\(\int \frac {x^2 \sin (c+d x)}{a+b x} \, dx\) [20]

Optimal result

Integrand size = 17, antiderivative size = 89 \[ \int \frac {x^2 \sin (c+d x)}{a+b x} \, dx=\frac {(a-b x) \cos (c+d x)}{b^2 d}+\frac {a^2 \operatorname {CosIntegral}\left (\frac {a d}{b}+d x\right ) \sin \left (c-\frac {a d}{b}\right )}{b^3}+\frac {\sin (c+d x)}{b d^2}+\frac {a^2 \cos \left (c-\frac {a d}{b}\right ) \text {Si}\left (\frac {a d}{b}+d x\right )}{b^3} \] Output:

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

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.98 \[ \int \frac {x^2 \sin (c+d x)}{a+b x} \, dx=\frac {a^2 d^2 \operatorname {CosIntegral}\left (d \left (\frac {a}{b}+x\right )\right ) \sin \left (c-\frac {a d}{b}\right )+b (d (a-b x) \cos (c+d x)+b \sin (c+d x))+a^2 d^2 \cos \left (c-\frac {a d}{b}\right ) \text {Si}\left (d \left (\frac {a}{b}+x\right )\right )}{b^3 d^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.11, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {7293, 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.

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 2009

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

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.96 (sec) , antiderivative size = 180, normalized size of antiderivative = 2.02

method	result	size
risch	\(-\frac {\left (d x b -a d \right ) \cos \left (d x +c \right )}{d^{2} b^{2}}+\frac {\sin \left (d x +c \right )}{b \,d^{2}}-\frac {i a^{2} \cos \left (\frac {a d -b c}{b}\right ) \operatorname {expIntegral}_{1}\left (\frac {i \left (b x +a \right ) d}{b}\right )}{2 b^{3}}+\frac {i a^{2} \cos \left (\frac {a d -b c}{b}\right ) \operatorname {expIntegral}_{1}\left (-\frac {i \left (b x +a \right ) d}{b}\right )}{2 b^{3}}+\frac {a^{2} \sin \left (\frac {a d -b c}{b}\right ) \operatorname {expIntegral}_{1}\left (\frac {i \left (b x +a \right ) d}{b}\right )}{2 b^{3}}+\frac {a^{2} \sin \left (\frac {a d -b c}{b}\right ) \operatorname {expIntegral}_{1}\left (-\frac {i \left (b x +a \right ) d}{b}\right )}{2 b^{3}}\)	\(180\)
derivativedivides	\(\frac {d \,c^{2} \left (\frac {\operatorname {Si}\left (d x +c +\frac {a d -b c}{b}\right ) \cos \left (\frac {a d -b c}{b}\right )}{b}-\frac {\operatorname {Ci}\left (d x +c +\frac {a d -b c}{b}\right ) \sin \left (\frac {a d -b c}{b}\right )}{b}\right )+\frac {2 \left (a d -b c \right ) d c \left (\frac {\operatorname {Si}\left (d x +c +\frac {a d -b c}{b}\right ) \cos \left (\frac {a d -b c}{b}\right )}{b}-\frac {\operatorname {Ci}\left (d x +c +\frac {a d -b c}{b}\right ) \sin \left (\frac {a d -b c}{b}\right )}{b}\right )}{b}+\frac {2 d c \cos \left (d x +c \right )}{b}+\frac {\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) d \left (\frac {\operatorname {Si}\left (d x +c +\frac {a d -b c}{b}\right ) \cos \left (\frac {a d -b c}{b}\right )}{b}-\frac {\operatorname {Ci}\left (d x +c +\frac {a d -b c}{b}\right ) \sin \left (\frac {a d -b c}{b}\right )}{b}\right )}{b^{2}}-\frac {d \left (a d -b c -b \right ) \left (\sin \left (d x +c \right )-\cos \left (d x +c \right ) \left (d x +c \right )\right )}{b^{2}}}{d^{3}}\)	\(318\)
default	\(\frac {d \,c^{2} \left (\frac {\operatorname {Si}\left (d x +c +\frac {a d -b c}{b}\right ) \cos \left (\frac {a d -b c}{b}\right )}{b}-\frac {\operatorname {Ci}\left (d x +c +\frac {a d -b c}{b}\right ) \sin \left (\frac {a d -b c}{b}\right )}{b}\right )+\frac {2 \left (a d -b c \right ) d c \left (\frac {\operatorname {Si}\left (d x +c +\frac {a d -b c}{b}\right ) \cos \left (\frac {a d -b c}{b}\right )}{b}-\frac {\operatorname {Ci}\left (d x +c +\frac {a d -b c}{b}\right ) \sin \left (\frac {a d -b c}{b}\right )}{b}\right )}{b}+\frac {2 d c \cos \left (d x +c \right )}{b}+\frac {\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) d \left (\frac {\operatorname {Si}\left (d x +c +\frac {a d -b c}{b}\right ) \cos \left (\frac {a d -b c}{b}\right )}{b}-\frac {\operatorname {Ci}\left (d x +c +\frac {a d -b c}{b}\right ) \sin \left (\frac {a d -b c}{b}\right )}{b}\right )}{b^{2}}-\frac {d \left (a d -b c -b \right ) \left (\sin \left (d x +c \right )-\cos \left (d x +c \right ) \left (d x +c \right )\right )}{b^{2}}}{d^{3}}\)	\(318\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.21 \[ \int \frac {x^2 \sin (c+d x)}{a+b x} \, dx=-\frac {a^{2} d^{2} \operatorname {Ci}\left (\frac {b d x + a d}{b}\right ) \sin \left (-\frac {b c - a d}{b}\right ) - a^{2} d^{2} \cos \left (-\frac {b c - a d}{b}\right ) \operatorname {Si}\left (\frac {b d x + a d}{b}\right ) - b^{2} \sin \left (d x + c\right ) + {\left (b^{2} d x - a b d\right )} \cos \left (d x + c\right )}{b^{3} d^{2}} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

-1/2*((b*cos(c)^2 + b*sin(c)^2)*d*x^2*cos(d*x + c) - ((a*(I*exp_integral_e 
(2, (I*b*d*x + I*a*d)/b) - I*exp_integral_e(2, -(I*b*d*x + I*a*d)/b))*cos( 
c)^2 + a*(I*exp_integral_e(2, (I*b*d*x + I*a*d)/b) - I*exp_integral_e(2, - 
(I*b*d*x + I*a*d)/b))*sin(c)^2)*cos(-(b*c - a*d)/b) - (a*(exp_integral_e(2 
, (I*b*d*x + I*a*d)/b) + exp_integral_e(2, -(I*b*d*x + I*a*d)/b))*cos(c)^2 
 + a*(exp_integral_e(2, (I*b*d*x + I*a*d)/b) + exp_integral_e(2, -(I*b*d*x 
 + I*a*d)/b))*sin(c)^2)*sin(-(b*c - a*d)/b))*cos(d*x + c)^2 - (b*cos(c)^2 
+ b*sin(c)^2)*x*sin(d*x + c) - ((a*(I*exp_integral_e(2, (I*b*d*x + I*a*d)/ 
b) - I*exp_integral_e(2, -(I*b*d*x + I*a*d)/b))*cos(c)^2 + a*(I*exp_integr 
al_e(2, (I*b*d*x + I*a*d)/b) - I*exp_integral_e(2, -(I*b*d*x + I*a*d)/b))* 
sin(c)^2)*cos(-(b*c - a*d)/b) - (a*(exp_integral_e(2, (I*b*d*x + I*a*d)/b) 
 + exp_integral_e(2, -(I*b*d*x + I*a*d)/b))*cos(c)^2 + a*(exp_integral_e(2 
, (I*b*d*x + I*a*d)/b) + exp_integral_e(2, -(I*b*d*x + I*a*d)/b))*sin(c)^2 
)*sin(-(b*c - a*d)/b))*sin(d*x + c)^2 + ((b*d*x^2*cos(c) + b*x*sin(c))*cos 
(d*x + c)^2 + (b*d*x^2*cos(c) + b*x*sin(c))*sin(d*x + c)^2)*cos(d*x + 2*c) 
 - 2*(((a*b^2*cos(c)^2 + a*b^2*sin(c)^2)*d^3*x + (a^2*b*cos(c)^2 + a^2*b*s 
in(c)^2)*d^3)*cos(d*x + c)^2 + ((a*b^2*cos(c)^2 + a*b^2*sin(c)^2)*d^3*x + 
(a^2*b*cos(c)^2 + a^2*b*sin(c)^2)*d^3)*sin(d*x + c)^2)*integrate(1/2*x*cos 
(d*x + c)/(b^2*d^2*x^2 + 2*a*b*d^2*x + a^2*d^2), x) - 2*(((a*b^2*cos(c)^2 
+ a*b^2*sin(c)^2)*d^3*x + (a^2*b*cos(c)^2 + a^2*b*sin(c)^2)*d^3)*cos(d*...
 

Giac [C] (verification not implemented)

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

Time = 0.16 (sec) , antiderivative size = 2205, normalized size of antiderivative = 24.78 \[ \int \frac {x^2 \sin (c+d x)}{a+b x} \, dx=\text {Too large to display} \] Input:

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

Output:

1/2*(a^2*d^2*imag_part(cos_integral(d*x + a*d/b))*tan(1/2*d*x + 1/2*c)^2*t 
an(1/2*c)^2*tan(1/2*a*d/b)^2 - a^2*d^2*imag_part(cos_integral(-d*x - a*d/b 
))*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c)^2*tan(1/2*a*d/b)^2 + 2*a^2*d^2*sin_in 
tegral((b*d*x + a*d)/b)*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c)^2*tan(1/2*a*d/b) 
^2 + 2*a^2*d^2*real_part(cos_integral(d*x + a*d/b))*tan(1/2*d*x + 1/2*c)^2 
*tan(1/2*c)^2*tan(1/2*a*d/b) + 2*a^2*d^2*real_part(cos_integral(-d*x - a*d 
/b))*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c)^2*tan(1/2*a*d/b) - 2*a^2*d^2*real_p 
art(cos_integral(d*x + a*d/b))*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c)*tan(1/2*a 
*d/b)^2 - 2*a^2*d^2*real_part(cos_integral(-d*x - a*d/b))*tan(1/2*d*x + 1/ 
2*c)^2*tan(1/2*c)*tan(1/2*a*d/b)^2 + 2*b^2*d*x*tan(1/2*d*x + 1/2*c)^2*tan( 
1/2*c)^2*tan(1/2*a*d/b)^2 - a^2*d^2*imag_part(cos_integral(d*x + a*d/b))*t 
an(1/2*d*x + 1/2*c)^2*tan(1/2*c)^2 + a^2*d^2*imag_part(cos_integral(-d*x - 
 a*d/b))*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c)^2 - 2*a^2*d^2*sin_integral((b*d 
*x + a*d)/b)*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c)^2 + 4*a^2*d^2*imag_part(cos 
_integral(d*x + a*d/b))*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c)*tan(1/2*a*d/b) - 
 4*a^2*d^2*imag_part(cos_integral(-d*x - a*d/b))*tan(1/2*d*x + 1/2*c)^2*ta 
n(1/2*c)*tan(1/2*a*d/b) + 8*a^2*d^2*sin_integral((b*d*x + a*d)/b)*tan(1/2* 
d*x + 1/2*c)^2*tan(1/2*c)*tan(1/2*a*d/b) - a^2*d^2*imag_part(cos_integral( 
d*x + a*d/b))*tan(1/2*d*x + 1/2*c)^2*tan(1/2*a*d/b)^2 + a^2*d^2*imag_part( 
cos_integral(-d*x - a*d/b))*tan(1/2*d*x + 1/2*c)^2*tan(1/2*a*d/b)^2 - 2...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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