3.1.12 \(\int x^5 (a+b \sin (c+d x^2))^2 \, dx\) [12]

3.1.12.1 Optimal result

Integrand size = 18, antiderivative size = 163 \[ \int x^5 \left (a+b \sin \left (c+d x^2\right )\right )^2 \, dx=-\frac {b^2 x^2}{8 d^2}+\frac {a^2 x^6}{6}+\frac {b^2 x^6}{12}+\frac {2 a b \cos \left (c+d x^2\right )}{d^3}-\frac {a b x^4 \cos \left (c+d x^2\right )}{d}+\frac {2 a b x^2 \sin \left (c+d x^2\right )}{d^2}+\frac {b^2 \cos \left (c+d x^2\right ) \sin \left (c+d x^2\right )}{8 d^3}-\frac {b^2 x^4 \cos \left (c+d x^2\right ) \sin \left (c+d x^2\right )}{4 d}+\frac {b^2 x^2 \sin ^2\left (c+d x^2\right )}{4 d^2} \]

output

-1/8*b^2*x^2/d^2+1/6*a^2*x^6+1/12*b^2*x^6+2*a*b*cos(d*x^2+c)/d^3-a*b*x^4*c 
os(d*x^2+c)/d+2*a*b*x^2*sin(d*x^2+c)/d^2+1/8*b^2*cos(d*x^2+c)*sin(d*x^2+c) 
/d^3-1/4*b^2*x^4*cos(d*x^2+c)*sin(d*x^2+c)/d+1/4*b^2*x^2*sin(d*x^2+c)^2/d^ 
2
 

3.1.12.2 Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.75 \[ \int x^5 \left (a+b \sin \left (c+d x^2\right )\right )^2 \, dx=\frac {8 a^2 d^3 x^6+4 b^2 d^3 x^6-48 a b \left (-2+d^2 x^4\right ) \cos \left (c+d x^2\right )-6 b^2 d x^2 \cos \left (2 \left (c+d x^2\right )\right )+96 a b d x^2 \sin \left (c+d x^2\right )+3 b^2 \sin \left (2 \left (c+d x^2\right )\right )-6 b^2 d^2 x^4 \sin \left (2 \left (c+d x^2\right )\right )}{48 d^3} \]

input

Integrate[x^5*(a + b*Sin[c + d*x^2])^2,x]
 

output

(8*a^2*d^3*x^6 + 4*b^2*d^3*x^6 - 48*a*b*(-2 + d^2*x^4)*Cos[c + d*x^2] - 6* 
b^2*d*x^2*Cos[2*(c + d*x^2)] + 96*a*b*d*x^2*Sin[c + d*x^2] + 3*b^2*Sin[2*( 
c + d*x^2)] - 6*b^2*d^2*x^4*Sin[2*(c + d*x^2)])/(48*d^3)
 

3.1.12.3 Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.02, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {3860, 3042, 3798, 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.

\(\displaystyle \int x^5 \left (a+b \sin \left (c+d x^2\right )\right )^2 \, dx\)

\(\Big \downarrow \) 3860

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3798

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} \left (\frac {a^2 x^6}{3}+\frac {4 a b \cos \left (c+d x^2\right )}{d^3}+\frac {4 a b x^2 \sin \left (c+d x^2\right )}{d^2}-\frac {2 a b x^4 \cos \left (c+d x^2\right )}{d}+\frac {b^2 \sin \left (c+d x^2\right ) \cos \left (c+d x^2\right )}{4 d^3}+\frac {b^2 x^2 \sin ^2\left (c+d x^2\right )}{2 d^2}-\frac {b^2 x^4 \sin \left (c+d x^2\right ) \cos \left (c+d x^2\right )}{2 d}-\frac {b^2 x^2}{4 d^2}+\frac {b^2 x^6}{6}\right )\)

input

Int[x^5*(a + b*Sin[c + d*x^2])^2,x]
 

output

(-1/4*(b^2*x^2)/d^2 + (a^2*x^6)/3 + (b^2*x^6)/6 + (4*a*b*Cos[c + d*x^2])/d 
^3 - (2*a*b*x^4*Cos[c + d*x^2])/d + (4*a*b*x^2*Sin[c + d*x^2])/d^2 + (b^2* 
Cos[c + d*x^2]*Sin[c + d*x^2])/(4*d^3) - (b^2*x^4*Cos[c + d*x^2]*Sin[c + d 
*x^2])/(2*d) + (b^2*x^2*Sin[c + d*x^2]^2)/(2*d^2))/2
 

3.1.12.3.1 Defintions of rubi rules used

rule 2009

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

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3798

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.) 
, x_Symbol] :> Int[ExpandIntegrand[(c + d*x)^m, (a + b*Sin[e + f*x])^n, x], 
 x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[n, 0] && (EqQ[n, 1] || IGtQ[ 
m, 0] || NeQ[a^2 - b^2, 0])
 

rule 3860

Int[(x_)^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol 
] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*Sin[c + d*x])^ 
p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IntegerQ[Simplify[ 
(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[ 
(m + 1)/n], 0]))
 

3.1.12.4 Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.68

method	result	size
risch	\(\frac {a^{2} x^{6}}{6}+\frac {b^{2} x^{6}}{12}-\frac {a b \left (x^{4} d^{2}-2\right ) \cos \left (d \,x^{2}+c \right )}{d^{3}}+\frac {2 a b \,x^{2} \sin \left (d \,x^{2}+c \right )}{d^{2}}-\frac {b^{2} x^{2} \cos \left (2 d \,x^{2}+2 c \right )}{8 d^{2}}-\frac {b^{2} \left (2 x^{4} d^{2}-1\right ) \sin \left (2 d \,x^{2}+2 c \right )}{16 d^{3}}\)	\(111\)
parallelrisch	\(\frac {\left (-6 x^{4} d^{2}+3\right ) b^{2} \sin \left (2 d \,x^{2}+2 c \right )-6 b^{2} x^{2} \cos \left (2 d \,x^{2}+2 c \right ) d -48 a b \left (x^{4} d^{2}-2\right ) \cos \left (d \,x^{2}+c \right )+8 a^{2} d^{3} x^{6}+4 b^{2} d^{3} x^{6}+96 a b \,x^{2} \sin \left (d \,x^{2}+c \right ) d +96 a b}{48 d^{3}}\)	\(115\)
parts	\(\frac {a^{2} x^{6}}{6}+b^{2} \left (\frac {x^{6}}{12}-\frac {x^{4} \sin \left (2 d \,x^{2}+2 c \right )}{8 d}+\frac {-\frac {x^{2} \cos \left (2 d \,x^{2}+2 c \right )}{4 d}+\frac {\sin \left (2 d \,x^{2}+2 c \right )}{8 d^{2}}}{2 d}\right )+2 a b \left (-\frac {x^{4} \cos \left (d \,x^{2}+c \right )}{2 d}+\frac {\frac {x^{2} \sin \left (d \,x^{2}+c \right )}{d}+\frac {\cos \left (d \,x^{2}+c \right )}{d^{2}}}{d}\right )\)	\(136\)
default	\(\frac {\left (a^{2}+\frac {b^{2}}{2}\right ) x^{6}}{6}-\frac {b^{2} \left (\frac {x^{4} \sin \left (2 d \,x^{2}+2 c \right )}{4 d}-\frac {-\frac {x^{2} \cos \left (2 d \,x^{2}+2 c \right )}{4 d}+\frac {\sin \left (2 d \,x^{2}+2 c \right )}{8 d^{2}}}{d}\right )}{2}+2 a b \left (-\frac {x^{4} \cos \left (d \,x^{2}+c \right )}{2 d}+\frac {\frac {x^{2} \sin \left (d \,x^{2}+c \right )}{d}+\frac {\cos \left (d \,x^{2}+c \right )}{d^{2}}}{d}\right )\)	\(138\)
norman	\(\frac {\left (\frac {a^{2}}{6}+\frac {b^{2}}{12}\right ) x^{6}+\left (\frac {a^{2}}{3}+\frac {b^{2}}{6}\right ) x^{6} \left (\tan ^{2}\left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )\right )+\left (\frac {a^{2}}{6}+\frac {b^{2}}{12}\right ) x^{6} \left (\tan ^{4}\left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )\right )+\frac {a b \,x^{4} \left (\tan ^{4}\left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )\right )}{d}+\frac {b^{2} \tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )}{4 d^{3}}-\frac {b^{2} \left (\tan ^{3}\left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )\right )}{4 d^{3}}-\frac {b^{2} x^{2}}{8 d^{2}}-\frac {4 a b \left (\tan ^{2}\left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )\right )}{d^{3}}-\frac {4 a b \left (\tan ^{4}\left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )\right )}{d^{3}}-\frac {b^{2} x^{4} \tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )}{2 d}+\frac {b^{2} x^{4} \left (\tan ^{3}\left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {3 b^{2} x^{2} \left (\tan ^{2}\left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )\right )}{4 d^{2}}-\frac {b^{2} x^{2} \left (\tan ^{4}\left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )\right )}{8 d^{2}}-\frac {a b \,x^{4}}{d}+\frac {4 a b \,x^{2} \tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )}{d^{2}}+\frac {4 a b \,x^{2} \left (\tan ^{3}\left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )\right )}{d^{2}}}{\left (1+\tan ^{2}\left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )\right )^{2}}\)	\(352\)

input

int(x^5*(a+b*sin(d*x^2+c))^2,x,method=_RETURNVERBOSE)
 

output

1/6*a^2*x^6+1/12*b^2*x^6-a*b*(d^2*x^4-2)/d^3*cos(d*x^2+c)+2*a*b*x^2*sin(d* 
x^2+c)/d^2-1/8*b^2/d^2*x^2*cos(2*d*x^2+2*c)-1/16*b^2*(2*d^2*x^4-1)/d^3*sin 
(2*d*x^2+2*c)
 

3.1.12.5 Fricas [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.74 \[ \int x^5 \left (a+b \sin \left (c+d x^2\right )\right )^2 \, dx=\frac {2 \, {\left (2 \, a^{2} + b^{2}\right )} d^{3} x^{6} - 6 \, b^{2} d x^{2} \cos \left (d x^{2} + c\right )^{2} + 3 \, b^{2} d x^{2} - 24 \, {\left (a b d^{2} x^{4} - 2 \, a b\right )} \cos \left (d x^{2} + c\right ) + 3 \, {\left (16 \, a b d x^{2} - {\left (2 \, b^{2} d^{2} x^{4} - b^{2}\right )} \cos \left (d x^{2} + c\right )\right )} \sin \left (d x^{2} + c\right )}{24 \, d^{3}} \]

input

integrate(x^5*(a+b*sin(d*x^2+c))^2,x, algorithm="fricas")
 

output

1/24*(2*(2*a^2 + b^2)*d^3*x^6 - 6*b^2*d*x^2*cos(d*x^2 + c)^2 + 3*b^2*d*x^2 
 - 24*(a*b*d^2*x^4 - 2*a*b)*cos(d*x^2 + c) + 3*(16*a*b*d*x^2 - (2*b^2*d^2* 
x^4 - b^2)*cos(d*x^2 + c))*sin(d*x^2 + c))/d^3
 

3.1.12.6 Sympy [A] (verification not implemented)

Time = 0.55 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.28 \[ \int x^5 \left (a+b \sin \left (c+d x^2\right )\right )^2 \, dx=\begin {cases} \frac {a^{2} x^{6}}{6} - \frac {a b x^{4} \cos {\left (c + d x^{2} \right )}}{d} + \frac {2 a b x^{2} \sin {\left (c + d x^{2} \right )}}{d^{2}} + \frac {2 a b \cos {\left (c + d x^{2} \right )}}{d^{3}} + \frac {b^{2} x^{6} \sin ^{2}{\left (c + d x^{2} \right )}}{12} + \frac {b^{2} x^{6} \cos ^{2}{\left (c + d x^{2} \right )}}{12} - \frac {b^{2} x^{4} \sin {\left (c + d x^{2} \right )} \cos {\left (c + d x^{2} \right )}}{4 d} + \frac {b^{2} x^{2} \sin ^{2}{\left (c + d x^{2} \right )}}{8 d^{2}} - \frac {b^{2} x^{2} \cos ^{2}{\left (c + d x^{2} \right )}}{8 d^{2}} + \frac {b^{2} \sin {\left (c + d x^{2} \right )} \cos {\left (c + d x^{2} \right )}}{8 d^{3}} & \text {for}\: d \neq 0 \\\frac {x^{6} \left (a + b \sin {\left (c \right )}\right )^{2}}{6} & \text {otherwise} \end {cases} \]

input

integrate(x**5*(a+b*sin(d*x**2+c))**2,x)
 

output

Piecewise((a**2*x**6/6 - a*b*x**4*cos(c + d*x**2)/d + 2*a*b*x**2*sin(c + d 
*x**2)/d**2 + 2*a*b*cos(c + d*x**2)/d**3 + b**2*x**6*sin(c + d*x**2)**2/12 
 + b**2*x**6*cos(c + d*x**2)**2/12 - b**2*x**4*sin(c + d*x**2)*cos(c + d*x 
**2)/(4*d) + b**2*x**2*sin(c + d*x**2)**2/(8*d**2) - b**2*x**2*cos(c + d*x 
**2)**2/(8*d**2) + b**2*sin(c + d*x**2)*cos(c + d*x**2)/(8*d**3), Ne(d, 0) 
), (x**6*(a + b*sin(c))**2/6, True))
 

3.1.12.7 Maxima [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.65 \[ \int x^5 \left (a+b \sin \left (c+d x^2\right )\right )^2 \, dx=\frac {1}{6} \, a^{2} x^{6} + \frac {{\left (2 \, d x^{2} \sin \left (d x^{2} + c\right ) - {\left (d^{2} x^{4} - 2\right )} \cos \left (d x^{2} + c\right )\right )} a b}{d^{3}} + \frac {{\left (4 \, d^{3} x^{6} - 6 \, d x^{2} \cos \left (2 \, d x^{2} + 2 \, c\right ) - 3 \, {\left (2 \, d^{2} x^{4} - 1\right )} \sin \left (2 \, d x^{2} + 2 \, c\right )\right )} b^{2}}{48 \, d^{3}} \]

input

integrate(x^5*(a+b*sin(d*x^2+c))^2,x, algorithm="maxima")
 

output

1/6*a^2*x^6 + (2*d*x^2*sin(d*x^2 + c) - (d^2*x^4 - 2)*cos(d*x^2 + c))*a*b/ 
d^3 + 1/48*(4*d^3*x^6 - 6*d*x^2*cos(2*d*x^2 + 2*c) - 3*(2*d^2*x^4 - 1)*sin 
(2*d*x^2 + 2*c))*b^2/d^3
 

3.1.12.8 Giac [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.74 \[ \int x^5 \left (a+b \sin \left (c+d x^2\right )\right )^2 \, dx=-\frac {{\left ({\left (d x^{2} + c\right )} b^{2} - b^{2} c\right )} \cos \left (2 \, d x^{2} + 2 \, c\right )}{8 \, d^{3}} - \frac {{\left ({\left (d x^{2} + c\right )}^{2} a b - 2 \, {\left (d x^{2} + c\right )} a b c - 2 \, a b\right )} \cos \left (d x^{2} + c\right )}{d^{3}} - \frac {{\left (2 \, {\left (d x^{2} + c\right )}^{2} b^{2} - 4 \, {\left (d x^{2} + c\right )} b^{2} c - b^{2}\right )} \sin \left (2 \, d x^{2} + 2 \, c\right )}{16 \, d^{3}} + \frac {2 \, {\left ({\left (d x^{2} + c\right )} a b - a b c\right )} \sin \left (d x^{2} + c\right )}{d^{3}} + \frac {2 \, {\left (d x^{2} + c\right )}^{3} a^{2} + {\left (d x^{2} + c\right )}^{3} b^{2} - 6 \, {\left (d x^{2} + c\right )}^{2} a^{2} c - 3 \, {\left (d x^{2} + c\right )}^{2} b^{2} c}{12 \, d^{3}} + \frac {4 \, {\left (d x^{2} + c\right )} a^{2} c^{2} + {\left (2 \, d x^{2} + 2 \, c - \sin \left (2 \, d x^{2} + 2 \, c\right )\right )} b^{2} c^{2} - 8 \, a b c^{2} \cos \left (d x^{2} + c\right )}{8 \, d^{3}} \]

input

integrate(x^5*(a+b*sin(d*x^2+c))^2,x, algorithm="giac")
 

output

-1/8*((d*x^2 + c)*b^2 - b^2*c)*cos(2*d*x^2 + 2*c)/d^3 - ((d*x^2 + c)^2*a*b 
 - 2*(d*x^2 + c)*a*b*c - 2*a*b)*cos(d*x^2 + c)/d^3 - 1/16*(2*(d*x^2 + c)^2 
*b^2 - 4*(d*x^2 + c)*b^2*c - b^2)*sin(2*d*x^2 + 2*c)/d^3 + 2*((d*x^2 + c)* 
a*b - a*b*c)*sin(d*x^2 + c)/d^3 + 1/12*(2*(d*x^2 + c)^3*a^2 + (d*x^2 + c)^ 
3*b^2 - 6*(d*x^2 + c)^2*a^2*c - 3*(d*x^2 + c)^2*b^2*c)/d^3 + 1/8*(4*(d*x^2 
 + c)*a^2*c^2 + (2*d*x^2 + 2*c - sin(2*d*x^2 + 2*c))*b^2*c^2 - 8*a*b*c^2*c 
os(d*x^2 + c))/d^3
 

3.1.12.9 Mupad [B] (verification not implemented)

Time = 0.48 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.91 \[ \int x^5 \left (a+b \sin \left (c+d x^2\right )\right )^2 \, dx=\frac {\frac {3\,b^2\,\sin \left (2\,d\,x^2+2\,c\right )}{2}-96\,a\,b\,{\sin \left (\frac {d\,x^2}{2}+\frac {c}{2}\right )}^2+4\,a^2\,d^3\,x^6+2\,b^2\,d^3\,x^6+3\,b^2\,d\,x^2\,\left (2\,{\sin \left (d\,x^2+c\right )}^2-1\right )-3\,b^2\,d^2\,x^4\,\sin \left (2\,d\,x^2+2\,c\right )+24\,a\,b\,d^2\,x^4\,\left (2\,{\sin \left (\frac {d\,x^2}{2}+\frac {c}{2}\right )}^2-1\right )+48\,a\,b\,d\,x^2\,\sin \left (d\,x^2+c\right )}{24\,d^3} \]

input

int(x^5*(a + b*sin(c + d*x^2))^2,x)
 

output

((3*b^2*sin(2*c + 2*d*x^2))/2 - 96*a*b*sin(c/2 + (d*x^2)/2)^2 + 4*a^2*d^3* 
x^6 + 2*b^2*d^3*x^6 + 3*b^2*d*x^2*(2*sin(c + d*x^2)^2 - 1) - 3*b^2*d^2*x^4 
*sin(2*c + 2*d*x^2) + 24*a*b*d^2*x^4*(2*sin(c/2 + (d*x^2)/2)^2 - 1) + 48*a 
*b*d*x^2*sin(c + d*x^2))/(24*d^3)