3.1.53 \(\int (e x)^m (a+b \sin (c+d x^2))^2 \, dx\) [53]

3.1.53.1 Optimal result

Integrand size = 20, antiderivative size = 279 \[ \int (e x)^m \left (a+b \sin \left (c+d x^2\right )\right )^2 \, dx=\frac {\left (2 a^2+b^2\right ) (e x)^{1+m}}{2 e (1+m)}+\frac {i a b e^{i c} (e x)^{1+m} \left (-i d x^2\right )^{\frac {1}{2} (-1-m)} \Gamma \left (\frac {1+m}{2},-i d x^2\right )}{2 e}-\frac {i a b e^{-i c} (e x)^{1+m} \left (i d x^2\right )^{\frac {1}{2} (-1-m)} \Gamma \left (\frac {1+m}{2},i d x^2\right )}{2 e}+\frac {2^{-\frac {7}{2}-\frac {m}{2}} b^2 e^{2 i c} (e x)^{1+m} \left (-i d x^2\right )^{\frac {1}{2} (-1-m)} \Gamma \left (\frac {1+m}{2},-2 i d x^2\right )}{e}+\frac {2^{-\frac {7}{2}-\frac {m}{2}} b^2 e^{-2 i c} (e x)^{1+m} \left (i d x^2\right )^{\frac {1}{2} (-1-m)} \Gamma \left (\frac {1+m}{2},2 i d x^2\right )}{e} \]

output

1/2*(2*a^2+b^2)*(e*x)^(1+m)/e/(1+m)+1/2*I*a*b*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-1/2*I*a*b*(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)+2^(-7/2-1/2*m)*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+ 
2^(-7/2-1/2*m)*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)
 

3.1.53.2 Mathematica [A] (verified)

Time = 1.90 (sec) , antiderivative size = 551, normalized size of antiderivative = 1.97 \[ \int (e x)^m \left (a+b \sin \left (c+d x^2\right )\right )^2 \, dx=\frac {2^{\frac {1}{2} (-7-m)} x (e x)^m \left (d^2 x^4\right )^{\frac {1}{2} (-1-m)} \left (2^{\frac {7+m}{2}} a^2 \left (d^2 x^4\right )^{\frac {1+m}{2}}+2^{\frac {5+m}{2}} b^2 \left (d^2 x^4\right )^{\frac {1+m}{2}}+b^2 \left (i d x^2\right )^{\frac {1+m}{2}} \cos (2 c) \Gamma \left (\frac {1+m}{2},-2 i d x^2\right )+b^2 m \left (i d x^2\right )^{\frac {1+m}{2}} \cos (2 c) \Gamma \left (\frac {1+m}{2},-2 i d x^2\right )+b^2 \left (-i d x^2\right )^{\frac {1+m}{2}} \cos (2 c) \Gamma \left (\frac {1+m}{2},2 i d x^2\right )+b^2 m \left (-i d x^2\right )^{\frac {1+m}{2}} \cos (2 c) \Gamma \left (\frac {1+m}{2},2 i d x^2\right )-i 2^{\frac {5+m}{2}} a b (1+m) \left (-i d x^2\right )^{\frac {1+m}{2}} \Gamma \left (\frac {1+m}{2},i d x^2\right ) (\cos (c)-i \sin (c))+i 2^{\frac {5+m}{2}} a b (1+m) \left (i d x^2\right )^{\frac {1+m}{2}} \Gamma \left (\frac {1+m}{2},-i d x^2\right ) (\cos (c)+i \sin (c))+i b^2 \left (i d x^2\right )^{\frac {1+m}{2}} \Gamma \left (\frac {1+m}{2},-2 i d x^2\right ) \sin (2 c)+i b^2 m \left (i d x^2\right )^{\frac {1+m}{2}} \Gamma \left (\frac {1+m}{2},-2 i d x^2\right ) \sin (2 c)-i b^2 \left (-i d x^2\right )^{\frac {1+m}{2}} \Gamma \left (\frac {1+m}{2},2 i d x^2\right ) \sin (2 c)-i b^2 m \left (-i d x^2\right )^{\frac {1+m}{2}} \Gamma \left (\frac {1+m}{2},2 i d x^2\right ) \sin (2 c)\right )}{1+m} \]

input

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

output

(2^((-7 - m)/2)*x*(e*x)^m*(d^2*x^4)^((-1 - m)/2)*(2^((7 + m)/2)*a^2*(d^2*x 
^4)^((1 + m)/2) + 2^((5 + m)/2)*b^2*(d^2*x^4)^((1 + m)/2) + b^2*(I*d*x^2)^ 
((1 + m)/2)*Cos[2*c]*Gamma[(1 + m)/2, (-2*I)*d*x^2] + b^2*m*(I*d*x^2)^((1 
+ m)/2)*Cos[2*c]*Gamma[(1 + m)/2, (-2*I)*d*x^2] + b^2*((-I)*d*x^2)^((1 + m 
)/2)*Cos[2*c]*Gamma[(1 + m)/2, (2*I)*d*x^2] + b^2*m*((-I)*d*x^2)^((1 + m)/ 
2)*Cos[2*c]*Gamma[(1 + m)/2, (2*I)*d*x^2] - I*2^((5 + m)/2)*a*b*(1 + m)*(( 
-I)*d*x^2)^((1 + m)/2)*Gamma[(1 + m)/2, I*d*x^2]*(Cos[c] - I*Sin[c]) + I*2 
^((5 + m)/2)*a*b*(1 + m)*(I*d*x^2)^((1 + m)/2)*Gamma[(1 + m)/2, (-I)*d*x^2 
]*(Cos[c] + I*Sin[c]) + I*b^2*(I*d*x^2)^((1 + m)/2)*Gamma[(1 + m)/2, (-2*I 
)*d*x^2]*Sin[2*c] + I*b^2*m*(I*d*x^2)^((1 + m)/2)*Gamma[(1 + m)/2, (-2*I)* 
d*x^2]*Sin[2*c] - I*b^2*((-I)*d*x^2)^((1 + m)/2)*Gamma[(1 + m)/2, (2*I)*d* 
x^2]*Sin[2*c] - I*b^2*m*((-I)*d*x^2)^((1 + m)/2)*Gamma[(1 + m)/2, (2*I)*d* 
x^2]*Sin[2*c]))/(1 + m)
 

3.1.53.3 Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 279, 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.

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

\(\Big \downarrow \) 3884

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

\(\Big \downarrow \) 6

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

\(\Big \downarrow \) 2009

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

input

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

output

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

3.1.53.3.1 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]
 

3.1.53.4 Maple [F]

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

input

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

output

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

3.1.53.5 Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.71 \[ \int (e x)^m \left (a+b \sin \left (c+d x^2\right )\right )^2 \, dx=\frac {8 \, {\left (2 \, a^{2} + b^{2}\right )} \left (e x\right )^{m} d x + {\left (-i \, b^{2} e m - i \, b^{2} e\right )} e^{\left (-\frac {1}{2} \, {\left (m - 1\right )} \log \left (\frac {2 i \, d}{e^{2}}\right ) - 2 i \, c\right )} \Gamma \left (\frac {1}{2} \, m + \frac {1}{2}, 2 i \, d x^{2}\right ) - 8 \, {\left (a b e m + a b e\right )} e^{\left (-\frac {1}{2} \, {\left (m - 1\right )} \log \left (\frac {i \, d}{e^{2}}\right ) - i \, c\right )} \Gamma \left (\frac {1}{2} \, m + \frac {1}{2}, i \, d x^{2}\right ) - 8 \, {\left (a b e m + a b e\right )} e^{\left (-\frac {1}{2} \, {\left (m - 1\right )} \log \left (-\frac {i \, d}{e^{2}}\right ) + i \, c\right )} \Gamma \left (\frac {1}{2} \, m + \frac {1}{2}, -i \, d x^{2}\right ) + {\left (i \, b^{2} e m + i \, b^{2} e\right )} e^{\left (-\frac {1}{2} \, {\left (m - 1\right )} \log \left (-\frac {2 i \, d}{e^{2}}\right ) + 2 i \, c\right )} \Gamma \left (\frac {1}{2} \, m + \frac {1}{2}, -2 i \, d x^{2}\right )}{16 \, {\left (d m + d\right )}} \]

input

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

output

1/16*(8*(2*a^2 + b^2)*(e*x)^m*d*x + (-I*b^2*e*m - I*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) - 8*(a*b*e*m + a*b* 
e)*e^(-1/2*(m - 1)*log(I*d/e^2) - I*c)*gamma(1/2*m + 1/2, I*d*x^2) - 8*(a* 
b*e*m + a*b*e)*e^(-1/2*(m - 1)*log(-I*d/e^2) + I*c)*gamma(1/2*m + 1/2, -I* 
d*x^2) + (I*b^2*e*m + I*b^2*e)*e^(-1/2*(m - 1)*log(-2*I*d/e^2) + 2*I*c)*ga 
mma(1/2*m + 1/2, -2*I*d*x^2))/(d*m + d)
 

3.1.53.6 Sympy [F]

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

input

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

output

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

3.1.53.7 Maxima [F]

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

input

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

3.1.53.8 Giac [F]

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

input

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

output

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

3.1.53.9 Mupad [F(-1)]

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

input

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

output

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