\(\int (e x)^m (a+b \sin (c+d x^3))^2 \, dx\) [99]

Optimal result

Integrand size = 20, antiderivative size = 285 \[ \int (e x)^m \left (a+b \sin \left (c+d x^3\right )\right )^2 \, dx=\frac {\left (2 a^2+b^2\right ) (e x)^{1+m}}{2 e (1+m)}+\frac {i a b e^{i c} (e x)^{1+m} \left (-i d x^3\right )^{\frac {1}{3} (-1-m)} \Gamma \left (\frac {1+m}{3},-i d x^3\right )}{3 e}-\frac {i a b e^{-i c} (e x)^{1+m} \left (i d x^3\right )^{\frac {1}{3} (-1-m)} \Gamma \left (\frac {1+m}{3},i d x^3\right )}{3 e}+\frac {2^{-\frac {7}{3}-\frac {m}{3}} b^2 e^{2 i c} (e x)^{1+m} \left (-i d x^3\right )^{\frac {1}{3} (-1-m)} \Gamma \left (\frac {1+m}{3},-2 i d x^3\right )}{3 e}+\frac {2^{-\frac {7}{3}-\frac {m}{3}} b^2 e^{-2 i c} (e x)^{1+m} \left (i d x^3\right )^{\frac {1}{3} (-1-m)} \Gamma \left (\frac {1+m}{3},2 i d x^3\right )}{3 e} \] Output:

1/2*(2*a^2+b^2)*(e*x)^(1+m)/e/(1+m)+1/3*I*a*b*exp(I*c)*(e*x)^(1+m)*(-I*d*x 
^3)^(-1/3-1/3*m)*GAMMA(1/3+1/3*m,-I*d*x^3)/e-1/3*I*a*b*(e*x)^(1+m)*(I*d*x^ 
3)^(-1/3-1/3*m)*GAMMA(1/3+1/3*m,I*d*x^3)/e/exp(I*c)+1/3*2^(-7/3-1/3*m)*b^2 
*exp(2*I*c)*(e*x)^(1+m)*(-I*d*x^3)^(-1/3-1/3*m)*GAMMA(1/3+1/3*m,-2*I*d*x^3 
)/e+1/3*2^(-7/3-1/3*m)*b^2*(e*x)^(1+m)*(I*d*x^3)^(-1/3-1/3*m)*GAMMA(1/3+1/ 
3*m,2*I*d*x^3)/e/exp(2*I*c)
 

Mathematica [A] (verified)

Time = 3.01 (sec) , antiderivative size = 556, normalized size of antiderivative = 1.95 \[ \int (e x)^m \left (a+b \sin \left (c+d x^3\right )\right )^2 \, dx=\frac {2^{\frac {1}{3} (-7-m)} x (e x)^m \left (d^2 x^6\right )^{\frac {1}{3} (-1-m)} \left (3\ 2^{\frac {7+m}{3}} a^2 \left (d^2 x^6\right )^{\frac {1+m}{3}}+3\ 2^{\frac {4+m}{3}} b^2 \left (d^2 x^6\right )^{\frac {1+m}{3}}+b^2 \left (i d x^3\right )^{\frac {1+m}{3}} \cos (2 c) \Gamma \left (\frac {1+m}{3},-2 i d x^3\right )+b^2 m \left (i d x^3\right )^{\frac {1+m}{3}} \cos (2 c) \Gamma \left (\frac {1+m}{3},-2 i d x^3\right )+b^2 \left (-i d x^3\right )^{\frac {1+m}{3}} \cos (2 c) \Gamma \left (\frac {1+m}{3},2 i d x^3\right )+b^2 m \left (-i d x^3\right )^{\frac {1+m}{3}} \cos (2 c) \Gamma \left (\frac {1+m}{3},2 i d x^3\right )-i 2^{\frac {7+m}{3}} a b (1+m) \left (-i d x^3\right )^{\frac {1+m}{3}} \Gamma \left (\frac {1+m}{3},i d x^3\right ) (\cos (c)-i \sin (c))+i 2^{\frac {7+m}{3}} a b (1+m) \left (i d x^3\right )^{\frac {1+m}{3}} \Gamma \left (\frac {1+m}{3},-i d x^3\right ) (\cos (c)+i \sin (c))+i b^2 \left (i d x^3\right )^{\frac {1+m}{3}} \Gamma \left (\frac {1+m}{3},-2 i d x^3\right ) \sin (2 c)+i b^2 m \left (i d x^3\right )^{\frac {1+m}{3}} \Gamma \left (\frac {1+m}{3},-2 i d x^3\right ) \sin (2 c)-i b^2 \left (-i d x^3\right )^{\frac {1+m}{3}} \Gamma \left (\frac {1+m}{3},2 i d x^3\right ) \sin (2 c)-i b^2 m \left (-i d x^3\right )^{\frac {1+m}{3}} \Gamma \left (\frac {1+m}{3},2 i d x^3\right ) \sin (2 c)\right )}{3 (1+m)} \] Input:

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

Output:

(2^((-7 - m)/3)*x*(e*x)^m*(d^2*x^6)^((-1 - m)/3)*(3*2^((7 + m)/3)*a^2*(d^2 
*x^6)^((1 + m)/3) + 3*2^((4 + m)/3)*b^2*(d^2*x^6)^((1 + m)/3) + b^2*(I*d*x 
^3)^((1 + m)/3)*Cos[2*c]*Gamma[(1 + m)/3, (-2*I)*d*x^3] + b^2*m*(I*d*x^3)^ 
((1 + m)/3)*Cos[2*c]*Gamma[(1 + m)/3, (-2*I)*d*x^3] + b^2*((-I)*d*x^3)^((1 
 + m)/3)*Cos[2*c]*Gamma[(1 + m)/3, (2*I)*d*x^3] + b^2*m*((-I)*d*x^3)^((1 + 
 m)/3)*Cos[2*c]*Gamma[(1 + m)/3, (2*I)*d*x^3] - I*2^((7 + m)/3)*a*b*(1 + m 
)*((-I)*d*x^3)^((1 + m)/3)*Gamma[(1 + m)/3, I*d*x^3]*(Cos[c] - I*Sin[c]) + 
 I*2^((7 + m)/3)*a*b*(1 + m)*(I*d*x^3)^((1 + m)/3)*Gamma[(1 + m)/3, (-I)*d 
*x^3]*(Cos[c] + I*Sin[c]) + I*b^2*(I*d*x^3)^((1 + m)/3)*Gamma[(1 + m)/3, ( 
-2*I)*d*x^3]*Sin[2*c] + I*b^2*m*(I*d*x^3)^((1 + m)/3)*Gamma[(1 + m)/3, (-2 
*I)*d*x^3]*Sin[2*c] - I*b^2*((-I)*d*x^3)^((1 + m)/3)*Gamma[(1 + m)/3, (2*I 
)*d*x^3]*Sin[2*c] - I*b^2*m*((-I)*d*x^3)^((1 + m)/3)*Gamma[(1 + m)/3, (2*I 
)*d*x^3]*Sin[2*c]))/(3*(1 + m))
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 285, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, 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.

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

\(\Big \downarrow \) 3884

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

\(\Big \downarrow \) 6

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\left (2 a^2+b^2\right ) (e x)^{m+1}}{2 e (m+1)}+\frac {i a b e^{i c} \left (-i d x^3\right )^{\frac {1}{3} (-m-1)} (e x)^{m+1} \Gamma \left (\frac {m+1}{3},-i d x^3\right )}{3 e}-\frac {i a b e^{-i c} \left (i d x^3\right )^{\frac {1}{3} (-m-1)} (e x)^{m+1} \Gamma \left (\frac {m+1}{3},i d x^3\right )}{3 e}+\frac {b^2 e^{2 i c} 2^{-\frac {m}{3}-\frac {7}{3}} \left (-i d x^3\right )^{\frac {1}{3} (-m-1)} (e x)^{m+1} \Gamma \left (\frac {m+1}{3},-2 i d x^3\right )}{3 e}+\frac {b^2 e^{-2 i c} 2^{-\frac {m}{3}-\frac {7}{3}} \left (i d x^3\right )^{\frac {1}{3} (-m-1)} (e x)^{m+1} \Gamma \left (\frac {m+1}{3},2 i d x^3\right )}{3 e}\)

Input:

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

Output:

((2*a^2 + b^2)*(e*x)^(1 + m))/(2*e*(1 + m)) + ((I/3)*a*b*E^(I*c)*(e*x)^(1 
+ m)*((-I)*d*x^3)^((-1 - m)/3)*Gamma[(1 + m)/3, (-I)*d*x^3])/e - ((I/3)*a* 
b*(e*x)^(1 + m)*(I*d*x^3)^((-1 - m)/3)*Gamma[(1 + m)/3, I*d*x^3])/(e*E^(I* 
c)) + (2^(-7/3 - m/3)*b^2*E^((2*I)*c)*(e*x)^(1 + m)*((-I)*d*x^3)^((-1 - m) 
/3)*Gamma[(1 + m)/3, (-2*I)*d*x^3])/(3*e) + (2^(-7/3 - m/3)*b^2*(e*x)^(1 + 
 m)*(I*d*x^3)^((-1 - m)/3)*Gamma[(1 + m)/3, (2*I)*d*x^3])/(3*e*E^((2*I)*c) 
)
 

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]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 214, normalized size of antiderivative = 0.75 \[ \int (e x)^m \left (a+b \sin \left (c+d x^3\right )\right )^2 \, dx=\frac {12 \, {\left (2 \, a^{2} + b^{2}\right )} \left (e x\right )^{m} d x + {\left (-i \, b^{2} e^{2} m - i \, b^{2} e^{2}\right )} e^{\left (-\frac {1}{3} \, {\left (m - 2\right )} \log \left (\frac {2 i \, d}{e^{3}}\right ) - 2 i \, c\right )} \Gamma \left (\frac {1}{3} \, m + \frac {1}{3}, 2 i \, d x^{3}\right ) - 8 \, {\left (a b e^{2} m + a b e^{2}\right )} e^{\left (-\frac {1}{3} \, {\left (m - 2\right )} \log \left (\frac {i \, d}{e^{3}}\right ) - i \, c\right )} \Gamma \left (\frac {1}{3} \, m + \frac {1}{3}, i \, d x^{3}\right ) - 8 \, {\left (a b e^{2} m + a b e^{2}\right )} e^{\left (-\frac {1}{3} \, {\left (m - 2\right )} \log \left (-\frac {i \, d}{e^{3}}\right ) + i \, c\right )} \Gamma \left (\frac {1}{3} \, m + \frac {1}{3}, -i \, d x^{3}\right ) + {\left (i \, b^{2} e^{2} m + i \, b^{2} e^{2}\right )} e^{\left (-\frac {1}{3} \, {\left (m - 2\right )} \log \left (-\frac {2 i \, d}{e^{3}}\right ) + 2 i \, c\right )} \Gamma \left (\frac {1}{3} \, m + \frac {1}{3}, -2 i \, d x^{3}\right )}{24 \, {\left (d m + d\right )}} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

(e*x)^(m + 1)*a^2/(e*(m + 1)) + 1/2*(b^2*e^m*x*x^m - (b^2*e^m*m + b^2*e^m) 
*integrate(x^m*cos(2*d*x^3 + 2*c), x) + 4*(a*b*e^m*m + a*b*e^m)*integrate( 
x^m*sin(d*x^3 + c), x))/(m + 1)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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