c+d x)^3 \cos ^2(a+b x) \sin (a+b x)}{3 b}-\frac {(2 d) \int (c+d x)^2 \sin (a+b x) \, dx}{b}-\frac {\left (4 d^2\right ) \int (c+d x) \cos (a+b x) \, dx}{9 b^2}\\ &=\frac {2 d (c+d x)^2 \cos (a+b x)}{b^2}-\frac {2 d^3 \cos ^3(a+b x)}{27 b^4}+\frac {d (c+d x)^2 \cos ^3(a+b x)}{3 b^2}-\frac {4 d^2 (c+d x) \sin (a+b x)}{9 b^3}+\frac {2 (c+d x)^3 \sin (a+b x)}{3 b}-\frac {2 d^2 (c+d x) \cos ^2(a+b x) \sin (a+b x)}{9 b^3}+\frac {(c+d x)^3 \cos ^2(a+b x) \sin (a+b x)}{3 b}-\frac {\left (4 d^2\right ) \int (c+d x) \cos (a+b x) \, dx}{b^2}+\frac {\left (4 d^3\right ) \int \sin (a+b x) \, dx}{9 b^3}\\ &=-\frac {4 d^3 \cos (a+b x)}{9 b^4}+\frac {2 d (c+d x)^2 \cos (a+b x)}{b^2}-\frac {2 d^3 \cos ^3(a+b x)}{27 b^4}+\frac {d (c+d x)^2 \cos ^3(a+b x)}{3 b^2}-\frac {40 d^2 (c+d x) \sin (a+b x)}{9 b^3}+\frac {2 (c+d x)^3 \sin (a+b x)}{3 b}-\frac {2 d^2 (c+d x) \cos ^2(a+b x) \sin (a+b x)}{9 b^3}+\frac {(c+d x)^3 \cos ^2(a+b x) \sin (a+b x)}{3 b}+\frac {\left (4 d^3\right ) \int \sin (a+b x) \, dx}{b^3}\\ &=-\frac {40 d^3 \cos (a+b x)}{9 b^4}+\frac {2 d (c+d x)^2 \cos (a+b x)}{b^2}-\frac {2 d^3 \cos ^3(a+b x)}{27 b^4}+\frac {d (c+d x)^2 \cos ^3(a+b x)}{3 b^2}-\frac {40 d^2 (c+d x) \sin (a+b x)}{9 b^3}+\frac {2 (c+d x)^3 \sin (a+b x)}{3 b}-\frac {2 d^2 (c+d x) \cos ^2(a+b x) \sin (a+b x)}{9 b^3}+\frac {(c+d x)^3 \cos ^2(a+b x) \sin (a+b x)}{3 b}\\ \end {align*}

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Mathematica [A]
time = 0.61, size = 121, normalized size = 0.69 \begin {gather*} \frac {243 d \left (-2 d^2+b^2 (c+d x)^2\right ) \cos (a+b x)+d \left (-2 d^2+9 b^2 (c+d x)^2\right ) \cos (3 (a+b x))+6 b (c+d x) \left (-82 d^2+15 b^2 (c+d x)^2+\left (-2 d^2+3 b^2 (c+d x)^2\right ) \cos (2 (a+b x))\right ) \sin (a+b x)}{108 b^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^3*Cos[a + b*x]^3,x]

[Out]

(243*d*(-2*d^2 + b^2*(c + d*x)^2)*Cos[a + b*x] + d*(-2*d^2 + 9*b^2*(c + d*x)^2)*Cos[3*(a + b*x)] + 6*b*(c + d*
x)*(-82*d^2 + 15*b^2*(c + d*x)^2 + (-2*d^2 + 3*b^2*(c + d*x)^2)*Cos[2*(a + b*x)])*Sin[a + b*x])/(108*b^4)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(559\) vs. \(2(161)=322\).
time = 0.18, size = 560, normalized size = 3.20 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^3*cos(b*x+a)^3,x,method=_RETURNVERBOSE)

[Out]

1/b*(-1/3/b^3*a^3*d^3*(2+cos(b*x+a)^2)*sin(b*x+a)+1/b^2*a^2*c*d^2*(2+cos(b*x+a)^2)*sin(b*x+a)+3/b^3*a^2*d^3*(1
/3*(b*x+a)*(2+cos(b*x+a)^2)*sin(b*x+a)+1/9*cos(b*x+a)^3+2/3*cos(b*x+a))-1/b*a*c^2*d*(2+cos(b*x+a)^2)*sin(b*x+a
)-6/b^2*a*c*d^2*(1/3*(b*x+a)*(2+cos(b*x+a)^2)*sin(b*x+a)+1/9*cos(b*x+a)^3+2/3*cos(b*x+a))-3/b^3*a*d^3*(1/3*(b*
x+a)^2*(2+cos(b*x+a)^2)*sin(b*x+a)-4/3*sin(b*x+a)+4/3*(b*x+a)*cos(b*x+a)+2/9*(b*x+a)*cos(b*x+a)^3-2/27*(2+cos(
b*x+a)^2)*sin(b*x+a))+1/3*c^3*(2+cos(b*x+a)^2)*sin(b*x+a)+3/b*c^2*d*(1/3*(b*x+a)*(2+cos(b*x+a)^2)*sin(b*x+a)+1
/9*cos(b*x+a)^3+2/3*cos(b*x+a))+3/b^2*c*d^2*(1/3*(b*x+a)^2*(2+cos(b*x+a)^2)*sin(b*x+a)-4/3*sin(b*x+a)+4/3*(b*x
+a)*cos(b*x+a)+2/9*(b*x+a)*cos(b*x+a)^3-2/27*(2+cos(b*x+a)^2)*sin(b*x+a))+1/b^3*d^3*(1/3*(b*x+a)^3*(2+cos(b*x+
a)^2)*sin(b*x+a)+2*(b*x+a)^2*cos(b*x+a)-40/9*cos(b*x+a)-4*(b*x+a)*sin(b*x+a)+1/3*(b*x+a)^2*cos(b*x+a)^3-2/9*(b
*x+a)*(2+cos(b*x+a)^2)*sin(b*x+a)-2/27*cos(b*x+a)^3))

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 535 vs. \(2 (161) = 322\).
time = 0.32, size = 535, normalized size = 3.06 \begin {gather*} -\frac {36 \, {\left (\sin \left (b x + a\right )^{3} - 3 \, \sin \left (b x + a\right )\right )} c^{3} - \frac {108 \, {\left (\sin \left (b x + a\right )^{3} - 3 \, \sin \left (b x + a\right )\right )} a c^{2} d}{b} + \frac {108 \, {\left (\sin \left (b x + a\right )^{3} - 3 \, \sin \left (b x + a\right )\right )} a^{2} c d^{2}}{b^{2}} - \frac {36 \, {\left (\sin \left (b x + a\right )^{3} - 3 \, \sin \left (b x + a\right )\right )} a^{3} d^{3}}{b^{3}} - \frac {9 \, {\left (3 \, {\left (b x + a\right )} \sin \left (3 \, b x + 3 \, a\right ) + 27 \, {\left (b x + a\right )} \sin \left (b x + a\right ) + \cos \left (3 \, b x + 3 \, a\right ) + 27 \, \cos \left (b x + a\right )\right )} c^{2} d}{b} + \frac {18 \, {\left (3 \, {\left (b x + a\right )} \sin \left (3 \, b x + 3 \, a\right ) + 27 \, {\left (b x + a\right )} \sin \left (b x + a\right ) + \cos \left (3 \, b x + 3 \, a\right ) + 27 \, \cos \left (b x + a\right )\right )} a c d^{2}}{b^{2}} - \frac {9 \, {\left (3 \, {\left (b x + a\right )} \sin \left (3 \, b x + 3 \, a\right ) + 27 \, {\left (b x + a\right )} \sin \left (b x + a\right ) + \cos \left (3 \, b x + 3 \, a\right ) + 27 \, \cos \left (b x + a\right )\right )} a^{2} d^{3}}{b^{3}} - \frac {3 \, {\left (6 \, {\left (b x + a\right )} \cos \left (3 \, b x + 3 \, a\right ) + 162 \, {\left (b x + a\right )} \cos \left (b x + a\right ) + {\left (9 \, {\left (b x + a\right )}^{2} - 2\right )} \sin \left (3 \, b x + 3 \, a\right ) + 81 \, {\left ({\left (b x + a\right )}^{2} - 2\right )} \sin \left (b x + a\right )\right )} c d^{2}}{b^{2}} + \frac {3 \, {\left (6 \, {\left (b x + a\right )} \cos \left (3 \, b x + 3 \, a\right ) + 162 \, {\left (b x + a\right )} \cos \left (b x + a\right ) + {\left (9 \, {\left (b x + a\right )}^{2} - 2\right )} \sin \left (3 \, b x + 3 \, a\right ) + 81 \, {\left ({\left (b x + a\right )}^{2} - 2\right )} \sin \left (b x + a\right )\right )} a d^{3}}{b^{3}} - \frac {{\left ({\left (9 \, {\left (b x + a\right )}^{2} - 2\right )} \cos \left (3 \, b x + 3 \, a\right ) + 243 \, {\left ({\left (b x + a\right )}^{2} - 2\right )} \cos \left (b x + a\right ) + 3 \, {\left (3 \, {\left (b x + a\right )}^{3} - 2 \, b x - 2 \, a\right )} \sin \left (3 \, b x + 3 \, a\right ) + 81 \, {\left ({\left (b x + a\right )}^{3} - 6 \, b x - 6 \, a\right )} \sin \left (b x + a\right )\right )} d^{3}}{b^{3}}}{108 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^3*cos(b*x+a)^3,x, algorithm="maxima")

[Out]

-1/108*(36*(sin(b*x + a)^3 - 3*sin(b*x + a))*c^3 - 108*(sin(b*x + a)^3 - 3*sin(b*x + a))*a*c^2*d/b + 108*(sin(
b*x + a)^3 - 3*sin(b*x + a))*a^2*c*d^2/b^2 - 36*(sin(b*x + a)^3 - 3*sin(b*x + a))*a^3*d^3/b^3 - 9*(3*(b*x + a)
*sin(3*b*x + 3*a) + 27*(b*x + a)*sin(b*x + a) + cos(3*b*x + 3*a) + 27*cos(b*x + a))*c^2*d/b + 18*(3*(b*x + a)*
sin(3*b*x + 3*a) + 27*(b*x + a)*sin(b*x + a) + cos(3*b*x + 3*a) + 27*cos(b*x + a))*a*c*d^2/b^2 - 9*(3*(b*x + a
)*sin(3*b*x + 3*a) + 27*(b*x + a)*sin(b*x + a) + cos(3*b*x + 3*a) + 27*cos(b*x + a))*a^2*d^3/b^3 - 3*(6*(b*x +
a)*cos(3*b*x + 3*a) + 162*(b*x + a)*cos(b*x + a) + (9*(b*x + a)^2 - 2)*sin(3*b*x + 3*a) + 81*((b*x + a)^2 - 2
)*sin(b*x + a))*c*d^2/b^2 + 3*(6*(b*x + a)*cos(3*b*x + 3*a) + 162*(b*x + a)*cos(b*x + a) + (9*(b*x + a)^2 - 2)
*sin(3*b*x + 3*a) + 81*((b*x + a)^2 - 2)*sin(b*x + a))*a*d^3/b^3 - ((9*(b*x + a)^2 - 2)*cos(3*b*x + 3*a) + 243
*((b*x + a)^2 - 2)*cos(b*x + a) + 3*(3*(b*x + a)^3 - 2*b*x - 2*a)*sin(3*b*x + 3*a) + 81*((b*x + a)^3 - 6*b*x -
6*a)*sin(b*x + a))*d^3/b^3)/b

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Fricas [A]
time = 0.38, size = 227, normalized size = 1.30 \begin {gather*} \frac {{\left (9 \, b^{2} d^{3} x^{2} + 18 \, b^{2} c d^{2} x + 9 \, b^{2} c^{2} d - 2 \, d^{3}\right )} \cos \left (b x + a\right )^{3} + 6 \, {\left (9 \, b^{2} d^{3} x^{2} + 18 \, b^{2} c d^{2} x + 9 \, b^{2} c^{2} d - 20 \, d^{3}\right )} \cos \left (b x + a\right ) + 3 \, {\left (6 \, b^{3} d^{3} x^{3} + 18 \, b^{3} c d^{2} x^{2} + 6 \, b^{3} c^{3} - 40 \, b c d^{2} + {\left (3 \, b^{3} d^{3} x^{3} + 9 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{3} - 2 \, b c d^{2} + {\left (9 \, b^{3} c^{2} d - 2 \, b d^{3}\right )} x\right )} \cos \left (b x + a\right )^{2} + 2 \, {\left (9 \, b^{3} c^{2} d - 20 \, b d^{3}\right )} x\right )} \sin \left (b x + a\right )}{27 \, b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^3*cos(b*x+a)^3,x, algorithm="fricas")

[Out]

1/27*((9*b^2*d^3*x^2 + 18*b^2*c*d^2*x + 9*b^2*c^2*d - 2*d^3)*cos(b*x + a)^3 + 6*(9*b^2*d^3*x^2 + 18*b^2*c*d^2*
x + 9*b^2*c^2*d - 20*d^3)*cos(b*x + a) + 3*(6*b^3*d^3*x^3 + 18*b^3*c*d^2*x^2 + 6*b^3*c^3 - 40*b*c*d^2 + (3*b^3
*d^3*x^3 + 9*b^3*c*d^2*x^2 + 3*b^3*c^3 - 2*b*c*d^2 + (9*b^3*c^2*d - 2*b*d^3)*x)*cos(b*x + a)^2 + 2*(9*b^3*c^2*
d - 20*b*d^3)*x)*sin(b*x + a))/b^4

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 495 vs. \(2 (173) = 346\).
time = 0.53, size = 495, normalized size = 2.83 \begin {gather*} \begin {cases} \frac {2 c^{3} \sin ^{3}{\left (a + b x \right )}}{3 b} + \frac {c^{3} \sin {\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{b} + \frac {2 c^{2} d x \sin ^{3}{\left (a + b x \right )}}{b} + \frac {3 c^{2} d x \sin {\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{b} + \frac {2 c d^{2} x^{2} \sin ^{3}{\left (a + b x \right )}}{b} + \frac {3 c d^{2} x^{2} \sin {\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{b} + \frac {2 d^{3} x^{3} \sin ^{3}{\left (a + b x \right )}}{3 b} + \frac {d^{3} x^{3} \sin {\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{b} + \frac {2 c^{2} d \sin ^{2}{\left (a + b x \right )} \cos {\left (a + b x \right )}}{b^{2}} + \frac {7 c^{2} d \cos ^{3}{\left (a + b x \right )}}{3 b^{2}} + \frac {4 c d^{2} x \sin ^{2}{\left (a + b x \right )} \cos {\left (a + b x \right )}}{b^{2}} + \frac {14 c d^{2} x \cos ^{3}{\left (a + b x \right )}}{3 b^{2}} + \frac {2 d^{3} x^{2} \sin ^{2}{\left (a + b x \right )} \cos {\left (a + b x \right )}}{b^{2}} + \frac {7 d^{3} x^{2} \cos ^{3}{\left (a + b x \right )}}{3 b^{2}} - \frac {40 c d^{2} \sin ^{3}{\left (a + b x \right )}}{9 b^{3}} - \frac {14 c d^{2} \sin {\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{3 b^{3}} - \frac {40 d^{3} x \sin ^{3}{\left (a + b x \right )}}{9 b^{3}} - \frac {14 d^{3} x \sin {\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{3 b^{3}} - \frac {40 d^{3} \sin ^{2}{\left (a + b x \right )} \cos {\left (a + b x \right )}}{9 b^{4}} - \frac {122 d^{3} \cos ^{3}{\left (a + b x \right )}}{27 b^{4}} & \text {for}\: b \neq 0 \\\left (c^{3} x + \frac {3 c^{2} d x^{2}}{2} + c d^{2} x^{3} + \frac {d^{3} x^{4}}{4}\right ) \cos ^{3}{\left (a \right )} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**3*cos(b*x+a)**3,x)

[Out]

Piecewise((2*c**3*sin(a + b*x)**3/(3*b) + c**3*sin(a + b*x)*cos(a + b*x)**2/b + 2*c**2*d*x*sin(a + b*x)**3/b +
3*c**2*d*x*sin(a + b*x)*cos(a + b*x)**2/b + 2*c*d**2*x**2*sin(a + b*x)**3/b + 3*c*d**2*x**2*sin(a + b*x)*cos(
a + b*x)**2/b + 2*d**3*x**3*sin(a + b*x)**3/(3*b) + d**3*x**3*sin(a + b*x)*cos(a + b*x)**2/b + 2*c**2*d*sin(a
+ b*x)**2*cos(a + b*x)/b**2 + 7*c**2*d*cos(a + b*x)**3/(3*b**2) + 4*c*d**2*x*sin(a + b*x)**2*cos(a + b*x)/b**2
+ 14*c*d**2*x*cos(a + b*x)**3/(3*b**2) + 2*d**3*x**2*sin(a + b*x)**2*cos(a + b*x)/b**2 + 7*d**3*x**2*cos(a +
b*x)**3/(3*b**2) - 40*c*d**2*sin(a + b*x)**3/(9*b**3) - 14*c*d**2*sin(a + b*x)*cos(a + b*x)**2/(3*b**3) - 40*d
**3*x*sin(a + b*x)**3/(9*b**3) - 14*d**3*x*sin(a + b*x)*cos(a + b*x)**2/(3*b**3) - 40*d**3*sin(a + b*x)**2*cos
(a + b*x)/(9*b**4) - 122*d**3*cos(a + b*x)**3/(27*b**4), Ne(b, 0)), ((c**3*x + 3*c**2*d*x**2/2 + c*d**2*x**3 +
d**3*x**4/4)*cos(a)**3, True))

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Giac [A]
time = 0.48, size = 231, normalized size = 1.32 \begin {gather*} \frac {{\left (9 \, b^{2} d^{3} x^{2} + 18 \, b^{2} c d^{2} x + 9 \, b^{2} c^{2} d - 2 \, d^{3}\right )} \cos \left (3 \, b x + 3 \, a\right )}{108 \, b^{4}} + \frac {9 \, {\left (b^{2} d^{3} x^{2} + 2 \, b^{2} c d^{2} x + b^{2} c^{2} d - 2 \, d^{3}\right )} \cos \left (b x + a\right )}{4 \, b^{4}} + \frac {{\left (3 \, b^{3} d^{3} x^{3} + 9 \, b^{3} c d^{2} x^{2} + 9 \, b^{3} c^{2} d x + 3 \, b^{3} c^{3} - 2 \, b d^{3} x - 2 \, b c d^{2}\right )} \sin \left (3 \, b x + 3 \, a\right )}{36 \, b^{4}} + \frac {3 \, {\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3} - 6 \, b d^{3} x - 6 \, b c d^{2}\right )} \sin \left (b x + a\right )}{4 \, b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^3*cos(b*x+a)^3,x, algorithm="giac")

[Out]

1/108*(9*b^2*d^3*x^2 + 18*b^2*c*d^2*x + 9*b^2*c^2*d - 2*d^3)*cos(3*b*x + 3*a)/b^4 + 9/4*(b^2*d^3*x^2 + 2*b^2*c
*d^2*x + b^2*c^2*d - 2*d^3)*cos(b*x + a)/b^4 + 1/36*(3*b^3*d^3*x^3 + 9*b^3*c*d^2*x^2 + 9*b^3*c^2*d*x + 3*b^3*c
^3 - 2*b*d^3*x - 2*b*c*d^2)*sin(3*b*x + 3*a)/b^4 + 3/4*(b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + b^3*c^
3 - 6*b*d^3*x - 6*b*c*d^2)*sin(b*x + a)/b^4

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Mupad [B]
time = 0.72, size = 364, normalized size = 2.08 \begin {gather*} \frac {7\,d^3\,x^2\,{\cos \left (a+b\,x\right )}^3}{3\,b^2}-\frac {2\,{\sin \left (a+b\,x\right )}^3\,\left (20\,c\,d^2-3\,b^2\,c^3\right )}{9\,b^3}-\frac {{\cos \left (a+b\,x\right )}^2\,\sin \left (a+b\,x\right )\,\left (14\,c\,d^2-3\,b^2\,c^3\right )}{3\,b^3}-\frac {2\,\cos \left (a+b\,x\right )\,{\sin \left (a+b\,x\right )}^2\,\left (20\,d^3-9\,b^2\,c^2\,d\right )}{9\,b^4}-\frac {2\,x\,{\sin \left (a+b\,x\right )}^3\,\left (20\,d^3-9\,b^2\,c^2\,d\right )}{9\,b^3}-\frac {{\cos \left (a+b\,x\right )}^3\,\left (122\,d^3-63\,b^2\,c^2\,d\right )}{27\,b^4}+\frac {2\,d^3\,x^3\,{\sin \left (a+b\,x\right )}^3}{3\,b}+\frac {14\,c\,d^2\,x\,{\cos \left (a+b\,x\right )}^3}{3\,b^2}-\frac {x\,{\cos \left (a+b\,x\right )}^2\,\sin \left (a+b\,x\right )\,\left (14\,d^3-9\,b^2\,c^2\,d\right )}{3\,b^3}+\frac {d^3\,x^3\,{\cos \left (a+b\,x\right )}^2\,\sin \left (a+b\,x\right )}{b}+\frac {2\,d^3\,x^2\,\cos \left (a+b\,x\right )\,{\sin \left (a+b\,x\right )}^2}{b^2}+\frac {2\,c\,d^2\,x^2\,{\sin \left (a+b\,x\right )}^3}{b}+\frac {3\,c\,d^2\,x^2\,{\cos \left (a+b\,x\right )}^2\,\sin \left (a+b\,x\right )}{b}+\frac {4\,c\,d^2\,x\,\cos \left (a+b\,x\right )\,{\sin \left (a+b\,x\right )}^2}{b^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)^3*(c + d*x)^3,x)

[Out]

(7*d^3*x^2*cos(a + b*x)^3)/(3*b^2) - (2*sin(a + b*x)^3*(20*c*d^2 - 3*b^2*c^3))/(9*b^3) - (cos(a + b*x)^2*sin(a
+ b*x)*(14*c*d^2 - 3*b^2*c^3))/(3*b^3) - (2*cos(a + b*x)*sin(a + b*x)^2*(20*d^3 - 9*b^2*c^2*d))/(9*b^4) - (2*
x*sin(a + b*x)^3*(20*d^3 - 9*b^2*c^2*d))/(9*b^3) - (cos(a + b*x)^3*(122*d^3 - 63*b^2*c^2*d))/(27*b^4) + (2*d^3
*x^3*sin(a + b*x)^3)/(3*b) + (14*c*d^2*x*cos(a + b*x)^3)/(3*b^2) - (x*cos(a + b*x)^2*sin(a + b*x)*(14*d^3 - 9*
b^2*c^2*d))/(3*b^3) + (d^3*x^3*cos(a + b*x)^2*sin(a + b*x))/b + (2*d^3*x^2*cos(a + b*x)*sin(a + b*x)^2)/b^2 +
(2*c*d^2*x^2*sin(a + b*x)^3)/b + (3*c*d^2*x^2*cos(a + b*x)^2*sin(a + b*x))/b + (4*c*d^2*x*cos(a + b*x)*sin(a +
b*x)^2)/b^2

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