Integrand size = 14, antiderivative size = 104 \[ \int \frac {\sin (c+d x)}{(a+b x)^3} \, dx=-\frac {d \cos (c+d x)}{2 b^2 (a+b x)}-\frac {d^2 \operatorname {CosIntegral}\left (\frac {a d}{b}+d x\right ) \sin \left (c-\frac {a d}{b}\right )}{2 b^3}-\frac {\sin (c+d x)}{2 b (a+b x)^2}-\frac {d^2 \cos \left (c-\frac {a d}{b}\right ) \text {Si}\left (\frac {a d}{b}+d x\right )}{2 b^3} \]
-1/2*d*cos(d*x+c)/b^2/(b*x+a)-1/2*d^2*cos(-c+a*d/b)*Si(a*d/b+d*x)/b^3+1/2* d^2*Ci(a*d/b+d*x)*sin(-c+a*d/b)/b^3-1/2*sin(d*x+c)/b/(b*x+a)^2
Time = 0.48 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.84 \[ \int \frac {\sin (c+d x)}{(a+b x)^3} \, dx=-\frac {d^2 \operatorname {CosIntegral}\left (d \left (\frac {a}{b}+x\right )\right ) \sin \left (c-\frac {a d}{b}\right )+\frac {b (d (a+b x) \cos (c+d x)+b \sin (c+d x))}{(a+b x)^2}+d^2 \cos \left (c-\frac {a d}{b}\right ) \text {Si}\left (d \left (\frac {a}{b}+x\right )\right )}{2 b^3} \]
-1/2*(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 + b*x)^2 + d^2*Cos[c - (a*d)/b]*SinIntegral [d*(a/b + x)])/b^3
Time = 0.59 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.01, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {3042, 3778, 3042, 3778, 25, 3042, 3784, 3042, 3780, 3783}
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.
\(\displaystyle \int \frac {\sin (c+d x)}{(a+b x)^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x)}{(a+b x)^3}dx\) |
\(\Big \downarrow \) 3778 |
\(\displaystyle \frac {d \int \frac {\cos (c+d x)}{(a+b x)^2}dx}{2 b}-\frac {\sin (c+d x)}{2 b (a+b x)^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {d \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )}{(a+b x)^2}dx}{2 b}-\frac {\sin (c+d x)}{2 b (a+b x)^2}\) |
\(\Big \downarrow \) 3778 |
\(\displaystyle \frac {d \left (\frac {d \int -\frac {\sin (c+d x)}{a+b x}dx}{b}-\frac {\cos (c+d x)}{b (a+b x)}\right )}{2 b}-\frac {\sin (c+d x)}{2 b (a+b x)^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {d \left (-\frac {d \int \frac {\sin (c+d x)}{a+b x}dx}{b}-\frac {\cos (c+d x)}{b (a+b x)}\right )}{2 b}-\frac {\sin (c+d x)}{2 b (a+b x)^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {d \left (-\frac {d \int \frac {\sin (c+d x)}{a+b x}dx}{b}-\frac {\cos (c+d x)}{b (a+b x)}\right )}{2 b}-\frac {\sin (c+d x)}{2 b (a+b x)^2}\) |
\(\Big \downarrow \) 3784 |
\(\displaystyle \frac {d \left (-\frac {d \left (\sin \left (c-\frac {a d}{b}\right ) \int \frac {\cos \left (x d+\frac {a d}{b}\right )}{a+b x}dx+\cos \left (c-\frac {a d}{b}\right ) \int \frac {\sin \left (x d+\frac {a d}{b}\right )}{a+b x}dx\right )}{b}-\frac {\cos (c+d x)}{b (a+b x)}\right )}{2 b}-\frac {\sin (c+d x)}{2 b (a+b x)^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {d \left (-\frac {d \left (\sin \left (c-\frac {a d}{b}\right ) \int \frac {\sin \left (x d+\frac {a d}{b}+\frac {\pi }{2}\right )}{a+b x}dx+\cos \left (c-\frac {a d}{b}\right ) \int \frac {\sin \left (x d+\frac {a d}{b}\right )}{a+b x}dx\right )}{b}-\frac {\cos (c+d x)}{b (a+b x)}\right )}{2 b}-\frac {\sin (c+d x)}{2 b (a+b x)^2}\) |
\(\Big \downarrow \) 3780 |
\(\displaystyle \frac {d \left (-\frac {d \left (\sin \left (c-\frac {a d}{b}\right ) \int \frac {\sin \left (x d+\frac {a d}{b}+\frac {\pi }{2}\right )}{a+b x}dx+\frac {\cos \left (c-\frac {a d}{b}\right ) \text {Si}\left (x d+\frac {a d}{b}\right )}{b}\right )}{b}-\frac {\cos (c+d x)}{b (a+b x)}\right )}{2 b}-\frac {\sin (c+d x)}{2 b (a+b x)^2}\) |
\(\Big \downarrow \) 3783 |
\(\displaystyle \frac {d \left (-\frac {d \left (\frac {\sin \left (c-\frac {a d}{b}\right ) \operatorname {CosIntegral}\left (x d+\frac {a d}{b}\right )}{b}+\frac {\cos \left (c-\frac {a d}{b}\right ) \text {Si}\left (x d+\frac {a d}{b}\right )}{b}\right )}{b}-\frac {\cos (c+d x)}{b (a+b x)}\right )}{2 b}-\frac {\sin (c+d x)}{2 b (a+b x)^2}\) |
-1/2*Sin[c + d*x]/(b*(a + b*x)^2) + (d*(-(Cos[c + d*x]/(b*(a + b*x))) - (d *((CosIntegral[(a*d)/b + d*x]*Sin[c - (a*d)/b])/b + (Cos[c - (a*d)/b]*SinI ntegral[(a*d)/b + d*x])/b))/b))/(2*b)
3.1.36.3.1 Defintions of rubi rules used
Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(c + d*x)^(m + 1)*(Sin[e + f*x]/(d*(m + 1))), x] - Simp[f/(d*(m + 1)) Int[( c + d*x)^(m + 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && LtQ[m, - 1]
Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinInte gral[e + f*x]/d, x] /; FreeQ[{c, d, e, f}, x] && EqQ[d*e - c*f, 0]
Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosInte gral[e - Pi/2 + f*x]/d, x] /; FreeQ[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]
Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[Cos[(d* e - c*f)/d] Int[Sin[c*(f/d) + f*x]/(c + d*x), x], x] + Simp[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]
Time = 0.32 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.39
method | result | size |
derivativedivides | \(d^{2} \left (-\frac {\sin \left (d x +c \right )}{2 \left (d a -c b +b \left (d x +c \right )\right )^{2} b}+\frac {-\frac {\cos \left (d x +c \right )}{\left (d a -c b +b \left (d x +c \right )\right ) b}-\frac {\frac {\operatorname {Si}\left (d x +c +\frac {d a -c b}{b}\right ) \cos \left (\frac {d a -c b}{b}\right )}{b}-\frac {\operatorname {Ci}\left (d x +c +\frac {d a -c b}{b}\right ) \sin \left (\frac {d a -c b}{b}\right )}{b}}{b}}{2 b}\right )\) | \(145\) |
default | \(d^{2} \left (-\frac {\sin \left (d x +c \right )}{2 \left (d a -c b +b \left (d x +c \right )\right )^{2} b}+\frac {-\frac {\cos \left (d x +c \right )}{\left (d a -c b +b \left (d x +c \right )\right ) b}-\frac {\frac {\operatorname {Si}\left (d x +c +\frac {d a -c b}{b}\right ) \cos \left (\frac {d a -c b}{b}\right )}{b}-\frac {\operatorname {Ci}\left (d x +c +\frac {d a -c b}{b}\right ) \sin \left (\frac {d a -c b}{b}\right )}{b}}{b}}{2 b}\right )\) | \(145\) |
risch | \(-\frac {i d^{2} {\mathrm e}^{-\frac {i \left (d a -c b \right )}{b}} \operatorname {Ei}_{1}\left (-i d x -i c -\frac {i a d -i c b}{b}\right )}{4 b^{3}}+\frac {i d^{2} {\mathrm e}^{\frac {i \left (d a -c b \right )}{b}} \operatorname {Ei}_{1}\left (i d x +i c +\frac {i \left (d a -c b \right )}{b}\right )}{4 b^{3}}+\frac {i \left (-2 i b^{3} d^{3} x^{3}-6 i a \,b^{2} d^{3} x^{2}-6 i a^{2} b \,d^{3} x -2 i a^{3} d^{3}\right ) \cos \left (d x +c \right )}{4 b^{2} \left (b x +a \right )^{2} \left (-d^{2} x^{2} b^{2}-2 a b \,d^{2} x -d^{2} a^{2}\right )}-\frac {\left (-2 d^{2} x^{2} b^{2}-4 a b \,d^{2} x -2 d^{2} a^{2}\right ) \sin \left (d x +c \right )}{4 b \left (b x +a \right )^{2} \left (-d^{2} x^{2} b^{2}-2 a b \,d^{2} x -d^{2} a^{2}\right )}\) | \(275\) |
d^2*(-1/2*sin(d*x+c)/(d*a-c*b+b*(d*x+c))^2/b+1/2*(-cos(d*x+c)/(d*a-c*b+b*( d*x+c))/b-(Si(d*x+c+(a*d-b*c)/b)*cos((a*d-b*c)/b)/b-Ci(d*x+c+(a*d-b*c)/b)* sin((a*d-b*c)/b)/b)/b)/b)
Time = 0.29 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.58 \[ \int \frac {\sin (c+d x)}{(a+b x)^3} \, dx=-\frac {b^{2} \sin \left (d x + c\right ) - {\left (b^{2} d^{2} x^{2} + 2 \, a b d^{2} x + a^{2} d^{2}\right )} \operatorname {Ci}\left (\frac {b d x + a d}{b}\right ) \sin \left (-\frac {b c - a d}{b}\right ) + {\left (b^{2} d^{2} x^{2} + 2 \, a b d^{2} x + a^{2} d^{2}\right )} \cos \left (-\frac {b c - a d}{b}\right ) \operatorname {Si}\left (\frac {b d x + a d}{b}\right ) + {\left (b^{2} d x + a b d\right )} \cos \left (d x + c\right )}{2 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}} \]
-1/2*(b^2*sin(d*x + c) - (b^2*d^2*x^2 + 2*a*b*d^2*x + a^2*d^2)*cos_integra l((b*d*x + a*d)/b)*sin(-(b*c - a*d)/b) + (b^2*d^2*x^2 + 2*a*b*d^2*x + a^2* d^2)*cos(-(b*c - a*d)/b)*sin_integral((b*d*x + a*d)/b) + (b^2*d*x + a*b*d) *cos(d*x + c))/(b^5*x^2 + 2*a*b^4*x + a^2*b^3)
\[ \int \frac {\sin (c+d x)}{(a+b x)^3} \, dx=\int \frac {\sin {\left (c + d x \right )}}{\left (a + b x\right )^{3}}\, dx \]
Result contains complex when optimal does not.
Time = 0.29 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.91 \[ \int \frac {\sin (c+d x)}{(a+b x)^3} \, dx=\frac {d^{3} {\left (-i \, E_{3}\left (\frac {i \, {\left (d x + c\right )} b - i \, b c + i \, a d}{b}\right ) + i \, E_{3}\left (-\frac {i \, {\left (d x + c\right )} b - i \, b c + i \, a d}{b}\right )\right )} \cos \left (-\frac {b c - a d}{b}\right ) + d^{3} {\left (E_{3}\left (\frac {i \, {\left (d x + c\right )} b - i \, b c + i \, a d}{b}\right ) + E_{3}\left (-\frac {i \, {\left (d x + c\right )} b - i \, b c + i \, a d}{b}\right )\right )} \sin \left (-\frac {b c - a d}{b}\right )}{2 \, {\left ({\left (d x + c\right )}^{2} b^{3} + b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2} - 2 \, {\left (b^{3} c - a b^{2} d\right )} {\left (d x + c\right )}\right )} d} \]
1/2*(d^3*(-I*exp_integral_e(3, (I*(d*x + c)*b - I*b*c + I*a*d)/b) + I*exp_ integral_e(3, -(I*(d*x + c)*b - I*b*c + I*a*d)/b))*cos(-(b*c - a*d)/b) + d ^3*(exp_integral_e(3, (I*(d*x + c)*b - I*b*c + I*a*d)/b) + exp_integral_e( 3, -(I*(d*x + c)*b - I*b*c + I*a*d)/b))*sin(-(b*c - a*d)/b))/(((d*x + c)^2 *b^3 + b^3*c^2 - 2*a*b^2*c*d + a^2*b*d^2 - 2*(b^3*c - a*b^2*d)*(d*x + c))* d)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.41 (sec) , antiderivative size = 5727, normalized size of antiderivative = 55.07 \[ \int \frac {\sin (c+d x)}{(a+b x)^3} \, dx=\text {Too large to display} \]
-1/4*(b^2*d^2*x^2*imag_part(cos_integral(d*x + a*d/b))*tan(1/2*d*x)^2*tan( 1/2*c)^2*tan(1/2*a*d/b)^2 - b^2*d^2*x^2*imag_part(cos_integral(-d*x - a*d/ b))*tan(1/2*d*x)^2*tan(1/2*c)^2*tan(1/2*a*d/b)^2 + 2*b^2*d^2*x^2*sin_integ ral((b*d*x + a*d)/b)*tan(1/2*d*x)^2*tan(1/2*c)^2*tan(1/2*a*d/b)^2 + 2*b^2* d^2*x^2*real_part(cos_integral(d*x + a*d/b))*tan(1/2*d*x)^2*tan(1/2*c)^2*t an(1/2*a*d/b) + 2*b^2*d^2*x^2*real_part(cos_integral(-d*x - a*d/b))*tan(1/ 2*d*x)^2*tan(1/2*c)^2*tan(1/2*a*d/b) - 2*b^2*d^2*x^2*real_part(cos_integra l(d*x + a*d/b))*tan(1/2*d*x)^2*tan(1/2*c)*tan(1/2*a*d/b)^2 - 2*b^2*d^2*x^2 *real_part(cos_integral(-d*x - a*d/b))*tan(1/2*d*x)^2*tan(1/2*c)*tan(1/2*a *d/b)^2 + 2*a*b*d^2*x*imag_part(cos_integral(d*x + a*d/b))*tan(1/2*d*x)^2* tan(1/2*c)^2*tan(1/2*a*d/b)^2 - 2*a*b*d^2*x*imag_part(cos_integral(-d*x - a*d/b))*tan(1/2*d*x)^2*tan(1/2*c)^2*tan(1/2*a*d/b)^2 + 4*a*b*d^2*x*sin_int egral((b*d*x + a*d)/b)*tan(1/2*d*x)^2*tan(1/2*c)^2*tan(1/2*a*d/b)^2 - b^2* d^2*x^2*imag_part(cos_integral(d*x + a*d/b))*tan(1/2*d*x)^2*tan(1/2*c)^2 + b^2*d^2*x^2*imag_part(cos_integral(-d*x - a*d/b))*tan(1/2*d*x)^2*tan(1/2* c)^2 - 2*b^2*d^2*x^2*sin_integral((b*d*x + a*d)/b)*tan(1/2*d*x)^2*tan(1/2* c)^2 + 4*b^2*d^2*x^2*imag_part(cos_integral(d*x + a*d/b))*tan(1/2*d*x)^2*t an(1/2*c)*tan(1/2*a*d/b) - 4*b^2*d^2*x^2*imag_part(cos_integral(-d*x - a*d /b))*tan(1/2*d*x)^2*tan(1/2*c)*tan(1/2*a*d/b) + 8*b^2*d^2*x^2*sin_integral ((b*d*x + a*d)/b)*tan(1/2*d*x)^2*tan(1/2*c)*tan(1/2*a*d/b) + 4*a*b*d^2*...
Timed out. \[ \int \frac {\sin (c+d x)}{(a+b x)^3} \, dx=\int \frac {\sin \left (c+d\,x\right )}{{\left (a+b\,x\right )}^3} \,d x \]