3.51 \(\int (a+b x^2)^2 \sin (c+d x) \, dx\)

Optimal. Leaf size=138 \[ -\frac {a^2 \cos (c+d x)}{d}+\frac {4 a b \cos (c+d x)}{d^3}+\frac {4 a b x \sin (c+d x)}{d^2}-\frac {2 a b x^2 \cos (c+d x)}{d}-\frac {24 b^2 \cos (c+d x)}{d^5}-\frac {24 b^2 x \sin (c+d x)}{d^4}+\frac {12 b^2 x^2 \cos (c+d x)}{d^3}+\frac {4 b^2 x^3 \sin (c+d x)}{d^2}-\frac {b^2 x^4 \cos (c+d x)}{d} \]

[Out]

-24*b^2*cos(d*x+c)/d^5+4*a*b*cos(d*x+c)/d^3-a^2*cos(d*x+c)/d+12*b^2*x^2*cos(d*x+c)/d^3-2*a*b*x^2*cos(d*x+c)/d-
b^2*x^4*cos(d*x+c)/d-24*b^2*x*sin(d*x+c)/d^4+4*a*b*x*sin(d*x+c)/d^2+4*b^2*x^3*sin(d*x+c)/d^2

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Rubi [A]  time = 0.16, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {3329, 2638, 3296} \[ -\frac {a^2 \cos (c+d x)}{d}+\frac {4 a b x \sin (c+d x)}{d^2}+\frac {4 a b \cos (c+d x)}{d^3}-\frac {2 a b x^2 \cos (c+d x)}{d}+\frac {4 b^2 x^3 \sin (c+d x)}{d^2}+\frac {12 b^2 x^2 \cos (c+d x)}{d^3}-\frac {24 b^2 x \sin (c+d x)}{d^4}-\frac {24 b^2 \cos (c+d x)}{d^5}-\frac {b^2 x^4 \cos (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^2*Sin[c + d*x],x]

[Out]

(-24*b^2*Cos[c + d*x])/d^5 + (4*a*b*Cos[c + d*x])/d^3 - (a^2*Cos[c + d*x])/d + (12*b^2*x^2*Cos[c + d*x])/d^3 -
(2*a*b*x^2*Cos[c + d*x])/d - (b^2*x^4*Cos[c + d*x])/d - (24*b^2*x*Sin[c + d*x])/d^4 + (4*a*b*x*Sin[c + d*x])/
d^2 + (4*b^2*x^3*Sin[c + d*x])/d^2

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[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 3329

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

Rubi steps

\begin {align*} \int \left (a+b x^2\right )^2 \sin (c+d x) \, dx &=\int \left (a^2 \sin (c+d x)+2 a b x^2 \sin (c+d x)+b^2 x^4 \sin (c+d x)\right ) \, dx\\ &=a^2 \int \sin (c+d x) \, dx+(2 a b) \int x^2 \sin (c+d x) \, dx+b^2 \int x^4 \sin (c+d x) \, dx\\ &=-\frac {a^2 \cos (c+d x)}{d}-\frac {2 a b x^2 \cos (c+d x)}{d}-\frac {b^2 x^4 \cos (c+d x)}{d}+\frac {(4 a b) \int x \cos (c+d x) \, dx}{d}+\frac {\left (4 b^2\right ) \int x^3 \cos (c+d x) \, dx}{d}\\ &=-\frac {a^2 \cos (c+d x)}{d}-\frac {2 a b x^2 \cos (c+d x)}{d}-\frac {b^2 x^4 \cos (c+d x)}{d}+\frac {4 a b x \sin (c+d x)}{d^2}+\frac {4 b^2 x^3 \sin (c+d x)}{d^2}-\frac {(4 a b) \int \sin (c+d x) \, dx}{d^2}-\frac {\left (12 b^2\right ) \int x^2 \sin (c+d x) \, dx}{d^2}\\ &=\frac {4 a b \cos (c+d x)}{d^3}-\frac {a^2 \cos (c+d x)}{d}+\frac {12 b^2 x^2 \cos (c+d x)}{d^3}-\frac {2 a b x^2 \cos (c+d x)}{d}-\frac {b^2 x^4 \cos (c+d x)}{d}+\frac {4 a b x \sin (c+d x)}{d^2}+\frac {4 b^2 x^3 \sin (c+d x)}{d^2}-\frac {\left (24 b^2\right ) \int x \cos (c+d x) \, dx}{d^3}\\ &=\frac {4 a b \cos (c+d x)}{d^3}-\frac {a^2 \cos (c+d x)}{d}+\frac {12 b^2 x^2 \cos (c+d x)}{d^3}-\frac {2 a b x^2 \cos (c+d x)}{d}-\frac {b^2 x^4 \cos (c+d x)}{d}-\frac {24 b^2 x \sin (c+d x)}{d^4}+\frac {4 a b x \sin (c+d x)}{d^2}+\frac {4 b^2 x^3 \sin (c+d x)}{d^2}+\frac {\left (24 b^2\right ) \int \sin (c+d x) \, dx}{d^4}\\ &=-\frac {24 b^2 \cos (c+d x)}{d^5}+\frac {4 a b \cos (c+d x)}{d^3}-\frac {a^2 \cos (c+d x)}{d}+\frac {12 b^2 x^2 \cos (c+d x)}{d^3}-\frac {2 a b x^2 \cos (c+d x)}{d}-\frac {b^2 x^4 \cos (c+d x)}{d}-\frac {24 b^2 x \sin (c+d x)}{d^4}+\frac {4 a b x \sin (c+d x)}{d^2}+\frac {4 b^2 x^3 \sin (c+d x)}{d^2}\\ \end {align*}

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Mathematica [A]  time = 0.21, size = 86, normalized size = 0.62 \[ \frac {4 b d x \left (a d^2+b \left (d^2 x^2-6\right )\right ) \sin (c+d x)-\left (a^2 d^4+2 a b d^2 \left (d^2 x^2-2\right )+b^2 \left (d^4 x^4-12 d^2 x^2+24\right )\right ) \cos (c+d x)}{d^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.77, size = 97, normalized size = 0.70 \[ -\frac {{\left (b^{2} d^{4} x^{4} + a^{2} d^{4} - 4 \, a b d^{2} + 2 \, {\left (a b d^{4} - 6 \, b^{2} d^{2}\right )} x^{2} + 24 \, b^{2}\right )} \cos \left (d x + c\right ) - 4 \, {\left (b^{2} d^{3} x^{3} + {\left (a b d^{3} - 6 \, b^{2} d\right )} x\right )} \sin \left (d x + c\right )}{d^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((b^2*d^4*x^4 + a^2*d^4 - 4*a*b*d^2 + 2*(a*b*d^4 - 6*b^2*d^2)*x^2 + 24*b^2)*cos(d*x + c) - 4*(b^2*d^3*x^3 + (
a*b*d^3 - 6*b^2*d)*x)*sin(d*x + c))/d^5

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giac [A]  time = 0.41, size = 99, normalized size = 0.72 \[ -\frac {{\left (b^{2} d^{4} x^{4} + 2 \, a b d^{4} x^{2} + a^{2} d^{4} - 12 \, b^{2} d^{2} x^{2} - 4 \, a b d^{2} + 24 \, b^{2}\right )} \cos \left (d x + c\right )}{d^{5}} + \frac {4 \, {\left (b^{2} d^{3} x^{3} + a b d^{3} x - 6 \, b^{2} d x\right )} \sin \left (d x + c\right )}{d^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.02, size = 336, normalized size = 2.43 \[ \frac {\frac {b^{2} \left (-\left (d x +c \right )^{4} \cos \left (d x +c \right )+4 \left (d x +c \right )^{3} \sin \left (d x +c \right )+12 \left (d x +c \right )^{2} \cos \left (d x +c \right )-24 \cos \left (d x +c \right )-24 \left (d x +c \right ) \sin \left (d x +c \right )\right )}{d^{4}}-\frac {4 b^{2} c \left (-\left (d x +c \right )^{3} \cos \left (d x +c \right )+3 \left (d x +c \right )^{2} \sin \left (d x +c \right )-6 \sin \left (d x +c \right )+6 \left (d x +c \right ) \cos \left (d x +c \right )\right )}{d^{4}}+\frac {2 a b \left (-\left (d x +c \right )^{2} \cos \left (d x +c \right )+2 \cos \left (d x +c \right )+2 \left (d x +c \right ) \sin \left (d x +c \right )\right )}{d^{2}}+\frac {6 b^{2} c^{2} \left (-\left (d x +c \right )^{2} \cos \left (d x +c \right )+2 \cos \left (d x +c \right )+2 \left (d x +c \right ) \sin \left (d x +c \right )\right )}{d^{4}}-\frac {4 a b c \left (\sin \left (d x +c \right )-\left (d x +c \right ) \cos \left (d x +c \right )\right )}{d^{2}}-\frac {4 b^{2} c^{3} \left (\sin \left (d x +c \right )-\left (d x +c \right ) \cos \left (d x +c \right )\right )}{d^{4}}-a^{2} \cos \left (d x +c \right )-\frac {2 a b \,c^{2} \cos \left (d x +c \right )}{d^{2}}-\frac {b^{2} c^{4} \cos \left (d x +c \right )}{d^{4}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.34, size = 292, normalized size = 2.12 \[ -\frac {a^{2} \cos \left (d x + c\right ) + \frac {b^{2} c^{4} \cos \left (d x + c\right )}{d^{4}} + \frac {2 \, a b c^{2} \cos \left (d x + c\right )}{d^{2}} - \frac {4 \, {\left ({\left (d x + c\right )} \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right )} b^{2} c^{3}}{d^{4}} - \frac {4 \, {\left ({\left (d x + c\right )} \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right )} a b c}{d^{2}} + \frac {6 \, {\left ({\left ({\left (d x + c\right )}^{2} - 2\right )} \cos \left (d x + c\right ) - 2 \, {\left (d x + c\right )} \sin \left (d x + c\right )\right )} b^{2} c^{2}}{d^{4}} + \frac {2 \, {\left ({\left ({\left (d x + c\right )}^{2} - 2\right )} \cos \left (d x + c\right ) - 2 \, {\left (d x + c\right )} \sin \left (d x + c\right )\right )} a b}{d^{2}} - \frac {4 \, {\left ({\left ({\left (d x + c\right )}^{3} - 6 \, d x - 6 \, c\right )} \cos \left (d x + c\right ) - 3 \, {\left ({\left (d x + c\right )}^{2} - 2\right )} \sin \left (d x + c\right )\right )} b^{2} c}{d^{4}} + \frac {{\left ({\left ({\left (d x + c\right )}^{4} - 12 \, {\left (d x + c\right )}^{2} + 24\right )} \cos \left (d x + c\right ) - 4 \, {\left ({\left (d x + c\right )}^{3} - 6 \, d x - 6 \, c\right )} \sin \left (d x + c\right )\right )} b^{2}}{d^{4}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.83, size = 118, normalized size = 0.86 \[ \frac {4\,b^2\,x^3\,\sin \left (c+d\,x\right )}{d^2}-\frac {b^2\,x^4\,\cos \left (c+d\,x\right )}{d}-\frac {\cos \left (c+d\,x\right )\,\left (a^2\,d^4-4\,a\,b\,d^2+24\,b^2\right )}{d^5}-\frac {4\,x\,\sin \left (c+d\,x\right )\,\left (6\,b^2-a\,b\,d^2\right )}{d^4}+\frac {2\,x^2\,\cos \left (c+d\,x\right )\,\left (6\,b^2-a\,b\,d^2\right )}{d^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)*(a + b*x^2)^2,x)

[Out]

(4*b^2*x^3*sin(c + d*x))/d^2 - (b^2*x^4*cos(c + d*x))/d - (cos(c + d*x)*(24*b^2 + a^2*d^4 - 4*a*b*d^2))/d^5 -
(4*x*sin(c + d*x)*(6*b^2 - a*b*d^2))/d^4 + (2*x^2*cos(c + d*x)*(6*b^2 - a*b*d^2))/d^3

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sympy [A]  time = 2.71, size = 172, normalized size = 1.25 \[ \begin {cases} - \frac {a^{2} \cos {\left (c + d x \right )}}{d} - \frac {2 a b x^{2} \cos {\left (c + d x \right )}}{d} + \frac {4 a b x \sin {\left (c + d x \right )}}{d^{2}} + \frac {4 a b \cos {\left (c + d x \right )}}{d^{3}} - \frac {b^{2} x^{4} \cos {\left (c + d x \right )}}{d} + \frac {4 b^{2} x^{3} \sin {\left (c + d x \right )}}{d^{2}} + \frac {12 b^{2} x^{2} \cos {\left (c + d x \right )}}{d^{3}} - \frac {24 b^{2} x \sin {\left (c + d x \right )}}{d^{4}} - \frac {24 b^{2} \cos {\left (c + d x \right )}}{d^{5}} & \text {for}\: d \neq 0 \\\left (a^{2} x + \frac {2 a b x^{3}}{3} + \frac {b^{2} x^{5}}{5}\right ) \sin {\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a**2*cos(c + d*x)/d - 2*a*b*x**2*cos(c + d*x)/d + 4*a*b*x*sin(c + d*x)/d**2 + 4*a*b*cos(c + d*x)/d
**3 - b**2*x**4*cos(c + d*x)/d + 4*b**2*x**3*sin(c + d*x)/d**2 + 12*b**2*x**2*cos(c + d*x)/d**3 - 24*b**2*x*si
n(c + d*x)/d**4 - 24*b**2*cos(c + d*x)/d**5, Ne(d, 0)), ((a**2*x + 2*a*b*x**3/3 + b**2*x**5/5)*sin(c), True))

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