\(\int (c e+d e x)^{4/3} \sin (a+b (c+d x)^{2/3}) \, dx\) [235]

Optimal result

Integrand size = 27, antiderivative size = 267 \[ \int (c e+d e x)^{4/3} \sin \left (a+b (c+d x)^{2/3}\right ) \, dx=\frac {45 e \sqrt [3]{e (c+d x)} \cos \left (a+b (c+d x)^{2/3}\right )}{8 b^3 d}-\frac {3 e (c+d x)^{4/3} \sqrt [3]{e (c+d x)} \cos \left (a+b (c+d x)^{2/3}\right )}{2 b d}-\frac {45 e \sqrt {\pi } \sqrt [3]{e (c+d x)} \cos (a) \operatorname {FresnelC}\left (\sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt [3]{c+d x}\right )}{8 \sqrt {2} b^{7/2} d \sqrt [3]{c+d x}}+\frac {45 e \sqrt {\pi } \sqrt [3]{e (c+d x)} \operatorname {FresnelS}\left (\sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt [3]{c+d x}\right ) \sin (a)}{8 \sqrt {2} b^{7/2} d \sqrt [3]{c+d x}}+\frac {15 e (c+d x)^{2/3} \sqrt [3]{e (c+d x)} \sin \left (a+b (c+d x)^{2/3}\right )}{4 b^2 d} \] Output:

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

Mathematica [A] (verified)

Time = 0.95 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.66 \[ \int (c e+d e x)^{4/3} \sin \left (a+b (c+d x)^{2/3}\right ) \, dx=-\frac {3 (e (c+d x))^{4/3} \left (15 \sqrt {2 \pi } \cos (a) \operatorname {FresnelC}\left (\sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt [3]{c+d x}\right )-15 \sqrt {2 \pi } \operatorname {FresnelS}\left (\sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt [3]{c+d x}\right ) \sin (a)+2 \sqrt {b} \left (\sqrt [3]{c+d x} \left (-15+4 b^2 (c+d x)^{4/3}\right ) \cos \left (a+b (c+d x)^{2/3}\right )-10 b (c+d x) \sin \left (a+b (c+d x)^{2/3}\right )\right )\right )}{16 b^{7/2} d (c+d x)^{4/3}} \] Input:

Integrate[(c*e + d*e*x)^(4/3)*Sin[a + b*(c + d*x)^(2/3)],x]
 

Output:

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

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 222, normalized size of antiderivative = 0.83, 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, 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.

\(\displaystyle \int (c e+d e x)^{4/3} \sin \left (a+b (c+d x)^{2/3}\right ) \, dx\)

\(\Big \downarrow \) 3916

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

\(\Big \downarrow \) 3898

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

\(\Big \downarrow \) 3896

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

\(\Big \downarrow \) 3866

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

\(\Big \downarrow \) 3867

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

\(\Big \downarrow \) 3866

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

\(\Big \downarrow \) 3835

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

\(\Big \downarrow \) 3832

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

\(\Big \downarrow \) 3833

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

Input:

Int[(c*e + d*e*x)^(4/3)*Sin[a + b*(c + d*x)^(2/3)],x]
 

Output:

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

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

Input:

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

Output:

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

Fricas [F]

\[ \int (c e+d e x)^{4/3} \sin \left (a+b (c+d x)^{2/3}\right ) \, dx=\int { {\left (d e x + c e\right )}^{\frac {4}{3}} \sin \left ({\left (d x + c\right )}^{\frac {2}{3}} b + a\right ) \,d x } \] Input:

integrate((d*e*x+c*e)^(4/3)*sin(a+b*(d*x+c)^(2/3)),x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int (c e+d e x)^{4/3} \sin \left (a+b (c+d x)^{2/3}\right ) \, dx=\text {Timed out} \] Input:

integrate((d*e*x+c*e)**(4/3)*sin(a+b*(d*x+c)**(2/3)),x)
 

Output:

Timed out
 

Maxima [C] (verification not implemented)

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

Time = 0.35 (sec) , antiderivative size = 386, normalized size of antiderivative = 1.45 \[ \int (c e+d e x)^{4/3} \sin \left (a+b (c+d x)^{2/3}\right ) \, dx =\text {Too large to display} \] Input:

integrate((d*e*x+c*e)^(4/3)*sin(a+b*(d*x+c)^(2/3)),x, algorithm="maxima")
 

Output:

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

Giac [F(-2)]

Exception generated. \[ \int (c e+d e x)^{4/3} \sin \left (a+b (c+d x)^{2/3}\right ) \, dx=\text {Exception raised: TypeError} \] Input:

integrate((d*e*x+c*e)^(4/3)*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. \[ \int (c e+d e x)^{4/3} \sin \left (a+b (c+d x)^{2/3}\right ) \, dx=\int \sin \left (a+b\,{\left (c+d\,x\right )}^{2/3}\right )\,{\left (c\,e+d\,e\,x\right )}^{4/3} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int (c e+d e x)^{4/3} \sin \left (a+b (c+d x)^{2/3}\right ) \, dx=\frac {e^{\frac {4}{3}} \left (-3 \left (d x +c \right )^{\frac {2}{3}} \cos \left (\left (d x +c \right )^{\frac {2}{3}} b +a \right ) b c +2 \left (\int \left (d x +c \right )^{\frac {1}{3}} \sin \left (\left (d x +c \right )^{\frac {2}{3}} b +a \right ) x d x \right ) b^{2} d^{2}+3 \sin \left (\left (d x +c \right )^{\frac {2}{3}} b +a \right ) c \right )}{2 b^{2} d} \] Input:

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

Output:

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