\(\int \frac {\sin (a+\frac {b}{(c+d x)^{2/3}})}{(c e+d e x)^{10/3}} \, dx\) [258]

Optimal result

Integrand size = 27, antiderivative size = 277 \[ \int \frac {\sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{(c e+d e x)^{10/3}} \, dx=-\frac {45 \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{8 b^3 d e^3 \sqrt [3]{e (c+d x)}}+\frac {3 \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{2 b d e^3 (c+d x)^{4/3} \sqrt [3]{e (c+d x)}}+\frac {45 \sqrt {\pi } \sqrt [3]{c+d x} \cos (a) \operatorname {FresnelC}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right )}{8 \sqrt {2} b^{7/2} d e^3 \sqrt [3]{e (c+d x)}}-\frac {45 \sqrt {\pi } \sqrt [3]{c+d x} \operatorname {FresnelS}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right ) \sin (a)}{8 \sqrt {2} b^{7/2} d e^3 \sqrt [3]{e (c+d x)}}-\frac {15 \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{4 b^2 d e^3 (c+d x)^{2/3} \sqrt [3]{e (c+d x)}} \] Output:

-45/8*cos(a+b/(d*x+c)^(2/3))/b^3/d/e^3/(e*(d*x+c))^(1/3)+3/2*cos(a+b/(d*x+ 
c)^(2/3))/b/d/e^3/(d*x+c)^(4/3)/(e*(d*x+c))^(1/3)+45/16*Pi^(1/2)*(d*x+c)^( 
1/3)*cos(a)*FresnelC(b^(1/2)*2^(1/2)/Pi^(1/2)/(d*x+c)^(1/3))*2^(1/2)/b^(7/ 
2)/d/e^3/(e*(d*x+c))^(1/3)-45/16*Pi^(1/2)*(d*x+c)^(1/3)*FresnelS(b^(1/2)*2 
^(1/2)/Pi^(1/2)/(d*x+c)^(1/3))*sin(a)*2^(1/2)/b^(7/2)/d/e^3/(e*(d*x+c))^(1 
/3)-15/4*sin(a+b/(d*x+c)^(2/3))/b^2/d/e^3/(d*x+c)^(2/3)/(e*(d*x+c))^(1/3)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 1.65 (sec) , antiderivative size = 192, normalized size of antiderivative = 0.69 \[ \int \frac {\sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{(c e+d e x)^{10/3}} \, dx=\frac {(e (c+d x))^{2/3} \left (45 \sqrt {2 \pi } (c+d x)^{5/3} \cos (a) \operatorname {FresnelC}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right )-45 \sqrt {2 \pi } (c+d x)^{5/3} \operatorname {FresnelS}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right ) \sin (a)-6 \sqrt {b} \left (\left (-4 b^2+15 (c+d x)^{4/3}\right ) \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )+10 b (c+d x)^{2/3} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )\right )\right )}{16 b^{7/2} d e^4 (c+d x)^{7/3}} \] Input:

Integrate[Sin[a + b/(c + d*x)^(2/3)]/(c*e + d*e*x)^(10/3),x]
 

Output:

((e*(c + d*x))^(2/3)*(45*Sqrt[2*Pi]*(c + d*x)^(5/3)*Cos[a]*FresnelC[(Sqrt[ 
b]*Sqrt[2/Pi])/(c + d*x)^(1/3)] - 45*Sqrt[2*Pi]*(c + d*x)^(5/3)*FresnelS[( 
Sqrt[b]*Sqrt[2/Pi])/(c + d*x)^(1/3)]*Sin[a] - 6*Sqrt[b]*((-4*b^2 + 15*(c + 
 d*x)^(4/3))*Cos[a + b/(c + d*x)^(2/3)] + 10*b*(c + d*x)^(2/3)*Sin[a + b/( 
c + d*x)^(2/3)])))/(16*b^(7/2)*d*e^4*(c + d*x)^(7/3))
 

Rubi [A] (warning: unable to verify)

Time = 0.73 (sec) , antiderivative size = 224, normalized size of antiderivative = 0.81, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.370, Rules used = {3916, 3898, 3896, 3890, 3866, 3867, 3866, 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.

\(\displaystyle \int \frac {\sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{(c e+d e x)^{10/3}} \, dx\)

\(\Big \downarrow \) 3916

\(\displaystyle \frac {\int \frac {\sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{(e (c+d x))^{10/3}}d(c+d x)}{d}\)

\(\Big \downarrow \) 3898

\(\displaystyle \frac {\sqrt [3]{c+d x} \int \frac {\sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{(c+d x)^{10/3}}d(c+d x)}{d e^3 \sqrt [3]{e (c+d x)}}\)

\(\Big \downarrow \) 3896

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

\(\Big \downarrow \) 3890

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

\(\Big \downarrow \) 3866

\(\displaystyle -\frac {3 \sqrt [3]{c+d x} \left (\frac {5 \int (c+d x)^{4/3} \cos \left (a+b (c+d x)^{2/3}\right )d\frac {1}{\sqrt [3]{c+d x}}}{2 b}-\frac {(c+d x)^{5/3} \cos \left (a+b (c+d x)^{2/3}\right )}{2 b}\right )}{d e^3 \sqrt [3]{e (c+d x)}}\)

\(\Big \downarrow \) 3867

\(\displaystyle -\frac {3 \sqrt [3]{c+d x} \left (\frac {5 \left (\frac {(c+d x) \sin \left (a+b (c+d x)^{2/3}\right )}{2 b}-\frac {3 \int (c+d x)^{2/3} \sin \left (a+b (c+d x)^{2/3}\right )d\frac {1}{\sqrt [3]{c+d x}}}{2 b}\right )}{2 b}-\frac {(c+d x)^{5/3} \cos \left (a+b (c+d x)^{2/3}\right )}{2 b}\right )}{d e^3 \sqrt [3]{e (c+d x)}}\)

\(\Big \downarrow \) 3866

\(\displaystyle -\frac {3 \sqrt [3]{c+d x} \left (\frac {5 \left (\frac {(c+d x) \sin \left (a+b (c+d x)^{2/3}\right )}{2 b}-\frac {3 \left (\frac {\int \cos \left (a+b (c+d x)^{2/3}\right )d\frac {1}{\sqrt [3]{c+d x}}}{2 b}-\frac {\cos \left (a+b (c+d x)^{2/3}\right )}{2 b \sqrt [3]{c+d x}}\right )}{2 b}\right )}{2 b}-\frac {(c+d x)^{5/3} \cos \left (a+b (c+d x)^{2/3}\right )}{2 b}\right )}{d e^3 \sqrt [3]{e (c+d x)}}\)

\(\Big \downarrow \) 3835

\(\displaystyle -\frac {3 \sqrt [3]{c+d x} \left (\frac {5 \left (\frac {(c+d x) \sin \left (a+b (c+d x)^{2/3}\right )}{2 b}-\frac {3 \left (\frac {\cos (a) \int \cos \left (b (c+d x)^{2/3}\right )d\frac {1}{\sqrt [3]{c+d x}}-\sin (a) \int \sin \left (b (c+d x)^{2/3}\right )d\frac {1}{\sqrt [3]{c+d x}}}{2 b}-\frac {\cos \left (a+b (c+d x)^{2/3}\right )}{2 b \sqrt [3]{c+d x}}\right )}{2 b}\right )}{2 b}-\frac {(c+d x)^{5/3} \cos \left (a+b (c+d x)^{2/3}\right )}{2 b}\right )}{d e^3 \sqrt [3]{e (c+d x)}}\)

\(\Big \downarrow \) 3832

\(\displaystyle -\frac {3 \sqrt [3]{c+d x} \left (\frac {5 \left (\frac {(c+d x) \sin \left (a+b (c+d x)^{2/3}\right )}{2 b}-\frac {3 \left (\frac {\cos (a) \int \cos \left (b (c+d x)^{2/3}\right )d\frac {1}{\sqrt [3]{c+d x}}-\frac {\sqrt {\frac {\pi }{2}} \sin (a) \operatorname {FresnelS}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right )}{\sqrt {b}}}{2 b}-\frac {\cos \left (a+b (c+d x)^{2/3}\right )}{2 b \sqrt [3]{c+d x}}\right )}{2 b}\right )}{2 b}-\frac {(c+d x)^{5/3} \cos \left (a+b (c+d x)^{2/3}\right )}{2 b}\right )}{d e^3 \sqrt [3]{e (c+d x)}}\)

\(\Big \downarrow \) 3833

\(\displaystyle -\frac {3 \sqrt [3]{c+d x} \left (\frac {5 \left (\frac {(c+d x) \sin \left (a+b (c+d x)^{2/3}\right )}{2 b}-\frac {3 \left (\frac {\frac {\sqrt {\frac {\pi }{2}} \cos (a) \operatorname {FresnelC}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right )}{\sqrt {b}}-\frac {\sqrt {\frac {\pi }{2}} \sin (a) \operatorname {FresnelS}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right )}{\sqrt {b}}}{2 b}-\frac {\cos \left (a+b (c+d x)^{2/3}\right )}{2 b \sqrt [3]{c+d x}}\right )}{2 b}\right )}{2 b}-\frac {(c+d x)^{5/3} \cos \left (a+b (c+d x)^{2/3}\right )}{2 b}\right )}{d e^3 \sqrt [3]{e (c+d x)}}\)

Input:

Int[Sin[a + b/(c + d*x)^(2/3)]/(c*e + d*e*x)^(10/3),x]
 

Output:

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

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 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]

\[\int \frac {\sin \left (a +\frac {b}{\left (d x +c \right )^{\frac {2}{3}}}\right )}{\left (d e x +c e \right )^{\frac {10}{3}}}d x\]

Input:

int(sin(a+b/(d*x+c)^(2/3))/(d*e*x+c*e)^(10/3),x)
 

Output:

int(sin(a+b/(d*x+c)^(2/3))/(d*e*x+c*e)^(10/3),x)
 

Fricas [F]

\[ \int \frac {\sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{(c e+d e x)^{10/3}} \, dx=\int { \frac {\sin \left (a + \frac {b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right )}{{\left (d e x + c e\right )}^{\frac {10}{3}}} \,d x } \] Input:

integrate(sin(a+b/(d*x+c)^(2/3))/(d*e*x+c*e)^(10/3),x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {\sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{(c e+d e x)^{10/3}} \, dx=\text {Timed out} \] Input:

integrate(sin(a+b/(d*x+c)**(2/3))/(d*e*x+c*e)**(10/3),x)
 

Output:

Timed out
 

Maxima [C] (verification not implemented)

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

Time = 0.37 (sec) , antiderivative size = 405, normalized size of antiderivative = 1.46 \[ \int \frac {\sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{(c e+d e x)^{10/3}} \, dx =\text {Too large to display} \] Input:

integrate(sin(a+b/(d*x+c)^(2/3))/(d*e*x+c*e)^(10/3),x, algorithm="maxima")
 

Output:

3/8*(((I*gamma(7/2, I*b*conjugate((d*x + c)^(-2/3))) - I*gamma(7/2, -I*b/( 
d*x + c)^(2/3)))*cos(7/4*pi + 7/3*arctan2(0, d*x + c)) + (-I*gamma(7/2, -I 
*b*conjugate((d*x + c)^(-2/3))) + I*gamma(7/2, I*b/(d*x + c)^(2/3)))*cos(- 
7/4*pi + 7/3*arctan2(0, d*x + c)) + (gamma(7/2, I*b*conjugate((d*x + c)^(- 
2/3))) + gamma(7/2, -I*b/(d*x + c)^(2/3)))*sin(7/4*pi + 7/3*arctan2(0, d*x 
 + c)) - (gamma(7/2, -I*b*conjugate((d*x + c)^(-2/3))) + gamma(7/2, I*b/(d 
*x + c)^(2/3)))*sin(-7/4*pi + 7/3*arctan2(0, d*x + c)))*cos(a) + ((gamma(7 
/2, I*b*conjugate((d*x + c)^(-2/3))) + gamma(7/2, -I*b/(d*x + c)^(2/3)))*c 
os(7/4*pi + 7/3*arctan2(0, d*x + c)) + (gamma(7/2, -I*b*conjugate((d*x + c 
)^(-2/3))) + gamma(7/2, I*b/(d*x + c)^(2/3)))*cos(-7/4*pi + 7/3*arctan2(0, 
 d*x + c)) + (-I*gamma(7/2, I*b*conjugate((d*x + c)^(-2/3))) + I*gamma(7/2 
, -I*b/(d*x + c)^(2/3)))*sin(7/4*pi + 7/3*arctan2(0, d*x + c)) + (-I*gamma 
(7/2, -I*b*conjugate((d*x + c)^(-2/3))) + I*gamma(7/2, I*b/(d*x + c)^(2/3) 
))*sin(-7/4*pi + 7/3*arctan2(0, d*x + c)))*sin(a))/((d^3*e^(10/3)*x^2 + 2* 
c*d^2*e^(10/3)*x + c^2*d*e^(10/3))*(d*x + c)^(1/3)*(b/(d*x + c)^(2/3))^(7/ 
2))
 

Giac [F]

\[ \int \frac {\sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{(c e+d e x)^{10/3}} \, dx=\int { \frac {\sin \left (a + \frac {b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right )}{{\left (d e x + c e\right )}^{\frac {10}{3}}} \,d x } \] Input:

integrate(sin(a+b/(d*x+c)^(2/3))/(d*e*x+c*e)^(10/3),x, algorithm="giac")
 

Output:

integrate(sin(a + b/(d*x + c)^(2/3))/(d*e*x + c*e)^(10/3), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{(c e+d e x)^{10/3}} \, dx=\int \frac {\sin \left (a+\frac {b}{{\left (c+d\,x\right )}^{2/3}}\right )}{{\left (c\,e+d\,e\,x\right )}^{10/3}} \,d x \] Input:

int(sin(a + b/(c + d*x)^(2/3))/(c*e + d*e*x)^(10/3),x)
 

Output:

int(sin(a + b/(c + d*x)^(2/3))/(c*e + d*e*x)^(10/3), x)
 

Reduce [F]

\[ \int \frac {\sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{(c e+d e x)^{10/3}} \, dx=\frac {\int \frac {\sin \left (\frac {\left (d x +c \right )^{\frac {2}{3}} a +b}{\left (d x +c \right )^{\frac {2}{3}}}\right )}{\left (d x +c \right )^{\frac {1}{3}} c^{3}+3 \left (d x +c \right )^{\frac {1}{3}} c^{2} d x +3 \left (d x +c \right )^{\frac {1}{3}} c \,d^{2} x^{2}+\left (d x +c \right )^{\frac {1}{3}} d^{3} x^{3}}d x}{e^{\frac {10}{3}}} \] Input:

int(sin(a+b/(d*x+c)^(2/3))/(d*e*x+c*e)^(10/3),x)
 

Output:

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