\(\int x^4 (a+b \sin (c+d x^2)) \, dx\) [7]

Optimal result

Integrand size = 16, antiderivative size = 121 \[ \int x^4 \left (a+b \sin \left (c+d x^2\right )\right ) \, dx=\frac {a x^5}{5}-\frac {b x^3 \cos \left (c+d x^2\right )}{2 d}-\frac {3 b \sqrt {\frac {\pi }{2}} \cos (c) \operatorname {FresnelS}\left (\sqrt {d} \sqrt {\frac {2}{\pi }} x\right )}{4 d^{5/2}}-\frac {3 b \sqrt {\frac {\pi }{2}} \operatorname {FresnelC}\left (\sqrt {d} \sqrt {\frac {2}{\pi }} x\right ) \sin (c)}{4 d^{5/2}}+\frac {3 b x \sin \left (c+d x^2\right )}{4 d^2} \] Output:

1/5*a*x^5-1/2*b*x^3*cos(d*x^2+c)/d-3/8*b*2^(1/2)*Pi^(1/2)*cos(c)*FresnelS( 
d^(1/2)*2^(1/2)/Pi^(1/2)*x)/d^(5/2)-3/8*b*2^(1/2)*Pi^(1/2)*FresnelC(d^(1/2 
)*2^(1/2)/Pi^(1/2)*x)*sin(c)/d^(5/2)+3/4*b*x*sin(d*x^2+c)/d^2
 

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.03 \[ \int x^4 \left (a+b \sin \left (c+d x^2\right )\right ) \, dx=\frac {a x^5}{5}-\frac {b x \cos \left (d x^2\right ) \left (2 d x^2 \cos (c)-3 \sin (c)\right )}{4 d^2}-\frac {3 b \sqrt {\frac {\pi }{2}} \left (\cos (c) \operatorname {FresnelS}\left (\sqrt {d} \sqrt {\frac {2}{\pi }} x\right )+\operatorname {FresnelC}\left (\sqrt {d} \sqrt {\frac {2}{\pi }} x\right ) \sin (c)\right )}{4 d^{5/2}}+\frac {b x \left (3 \cos (c)+2 d x^2 \sin (c)\right ) \sin \left (d x^2\right )}{4 d^2} \] Input:

Integrate[x^4*(a + b*Sin[c + d*x^2]),x]
 

Output:

(a*x^5)/5 - (b*x*Cos[d*x^2]*(2*d*x^2*Cos[c] - 3*Sin[c]))/(4*d^2) - (3*b*Sq 
rt[Pi/2]*(Cos[c]*FresnelS[Sqrt[d]*Sqrt[2/Pi]*x] + FresnelC[Sqrt[d]*Sqrt[2/ 
Pi]*x]*Sin[c]))/(4*d^(5/2)) + (b*x*(3*Cos[c] + 2*d*x^2*Sin[c])*Sin[d*x^2]) 
/(4*d^2)
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2010, 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.

\(\displaystyle \int x^4 \left (a+b \sin \left (c+d x^2\right )\right ) \, dx\)

\(\Big \downarrow \) 2010

\(\displaystyle \int \left (a x^4+b x^4 \sin \left (c+d x^2\right )\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {a x^5}{5}-\frac {3 \sqrt {\frac {\pi }{2}} b \sin (c) \operatorname {FresnelC}\left (\sqrt {d} \sqrt {\frac {2}{\pi }} x\right )}{4 d^{5/2}}-\frac {3 \sqrt {\frac {\pi }{2}} b \cos (c) \operatorname {FresnelS}\left (\sqrt {d} \sqrt {\frac {2}{\pi }} x\right )}{4 d^{5/2}}+\frac {3 b x \sin \left (c+d x^2\right )}{4 d^2}-\frac {b x^3 \cos \left (c+d x^2\right )}{2 d}\)

Input:

Int[x^4*(a + b*Sin[c + d*x^2]),x]
 

Output:

(a*x^5)/5 - (b*x^3*Cos[c + d*x^2])/(2*d) - (3*b*Sqrt[Pi/2]*Cos[c]*FresnelS 
[Sqrt[d]*Sqrt[2/Pi]*x])/(4*d^(5/2)) - (3*b*Sqrt[Pi/2]*FresnelC[Sqrt[d]*Sqr 
t[2/Pi]*x]*Sin[c])/(4*d^(5/2)) + (3*b*x*Sin[c + d*x^2])/(4*d^2)
 

Defintions of rubi rules used

rule 2009

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

rule 2010

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] 
, x] /; FreeQ[{c, m}, x] && SumQ[u] &&  !LinearQ[u, x] &&  !MatchQ[u, (a_) 
+ (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
 

Maple [A] (verified)

Time = 0.77 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.74

method	result	size
default	\(\frac {a \,x^{5}}{5}+b \left (-\frac {x^{3} \cos \left (d \,x^{2}+c \right )}{2 d}+\frac {\frac {3 x \sin \left (d \,x^{2}+c \right )}{4 d}-\frac {3 \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (c \right ) \operatorname {FresnelS}\left (\frac {\sqrt {d}\, \sqrt {2}\, x}{\sqrt {\pi }}\right )+\sin \left (c \right ) \operatorname {FresnelC}\left (\frac {\sqrt {d}\, \sqrt {2}\, x}{\sqrt {\pi }}\right )\right )}{8 d^{\frac {3}{2}}}}{d}\right )\)	\(89\)
parts	\(\frac {a \,x^{5}}{5}+b \left (-\frac {x^{3} \cos \left (d \,x^{2}+c \right )}{2 d}+\frac {\frac {3 x \sin \left (d \,x^{2}+c \right )}{4 d}-\frac {3 \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (c \right ) \operatorname {FresnelS}\left (\frac {\sqrt {d}\, \sqrt {2}\, x}{\sqrt {\pi }}\right )+\sin \left (c \right ) \operatorname {FresnelC}\left (\frac {\sqrt {d}\, \sqrt {2}\, x}{\sqrt {\pi }}\right )\right )}{8 d^{\frac {3}{2}}}}{d}\right )\)	\(89\)
risch	\(\frac {a \,x^{5}}{5}-\frac {3 i b \sqrt {\pi }\, \operatorname {erf}\left (\sqrt {i d}\, x \right ) {\mathrm e}^{-i c}}{16 d^{2} \sqrt {i d}}+\frac {3 i b \sqrt {\pi }\, \operatorname {erf}\left (\sqrt {-i d}\, x \right ) {\mathrm e}^{i c}}{16 d^{2} \sqrt {-i d}}-\frac {b \,x^{3} \cos \left (d \,x^{2}+c \right )}{2 d}+\frac {3 b x \sin \left (d \,x^{2}+c \right )}{4 d^{2}}\)	\(100\)

Input:

int(x^4*(a+b*sin(d*x^2+c)),x,method=_RETURNVERBOSE)
 

Output:

1/5*a*x^5+b*(-1/2/d*x^3*cos(d*x^2+c)+3/2/d*(1/2/d*x*sin(d*x^2+c)-1/4/d^(3/ 
2)*2^(1/2)*Pi^(1/2)*(cos(c)*FresnelS(d^(1/2)*2^(1/2)/Pi^(1/2)*x)+sin(c)*Fr 
esnelC(d^(1/2)*2^(1/2)/Pi^(1/2)*x))))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.85 \[ \int x^4 \left (a+b \sin \left (c+d x^2\right )\right ) \, dx=\frac {8 \, a d^{3} x^{5} - 20 \, b d^{2} x^{3} \cos \left (d x^{2} + c\right ) - 15 \, \sqrt {2} \pi b \sqrt {\frac {d}{\pi }} \cos \left (c\right ) \operatorname {S}\left (\sqrt {2} x \sqrt {\frac {d}{\pi }}\right ) - 15 \, \sqrt {2} \pi b \sqrt {\frac {d}{\pi }} \operatorname {C}\left (\sqrt {2} x \sqrt {\frac {d}{\pi }}\right ) \sin \left (c\right ) + 30 \, b d x \sin \left (d x^{2} + c\right )}{40 \, d^{3}} \] Input:

integrate(x^4*(a+b*sin(d*x^2+c)),x, algorithm="fricas")
 

Output:

1/40*(8*a*d^3*x^5 - 20*b*d^2*x^3*cos(d*x^2 + c) - 15*sqrt(2)*pi*b*sqrt(d/p 
i)*cos(c)*fresnel_sin(sqrt(2)*x*sqrt(d/pi)) - 15*sqrt(2)*pi*b*sqrt(d/pi)*f 
resnel_cos(sqrt(2)*x*sqrt(d/pi))*sin(c) + 30*b*d*x*sin(d*x^2 + c))/d^3
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 488 vs. \(2 (126) = 252\).

Time = 2.29 (sec) , antiderivative size = 488, normalized size of antiderivative = 4.03 \[ \int x^4 \left (a+b \sin \left (c+d x^2\right )\right ) \, dx=\frac {a x^{5}}{5} - \frac {5 \sqrt {2} \sqrt {\pi } b x^{4} \sqrt {\frac {1}{d}} \sin {\left (c \right )} C\left (\frac {\sqrt {2} \sqrt {d} x}{\sqrt {\pi }}\right ) \Gamma \left (\frac {1}{4}\right )}{32 \Gamma \left (\frac {9}{4}\right )} + \frac {\sqrt {2} \sqrt {\pi } b x^{4} \sqrt {\frac {1}{d}} \sin {\left (c \right )} C\left (\frac {\sqrt {2} \sqrt {d} x}{\sqrt {\pi }}\right )}{2} - \frac {21 \sqrt {2} \sqrt {\pi } b x^{4} \sqrt {\frac {1}{d}} \cos {\left (c \right )} S\left (\frac {\sqrt {2} \sqrt {d} x}{\sqrt {\pi }}\right ) \Gamma \left (\frac {3}{4}\right )}{32 \Gamma \left (\frac {11}{4}\right )} + \frac {\sqrt {2} \sqrt {\pi } b x^{4} \sqrt {\frac {1}{d}} \cos {\left (c \right )} S\left (\frac {\sqrt {2} \sqrt {d} x}{\sqrt {\pi }}\right )}{2} - \frac {15 \sqrt {2} \sqrt {\pi } b \sqrt {\frac {1}{d}} \sin {\left (c \right )} C\left (\frac {\sqrt {2} \sqrt {d} x}{\sqrt {\pi }}\right ) \Gamma \left (\frac {1}{4}\right )}{128 d^{2} \Gamma \left (\frac {9}{4}\right )} - \frac {63 \sqrt {2} \sqrt {\pi } b \sqrt {\frac {1}{d}} \cos {\left (c \right )} S\left (\frac {\sqrt {2} \sqrt {d} x}{\sqrt {\pi }}\right ) \Gamma \left (\frac {3}{4}\right )}{128 d^{2} \Gamma \left (\frac {11}{4}\right )} + \frac {5 b x^{3} \sqrt {\frac {1}{d}} \sin {\left (c \right )} \sin {\left (d x^{2} \right )} \Gamma \left (\frac {1}{4}\right )}{32 \sqrt {d} \Gamma \left (\frac {9}{4}\right )} - \frac {21 b x^{3} \sqrt {\frac {1}{d}} \cos {\left (c \right )} \cos {\left (d x^{2} \right )} \Gamma \left (\frac {3}{4}\right )}{32 \sqrt {d} \Gamma \left (\frac {11}{4}\right )} + \frac {15 b x \sqrt {\frac {1}{d}} \sin {\left (c \right )} \cos {\left (d x^{2} \right )} \Gamma \left (\frac {1}{4}\right )}{64 d^{\frac {3}{2}} \Gamma \left (\frac {9}{4}\right )} + \frac {63 b x \sqrt {\frac {1}{d}} \sin {\left (d x^{2} \right )} \cos {\left (c \right )} \Gamma \left (\frac {3}{4}\right )}{64 d^{\frac {3}{2}} \Gamma \left (\frac {11}{4}\right )} \] Input:

integrate(x**4*(a+b*sin(d*x**2+c)),x)
 

Output:

a*x**5/5 - 5*sqrt(2)*sqrt(pi)*b*x**4*sqrt(1/d)*sin(c)*fresnelc(sqrt(2)*sqr 
t(d)*x/sqrt(pi))*gamma(1/4)/(32*gamma(9/4)) + sqrt(2)*sqrt(pi)*b*x**4*sqrt 
(1/d)*sin(c)*fresnelc(sqrt(2)*sqrt(d)*x/sqrt(pi))/2 - 21*sqrt(2)*sqrt(pi)* 
b*x**4*sqrt(1/d)*cos(c)*fresnels(sqrt(2)*sqrt(d)*x/sqrt(pi))*gamma(3/4)/(3 
2*gamma(11/4)) + sqrt(2)*sqrt(pi)*b*x**4*sqrt(1/d)*cos(c)*fresnels(sqrt(2) 
*sqrt(d)*x/sqrt(pi))/2 - 15*sqrt(2)*sqrt(pi)*b*sqrt(1/d)*sin(c)*fresnelc(s 
qrt(2)*sqrt(d)*x/sqrt(pi))*gamma(1/4)/(128*d**2*gamma(9/4)) - 63*sqrt(2)*s 
qrt(pi)*b*sqrt(1/d)*cos(c)*fresnels(sqrt(2)*sqrt(d)*x/sqrt(pi))*gamma(3/4) 
/(128*d**2*gamma(11/4)) + 5*b*x**3*sqrt(1/d)*sin(c)*sin(d*x**2)*gamma(1/4) 
/(32*sqrt(d)*gamma(9/4)) - 21*b*x**3*sqrt(1/d)*cos(c)*cos(d*x**2)*gamma(3/ 
4)/(32*sqrt(d)*gamma(11/4)) + 15*b*x*sqrt(1/d)*sin(c)*cos(d*x**2)*gamma(1/ 
4)/(64*d**(3/2)*gamma(9/4)) + 63*b*x*sqrt(1/d)*sin(d*x**2)*cos(c)*gamma(3/ 
4)/(64*d**(3/2)*gamma(11/4))
 

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.06 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.76 \[ \int x^4 \left (a+b \sin \left (c+d x^2\right )\right ) \, dx=\frac {1}{5} \, a x^{5} - \frac {{\left (16 \, d^{3} x^{3} \cos \left (d x^{2} + c\right ) - 24 \, d^{2} x \sin \left (d x^{2} + c\right ) + 3 \, \sqrt {2} \sqrt {\pi } {\left ({\left (\left (i + 1\right ) \, \cos \left (c\right ) - \left (i - 1\right ) \, \sin \left (c\right )\right )} \operatorname {erf}\left (\sqrt {i \, d} x\right ) + {\left (-\left (i - 1\right ) \, \cos \left (c\right ) + \left (i + 1\right ) \, \sin \left (c\right )\right )} \operatorname {erf}\left (\sqrt {-i \, d} x\right )\right )} d^{\frac {3}{2}}\right )} b}{32 \, d^{4}} \] Input:

integrate(x^4*(a+b*sin(d*x^2+c)),x, algorithm="maxima")
 

Output:

1/5*a*x^5 - 1/32*(16*d^3*x^3*cos(d*x^2 + c) - 24*d^2*x*sin(d*x^2 + c) + 3* 
sqrt(2)*sqrt(pi)*(((I + 1)*cos(c) - (I - 1)*sin(c))*erf(sqrt(I*d)*x) + (-( 
I - 1)*cos(c) + (I + 1)*sin(c))*erf(sqrt(-I*d)*x))*d^(3/2))*b/d^4
 

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.12 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.36 \[ \int x^4 \left (a+b \sin \left (c+d x^2\right )\right ) \, dx=\frac {1}{5} \, a x^{5} - \frac {3 i \, \sqrt {2} \sqrt {\pi } b \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {2} x {\left (-\frac {i \, d}{{\left | d \right |}} + 1\right )} \sqrt {{\left | d \right |}}\right ) e^{\left (i \, c\right )}}{16 \, d^{2} {\left (-\frac {i \, d}{{\left | d \right |}} + 1\right )} \sqrt {{\left | d \right |}}} + \frac {3 i \, \sqrt {2} \sqrt {\pi } b \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {2} x {\left (\frac {i \, d}{{\left | d \right |}} + 1\right )} \sqrt {{\left | d \right |}}\right ) e^{\left (-i \, c\right )}}{16 \, d^{2} {\left (\frac {i \, d}{{\left | d \right |}} + 1\right )} \sqrt {{\left | d \right |}}} + \frac {i \, {\left (2 i \, b d x^{3} - 3 \, b x\right )} e^{\left (i \, d x^{2} + i \, c\right )}}{8 \, d^{2}} + \frac {i \, {\left (2 i \, b d x^{3} + 3 \, b x\right )} e^{\left (-i \, d x^{2} - i \, c\right )}}{8 \, d^{2}} \] Input:

integrate(x^4*(a+b*sin(d*x^2+c)),x, algorithm="giac")
 

Output:

1/5*a*x^5 - 3/16*I*sqrt(2)*sqrt(pi)*b*erf(-1/2*sqrt(2)*x*(-I*d/abs(d) + 1) 
*sqrt(abs(d)))*e^(I*c)/(d^2*(-I*d/abs(d) + 1)*sqrt(abs(d))) + 3/16*I*sqrt( 
2)*sqrt(pi)*b*erf(-1/2*sqrt(2)*x*(I*d/abs(d) + 1)*sqrt(abs(d)))*e^(-I*c)/( 
d^2*(I*d/abs(d) + 1)*sqrt(abs(d))) + 1/8*I*(2*I*b*d*x^3 - 3*b*x)*e^(I*d*x^ 
2 + I*c)/d^2 + 1/8*I*(2*I*b*d*x^3 + 3*b*x)*e^(-I*d*x^2 - I*c)/d^2
 

Mupad [F(-1)]

Timed out. \[ \int x^4 \left (a+b \sin \left (c+d x^2\right )\right ) \, dx=\int x^4\,\left (a+b\,\sin \left (d\,x^2+c\right )\right ) \,d x \] Input:

int(x^4*(a + b*sin(c + d*x^2)),x)
 

Output:

int(x^4*(a + b*sin(c + d*x^2)), x)
 

Reduce [F]

\[ \int x^4 \left (a+b \sin \left (c+d x^2\right )\right ) \, dx=\frac {-10 \cos \left (d \,x^{2}+c \right ) b d \,x^{3}-15 \left (\int \sin \left (d \,x^{2}+c \right )d x \right ) b +15 \sin \left (d \,x^{2}+c \right ) b x +4 a \,d^{2} x^{5}}{20 d^{2}} \] Input:

int(x^4*(a+b*sin(d*x^2+c)),x)
 

Output:

( - 10*cos(c + d*x**2)*b*d*x**3 - 15*int(sin(c + d*x**2),x)*b + 15*sin(c + 
 d*x**2)*b*x + 4*a*d**2*x**5)/(20*d**2)