\(\int \frac {(a+b x^2)^2 \sin (c+d x)}{x} \, dx\) [52]

Optimal result

Integrand size = 19, antiderivative size = 111 \[ \int \frac {\left (a+b x^2\right )^2 \sin (c+d x)}{x} \, dx=\frac {6 b^2 x \cos (c+d x)}{d^3}-\frac {2 a b x \cos (c+d x)}{d}-\frac {b^2 x^3 \cos (c+d x)}{d}+a^2 \operatorname {CosIntegral}(d x) \sin (c)-\frac {6 b^2 \sin (c+d x)}{d^4}+\frac {2 a b \sin (c+d x)}{d^2}+\frac {3 b^2 x^2 \sin (c+d x)}{d^2}+a^2 \cos (c) \text {Si}(d x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {3420, 3384, 3380, 3383, 3377, 2717} \[ \int \frac {\left (a+b x^2\right )^2 \sin (c+d x)}{x} \, dx=a^2 \sin (c) \operatorname {CosIntegral}(d x)+a^2 \cos (c) \text {Si}(d x)+\frac {2 a b \sin (c+d x)}{d^2}-\frac {2 a b x \cos (c+d x)}{d}-\frac {6 b^2 \sin (c+d x)}{d^4}+\frac {6 b^2 x \cos (c+d x)}{d^3}+\frac {3 b^2 x^2 \sin (c+d x)}{d^2}-\frac {b^2 x^3 \cos (c+d x)}{d} \]

[In]

Int[((a + b*x^2)^2*Sin[c + d*x])/x,x]

[Out]

(6*b^2*x*Cos[c + d*x])/d^3 - (2*a*b*x*Cos[c + d*x])/d - (b^2*x^3*Cos[c + d*x])/d + a^2*CosIntegral[d*x]*Sin[c]
- (6*b^2*Sin[c + d*x])/d^4 + (2*a*b*Sin[c + d*x])/d^2 + (3*b^2*x^2*Sin[c + d*x])/d^2 + a^2*Cos[c]*SinIntegral
[d*x]

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 3380

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]

Rule 3383

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]

Rule 3384

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[c*(f/d) + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[c*(f/d) + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 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*} \text {integral}& = \int \left (\frac {a^2 \sin (c+d x)}{x}+2 a b x \sin (c+d x)+b^2 x^3 \sin (c+d x)\right ) \, dx \\ & = a^2 \int \frac {\sin (c+d x)}{x} \, dx+(2 a b) \int x \sin (c+d x) \, dx+b^2 \int x^3 \sin (c+d x) \, dx \\ & = -\frac {2 a b x \cos (c+d x)}{d}-\frac {b^2 x^3 \cos (c+d x)}{d}+\frac {(2 a b) \int \cos (c+d x) \, dx}{d}+\frac {\left (3 b^2\right ) \int x^2 \cos (c+d x) \, dx}{d}+\left (a^2 \cos (c)\right ) \int \frac {\sin (d x)}{x} \, dx+\left (a^2 \sin (c)\right ) \int \frac {\cos (d x)}{x} \, dx \\ & = -\frac {2 a b x \cos (c+d x)}{d}-\frac {b^2 x^3 \cos (c+d x)}{d}+a^2 \operatorname {CosIntegral}(d x) \sin (c)+\frac {2 a b \sin (c+d x)}{d^2}+\frac {3 b^2 x^2 \sin (c+d x)}{d^2}+a^2 \cos (c) \text {Si}(d x)-\frac {\left (6 b^2\right ) \int x \sin (c+d x) \, dx}{d^2} \\ & = \frac {6 b^2 x \cos (c+d x)}{d^3}-\frac {2 a b x \cos (c+d x)}{d}-\frac {b^2 x^3 \cos (c+d x)}{d}+a^2 \operatorname {CosIntegral}(d x) \sin (c)+\frac {2 a b \sin (c+d x)}{d^2}+\frac {3 b^2 x^2 \sin (c+d x)}{d^2}+a^2 \cos (c) \text {Si}(d x)-\frac {\left (6 b^2\right ) \int \cos (c+d x) \, dx}{d^3} \\ & = \frac {6 b^2 x \cos (c+d x)}{d^3}-\frac {2 a b x \cos (c+d x)}{d}-\frac {b^2 x^3 \cos (c+d x)}{d}+a^2 \operatorname {CosIntegral}(d x) \sin (c)-\frac {6 b^2 \sin (c+d x)}{d^4}+\frac {2 a b \sin (c+d x)}{d^2}+\frac {3 b^2 x^2 \sin (c+d x)}{d^2}+a^2 \cos (c) \text {Si}(d x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.74 \[ \int \frac {\left (a+b x^2\right )^2 \sin (c+d x)}{x} \, dx=-\frac {b x \left (2 a d^2+b \left (-6+d^2 x^2\right )\right ) \cos (c+d x)}{d^3}+a^2 \operatorname {CosIntegral}(d x) \sin (c)+\frac {b \left (2 a d^2+3 b \left (-2+d^2 x^2\right )\right ) \sin (c+d x)}{d^4}+a^2 \cos (c) \text {Si}(d x) \]

[In]

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

[Out]

-((b*x*(2*a*d^2 + b*(-6 + d^2*x^2))*Cos[c + d*x])/d^3) + a^2*CosIntegral[d*x]*Sin[c] + (b*(2*a*d^2 + 3*b*(-2 +
d^2*x^2))*Sin[c + d*x])/d^4 + a^2*Cos[c]*SinIntegral[d*x]

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.32 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.15

method	result	size
risch	\(-\frac {b^{2} x^{3} \cos \left (d x +c \right )}{d}+\frac {i a^{2} {\mathrm e}^{i c} \operatorname {Ei}_{1}\left (-i d x \right )}{2}-\frac {i a^{2} {\mathrm e}^{-i c} \operatorname {Ei}_{1}\left (i d x \right )}{2}+\frac {3 b^{2} x^{2} \sin \left (d x +c \right )}{d^{2}}-\frac {2 a b x \cos \left (d x +c \right )}{d}+\frac {2 a b \sin \left (d x +c \right )}{d^{2}}+\frac {6 b^{2} x \cos \left (d x +c \right )}{d^{3}}-\frac {6 b^{2} \sin \left (d x +c \right )}{d^{4}}\)	\(128\)
derivativedivides	\(a^{2} \left (\operatorname {Si}\left (d x \right ) \cos \left (c \right )+\operatorname {Ci}\left (d x \right ) \sin \left (c \right )\right )+\frac {4 a b c \cos \left (d x +c \right )}{d^{2}}+\frac {2 \left (c +1\right ) a b \left (\sin \left (d x +c \right )-\cos \left (d x +c \right ) \left (d x +c \right )\right )}{d^{2}}+\frac {4 b^{2} c^{3} \cos \left (d x +c \right )}{d^{4}}+\frac {6 \left (c +1\right ) b^{2} c^{2} \left (\sin \left (d x +c \right )-\cos \left (d x +c \right ) \left (d x +c \right )\right )}{d^{4}}-\frac {4 b^{2} c \left (c^{2}+c +1\right ) \left (-\left (d x +c \right )^{2} \cos \left (d x +c \right )+2 \cos \left (d x +c \right )+2 \left (d x +c \right ) \sin \left (d x +c \right )\right )}{d^{4}}+\frac {\left (c^{3}+c^{2}+c +1\right ) b^{2} \left (-\left (d x +c \right )^{3} \cos \left (d x +c \right )+3 \left (d x +c \right )^{2} \sin \left (d x +c \right )-6 \sin \left (d x +c \right )+6 \cos \left (d x +c \right ) \left (d x +c \right )\right )}{d^{4}}\)	\(236\)
default	\(a^{2} \left (\operatorname {Si}\left (d x \right ) \cos \left (c \right )+\operatorname {Ci}\left (d x \right ) \sin \left (c \right )\right )+\frac {4 a b c \cos \left (d x +c \right )}{d^{2}}+\frac {2 \left (c +1\right ) a b \left (\sin \left (d x +c \right )-\cos \left (d x +c \right ) \left (d x +c \right )\right )}{d^{2}}+\frac {4 b^{2} c^{3} \cos \left (d x +c \right )}{d^{4}}+\frac {6 \left (c +1\right ) b^{2} c^{2} \left (\sin \left (d x +c \right )-\cos \left (d x +c \right ) \left (d x +c \right )\right )}{d^{4}}-\frac {4 b^{2} c \left (c^{2}+c +1\right ) \left (-\left (d x +c \right )^{2} \cos \left (d x +c \right )+2 \cos \left (d x +c \right )+2 \left (d x +c \right ) \sin \left (d x +c \right )\right )}{d^{4}}+\frac {\left (c^{3}+c^{2}+c +1\right ) b^{2} \left (-\left (d x +c \right )^{3} \cos \left (d x +c \right )+3 \left (d x +c \right )^{2} \sin \left (d x +c \right )-6 \sin \left (d x +c \right )+6 \cos \left (d x +c \right ) \left (d x +c \right )\right )}{d^{4}}\)	\(236\)
meijerg	\(\frac {8 b^{2} \sqrt {\pi }\, \sin \left (c \right ) \left (\frac {3}{4 \sqrt {\pi }}-\frac {\left (-\frac {3 d^{2} x^{2}}{2}+3\right ) \cos \left (d x \right )}{4 \sqrt {\pi }}-\frac {d x \left (-\frac {d^{2} x^{2}}{2}+3\right ) \sin \left (d x \right )}{4 \sqrt {\pi }}\right )}{d^{4}}+\frac {8 b^{2} \sqrt {\pi }\, \cos \left (c \right ) \left (\frac {x d \left (-\frac {5 d^{2} x^{2}}{2}+15\right ) \cos \left (d x \right )}{20 \sqrt {\pi }}-\frac {\left (-\frac {15 d^{2} x^{2}}{2}+15\right ) \sin \left (d x \right )}{20 \sqrt {\pi }}\right )}{d^{4}}+\frac {4 a b \sqrt {\pi }\, \sin \left (c \right ) \left (-\frac {1}{2 \sqrt {\pi }}+\frac {\cos \left (d x \right )}{2 \sqrt {\pi }}+\frac {d x \sin \left (d x \right )}{2 \sqrt {\pi }}\right )}{d^{2}}+\frac {4 a b \sqrt {\pi }\, \cos \left (c \right ) \left (-\frac {d x \cos \left (d x \right )}{2 \sqrt {\pi }}+\frac {\sin \left (d x \right )}{2 \sqrt {\pi }}\right )}{d^{2}}+\frac {a^{2} \sqrt {\pi }\, \sin \left (c \right ) \left (\frac {2 \gamma +2 \ln \left (x \right )+\ln \left (d^{2}\right )}{\sqrt {\pi }}-\frac {2 \gamma }{\sqrt {\pi }}-\frac {2 \ln \left (2\right )}{\sqrt {\pi }}-\frac {2 \ln \left (\frac {d x}{2}\right )}{\sqrt {\pi }}+\frac {2 \,\operatorname {Ci}\left (d x \right )}{\sqrt {\pi }}\right )}{2}+a^{2} \cos \left (c \right ) \operatorname {Si}\left (d x \right )\)	\(255\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.87 \[ \int \frac {\left (a+b x^2\right )^2 \sin (c+d x)}{x} \, dx=\frac {a^{2} d^{4} \operatorname {Ci}\left (d x\right ) \sin \left (c\right ) + a^{2} d^{4} \cos \left (c\right ) \operatorname {Si}\left (d x\right ) - {\left (b^{2} d^{3} x^{3} + 2 \, {\left (a b d^{3} - 3 \, b^{2} d\right )} x\right )} \cos \left (d x + c\right ) + {\left (3 \, b^{2} d^{2} x^{2} + 2 \, a b d^{2} - 6 \, b^{2}\right )} \sin \left (d x + c\right )}{d^{4}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 2.49 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.44 \[ \int \frac {\left (a+b x^2\right )^2 \sin (c+d x)}{x} \, dx=a^{2} \sin {\left (c \right )} \operatorname {Ci}{\left (d x \right )} + a^{2} \cos {\left (c \right )} \operatorname {Si}{\left (d x \right )} + 2 a b x \left (\begin {cases} x \sin {\left (c \right )} & \text {for}\: d = 0 \\- \frac {\cos {\left (c + d x \right )}}{d} & \text {otherwise} \end {cases}\right ) - 2 a b \left (\begin {cases} \frac {x^{2} \sin {\left (c \right )}}{2} & \text {for}\: d = 0 \\- \frac {\begin {cases} \frac {\sin {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \cos {\left (c \right )} & \text {otherwise} \end {cases}}{d} & \text {otherwise} \end {cases}\right ) + b^{2} x^{3} \left (\begin {cases} x \sin {\left (c \right )} & \text {for}\: d = 0 \\- \frac {\cos {\left (c + d x \right )}}{d} & \text {otherwise} \end {cases}\right ) - 3 b^{2} \left (\begin {cases} \frac {x^{4} \sin {\left (c \right )}}{4} & \text {for}\: d = 0 \\- \frac {\begin {cases} \frac {x^{2} \sin {\left (c + d x \right )}}{d} + \frac {2 x \cos {\left (c + d x \right )}}{d^{2}} - \frac {2 \sin {\left (c + d x \right )}}{d^{3}} & \text {for}\: d \neq 0 \\\frac {x^{3} \cos {\left (c \right )}}{3} & \text {otherwise} \end {cases}}{d} & \text {otherwise} \end {cases}\right ) \]

[In]

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

[Out]

a**2*sin(c)*Ci(d*x) + a**2*cos(c)*Si(d*x) + 2*a*b*x*Piecewise((x*sin(c), Eq(d, 0)), (-cos(c + d*x)/d, True)) -
2*a*b*Piecewise((x**2*sin(c)/2, Eq(d, 0)), (-Piecewise((sin(c + d*x)/d, Ne(d, 0)), (x*cos(c), True))/d, True)
) + b**2*x**3*Piecewise((x*sin(c), Eq(d, 0)), (-cos(c + d*x)/d, True)) - 3*b**2*Piecewise((x**4*sin(c)/4, Eq(d
, 0)), (-Piecewise((x**2*sin(c + d*x)/d + 2*x*cos(c + d*x)/d**2 - 2*sin(c + d*x)/d**3, Ne(d, 0)), (x**3*cos(c)
/3, True))/d, True))

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.38 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.05 \[ \int \frac {\left (a+b x^2\right )^2 \sin (c+d x)}{x} \, dx=\frac {{\left (a^{2} {\left (-i \, {\rm Ei}\left (i \, d x\right ) + i \, {\rm Ei}\left (-i \, d x\right )\right )} \cos \left (c\right ) + a^{2} {\left ({\rm Ei}\left (i \, d x\right ) + {\rm Ei}\left (-i \, d x\right )\right )} \sin \left (c\right )\right )} d^{4} - 2 \, {\left (b^{2} d^{3} x^{3} + 2 \, {\left (a b d^{3} - 3 \, b^{2} d\right )} x\right )} \cos \left (d x + c\right ) + 2 \, {\left (3 \, b^{2} d^{2} x^{2} + 2 \, a b d^{2} - 6 \, b^{2}\right )} \sin \left (d x + c\right )}{2 \, d^{4}} \]

[In]

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

[Out]

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

Giac [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.31 (sec) , antiderivative size = 725, normalized size of antiderivative = 6.53 \[ \int \frac {\left (a+b x^2\right )^2 \sin (c+d x)}{x} \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/2*(2*b^2*d^3*x^3*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c)^2 - a^2*d^4*imag_part(cos_integral(d*x))*tan(1/2*d*x + 1/
2*c)^2*tan(1/2*c)^2 + a^2*d^4*imag_part(cos_integral(-d*x))*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c)^2 - 2*a^2*d^4*si
n_integral(d*x)*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c)^2 + 2*b^2*d^3*x^3*tan(1/2*d*x + 1/2*c)^2 + 2*a^2*d^4*real_pa
rt(cos_integral(d*x))*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c) + 2*a^2*d^4*real_part(cos_integral(-d*x))*tan(1/2*d*x
+ 1/2*c)^2*tan(1/2*c) - 2*b^2*d^3*x^3*tan(1/2*c)^2 + 4*a*b*d^3*x*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c)^2 + a^2*d^4
*imag_part(cos_integral(d*x))*tan(1/2*d*x + 1/2*c)^2 - a^2*d^4*imag_part(cos_integral(-d*x))*tan(1/2*d*x + 1/2
*c)^2 + 2*a^2*d^4*sin_integral(d*x)*tan(1/2*d*x + 1/2*c)^2 - a^2*d^4*imag_part(cos_integral(d*x))*tan(1/2*c)^2
+ a^2*d^4*imag_part(cos_integral(-d*x))*tan(1/2*c)^2 - 2*a^2*d^4*sin_integral(d*x)*tan(1/2*c)^2 + 12*b^2*d^2*
x^2*tan(1/2*d*x + 1/2*c)*tan(1/2*c)^2 - 2*b^2*d^3*x^3 + 4*a*b*d^3*x*tan(1/2*d*x + 1/2*c)^2 + 2*a^2*d^4*real_pa
rt(cos_integral(d*x))*tan(1/2*c) + 2*a^2*d^4*real_part(cos_integral(-d*x))*tan(1/2*c) - 4*a*b*d^3*x*tan(1/2*c)
^2 - 12*b^2*d*x*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c)^2 + a^2*d^4*imag_part(cos_integral(d*x)) - a^2*d^4*imag_part
(cos_integral(-d*x)) + 2*a^2*d^4*sin_integral(d*x) + 12*b^2*d^2*x^2*tan(1/2*d*x + 1/2*c) + 8*a*b*d^2*tan(1/2*d
*x + 1/2*c)*tan(1/2*c)^2 - 4*a*b*d^3*x - 12*b^2*d*x*tan(1/2*d*x + 1/2*c)^2 + 12*b^2*d*x*tan(1/2*c)^2 + 8*a*b*d
^2*tan(1/2*d*x + 1/2*c) - 24*b^2*tan(1/2*d*x + 1/2*c)*tan(1/2*c)^2 + 12*b^2*d*x - 24*b^2*tan(1/2*d*x + 1/2*c))
/(d^4*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c)^2 + d^4*tan(1/2*d*x + 1/2*c)^2 + d^4*tan(1/2*c)^2 + d^4)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^2\right )^2 \sin (c+d x)}{x} \, dx=\int \frac {\sin \left (c+d\,x\right )\,{\left (b\,x^2+a\right )}^2}{x} \,d x \]

[In]

int((sin(c + d*x)*(a + b*x^2)^2)/x,x)

[Out]

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