\(\int \frac {a+b \sin (e+f x)}{(c+d x)^3} \, dx\) [156]

Optimal result

Integrand size = 18, antiderivative size = 123 \[ \int \frac {a+b \sin (e+f x)}{(c+d x)^3} \, dx=-\frac {a}{2 d (c+d x)^2}-\frac {b f \cos (e+f x)}{2 d^2 (c+d x)}-\frac {b f^2 \operatorname {CosIntegral}\left (\frac {c f}{d}+f x\right ) \sin \left (e-\frac {c f}{d}\right )}{2 d^3}-\frac {b \sin (e+f x)}{2 d (c+d x)^2}-\frac {b f^2 \cos \left (e-\frac {c f}{d}\right ) \text {Si}\left (\frac {c f}{d}+f x\right )}{2 d^3} \] Output:

-1/2*a/d/(d*x+c)^2-1/2*b*f*cos(f*x+e)/d^2/(d*x+c)+1/2*b*f^2*Ci(c*f/d+f*x)* 
sin(-e+c*f/d)/d^3-1/2*b*sin(f*x+e)/d/(d*x+c)^2-1/2*b*f^2*cos(-e+c*f/d)*Si( 
c*f/d+f*x)/d^3
 

Mathematica [A] (verified)

Time = 0.95 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.76 \[ \int \frac {a+b \sin (e+f x)}{(c+d x)^3} \, dx=-\frac {b f^2 \operatorname {CosIntegral}\left (f \left (\frac {c}{d}+x\right )\right ) \sin \left (e-\frac {c f}{d}\right )+\frac {d (b f (c+d x) \cos (e+f x)+d (a+b \sin (e+f x)))}{(c+d x)^2}+b f^2 \cos \left (e-\frac {c f}{d}\right ) \text {Si}\left (f \left (\frac {c}{d}+x\right )\right )}{2 d^3} \] Input:

Integrate[(a + b*Sin[e + f*x])/(c + d*x)^3,x]
 

Output:

-1/2*(b*f^2*CosIntegral[f*(c/d + x)]*Sin[e - (c*f)/d] + (d*(b*f*(c + d*x)* 
Cos[e + f*x] + d*(a + b*Sin[e + f*x])))/(c + d*x)^2 + b*f^2*Cos[e - (c*f)/ 
d]*SinIntegral[f*(c/d + x)])/d^3
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3042, 3798, 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.

\(\displaystyle \int \frac {a+b \sin (e+f x)}{(c+d x)^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {a+b \sin (e+f x)}{(c+d x)^3}dx\)

\(\Big \downarrow \) 3798

\(\displaystyle \int \left (\frac {a}{(c+d x)^3}+\frac {b \sin (e+f x)}{(c+d x)^3}\right )dx\)

\(\Big \downarrow \) 2009

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

Input:

Int[(a + b*Sin[e + f*x])/(c + d*x)^3,x]
 

Output:

-1/2*a/(d*(c + d*x)^2) - (b*f*Cos[e + f*x])/(2*d^2*(c + d*x)) - (b*f^2*Cos 
Integral[(c*f)/d + f*x]*Sin[e - (c*f)/d])/(2*d^3) - (b*Sin[e + f*x])/(2*d* 
(c + d*x)^2) - (b*f^2*Cos[e - (c*f)/d]*SinIntegral[(c*f)/d + f*x])/(2*d^3)
 

Defintions of rubi rules used

rule 2009

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

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3798

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.) 
, x_Symbol] :> Int[ExpandIntegrand[(c + d*x)^m, (a + b*Sin[e + f*x])^n, x], 
 x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[n, 0] && (EqQ[n, 1] || IGtQ[ 
m, 0] || NeQ[a^2 - b^2, 0])
 

Maple [A] (verified)

Time = 1.12 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.30

method	result	size
parts	\(-\frac {a}{2 d \left (d x +c \right )^{2}}+b \,f^{2} \left (-\frac {\sin \left (f x +e \right )}{2 \left (c f -d e +d \left (f x +e \right )\right )^{2} d}+\frac {-\frac {\cos \left (f x +e \right )}{\left (c f -d e +d \left (f x +e \right )\right ) d}-\frac {\frac {\operatorname {Si}\left (f x +e +\frac {c f -d e}{d}\right ) \cos \left (\frac {c f -d e}{d}\right )}{d}-\frac {\operatorname {Ci}\left (f x +e +\frac {c f -d e}{d}\right ) \sin \left (\frac {c f -d e}{d}\right )}{d}}{d}}{2 d}\right )\)	\(160\)
derivativedivides	\(\frac {-\frac {f^{3} a}{2 \left (c f -d e +d \left (f x +e \right )\right )^{2} d}+f^{3} b \left (-\frac {\sin \left (f x +e \right )}{2 \left (c f -d e +d \left (f x +e \right )\right )^{2} d}+\frac {-\frac {\cos \left (f x +e \right )}{\left (c f -d e +d \left (f x +e \right )\right ) d}-\frac {\frac {\operatorname {Si}\left (f x +e +\frac {c f -d e}{d}\right ) \cos \left (\frac {c f -d e}{d}\right )}{d}-\frac {\operatorname {Ci}\left (f x +e +\frac {c f -d e}{d}\right ) \sin \left (\frac {c f -d e}{d}\right )}{d}}{d}}{2 d}\right )}{f}\)	\(177\)
default	\(\frac {-\frac {f^{3} a}{2 \left (c f -d e +d \left (f x +e \right )\right )^{2} d}+f^{3} b \left (-\frac {\sin \left (f x +e \right )}{2 \left (c f -d e +d \left (f x +e \right )\right )^{2} d}+\frac {-\frac {\cos \left (f x +e \right )}{\left (c f -d e +d \left (f x +e \right )\right ) d}-\frac {\frac {\operatorname {Si}\left (f x +e +\frac {c f -d e}{d}\right ) \cos \left (\frac {c f -d e}{d}\right )}{d}-\frac {\operatorname {Ci}\left (f x +e +\frac {c f -d e}{d}\right ) \sin \left (\frac {c f -d e}{d}\right )}{d}}{d}}{2 d}\right )}{f}\)	\(177\)
risch	\(-\frac {a}{2 d \left (d x +c \right )^{2}}+\frac {i f^{2} b \,{\mathrm e}^{\frac {i \left (c f -d e \right )}{d}} \operatorname {expIntegral}_{1}\left (i f x +i e +\frac {i \left (c f -d e \right )}{d}\right )}{4 d^{3}}-\frac {i f^{2} b \,{\mathrm e}^{-\frac {i \left (c f -d e \right )}{d}} \operatorname {expIntegral}_{1}\left (-i f x -i e -\frac {i c f -i d e}{d}\right )}{4 d^{3}}+\frac {i b \left (-2 i d^{3} f^{3} x^{3}-6 i c \,d^{2} f^{3} x^{2}-6 i c^{2} d \,f^{3} x -2 i c^{3} f^{3}\right ) \cos \left (f x +e \right )}{4 d^{2} \left (d x +c \right )^{2} \left (-d^{2} x^{2} f^{2}-2 c d \,f^{2} x -c^{2} f^{2}\right )}-\frac {b \left (-2 d^{2} x^{2} f^{2}-4 c d \,f^{2} x -2 c^{2} f^{2}\right ) \sin \left (f x +e \right )}{4 d \left (d x +c \right )^{2} \left (-d^{2} x^{2} f^{2}-2 c d \,f^{2} x -c^{2} f^{2}\right )}\)	\(292\)

Input:

int((a+b*sin(f*x+e))/(d*x+c)^3,x,method=_RETURNVERBOSE)
 

Output:

-1/2*a/d/(d*x+c)^2+b*f^2*(-1/2*sin(f*x+e)/(c*f-d*e+d*(f*x+e))^2/d+1/2*(-co 
s(f*x+e)/(c*f-d*e+d*(f*x+e))/d-(Si(f*x+e+(c*f-d*e)/d)*cos((c*f-d*e)/d)/d-C 
i(f*x+e+(c*f-d*e)/d)*sin((c*f-d*e)/d)/d)/d)/d)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.45 \[ \int \frac {a+b \sin (e+f x)}{(c+d x)^3} \, dx=-\frac {b d^{2} \sin \left (f x + e\right ) + a d^{2} - {\left (b d^{2} f^{2} x^{2} + 2 \, b c d f^{2} x + b c^{2} f^{2}\right )} \operatorname {Ci}\left (\frac {d f x + c f}{d}\right ) \sin \left (-\frac {d e - c f}{d}\right ) + {\left (b d^{2} f^{2} x^{2} + 2 \, b c d f^{2} x + b c^{2} f^{2}\right )} \cos \left (-\frac {d e - c f}{d}\right ) \operatorname {Si}\left (\frac {d f x + c f}{d}\right ) + {\left (b d^{2} f x + b c d f\right )} \cos \left (f x + e\right )}{2 \, {\left (d^{5} x^{2} + 2 \, c d^{4} x + c^{2} d^{3}\right )}} \] Input:

integrate((a+b*sin(f*x+e))/(d*x+c)^3,x, algorithm="fricas")
 

Output:

-1/2*(b*d^2*sin(f*x + e) + a*d^2 - (b*d^2*f^2*x^2 + 2*b*c*d*f^2*x + b*c^2* 
f^2)*cos_integral((d*f*x + c*f)/d)*sin(-(d*e - c*f)/d) + (b*d^2*f^2*x^2 + 
2*b*c*d*f^2*x + b*c^2*f^2)*cos(-(d*e - c*f)/d)*sin_integral((d*f*x + c*f)/ 
d) + (b*d^2*f*x + b*c*d*f)*cos(f*x + e))/(d^5*x^2 + 2*c*d^4*x + c^2*d^3)
 

Sympy [F]

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

integrate((a+b*sin(f*x+e))/(d*x+c)**3,x)
 

Output:

Integral((a + b*sin(e + f*x))/(c + d*x)**3, x)
 

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.15 (sec) , antiderivative size = 265, normalized size of antiderivative = 2.15 \[ \int \frac {a+b \sin (e+f x)}{(c+d x)^3} \, dx=-\frac {\frac {a f^{3}}{{\left (f x + e\right )}^{2} d^{3} + d^{3} e^{2} - 2 \, c d^{2} e f + c^{2} d f^{2} - 2 \, {\left (d^{3} e - c d^{2} f\right )} {\left (f x + e\right )}} - \frac {{\left (f^{3} {\left (-i \, E_{3}\left (\frac {i \, {\left (f x + e\right )} d - i \, d e + i \, c f}{d}\right ) + i \, E_{3}\left (-\frac {i \, {\left (f x + e\right )} d - i \, d e + i \, c f}{d}\right )\right )} \cos \left (-\frac {d e - c f}{d}\right ) + f^{3} {\left (E_{3}\left (\frac {i \, {\left (f x + e\right )} d - i \, d e + i \, c f}{d}\right ) + E_{3}\left (-\frac {i \, {\left (f x + e\right )} d - i \, d e + i \, c f}{d}\right )\right )} \sin \left (-\frac {d e - c f}{d}\right )\right )} b}{{\left (f x + e\right )}^{2} d^{3} + d^{3} e^{2} - 2 \, c d^{2} e f + c^{2} d f^{2} - 2 \, {\left (d^{3} e - c d^{2} f\right )} {\left (f x + e\right )}}}{2 \, f} \] Input:

integrate((a+b*sin(f*x+e))/(d*x+c)^3,x, algorithm="maxima")
 

Output:

-1/2*(a*f^3/((f*x + e)^2*d^3 + d^3*e^2 - 2*c*d^2*e*f + c^2*d*f^2 - 2*(d^3* 
e - c*d^2*f)*(f*x + e)) - (f^3*(-I*exp_integral_e(3, (I*(f*x + e)*d - I*d* 
e + I*c*f)/d) + I*exp_integral_e(3, -(I*(f*x + e)*d - I*d*e + I*c*f)/d))*c 
os(-(d*e - c*f)/d) + f^3*(exp_integral_e(3, (I*(f*x + e)*d - I*d*e + I*c*f 
)/d) + exp_integral_e(3, -(I*(f*x + e)*d - I*d*e + I*c*f)/d))*sin(-(d*e - 
c*f)/d))*b/((f*x + e)^2*d^3 + d^3*e^2 - 2*c*d^2*e*f + c^2*d*f^2 - 2*(d^3*e 
 - c*d^2*f)*(f*x + e)))/f
 

Giac [C] (verification not implemented)

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

Time = 0.37 (sec) , antiderivative size = 6033, normalized size of antiderivative = 49.05 \[ \int \frac {a+b \sin (e+f x)}{(c+d x)^3} \, dx=\text {Too large to display} \] Input:

integrate((a+b*sin(f*x+e))/(d*x+c)^3,x, algorithm="giac")
 

Output:

-1/4*(b*d^2*f^2*x^2*imag_part(cos_integral(f*x + c*f/d))*tan(1/2*f*x)^2*ta 
n(1/2*e)^2*tan(1/2*c*f/d)^2 - b*d^2*f^2*x^2*imag_part(cos_integral(-f*x - 
c*f/d))*tan(1/2*f*x)^2*tan(1/2*e)^2*tan(1/2*c*f/d)^2 + 2*b*d^2*f^2*x^2*sin 
_integral((d*f*x + c*f)/d)*tan(1/2*f*x)^2*tan(1/2*e)^2*tan(1/2*c*f/d)^2 + 
2*b*d^2*f^2*x^2*real_part(cos_integral(f*x + c*f/d))*tan(1/2*f*x)^2*tan(1/ 
2*e)^2*tan(1/2*c*f/d) + 2*b*d^2*f^2*x^2*real_part(cos_integral(-f*x - c*f/ 
d))*tan(1/2*f*x)^2*tan(1/2*e)^2*tan(1/2*c*f/d) - 2*b*d^2*f^2*x^2*real_part 
(cos_integral(f*x + c*f/d))*tan(1/2*f*x)^2*tan(1/2*e)*tan(1/2*c*f/d)^2 - 2 
*b*d^2*f^2*x^2*real_part(cos_integral(-f*x - c*f/d))*tan(1/2*f*x)^2*tan(1/ 
2*e)*tan(1/2*c*f/d)^2 + 2*b*c*d*f^2*x*imag_part(cos_integral(f*x + c*f/d)) 
*tan(1/2*f*x)^2*tan(1/2*e)^2*tan(1/2*c*f/d)^2 - 2*b*c*d*f^2*x*imag_part(co 
s_integral(-f*x - c*f/d))*tan(1/2*f*x)^2*tan(1/2*e)^2*tan(1/2*c*f/d)^2 + 4 
*b*c*d*f^2*x*sin_integral((d*f*x + c*f)/d)*tan(1/2*f*x)^2*tan(1/2*e)^2*tan 
(1/2*c*f/d)^2 - b*d^2*f^2*x^2*imag_part(cos_integral(f*x + c*f/d))*tan(1/2 
*f*x)^2*tan(1/2*e)^2 + b*d^2*f^2*x^2*imag_part(cos_integral(-f*x - c*f/d)) 
*tan(1/2*f*x)^2*tan(1/2*e)^2 - 2*b*d^2*f^2*x^2*sin_integral((d*f*x + c*f)/ 
d)*tan(1/2*f*x)^2*tan(1/2*e)^2 + 4*b*d^2*f^2*x^2*imag_part(cos_integral(f* 
x + c*f/d))*tan(1/2*f*x)^2*tan(1/2*e)*tan(1/2*c*f/d) - 4*b*d^2*f^2*x^2*ima 
g_part(cos_integral(-f*x - c*f/d))*tan(1/2*f*x)^2*tan(1/2*e)*tan(1/2*c*f/d 
) + 8*b*d^2*f^2*x^2*sin_integral((d*f*x + c*f)/d)*tan(1/2*f*x)^2*tan(1/...
 

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b \sin (e+f x)}{(c+d x)^3} \, dx=\int \frac {a+b\,\sin \left (e+f\,x\right )}{{\left (c+d\,x\right )}^3} \,d x \] Input:

int((a + b*sin(e + f*x))/(c + d*x)^3,x)
 

Output:

int((a + b*sin(e + f*x))/(c + d*x)^3, x)
 

Reduce [F]

\[ \int \frac {a+b \sin (e+f x)}{(c+d x)^3} \, dx=\frac {2 \left (\int \frac {\sin \left (f x +e \right )}{d^{3} x^{3}+3 c \,d^{2} x^{2}+3 c^{2} d x +c^{3}}d x \right ) b \,c^{2} d +4 \left (\int \frac {\sin \left (f x +e \right )}{d^{3} x^{3}+3 c \,d^{2} x^{2}+3 c^{2} d x +c^{3}}d x \right ) b c \,d^{2} x +2 \left (\int \frac {\sin \left (f x +e \right )}{d^{3} x^{3}+3 c \,d^{2} x^{2}+3 c^{2} d x +c^{3}}d x \right ) b \,d^{3} x^{2}-a}{2 d \left (d^{2} x^{2}+2 c d x +c^{2}\right )} \] Input:

int((a+b*sin(f*x+e))/(d*x+c)^3,x)
 

Output:

(2*int(sin(e + f*x)/(c**3 + 3*c**2*d*x + 3*c*d**2*x**2 + d**3*x**3),x)*b*c 
**2*d + 4*int(sin(e + f*x)/(c**3 + 3*c**2*d*x + 3*c*d**2*x**2 + d**3*x**3) 
,x)*b*c*d**2*x + 2*int(sin(e + f*x)/(c**3 + 3*c**2*d*x + 3*c*d**2*x**2 + d 
**3*x**3),x)*b*d**3*x**2 - a)/(2*d*(c**2 + 2*c*d*x + d**2*x**2))