3.9 \(\int (c+d x)^4 \cos ^2(a+b x) \, dx\)

Optimal. Leaf size=161 \[ \frac {3 d^4 \sin (a+b x) \cos (a+b x)}{4 b^5}-\frac {3 d^3 (c+d x) \cos ^2(a+b x)}{2 b^4}-\frac {3 d^2 (c+d x)^2 \sin (a+b x) \cos (a+b x)}{2 b^3}+\frac {d (c+d x)^3 \cos ^2(a+b x)}{b^2}+\frac {(c+d x)^4 \sin (a+b x) \cos (a+b x)}{2 b}+\frac {3 d^4 x}{4 b^4}-\frac {d (c+d x)^3}{2 b^2}+\frac {(c+d x)^5}{10 d} \]

[Out]

3/4*d^4*x/b^4-1/2*d*(d*x+c)^3/b^2+1/10*(d*x+c)^5/d-3/2*d^3*(d*x+c)*cos(b*x+a)^2/b^4+d*(d*x+c)^3*cos(b*x+a)^2/b
^2+3/4*d^4*cos(b*x+a)*sin(b*x+a)/b^5-3/2*d^2*(d*x+c)^2*cos(b*x+a)*sin(b*x+a)/b^3+1/2*(d*x+c)^4*cos(b*x+a)*sin(
b*x+a)/b

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Rubi [A]  time = 0.10, antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3311, 32, 2635, 8} \[ -\frac {3 d^3 (c+d x) \cos ^2(a+b x)}{2 b^4}-\frac {3 d^2 (c+d x)^2 \sin (a+b x) \cos (a+b x)}{2 b^3}+\frac {d (c+d x)^3 \cos ^2(a+b x)}{b^2}+\frac {3 d^4 \sin (a+b x) \cos (a+b x)}{4 b^5}+\frac {(c+d x)^4 \sin (a+b x) \cos (a+b x)}{2 b}-\frac {d (c+d x)^3}{2 b^2}+\frac {3 d^4 x}{4 b^4}+\frac {(c+d x)^5}{10 d} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^4*Cos[a + b*x]^2,x]

[Out]

(3*d^4*x)/(4*b^4) - (d*(c + d*x)^3)/(2*b^2) + (c + d*x)^5/(10*d) - (3*d^3*(c + d*x)*Cos[a + b*x]^2)/(2*b^4) +
(d*(c + d*x)^3*Cos[a + b*x]^2)/b^2 + (3*d^4*Cos[a + b*x]*Sin[a + b*x])/(4*b^5) - (3*d^2*(c + d*x)^2*Cos[a + b*
x]*Sin[a + b*x])/(2*b^3) + ((c + d*x)^4*Cos[a + b*x]*Sin[a + b*x])/(2*b)

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 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]

Rubi steps

\begin {align*} \int (c+d x)^4 \cos ^2(a+b x) \, dx &=\frac {d (c+d x)^3 \cos ^2(a+b x)}{b^2}+\frac {(c+d x)^4 \cos (a+b x) \sin (a+b x)}{2 b}+\frac {1}{2} \int (c+d x)^4 \, dx-\frac {\left (3 d^2\right ) \int (c+d x)^2 \cos ^2(a+b x) \, dx}{b^2}\\ &=\frac {(c+d x)^5}{10 d}-\frac {3 d^3 (c+d x) \cos ^2(a+b x)}{2 b^4}+\frac {d (c+d x)^3 \cos ^2(a+b x)}{b^2}-\frac {3 d^2 (c+d x)^2 \cos (a+b x) \sin (a+b x)}{2 b^3}+\frac {(c+d x)^4 \cos (a+b x) \sin (a+b x)}{2 b}-\frac {\left (3 d^2\right ) \int (c+d x)^2 \, dx}{2 b^2}+\frac {\left (3 d^4\right ) \int \cos ^2(a+b x) \, dx}{2 b^4}\\ &=-\frac {d (c+d x)^3}{2 b^2}+\frac {(c+d x)^5}{10 d}-\frac {3 d^3 (c+d x) \cos ^2(a+b x)}{2 b^4}+\frac {d (c+d x)^3 \cos ^2(a+b x)}{b^2}+\frac {3 d^4 \cos (a+b x) \sin (a+b x)}{4 b^5}-\frac {3 d^2 (c+d x)^2 \cos (a+b x) \sin (a+b x)}{2 b^3}+\frac {(c+d x)^4 \cos (a+b x) \sin (a+b x)}{2 b}+\frac {\left (3 d^4\right ) \int 1 \, dx}{4 b^4}\\ &=\frac {3 d^4 x}{4 b^4}-\frac {d (c+d x)^3}{2 b^2}+\frac {(c+d x)^5}{10 d}-\frac {3 d^3 (c+d x) \cos ^2(a+b x)}{2 b^4}+\frac {d (c+d x)^3 \cos ^2(a+b x)}{b^2}+\frac {3 d^4 \cos (a+b x) \sin (a+b x)}{4 b^5}-\frac {3 d^2 (c+d x)^2 \cos (a+b x) \sin (a+b x)}{2 b^3}+\frac {(c+d x)^4 \cos (a+b x) \sin (a+b x)}{2 b}\\ \end {align*}

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Mathematica [A]  time = 0.68, size = 132, normalized size = 0.82 \[ \frac {20 b d (c+d x) \cos (2 (a+b x)) \left (2 b^2 (c+d x)^2-3 d^2\right )+10 \sin (2 (a+b x)) \left (2 b^4 (c+d x)^4-6 b^2 d^2 (c+d x)^2+3 d^4\right )+8 b^5 x \left (5 c^4+10 c^3 d x+10 c^2 d^2 x^2+5 c d^3 x^3+d^4 x^4\right )}{80 b^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^4*Cos[a + b*x]^2,x]

[Out]

(8*b^5*x*(5*c^4 + 10*c^3*d*x + 10*c^2*d^2*x^2 + 5*c*d^3*x^3 + d^4*x^4) + 20*b*d*(c + d*x)*(-3*d^2 + 2*b^2*(c +
d*x)^2)*Cos[2*(a + b*x)] + 10*(3*d^4 - 6*b^2*d^2*(c + d*x)^2 + 2*b^4*(c + d*x)^4)*Sin[2*(a + b*x)])/(80*b^5)

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fricas [A]  time = 0.78, size = 287, normalized size = 1.78 \[ \frac {2 \, b^{5} d^{4} x^{5} + 10 \, b^{5} c d^{3} x^{4} + 10 \, {\left (2 \, b^{5} c^{2} d^{2} - b^{3} d^{4}\right )} x^{3} + 10 \, {\left (2 \, b^{5} c^{3} d - 3 \, b^{3} c d^{3}\right )} x^{2} + 10 \, {\left (2 \, b^{3} d^{4} x^{3} + 6 \, b^{3} c d^{3} x^{2} + 2 \, b^{3} c^{3} d - 3 \, b c d^{3} + 3 \, {\left (2 \, b^{3} c^{2} d^{2} - b d^{4}\right )} x\right )} \cos \left (b x + a\right )^{2} + 5 \, {\left (2 \, b^{4} d^{4} x^{4} + 8 \, b^{4} c d^{3} x^{3} + 2 \, b^{4} c^{4} - 6 \, b^{2} c^{2} d^{2} + 3 \, d^{4} + 6 \, {\left (2 \, b^{4} c^{2} d^{2} - b^{2} d^{4}\right )} x^{2} + 4 \, {\left (2 \, b^{4} c^{3} d - 3 \, b^{2} c d^{3}\right )} x\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 5 \, {\left (2 \, b^{5} c^{4} - 6 \, b^{3} c^{2} d^{2} + 3 \, b d^{4}\right )} x}{20 \, b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^4*cos(b*x+a)^2,x, algorithm="fricas")

[Out]

1/20*(2*b^5*d^4*x^5 + 10*b^5*c*d^3*x^4 + 10*(2*b^5*c^2*d^2 - b^3*d^4)*x^3 + 10*(2*b^5*c^3*d - 3*b^3*c*d^3)*x^2
+ 10*(2*b^3*d^4*x^3 + 6*b^3*c*d^3*x^2 + 2*b^3*c^3*d - 3*b*c*d^3 + 3*(2*b^3*c^2*d^2 - b*d^4)*x)*cos(b*x + a)^2
+ 5*(2*b^4*d^4*x^4 + 8*b^4*c*d^3*x^3 + 2*b^4*c^4 - 6*b^2*c^2*d^2 + 3*d^4 + 6*(2*b^4*c^2*d^2 - b^2*d^4)*x^2 +
4*(2*b^4*c^3*d - 3*b^2*c*d^3)*x)*cos(b*x + a)*sin(b*x + a) + 5*(2*b^5*c^4 - 6*b^3*c^2*d^2 + 3*b*d^4)*x)/b^5

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giac [A]  time = 0.53, size = 222, normalized size = 1.38 \[ \frac {1}{10} \, d^{4} x^{5} + \frac {1}{2} \, c d^{3} x^{4} + c^{2} d^{2} x^{3} + c^{3} d x^{2} + \frac {1}{2} \, c^{4} x + \frac {{\left (2 \, b^{3} d^{4} x^{3} + 6 \, b^{3} c d^{3} x^{2} + 6 \, b^{3} c^{2} d^{2} x + 2 \, b^{3} c^{3} d - 3 \, b d^{4} x - 3 \, b c d^{3}\right )} \cos \left (2 \, b x + 2 \, a\right )}{4 \, b^{5}} + \frac {{\left (2 \, b^{4} d^{4} x^{4} + 8 \, b^{4} c d^{3} x^{3} + 12 \, b^{4} c^{2} d^{2} x^{2} + 8 \, b^{4} c^{3} d x + 2 \, b^{4} c^{4} - 6 \, b^{2} d^{4} x^{2} - 12 \, b^{2} c d^{3} x - 6 \, b^{2} c^{2} d^{2} + 3 \, d^{4}\right )} \sin \left (2 \, b x + 2 \, a\right )}{8 \, b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^4*cos(b*x+a)^2,x, algorithm="giac")

[Out]

1/10*d^4*x^5 + 1/2*c*d^3*x^4 + c^2*d^2*x^3 + c^3*d*x^2 + 1/2*c^4*x + 1/4*(2*b^3*d^4*x^3 + 6*b^3*c*d^3*x^2 + 6*
b^3*c^2*d^2*x + 2*b^3*c^3*d - 3*b*d^4*x - 3*b*c*d^3)*cos(2*b*x + 2*a)/b^5 + 1/8*(2*b^4*d^4*x^4 + 8*b^4*c*d^3*x
^3 + 12*b^4*c^2*d^2*x^2 + 8*b^4*c^3*d*x + 2*b^4*c^4 - 6*b^2*d^4*x^2 - 12*b^2*c*d^3*x - 6*b^2*c^2*d^2 + 3*d^4)*
sin(2*b*x + 2*a)/b^5

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maple [B]  time = 0.08, size = 1027, normalized size = 6.38 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^4*cos(b*x+a)^2,x)

[Out]

1/b*(1/b^4*d^4*((b*x+a)^4*(1/2*cos(b*x+a)*sin(b*x+a)+1/2*b*x+1/2*a)+(b*x+a)^3*cos(b*x+a)^2-3*(b*x+a)^2*(1/2*co
s(b*x+a)*sin(b*x+a)+1/2*b*x+1/2*a)-3/2*(b*x+a)*cos(b*x+a)^2+3/4*cos(b*x+a)*sin(b*x+a)+3/4*b*x+3/4*a+(b*x+a)^3-
2/5*(b*x+a)^5)-4/b^4*a*d^4*((b*x+a)^3*(1/2*cos(b*x+a)*sin(b*x+a)+1/2*b*x+1/2*a)+3/4*(b*x+a)^2*cos(b*x+a)^2-3/2
*(b*x+a)*(1/2*cos(b*x+a)*sin(b*x+a)+1/2*b*x+1/2*a)+3/8*(b*x+a)^2+3/8*sin(b*x+a)^2-3/8*(b*x+a)^4)+4/b^3*c*d^3*(
(b*x+a)^3*(1/2*cos(b*x+a)*sin(b*x+a)+1/2*b*x+1/2*a)+3/4*(b*x+a)^2*cos(b*x+a)^2-3/2*(b*x+a)*(1/2*cos(b*x+a)*sin
(b*x+a)+1/2*b*x+1/2*a)+3/8*(b*x+a)^2+3/8*sin(b*x+a)^2-3/8*(b*x+a)^4)+6/b^4*a^2*d^4*((b*x+a)^2*(1/2*cos(b*x+a)*
sin(b*x+a)+1/2*b*x+1/2*a)+1/2*(b*x+a)*cos(b*x+a)^2-1/4*cos(b*x+a)*sin(b*x+a)-1/4*b*x-1/4*a-1/3*(b*x+a)^3)-12/b
^3*a*c*d^3*((b*x+a)^2*(1/2*cos(b*x+a)*sin(b*x+a)+1/2*b*x+1/2*a)+1/2*(b*x+a)*cos(b*x+a)^2-1/4*cos(b*x+a)*sin(b*
x+a)-1/4*b*x-1/4*a-1/3*(b*x+a)^3)+6/b^2*c^2*d^2*((b*x+a)^2*(1/2*cos(b*x+a)*sin(b*x+a)+1/2*b*x+1/2*a)+1/2*(b*x+
a)*cos(b*x+a)^2-1/4*cos(b*x+a)*sin(b*x+a)-1/4*b*x-1/4*a-1/3*(b*x+a)^3)-4/b^4*a^3*d^4*((b*x+a)*(1/2*cos(b*x+a)*
sin(b*x+a)+1/2*b*x+1/2*a)-1/4*(b*x+a)^2-1/4*sin(b*x+a)^2)+12/b^3*a^2*c*d^3*((b*x+a)*(1/2*cos(b*x+a)*sin(b*x+a)
+1/2*b*x+1/2*a)-1/4*(b*x+a)^2-1/4*sin(b*x+a)^2)-12/b^2*a*c^2*d^2*((b*x+a)*(1/2*cos(b*x+a)*sin(b*x+a)+1/2*b*x+1
/2*a)-1/4*(b*x+a)^2-1/4*sin(b*x+a)^2)+4/b*c^3*d*((b*x+a)*(1/2*cos(b*x+a)*sin(b*x+a)+1/2*b*x+1/2*a)-1/4*(b*x+a)
^2-1/4*sin(b*x+a)^2)+1/b^4*a^4*d^4*(1/2*cos(b*x+a)*sin(b*x+a)+1/2*b*x+1/2*a)-4/b^3*a^3*c*d^3*(1/2*cos(b*x+a)*s
in(b*x+a)+1/2*b*x+1/2*a)+6/b^2*a^2*c^2*d^2*(1/2*cos(b*x+a)*sin(b*x+a)+1/2*b*x+1/2*a)-4/b*a*c^3*d*(1/2*cos(b*x+
a)*sin(b*x+a)+1/2*b*x+1/2*a)+c^4*(1/2*cos(b*x+a)*sin(b*x+a)+1/2*b*x+1/2*a))

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maxima [B]  time = 0.77, size = 717, normalized size = 4.45 \[ \frac {10 \, {\left (2 \, b x + 2 \, a + \sin \left (2 \, b x + 2 \, a\right )\right )} c^{4} - \frac {40 \, {\left (2 \, b x + 2 \, a + \sin \left (2 \, b x + 2 \, a\right )\right )} a c^{3} d}{b} + \frac {60 \, {\left (2 \, b x + 2 \, a + \sin \left (2 \, b x + 2 \, a\right )\right )} a^{2} c^{2} d^{2}}{b^{2}} - \frac {40 \, {\left (2 \, b x + 2 \, a + \sin \left (2 \, b x + 2 \, a\right )\right )} a^{3} c d^{3}}{b^{3}} + \frac {10 \, {\left (2 \, b x + 2 \, a + \sin \left (2 \, b x + 2 \, a\right )\right )} a^{4} d^{4}}{b^{4}} + \frac {20 \, {\left (2 \, {\left (b x + a\right )}^{2} + 2 \, {\left (b x + a\right )} \sin \left (2 \, b x + 2 \, a\right ) + \cos \left (2 \, b x + 2 \, a\right )\right )} c^{3} d}{b} - \frac {60 \, {\left (2 \, {\left (b x + a\right )}^{2} + 2 \, {\left (b x + a\right )} \sin \left (2 \, b x + 2 \, a\right ) + \cos \left (2 \, b x + 2 \, a\right )\right )} a c^{2} d^{2}}{b^{2}} + \frac {60 \, {\left (2 \, {\left (b x + a\right )}^{2} + 2 \, {\left (b x + a\right )} \sin \left (2 \, b x + 2 \, a\right ) + \cos \left (2 \, b x + 2 \, a\right )\right )} a^{2} c d^{3}}{b^{3}} - \frac {20 \, {\left (2 \, {\left (b x + a\right )}^{2} + 2 \, {\left (b x + a\right )} \sin \left (2 \, b x + 2 \, a\right ) + \cos \left (2 \, b x + 2 \, a\right )\right )} a^{3} d^{4}}{b^{4}} + \frac {10 \, {\left (4 \, {\left (b x + a\right )}^{3} + 6 \, {\left (b x + a\right )} \cos \left (2 \, b x + 2 \, a\right ) + 3 \, {\left (2 \, {\left (b x + a\right )}^{2} - 1\right )} \sin \left (2 \, b x + 2 \, a\right )\right )} c^{2} d^{2}}{b^{2}} - \frac {20 \, {\left (4 \, {\left (b x + a\right )}^{3} + 6 \, {\left (b x + a\right )} \cos \left (2 \, b x + 2 \, a\right ) + 3 \, {\left (2 \, {\left (b x + a\right )}^{2} - 1\right )} \sin \left (2 \, b x + 2 \, a\right )\right )} a c d^{3}}{b^{3}} + \frac {10 \, {\left (4 \, {\left (b x + a\right )}^{3} + 6 \, {\left (b x + a\right )} \cos \left (2 \, b x + 2 \, a\right ) + 3 \, {\left (2 \, {\left (b x + a\right )}^{2} - 1\right )} \sin \left (2 \, b x + 2 \, a\right )\right )} a^{2} d^{4}}{b^{4}} + \frac {10 \, {\left (2 \, {\left (b x + a\right )}^{4} + 3 \, {\left (2 \, {\left (b x + a\right )}^{2} - 1\right )} \cos \left (2 \, b x + 2 \, a\right ) + 2 \, {\left (2 \, {\left (b x + a\right )}^{3} - 3 \, b x - 3 \, a\right )} \sin \left (2 \, b x + 2 \, a\right )\right )} c d^{3}}{b^{3}} - \frac {10 \, {\left (2 \, {\left (b x + a\right )}^{4} + 3 \, {\left (2 \, {\left (b x + a\right )}^{2} - 1\right )} \cos \left (2 \, b x + 2 \, a\right ) + 2 \, {\left (2 \, {\left (b x + a\right )}^{3} - 3 \, b x - 3 \, a\right )} \sin \left (2 \, b x + 2 \, a\right )\right )} a d^{4}}{b^{4}} + \frac {{\left (4 \, {\left (b x + a\right )}^{5} + 10 \, {\left (2 \, {\left (b x + a\right )}^{3} - 3 \, b x - 3 \, a\right )} \cos \left (2 \, b x + 2 \, a\right ) + 5 \, {\left (2 \, {\left (b x + a\right )}^{4} - 6 \, {\left (b x + a\right )}^{2} + 3\right )} \sin \left (2 \, b x + 2 \, a\right )\right )} d^{4}}{b^{4}}}{40 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^4*cos(b*x+a)^2,x, algorithm="maxima")

[Out]

1/40*(10*(2*b*x + 2*a + sin(2*b*x + 2*a))*c^4 - 40*(2*b*x + 2*a + sin(2*b*x + 2*a))*a*c^3*d/b + 60*(2*b*x + 2*
a + sin(2*b*x + 2*a))*a^2*c^2*d^2/b^2 - 40*(2*b*x + 2*a + sin(2*b*x + 2*a))*a^3*c*d^3/b^3 + 10*(2*b*x + 2*a +
sin(2*b*x + 2*a))*a^4*d^4/b^4 + 20*(2*(b*x + a)^2 + 2*(b*x + a)*sin(2*b*x + 2*a) + cos(2*b*x + 2*a))*c^3*d/b -
60*(2*(b*x + a)^2 + 2*(b*x + a)*sin(2*b*x + 2*a) + cos(2*b*x + 2*a))*a*c^2*d^2/b^2 + 60*(2*(b*x + a)^2 + 2*(b
*x + a)*sin(2*b*x + 2*a) + cos(2*b*x + 2*a))*a^2*c*d^3/b^3 - 20*(2*(b*x + a)^2 + 2*(b*x + a)*sin(2*b*x + 2*a)
+ cos(2*b*x + 2*a))*a^3*d^4/b^4 + 10*(4*(b*x + a)^3 + 6*(b*x + a)*cos(2*b*x + 2*a) + 3*(2*(b*x + a)^2 - 1)*sin
(2*b*x + 2*a))*c^2*d^2/b^2 - 20*(4*(b*x + a)^3 + 6*(b*x + a)*cos(2*b*x + 2*a) + 3*(2*(b*x + a)^2 - 1)*sin(2*b*
x + 2*a))*a*c*d^3/b^3 + 10*(4*(b*x + a)^3 + 6*(b*x + a)*cos(2*b*x + 2*a) + 3*(2*(b*x + a)^2 - 1)*sin(2*b*x + 2
*a))*a^2*d^4/b^4 + 10*(2*(b*x + a)^4 + 3*(2*(b*x + a)^2 - 1)*cos(2*b*x + 2*a) + 2*(2*(b*x + a)^3 - 3*b*x - 3*a
)*sin(2*b*x + 2*a))*c*d^3/b^3 - 10*(2*(b*x + a)^4 + 3*(2*(b*x + a)^2 - 1)*cos(2*b*x + 2*a) + 2*(2*(b*x + a)^3
- 3*b*x - 3*a)*sin(2*b*x + 2*a))*a*d^4/b^4 + (4*(b*x + a)^5 + 10*(2*(b*x + a)^3 - 3*b*x - 3*a)*cos(2*b*x + 2*a
) + 5*(2*(b*x + a)^4 - 6*(b*x + a)^2 + 3)*sin(2*b*x + 2*a))*d^4/b^4)/b

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mupad [B]  time = 0.61, size = 349, normalized size = 2.17 \[ \frac {\frac {15\,d^4\,\sin \left (2\,a+2\,b\,x\right )}{2}+10\,b^5\,c^4\,x+5\,b^4\,c^4\,\sin \left (2\,a+2\,b\,x\right )+2\,b^5\,d^4\,x^5+10\,b^3\,c^3\,d\,\cos \left (2\,a+2\,b\,x\right )+20\,b^5\,c^3\,d\,x^2+10\,b^5\,c\,d^3\,x^4-15\,b^2\,c^2\,d^2\,\sin \left (2\,a+2\,b\,x\right )+10\,b^3\,d^4\,x^3\,\cos \left (2\,a+2\,b\,x\right )+20\,b^5\,c^2\,d^2\,x^3-15\,b^2\,d^4\,x^2\,\sin \left (2\,a+2\,b\,x\right )+5\,b^4\,d^4\,x^4\,\sin \left (2\,a+2\,b\,x\right )-15\,b\,c\,d^3\,\cos \left (2\,a+2\,b\,x\right )-15\,b\,d^4\,x\,\cos \left (2\,a+2\,b\,x\right )+30\,b^4\,c^2\,d^2\,x^2\,\sin \left (2\,a+2\,b\,x\right )-30\,b^2\,c\,d^3\,x\,\sin \left (2\,a+2\,b\,x\right )+20\,b^4\,c^3\,d\,x\,\sin \left (2\,a+2\,b\,x\right )+30\,b^3\,c^2\,d^2\,x\,\cos \left (2\,a+2\,b\,x\right )+30\,b^3\,c\,d^3\,x^2\,\cos \left (2\,a+2\,b\,x\right )+20\,b^4\,c\,d^3\,x^3\,\sin \left (2\,a+2\,b\,x\right )}{20\,b^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)^2*(c + d*x)^4,x)

[Out]

((15*d^4*sin(2*a + 2*b*x))/2 + 10*b^5*c^4*x + 5*b^4*c^4*sin(2*a + 2*b*x) + 2*b^5*d^4*x^5 + 10*b^3*c^3*d*cos(2*
a + 2*b*x) + 20*b^5*c^3*d*x^2 + 10*b^5*c*d^3*x^4 - 15*b^2*c^2*d^2*sin(2*a + 2*b*x) + 10*b^3*d^4*x^3*cos(2*a +
2*b*x) + 20*b^5*c^2*d^2*x^3 - 15*b^2*d^4*x^2*sin(2*a + 2*b*x) + 5*b^4*d^4*x^4*sin(2*a + 2*b*x) - 15*b*c*d^3*co
s(2*a + 2*b*x) - 15*b*d^4*x*cos(2*a + 2*b*x) + 30*b^4*c^2*d^2*x^2*sin(2*a + 2*b*x) - 30*b^2*c*d^3*x*sin(2*a +
2*b*x) + 20*b^4*c^3*d*x*sin(2*a + 2*b*x) + 30*b^3*c^2*d^2*x*cos(2*a + 2*b*x) + 30*b^3*c*d^3*x^2*cos(2*a + 2*b*
x) + 20*b^4*c*d^3*x^3*sin(2*a + 2*b*x))/(20*b^5)

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sympy [A]  time = 4.60, size = 660, normalized size = 4.10 \[ \begin {cases} \frac {c^{4} x \sin ^{2}{\left (a + b x \right )}}{2} + \frac {c^{4} x \cos ^{2}{\left (a + b x \right )}}{2} + c^{3} d x^{2} \sin ^{2}{\left (a + b x \right )} + c^{3} d x^{2} \cos ^{2}{\left (a + b x \right )} + c^{2} d^{2} x^{3} \sin ^{2}{\left (a + b x \right )} + c^{2} d^{2} x^{3} \cos ^{2}{\left (a + b x \right )} + \frac {c d^{3} x^{4} \sin ^{2}{\left (a + b x \right )}}{2} + \frac {c d^{3} x^{4} \cos ^{2}{\left (a + b x \right )}}{2} + \frac {d^{4} x^{5} \sin ^{2}{\left (a + b x \right )}}{10} + \frac {d^{4} x^{5} \cos ^{2}{\left (a + b x \right )}}{10} + \frac {c^{4} \sin {\left (a + b x \right )} \cos {\left (a + b x \right )}}{2 b} + \frac {2 c^{3} d x \sin {\left (a + b x \right )} \cos {\left (a + b x \right )}}{b} + \frac {3 c^{2} d^{2} x^{2} \sin {\left (a + b x \right )} \cos {\left (a + b x \right )}}{b} + \frac {2 c d^{3} x^{3} \sin {\left (a + b x \right )} \cos {\left (a + b x \right )}}{b} + \frac {d^{4} x^{4} \sin {\left (a + b x \right )} \cos {\left (a + b x \right )}}{2 b} - \frac {c^{3} d \sin ^{2}{\left (a + b x \right )}}{b^{2}} - \frac {3 c^{2} d^{2} x \sin ^{2}{\left (a + b x \right )}}{2 b^{2}} + \frac {3 c^{2} d^{2} x \cos ^{2}{\left (a + b x \right )}}{2 b^{2}} - \frac {3 c d^{3} x^{2} \sin ^{2}{\left (a + b x \right )}}{2 b^{2}} + \frac {3 c d^{3} x^{2} \cos ^{2}{\left (a + b x \right )}}{2 b^{2}} - \frac {d^{4} x^{3} \sin ^{2}{\left (a + b x \right )}}{2 b^{2}} + \frac {d^{4} x^{3} \cos ^{2}{\left (a + b x \right )}}{2 b^{2}} - \frac {3 c^{2} d^{2} \sin {\left (a + b x \right )} \cos {\left (a + b x \right )}}{2 b^{3}} - \frac {3 c d^{3} x \sin {\left (a + b x \right )} \cos {\left (a + b x \right )}}{b^{3}} - \frac {3 d^{4} x^{2} \sin {\left (a + b x \right )} \cos {\left (a + b x \right )}}{2 b^{3}} + \frac {3 c d^{3} \sin ^{2}{\left (a + b x \right )}}{2 b^{4}} + \frac {3 d^{4} x \sin ^{2}{\left (a + b x \right )}}{4 b^{4}} - \frac {3 d^{4} x \cos ^{2}{\left (a + b x \right )}}{4 b^{4}} + \frac {3 d^{4} \sin {\left (a + b x \right )} \cos {\left (a + b x \right )}}{4 b^{5}} & \text {for}\: b \neq 0 \\\left (c^{4} x + 2 c^{3} d x^{2} + 2 c^{2} d^{2} x^{3} + c d^{3} x^{4} + \frac {d^{4} x^{5}}{5}\right ) \cos ^{2}{\relax (a )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**4*cos(b*x+a)**2,x)

[Out]

Piecewise((c**4*x*sin(a + b*x)**2/2 + c**4*x*cos(a + b*x)**2/2 + c**3*d*x**2*sin(a + b*x)**2 + c**3*d*x**2*cos
(a + b*x)**2 + c**2*d**2*x**3*sin(a + b*x)**2 + c**2*d**2*x**3*cos(a + b*x)**2 + c*d**3*x**4*sin(a + b*x)**2/2
+ c*d**3*x**4*cos(a + b*x)**2/2 + d**4*x**5*sin(a + b*x)**2/10 + d**4*x**5*cos(a + b*x)**2/10 + c**4*sin(a +
b*x)*cos(a + b*x)/(2*b) + 2*c**3*d*x*sin(a + b*x)*cos(a + b*x)/b + 3*c**2*d**2*x**2*sin(a + b*x)*cos(a + b*x)/
b + 2*c*d**3*x**3*sin(a + b*x)*cos(a + b*x)/b + d**4*x**4*sin(a + b*x)*cos(a + b*x)/(2*b) - c**3*d*sin(a + b*x
)**2/b**2 - 3*c**2*d**2*x*sin(a + b*x)**2/(2*b**2) + 3*c**2*d**2*x*cos(a + b*x)**2/(2*b**2) - 3*c*d**3*x**2*si
n(a + b*x)**2/(2*b**2) + 3*c*d**3*x**2*cos(a + b*x)**2/(2*b**2) - d**4*x**3*sin(a + b*x)**2/(2*b**2) + d**4*x*
*3*cos(a + b*x)**2/(2*b**2) - 3*c**2*d**2*sin(a + b*x)*cos(a + b*x)/(2*b**3) - 3*c*d**3*x*sin(a + b*x)*cos(a +
b*x)/b**3 - 3*d**4*x**2*sin(a + b*x)*cos(a + b*x)/(2*b**3) + 3*c*d**3*sin(a + b*x)**2/(2*b**4) + 3*d**4*x*sin
(a + b*x)**2/(4*b**4) - 3*d**4*x*cos(a + b*x)**2/(4*b**4) + 3*d**4*sin(a + b*x)*cos(a + b*x)/(4*b**5), Ne(b, 0
)), ((c**4*x + 2*c**3*d*x**2 + 2*c**2*d**2*x**3 + c*d**3*x**4 + d**4*x**5/5)*cos(a)**2, True))

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