\(\int \frac {\sin (a+\frac {b}{(c+d x)^{2/3}})}{(c e+d e x)^{8/3}} \, dx\) [257]

Optimal result

Integrand size = 27, antiderivative size = 237 \[ \int \frac {\sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{(c e+d e x)^{8/3}} \, dx=\frac {3 \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{2 b d e^2 \sqrt [3]{c+d x} (e (c+d x))^{2/3}}+\frac {9 \sqrt {\frac {\pi }{2}} (c+d x)^{2/3} \cos (a) \operatorname {FresnelS}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right )}{4 b^{5/2} d e^2 (e (c+d x))^{2/3}}+\frac {9 \sqrt {\frac {\pi }{2}} (c+d x)^{2/3} \operatorname {FresnelC}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right ) \sin (a)}{4 b^{5/2} d e^2 (e (c+d x))^{2/3}}-\frac {9 \sqrt [3]{c+d x} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{4 b^2 d e^2 (e (c+d x))^{2/3}} \] Output:

3/2*cos(a+b/(d*x+c)^(2/3))/b/d/e^2/(d*x+c)^(1/3)/(e*(d*x+c))^(2/3)+9/8*2^( 
1/2)*Pi^(1/2)*(d*x+c)^(2/3)*cos(a)*FresnelS(b^(1/2)*2^(1/2)/Pi^(1/2)/(d*x+ 
c)^(1/3))/b^(5/2)/d/e^2/(e*(d*x+c))^(2/3)+9/8*2^(1/2)*Pi^(1/2)*(d*x+c)^(2/ 
3)*FresnelC(b^(1/2)*2^(1/2)/Pi^(1/2)/(d*x+c)^(1/3))*sin(a)/b^(5/2)/d/e^2/( 
e*(d*x+c))^(2/3)-9/4*(d*x+c)^(1/3)*sin(a+b/(d*x+c)^(2/3))/b^2/d/e^2/(e*(d* 
x+c))^(2/3)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 1.21 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.70 \[ \int \frac {\sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{(c e+d e x)^{8/3}} \, dx=\frac {(c+d x)^{5/3} \left (9 \sqrt {2 \pi } (c+d x) \cos (a) \operatorname {FresnelS}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right )+9 \sqrt {2 \pi } (c+d x) \operatorname {FresnelC}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right ) \sin (a)+6 \sqrt {b} \left (2 b \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )-3 (c+d x)^{2/3} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )\right )\right )}{8 b^{5/2} d (e (c+d x))^{8/3}} \] Input:

Integrate[Sin[a + b/(c + d*x)^(2/3)]/(c*e + d*e*x)^(8/3),x]
 

Output:

((c + d*x)^(5/3)*(9*Sqrt[2*Pi]*(c + d*x)*Cos[a]*FresnelS[(Sqrt[b]*Sqrt[2/P 
i])/(c + d*x)^(1/3)] + 9*Sqrt[2*Pi]*(c + d*x)*FresnelC[(Sqrt[b]*Sqrt[2/Pi] 
)/(c + d*x)^(1/3)]*Sin[a] + 6*Sqrt[b]*(2*b*Cos[a + b/(c + d*x)^(2/3)] - 3* 
(c + d*x)^(2/3)*Sin[a + b/(c + d*x)^(2/3)])))/(8*b^(5/2)*d*(e*(c + d*x))^( 
8/3))
 

Rubi [A] (warning: unable to verify)

Time = 0.63 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.78, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3916, 3898, 3896, 3890, 3866, 3867, 3834, 3832, 3833}

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.

\(\displaystyle \int \frac {\sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{(c e+d e x)^{8/3}} \, dx\)

\(\Big \downarrow \) 3916

\(\displaystyle \frac {\int \frac {\sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{(e (c+d x))^{8/3}}d(c+d x)}{d}\)

\(\Big \downarrow \) 3898

\(\displaystyle \frac {(c+d x)^{2/3} \int \frac {\sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{(c+d x)^{8/3}}d(c+d x)}{d e^2 (e (c+d x))^{2/3}}\)

\(\Big \downarrow \) 3896

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

\(\Big \downarrow \) 3890

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

\(\Big \downarrow \) 3866

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

\(\Big \downarrow \) 3867

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

\(\Big \downarrow \) 3834

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

\(\Big \downarrow \) 3832

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

\(\Big \downarrow \) 3833

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

Input:

Int[Sin[a + b/(c + d*x)^(2/3)]/(c*e + d*e*x)^(8/3),x]
 

Output:

(-3*(c + d*x)^(2/3)*(-1/2*((c + d*x)*Cos[a + b*(c + d*x)^(2/3)])/b + (3*(- 
1/2*((Sqrt[Pi/2]*Cos[a]*FresnelS[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)^(1/3)])/Sq 
rt[b] + (Sqrt[Pi/2]*FresnelC[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)^(1/3)]*Sin[a]) 
/Sqrt[b])/b + Sin[a + b*(c + d*x)^(2/3)]/(2*b*(c + d*x)^(1/3))))/(2*b)))/( 
d*e^2*(e*(c + d*x))^(2/3))
 

Defintions of rubi rules used

rule 3832

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[ 
d, 2]))*FresnelS[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]
 

rule 3833

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[ 
d, 2]))*FresnelC[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]
 

rule 3834

Int[Sin[(c_) + (d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[Sin[c]   In 
t[Cos[d*(e + f*x)^2], x], x] + Simp[Cos[c]   Int[Sin[d*(e + f*x)^2], x], x] 
 /; FreeQ[{c, d, e, f}, x]
 

rule 3866

Int[((e_.)*(x_))^(m_.)*Sin[(c_.) + (d_.)*(x_)^(n_)], x_Symbol] :> Simp[(-e^ 
(n - 1))*(e*x)^(m - n + 1)*(Cos[c + d*x^n]/(d*n)), x] + Simp[e^n*((m - n + 
1)/(d*n))   Int[(e*x)^(m - n)*Cos[c + d*x^n], x], x] /; FreeQ[{c, d, e}, x] 
 && IGtQ[n, 0] && LtQ[n, m + 1]
 

rule 3867

Int[Cos[(c_.) + (d_.)*(x_)^(n_)]*((e_.)*(x_))^(m_.), x_Symbol] :> Simp[e^(n 
 - 1)*(e*x)^(m - n + 1)*(Sin[c + d*x^n]/(d*n)), x] - Simp[e^n*((m - n + 1)/ 
(d*n))   Int[(e*x)^(m - n)*Sin[c + d*x^n], x], x] /; FreeQ[{c, d, e}, x] && 
 IGtQ[n, 0] && LtQ[n, m + 1]
 

rule 3890

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

rule 3896

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

rule 3898

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}, x] && IntegerQ[ 
p] && FractionQ[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]

\[\int \frac {\sin \left (a +\frac {b}{\left (d x +c \right )^{\frac {2}{3}}}\right )}{\left (d e x +c e \right )^{\frac {8}{3}}}d x\]

Input:

int(sin(a+b/(d*x+c)^(2/3))/(d*e*x+c*e)^(8/3),x)
 

Output:

int(sin(a+b/(d*x+c)^(2/3))/(d*e*x+c*e)^(8/3),x)
 

Fricas [F]

\[ \int \frac {\sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{(c e+d e x)^{8/3}} \, dx=\int { \frac {\sin \left (a + \frac {b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right )}{{\left (d e x + c e\right )}^{\frac {8}{3}}} \,d x } \] Input:

integrate(sin(a+b/(d*x+c)^(2/3))/(d*e*x+c*e)^(8/3),x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {\sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{(c e+d e x)^{8/3}} \, dx=\text {Timed out} \] Input:

integrate(sin(a+b/(d*x+c)**(2/3))/(d*e*x+c*e)**(8/3),x)
 

Output:

Timed out
 

Maxima [C] (verification not implemented)

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

Time = 0.38 (sec) , antiderivative size = 408, normalized size of antiderivative = 1.72 \[ \int \frac {\sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{(c e+d e x)^{8/3}} \, dx =\text {Too large to display} \] Input:

integrate(sin(a+b/(d*x+c)^(2/3))/(d*e*x+c*e)^(8/3),x, algorithm="maxima")
 

Output:

3/8*(d*x + c)^(1/3)*(((I*gamma(5/2, I*b*conjugate((d*x + c)^(-2/3))) - I*g 
amma(5/2, -I*b/(d*x + c)^(2/3)))*cos(5/4*pi + 5/3*arctan2(0, d*x + c)) + ( 
-I*gamma(5/2, -I*b*conjugate((d*x + c)^(-2/3))) + I*gamma(5/2, I*b/(d*x + 
c)^(2/3)))*cos(-5/4*pi + 5/3*arctan2(0, d*x + c)) + (gamma(5/2, I*b*conjug 
ate((d*x + c)^(-2/3))) + gamma(5/2, -I*b/(d*x + c)^(2/3)))*sin(5/4*pi + 5/ 
3*arctan2(0, d*x + c)) - (gamma(5/2, -I*b*conjugate((d*x + c)^(-2/3))) + g 
amma(5/2, I*b/(d*x + c)^(2/3)))*sin(-5/4*pi + 5/3*arctan2(0, d*x + c)))*co 
s(a) + ((gamma(5/2, I*b*conjugate((d*x + c)^(-2/3))) + gamma(5/2, -I*b/(d* 
x + c)^(2/3)))*cos(5/4*pi + 5/3*arctan2(0, d*x + c)) + (gamma(5/2, -I*b*co 
njugate((d*x + c)^(-2/3))) + gamma(5/2, I*b/(d*x + c)^(2/3)))*cos(-5/4*pi 
+ 5/3*arctan2(0, d*x + c)) + (-I*gamma(5/2, I*b*conjugate((d*x + c)^(-2/3) 
)) + I*gamma(5/2, -I*b/(d*x + c)^(2/3)))*sin(5/4*pi + 5/3*arctan2(0, d*x + 
 c)) + (-I*gamma(5/2, -I*b*conjugate((d*x + c)^(-2/3))) + I*gamma(5/2, I*b 
/(d*x + c)^(2/3)))*sin(-5/4*pi + 5/3*arctan2(0, d*x + c)))*sin(a))*e^(1/3) 
/((d^3*e^3*x^2 + 2*c*d^2*e^3*x + c^2*d*e^3)*(b/(d*x + c)^(2/3))^(5/2))
 

Giac [F]

\[ \int \frac {\sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{(c e+d e x)^{8/3}} \, dx=\int { \frac {\sin \left (a + \frac {b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right )}{{\left (d e x + c e\right )}^{\frac {8}{3}}} \,d x } \] Input:

integrate(sin(a+b/(d*x+c)^(2/3))/(d*e*x+c*e)^(8/3),x, algorithm="giac")
 

Output:

integrate(sin(a + b/(d*x + c)^(2/3))/(d*e*x + c*e)^(8/3), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{(c e+d e x)^{8/3}} \, dx=\int \frac {\sin \left (a+\frac {b}{{\left (c+d\,x\right )}^{2/3}}\right )}{{\left (c\,e+d\,e\,x\right )}^{8/3}} \,d x \] Input:

int(sin(a + b/(c + d*x)^(2/3))/(c*e + d*e*x)^(8/3),x)
 

Output:

int(sin(a + b/(c + d*x)^(2/3))/(c*e + d*e*x)^(8/3), x)
 

Reduce [F]

\[ \int \frac {\sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{(c e+d e x)^{8/3}} \, dx=\frac {\int \frac {\sin \left (\frac {\left (d x +c \right )^{\frac {2}{3}} a +b}{\left (d x +c \right )^{\frac {2}{3}}}\right )}{\left (d x +c \right )^{\frac {2}{3}} c^{2}+2 \left (d x +c \right )^{\frac {2}{3}} c d x +\left (d x +c \right )^{\frac {2}{3}} d^{2} x^{2}}d x}{e^{\frac {8}{3}}} \] Input:

int(sin(a+b/(d*x+c)^(2/3))/(d*e*x+c*e)^(8/3),x)
 

Output:

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