\(\int \frac {(a+b x^2) \sin (c+d x)}{x^2} \, dx\) [45]

Optimal result

Integrand size = 17, antiderivative size = 44 \[ \int \frac {\left (a+b x^2\right ) \sin (c+d x)}{x^2} \, dx=-\frac {b \cos (c+d x)}{d}+a d \cos (c) \operatorname {CosIntegral}(d x)-\frac {a \sin (c+d x)}{x}-a d \sin (c) \text {Si}(d x) \] Output:

-b*cos(d*x+c)/d+a*d*cos(c)*Ci(d*x)-a*sin(d*x+c)/x-a*d*sin(c)*Si(d*x)
 

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^2\right ) \sin (c+d x)}{x^2} \, dx=-\frac {b \cos (c+d x)}{d}+a d \cos (c) \operatorname {CosIntegral}(d x)-\frac {a \sin (c+d x)}{x}-a d \sin (c) \text {Si}(d x) \] Input:

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

Output:

-((b*Cos[c + d*x])/d) + a*d*Cos[c]*CosIntegral[d*x] - (a*Sin[c + d*x])/x - 
 a*d*Sin[c]*SinIntegral[d*x]
 

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {3820, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\left (a+b x^2\right ) \sin (c+d x)}{x^2} \, dx\)

\(\Big \downarrow \) 3820

\(\displaystyle \int \left (\frac {a \sin (c+d x)}{x^2}+b \sin (c+d x)\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle a d \cos (c) \operatorname {CosIntegral}(d x)-a d \sin (c) \text {Si}(d x)-\frac {a \sin (c+d x)}{x}-\frac {b \cos (c+d x)}{d}\)

Input:

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

Output:

-((b*Cos[c + d*x])/d) + a*d*Cos[c]*CosIntegral[d*x] - (a*Sin[c + d*x])/x - 
 a*d*Sin[c]*SinIntegral[d*x]
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3820

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*Sin[(c_.) + (d_.)*(x_ 
)], x_Symbol] :> Int[ExpandIntegrand[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]
 

Maple [A] (verified)

Time = 0.81 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.09

method	result	size
derivativedivides	\(d \left (a \left (-\frac {\sin \left (d x +c \right )}{d x}-\operatorname {Si}\left (d x \right ) \sin \left (c \right )+\operatorname {Ci}\left (d x \right ) \cos \left (c \right )\right )-\frac {b \cos \left (d x +c \right )}{d^{2}}\right )\)	\(48\)
default	\(d \left (a \left (-\frac {\sin \left (d x +c \right )}{d x}-\operatorname {Si}\left (d x \right ) \sin \left (c \right )+\operatorname {Ci}\left (d x \right ) \cos \left (c \right )\right )-\frac {b \cos \left (d x +c \right )}{d^{2}}\right )\)	\(48\)
risch	\(-\frac {d \cos \left (c \right ) a \,\operatorname {expIntegral}_{1}\left (-i d x \right )}{2}-\frac {d \cos \left (c \right ) a \,\operatorname {expIntegral}_{1}\left (i d x \right )}{2}-\frac {i d \sin \left (c \right ) a \,\operatorname {expIntegral}_{1}\left (-i d x \right )}{2}+\frac {i d \sin \left (c \right ) a \,\operatorname {expIntegral}_{1}\left (i d x \right )}{2}-\frac {b \cos \left (d x +c \right )}{d}-\frac {a \sin \left (d x +c \right )}{x}\)	\(80\)
meijerg	\(\frac {b \sin \left (c \right ) \sin \left (d x \right )}{d}+\frac {b \cos \left (c \right ) \sqrt {\pi }\, \left (\frac {1}{\sqrt {\pi }}-\frac {\cos \left (d x \right )}{\sqrt {\pi }}\right )}{d}+\frac {a \sin \left (c \right ) \sqrt {\pi }\, d^{2} \left (-\frac {4 d^{2} \cos \left (x \sqrt {d^{2}}\right )}{x \left (d^{2}\right )^{\frac {3}{2}} \sqrt {\pi }}-\frac {4 \,\operatorname {Si}\left (x \sqrt {d^{2}}\right )}{\sqrt {\pi }}\right )}{4 \sqrt {d^{2}}}+\frac {a \cos \left (c \right ) \sqrt {\pi }\, d \left (\frac {4 \gamma -4+4 \ln \left (x \right )+4 \ln \left (d \right )}{\sqrt {\pi }}+\frac {4}{\sqrt {\pi }}-\frac {4 \gamma }{\sqrt {\pi }}-\frac {4 \ln \left (2\right )}{\sqrt {\pi }}-\frac {4 \ln \left (\frac {d x}{2}\right )}{\sqrt {\pi }}-\frac {4 \sin \left (d x \right )}{\sqrt {\pi }\, x d}+\frac {4 \,\operatorname {Ci}\left (d x \right )}{\sqrt {\pi }}\right )}{4}\)	\(170\)

Input:

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

Output:

d*(a*(-sin(d*x+c)/d/x-Si(d*x)*sin(c)+Ci(d*x)*cos(c))-1/d^2*b*cos(d*x+c))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.20 \[ \int \frac {\left (a+b x^2\right ) \sin (c+d x)}{x^2} \, dx=\frac {a d^{2} x \cos \left (c\right ) \operatorname {Ci}\left (d x\right ) - a d^{2} x \sin \left (c\right ) \operatorname {Si}\left (d x\right ) - b x \cos \left (d x + c\right ) - a d \sin \left (d x + c\right )}{d x} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.24 (sec) , antiderivative size = 937, normalized size of antiderivative = 21.30 \[ \int \frac {\left (a+b x^2\right ) \sin (c+d x)}{x^2} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/4*(((I*exp_integral_e(2, I*d*x) - I*exp_integral_e(2, -I*d*x))*cos(c)^3 
 + (I*exp_integral_e(2, I*d*x) - I*exp_integral_e(2, -I*d*x))*cos(c)*sin(c 
)^2 + (exp_integral_e(2, I*d*x) + exp_integral_e(2, -I*d*x))*sin(c)^3 + (I 
*exp_integral_e(2, I*d*x) - I*exp_integral_e(2, -I*d*x))*cos(c) + ((exp_in 
tegral_e(2, I*d*x) + exp_integral_e(2, -I*d*x))*cos(c)^2 + exp_integral_e( 
2, I*d*x) + exp_integral_e(2, -I*d*x))*sin(c))*b*c^2/((d*x + c)*(cos(c)^2 
+ sin(c)^2)*d^2 - (c*cos(c)^2 + c*sin(c)^2)*d^2) - ((I*exp_integral_e(2, I 
*d*x) - I*exp_integral_e(2, -I*d*x))*cos(c)^3 + (I*exp_integral_e(2, I*d*x 
) - I*exp_integral_e(2, -I*d*x))*cos(c)*sin(c)^2 + (exp_integral_e(2, I*d* 
x) + exp_integral_e(2, -I*d*x))*sin(c)^3 + (I*exp_integral_e(2, I*d*x) - I 
*exp_integral_e(2, -I*d*x))*cos(c) + ((exp_integral_e(2, I*d*x) + exp_inte 
gral_e(2, -I*d*x))*cos(c)^2 + exp_integral_e(2, I*d*x) + exp_integral_e(2, 
 -I*d*x))*sin(c))*a/(c*cos(c)^2 + c*sin(c)^2 - (d*x + c)*(cos(c)^2 + sin(c 
)^2)) + 2*(((b*cos(c)^2 + b*sin(c)^2)*(d*x + c)^2 - 2*(b*c*cos(c)^2 + b*c* 
sin(c)^2)*(d*x + c))*cos(d*x + c)^3 + (b*c^2*(exp_integral_e(3, I*d*x) + e 
xp_integral_e(3, -I*d*x))*cos(c)^3 + b*c^2*(exp_integral_e(3, I*d*x) + exp 
_integral_e(3, -I*d*x))*cos(c)*sin(c)^2 + b*c^2*(-I*exp_integral_e(3, I*d* 
x) + I*exp_integral_e(3, -I*d*x))*sin(c)^3 + b*c^2*(exp_integral_e(3, I*d* 
x) + exp_integral_e(3, -I*d*x))*cos(c) + (b*c^2*(-I*exp_integral_e(3, I*d* 
x) + I*exp_integral_e(3, -I*d*x))*cos(c)^2 + b*c^2*(-I*exp_integral_e(3...
 

Giac [C] (verification not implemented)

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

Time = 0.14 (sec) , antiderivative size = 411, normalized size of antiderivative = 9.34 \[ \int \frac {\left (a+b x^2\right ) \sin (c+d x)}{x^2} \, dx =\text {Too large to display} \] Input:

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

Output:

-1/2*(a*d^2*x*real_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 + a 
*d^2*x*real_part(cos_integral(-d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*a*d^2 
*x*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c) - 2*a*d^2*x*imag 
_part(cos_integral(-d*x))*tan(1/2*d*x)^2*tan(1/2*c) + 4*a*d^2*x*sin_integr 
al(d*x)*tan(1/2*d*x)^2*tan(1/2*c) - a*d^2*x*real_part(cos_integral(d*x))*t 
an(1/2*d*x)^2 - a*d^2*x*real_part(cos_integral(-d*x))*tan(1/2*d*x)^2 + a*d 
^2*x*real_part(cos_integral(d*x))*tan(1/2*c)^2 + a*d^2*x*real_part(cos_int 
egral(-d*x))*tan(1/2*c)^2 + 2*a*d^2*x*imag_part(cos_integral(d*x))*tan(1/2 
*c) - 2*a*d^2*x*imag_part(cos_integral(-d*x))*tan(1/2*c) + 4*a*d^2*x*sin_i 
ntegral(d*x)*tan(1/2*c) + 2*b*x*tan(1/2*d*x)^2*tan(1/2*c)^2 - a*d^2*x*real 
_part(cos_integral(d*x)) - a*d^2*x*real_part(cos_integral(-d*x)) - 4*a*d*t 
an(1/2*d*x)^2*tan(1/2*c) - 4*a*d*tan(1/2*d*x)*tan(1/2*c)^2 - 2*b*x*tan(1/2 
*d*x)^2 - 8*b*x*tan(1/2*d*x)*tan(1/2*c) - 2*b*x*tan(1/2*c)^2 + 4*a*d*tan(1 
/2*d*x) + 4*a*d*tan(1/2*c) + 2*b*x)/(d*x*tan(1/2*d*x)^2*tan(1/2*c)^2 + d*x 
*tan(1/2*d*x)^2 + d*x*tan(1/2*c)^2 + d*x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (a+b x^2\right ) \sin (c+d x)}{x^2} \, dx=\frac {-\cos \left (d x +c \right ) b +\left (\int \frac {\sin \left (d x +c \right )}{x^{2}}d x \right ) a d}{d} \] Input:

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

Output:

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