\(\int \frac {\sin (a+\frac {b}{(c+d x)^{2/3}})}{\sqrt [3]{c e+d e x}} \, dx\) [252]

Optimal result

Integrand size = 27, antiderivative size = 122 \[ \int \frac {\sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{\sqrt [3]{c e+d e x}} \, dx=-\frac {3 b \sqrt [3]{c+d x} \cos (a) \operatorname {CosIntegral}\left (\frac {b}{(c+d x)^{2/3}}\right )}{2 d \sqrt [3]{e (c+d x)}}+\frac {3 (c+d x) \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{2 d \sqrt [3]{e (c+d x)}}+\frac {3 b \sqrt [3]{c+d x} \sin (a) \text {Si}\left (\frac {b}{(c+d x)^{2/3}}\right )}{2 d \sqrt [3]{e (c+d x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.74 \[ \int \frac {\sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{\sqrt [3]{c e+d e x}} \, dx=\frac {3 \left (-b \sqrt [3]{c+d x} \cos (a) \operatorname {CosIntegral}\left (\frac {b}{(c+d x)^{2/3}}\right )+(c+d x) \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )+b \sqrt [3]{c+d x} \sin (a) \text {Si}\left (\frac {b}{(c+d x)^{2/3}}\right )\right )}{2 d \sqrt [3]{e (c+d x)}} \] Input:

Integrate[Sin[a + b/(c + d*x)^(2/3)]/(c*e + d*e*x)^(1/3),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.71, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.370, Rules used = {3916, 3862, 3860, 3042, 3778, 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.

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

\(\Big \downarrow \) 3916

\(\displaystyle \frac {\int \frac {\sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{\sqrt [3]{e (c+d x)}}d(c+d x)}{d}\)

\(\Big \downarrow \) 3862

\(\displaystyle \frac {\sqrt [3]{c+d x} \int \frac {\sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{\sqrt [3]{c+d x}}d(c+d x)}{d \sqrt [3]{e (c+d x)}}\)

\(\Big \downarrow \) 3860

\(\displaystyle -\frac {3 \sqrt [3]{c+d x} \int (c+d x)^{4/3} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )d\frac {1}{(c+d x)^{2/3}}}{2 d \sqrt [3]{e (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {3 \sqrt [3]{c+d x} \int (c+d x)^{4/3} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )d\frac {1}{(c+d x)^{2/3}}}{2 d \sqrt [3]{e (c+d x)}}\)

\(\Big \downarrow \) 3778

\(\displaystyle -\frac {3 \sqrt [3]{c+d x} \left (b \int (c+d x)^{2/3} \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )d\frac {1}{(c+d x)^{2/3}}-(c+d x)^{2/3} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )\right )}{2 d \sqrt [3]{e (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {3 \sqrt [3]{c+d x} \left (b \int (c+d x)^{2/3} \sin \left (a+\frac {b}{(c+d x)^{2/3}}+\frac {\pi }{2}\right )d\frac {1}{(c+d x)^{2/3}}-(c+d x)^{2/3} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )\right )}{2 d \sqrt [3]{e (c+d x)}}\)

\(\Big \downarrow \) 3784

\(\displaystyle -\frac {3 \sqrt [3]{c+d x} \left (b \left (\cos (a) \int (c+d x)^{2/3} \cos \left (\frac {b}{(c+d x)^{2/3}}\right )d\frac {1}{(c+d x)^{2/3}}-\sin (a) \int (c+d x)^{2/3} \sin \left (\frac {b}{(c+d x)^{2/3}}\right )d\frac {1}{(c+d x)^{2/3}}\right )-(c+d x)^{2/3} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )\right )}{2 d \sqrt [3]{e (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {3 \sqrt [3]{c+d x} \left (b \left (\cos (a) \int (c+d x)^{2/3} \sin \left (\frac {b}{(c+d x)^{2/3}}+\frac {\pi }{2}\right )d\frac {1}{(c+d x)^{2/3}}-\sin (a) \int (c+d x)^{2/3} \sin \left (\frac {b}{(c+d x)^{2/3}}\right )d\frac {1}{(c+d x)^{2/3}}\right )-(c+d x)^{2/3} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )\right )}{2 d \sqrt [3]{e (c+d x)}}\)

\(\Big \downarrow \) 3780

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

\(\Big \downarrow \) 3783

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

Input:

Int[Sin[a + b/(c + d*x)^(2/3)]/(c*e + d*e*x)^(1/3),x]
 

Output:

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

Defintions of rubi rules used

rule 3042

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

rule 3778

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]
 

rule 3780

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]
 

rule 3783

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]
 

rule 3784

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]
 

rule 3860

Int[(x_)^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol 
] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*Sin[c + d*x])^ 
p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IntegerQ[Simplify[ 
(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[ 
(m + 1)/n], 0]))
 

rule 3862

Int[((e_)*(x_))^(m_)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_ 
Symbol] :> Simp[e^IntPart[m]*((e*x)^FracPart[m]/x^FracPart[m])   Int[x^m*(a 
 + b*Sin[c + d*x^n])^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && Int 
egerQ[Simplify[(m + 1)/n]]
 

rule 3916

Int[((g_.) + (h_.)*(x_))^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*((e_.) + (f 
_.)*(x_))^(n_)])^(p_.), x_Symbol] :> Simp[1/f   Subst[Int[(h*(x/f))^m*(a + 
b*Sin[c + d*x^n])^p, x], x, e + f*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, 
m}, x] && IGtQ[p, 0] && EqQ[f*g - e*h, 0]
 

Maple [F]

\[\int \frac {\sin \left (a +\frac {b}{\left (d x +c \right )^{\frac {2}{3}}}\right )}{\left (d e x +c e \right )^{\frac {1}{3}}}d x\]

Input:

int(sin(a+b/(d*x+c)^(2/3))/(d*e*x+c*e)^(1/3),x)
 

Output:

int(sin(a+b/(d*x+c)^(2/3))/(d*e*x+c*e)^(1/3),x)
 

Fricas [F]

\[ \int \frac {\sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{\sqrt [3]{c e+d e x}} \, dx=\int { \frac {\sin \left (a + \frac {b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right )}{{\left (d e x + c e\right )}^{\frac {1}{3}}} \,d x } \] Input:

integrate(sin(a+b/(d*x+c)^(2/3))/(d*e*x+c*e)^(1/3),x, algorithm="fricas")
 

Output:

integral(sin((a*d*x + a*c + (d*x + c)^(1/3)*b)/(d*x + c))/(d*e*x + c*e)^(1 
/3), x)
 

Sympy [F]

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

integrate(sin(a+b/(d*x+c)**(2/3))/(d*e*x+c*e)**(1/3),x)
 

Output:

Integral(sin(a + b/(c + d*x)**(2/3))/(e*(c + d*x))**(1/3), x)
 

Maxima [C] (verification not implemented)

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

Time = 0.23 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.03 \[ \int \frac {\sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{\sqrt [3]{c e+d e x}} \, dx=-\frac {3 \, {\left ({\left (\Gamma \left (-1, i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {2}{3}}}}\right ) + \Gamma \left (-1, -i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {2}{3}}}}\right ) + \Gamma \left (-1, \frac {i \, b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right ) + \Gamma \left (-1, -\frac {i \, b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right )\right )} \cos \left (a\right ) + {\left (-i \, \Gamma \left (-1, i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {2}{3}}}}\right ) + i \, \Gamma \left (-1, -i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {2}{3}}}}\right ) - i \, \Gamma \left (-1, \frac {i \, b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right ) + i \, \Gamma \left (-1, -\frac {i \, b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right )\right )} \sin \left (a\right )\right )} b}{8 \, d e^{\frac {1}{3}}} \] Input:

integrate(sin(a+b/(d*x+c)^(2/3))/(d*e*x+c*e)^(1/3),x, algorithm="maxima")
 

Output:

-3/8*((gamma(-1, I*b*conjugate((d*x + c)^(-2/3))) + gamma(-1, -I*b*conjuga 
te((d*x + c)^(-2/3))) + gamma(-1, I*b/(d*x + c)^(2/3)) + gamma(-1, -I*b/(d 
*x + c)^(2/3)))*cos(a) + (-I*gamma(-1, I*b*conjugate((d*x + c)^(-2/3))) + 
I*gamma(-1, -I*b*conjugate((d*x + c)^(-2/3))) - I*gamma(-1, I*b/(d*x + c)^ 
(2/3)) + I*gamma(-1, -I*b/(d*x + c)^(2/3)))*sin(a))*b/(d*e^(1/3))
 

Giac [F]

\[ \int \frac {\sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{\sqrt [3]{c e+d e x}} \, dx=\int { \frac {\sin \left (a + \frac {b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right )}{{\left (d e x + c e\right )}^{\frac {1}{3}}} \,d x } \] Input:

integrate(sin(a+b/(d*x+c)^(2/3))/(d*e*x+c*e)^(1/3),x, algorithm="giac")
 

Output:

integrate(sin(a + b/(d*x + c)^(2/3))/(d*e*x + c*e)^(1/3), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{\sqrt [3]{c e+d e x}} \, dx=\int \frac {\sin \left (a+\frac {b}{{\left (c+d\,x\right )}^{2/3}}\right )}{{\left (c\,e+d\,e\,x\right )}^{1/3}} \,d x \] Input:

int(sin(a + b/(c + d*x)^(2/3))/(c*e + d*e*x)^(1/3),x)
 

Output:

int(sin(a + b/(c + d*x)^(2/3))/(c*e + d*e*x)^(1/3), x)
 

Reduce [F]

\[ \int \frac {\sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{\sqrt [3]{c e+d e x}} \, dx=\frac {\int \frac {\sin \left (\frac {\left (d x +c \right )^{\frac {2}{3}} a +b}{\left (d x +c \right )^{\frac {2}{3}}}\right )}{\left (d x +c \right )^{\frac {1}{3}}}d x}{e^{\frac {1}{3}}} \] Input:

int(sin(a+b/(d*x+c)^(2/3))/(d*e*x+c*e)^(1/3),x)
 

Output:

int(sin(((c + d*x)**(2/3)*a + b)/(c + d*x)**(2/3))/(c + d*x)**(1/3),x)/e** 
(1/3)