\(\int \frac {(a+b x^2)^2 \sin (c+d x)}{x^2} \, dx\) [53]

Optimal result

Integrand size = 19, antiderivative size = 97 \[ \int \frac {\left (a+b x^2\right )^2 \sin (c+d x)}{x^2} \, dx=\frac {2 b^2 \cos (c+d x)}{d^3}-\frac {2 a b \cos (c+d x)}{d}-\frac {b^2 x^2 \cos (c+d x)}{d}+a^2 d \cos (c) \operatorname {CosIntegral}(d x)-\frac {a^2 \sin (c+d x)}{x}+\frac {2 b^2 x \sin (c+d x)}{d^2}-a^2 d \sin (c) \text {Si}(d x) \] Output:

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

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^2\right )^2 \sin (c+d x)}{x^2} \, dx=\frac {2 b^2 \cos (c+d x)}{d^3}-\frac {2 a b \cos (c+d x)}{d}-\frac {b^2 x^2 \cos (c+d x)}{d}+a^2 d \cos (c) \operatorname {CosIntegral}(d x)-\frac {a^2 \sin (c+d x)}{x}+\frac {2 b^2 x \sin (c+d x)}{d^2}-a^2 d \sin (c) \text {Si}(d x) \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, 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.

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

\(\Big \downarrow \) 3820

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

\(\Big \downarrow \) 2009

\(\displaystyle a^2 d \cos (c) \operatorname {CosIntegral}(d x)-a^2 d \sin (c) \text {Si}(d x)-\frac {a^2 \sin (c+d x)}{x}-\frac {2 a b \cos (c+d x)}{d}+\frac {2 b^2 \cos (c+d x)}{d^3}+\frac {2 b^2 x \sin (c+d x)}{d^2}-\frac {b^2 x^2 \cos (c+d x)}{d}\)

Input:

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

Output:

(2*b^2*Cos[c + d*x])/d^3 - (2*a*b*Cos[c + d*x])/d - (b^2*x^2*Cos[c + d*x]) 
/d + a^2*d*Cos[c]*CosIntegral[d*x] - (a^2*Sin[c + d*x])/x + (2*b^2*x*Sin[c 
 + d*x])/d^2 - a^2*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 = 1.14 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.61

method	result	size
derivativedivides	\(d \left (a^{2} \left (-\frac {\sin \left (d x +c \right )}{d x}-\operatorname {Si}\left (d x \right ) \sin \left (c \right )+\operatorname {Ci}\left (d x \right ) \cos \left (c \right )\right )-\frac {2 a b \cos \left (d x +c \right )}{d^{2}}-\frac {6 b^{2} c^{2} \cos \left (d x +c \right )}{d^{4}}-\frac {4 c \,b^{2} \left (2 c +1\right ) \left (\sin \left (d x +c \right )-\cos \left (d x +c \right ) \left (d x +c \right )\right )}{d^{4}}+\frac {\left (3 c^{2}+2 c +1\right ) b^{2} \left (-\left (d x +c \right )^{2} \cos \left (d x +c \right )+2 \cos \left (d x +c \right )+2 \left (d x +c \right ) \sin \left (d x +c \right )\right )}{d^{4}}\right )\)	\(156\)
default	\(d \left (a^{2} \left (-\frac {\sin \left (d x +c \right )}{d x}-\operatorname {Si}\left (d x \right ) \sin \left (c \right )+\operatorname {Ci}\left (d x \right ) \cos \left (c \right )\right )-\frac {2 a b \cos \left (d x +c \right )}{d^{2}}-\frac {6 b^{2} c^{2} \cos \left (d x +c \right )}{d^{4}}-\frac {4 c \,b^{2} \left (2 c +1\right ) \left (\sin \left (d x +c \right )-\cos \left (d x +c \right ) \left (d x +c \right )\right )}{d^{4}}+\frac {\left (3 c^{2}+2 c +1\right ) b^{2} \left (-\left (d x +c \right )^{2} \cos \left (d x +c \right )+2 \cos \left (d x +c \right )+2 \left (d x +c \right ) \sin \left (d x +c \right )\right )}{d^{4}}\right )\)	\(156\)
risch	\(-\frac {-\pi \,\operatorname {csgn}\left (d x \right ) \sin \left (c \right ) a^{2} d^{4} x +2 \,\operatorname {Si}\left (d x \right ) \sin \left (c \right ) a^{2} d^{4} x -i \pi \,\operatorname {csgn}\left (d x \right ) \cos \left (c \right ) a^{2} d^{4} x +2 i \operatorname {Si}\left (d x \right ) \cos \left (c \right ) a^{2} d^{4} x +2 \,\operatorname {expIntegral}_{1}\left (-i d x \right ) \cos \left (c \right ) a^{2} d^{4} x +2 \cos \left (d x +c \right ) b^{2} d^{2} x^{3}+2 \sin \left (d x +c \right ) a^{2} d^{3}-4 \sin \left (d x +c \right ) b^{2} d \,x^{2}+4 \cos \left (d x +c \right ) a b \,d^{2} x -4 \cos \left (d x +c \right ) b^{2} x}{2 d^{3} x}\)	\(164\)
meijerg	\(\frac {4 b^{2} \sin \left (c \right ) \sqrt {\pi }\, \left (\frac {x \left (d^{2}\right )^{\frac {3}{2}} \cos \left (d x \right )}{2 \sqrt {\pi }\, d^{2}}-\frac {\left (d^{2}\right )^{\frac {3}{2}} \left (-\frac {3 x^{2} d^{2}}{2}+3\right ) \sin \left (d x \right )}{6 \sqrt {\pi }\, d^{3}}\right )}{d^{2} \sqrt {d^{2}}}+\frac {4 b^{2} \cos \left (c \right ) \sqrt {\pi }\, \left (-\frac {1}{2 \sqrt {\pi }}+\frac {\left (-\frac {x^{2} d^{2}}{2}+1\right ) \cos \left (d x \right )}{2 \sqrt {\pi }}+\frac {x d \sin \left (d x \right )}{2 \sqrt {\pi }}\right )}{d^{3}}+\frac {2 a b \sin \left (c \right ) \sin \left (d x \right )}{d}+\frac {2 a b \cos \left (c \right ) \sqrt {\pi }\, \left (\frac {1}{\sqrt {\pi }}-\frac {\cos \left (d x \right )}{\sqrt {\pi }}\right )}{d}+\frac {a^{2} \sin \left (c \right ) \sqrt {\pi }\, d^{2} \left (-\frac {4 d^{2} \cos \left (x \sqrt {d^{2}}\right )}{x \left (d^{2}\right )^{\frac {3}{2}} \sqrt {\pi }}-\frac {4 \,\operatorname {Si}\left (x \sqrt {d^{2}}\right )}{\sqrt {\pi }}\right )}{4 \sqrt {d^{2}}}+\frac {a^{2} \cos \left (c \right ) \sqrt {\pi }\, d \left (\frac {4 \gamma -4+4 \ln \left (x \right )+4 \ln \left (d \right )}{\sqrt {\pi }}+\frac {4}{\sqrt {\pi }}-\frac {4 \gamma }{\sqrt {\pi }}-\frac {4 \ln \left (2\right )}{\sqrt {\pi }}-\frac {4 \ln \left (\frac {d x}{2}\right )}{\sqrt {\pi }}-\frac {4 \sin \left (d x \right )}{\sqrt {\pi }\, x d}+\frac {4 \,\operatorname {Ci}\left (d x \right )}{\sqrt {\pi }}\right )}{4}\)	\(291\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.99 \[ \int \frac {\left (a+b x^2\right )^2 \sin (c+d x)}{x^2} \, dx=\frac {a^{2} d^{4} x \cos \left (c\right ) \operatorname {Ci}\left (d x\right ) - a^{2} d^{4} x \sin \left (c\right ) \operatorname {Si}\left (d x\right ) - {\left (b^{2} d^{2} x^{3} + 2 \, {\left (a b d^{2} - b^{2}\right )} x\right )} \cos \left (d x + c\right ) - {\left (a^{2} d^{3} - 2 \, b^{2} d x^{2}\right )} \sin \left (d x + c\right )}{d^{3} x} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.99 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^2\right )^2 \sin (c+d x)}{x^2} \, dx=\frac {{\left (a^{2} {\left (\Gamma \left (-1, i \, d x\right ) + \Gamma \left (-1, -i \, d x\right )\right )} \cos \left (c\right ) + a^{2} {\left (-i \, \Gamma \left (-1, i \, d x\right ) + i \, \Gamma \left (-1, -i \, d x\right )\right )} \sin \left (c\right )\right )} d^{4} + 4 \, b^{2} d x \sin \left (d x + c\right ) - 2 \, {\left (b^{2} d^{2} x^{2} + 2 \, a b d^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )}{2 \, d^{3}} \] Input:

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

Output:

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

Giac [C] (verification not implemented)

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

Time = 0.15 (sec) , antiderivative size = 1638, normalized size of antiderivative = 16.89 \[ \int \frac {\left (a+b x^2\right )^2 \sin (c+d x)}{x^2} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/2*(a^2*d^4*x*real_part(cos_integral(d*x))*tan(1/2*d*x + 1/2*c)^2*tan(1/ 
2*d*x)^2*tan(1/2*c)^2 + a^2*d^4*x*real_part(cos_integral(-d*x))*tan(1/2*d* 
x + 1/2*c)^2*tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*a^2*d^4*x*imag_part(cos_integ 
ral(d*x))*tan(1/2*d*x + 1/2*c)^2*tan(1/2*d*x)^2*tan(1/2*c) - 2*a^2*d^4*x*i 
mag_part(cos_integral(-d*x))*tan(1/2*d*x + 1/2*c)^2*tan(1/2*d*x)^2*tan(1/2 
*c) + 4*a^2*d^4*x*sin_integral(d*x)*tan(1/2*d*x + 1/2*c)^2*tan(1/2*d*x)^2* 
tan(1/2*c) - 2*b^2*d^2*x^3*tan(1/2*d*x + 1/2*c)^2*tan(1/2*d*x)^2*tan(1/2*c 
)^2 - a^2*d^4*x*real_part(cos_integral(d*x))*tan(1/2*d*x + 1/2*c)^2*tan(1/ 
2*d*x)^2 - a^2*d^4*x*real_part(cos_integral(-d*x))*tan(1/2*d*x + 1/2*c)^2* 
tan(1/2*d*x)^2 + a^2*d^4*x*real_part(cos_integral(d*x))*tan(1/2*d*x + 1/2* 
c)^2*tan(1/2*c)^2 + a^2*d^4*x*real_part(cos_integral(-d*x))*tan(1/2*d*x + 
1/2*c)^2*tan(1/2*c)^2 + a^2*d^4*x*real_part(cos_integral(d*x))*tan(1/2*d*x 
)^2*tan(1/2*c)^2 + a^2*d^4*x*real_part(cos_integral(-d*x))*tan(1/2*d*x)^2* 
tan(1/2*c)^2 - 2*b^2*d^2*x^3*tan(1/2*d*x + 1/2*c)^2*tan(1/2*d*x)^2 + 2*a^2 
*d^4*x*imag_part(cos_integral(d*x))*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c) - 2* 
a^2*d^4*x*imag_part(cos_integral(-d*x))*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c) 
+ 4*a^2*d^4*x*sin_integral(d*x)*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c) + 2*a^2* 
d^4*x*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c) - 2*a^2*d^4*x 
*imag_part(cos_integral(-d*x))*tan(1/2*d*x)^2*tan(1/2*c) + 4*a^2*d^4*x*sin 
_integral(d*x)*tan(1/2*d*x)^2*tan(1/2*c) - 2*b^2*d^2*x^3*tan(1/2*d*x + ...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (a+b x^2\right )^2 \sin (c+d x)}{x^2} \, dx=\frac {-2 \cos \left (d x +c \right ) a b \,d^{2}-\cos \left (d x +c \right ) b^{2} d^{2} x^{2}+2 \cos \left (d x +c \right ) b^{2}+\left (\int \frac {\sin \left (d x +c \right )}{x^{2}}d x \right ) a^{2} d^{3}+2 \sin \left (d x +c \right ) b^{2} d x}{d^{3}} \] Input:

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

Output:

( - 2*cos(c + d*x)*a*b*d**2 - cos(c + d*x)*b**2*d**2*x**2 + 2*cos(c + d*x) 
*b**2 + int(sin(c + d*x)/x**2,x)*a**2*d**3 + 2*sin(c + d*x)*b**2*d*x)/d**3