\(\int (a+b \sin (c+d x^3))^2 \, dx\) [78]

Optimal result

Integrand size = 14, antiderivative size = 183 \[ \int \left (a+b \sin \left (c+d x^3\right )\right )^2 \, dx=\frac {1}{2} \left (2 a^2+b^2\right ) x+\frac {i a b e^{i c} x \Gamma \left (\frac {1}{3},-i d x^3\right )}{3 \sqrt [3]{-i d x^3}}-\frac {i a b e^{-i c} x \Gamma \left (\frac {1}{3},i d x^3\right )}{3 \sqrt [3]{i d x^3}}+\frac {b^2 e^{2 i c} x \Gamma \left (\frac {1}{3},-2 i d x^3\right )}{12 \sqrt [3]{2} \sqrt [3]{-i d x^3}}+\frac {b^2 e^{-2 i c} x \Gamma \left (\frac {1}{3},2 i d x^3\right )}{12 \sqrt [3]{2} \sqrt [3]{i d x^3}} \] Output:

1/2*(2*a^2+b^2)*x+1/3*I*a*b*exp(I*c)*x*GAMMA(1/3,-I*d*x^3)/(-I*d*x^3)^(1/3 
)-1/3*I*a*b*x*GAMMA(1/3,I*d*x^3)/exp(I*c)/(I*d*x^3)^(1/3)+1/24*b^2*exp(2*I 
*c)*x*GAMMA(1/3,-2*I*d*x^3)*2^(2/3)/(-I*d*x^3)^(1/3)+1/24*b^2*x*GAMMA(1/3, 
2*I*d*x^3)*2^(2/3)/exp(2*I*c)/(I*d*x^3)^(1/3)
 

Mathematica [A] (verified)

Time = 1.04 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.25 \[ \int \left (a+b \sin \left (c+d x^3\right )\right )^2 \, dx=\frac {1}{24} x \left (12 \left (2 a^2+b^2\right )-8 i a b \cos (c) \left (-\frac {\Gamma \left (\frac {1}{3},-i d x^3\right )}{\sqrt [3]{-i d x^3}}+\frac {\Gamma \left (\frac {1}{3},i d x^3\right )}{\sqrt [3]{i d x^3}}\right )+8 a b \left (-\frac {\Gamma \left (\frac {1}{3},-i d x^3\right )}{\sqrt [3]{-i d x^3}}-\frac {\Gamma \left (\frac {1}{3},i d x^3\right )}{\sqrt [3]{i d x^3}}\right ) \sin (c)+\frac {2^{2/3} b^2 \Gamma \left (\frac {1}{3},2 i d x^3\right ) (\cos (2 c)-i \sin (2 c))}{\sqrt [3]{i d x^3}}+\frac {2^{2/3} b^2 \Gamma \left (\frac {1}{3},-2 i d x^3\right ) (\cos (2 c)+i \sin (2 c))}{\sqrt [3]{-i d x^3}}\right ) \] Input:

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

Output:

(x*(12*(2*a^2 + b^2) - (8*I)*a*b*Cos[c]*(-(Gamma[1/3, (-I)*d*x^3]/((-I)*d* 
x^3)^(1/3)) + Gamma[1/3, I*d*x^3]/(I*d*x^3)^(1/3)) + 8*a*b*(-(Gamma[1/3, ( 
-I)*d*x^3]/((-I)*d*x^3)^(1/3)) - Gamma[1/3, I*d*x^3]/(I*d*x^3)^(1/3))*Sin[ 
c] + (2^(2/3)*b^2*Gamma[1/3, (2*I)*d*x^3]*(Cos[2*c] - I*Sin[2*c]))/(I*d*x^ 
3)^(1/3) + (2^(2/3)*b^2*Gamma[1/3, (-2*I)*d*x^3]*(Cos[2*c] + I*Sin[2*c]))/ 
((-I)*d*x^3)^(1/3)))/24
 

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3838, 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.

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

\(\Big \downarrow \) 3838

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} x \left (2 a^2+b^2\right )+\frac {i a b e^{i c} x \Gamma \left (\frac {1}{3},-i d x^3\right )}{3 \sqrt [3]{-i d x^3}}-\frac {i a b e^{-i c} x \Gamma \left (\frac {1}{3},i d x^3\right )}{3 \sqrt [3]{i d x^3}}+\frac {b^2 e^{2 i c} x \Gamma \left (\frac {1}{3},-2 i d x^3\right )}{12 \sqrt [3]{2} \sqrt [3]{-i d x^3}}+\frac {b^2 e^{-2 i c} x \Gamma \left (\frac {1}{3},2 i d x^3\right )}{12 \sqrt [3]{2} \sqrt [3]{i d x^3}}\)

Input:

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

Output:

((2*a^2 + b^2)*x)/2 + ((I/3)*a*b*E^(I*c)*x*Gamma[1/3, (-I)*d*x^3])/((-I)*d 
*x^3)^(1/3) - ((I/3)*a*b*x*Gamma[1/3, I*d*x^3])/(E^(I*c)*(I*d*x^3)^(1/3)) 
+ (b^2*E^((2*I)*c)*x*Gamma[1/3, (-2*I)*d*x^3])/(12*2^(1/3)*((-I)*d*x^3)^(1 
/3)) + (b^2*x*Gamma[1/3, (2*I)*d*x^3])/(12*2^(1/3)*E^((2*I)*c)*(I*d*x^3)^( 
1/3))
 

Defintions of rubi rules used

rule 2009

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

rule 3838

Int[((a_.) + (b_.)*Sin[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)])^(p_), x_Sy 
mbol] :> Int[ExpandTrigReduce[(a + b*Sin[c + d*(e + f*x)^n])^p, x], x] /; F 
reeQ[{a, b, c, d, e, f}, x] && IGtQ[p, 1] && IGtQ[n, 1]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.76 \[ \int \left (a+b \sin \left (c+d x^3\right )\right )^2 \, dx=\frac {12 \, {\left (2 \, a^{2} + b^{2}\right )} d x + {\left (-i \, b^{2} \cos \left (2 \, c\right ) - b^{2} \sin \left (2 \, c\right )\right )} \left (2 i \, d\right )^{\frac {2}{3}} \Gamma \left (\frac {1}{3}, 2 i \, d x^{3}\right ) - 8 \, {\left (a b \cos \left (c\right ) - i \, a b \sin \left (c\right )\right )} \left (i \, d\right )^{\frac {2}{3}} \Gamma \left (\frac {1}{3}, i \, d x^{3}\right ) - 8 \, {\left (a b \cos \left (c\right ) + i \, a b \sin \left (c\right )\right )} \left (-i \, d\right )^{\frac {2}{3}} \Gamma \left (\frac {1}{3}, -i \, d x^{3}\right ) + {\left (i \, b^{2} \cos \left (2 \, c\right ) - b^{2} \sin \left (2 \, c\right )\right )} \left (-2 i \, d\right )^{\frac {2}{3}} \Gamma \left (\frac {1}{3}, -2 i \, d x^{3}\right )}{24 \, d} \] Input:

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

Output:

1/24*(12*(2*a^2 + b^2)*d*x + (-I*b^2*cos(2*c) - b^2*sin(2*c))*(2*I*d)^(2/3 
)*gamma(1/3, 2*I*d*x^3) - 8*(a*b*cos(c) - I*a*b*sin(c))*(I*d)^(2/3)*gamma( 
1/3, I*d*x^3) - 8*(a*b*cos(c) + I*a*b*sin(c))*(-I*d)^(2/3)*gamma(1/3, -I*d 
*x^3) + (I*b^2*cos(2*c) - b^2*sin(2*c))*(-2*I*d)^(2/3)*gamma(1/3, -2*I*d*x 
^3))/d
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.05 \[ \int \left (a+b \sin \left (c+d x^3\right )\right )^2 \, dx=\frac {{\left ({\left ({\left (-i \, \sqrt {3} - 1\right )} \Gamma \left (\frac {1}{3}, i \, d x^{3}\right ) + {\left (i \, \sqrt {3} - 1\right )} \Gamma \left (\frac {1}{3}, -i \, d x^{3}\right )\right )} \cos \left (c\right ) - {\left ({\left (\sqrt {3} - i\right )} \Gamma \left (\frac {1}{3}, i \, d x^{3}\right ) + {\left (\sqrt {3} + i\right )} \Gamma \left (\frac {1}{3}, -i \, d x^{3}\right )\right )} \sin \left (c\right )\right )} a b x}{6 \, \left (d x^{3}\right )^{\frac {1}{3}}} + \frac {2^{\frac {2}{3}} {\left ({\left ({\left ({\left (\sqrt {3} - i\right )} \Gamma \left (\frac {1}{3}, 2 i \, d x^{3}\right ) + {\left (\sqrt {3} + i\right )} \Gamma \left (\frac {1}{3}, -2 i \, d x^{3}\right )\right )} \cos \left (2 \, c\right ) + {\left ({\left (-i \, \sqrt {3} - 1\right )} \Gamma \left (\frac {1}{3}, 2 i \, d x^{3}\right ) + {\left (i \, \sqrt {3} - 1\right )} \Gamma \left (\frac {1}{3}, -2 i \, d x^{3}\right )\right )} \sin \left (2 \, c\right )\right )} x + 12 \cdot 2^{\frac {1}{3}} \left (d x^{3}\right )^{\frac {1}{3}} x\right )} b^{2}}{48 \, \left (d x^{3}\right )^{\frac {1}{3}}} + a^{2} x \] Input:

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

Output:

1/6*(((-I*sqrt(3) - 1)*gamma(1/3, I*d*x^3) + (I*sqrt(3) - 1)*gamma(1/3, -I 
*d*x^3))*cos(c) - ((sqrt(3) - I)*gamma(1/3, I*d*x^3) + (sqrt(3) + I)*gamma 
(1/3, -I*d*x^3))*sin(c))*a*b*x/(d*x^3)^(1/3) + 1/48*2^(2/3)*((((sqrt(3) - 
I)*gamma(1/3, 2*I*d*x^3) + (sqrt(3) + I)*gamma(1/3, -2*I*d*x^3))*cos(2*c) 
+ ((-I*sqrt(3) - 1)*gamma(1/3, 2*I*d*x^3) + (I*sqrt(3) - 1)*gamma(1/3, -2* 
I*d*x^3))*sin(2*c))*x + 12*2^(1/3)*(d*x^3)^(1/3)*x)*b^2/(d*x^3)^(1/3) + a^ 
2*x
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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