\(\int \frac {\sin (a+\frac {b}{\sqrt [3]{c+d x}})}{(c e+d e x)^{5/3}} \, dx\) [246]

Optimal result

Integrand size = 27, antiderivative size = 91 \[ \int \frac {\sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{(c e+d e x)^{5/3}} \, dx=\frac {3 \sqrt [3]{c+d x} \cos \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{b d e (e (c+d x))^{2/3}}-\frac {3 (c+d x)^{2/3} \sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{b^2 d e (e (c+d x))^{2/3}} \] Output:

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.79 \[ \int \frac {\sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{(c e+d e x)^{5/3}} \, dx=\frac {3 (c+d x)^{5/3} \left (\frac {\cos \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{b \sqrt [3]{c+d x}}-\frac {\sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{b^2}\right )}{d (e (c+d x))^{5/3}} \] Input:

Integrate[Sin[a + b/(c + d*x)^(1/3)]/(c*e + d*e*x)^(5/3),x]
 

Output:

(3*(c + d*x)^(5/3)*(Cos[a + b/(c + d*x)^(1/3)]/(b*(c + d*x)^(1/3)) - Sin[a 
 + b/(c + d*x)^(1/3)]/b^2))/(d*(e*(c + d*x))^(5/3))
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.82, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {3912, 30, 3042, 3777, 3042, 3117}

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.

\(\displaystyle \int \frac {\sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{(c e+d e x)^{5/3}} \, dx\)

\(\Big \downarrow \) 3912

\(\displaystyle -\frac {3 \int \frac {(c+d x)^{4/3} \sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{(e (c+d x))^{5/3}}d\frac {1}{\sqrt [3]{c+d x}}}{d}\)

\(\Big \downarrow \) 30

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3777

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3117

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

Input:

Int[Sin[a + b/(c + d*x)^(1/3)]/(c*e + d*e*x)^(5/3),x]
 

Output:

(-3*(c + d*x)^(2/3)*(-(Cos[a + b/(c + d*x)^(1/3)]/(b*(c + d*x)^(1/3))) + S 
in[a + b/(c + d*x)^(1/3)]/b^2))/(d*e*(e*(c + d*x))^(2/3))
 

Defintions of rubi rules used

rule 30

Int[(u_.)*((a_.)*(x_))^(m_.)*((b_.)*(x_)^(i_.))^(p_), x_Symbol] :> Simp[b^I 
ntPart[p]*((b*x^i)^FracPart[p]/(a^(i*IntPart[p])*(a*x)^(i*FracPart[p]))) 
Int[u*(a*x)^(m + i*p), x], x] /; FreeQ[{a, b, i, m, p}, x] && IntegerQ[i] & 
&  !IntegerQ[p]
 

rule 3042

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

rule 3117

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; 
 FreeQ[{c, d}, x]
 

rule 3777

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[( 
-(c + d*x)^m)*(Cos[e + f*x]/f), x] + Simp[d*(m/f)   Int[(c + d*x)^(m - 1)*C 
os[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]
 

rule 3912

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

Maple [F]

\[\int \frac {\sin \left (a +\frac {b}{\left (d x +c \right )^{\frac {1}{3}}}\right )}{\left (d e x +c e \right )^{\frac {5}{3}}}d x\]

Input:

int(sin(a+b/(d*x+c)^(1/3))/(d*e*x+c*e)^(5/3),x)
 

Output:

int(sin(a+b/(d*x+c)^(1/3))/(d*e*x+c*e)^(5/3),x)
 

Fricas [A] (verification not implemented)

Time = 0.46 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.27 \[ \int \frac {\sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{(c e+d e x)^{5/3}} \, dx=\frac {3 \, {\left ({\left (d e x + c e\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {1}{3}} b \cos \left (\frac {a d x + a c + {\left (d x + c\right )}^{\frac {2}{3}} b}{d x + c}\right ) - {\left (d e x + c e\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}} \sin \left (\frac {a d x + a c + {\left (d x + c\right )}^{\frac {2}{3}} b}{d x + c}\right )\right )}}{b^{2} d^{2} e^{2} x + b^{2} c d e^{2}} \] Input:

integrate(sin(a+b/(d*x+c)^(1/3))/(d*e*x+c*e)^(5/3),x, algorithm="fricas")
 

Output:

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

Sympy [F]

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

integrate(sin(a+b/(d*x+c)**(1/3))/(d*e*x+c*e)**(5/3),x)
 

Output:

Integral(sin(a + b/(c + d*x)**(1/3))/(e*(c + d*x))**(5/3), x)
 

Maxima [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.25 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.88 \[ \int \frac {\sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{(c e+d e x)^{5/3}} \, dx=-\frac {3 \, {\left (4 \, b^{2} \sin \left (\frac {{\left (d x + c\right )}^{\frac {1}{3}} a + b}{{\left (d x + c\right )}^{\frac {1}{3}}}\right ) - {\left (d x + c\right )}^{\frac {2}{3}} {\left ({\left (-i \, \Gamma \left (3, i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {1}{3}}}}\right ) + i \, \Gamma \left (3, -i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {1}{3}}}}\right ) - i \, \Gamma \left (3, \frac {i \, b}{{\left (d x + c\right )}^{\frac {1}{3}}}\right ) + i \, \Gamma \left (3, -\frac {i \, b}{{\left (d x + c\right )}^{\frac {1}{3}}}\right )\right )} \cos \left (a\right ) - {\left (\Gamma \left (3, i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {1}{3}}}}\right ) + \Gamma \left (3, -i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {1}{3}}}}\right ) + \Gamma \left (3, \frac {i \, b}{{\left (d x + c\right )}^{\frac {1}{3}}}\right ) + \Gamma \left (3, -\frac {i \, b}{{\left (d x + c\right )}^{\frac {1}{3}}}\right )\right )} \sin \left (a\right )\right )}\right )}}{8 \, {\left (d x + c\right )}^{\frac {2}{3}} b^{2} d e^{\frac {5}{3}}} \] Input:

integrate(sin(a+b/(d*x+c)^(1/3))/(d*e*x+c*e)^(5/3),x, algorithm="maxima")
 

Output:

-3/8*(4*b^2*sin(((d*x + c)^(1/3)*a + b)/(d*x + c)^(1/3)) - (d*x + c)^(2/3) 
*((-I*gamma(3, I*b*conjugate((d*x + c)^(-1/3))) + I*gamma(3, -I*b*conjugat 
e((d*x + c)^(-1/3))) - I*gamma(3, I*b/(d*x + c)^(1/3)) + I*gamma(3, -I*b/( 
d*x + c)^(1/3)))*cos(a) - (gamma(3, I*b*conjugate((d*x + c)^(-1/3))) + gam 
ma(3, -I*b*conjugate((d*x + c)^(-1/3))) + gamma(3, I*b/(d*x + c)^(1/3)) + 
gamma(3, -I*b/(d*x + c)^(1/3)))*sin(a)))/((d*x + c)^(2/3)*b^2*d*e^(5/3))
 

Giac [F]

\[ \int \frac {\sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{(c e+d e x)^{5/3}} \, dx=\int { \frac {\sin \left (a + \frac {b}{{\left (d x + c\right )}^{\frac {1}{3}}}\right )}{{\left (d e x + c e\right )}^{\frac {5}{3}}} \,d x } \] Input:

integrate(sin(a+b/(d*x+c)^(1/3))/(d*e*x+c*e)^(5/3),x, algorithm="giac")
                                                                                    
                                                                                    
 

Output:

integrate(sin(a + b/(d*x + c)^(1/3))/(d*e*x + c*e)^(5/3), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{(c e+d e x)^{5/3}} \, dx=\int \frac {\sin \left (a+\frac {b}{{\left (c+d\,x\right )}^{1/3}}\right )}{{\left (c\,e+d\,e\,x\right )}^{5/3}} \,d x \] Input:

int(sin(a + b/(c + d*x)^(1/3))/(c*e + d*e*x)^(5/3),x)
 

Output:

int(sin(a + b/(c + d*x)^(1/3))/(c*e + d*e*x)^(5/3), x)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.77 \[ \int \frac {\sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{(c e+d e x)^{5/3}} \, dx=\frac {3 \cos \left (\frac {\left (d x +c \right )^{\frac {1}{3}} a +b}{\left (d x +c \right )^{\frac {1}{3}}}\right ) b -3 \left (d x +c \right )^{\frac {1}{3}} \sin \left (\frac {\left (d x +c \right )^{\frac {1}{3}} a +b}{\left (d x +c \right )^{\frac {1}{3}}}\right )}{e^{\frac {5}{3}} \left (d x +c \right )^{\frac {1}{3}} b^{2} d} \] Input:

int(sin(a+b/(d*x+c)^(1/3))/(d*e*x+c*e)^(5/3),x)
 

Output:

(3*(cos(((c + d*x)**(1/3)*a + b)/(c + d*x)**(1/3))*b - (c + d*x)**(1/3)*si 
n(((c + d*x)**(1/3)*a + b)/(c + d*x)**(1/3))))/(e**(2/3)*(c + d*x)**(1/3)* 
b**2*d*e)