\(\int \frac {\sin (a+b \sqrt [3]{c+d x})}{(c e+d e x)^{5/3}} \, dx\) [233]

Optimal result

Integrand size = 27, antiderivative size = 175 \[ \int \frac {\sin \left (a+b \sqrt [3]{c+d x}\right )}{(c e+d e x)^{5/3}} \, dx=-\frac {3 b \sqrt [3]{c+d x} \cos \left (a+b \sqrt [3]{c+d x}\right )}{2 d e (e (c+d x))^{2/3}}-\frac {3 b^2 (c+d x)^{2/3} \operatorname {CosIntegral}\left (b \sqrt [3]{c+d x}\right ) \sin (a)}{2 d e (e (c+d x))^{2/3}}-\frac {3 \sin \left (a+b \sqrt [3]{c+d x}\right )}{2 d e (e (c+d x))^{2/3}}-\frac {3 b^2 (c+d x)^{2/3} \cos (a) \text {Si}\left (b \sqrt [3]{c+d x}\right )}{2 d e (e (c+d x))^{2/3}} \] Output:

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

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.66 \[ \int \frac {\sin \left (a+b \sqrt [3]{c+d x}\right )}{(c e+d e x)^{5/3}} \, dx=-\frac {3 \left (b \sqrt [3]{c+d x} \cos \left (a+b \sqrt [3]{c+d x}\right )+b^2 (c+d x)^{2/3} \operatorname {CosIntegral}\left (b \sqrt [3]{c+d x}\right ) \sin (a)+\sin \left (a+b \sqrt [3]{c+d x}\right )+b^2 (c+d x)^{2/3} \cos (a) \text {Si}\left (b \sqrt [3]{c+d x}\right )\right )}{2 d e (e (c+d x))^{2/3}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.69, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {3912, 30, 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.

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

\(\Big \downarrow \) 3912

\(\displaystyle \frac {3 \int \frac {(c+d x)^{2/3} \sin \left (a+b \sqrt [3]{c+d x}\right )}{(e (c+d x))^{5/3}}d\sqrt [3]{c+d x}}{d}\)

\(\Big \downarrow \) 30

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3778

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3778

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3784

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3780

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

\(\Big \downarrow \) 3783

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

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

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 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 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]

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

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 [F]

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

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

Output:

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

Sympy [F]

\[ \int \frac {\sin \left (a+b \sqrt [3]{c+d x}\right )}{(c e+d e x)^{5/3}} \, dx=\int \frac {\sin {\left (a + b \sqrt [3]{c + d x} \right )}}{\left (e \left (c + d x\right )\right )^{\frac {5}{3}}}\, dx \] 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 4.

Time = 0.35 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.74 \[ \int \frac {\sin \left (a+b \sqrt [3]{c+d x}\right )}{(c e+d e x)^{5/3}} \, dx=-\frac {3 \, {\left ({\left (-i \, \Gamma \left (-2, i \, b \overline {{\left (d x + c\right )}^{\frac {1}{3}}}\right ) + i \, \Gamma \left (-2, -i \, b \overline {{\left (d x + c\right )}^{\frac {1}{3}}}\right ) - i \, \Gamma \left (-2, i \, {\left (d x + c\right )}^{\frac {1}{3}} b\right ) + i \, \Gamma \left (-2, -i \, {\left (d x + c\right )}^{\frac {1}{3}} b\right )\right )} \cos \left (a\right ) - {\left (\Gamma \left (-2, i \, b \overline {{\left (d x + c\right )}^{\frac {1}{3}}}\right ) + \Gamma \left (-2, -i \, b \overline {{\left (d x + c\right )}^{\frac {1}{3}}}\right ) + \Gamma \left (-2, i \, {\left (d x + c\right )}^{\frac {1}{3}} b\right ) + \Gamma \left (-2, -i \, {\left (d x + c\right )}^{\frac {1}{3}} b\right )\right )} \sin \left (a\right )\right )} b^{2}}{4 \, d e^{\frac {5}{3}}} \] Input:

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

\[ \int \frac {\sin \left (a+b \sqrt [3]{c+d x}\right )}{(c e+d e x)^{5/3}} \, dx=\frac {\int \frac {\sin \left (\left (d x +c \right )^{\frac {1}{3}} b +a \right )}{\left (d x +c \right )^{\frac {2}{3}} c +\left (d x +c \right )^{\frac {2}{3}} d x}d x}{e^{\frac {5}{3}}} \] Input:

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

Output:

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