\(\int \frac {\sin (c+d x)}{x^2 (a+b x^3)} \, dx\) [100]

Optimal result

Integrand size = 19, antiderivative size = 380 \[ \int \frac {\sin (c+d x)}{x^2 \left (a+b x^3\right )} \, dx=\frac {d \cos (c) \operatorname {CosIntegral}(d x)}{a}+\frac {\sqrt [3]{b} \operatorname {CosIntegral}\left (\frac {\sqrt [3]{a} d}{\sqrt [3]{b}}+d x\right ) \sin \left (c-\frac {\sqrt [3]{a} d}{\sqrt [3]{b}}\right )}{3 a^{4/3}}+\frac {(-1)^{2/3} \sqrt [3]{b} \operatorname {CosIntegral}\left (\frac {\sqrt [3]{-1} \sqrt [3]{a} d}{\sqrt [3]{b}}-d x\right ) \sin \left (c+\frac {\sqrt [3]{-1} \sqrt [3]{a} d}{\sqrt [3]{b}}\right )}{3 a^{4/3}}-\frac {\sqrt [3]{-1} \sqrt [3]{b} \operatorname {CosIntegral}\left (\frac {(-1)^{2/3} \sqrt [3]{a} d}{\sqrt [3]{b}}+d x\right ) \sin \left (c-\frac {(-1)^{2/3} \sqrt [3]{a} d}{\sqrt [3]{b}}\right )}{3 a^{4/3}}-\frac {\sin (c+d x)}{a x}-\frac {d \sin (c) \text {Si}(d x)}{a}-\frac {(-1)^{2/3} \sqrt [3]{b} \cos \left (c+\frac {\sqrt [3]{-1} \sqrt [3]{a} d}{\sqrt [3]{b}}\right ) \text {Si}\left (\frac {\sqrt [3]{-1} \sqrt [3]{a} d}{\sqrt [3]{b}}-d x\right )}{3 a^{4/3}}+\frac {\sqrt [3]{b} \cos \left (c-\frac {\sqrt [3]{a} d}{\sqrt [3]{b}}\right ) \text {Si}\left (\frac {\sqrt [3]{a} d}{\sqrt [3]{b}}+d x\right )}{3 a^{4/3}}-\frac {\sqrt [3]{-1} \sqrt [3]{b} \cos \left (c-\frac {(-1)^{2/3} \sqrt [3]{a} d}{\sqrt [3]{b}}\right ) \text {Si}\left (\frac {(-1)^{2/3} \sqrt [3]{a} d}{\sqrt [3]{b}}+d x\right )}{3 a^{4/3}} \] Output:

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

Mathematica [C] (verified)

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

Time = 0.27 (sec) , antiderivative size = 233, normalized size of antiderivative = 0.61 \[ \int \frac {\sin (c+d x)}{x^2 \left (a+b x^3\right )} \, dx=\frac {6 d x \cos (c) \operatorname {CosIntegral}(d x)-i x \text {RootSum}\left [a+b \text {$\#$1}^3\&,\frac {\cos (c+d \text {$\#$1}) \operatorname {CosIntegral}(d (x-\text {$\#$1}))-i \operatorname {CosIntegral}(d (x-\text {$\#$1})) \sin (c+d \text {$\#$1})-i \cos (c+d \text {$\#$1}) \text {Si}(d (x-\text {$\#$1}))-\sin (c+d \text {$\#$1}) \text {Si}(d (x-\text {$\#$1}))}{\text {$\#$1}}\&\right ]+i x \text {RootSum}\left [a+b \text {$\#$1}^3\&,\frac {\cos (c+d \text {$\#$1}) \operatorname {CosIntegral}(d (x-\text {$\#$1}))+i \operatorname {CosIntegral}(d (x-\text {$\#$1})) \sin (c+d \text {$\#$1})+i \cos (c+d \text {$\#$1}) \text {Si}(d (x-\text {$\#$1}))-\sin (c+d \text {$\#$1}) \text {Si}(d (x-\text {$\#$1}))}{\text {$\#$1}}\&\right ]-6 \sin (c+d x)-6 d x \sin (c) \text {Si}(d x)}{6 a x} \] Input:

Integrate[Sin[c + d*x]/(x^2*(a + b*x^3)),x]
 

Output:

(6*d*x*Cos[c]*CosIntegral[d*x] - I*x*RootSum[a + b*#1^3 & , (Cos[c + d*#1] 
*CosIntegral[d*(x - #1)] - I*CosIntegral[d*(x - #1)]*Sin[c + d*#1] - I*Cos 
[c + d*#1]*SinIntegral[d*(x - #1)] - Sin[c + d*#1]*SinIntegral[d*(x - #1)] 
)/#1 & ] + I*x*RootSum[a + b*#1^3 & , (Cos[c + d*#1]*CosIntegral[d*(x - #1 
)] + I*CosIntegral[d*(x - #1)]*Sin[c + d*#1] + I*Cos[c + d*#1]*SinIntegral 
[d*(x - #1)] - Sin[c + d*#1]*SinIntegral[d*(x - #1)])/#1 & ] - 6*Sin[c + d 
*x] - 6*d*x*Sin[c]*SinIntegral[d*x])/(6*a*x)
 

Rubi [A] (verified)

Time = 0.83 (sec) , antiderivative size = 380, 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 = {3826, 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.

\(\displaystyle \int \frac {\sin (c+d x)}{x^2 \left (a+b x^3\right )} \, dx\)

\(\Big \downarrow \) 3826

\(\displaystyle \int \left (\frac {\sin (c+d x)}{a x^2}-\frac {b x \sin (c+d x)}{a \left (a+b x^3\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\sqrt [3]{b} \sin \left (c-\frac {\sqrt [3]{a} d}{\sqrt [3]{b}}\right ) \operatorname {CosIntegral}\left (x d+\frac {\sqrt [3]{a} d}{\sqrt [3]{b}}\right )}{3 a^{4/3}}+\frac {(-1)^{2/3} \sqrt [3]{b} \sin \left (\frac {\sqrt [3]{-1} \sqrt [3]{a} d}{\sqrt [3]{b}}+c\right ) \operatorname {CosIntegral}\left (\frac {\sqrt [3]{-1} \sqrt [3]{a} d}{\sqrt [3]{b}}-d x\right )}{3 a^{4/3}}-\frac {\sqrt [3]{-1} \sqrt [3]{b} \sin \left (c-\frac {(-1)^{2/3} \sqrt [3]{a} d}{\sqrt [3]{b}}\right ) \operatorname {CosIntegral}\left (x d+\frac {(-1)^{2/3} \sqrt [3]{a} d}{\sqrt [3]{b}}\right )}{3 a^{4/3}}-\frac {(-1)^{2/3} \sqrt [3]{b} \cos \left (\frac {\sqrt [3]{-1} \sqrt [3]{a} d}{\sqrt [3]{b}}+c\right ) \text {Si}\left (\frac {\sqrt [3]{-1} \sqrt [3]{a} d}{\sqrt [3]{b}}-d x\right )}{3 a^{4/3}}+\frac {\sqrt [3]{b} \cos \left (c-\frac {\sqrt [3]{a} d}{\sqrt [3]{b}}\right ) \text {Si}\left (x d+\frac {\sqrt [3]{a} d}{\sqrt [3]{b}}\right )}{3 a^{4/3}}-\frac {\sqrt [3]{-1} \sqrt [3]{b} \cos \left (c-\frac {(-1)^{2/3} \sqrt [3]{a} d}{\sqrt [3]{b}}\right ) \text {Si}\left (x d+\frac {(-1)^{2/3} \sqrt [3]{a} d}{\sqrt [3]{b}}\right )}{3 a^{4/3}}+\frac {d \cos (c) \operatorname {CosIntegral}(d x)}{a}-\frac {d \sin (c) \text {Si}(d x)}{a}-\frac {\sin (c+d x)}{a x}\)

Input:

Int[Sin[c + d*x]/(x^2*(a + b*x^3)),x]
 

Output:

(d*Cos[c]*CosIntegral[d*x])/a + (b^(1/3)*CosIntegral[(a^(1/3)*d)/b^(1/3) + 
 d*x]*Sin[c - (a^(1/3)*d)/b^(1/3)])/(3*a^(4/3)) + ((-1)^(2/3)*b^(1/3)*CosI 
ntegral[((-1)^(1/3)*a^(1/3)*d)/b^(1/3) - d*x]*Sin[c + ((-1)^(1/3)*a^(1/3)* 
d)/b^(1/3)])/(3*a^(4/3)) - ((-1)^(1/3)*b^(1/3)*CosIntegral[((-1)^(2/3)*a^( 
1/3)*d)/b^(1/3) + d*x]*Sin[c - ((-1)^(2/3)*a^(1/3)*d)/b^(1/3)])/(3*a^(4/3) 
) - Sin[c + d*x]/(a*x) - (d*Sin[c]*SinIntegral[d*x])/a - ((-1)^(2/3)*b^(1/ 
3)*Cos[c + ((-1)^(1/3)*a^(1/3)*d)/b^(1/3)]*SinIntegral[((-1)^(1/3)*a^(1/3) 
*d)/b^(1/3) - d*x])/(3*a^(4/3)) + (b^(1/3)*Cos[c - (a^(1/3)*d)/b^(1/3)]*Si 
nIntegral[(a^(1/3)*d)/b^(1/3) + d*x])/(3*a^(4/3)) - ((-1)^(1/3)*b^(1/3)*Co 
s[c - ((-1)^(2/3)*a^(1/3)*d)/b^(1/3)]*SinIntegral[((-1)^(2/3)*a^(1/3)*d)/b 
^(1/3) + d*x])/(3*a^(4/3))
 

Defintions of rubi rules used

rule 2009

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

rule 3826

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*Sin[(c_.) + (d_.)*(x_)], x_Sym 
bol] :> Int[ExpandIntegrand[Sin[c + d*x], x^m*(a + b*x^n)^p, x], x] /; Free 
Q[{a, b, c, d, m}, x] && ILtQ[p, 0] && IGtQ[n, 0] && (EqQ[n, 2] || EqQ[p, - 
1]) && IntegerQ[m]
 

Maple [C] (verified)

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

Time = 1.19 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.31

method	result	size
derivativedivides	\(d \left (-\frac {\sin \left (d x +c \right )}{a d x}+\frac {\munderset {\textit {\_R1} =\operatorname {RootOf}\left (b \,\textit {\_Z}^{3}-3 b c \,\textit {\_Z}^{2}+3 \textit {\_Z} b \,c^{2}+a \,d^{3}-b \,c^{3}\right )}{\sum }\frac {-\operatorname {Si}\left (-d x +\textit {\_R1} -c \right ) \cos \left (\textit {\_R1} \right )+\operatorname {Ci}\left (d x -\textit {\_R1} +c \right ) \sin \left (\textit {\_R1} \right )}{-\textit {\_R1} +c}}{3 a}+\frac {-\operatorname {Si}\left (d x \right ) \sin \left (c \right )+\operatorname {Ci}\left (d x \right ) \cos \left (c \right )}{a}\right )\)	\(116\)
default	\(d \left (-\frac {\sin \left (d x +c \right )}{a d x}+\frac {\munderset {\textit {\_R1} =\operatorname {RootOf}\left (b \,\textit {\_Z}^{3}-3 b c \,\textit {\_Z}^{2}+3 \textit {\_Z} b \,c^{2}+a \,d^{3}-b \,c^{3}\right )}{\sum }\frac {-\operatorname {Si}\left (-d x +\textit {\_R1} -c \right ) \cos \left (\textit {\_R1} \right )+\operatorname {Ci}\left (d x -\textit {\_R1} +c \right ) \sin \left (\textit {\_R1} \right )}{-\textit {\_R1} +c}}{3 a}+\frac {-\operatorname {Si}\left (d x \right ) \sin \left (c \right )+\operatorname {Ci}\left (d x \right ) \cos \left (c \right )}{a}\right )\)	\(116\)
risch	\(-\frac {d \,\operatorname {expIntegral}_{1}\left (-i d x \right ) {\mathrm e}^{i c}}{2 a}+\frac {d \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (-3 i \textit {\_Z}^{2} b c -i d^{3} a +i b \,c^{3}+b \,\textit {\_Z}^{3}-3 \textit {\_Z} b \,c^{2}\right )}{\sum }\frac {{\mathrm e}^{\textit {\_R1}} \operatorname {expIntegral}_{1}\left (-i d x -i c +\textit {\_R1} \right )}{-i c +\textit {\_R1}}\right )}{6 a}-\frac {d \,\operatorname {expIntegral}_{1}\left (i d x \right ) {\mathrm e}^{-i c}}{2 a}-\frac {d \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (-3 i \textit {\_Z}^{2} b c -i d^{3} a +i b \,c^{3}+b \,\textit {\_Z}^{3}-3 \textit {\_Z} b \,c^{2}\right )}{\sum }\frac {{\mathrm e}^{-\textit {\_R1}} \operatorname {expIntegral}_{1}\left (i d x +i c -\textit {\_R1} \right )}{-i c +\textit {\_R1}}\right )}{6 a}-\frac {\sin \left (d x +c \right )}{a x}\)	\(194\)

Input:

int(sin(d*x+c)/x^2/(b*x^3+a),x,method=_RETURNVERBOSE)
 

Output:

d*(-sin(d*x+c)/a/d/x+1/3/a*sum(1/(-_R1+c)*(-Si(-d*x+_R1-c)*cos(_R1)+Ci(d*x 
-_R1+c)*sin(_R1)),_R1=RootOf(_Z^3*b-3*_Z^2*b*c+3*_Z*b*c^2+a*d^3-b*c^3))+1/ 
a*(-Si(d*x)*sin(c)+Ci(d*x)*cos(c)))
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.12 (sec) , antiderivative size = 448, normalized size of antiderivative = 1.18 \[ \int \frac {\sin (c+d x)}{x^2 \left (a+b x^3\right )} \, dx =\text {Too large to display} \] Input:

integrate(sin(d*x+c)/x^2/(b*x^3+a),x, algorithm="fricas")
 

Output:

1/12*(12*a*d^3*x*cos(c)*cos_integral(d*x) - 12*a*d^3*x*sin(c)*sin_integral 
(d*x) + 2*I*(-I*a*d^3/b)^(2/3)*b*x*Ei(I*d*x + (-I*a*d^3/b)^(1/3))*e^(I*c - 
 (-I*a*d^3/b)^(1/3)) - 2*I*(I*a*d^3/b)^(2/3)*b*x*Ei(-I*d*x + (I*a*d^3/b)^( 
1/3))*e^(-I*c - (I*a*d^3/b)^(1/3)) - 12*a*d^2*sin(d*x + c) + (I*a*d^3/b)^( 
2/3)*(sqrt(3)*b*x + I*b*x)*Ei(-I*d*x + 1/2*(I*a*d^3/b)^(1/3)*(-I*sqrt(3) - 
 1))*e^(1/2*(I*a*d^3/b)^(1/3)*(I*sqrt(3) + 1) - I*c) - (-I*a*d^3/b)^(2/3)* 
(sqrt(3)*b*x + I*b*x)*Ei(I*d*x + 1/2*(-I*a*d^3/b)^(1/3)*(-I*sqrt(3) - 1))* 
e^(1/2*(-I*a*d^3/b)^(1/3)*(I*sqrt(3) + 1) + I*c) - (I*a*d^3/b)^(2/3)*(sqrt 
(3)*b*x - I*b*x)*Ei(-I*d*x + 1/2*(I*a*d^3/b)^(1/3)*(I*sqrt(3) - 1))*e^(1/2 
*(I*a*d^3/b)^(1/3)*(-I*sqrt(3) + 1) - I*c) + (-I*a*d^3/b)^(2/3)*(sqrt(3)*b 
*x - I*b*x)*Ei(I*d*x + 1/2*(-I*a*d^3/b)^(1/3)*(I*sqrt(3) - 1))*e^(1/2*(-I* 
a*d^3/b)^(1/3)*(-I*sqrt(3) + 1) + I*c))/(a^2*d^2*x)
 

Sympy [F]

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

integrate(sin(d*x+c)/x**2/(b*x**3+a),x)
                                                                                    
                                                                                    
 

Output:

Integral(sin(c + d*x)/(x**2*(a + b*x**3)), x)
 

Maxima [F]

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

integrate(sin(d*x+c)/x^2/(b*x^3+a),x, algorithm="maxima")
 

Output:

integrate(sin(d*x + c)/((b*x^3 + a)*x^2), x)
 

Giac [F]

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

integrate(sin(d*x+c)/x^2/(b*x^3+a),x, algorithm="giac")
 

Output:

integrate(sin(d*x + c)/((b*x^3 + a)*x^2), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\sin (c+d x)}{x^2 \left (a+b x^3\right )} \, dx=\int \frac {\sin \left (d x +c \right )}{b \,x^{5}+a \,x^{2}}d x \] Input:

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

Output:

int(sin(c + d*x)/(a*x**2 + b*x**5),x)