3.10.26 \(\int x^5 \cos ^7(a+b x^3) \sin (a+b x^3) \, dx\) [926]

3.10.26.1 Optimal result

Integrand size = 22, antiderivative size = 129 \[ \int x^5 \cos ^7\left (a+b x^3\right ) \sin \left (a+b x^3\right ) \, dx=\frac {35 x^3}{3072 b}-\frac {x^3 \cos ^8\left (a+b x^3\right )}{24 b}+\frac {35 \cos \left (a+b x^3\right ) \sin \left (a+b x^3\right )}{3072 b^2}+\frac {35 \cos ^3\left (a+b x^3\right ) \sin \left (a+b x^3\right )}{4608 b^2}+\frac {7 \cos ^5\left (a+b x^3\right ) \sin \left (a+b x^3\right )}{1152 b^2}+\frac {\cos ^7\left (a+b x^3\right ) \sin \left (a+b x^3\right )}{192 b^2} \]

output

35/3072*x^3/b-1/24*x^3*cos(b*x^3+a)^8/b+35/3072*cos(b*x^3+a)*sin(b*x^3+a)/ 
b^2+35/4608*cos(b*x^3+a)^3*sin(b*x^3+a)/b^2+7/1152*cos(b*x^3+a)^5*sin(b*x^ 
3+a)/b^2+1/192*cos(b*x^3+a)^7*sin(b*x^3+a)/b^2
 

3.10.26.2 Mathematica [A] (verified)

Time = 0.39 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.93 \[ \int x^5 \cos ^7\left (a+b x^3\right ) \sin \left (a+b x^3\right ) \, dx=\frac {-1344 b x^3 \cos \left (2 \left (a+b x^3\right )\right )-672 b x^3 \cos \left (4 \left (a+b x^3\right )\right )-192 b x^3 \cos \left (6 \left (a+b x^3\right )\right )-24 b x^3 \cos \left (8 \left (a+b x^3\right )\right )+672 \sin \left (2 \left (a+b x^3\right )\right )+168 \sin \left (4 \left (a+b x^3\right )\right )+32 \sin \left (6 \left (a+b x^3\right )\right )+3 \sin \left (8 \left (a+b x^3\right )\right )}{73728 b^2} \]

input

Integrate[x^5*Cos[a + b*x^3]^7*Sin[a + b*x^3],x]
 

output

(-1344*b*x^3*Cos[2*(a + b*x^3)] - 672*b*x^3*Cos[4*(a + b*x^3)] - 192*b*x^3 
*Cos[6*(a + b*x^3)] - 24*b*x^3*Cos[8*(a + b*x^3)] + 672*Sin[2*(a + b*x^3)] 
 + 168*Sin[4*(a + b*x^3)] + 32*Sin[6*(a + b*x^3)] + 3*Sin[8*(a + b*x^3)])/ 
(73728*b^2)
 

3.10.26.3 Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.16, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3925, 3861, 3042, 3115, 3042, 3115, 3042, 3115, 3042, 3115, 24}

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.

\(\displaystyle \int x^5 \sin \left (a+b x^3\right ) \cos ^7\left (a+b x^3\right ) \, dx\)

\(\Big \downarrow \) 3925

\(\displaystyle \frac {\int x^2 \cos ^8\left (b x^3+a\right )dx}{8 b}-\frac {x^3 \cos ^8\left (a+b x^3\right )}{24 b}\)

\(\Big \downarrow \) 3861

\(\displaystyle \frac {\int \cos ^8\left (b x^3+a\right )dx^3}{24 b}-\frac {x^3 \cos ^8\left (a+b x^3\right )}{24 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \sin \left (b x^3+a+\frac {\pi }{2}\right )^8dx^3}{24 b}-\frac {x^3 \cos ^8\left (a+b x^3\right )}{24 b}\)

\(\Big \downarrow \) 3115

\(\displaystyle \frac {\frac {7}{8} \int \cos ^6\left (b x^3+a\right )dx^3+\frac {\sin \left (a+b x^3\right ) \cos ^7\left (a+b x^3\right )}{8 b}}{24 b}-\frac {x^3 \cos ^8\left (a+b x^3\right )}{24 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {7}{8} \int \sin \left (b x^3+a+\frac {\pi }{2}\right )^6dx^3+\frac {\sin \left (a+b x^3\right ) \cos ^7\left (a+b x^3\right )}{8 b}}{24 b}-\frac {x^3 \cos ^8\left (a+b x^3\right )}{24 b}\)

\(\Big \downarrow \) 3115

\(\displaystyle \frac {\frac {7}{8} \left (\frac {5}{6} \int \cos ^4\left (b x^3+a\right )dx^3+\frac {\sin \left (a+b x^3\right ) \cos ^5\left (a+b x^3\right )}{6 b}\right )+\frac {\sin \left (a+b x^3\right ) \cos ^7\left (a+b x^3\right )}{8 b}}{24 b}-\frac {x^3 \cos ^8\left (a+b x^3\right )}{24 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {7}{8} \left (\frac {5}{6} \int \sin \left (b x^3+a+\frac {\pi }{2}\right )^4dx^3+\frac {\sin \left (a+b x^3\right ) \cos ^5\left (a+b x^3\right )}{6 b}\right )+\frac {\sin \left (a+b x^3\right ) \cos ^7\left (a+b x^3\right )}{8 b}}{24 b}-\frac {x^3 \cos ^8\left (a+b x^3\right )}{24 b}\)

\(\Big \downarrow \) 3115

\(\displaystyle \frac {\frac {7}{8} \left (\frac {5}{6} \left (\frac {3}{4} \int \cos ^2\left (b x^3+a\right )dx^3+\frac {\sin \left (a+b x^3\right ) \cos ^3\left (a+b x^3\right )}{4 b}\right )+\frac {\sin \left (a+b x^3\right ) \cos ^5\left (a+b x^3\right )}{6 b}\right )+\frac {\sin \left (a+b x^3\right ) \cos ^7\left (a+b x^3\right )}{8 b}}{24 b}-\frac {x^3 \cos ^8\left (a+b x^3\right )}{24 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {7}{8} \left (\frac {5}{6} \left (\frac {3}{4} \int \sin \left (b x^3+a+\frac {\pi }{2}\right )^2dx^3+\frac {\sin \left (a+b x^3\right ) \cos ^3\left (a+b x^3\right )}{4 b}\right )+\frac {\sin \left (a+b x^3\right ) \cos ^5\left (a+b x^3\right )}{6 b}\right )+\frac {\sin \left (a+b x^3\right ) \cos ^7\left (a+b x^3\right )}{8 b}}{24 b}-\frac {x^3 \cos ^8\left (a+b x^3\right )}{24 b}\)

\(\Big \downarrow \) 3115

\(\displaystyle \frac {\frac {7}{8} \left (\frac {5}{6} \left (\frac {3}{4} \left (\frac {\int 1dx^3}{2}+\frac {\sin \left (a+b x^3\right ) \cos \left (a+b x^3\right )}{2 b}\right )+\frac {\sin \left (a+b x^3\right ) \cos ^3\left (a+b x^3\right )}{4 b}\right )+\frac {\sin \left (a+b x^3\right ) \cos ^5\left (a+b x^3\right )}{6 b}\right )+\frac {\sin \left (a+b x^3\right ) \cos ^7\left (a+b x^3\right )}{8 b}}{24 b}-\frac {x^3 \cos ^8\left (a+b x^3\right )}{24 b}\)

\(\Big \downarrow \) 24

\(\displaystyle \frac {\frac {\sin \left (a+b x^3\right ) \cos ^7\left (a+b x^3\right )}{8 b}+\frac {7}{8} \left (\frac {\sin \left (a+b x^3\right ) \cos ^5\left (a+b x^3\right )}{6 b}+\frac {5}{6} \left (\frac {\sin \left (a+b x^3\right ) \cos ^3\left (a+b x^3\right )}{4 b}+\frac {3}{4} \left (\frac {\sin \left (a+b x^3\right ) \cos \left (a+b x^3\right )}{2 b}+\frac {x^3}{2}\right )\right )\right )}{24 b}-\frac {x^3 \cos ^8\left (a+b x^3\right )}{24 b}\)

input

Int[x^5*Cos[a + b*x^3]^7*Sin[a + b*x^3],x]
 

output

-1/24*(x^3*Cos[a + b*x^3]^8)/b + ((Cos[a + b*x^3]^7*Sin[a + b*x^3])/(8*b) 
+ (7*((Cos[a + b*x^3]^5*Sin[a + b*x^3])/(6*b) + (5*((Cos[a + b*x^3]^3*Sin[ 
a + b*x^3])/(4*b) + (3*(x^3/2 + (Cos[a + b*x^3]*Sin[a + b*x^3])/(2*b)))/4) 
)/6))/8)/(24*b)
 

3.10.26.3.1 Defintions of rubi rules used

rule 24

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

rule 3042

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

rule 3115

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n)   Int[(b*Sin 
[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 
2*n]
 

rule 3861

Int[((a_.) + Cos[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol 
] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*Cos[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]))
 

rule 3925

Int[Cos[(a_.) + (b_.)*(x_)^(n_.)]^(p_.)*(x_)^(m_.)*Sin[(a_.) + (b_.)*(x_)^( 
n_.)], x_Symbol] :> Simp[(-x^(m - n + 1))*(Cos[a + b*x^n]^(p + 1)/(b*n*(p + 
 1))), x] + Simp[(m - n + 1)/(b*n*(p + 1))   Int[x^(m - n)*Cos[a + b*x^n]^( 
p + 1), x], x] /; FreeQ[{a, b, p}, x] && LtQ[0, n, m + 1] && NeQ[p, -1]
 

3.10.26.4 Maple [A] (verified)

Time = 1.99 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.98

method	result	size
parallelrisch	\(\frac {-24 x^{3} \cos \left (8 b \,x^{3}+8 a \right ) b -192 x^{3} \cos \left (6 b \,x^{3}+6 a \right ) b -672 x^{3} \cos \left (4 b \,x^{3}+4 a \right ) b -1344 \cos \left (2 b \,x^{3}+2 a \right ) x^{3} b +3 \sin \left (8 b \,x^{3}+8 a \right )+32 \sin \left (6 b \,x^{3}+6 a \right )+168 \sin \left (4 b \,x^{3}+4 a \right )+672 \sin \left (2 b \,x^{3}+2 a \right )}{73728 b^{2}}\)	\(127\)
risch	\(-\frac {x^{3} \cos \left (8 b \,x^{3}+8 a \right )}{3072 b}+\frac {\sin \left (8 b \,x^{3}+8 a \right )}{24576 b^{2}}-\frac {x^{3} \cos \left (6 b \,x^{3}+6 a \right )}{384 b}+\frac {\sin \left (6 b \,x^{3}+6 a \right )}{2304 b^{2}}-\frac {7 x^{3} \cos \left (4 b \,x^{3}+4 a \right )}{768 b}+\frac {7 \sin \left (4 b \,x^{3}+4 a \right )}{3072 b^{2}}-\frac {7 x^{3} \cos \left (2 b \,x^{3}+2 a \right )}{384 b}+\frac {7 \sin \left (2 b \,x^{3}+2 a \right )}{768 b^{2}}\)	\(142\)
default	\(\frac {-\frac {x^{3}}{24 b}+\frac {\tan \left (4 b \,x^{3}+4 a \right )}{96 b^{2}}+\frac {x^{3} \tan \left (4 b \,x^{3}+4 a \right )^{2}}{24 b}}{128+128 \tan \left (4 b \,x^{3}+4 a \right )^{2}}+\frac {-\frac {7 x^{3}}{3 b}+\frac {7 \tan \left (b \,x^{3}+a \right )}{3 b^{2}}+\frac {7 x^{3} \tan \left (b \,x^{3}+a \right )^{2}}{3 b}}{128+128 \tan \left (b \,x^{3}+a \right )^{2}}+\frac {-\frac {7 x^{3}}{6 b}+\frac {7 \tan \left (2 b \,x^{3}+2 a \right )}{12 b^{2}}+\frac {7 x^{3} \tan \left (2 b \,x^{3}+2 a \right )^{2}}{6 b}}{128+128 \tan \left (2 b \,x^{3}+2 a \right )^{2}}+\frac {-\frac {x^{3}}{3 b}+\frac {\tan \left (3 b \,x^{3}+3 a \right )}{9 b^{2}}+\frac {x^{3} \tan \left (3 b \,x^{3}+3 a \right )^{2}}{3 b}}{128+128 \tan \left (3 b \,x^{3}+3 a \right )^{2}}\)	\(253\)

input

int(x^5*cos(b*x^3+a)^7*sin(b*x^3+a),x,method=_RETURNVERBOSE)
 

output

1/73728*(-24*x^3*cos(8*b*x^3+8*a)*b-192*x^3*cos(6*b*x^3+6*a)*b-672*x^3*cos 
(4*b*x^3+4*a)*b-1344*cos(2*b*x^3+2*a)*x^3*b+3*sin(8*b*x^3+8*a)+32*sin(6*b* 
x^3+6*a)+168*sin(4*b*x^3+4*a)+672*sin(2*b*x^3+2*a))/b^2
 

3.10.26.5 Fricas [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.66 \[ \int x^5 \cos ^7\left (a+b x^3\right ) \sin \left (a+b x^3\right ) \, dx=-\frac {384 \, b x^{3} \cos \left (b x^{3} + a\right )^{8} - 105 \, b x^{3} - {\left (48 \, \cos \left (b x^{3} + a\right )^{7} + 56 \, \cos \left (b x^{3} + a\right )^{5} + 70 \, \cos \left (b x^{3} + a\right )^{3} + 105 \, \cos \left (b x^{3} + a\right )\right )} \sin \left (b x^{3} + a\right )}{9216 \, b^{2}} \]

input

integrate(x^5*cos(b*x^3+a)^7*sin(b*x^3+a),x, algorithm="fricas")
 

output

-1/9216*(384*b*x^3*cos(b*x^3 + a)^8 - 105*b*x^3 - (48*cos(b*x^3 + a)^7 + 5 
6*cos(b*x^3 + a)^5 + 70*cos(b*x^3 + a)^3 + 105*cos(b*x^3 + a))*sin(b*x^3 + 
 a))/b^2
 

3.10.26.6 Sympy [A] (verification not implemented)

Time = 3.41 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.87 \[ \int x^5 \cos ^7\left (a+b x^3\right ) \sin \left (a+b x^3\right ) \, dx=\begin {cases} \frac {35 x^{3} \sin ^{8}{\left (a + b x^{3} \right )}}{3072 b} + \frac {35 x^{3} \sin ^{6}{\left (a + b x^{3} \right )} \cos ^{2}{\left (a + b x^{3} \right )}}{768 b} + \frac {35 x^{3} \sin ^{4}{\left (a + b x^{3} \right )} \cos ^{4}{\left (a + b x^{3} \right )}}{512 b} + \frac {35 x^{3} \sin ^{2}{\left (a + b x^{3} \right )} \cos ^{6}{\left (a + b x^{3} \right )}}{768 b} - \frac {31 x^{3} \cos ^{8}{\left (a + b x^{3} \right )}}{1024 b} + \frac {35 \sin ^{7}{\left (a + b x^{3} \right )} \cos {\left (a + b x^{3} \right )}}{3072 b^{2}} + \frac {385 \sin ^{5}{\left (a + b x^{3} \right )} \cos ^{3}{\left (a + b x^{3} \right )}}{9216 b^{2}} + \frac {511 \sin ^{3}{\left (a + b x^{3} \right )} \cos ^{5}{\left (a + b x^{3} \right )}}{9216 b^{2}} + \frac {31 \sin {\left (a + b x^{3} \right )} \cos ^{7}{\left (a + b x^{3} \right )}}{1024 b^{2}} & \text {for}\: b \neq 0 \\\frac {x^{6} \sin {\left (a \right )} \cos ^{7}{\left (a \right )}}{6} & \text {otherwise} \end {cases} \]

input

integrate(x**5*cos(b*x**3+a)**7*sin(b*x**3+a),x)
 

output

Piecewise((35*x**3*sin(a + b*x**3)**8/(3072*b) + 35*x**3*sin(a + b*x**3)** 
6*cos(a + b*x**3)**2/(768*b) + 35*x**3*sin(a + b*x**3)**4*cos(a + b*x**3)* 
*4/(512*b) + 35*x**3*sin(a + b*x**3)**2*cos(a + b*x**3)**6/(768*b) - 31*x* 
*3*cos(a + b*x**3)**8/(1024*b) + 35*sin(a + b*x**3)**7*cos(a + b*x**3)/(30 
72*b**2) + 385*sin(a + b*x**3)**5*cos(a + b*x**3)**3/(9216*b**2) + 511*sin 
(a + b*x**3)**3*cos(a + b*x**3)**5/(9216*b**2) + 31*sin(a + b*x**3)*cos(a 
+ b*x**3)**7/(1024*b**2), Ne(b, 0)), (x**6*sin(a)*cos(a)**7/6, True))
 

3.10.26.7 Maxima [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.98 \[ \int x^5 \cos ^7\left (a+b x^3\right ) \sin \left (a+b x^3\right ) \, dx=-\frac {24 \, b x^{3} \cos \left (8 \, b x^{3} + 8 \, a\right ) + 192 \, b x^{3} \cos \left (6 \, b x^{3} + 6 \, a\right ) + 672 \, b x^{3} \cos \left (4 \, b x^{3} + 4 \, a\right ) + 1344 \, b x^{3} \cos \left (2 \, b x^{3} + 2 \, a\right ) - 3 \, \sin \left (8 \, b x^{3} + 8 \, a\right ) - 32 \, \sin \left (6 \, b x^{3} + 6 \, a\right ) - 168 \, \sin \left (4 \, b x^{3} + 4 \, a\right ) - 672 \, \sin \left (2 \, b x^{3} + 2 \, a\right )}{73728 \, b^{2}} \]

input

integrate(x^5*cos(b*x^3+a)^7*sin(b*x^3+a),x, algorithm="maxima")
 

output

-1/73728*(24*b*x^3*cos(8*b*x^3 + 8*a) + 192*b*x^3*cos(6*b*x^3 + 6*a) + 672 
*b*x^3*cos(4*b*x^3 + 4*a) + 1344*b*x^3*cos(2*b*x^3 + 2*a) - 3*sin(8*b*x^3 
+ 8*a) - 32*sin(6*b*x^3 + 6*a) - 168*sin(4*b*x^3 + 4*a) - 672*sin(2*b*x^3 
+ 2*a))/b^2
 

3.10.26.8 Giac [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.20 \[ \int x^5 \cos ^7\left (a+b x^3\right ) \sin \left (a+b x^3\right ) \, dx=\frac {a \cos \left (b x^{3} + a\right )^{8}}{24 \, b^{2}} - \frac {24 \, {\left (b x^{3} + a\right )} \cos \left (8 \, b x^{3} + 8 \, a\right ) + 192 \, {\left (b x^{3} + a\right )} \cos \left (6 \, b x^{3} + 6 \, a\right ) + 672 \, {\left (b x^{3} + a\right )} \cos \left (4 \, b x^{3} + 4 \, a\right ) + 1344 \, {\left (b x^{3} + a\right )} \cos \left (2 \, b x^{3} + 2 \, a\right ) - 3 \, \sin \left (8 \, b x^{3} + 8 \, a\right ) - 32 \, \sin \left (6 \, b x^{3} + 6 \, a\right ) - 168 \, \sin \left (4 \, b x^{3} + 4 \, a\right ) - 672 \, \sin \left (2 \, b x^{3} + 2 \, a\right )}{73728 \, b^{2}} \]

input

integrate(x^5*cos(b*x^3+a)^7*sin(b*x^3+a),x, algorithm="giac")
 

output

1/24*a*cos(b*x^3 + a)^8/b^2 - 1/73728*(24*(b*x^3 + a)*cos(8*b*x^3 + 8*a) + 
 192*(b*x^3 + a)*cos(6*b*x^3 + 6*a) + 672*(b*x^3 + a)*cos(4*b*x^3 + 4*a) + 
 1344*(b*x^3 + a)*cos(2*b*x^3 + 2*a) - 3*sin(8*b*x^3 + 8*a) - 32*sin(6*b*x 
^3 + 6*a) - 168*sin(4*b*x^3 + 4*a) - 672*sin(2*b*x^3 + 2*a))/b^2
 

3.10.26.9 Mupad [B] (verification not implemented)

Time = 27.28 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.14 \[ \int x^5 \cos ^7\left (a+b x^3\right ) \sin \left (a+b x^3\right ) \, dx=\frac {168\,\sin \left (2\,b\,x^3+2\,a\right )+42\,\sin \left (4\,b\,x^3+4\,a\right )+8\,\sin \left (6\,b\,x^3+6\,a\right )+\frac {3\,\sin \left (8\,b\,x^3+8\,a\right )}{4}+336\,b\,x^3\,\left (2\,{\sin \left (b\,x^3+a\right )}^2-1\right )+168\,b\,x^3\,\left (2\,{\sin \left (2\,b\,x^3+2\,a\right )}^2-1\right )+48\,b\,x^3\,\left (2\,{\sin \left (3\,b\,x^3+3\,a\right )}^2-1\right )+6\,b\,x^3\,\left (2\,{\sin \left (4\,b\,x^3+4\,a\right )}^2-1\right )}{18432\,b^2} \]

input

int(x^5*cos(a + b*x^3)^7*sin(a + b*x^3),x)
 

output

(168*sin(2*a + 2*b*x^3) + 42*sin(4*a + 4*b*x^3) + 8*sin(6*a + 6*b*x^3) + ( 
3*sin(8*a + 8*b*x^3))/4 + 336*b*x^3*(2*sin(a + b*x^3)^2 - 1) + 168*b*x^3*( 
2*sin(2*a + 2*b*x^3)^2 - 1) + 48*b*x^3*(2*sin(3*a + 3*b*x^3)^2 - 1) + 6*b* 
x^3*(2*sin(4*a + 4*b*x^3)^2 - 1))/(18432*b^2)