\(\int (e x)^m (a+b \sin (c+d x^3))^3 \, dx\) [98]

Optimal result

Integrand size = 20, antiderivative size = 442 \[ \int (e x)^m \left (a+b \sin \left (c+d x^3\right )\right )^3 \, dx=\frac {a \left (2 a^2+3 b^2\right ) (e x)^{1+m}}{2 e (1+m)}+\frac {i b \left (4 a^2+b^2\right ) e^{i c} (e x)^{1+m} \left (-i d x^3\right )^{\frac {1}{3} (-1-m)} \Gamma \left (\frac {1+m}{3},-i d x^3\right )}{8 e}-\frac {i b \left (4 a^2+b^2\right ) e^{-i c} (e x)^{1+m} \left (i d x^3\right )^{\frac {1}{3} (-1-m)} \Gamma \left (\frac {1+m}{3},i d x^3\right )}{8 e}+\frac {2^{-\frac {7}{3}-\frac {m}{3}} a b^2 e^{2 i c} (e x)^{1+m} \left (-i d x^3\right )^{\frac {1}{3} (-1-m)} \Gamma \left (\frac {1+m}{3},-2 i d x^3\right )}{e}+\frac {2^{-\frac {7}{3}-\frac {m}{3}} a b^2 e^{-2 i c} (e x)^{1+m} \left (i d x^3\right )^{\frac {1}{3} (-1-m)} \Gamma \left (\frac {1+m}{3},2 i d x^3\right )}{e}-\frac {i 3^{-\frac {4}{3}-\frac {m}{3}} b^3 e^{3 i c} (e x)^{1+m} \left (-i d x^3\right )^{\frac {1}{3} (-1-m)} \Gamma \left (\frac {1+m}{3},-3 i d x^3\right )}{8 e}+\frac {i 3^{-\frac {4}{3}-\frac {m}{3}} b^3 e^{-3 i c} (e x)^{1+m} \left (i d x^3\right )^{\frac {1}{3} (-1-m)} \Gamma \left (\frac {1+m}{3},3 i d x^3\right )}{8 e} \] Output:

1/2*a*(2*a^2+3*b^2)*(e*x)^(1+m)/e/(1+m)+1/8*I*b*(4*a^2+b^2)*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-1/8*I*b*(4*a^2+ 
b^2)*(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 
)+2^(-7/3-1/3*m)*a*b^2*exp(2*I*c)*(e*x)^(1+m)*(-I*d*x^3)^(-1/3-1/3*m)*GAMM 
A(1/3+1/3*m,-2*I*d*x^3)/e+2^(-7/3-1/3*m)*a*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)-1/8*I*3^(-4/3-1/3*m)*b^3*e 
xp(3*I*c)*(e*x)^(1+m)*(-I*d*x^3)^(-1/3-1/3*m)*GAMMA(1/3+1/3*m,-3*I*d*x^3)/ 
e+1/8*I*3^(-4/3-1/3*m)*b^3*(e*x)^(1+m)*(I*d*x^3)^(-1/3-1/3*m)*GAMMA(1/3+1/ 
3*m,3*I*d*x^3)/e/exp(3*I*c)
 

Mathematica [A] (verified)

Time = 2.77 (sec) , antiderivative size = 373, normalized size of antiderivative = 0.84 \[ \int (e x)^m \left (a+b \sin \left (c+d x^3\right )\right )^3 \, dx=\frac {1}{24} i x (e x)^m \left (-\frac {12 i a \left (2 a^2+3 b^2\right )}{1+m}+3 b \left (4 a^2+b^2\right ) e^{i c} \left (-i d x^3\right )^{-\frac {1}{3}-\frac {m}{3}} \Gamma \left (\frac {1+m}{3},-i d x^3\right )-3 b \left (4 a^2+b^2\right ) e^{-i c} \left (i d x^3\right )^{-\frac {1}{3}-\frac {m}{3}} \Gamma \left (\frac {1+m}{3},i d x^3\right )-3 i 2^{\frac {2}{3}-\frac {m}{3}} a b^2 e^{2 i c} \left (-i d x^3\right )^{-\frac {1}{3}-\frac {m}{3}} \Gamma \left (\frac {1+m}{3},-2 i d x^3\right )-3 i 2^{\frac {2}{3}-\frac {m}{3}} a b^2 e^{-2 i c} \left (i d x^3\right )^{-\frac {1}{3}-\frac {m}{3}} \Gamma \left (\frac {1+m}{3},2 i d x^3\right )-3^{-\frac {1}{3}-\frac {m}{3}} b^3 e^{3 i c} \left (-i d x^3\right )^{-\frac {1}{3}-\frac {m}{3}} \Gamma \left (\frac {1+m}{3},-3 i d x^3\right )+3^{-\frac {1}{3}-\frac {m}{3}} b^3 e^{-3 i c} \left (i d x^3\right )^{-\frac {1}{3}-\frac {m}{3}} \Gamma \left (\frac {1+m}{3},3 i d x^3\right )\right ) \] Input:

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

Output:

(I/24)*x*(e*x)^m*(((-12*I)*a*(2*a^2 + 3*b^2))/(1 + m) + 3*b*(4*a^2 + b^2)* 
E^(I*c)*((-I)*d*x^3)^(-1/3 - m/3)*Gamma[(1 + m)/3, (-I)*d*x^3] - (3*b*(4*a 
^2 + b^2)*(I*d*x^3)^(-1/3 - m/3)*Gamma[(1 + m)/3, I*d*x^3])/E^(I*c) - (3*I 
)*2^(2/3 - m/3)*a*b^2*E^((2*I)*c)*((-I)*d*x^3)^(-1/3 - m/3)*Gamma[(1 + m)/ 
3, (-2*I)*d*x^3] - ((3*I)*2^(2/3 - m/3)*a*b^2*(I*d*x^3)^(-1/3 - m/3)*Gamma 
[(1 + m)/3, (2*I)*d*x^3])/E^((2*I)*c) - 3^(-1/3 - m/3)*b^3*E^((3*I)*c)*((- 
I)*d*x^3)^(-1/3 - m/3)*Gamma[(1 + m)/3, (-3*I)*d*x^3] + (3^(-1/3 - m/3)*b^ 
3*(I*d*x^3)^(-1/3 - m/3)*Gamma[(1 + m)/3, (3*I)*d*x^3])/E^((3*I)*c))
 

Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 442, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3884, 6, 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.

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

\(\Big \downarrow \) 3884

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

\(\Big \downarrow \) 6

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

\(\Big \downarrow \) 6

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {i b e^{i c} \left (4 a^2+b^2\right ) \left (-i d x^3\right )^{\frac {1}{3} (-m-1)} (e x)^{m+1} \Gamma \left (\frac {m+1}{3},-i d x^3\right )}{8 e}-\frac {i b e^{-i c} \left (4 a^2+b^2\right ) \left (i d x^3\right )^{\frac {1}{3} (-m-1)} (e x)^{m+1} \Gamma \left (\frac {m+1}{3},i d x^3\right )}{8 e}+\frac {a \left (2 a^2+3 b^2\right ) (e x)^{m+1}}{2 e (m+1)}+\frac {a b^2 e^{2 i c} 2^{-\frac {m}{3}-\frac {7}{3}} \left (-i d x^3\right )^{\frac {1}{3} (-m-1)} (e x)^{m+1} \Gamma \left (\frac {m+1}{3},-2 i d x^3\right )}{e}+\frac {a b^2 e^{-2 i c} 2^{-\frac {m}{3}-\frac {7}{3}} \left (i d x^3\right )^{\frac {1}{3} (-m-1)} (e x)^{m+1} \Gamma \left (\frac {m+1}{3},2 i d x^3\right )}{e}-\frac {i b^3 e^{3 i c} 3^{-\frac {m}{3}-\frac {4}{3}} \left (-i d x^3\right )^{\frac {1}{3} (-m-1)} (e x)^{m+1} \Gamma \left (\frac {m+1}{3},-3 i d x^3\right )}{8 e}+\frac {i b^3 e^{-3 i c} 3^{-\frac {m}{3}-\frac {4}{3}} \left (i d x^3\right )^{\frac {1}{3} (-m-1)} (e x)^{m+1} \Gamma \left (\frac {m+1}{3},3 i d x^3\right )}{8 e}\)

Input:

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

Output:

(a*(2*a^2 + 3*b^2)*(e*x)^(1 + m))/(2*e*(1 + m)) + ((I/8)*b*(4*a^2 + b^2)*E 
^(I*c)*(e*x)^(1 + m)*((-I)*d*x^3)^((-1 - m)/3)*Gamma[(1 + m)/3, (-I)*d*x^3 
])/e - ((I/8)*b*(4*a^2 + b^2)*(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)*a*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])/e + (2^( 
-7/3 - m/3)*a*b^2*(e*x)^(1 + m)*(I*d*x^3)^((-1 - m)/3)*Gamma[(1 + m)/3, (2 
*I)*d*x^3])/(e*E^((2*I)*c)) - ((I/8)*3^(-4/3 - m/3)*b^3*E^((3*I)*c)*(e*x)^ 
(1 + m)*((-I)*d*x^3)^((-1 - m)/3)*Gamma[(1 + m)/3, (-3*I)*d*x^3])/e + ((I/ 
8)*3^(-4/3 - m/3)*b^3*(e*x)^(1 + m)*(I*d*x^3)^((-1 - m)/3)*Gamma[(1 + m)/3 
, (3*I)*d*x^3])/(e*E^((3*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]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 347, normalized size of antiderivative = 0.79 \[ \int (e x)^m \left (a+b \sin \left (c+d x^3\right )\right )^3 \, dx=\frac {36 \, {\left (2 \, a^{3} + 3 \, a b^{2}\right )} \left (e x\right )^{m} d x + {\left (b^{3} e^{2} m + b^{3} e^{2}\right )} e^{\left (-\frac {1}{3} \, {\left (m - 2\right )} \log \left (\frac {3 i \, d}{e^{3}}\right ) - 3 i \, c\right )} \Gamma \left (\frac {1}{3} \, m + \frac {1}{3}, 3 i \, d x^{3}\right ) - 9 \, {\left (i \, a b^{2} e^{2} m + i \, a b^{2} e^{2}\right )} e^{\left (-\frac {1}{3} \, {\left (m - 2\right )} \log \left (\frac {2 i \, d}{e^{3}}\right ) - 2 i \, c\right )} \Gamma \left (\frac {1}{3} \, m + \frac {1}{3}, 2 i \, d x^{3}\right ) - 9 \, {\left ({\left (4 \, a^{2} b + b^{3}\right )} e^{2} m + {\left (4 \, a^{2} b + b^{3}\right )} e^{2}\right )} e^{\left (-\frac {1}{3} \, {\left (m - 2\right )} \log \left (\frac {i \, d}{e^{3}}\right ) - i \, c\right )} \Gamma \left (\frac {1}{3} \, m + \frac {1}{3}, i \, d x^{3}\right ) - 9 \, {\left ({\left (4 \, a^{2} b + b^{3}\right )} e^{2} m + {\left (4 \, a^{2} b + b^{3}\right )} e^{2}\right )} e^{\left (-\frac {1}{3} \, {\left (m - 2\right )} \log \left (-\frac {i \, d}{e^{3}}\right ) + i \, c\right )} \Gamma \left (\frac {1}{3} \, m + \frac {1}{3}, -i \, d x^{3}\right ) - 9 \, {\left (-i \, a b^{2} e^{2} m - i \, a b^{2} e^{2}\right )} e^{\left (-\frac {1}{3} \, {\left (m - 2\right )} \log \left (-\frac {2 i \, d}{e^{3}}\right ) + 2 i \, c\right )} \Gamma \left (\frac {1}{3} \, m + \frac {1}{3}, -2 i \, d x^{3}\right ) + {\left (b^{3} e^{2} m + b^{3} e^{2}\right )} e^{\left (-\frac {1}{3} \, {\left (m - 2\right )} \log \left (-\frac {3 i \, d}{e^{3}}\right ) + 3 i \, c\right )} \Gamma \left (\frac {1}{3} \, m + \frac {1}{3}, -3 i \, d x^{3}\right )}{72 \, {\left (d m + d\right )}} \] Input:

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

Output:

1/72*(36*(2*a^3 + 3*a*b^2)*(e*x)^m*d*x + (b^3*e^2*m + b^3*e^2)*e^(-1/3*(m 
- 2)*log(3*I*d/e^3) - 3*I*c)*gamma(1/3*m + 1/3, 3*I*d*x^3) - 9*(I*a*b^2*e^ 
2*m + I*a*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) - 9*((4*a^2*b + b^3)*e^2*m + (4*a^2*b + b^3)*e^2)*e^(-1/3*( 
m - 2)*log(I*d/e^3) - I*c)*gamma(1/3*m + 1/3, I*d*x^3) - 9*((4*a^2*b + b^3 
)*e^2*m + (4*a^2*b + b^3)*e^2)*e^(-1/3*(m - 2)*log(-I*d/e^3) + I*c)*gamma( 
1/3*m + 1/3, -I*d*x^3) - 9*(-I*a*b^2*e^2*m - I*a*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) + (b^3*e^2*m + b^3 
*e^2)*e^(-1/3*(m - 2)*log(-3*I*d/e^3) + 3*I*c)*gamma(1/3*m + 1/3, -3*I*d*x 
^3))/(d*m + d)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

(e*x)^(m + 1)*a^3/(e*(m + 1)) + 1/8*(12*a*b^2*e^m*x*x^m - 12*(a*b^2*e^m*m 
+ a*b^2*e^m)*integrate(x^m*cos(2*d*x^3 + 2*c), x) + 3*((4*a^2*b + b^3)*e^m 
*m*sin(c) + (4*a^2*b + b^3)*e^m*sin(c))*integrate(x^m*cos(d*x^3), x) - 2*( 
b^3*e^m*m + b^3*e^m)*integrate(x^m*sin(3*d*x^3 + 3*c), x) + 3*((4*a^2*b + 
b^3)*e^m*m + (4*a^2*b + b^3)*e^m)*integrate(x^m*sin(d*x^3 + c), x) + 3*((4 
*a^2*b + b^3)*e^m*m*cos(c) + (4*a^2*b + b^3)*e^m*cos(c))*integrate(x^m*sin 
(d*x^3), x))/(m + 1)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int (e x)^m \left (a+b \sin \left (c+d x^3\right )\right )^3 \, dx=\frac {e^{m} \left (x^{m} a^{3} x +6 x^{m} a \,b^{2} x -6 \left (\int x^{m}d x \right ) a \,b^{2} m -6 \left (\int x^{m}d x \right ) a \,b^{2}+\left (\int x^{m} \sin \left (d \,x^{3}+c \right )^{3}d x \right ) b^{3} m +\left (\int x^{m} \sin \left (d \,x^{3}+c \right )^{3}d x \right ) b^{3}+3 \left (\int x^{m} \sin \left (d \,x^{3}+c \right )^{2}d x \right ) a \,b^{2} m +3 \left (\int x^{m} \sin \left (d \,x^{3}+c \right )^{2}d x \right ) a \,b^{2}+3 \left (\int x^{m} \sin \left (d \,x^{3}+c \right )d x \right ) a^{2} b m +3 \left (\int x^{m} \sin \left (d \,x^{3}+c \right )d x \right ) a^{2} b \right )}{m +1} \] Input:

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

Output:

(e**m*(x**m*a**3*x + 6*x**m*a*b**2*x - 6*int(x**m,x)*a*b**2*m - 6*int(x**m 
,x)*a*b**2 + int(x**m*sin(c + d*x**3)**3,x)*b**3*m + int(x**m*sin(c + d*x* 
*3)**3,x)*b**3 + 3*int(x**m*sin(c + d*x**3)**2,x)*a*b**2*m + 3*int(x**m*si 
n(c + d*x**3)**2,x)*a*b**2 + 3*int(x**m*sin(c + d*x**3),x)*a**2*b*m + 3*in 
t(x**m*sin(c + d*x**3),x)*a**2*b))/(m + 1)