\(\int (c e+d e x)^{2/3} \sin (a+\frac {b}{(c+d x)^{2/3}}) \, dx\) [250]

Optimal result

Integrand size = 27, antiderivative size = 262 \[ \int (c e+d e x)^{2/3} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right ) \, dx=\frac {2 b \sqrt [3]{c+d x} (e (c+d x))^{2/3} \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{5 d}+\frac {4 \sqrt {2} b^{5/2} \sqrt {\pi } (e (c+d x))^{2/3} \cos (a) \operatorname {FresnelC}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right )}{5 d (c+d x)^{2/3}}-\frac {4 \sqrt {2} b^{5/2} \sqrt {\pi } (e (c+d x))^{2/3} \operatorname {FresnelS}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right ) \sin (a)}{5 d (c+d x)^{2/3}}-\frac {4 b^2 (e (c+d x))^{2/3} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{5 d \sqrt [3]{c+d x}}+\frac {3 (c+d x) (e (c+d x))^{2/3} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{5 d} \] Output:

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

Mathematica [A] (verified)

Time = 0.57 (sec) , antiderivative size = 228, normalized size of antiderivative = 0.87 \[ \int (c e+d e x)^{2/3} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right ) \, dx=\frac {(e (c+d x))^{2/3} \left (2 b c \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )+2 b d x \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )+4 b^{5/2} \sqrt {2 \pi } \cos (a) \operatorname {FresnelC}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right )-4 b^{5/2} \sqrt {2 \pi } \operatorname {FresnelS}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right ) \sin (a)-4 b^2 \sqrt [3]{c+d x} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )+3 c (c+d x)^{2/3} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )+3 d x (c+d x)^{2/3} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )\right )}{5 d (c+d x)^{2/3}} \] Input:

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

Output:

((e*(c + d*x))^(2/3)*(2*b*c*Cos[a + b/(c + d*x)^(2/3)] + 2*b*d*x*Cos[a + b 
/(c + d*x)^(2/3)] + 4*b^(5/2)*Sqrt[2*Pi]*Cos[a]*FresnelC[(Sqrt[b]*Sqrt[2/P 
i])/(c + d*x)^(1/3)] - 4*b^(5/2)*Sqrt[2*Pi]*FresnelS[(Sqrt[b]*Sqrt[2/Pi])/ 
(c + d*x)^(1/3)]*Sin[a] - 4*b^2*(c + d*x)^(1/3)*Sin[a + b/(c + d*x)^(2/3)] 
 + 3*c*(c + d*x)^(2/3)*Sin[a + b/(c + d*x)^(2/3)] + 3*d*x*(c + d*x)^(2/3)* 
Sin[a + b/(c + d*x)^(2/3)]))/(5*d*(c + d*x)^(2/3))
 

Rubi [A] (warning: unable to verify)

Time = 0.69 (sec) , antiderivative size = 204, normalized size of antiderivative = 0.78, 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, 3868, 3869, 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.

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

\(\Big \downarrow \) 3916

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

\(\Big \downarrow \) 3898

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

\(\Big \downarrow \) 3896

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

\(\Big \downarrow \) 3890

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

\(\Big \downarrow \) 3868

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

\(\Big \downarrow \) 3869

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

\(\Big \downarrow \) 3868

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

\(\Big \downarrow \) 3835

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

\(\Big \downarrow \) 3832

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

\(\Big \downarrow \) 3833

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

Input:

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

Output:

(-3*(e*(c + d*x))^(2/3)*(-1/5*Sin[a + b*(c + d*x)^(2/3)]/(c + d*x)^(5/3) + 
 (2*b*(-1/3*Cos[a + b*(c + d*x)^(2/3)]/(c + d*x) - (2*b*(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)))/3))/5))/(d*(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 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 3869

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

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

Input:

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

Output:

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

Fricas [F]

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

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

Sympy [F]

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

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

Output:

Integral((e*(c + d*x))**(2/3)*sin(a + b/(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 = 0.61 (sec) , antiderivative size = 749, normalized size of antiderivative = 2.86 \[ \int (c e+d e x)^{2/3} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right ) \, dx=\text {Too large to display} \] Input:

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

Output:

-3/8*((((-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)) + (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*conjugate((d*x 
+ c)^(-2/3))) + gamma(-5/2, -I*b/(d*x + c)^(2/3)))*sin(5/4*pi + 5/3*arctan 
2(0, d*x + c)) - (gamma(-5/2, -I*b*conjugate((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)))*cos(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*conju 
gate((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))*d*e^( 
2/3)*x + (((-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)) + (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*conjugate((d 
*x + c)^(-2/3))) + gamma(-5/2, -I*b/(d*x + c)^(2/3)))*sin(5/4*pi + 5/3*arc 
tan2(0, d*x + c)) - (gamma(-5/2, -I*b*conjugate((d*x + c)^(-2/3))) + ga...
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

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

Output:

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