\(\int (e x)^m (a+b \sin (c+d x^2))^3 \, dx\) [52]

Optimal result

Integrand size = 20, antiderivative size = 444 \[ \int (e x)^m \left (a+b \sin \left (c+d x^2\right )\right )^3 \, dx=\frac {a \left (2 a^2+3 b^2\right ) (e x)^{1+m}}{2 e (1+m)}+\frac {3 i b \left (4 a^2+b^2\right ) e^{i c} (e x)^{1+m} \left (-i d x^2\right )^{\frac {1}{2} (-1-m)} \Gamma \left (\frac {1+m}{2},-i d x^2\right )}{16 e}-\frac {3 i b \left (4 a^2+b^2\right ) e^{-i c} (e x)^{1+m} \left (i d x^2\right )^{\frac {1}{2} (-1-m)} \Gamma \left (\frac {1+m}{2},i d x^2\right )}{16 e}+\frac {3\ 2^{-\frac {7}{2}-\frac {m}{2}} a b^2 e^{2 i c} (e x)^{1+m} \left (-i d x^2\right )^{\frac {1}{2} (-1-m)} \Gamma \left (\frac {1+m}{2},-2 i d x^2\right )}{e}+\frac {3\ 2^{-\frac {7}{2}-\frac {m}{2}} a b^2 e^{-2 i c} (e x)^{1+m} \left (i d x^2\right )^{\frac {1}{2} (-1-m)} \Gamma \left (\frac {1+m}{2},2 i d x^2\right )}{e}-\frac {i 3^{-\frac {1}{2}-\frac {m}{2}} b^3 e^{3 i c} (e x)^{1+m} \left (-i d x^2\right )^{\frac {1}{2} (-1-m)} \Gamma \left (\frac {1+m}{2},-3 i d x^2\right )}{16 e}+\frac {i 3^{-\frac {1}{2}-\frac {m}{2}} b^3 e^{-3 i c} (e x)^{1+m} \left (i d x^2\right )^{\frac {1}{2} (-1-m)} \Gamma \left (\frac {1+m}{2},3 i d x^2\right )}{16 e} \] Output:

1/2*a*(2*a^2+3*b^2)*(e*x)^(1+m)/e/(1+m)+3/16*I*b*(4*a^2+b^2)*exp(I*c)*(e*x 
)^(1+m)*(-I*d*x^2)^(-1/2-1/2*m)*GAMMA(1/2+1/2*m,-I*d*x^2)/e-3/16*I*b*(4*a^ 
2+b^2)*(e*x)^(1+m)*(I*d*x^2)^(-1/2-1/2*m)*GAMMA(1/2+1/2*m,I*d*x^2)/e/exp(I 
*c)+3*2^(-7/2-1/2*m)*a*b^2*exp(2*I*c)*(e*x)^(1+m)*(-I*d*x^2)^(-1/2-1/2*m)* 
GAMMA(1/2+1/2*m,-2*I*d*x^2)/e+3*2^(-7/2-1/2*m)*a*b^2*(e*x)^(1+m)*(I*d*x^2) 
^(-1/2-1/2*m)*GAMMA(1/2+1/2*m,2*I*d*x^2)/e/exp(2*I*c)-1/16*I*3^(-1/2-1/2*m 
)*b^3*exp(3*I*c)*(e*x)^(1+m)*(-I*d*x^2)^(-1/2-1/2*m)*GAMMA(1/2+1/2*m,-3*I* 
d*x^2)/e+1/16*I*3^(-1/2-1/2*m)*b^3*(e*x)^(1+m)*(I*d*x^2)^(-1/2-1/2*m)*GAMM 
A(1/2+1/2*m,3*I*d*x^2)/e/exp(3*I*c)
 

Mathematica [A] (verified)

Time = 2.38 (sec) , antiderivative size = 373, normalized size of antiderivative = 0.84 \[ \int (e x)^m \left (a+b \sin \left (c+d x^2\right )\right )^3 \, dx=\frac {1}{16} i x (e x)^m \left (-\frac {8 i a \left (2 a^2+3 b^2\right )}{1+m}+3 b \left (4 a^2+b^2\right ) e^{i c} \left (-i d x^2\right )^{-\frac {1}{2}-\frac {m}{2}} \Gamma \left (\frac {1+m}{2},-i d x^2\right )-3 b \left (4 a^2+b^2\right ) e^{-i c} \left (i d x^2\right )^{-\frac {1}{2}-\frac {m}{2}} \Gamma \left (\frac {1+m}{2},i d x^2\right )-3 i 2^{\frac {1}{2}-\frac {m}{2}} a b^2 e^{2 i c} \left (-i d x^2\right )^{-\frac {1}{2}-\frac {m}{2}} \Gamma \left (\frac {1+m}{2},-2 i d x^2\right )-3 i 2^{\frac {1}{2}-\frac {m}{2}} a b^2 e^{-2 i c} \left (i d x^2\right )^{-\frac {1}{2}-\frac {m}{2}} \Gamma \left (\frac {1+m}{2},2 i d x^2\right )-3^{-\frac {1}{2}-\frac {m}{2}} b^3 e^{3 i c} \left (-i d x^2\right )^{-\frac {1}{2}-\frac {m}{2}} \Gamma \left (\frac {1+m}{2},-3 i d x^2\right )+3^{-\frac {1}{2}-\frac {m}{2}} b^3 e^{-3 i c} \left (i d x^2\right )^{-\frac {1}{2}-\frac {m}{2}} \Gamma \left (\frac {1+m}{2},3 i d x^2\right )\right ) \] Input:

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

Output:

(I/16)*x*(e*x)^m*(((-8*I)*a*(2*a^2 + 3*b^2))/(1 + m) + 3*b*(4*a^2 + b^2)*E 
^(I*c)*((-I)*d*x^2)^(-1/2 - m/2)*Gamma[(1 + m)/2, (-I)*d*x^2] - (3*b*(4*a^ 
2 + b^2)*(I*d*x^2)^(-1/2 - m/2)*Gamma[(1 + m)/2, I*d*x^2])/E^(I*c) - (3*I) 
*2^(1/2 - m/2)*a*b^2*E^((2*I)*c)*((-I)*d*x^2)^(-1/2 - m/2)*Gamma[(1 + m)/2 
, (-2*I)*d*x^2] - ((3*I)*2^(1/2 - m/2)*a*b^2*(I*d*x^2)^(-1/2 - m/2)*Gamma[ 
(1 + m)/2, (2*I)*d*x^2])/E^((2*I)*c) - 3^(-1/2 - m/2)*b^3*E^((3*I)*c)*((-I 
)*d*x^2)^(-1/2 - m/2)*Gamma[(1 + m)/2, (-3*I)*d*x^2] + (3^(-1/2 - m/2)*b^3 
*(I*d*x^2)^(-1/2 - m/2)*Gamma[(1 + m)/2, (3*I)*d*x^2])/E^((3*I)*c))
 

Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 444, 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.

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

\(\Big \downarrow \) 3884

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

\(\Big \downarrow \) 6

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

\(\Big \downarrow \) 6

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {3 i b e^{i c} \left (4 a^2+b^2\right ) \left (-i d x^2\right )^{\frac {1}{2} (-m-1)} (e x)^{m+1} \Gamma \left (\frac {m+1}{2},-i d x^2\right )}{16 e}-\frac {3 i b e^{-i c} \left (4 a^2+b^2\right ) \left (i d x^2\right )^{\frac {1}{2} (-m-1)} (e x)^{m+1} \Gamma \left (\frac {m+1}{2},i d x^2\right )}{16 e}+\frac {a \left (2 a^2+3 b^2\right ) (e x)^{m+1}}{2 e (m+1)}+\frac {3 a b^2 e^{2 i c} 2^{-\frac {m}{2}-\frac {7}{2}} \left (-i d x^2\right )^{\frac {1}{2} (-m-1)} (e x)^{m+1} \Gamma \left (\frac {m+1}{2},-2 i d x^2\right )}{e}+\frac {3 a b^2 e^{-2 i c} 2^{-\frac {m}{2}-\frac {7}{2}} \left (i d x^2\right )^{\frac {1}{2} (-m-1)} (e x)^{m+1} \Gamma \left (\frac {m+1}{2},2 i d x^2\right )}{e}-\frac {i b^3 e^{3 i c} 3^{-\frac {m}{2}-\frac {1}{2}} \left (-i d x^2\right )^{\frac {1}{2} (-m-1)} (e x)^{m+1} \Gamma \left (\frac {m+1}{2},-3 i d x^2\right )}{16 e}+\frac {i b^3 e^{-3 i c} 3^{-\frac {m}{2}-\frac {1}{2}} \left (i d x^2\right )^{\frac {1}{2} (-m-1)} (e x)^{m+1} \Gamma \left (\frac {m+1}{2},3 i d x^2\right )}{16 e}\)

Input:

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

Output:

(a*(2*a^2 + 3*b^2)*(e*x)^(1 + m))/(2*e*(1 + m)) + (((3*I)/16)*b*(4*a^2 + b 
^2)*E^(I*c)*(e*x)^(1 + m)*((-I)*d*x^2)^((-1 - m)/2)*Gamma[(1 + m)/2, (-I)* 
d*x^2])/e - (((3*I)/16)*b*(4*a^2 + b^2)*(e*x)^(1 + m)*(I*d*x^2)^((-1 - m)/ 
2)*Gamma[(1 + m)/2, I*d*x^2])/(e*E^(I*c)) + (3*2^(-7/2 - m/2)*a*b^2*E^((2* 
I)*c)*(e*x)^(1 + m)*((-I)*d*x^2)^((-1 - m)/2)*Gamma[(1 + m)/2, (-2*I)*d*x^ 
2])/e + (3*2^(-7/2 - m/2)*a*b^2*(e*x)^(1 + m)*(I*d*x^2)^((-1 - m)/2)*Gamma 
[(1 + m)/2, (2*I)*d*x^2])/(e*E^((2*I)*c)) - ((I/16)*3^(-1/2 - m/2)*b^3*E^( 
(3*I)*c)*(e*x)^(1 + m)*((-I)*d*x^2)^((-1 - m)/2)*Gamma[(1 + m)/2, (-3*I)*d 
*x^2])/e + ((I/16)*3^(-1/2 - m/2)*b^3*(e*x)^(1 + m)*(I*d*x^2)^((-1 - m)/2) 
*Gamma[(1 + m)/2, (3*I)*d*x^2])/(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]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 323, normalized size of antiderivative = 0.73 \[ \int (e x)^m \left (a+b \sin \left (c+d x^2\right )\right )^3 \, dx=\frac {24 \, {\left (2 \, a^{3} + 3 \, a b^{2}\right )} \left (e x\right )^{m} d x + {\left (b^{3} e m + b^{3} e\right )} e^{\left (-\frac {1}{2} \, {\left (m - 1\right )} \log \left (\frac {3 i \, d}{e^{2}}\right ) - 3 i \, c\right )} \Gamma \left (\frac {1}{2} \, m + \frac {1}{2}, 3 i \, d x^{2}\right ) - 9 \, {\left (i \, a b^{2} e m + i \, a b^{2} e\right )} e^{\left (-\frac {1}{2} \, {\left (m - 1\right )} \log \left (\frac {2 i \, d}{e^{2}}\right ) - 2 i \, c\right )} \Gamma \left (\frac {1}{2} \, m + \frac {1}{2}, 2 i \, d x^{2}\right ) - 9 \, {\left ({\left (4 \, a^{2} b + b^{3}\right )} e m + {\left (4 \, a^{2} b + b^{3}\right )} e\right )} e^{\left (-\frac {1}{2} \, {\left (m - 1\right )} \log \left (\frac {i \, d}{e^{2}}\right ) - i \, c\right )} \Gamma \left (\frac {1}{2} \, m + \frac {1}{2}, i \, d x^{2}\right ) - 9 \, {\left ({\left (4 \, a^{2} b + b^{3}\right )} e m + {\left (4 \, a^{2} b + b^{3}\right )} e\right )} e^{\left (-\frac {1}{2} \, {\left (m - 1\right )} \log \left (-\frac {i \, d}{e^{2}}\right ) + i \, c\right )} \Gamma \left (\frac {1}{2} \, m + \frac {1}{2}, -i \, d x^{2}\right ) - 9 \, {\left (-i \, a b^{2} e m - i \, a b^{2} e\right )} e^{\left (-\frac {1}{2} \, {\left (m - 1\right )} \log \left (-\frac {2 i \, d}{e^{2}}\right ) + 2 i \, c\right )} \Gamma \left (\frac {1}{2} \, m + \frac {1}{2}, -2 i \, d x^{2}\right ) + {\left (b^{3} e m + b^{3} e\right )} e^{\left (-\frac {1}{2} \, {\left (m - 1\right )} \log \left (-\frac {3 i \, d}{e^{2}}\right ) + 3 i \, c\right )} \Gamma \left (\frac {1}{2} \, m + \frac {1}{2}, -3 i \, d x^{2}\right )}{48 \, {\left (d m + d\right )}} \] Input:

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

Output:

1/48*(24*(2*a^3 + 3*a*b^2)*(e*x)^m*d*x + (b^3*e*m + b^3*e)*e^(-1/2*(m - 1) 
*log(3*I*d/e^2) - 3*I*c)*gamma(1/2*m + 1/2, 3*I*d*x^2) - 9*(I*a*b^2*e*m + 
I*a*b^2*e)*e^(-1/2*(m - 1)*log(2*I*d/e^2) - 2*I*c)*gamma(1/2*m + 1/2, 2*I* 
d*x^2) - 9*((4*a^2*b + b^3)*e*m + (4*a^2*b + b^3)*e)*e^(-1/2*(m - 1)*log(I 
*d/e^2) - I*c)*gamma(1/2*m + 1/2, I*d*x^2) - 9*((4*a^2*b + b^3)*e*m + (4*a 
^2*b + b^3)*e)*e^(-1/2*(m - 1)*log(-I*d/e^2) + I*c)*gamma(1/2*m + 1/2, -I* 
d*x^2) - 9*(-I*a*b^2*e*m - I*a*b^2*e)*e^(-1/2*(m - 1)*log(-2*I*d/e^2) + 2* 
I*c)*gamma(1/2*m + 1/2, -2*I*d*x^2) + (b^3*e*m + b^3*e)*e^(-1/2*(m - 1)*lo 
g(-3*I*d/e^2) + 3*I*c)*gamma(1/2*m + 1/2, -3*I*d*x^2))/(d*m + d)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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