3.1.50 \(\int x (a+b x^2)^2 \sin (c+d x) \, dx\) [50]

Optimal. Leaf size=185 \[ -\frac {120 b^2 x \cos (c+d x)}{d^5}+\frac {12 a b x \cos (c+d x)}{d^3}-\frac {a^2 x \cos (c+d x)}{d}+\frac {20 b^2 x^3 \cos (c+d x)}{d^3}-\frac {2 a b x^3 \cos (c+d x)}{d}-\frac {b^2 x^5 \cos (c+d x)}{d}+\frac {120 b^2 \sin (c+d x)}{d^6}-\frac {12 a b \sin (c+d x)}{d^4}+\frac {a^2 \sin (c+d x)}{d^2}-\frac {60 b^2 x^2 \sin (c+d x)}{d^4}+\frac {6 a b x^2 \sin (c+d x)}{d^2}+\frac {5 b^2 x^4 \sin (c+d x)}{d^2} \]

[Out]

-120*b^2*x*cos(d*x+c)/d^5+12*a*b*x*cos(d*x+c)/d^3-a^2*x*cos(d*x+c)/d+20*b^2*x^3*cos(d*x+c)/d^3-2*a*b*x^3*cos(d
*x+c)/d-b^2*x^5*cos(d*x+c)/d+120*b^2*sin(d*x+c)/d^6-12*a*b*sin(d*x+c)/d^4+a^2*sin(d*x+c)/d^2-60*b^2*x^2*sin(d*
x+c)/d^4+6*a*b*x^2*sin(d*x+c)/d^2+5*b^2*x^4*sin(d*x+c)/d^2

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Rubi [A]
time = 0.17, antiderivative size = 185, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {3420, 3377, 2717} \begin {gather*} \frac {a^2 \sin (c+d x)}{d^2}-\frac {a^2 x \cos (c+d x)}{d}-\frac {12 a b \sin (c+d x)}{d^4}+\frac {12 a b x \cos (c+d x)}{d^3}+\frac {6 a b x^2 \sin (c+d x)}{d^2}-\frac {2 a b x^3 \cos (c+d x)}{d}+\frac {120 b^2 \sin (c+d x)}{d^6}-\frac {120 b^2 x \cos (c+d x)}{d^5}-\frac {60 b^2 x^2 \sin (c+d x)}{d^4}+\frac {20 b^2 x^3 \cos (c+d x)}{d^3}+\frac {5 b^2 x^4 \sin (c+d x)}{d^2}-\frac {b^2 x^5 \cos (c+d x)}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*(a + b*x^2)^2*Sin[c + d*x],x]

[Out]

(-120*b^2*x*Cos[c + d*x])/d^5 + (12*a*b*x*Cos[c + d*x])/d^3 - (a^2*x*Cos[c + d*x])/d + (20*b^2*x^3*Cos[c + d*x
])/d^3 - (2*a*b*x^3*Cos[c + d*x])/d - (b^2*x^5*Cos[c + d*x])/d + (120*b^2*Sin[c + d*x])/d^6 - (12*a*b*Sin[c +
d*x])/d^4 + (a^2*Sin[c + d*x])/d^2 - (60*b^2*x^2*Sin[c + d*x])/d^4 + (6*a*b*x^2*Sin[c + d*x])/d^2 + (5*b^2*x^4
*Sin[c + d*x])/d^2

Rule 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[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 3420

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*Sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Int[ExpandIntegran
d[Sin[c + d*x], (e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]

Rubi steps

\begin {align*} \int x \left (a+b x^2\right )^2 \sin (c+d x) \, dx &=\int \left (a^2 x \sin (c+d x)+2 a b x^3 \sin (c+d x)+b^2 x^5 \sin (c+d x)\right ) \, dx\\ &=a^2 \int x \sin (c+d x) \, dx+(2 a b) \int x^3 \sin (c+d x) \, dx+b^2 \int x^5 \sin (c+d x) \, dx\\ &=-\frac {a^2 x \cos (c+d x)}{d}-\frac {2 a b x^3 \cos (c+d x)}{d}-\frac {b^2 x^5 \cos (c+d x)}{d}+\frac {a^2 \int \cos (c+d x) \, dx}{d}+\frac {(6 a b) \int x^2 \cos (c+d x) \, dx}{d}+\frac {\left (5 b^2\right ) \int x^4 \cos (c+d x) \, dx}{d}\\ &=-\frac {a^2 x \cos (c+d x)}{d}-\frac {2 a b x^3 \cos (c+d x)}{d}-\frac {b^2 x^5 \cos (c+d x)}{d}+\frac {a^2 \sin (c+d x)}{d^2}+\frac {6 a b x^2 \sin (c+d x)}{d^2}+\frac {5 b^2 x^4 \sin (c+d x)}{d^2}-\frac {(12 a b) \int x \sin (c+d x) \, dx}{d^2}-\frac {\left (20 b^2\right ) \int x^3 \sin (c+d x) \, dx}{d^2}\\ &=\frac {12 a b x \cos (c+d x)}{d^3}-\frac {a^2 x \cos (c+d x)}{d}+\frac {20 b^2 x^3 \cos (c+d x)}{d^3}-\frac {2 a b x^3 \cos (c+d x)}{d}-\frac {b^2 x^5 \cos (c+d x)}{d}+\frac {a^2 \sin (c+d x)}{d^2}+\frac {6 a b x^2 \sin (c+d x)}{d^2}+\frac {5 b^2 x^4 \sin (c+d x)}{d^2}-\frac {(12 a b) \int \cos (c+d x) \, dx}{d^3}-\frac {\left (60 b^2\right ) \int x^2 \cos (c+d x) \, dx}{d^3}\\ &=\frac {12 a b x \cos (c+d x)}{d^3}-\frac {a^2 x \cos (c+d x)}{d}+\frac {20 b^2 x^3 \cos (c+d x)}{d^3}-\frac {2 a b x^3 \cos (c+d x)}{d}-\frac {b^2 x^5 \cos (c+d x)}{d}-\frac {12 a b \sin (c+d x)}{d^4}+\frac {a^2 \sin (c+d x)}{d^2}-\frac {60 b^2 x^2 \sin (c+d x)}{d^4}+\frac {6 a b x^2 \sin (c+d x)}{d^2}+\frac {5 b^2 x^4 \sin (c+d x)}{d^2}+\frac {\left (120 b^2\right ) \int x \sin (c+d x) \, dx}{d^4}\\ &=-\frac {120 b^2 x \cos (c+d x)}{d^5}+\frac {12 a b x \cos (c+d x)}{d^3}-\frac {a^2 x \cos (c+d x)}{d}+\frac {20 b^2 x^3 \cos (c+d x)}{d^3}-\frac {2 a b x^3 \cos (c+d x)}{d}-\frac {b^2 x^5 \cos (c+d x)}{d}-\frac {12 a b \sin (c+d x)}{d^4}+\frac {a^2 \sin (c+d x)}{d^2}-\frac {60 b^2 x^2 \sin (c+d x)}{d^4}+\frac {6 a b x^2 \sin (c+d x)}{d^2}+\frac {5 b^2 x^4 \sin (c+d x)}{d^2}+\frac {\left (120 b^2\right ) \int \cos (c+d x) \, dx}{d^5}\\ &=-\frac {120 b^2 x \cos (c+d x)}{d^5}+\frac {12 a b x \cos (c+d x)}{d^3}-\frac {a^2 x \cos (c+d x)}{d}+\frac {20 b^2 x^3 \cos (c+d x)}{d^3}-\frac {2 a b x^3 \cos (c+d x)}{d}-\frac {b^2 x^5 \cos (c+d x)}{d}+\frac {120 b^2 \sin (c+d x)}{d^6}-\frac {12 a b \sin (c+d x)}{d^4}+\frac {a^2 \sin (c+d x)}{d^2}-\frac {60 b^2 x^2 \sin (c+d x)}{d^4}+\frac {6 a b x^2 \sin (c+d x)}{d^2}+\frac {5 b^2 x^4 \sin (c+d x)}{d^2}\\ \end {align*}

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Mathematica [A]
time = 0.15, size = 113, normalized size = 0.61 \begin {gather*} \frac {-d x \left (a^2 d^4+2 a b d^2 \left (-6+d^2 x^2\right )+b^2 \left (120-20 d^2 x^2+d^4 x^4\right )\right ) \cos (c+d x)+\left (a^2 d^4+6 a b d^2 \left (-2+d^2 x^2\right )+5 b^2 \left (24-12 d^2 x^2+d^4 x^4\right )\right ) \sin (c+d x)}{d^6} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-(d*x*(a^2*d^4 + 2*a*b*d^2*(-6 + d^2*x^2) + b^2*(120 - 20*d^2*x^2 + d^4*x^4))*Cos[c + d*x]) + (a^2*d^4 + 6*a*
b*d^2*(-2 + d^2*x^2) + 5*b^2*(24 - 12*d^2*x^2 + d^4*x^4))*Sin[c + d*x])/d^6

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(513\) vs. \(2(185)=370\).
time = 0.07, size = 514, normalized size = 2.78 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d^2*(a^2*c*cos(d*x+c)+a^2*(sin(d*x+c)-(d*x+c)*cos(d*x+c))+2/d^2*a*b*c^3*cos(d*x+c)+6/d^2*a*b*c^2*(sin(d*x+c)
-(d*x+c)*cos(d*x+c))-6/d^2*a*b*c*(-(d*x+c)^2*cos(d*x+c)+2*cos(d*x+c)+2*(d*x+c)*sin(d*x+c))+2/d^2*a*b*(-(d*x+c)
^3*cos(d*x+c)+3*(d*x+c)^2*sin(d*x+c)-6*sin(d*x+c)+6*(d*x+c)*cos(d*x+c))+1/d^4*b^2*c^5*cos(d*x+c)+5/d^4*b^2*c^4
*(sin(d*x+c)-(d*x+c)*cos(d*x+c))-10/d^4*b^2*c^3*(-(d*x+c)^2*cos(d*x+c)+2*cos(d*x+c)+2*(d*x+c)*sin(d*x+c))+10/d
^4*b^2*c^2*(-(d*x+c)^3*cos(d*x+c)+3*(d*x+c)^2*sin(d*x+c)-6*sin(d*x+c)+6*(d*x+c)*cos(d*x+c))-5/d^4*b^2*c*(-(d*x
+c)^4*cos(d*x+c)+4*(d*x+c)^3*sin(d*x+c)+12*(d*x+c)^2*cos(d*x+c)-24*cos(d*x+c)-24*(d*x+c)*sin(d*x+c))+1/d^4*b^2
*(-(d*x+c)^5*cos(d*x+c)+5*(d*x+c)^4*sin(d*x+c)+20*(d*x+c)^3*cos(d*x+c)-60*(d*x+c)^2*sin(d*x+c)+120*sin(d*x+c)-
120*(d*x+c)*cos(d*x+c)))

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 438 vs. \(2 (185) = 370\).
time = 0.30, size = 438, normalized size = 2.37 \begin {gather*} \frac {a^{2} c \cos \left (d x + c\right ) + \frac {b^{2} c^{5} \cos \left (d x + c\right )}{d^{4}} + \frac {2 \, a b c^{3} \cos \left (d x + c\right )}{d^{2}} - {\left ({\left (d x + c\right )} \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right )} a^{2} - \frac {5 \, {\left ({\left (d x + c\right )} \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right )} b^{2} c^{4}}{d^{4}} - \frac {6 \, {\left ({\left (d x + c\right )} \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right )} a b c^{2}}{d^{2}} + \frac {10 \, {\left ({\left ({\left (d x + c\right )}^{2} - 2\right )} \cos \left (d x + c\right ) - 2 \, {\left (d x + c\right )} \sin \left (d x + c\right )\right )} b^{2} c^{3}}{d^{4}} + \frac {6 \, {\left ({\left ({\left (d x + c\right )}^{2} - 2\right )} \cos \left (d x + c\right ) - 2 \, {\left (d x + c\right )} \sin \left (d x + c\right )\right )} a b c}{d^{2}} - \frac {10 \, {\left ({\left ({\left (d x + c\right )}^{3} - 6 \, d x - 6 \, c\right )} \cos \left (d x + c\right ) - 3 \, {\left ({\left (d x + c\right )}^{2} - 2\right )} \sin \left (d x + c\right )\right )} b^{2} c^{2}}{d^{4}} - \frac {2 \, {\left ({\left ({\left (d x + c\right )}^{3} - 6 \, d x - 6 \, c\right )} \cos \left (d x + c\right ) - 3 \, {\left ({\left (d x + c\right )}^{2} - 2\right )} \sin \left (d x + c\right )\right )} a b}{d^{2}} + \frac {5 \, {\left ({\left ({\left (d x + c\right )}^{4} - 12 \, {\left (d x + c\right )}^{2} + 24\right )} \cos \left (d x + c\right ) - 4 \, {\left ({\left (d x + c\right )}^{3} - 6 \, d x - 6 \, c\right )} \sin \left (d x + c\right )\right )} b^{2} c}{d^{4}} - \frac {{\left ({\left ({\left (d x + c\right )}^{5} - 20 \, {\left (d x + c\right )}^{3} + 120 \, d x + 120 \, c\right )} \cos \left (d x + c\right ) - 5 \, {\left ({\left (d x + c\right )}^{4} - 12 \, {\left (d x + c\right )}^{2} + 24\right )} \sin \left (d x + c\right )\right )} b^{2}}{d^{4}}}{d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^2*c*cos(d*x + c) + b^2*c^5*cos(d*x + c)/d^4 + 2*a*b*c^3*cos(d*x + c)/d^2 - ((d*x + c)*cos(d*x + c) - sin(d*
x + c))*a^2 - 5*((d*x + c)*cos(d*x + c) - sin(d*x + c))*b^2*c^4/d^4 - 6*((d*x + c)*cos(d*x + c) - sin(d*x + c)
)*a*b*c^2/d^2 + 10*(((d*x + c)^2 - 2)*cos(d*x + c) - 2*(d*x + c)*sin(d*x + c))*b^2*c^3/d^4 + 6*(((d*x + c)^2 -
2)*cos(d*x + c) - 2*(d*x + c)*sin(d*x + c))*a*b*c/d^2 - 10*(((d*x + c)^3 - 6*d*x - 6*c)*cos(d*x + c) - 3*((d*
x + c)^2 - 2)*sin(d*x + c))*b^2*c^2/d^4 - 2*(((d*x + c)^3 - 6*d*x - 6*c)*cos(d*x + c) - 3*((d*x + c)^2 - 2)*si
n(d*x + c))*a*b/d^2 + 5*(((d*x + c)^4 - 12*(d*x + c)^2 + 24)*cos(d*x + c) - 4*((d*x + c)^3 - 6*d*x - 6*c)*sin(
d*x + c))*b^2*c/d^4 - (((d*x + c)^5 - 20*(d*x + c)^3 + 120*d*x + 120*c)*cos(d*x + c) - 5*((d*x + c)^4 - 12*(d*
x + c)^2 + 24)*sin(d*x + c))*b^2/d^4)/d^2

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Fricas [A]
time = 0.37, size = 126, normalized size = 0.68 \begin {gather*} -\frac {{\left (b^{2} d^{5} x^{5} + 2 \, {\left (a b d^{5} - 10 \, b^{2} d^{3}\right )} x^{3} + {\left (a^{2} d^{5} - 12 \, a b d^{3} + 120 \, b^{2} d\right )} x\right )} \cos \left (d x + c\right ) - {\left (5 \, b^{2} d^{4} x^{4} + a^{2} d^{4} - 12 \, a b d^{2} + 6 \, {\left (a b d^{4} - 10 \, b^{2} d^{2}\right )} x^{2} + 120 \, b^{2}\right )} \sin \left (d x + c\right )}{d^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((b^2*d^5*x^5 + 2*(a*b*d^5 - 10*b^2*d^3)*x^3 + (a^2*d^5 - 12*a*b*d^3 + 120*b^2*d)*x)*cos(d*x + c) - (5*b^2*d^
4*x^4 + a^2*d^4 - 12*a*b*d^2 + 6*(a*b*d^4 - 10*b^2*d^2)*x^2 + 120*b^2)*sin(d*x + c))/d^6

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Sympy [A]
time = 0.49, size = 226, normalized size = 1.22 \begin {gather*} \begin {cases} - \frac {a^{2} x \cos {\left (c + d x \right )}}{d} + \frac {a^{2} \sin {\left (c + d x \right )}}{d^{2}} - \frac {2 a b x^{3} \cos {\left (c + d x \right )}}{d} + \frac {6 a b x^{2} \sin {\left (c + d x \right )}}{d^{2}} + \frac {12 a b x \cos {\left (c + d x \right )}}{d^{3}} - \frac {12 a b \sin {\left (c + d x \right )}}{d^{4}} - \frac {b^{2} x^{5} \cos {\left (c + d x \right )}}{d} + \frac {5 b^{2} x^{4} \sin {\left (c + d x \right )}}{d^{2}} + \frac {20 b^{2} x^{3} \cos {\left (c + d x \right )}}{d^{3}} - \frac {60 b^{2} x^{2} \sin {\left (c + d x \right )}}{d^{4}} - \frac {120 b^{2} x \cos {\left (c + d x \right )}}{d^{5}} + \frac {120 b^{2} \sin {\left (c + d x \right )}}{d^{6}} & \text {for}\: d \neq 0 \\\left (\frac {a^{2} x^{2}}{2} + \frac {a b x^{4}}{2} + \frac {b^{2} x^{6}}{6}\right ) \sin {\left (c \right )} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a**2*x*cos(c + d*x)/d + a**2*sin(c + d*x)/d**2 - 2*a*b*x**3*cos(c + d*x)/d + 6*a*b*x**2*sin(c + d*
x)/d**2 + 12*a*b*x*cos(c + d*x)/d**3 - 12*a*b*sin(c + d*x)/d**4 - b**2*x**5*cos(c + d*x)/d + 5*b**2*x**4*sin(c
+ d*x)/d**2 + 20*b**2*x**3*cos(c + d*x)/d**3 - 60*b**2*x**2*sin(c + d*x)/d**4 - 120*b**2*x*cos(c + d*x)/d**5
+ 120*b**2*sin(c + d*x)/d**6, Ne(d, 0)), ((a**2*x**2/2 + a*b*x**4/2 + b**2*x**6/6)*sin(c), True))

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Giac [A]
time = 4.87, size = 129, normalized size = 0.70 \begin {gather*} -\frac {{\left (b^{2} d^{5} x^{5} + 2 \, a b d^{5} x^{3} + a^{2} d^{5} x - 20 \, b^{2} d^{3} x^{3} - 12 \, a b d^{3} x + 120 \, b^{2} d x\right )} \cos \left (d x + c\right )}{d^{6}} + \frac {{\left (5 \, b^{2} d^{4} x^{4} + 6 \, a b d^{4} x^{2} + a^{2} d^{4} - 60 \, b^{2} d^{2} x^{2} - 12 \, a b d^{2} + 120 \, b^{2}\right )} \sin \left (d x + c\right )}{d^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(b^2*d^5*x^5 + 2*a*b*d^5*x^3 + a^2*d^5*x - 20*b^2*d^3*x^3 - 12*a*b*d^3*x + 120*b^2*d*x)*cos(d*x + c)/d^6 + (5
*b^2*d^4*x^4 + 6*a*b*d^4*x^2 + a^2*d^4 - 60*b^2*d^2*x^2 - 12*a*b*d^2 + 120*b^2)*sin(d*x + c)/d^6

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Mupad [B]
time = 4.93, size = 151, normalized size = 0.82 \begin {gather*} \frac {\sin \left (c+d\,x\right )\,\left (a^2\,d^4-12\,a\,b\,d^2+120\,b^2\right )}{d^6}-\frac {b^2\,x^5\,\cos \left (c+d\,x\right )}{d}+\frac {5\,b^2\,x^4\,\sin \left (c+d\,x\right )}{d^2}-\frac {x\,\cos \left (c+d\,x\right )\,\left (a^2\,d^4-12\,a\,b\,d^2+120\,b^2\right )}{d^5}+\frac {2\,x^3\,\cos \left (c+d\,x\right )\,\left (10\,b^2-a\,b\,d^2\right )}{d^3}-\frac {6\,x^2\,\sin \left (c+d\,x\right )\,\left (10\,b^2-a\,b\,d^2\right )}{d^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*sin(c + d*x)*(a + b*x^2)^2,x)

[Out]

(sin(c + d*x)*(120*b^2 + a^2*d^4 - 12*a*b*d^2))/d^6 - (b^2*x^5*cos(c + d*x))/d + (5*b^2*x^4*sin(c + d*x))/d^2
- (x*cos(c + d*x)*(120*b^2 + a^2*d^4 - 12*a*b*d^2))/d^5 + (2*x^3*cos(c + d*x)*(10*b^2 - a*b*d^2))/d^3 - (6*x^2
*sin(c + d*x)*(10*b^2 - a*b*d^2))/d^4

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