3.1.73 \(\int x^4 (a+b \sin (c+d x^3))^2 \, dx\) [73]

3.1.73.1 Optimal result

Integrand size = 18, antiderivative size = 249 \[ \int x^4 \left (a+b \sin \left (c+d x^3\right )\right )^2 \, dx=\frac {1}{10} \left (2 a^2+b^2\right ) x^5-\frac {2 a b x^2 \cos \left (c+d x^3\right )}{3 d}-\frac {2 a b e^{i c} x^2 \Gamma \left (\frac {2}{3},-i d x^3\right )}{9 d \left (-i d x^3\right )^{2/3}}-\frac {2 a b e^{-i c} x^2 \Gamma \left (\frac {2}{3},i d x^3\right )}{9 d \left (i d x^3\right )^{2/3}}+\frac {i b^2 e^{2 i c} x^2 \Gamma \left (\frac {2}{3},-2 i d x^3\right )}{36\ 2^{2/3} d \left (-i d x^3\right )^{2/3}}-\frac {i b^2 e^{-2 i c} x^2 \Gamma \left (\frac {2}{3},2 i d x^3\right )}{36\ 2^{2/3} d \left (i d x^3\right )^{2/3}}-\frac {b^2 x^2 \sin \left (2 c+2 d x^3\right )}{12 d} \]

output

1/10*(2*a^2+b^2)*x^5-2/3*a*b*x^2*cos(d*x^3+c)/d-2/9*a*b*exp(I*c)*x^2*GAMMA 
(2/3,-I*d*x^3)/d/(-I*d*x^3)^(2/3)-2/9*a*b*x^2*GAMMA(2/3,I*d*x^3)/d/exp(I*c 
)/(I*d*x^3)^(2/3)+1/72*I*b^2*exp(2*I*c)*x^2*GAMMA(2/3,-2*I*d*x^3)*2^(1/3)/ 
d/(-I*d*x^3)^(2/3)-1/72*I*b^2*x^2*GAMMA(2/3,2*I*d*x^3)*2^(1/3)/d/exp(2*I*c 
)/(I*d*x^3)^(2/3)-1/12*b^2*x^2*sin(2*d*x^3+2*c)/d
 

3.1.73.2 Mathematica [A] (verified)

Time = 0.61 (sec) , antiderivative size = 339, normalized size of antiderivative = 1.36 \[ \int x^4 \left (a+b \sin \left (c+d x^3\right )\right )^2 \, dx=\frac {d x^8 \left (72 a^2 d x^3 \left (d^2 x^6\right )^{2/3}+36 b^2 d x^3 \left (d^2 x^6\right )^{2/3}-240 a b \left (d^2 x^6\right )^{2/3} \cos \left (c+d x^3\right )+5 i \sqrt [3]{2} b^2 \left (i d x^3\right )^{2/3} \cos (2 c) \Gamma \left (\frac {2}{3},-2 i d x^3\right )-5 i \sqrt [3]{2} b^2 \left (-i d x^3\right )^{2/3} \cos (2 c) \Gamma \left (\frac {2}{3},2 i d x^3\right )-80 a b \left (-i d x^3\right )^{2/3} \Gamma \left (\frac {2}{3},i d x^3\right ) (\cos (c)-i \sin (c))-80 a b \left (i d x^3\right )^{2/3} \Gamma \left (\frac {2}{3},-i d x^3\right ) (\cos (c)+i \sin (c))-5 \sqrt [3]{2} b^2 \left (i d x^3\right )^{2/3} \Gamma \left (\frac {2}{3},-2 i d x^3\right ) \sin (2 c)-5 \sqrt [3]{2} b^2 \left (-i d x^3\right )^{2/3} \Gamma \left (\frac {2}{3},2 i d x^3\right ) \sin (2 c)-30 b^2 \left (d^2 x^6\right )^{2/3} \sin \left (2 \left (c+d x^3\right )\right )\right )}{360 \left (d^2 x^6\right )^{5/3}} \]

input

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

output

(d*x^8*(72*a^2*d*x^3*(d^2*x^6)^(2/3) + 36*b^2*d*x^3*(d^2*x^6)^(2/3) - 240* 
a*b*(d^2*x^6)^(2/3)*Cos[c + d*x^3] + (5*I)*2^(1/3)*b^2*(I*d*x^3)^(2/3)*Cos 
[2*c]*Gamma[2/3, (-2*I)*d*x^3] - (5*I)*2^(1/3)*b^2*((-I)*d*x^3)^(2/3)*Cos[ 
2*c]*Gamma[2/3, (2*I)*d*x^3] - 80*a*b*((-I)*d*x^3)^(2/3)*Gamma[2/3, I*d*x^ 
3]*(Cos[c] - I*Sin[c]) - 80*a*b*(I*d*x^3)^(2/3)*Gamma[2/3, (-I)*d*x^3]*(Co 
s[c] + I*Sin[c]) - 5*2^(1/3)*b^2*(I*d*x^3)^(2/3)*Gamma[2/3, (-2*I)*d*x^3]* 
Sin[2*c] - 5*2^(1/3)*b^2*((-I)*d*x^3)^(2/3)*Gamma[2/3, (2*I)*d*x^3]*Sin[2* 
c] - 30*b^2*(d^2*x^6)^(2/3)*Sin[2*(c + d*x^3)]))/(360*(d^2*x^6)^(5/3))
 

3.1.73.3 Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3884, 6, 2009}

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.

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

\(\Big \downarrow \) 3884

\(\displaystyle \int \left (a^2 x^4+2 a b x^4 \sin \left (c+d x^3\right )-\frac {1}{2} b^2 x^4 \cos \left (2 c+2 d x^3\right )+\frac {b^2 x^4}{2}\right )dx\)

\(\Big \downarrow \) 6

\(\displaystyle \int \left (x^4 \left (a^2+\frac {b^2}{2}\right )+2 a b x^4 \sin \left (c+d x^3\right )-\frac {1}{2} b^2 x^4 \cos \left (2 c+2 d x^3\right )\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{10} x^5 \left (2 a^2+b^2\right )-\frac {2 a b x^2 \cos \left (c+d x^3\right )}{3 d}-\frac {2 a b e^{i c} x^2 \Gamma \left (\frac {2}{3},-i d x^3\right )}{9 d \left (-i d x^3\right )^{2/3}}-\frac {2 a b e^{-i c} x^2 \Gamma \left (\frac {2}{3},i d x^3\right )}{9 d \left (i d x^3\right )^{2/3}}-\frac {b^2 x^2 \sin \left (2 c+2 d x^3\right )}{12 d}+\frac {i b^2 e^{2 i c} x^2 \Gamma \left (\frac {2}{3},-2 i d x^3\right )}{36\ 2^{2/3} d \left (-i d x^3\right )^{2/3}}-\frac {i b^2 e^{-2 i c} x^2 \Gamma \left (\frac {2}{3},2 i d x^3\right )}{36\ 2^{2/3} d \left (i d x^3\right )^{2/3}}\)

input

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

output

((2*a^2 + b^2)*x^5)/10 - (2*a*b*x^2*Cos[c + d*x^3])/(3*d) - (2*a*b*E^(I*c) 
*x^2*Gamma[2/3, (-I)*d*x^3])/(9*d*((-I)*d*x^3)^(2/3)) - (2*a*b*x^2*Gamma[2 
/3, I*d*x^3])/(9*d*E^(I*c)*(I*d*x^3)^(2/3)) + ((I/36)*b^2*E^((2*I)*c)*x^2* 
Gamma[2/3, (-2*I)*d*x^3])/(2^(2/3)*d*((-I)*d*x^3)^(2/3)) - ((I/36)*b^2*x^2 
*Gamma[2/3, (2*I)*d*x^3])/(2^(2/3)*d*E^((2*I)*c)*(I*d*x^3)^(2/3)) - (b^2*x 
^2*Sin[2*c + 2*d*x^3])/(12*d)
 

3.1.73.3.1 Defintions of rubi rules used

rule 6

Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v 
+ (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] &&  !FreeQ[Fx, x]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3884

Int[((e_.)*(x_))^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_), x 
_Symbol] :> Int[ExpandTrigReduce[(e*x)^m, (a + b*Sin[c + d*x^n])^p, x], x] 
/; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[p, 1] && IGtQ[n, 0]
 

3.1.73.4 Maple [F]

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

input

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

output

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

3.1.73.5 Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.75 \[ \int x^4 \left (a+b \sin \left (c+d x^3\right )\right )^2 \, dx=\frac {36 \, {\left (2 \, a^{2} + b^{2}\right )} d^{2} x^{5} - 60 \, b^{2} d x^{2} \cos \left (d x^{3} + c\right ) \sin \left (d x^{3} + c\right ) - 240 \, a b d x^{2} \cos \left (d x^{3} + c\right ) - 5 \, {\left (b^{2} \cos \left (2 \, c\right ) - i \, b^{2} \sin \left (2 \, c\right )\right )} \left (2 i \, d\right )^{\frac {1}{3}} \Gamma \left (\frac {2}{3}, 2 i \, d x^{3}\right ) - 80 \, {\left (-i \, a b \cos \left (c\right ) - a b \sin \left (c\right )\right )} \left (i \, d\right )^{\frac {1}{3}} \Gamma \left (\frac {2}{3}, i \, d x^{3}\right ) - 80 \, {\left (i \, a b \cos \left (c\right ) - a b \sin \left (c\right )\right )} \left (-i \, d\right )^{\frac {1}{3}} \Gamma \left (\frac {2}{3}, -i \, d x^{3}\right ) - 5 \, {\left (b^{2} \cos \left (2 \, c\right ) + i \, b^{2} \sin \left (2 \, c\right )\right )} \left (-2 i \, d\right )^{\frac {1}{3}} \Gamma \left (\frac {2}{3}, -2 i \, d x^{3}\right )}{360 \, d^{2}} \]

input

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

output

1/360*(36*(2*a^2 + b^2)*d^2*x^5 - 60*b^2*d*x^2*cos(d*x^3 + c)*sin(d*x^3 + 
c) - 240*a*b*d*x^2*cos(d*x^3 + c) - 5*(b^2*cos(2*c) - I*b^2*sin(2*c))*(2*I 
*d)^(1/3)*gamma(2/3, 2*I*d*x^3) - 80*(-I*a*b*cos(c) - a*b*sin(c))*(I*d)^(1 
/3)*gamma(2/3, I*d*x^3) - 80*(I*a*b*cos(c) - a*b*sin(c))*(-I*d)^(1/3)*gamm 
a(2/3, -I*d*x^3) - 5*(b^2*cos(2*c) + I*b^2*sin(2*c))*(-2*I*d)^(1/3)*gamma( 
2/3, -2*I*d*x^3))/d^2
 

3.1.73.6 Sympy [F]

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

input

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

output

Integral(x**4*(a + b*sin(c + d*x**3))**2, x)
 

3.1.73.7 Maxima [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 234, normalized size of antiderivative = 0.94 \[ \int x^4 \left (a+b \sin \left (c+d x^3\right )\right )^2 \, dx=\frac {1}{5} \, a^{2} x^{5} - \frac {{\left (6 \, d x^{3} \cos \left (d x^{3} + c\right ) - \left (d x^{3}\right )^{\frac {1}{3}} {\left ({\left ({\left (i \, \sqrt {3} - 1\right )} \Gamma \left (\frac {2}{3}, i \, d x^{3}\right ) + {\left (-i \, \sqrt {3} - 1\right )} \Gamma \left (\frac {2}{3}, -i \, d x^{3}\right )\right )} \cos \left (c\right ) + {\left ({\left (\sqrt {3} + i\right )} \Gamma \left (\frac {2}{3}, i \, d x^{3}\right ) + {\left (\sqrt {3} - i\right )} \Gamma \left (\frac {2}{3}, -i \, d x^{3}\right )\right )} \sin \left (c\right )\right )}\right )} a b}{9 \, d^{2} x} + \frac {{\left (72 \, d^{2} x^{6} - 60 \, d x^{3} \sin \left (2 \, d x^{3} + 2 \, c\right ) - 5 \cdot 2^{\frac {1}{3}} \left (d x^{3}\right )^{\frac {1}{3}} {\left ({\left ({\left (\sqrt {3} + i\right )} \Gamma \left (\frac {2}{3}, 2 i \, d x^{3}\right ) + {\left (\sqrt {3} - i\right )} \Gamma \left (\frac {2}{3}, -2 i \, d x^{3}\right )\right )} \cos \left (2 \, c\right ) + {\left ({\left (-i \, \sqrt {3} + 1\right )} \Gamma \left (\frac {2}{3}, 2 i \, d x^{3}\right ) + {\left (i \, \sqrt {3} + 1\right )} \Gamma \left (\frac {2}{3}, -2 i \, d x^{3}\right )\right )} \sin \left (2 \, c\right )\right )}\right )} b^{2}}{720 \, d^{2} x} \]

input

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

output

1/5*a^2*x^5 - 1/9*(6*d*x^3*cos(d*x^3 + c) - (d*x^3)^(1/3)*(((I*sqrt(3) - 1 
)*gamma(2/3, I*d*x^3) + (-I*sqrt(3) - 1)*gamma(2/3, -I*d*x^3))*cos(c) + (( 
sqrt(3) + I)*gamma(2/3, I*d*x^3) + (sqrt(3) - I)*gamma(2/3, -I*d*x^3))*sin 
(c)))*a*b/(d^2*x) + 1/720*(72*d^2*x^6 - 60*d*x^3*sin(2*d*x^3 + 2*c) - 5*2^ 
(1/3)*(d*x^3)^(1/3)*(((sqrt(3) + I)*gamma(2/3, 2*I*d*x^3) + (sqrt(3) - I)* 
gamma(2/3, -2*I*d*x^3))*cos(2*c) + ((-I*sqrt(3) + 1)*gamma(2/3, 2*I*d*x^3) 
 + (I*sqrt(3) + 1)*gamma(2/3, -2*I*d*x^3))*sin(2*c)))*b^2/(d^2*x)
 

3.1.73.8 Giac [F]

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

input

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

output

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

3.1.73.9 Mupad [F(-1)]

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

input

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

output

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