\(\int \frac {\sin (a+\frac {b}{(c+d x)^{2/3}})}{(c e+d e x)^{2/3}} \, dx\) [253]

Optimal result

Integrand size = 27, antiderivative size = 164 \[ \int \frac {\sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{(c e+d e x)^{2/3}} \, dx=-\frac {3 \sqrt {b} \sqrt {2 \pi } (c+d x)^{2/3} \cos (a) \operatorname {FresnelC}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right )}{d (e (c+d x))^{2/3}}+\frac {3 \sqrt {b} \sqrt {2 \pi } (c+d x)^{2/3} \operatorname {FresnelS}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right ) \sin (a)}{d (e (c+d x))^{2/3}}+\frac {3 (c+d x) \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{d (e (c+d x))^{2/3}} \] Output:

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

Mathematica [A] (verified)

Time = 0.43 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.83 \[ \int \frac {\sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{(c e+d e x)^{2/3}} \, dx=\frac {3 \left (-\sqrt {b} \sqrt {2 \pi } (c+d x)^{2/3} \cos (a) \operatorname {FresnelC}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right )+\sqrt {b} \sqrt {2 \pi } (c+d x)^{2/3} \operatorname {FresnelS}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right ) \sin (a)+(c+d x) \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )\right )}{d (e (c+d x))^{2/3}} \] Input:

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.54 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.85, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {3916, 3898, 3896, 3840, 3868, 3835, 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.

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

\(\Big \downarrow \) 3916

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

\(\Big \downarrow \) 3898

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

\(\Big \downarrow \) 3896

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

\(\Big \downarrow \) 3840

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

\(\Big \downarrow \) 3868

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

\(\Big \downarrow \) 3835

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

\(\Big \downarrow \) 3832

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

\(\Big \downarrow \) 3833

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

Input:

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

Output:

(-3*(c + d*x)^(2/3)*(2*b*((Sqrt[Pi/2]*Cos[a]*FresnelC[(Sqrt[b]*Sqrt[2/Pi]) 
/(c + d*x)^(1/3)])/Sqrt[b] - (Sqrt[Pi/2]*FresnelS[(Sqrt[b]*Sqrt[2/Pi])/(c 
+ d*x)^(1/3)]*Sin[a])/Sqrt[b]) - Sin[a + b*(c + d*x)^(2/3)]/(c + d*x)^(1/3 
)))/(d*(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 3835

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

rule 3840

Int[((a_.) + (b_.)*Sin[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)])^(p_.), x_S 
ymbol] :> Simp[-f^(-1)   Subst[Int[(a + b*Sin[c + d/x^n])^p/x^2, x], x, 1/( 
e + f*x)], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[p, 0] && ILtQ[n, 0] & 
& EqQ[n, -2]
 

rule 3868

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

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Sympy [F]

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

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

Output:

Integral(sin(a + b/(c + d*x)**(2/3))/(e*(c + d*x))**(2/3), 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 = 383, normalized size of antiderivative = 2.34 \[ \int \frac {\sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{(c e+d e x)^{2/3}} \, dx =\text {Too large to display} \] Input:

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

Output:

-3/8*(d*x + c)^(1/3)*(((-I*gamma(-1/2, I*b*conjugate((d*x + c)^(-2/3))) + 
I*gamma(-1/2, -I*b/(d*x + c)^(2/3)))*cos(1/4*pi + 1/3*arctan2(0, d*x + c)) 
 + (I*gamma(-1/2, -I*b*conjugate((d*x + c)^(-2/3))) - I*gamma(-1/2, I*b/(d 
*x + c)^(2/3)))*cos(-1/4*pi + 1/3*arctan2(0, d*x + c)) + (gamma(-1/2, I*b* 
conjugate((d*x + c)^(-2/3))) + gamma(-1/2, -I*b/(d*x + c)^(2/3)))*sin(1/4* 
pi + 1/3*arctan2(0, d*x + c)) - (gamma(-1/2, -I*b*conjugate((d*x + c)^(-2/ 
3))) + gamma(-1/2, I*b/(d*x + c)^(2/3)))*sin(-1/4*pi + 1/3*arctan2(0, d*x 
+ c)))*cos(a) - ((gamma(-1/2, I*b*conjugate((d*x + c)^(-2/3))) + gamma(-1/ 
2, -I*b/(d*x + c)^(2/3)))*cos(1/4*pi + 1/3*arctan2(0, d*x + c)) + (gamma(- 
1/2, -I*b*conjugate((d*x + c)^(-2/3))) + gamma(-1/2, I*b/(d*x + c)^(2/3))) 
*cos(-1/4*pi + 1/3*arctan2(0, d*x + c)) - (-I*gamma(-1/2, I*b*conjugate((d 
*x + c)^(-2/3))) + I*gamma(-1/2, -I*b/(d*x + c)^(2/3)))*sin(1/4*pi + 1/3*a 
rctan2(0, d*x + c)) - (-I*gamma(-1/2, -I*b*conjugate((d*x + c)^(-2/3))) + 
I*gamma(-1/2, I*b/(d*x + c)^(2/3)))*sin(-1/4*pi + 1/3*arctan2(0, d*x + c)) 
)*sin(a))*sqrt(b/(d*x + c)^(2/3))/(d*e^(2/3))
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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