\(\int \frac {\sin (a+\frac {b}{\sqrt [3]{c+d x}})}{(c e+d e x)^{7/3}} \, dx\) [247]

Optimal result

Integrand size = 27, antiderivative size = 172 \[ \int \frac {\sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{(c e+d e x)^{7/3}} \, dx=-\frac {18 \cos \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{b^3 d e^2 \sqrt [3]{e (c+d x)}}+\frac {3 \cos \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{b d e^2 (c+d x)^{2/3} \sqrt [3]{e (c+d x)}}-\frac {9 \sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{b^2 d e^2 \sqrt [3]{c+d x} \sqrt [3]{e (c+d x)}}+\frac {18 \sqrt [3]{c+d x} \sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{b^4 d e^2 \sqrt [3]{e (c+d x)}} \] Output:

-18*cos(a+b/(d*x+c)^(1/3))/b^3/d/e^2/(e*(d*x+c))^(1/3)+3*cos(a+b/(d*x+c)^( 
1/3))/b/d/e^2/(d*x+c)^(2/3)/(e*(d*x+c))^(1/3)-9*sin(a+b/(d*x+c)^(1/3))/b^2 
/d/e^2/(d*x+c)^(1/3)/(e*(d*x+c))^(1/3)+18*(d*x+c)^(1/3)*sin(a+b/(d*x+c)^(1 
/3))/b^4/d/e^2/(e*(d*x+c))^(1/3)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.66 \[ \int \frac {\sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{(c e+d e x)^{7/3}} \, dx=\frac {(c+d x)^{2/3} \left (3 b \sqrt [3]{c+d x} \left (-6 c-6 d x+b^2 \sqrt [3]{c+d x}\right ) \cos \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )+9 (c+d x) \left (-b^2+2 (c+d x)^{2/3}\right ) \sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )\right )}{b^4 d (e (c+d x))^{7/3}} \] Input:

Integrate[Sin[a + b/(c + d*x)^(1/3)]/(c*e + d*e*x)^(7/3),x]
 

Output:

((c + d*x)^(2/3)*(3*b*(c + d*x)^(1/3)*(-6*c - 6*d*x + b^2*(c + d*x)^(1/3)) 
*Cos[a + b/(c + d*x)^(1/3)] + 9*(c + d*x)*(-b^2 + 2*(c + d*x)^(2/3))*Sin[a 
 + b/(c + d*x)^(1/3)]))/(b^4*d*(e*(c + d*x))^(7/3))
 

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.81, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.407, Rules used = {3912, 30, 3042, 3777, 3042, 3777, 25, 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.

\(\displaystyle \int \frac {\sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{(c e+d e x)^{7/3}} \, dx\)

\(\Big \downarrow \) 3912

\(\displaystyle -\frac {3 \int \frac {(c+d x)^{4/3} \sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{(e (c+d x))^{7/3}}d\frac {1}{\sqrt [3]{c+d x}}}{d}\)

\(\Big \downarrow \) 30

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3777

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3777

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3777

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3117

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

Input:

Int[Sin[a + b/(c + d*x)^(1/3)]/(c*e + d*e*x)^(7/3),x]
 

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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]

\[\int \frac {\sin \left (a +\frac {b}{\left (d x +c \right )^{\frac {1}{3}}}\right )}{\left (d e x +c e \right )^{\frac {7}{3}}}d x\]

Input:

int(sin(a+b/(d*x+c)^(1/3))/(d*e*x+c*e)^(7/3),x)
 

Output:

int(sin(a+b/(d*x+c)^(1/3))/(d*e*x+c*e)^(7/3),x)
 

Fricas [A] (verification not implemented)

Time = 0.54 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.93 \[ \int \frac {\sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{(c e+d e x)^{7/3}} \, dx=\frac {3 \, {\left ({\left ({\left (d x + c\right )}^{\frac {1}{3}} b^{3} - 6 \, b d x - 6 \, b c\right )} {\left (d e x + c e\right )}^{\frac {2}{3}} \cos \left (\frac {a d x + a c + {\left (d x + c\right )}^{\frac {2}{3}} b}{d x + c}\right ) - 3 \, {\left (d e x + c e\right )}^{\frac {2}{3}} {\left ({\left (d x + c\right )}^{\frac {2}{3}} b^{2} - 2 \, {\left (d x + c\right )}^{\frac {4}{3}}\right )} \sin \left (\frac {a d x + a c + {\left (d x + c\right )}^{\frac {2}{3}} b}{d x + c}\right )\right )}}{b^{4} d^{3} e^{3} x^{2} + 2 \, b^{4} c d^{2} e^{3} x + b^{4} c^{2} d e^{3}} \] Input:

integrate(sin(a+b/(d*x+c)^(1/3))/(d*e*x+c*e)^(7/3),x, algorithm="fricas")
 

Output:

3*(((d*x + c)^(1/3)*b^3 - 6*b*d*x - 6*b*c)*(d*e*x + c*e)^(2/3)*cos((a*d*x 
+ a*c + (d*x + c)^(2/3)*b)/(d*x + c)) - 3*(d*e*x + c*e)^(2/3)*((d*x + c)^( 
2/3)*b^2 - 2*(d*x + c)^(4/3))*sin((a*d*x + a*c + (d*x + c)^(2/3)*b)/(d*x + 
 c)))/(b^4*d^3*e^3*x^2 + 2*b^4*c*d^2*e^3*x + b^4*c^2*d*e^3)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{(c e+d e x)^{7/3}} \, dx=\text {Timed out} \] Input:

integrate(sin(a+b/(d*x+c)**(1/3))/(d*e*x+c*e)**(7/3),x)
 

Output:

Timed out
 

Maxima [C] (verification not implemented)

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

Time = 1.11 (sec) , antiderivative size = 1389, normalized size of antiderivative = 8.08 \[ \int \frac {\sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{(c e+d e x)^{7/3}} \, dx=\text {Too large to display} \] Input:

integrate(sin(a+b/(d*x+c)^(1/3))/(d*e*x+c*e)^(7/3),x, algorithm="maxima")
 

Output:

-3/16*(2*(cos(a)^2 + sin(a)^2)*b^4*sin(((d*x + c)^(1/3)*a + b)/(d*x + c)^( 
1/3)) - 2*(b^4*cos(((d*x + c)^(1/3)*a + b)/(d*x + c)^(1/3))^2*sin(a) + b^4 
*sin(a)*sin(((d*x + c)^(1/3)*a + b)/(d*x + c)^(1/3))^2)*cos((2*(d*x + c)^( 
1/3)*a + b)/(d*x + c)^(1/3)) + 2*(b^4*cos(a)*cos(((d*x + c)^(1/3)*a + b)/( 
d*x + c)^(1/3))^2 + b^4*cos(a)*sin(((d*x + c)^(1/3)*a + b)/(d*x + c)^(1/3) 
)^2)*sin((2*(d*x + c)^(1/3)*a + b)/(d*x + c)^(1/3)) + ((((-I*gamma(5, I*b* 
conjugate((d*x + c)^(-1/3))) + I*gamma(5, -I*b*conjugate((d*x + c)^(-1/3)) 
) - I*gamma(5, I*b/(d*x + c)^(1/3)) + I*gamma(5, -I*b/(d*x + c)^(1/3)))*co 
s(a)^3 - (gamma(5, I*b*conjugate((d*x + c)^(-1/3))) + gamma(5, -I*b*conjug 
ate((d*x + c)^(-1/3))) + gamma(5, I*b/(d*x + c)^(1/3)) + gamma(5, -I*b/(d* 
x + c)^(1/3)))*cos(a)^2*sin(a) + (-I*gamma(5, I*b*conjugate((d*x + c)^(-1/ 
3))) + I*gamma(5, -I*b*conjugate((d*x + c)^(-1/3))) - I*gamma(5, I*b/(d*x 
+ c)^(1/3)) + I*gamma(5, -I*b/(d*x + c)^(1/3)))*cos(a)*sin(a)^2 - (gamma(5 
, I*b*conjugate((d*x + c)^(-1/3))) + gamma(5, -I*b*conjugate((d*x + c)^(-1 
/3))) + gamma(5, I*b/(d*x + c)^(1/3)) + gamma(5, -I*b/(d*x + c)^(1/3)))*si 
n(a)^3)*d*x + ((-I*gamma(5, I*b*conjugate((d*x + c)^(-1/3))) + I*gamma(5, 
-I*b*conjugate((d*x + c)^(-1/3))) - I*gamma(5, I*b/(d*x + c)^(1/3)) + I*ga 
mma(5, -I*b/(d*x + c)^(1/3)))*cos(a)^3 - (gamma(5, I*b*conjugate((d*x + c) 
^(-1/3))) + gamma(5, -I*b*conjugate((d*x + c)^(-1/3))) + gamma(5, I*b/(d*x 
 + c)^(1/3)) + gamma(5, -I*b/(d*x + c)^(1/3)))*cos(a)^2*sin(a) + (-I*ga...
 

Giac [F]

\[ \int \frac {\sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{(c e+d e x)^{7/3}} \, dx=\int { \frac {\sin \left (a + \frac {b}{{\left (d x + c\right )}^{\frac {1}{3}}}\right )}{{\left (d e x + c e\right )}^{\frac {7}{3}}} \,d x } \] Input:

integrate(sin(a+b/(d*x+c)^(1/3))/(d*e*x+c*e)^(7/3),x, algorithm="giac")
 

Output:

integrate(sin(a + b/(d*x + c)^(1/3))/(d*e*x + c*e)^(7/3), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{(c e+d e x)^{7/3}} \, dx=\int \frac {\sin \left (a+\frac {b}{{\left (c+d\,x\right )}^{1/3}}\right )}{{\left (c\,e+d\,e\,x\right )}^{7/3}} \,d x \] Input:

int(sin(a + b/(c + d*x)^(1/3))/(c*e + d*e*x)^(7/3),x)
 

Output:

int(sin(a + b/(c + d*x)^(1/3))/(c*e + d*e*x)^(7/3), x)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.94 \[ \int \frac {\sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{(c e+d e x)^{7/3}} \, dx=\frac {-18 \left (d x +c \right )^{\frac {2}{3}} \cos \left (\frac {\left (d x +c \right )^{\frac {1}{3}} a +b}{\left (d x +c \right )^{\frac {1}{3}}}\right ) b +3 \cos \left (\frac {\left (d x +c \right )^{\frac {1}{3}} a +b}{\left (d x +c \right )^{\frac {1}{3}}}\right ) b^{3}-9 \left (d x +c \right )^{\frac {1}{3}} \sin \left (\frac {\left (d x +c \right )^{\frac {1}{3}} a +b}{\left (d x +c \right )^{\frac {1}{3}}}\right ) b^{2}+18 \sin \left (\frac {\left (d x +c \right )^{\frac {1}{3}} a +b}{\left (d x +c \right )^{\frac {1}{3}}}\right ) c +18 \sin \left (\frac {\left (d x +c \right )^{\frac {1}{3}} a +b}{\left (d x +c \right )^{\frac {1}{3}}}\right ) d x -18 a c -18 a d x}{e^{\frac {7}{3}} b^{4} d \left (d x +c \right )} \] Input:

int(sin(a+b/(d*x+c)^(1/3))/(d*e*x+c*e)^(7/3),x)
 

Output:

(3*( - 6*(c + d*x)**(2/3)*cos(((c + d*x)**(1/3)*a + b)/(c + d*x)**(1/3))*b 
 + cos(((c + d*x)**(1/3)*a + b)/(c + d*x)**(1/3))*b**3 - 3*(c + d*x)**(1/3 
)*sin(((c + d*x)**(1/3)*a + b)/(c + d*x)**(1/3))*b**2 + 6*sin(((c + d*x)** 
(1/3)*a + b)/(c + d*x)**(1/3))*c + 6*sin(((c + d*x)**(1/3)*a + b)/(c + d*x 
)**(1/3))*d*x - 6*a*c - 6*a*d*x))/(e**(1/3)*b**4*d*e**2*(c + d*x))