\(\int (c e+d e x)^{2/3} \sin (a+b (c+d x)^{2/3}) \, dx\) [236]

Optimal result

Integrand size = 27, antiderivative size = 227 \[ \int (c e+d e x)^{2/3} \sin \left (a+b (c+d x)^{2/3}\right ) \, dx=-\frac {3 \sqrt [3]{c+d x} (e (c+d x))^{2/3} \cos \left (a+b (c+d x)^{2/3}\right )}{2 b d}-\frac {9 \sqrt {\pi } (e (c+d x))^{2/3} \cos (a) \operatorname {FresnelS}\left (\sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt [3]{c+d x}\right )}{4 \sqrt {2} b^{5/2} d (c+d x)^{2/3}}-\frac {9 \sqrt {\pi } (e (c+d x))^{2/3} \operatorname {FresnelC}\left (\sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt [3]{c+d x}\right ) \sin (a)}{4 \sqrt {2} b^{5/2} d (c+d x)^{2/3}}+\frac {9 (e (c+d x))^{2/3} \sin \left (a+b (c+d x)^{2/3}\right )}{4 b^2 d \sqrt [3]{c+d x}} \] Output:

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

Mathematica [A] (verified)

Time = 0.61 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.70 \[ \int (c e+d e x)^{2/3} \sin \left (a+b (c+d x)^{2/3}\right ) \, dx=-\frac {3 (e (c+d x))^{2/3} \left (3 \sqrt {2 \pi } \cos (a) \operatorname {FresnelS}\left (\sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt [3]{c+d x}\right )+3 \sqrt {2 \pi } \operatorname {FresnelC}\left (\sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt [3]{c+d x}\right ) \sin (a)+2 \sqrt {b} \left (2 b (c+d x) \cos \left (a+b (c+d x)^{2/3}\right )-3 \sqrt [3]{c+d x} \sin \left (a+b (c+d x)^{2/3}\right )\right )\right )}{8 b^{5/2} d (c+d x)^{2/3}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.80, 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, 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.

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

\(\Big \downarrow \) 3916

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

\(\Big \downarrow \) 3898

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

\(\Big \downarrow \) 3896

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

\(\Big \downarrow \) 3866

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

\(\Big \downarrow \) 3867

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

\(\Big \downarrow \) 3834

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

\(\Big \downarrow \) 3832

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

\(\Big \downarrow \) 3833

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

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

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

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]

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

Sympy [F]

\[ \int (c e+d e x)^{2/3} \sin \left (a+b (c+d x)^{2/3}\right ) \, dx=\int \left (e \left (c + d x\right )\right )^{\frac {2}{3}} \sin {\left (a + b \left (c + d x\right )^{\frac {2}{3}} \right )}\, dx \] 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 = 424, normalized size of antiderivative = 1.87 \[ \int (c e+d e x)^{2/3} \sin \left (a+b (c+d x)^{2/3}\right ) \, dx =\text {Too large to display} \] Input:

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

Output:

-3/16*(3*(d*x + c)^(2/3)*(((gamma(3/2, -I*b*conjugate((d*x + c)^(2/3))) + 
gamma(3/2, I*(d*x + c)^(2/3)*b))*cos(3/4*pi + arctan2(0, d*x + c)) + (gamm 
a(3/2, I*b*conjugate((d*x + c)^(2/3))) + gamma(3/2, -I*(d*x + c)^(2/3)*b)) 
*cos(-3/4*pi + arctan2(0, d*x + c)) + (I*gamma(3/2, -I*b*conjugate((d*x + 
c)^(2/3))) - I*gamma(3/2, I*(d*x + c)^(2/3)*b))*sin(3/4*pi + arctan2(0, d* 
x + c)) + (I*gamma(3/2, I*b*conjugate((d*x + c)^(2/3))) - I*gamma(3/2, -I* 
(d*x + c)^(2/3)*b))*sin(-3/4*pi + arctan2(0, d*x + c)))*cos(a) + ((I*gamma 
(3/2, -I*b*conjugate((d*x + c)^(2/3))) - I*gamma(3/2, I*(d*x + c)^(2/3)*b) 
)*cos(3/4*pi + arctan2(0, d*x + c)) + (-I*gamma(3/2, I*b*conjugate((d*x + 
c)^(2/3))) + I*gamma(3/2, -I*(d*x + c)^(2/3)*b))*cos(-3/4*pi + arctan2(0, 
d*x + c)) - (gamma(3/2, -I*b*conjugate((d*x + c)^(2/3))) + gamma(3/2, I*(d 
*x + c)^(2/3)*b))*sin(3/4*pi + arctan2(0, d*x + c)) + (gamma(3/2, I*b*conj 
ugate((d*x + c)^(2/3))) + gamma(3/2, -I*(d*x + c)^(2/3)*b))*sin(-3/4*pi + 
arctan2(0, d*x + c)))*sin(a))*sqrt((d*x + c)^(2/3)*b)*e^(2/3) + 8*(b^2*d^2 
*x^2 + 2*b^2*c*d*x + b^2*c^2)*e^(2/3)*cos((d*x + c)^(2/3)*b + a))/(b^3*d^2 
*x + b^3*c*d)
 

Giac [F(-2)]

Exception generated. \[ \int (c e+d e x)^{2/3} \sin \left (a+b (c+d x)^{2/3}\right ) \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument Type
 

Mupad [F(-1)]

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

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