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

Optimal. Leaf size=129 \[ \frac {\sin \left (a+b x^3\right ) \cos ^7\left (a+b x^3\right )}{192 b^2}+\frac {7 \sin \left (a+b x^3\right ) \cos ^5\left (a+b x^3\right )}{1152 b^2}+\frac {35 \sin \left (a+b x^3\right ) \cos ^3\left (a+b x^3\right )}{4608 b^2}+\frac {35 \sin \left (a+b x^3\right ) \cos \left (a+b x^3\right )}{3072 b^2}-\frac {x^3 \cos ^8\left (a+b x^3\right )}{24 b}+\frac {35 x^3}{3072 b} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.14, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3444, 3380, 2635, 8} \[ \frac {\sin \left (a+b x^3\right ) \cos ^7\left (a+b x^3\right )}{192 b^2}+\frac {7 \sin \left (a+b x^3\right ) \cos ^5\left (a+b x^3\right )}{1152 b^2}+\frac {35 \sin \left (a+b x^3\right ) \cos ^3\left (a+b x^3\right )}{4608 b^2}+\frac {35 \sin \left (a+b x^3\right ) \cos \left (a+b x^3\right )}{3072 b^2}-\frac {x^3 \cos ^8\left (a+b x^3\right )}{24 b}+\frac {35 x^3}{3072 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(35*x^3)/(3072*b) - (x^3*Cos[a + b*x^3]^8)/(24*b) + (35*Cos[a + b*x^3]*Sin[a + b*x^3])/(3072*b^2) + (35*Cos[a
+ b*x^3]^3*Sin[a + b*x^3])/(4608*b^2) + (7*Cos[a + b*x^3]^5*Sin[a + b*x^3])/(1152*b^2) + (Cos[a + b*x^3]^7*Sin
[a + b*x^3])/(192*b^2)

Rule 8

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

Rule 2635

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] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 3380

Int[((a_.) + Cos[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(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[Simpl
ify[(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[(m + 1)/n], 0]))

Rule 3444

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] + Dist[(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]

Rubi steps

\begin {align*} \int x^5 \cos ^7\left (a+b x^3\right ) \sin \left (a+b x^3\right ) \, dx &=-\frac {x^3 \cos ^8\left (a+b x^3\right )}{24 b}+\frac {\int x^2 \cos ^8\left (a+b x^3\right ) \, dx}{8 b}\\ &=-\frac {x^3 \cos ^8\left (a+b x^3\right )}{24 b}+\frac {\operatorname {Subst}\left (\int \cos ^8(a+b x) \, dx,x,x^3\right )}{24 b}\\ &=-\frac {x^3 \cos ^8\left (a+b x^3\right )}{24 b}+\frac {\cos ^7\left (a+b x^3\right ) \sin \left (a+b x^3\right )}{192 b^2}+\frac {7 \operatorname {Subst}\left (\int \cos ^6(a+b x) \, dx,x,x^3\right )}{192 b}\\ &=-\frac {x^3 \cos ^8\left (a+b x^3\right )}{24 b}+\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}+\frac {35 \operatorname {Subst}\left (\int \cos ^4(a+b x) \, dx,x,x^3\right )}{1152 b}\\ &=-\frac {x^3 \cos ^8\left (a+b x^3\right )}{24 b}+\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}+\frac {35 \operatorname {Subst}\left (\int \cos ^2(a+b x) \, dx,x,x^3\right )}{1536 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}+\frac {35 \operatorname {Subst}\left (\int 1 \, dx,x,x^3\right )}{3072 b}\\ &=\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}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.57, size = 120, normalized size = 0.93 \[ \frac {672 \sin \left (2 \left (a+b x^3\right )\right )+168 \sin \left (4 \left (a+b x^3\right )\right )+32 \sin \left (6 \left (a+b x^3\right )\right )+3 \sin \left (8 \left (a+b x^3\right )\right )-1344 b x^3 \cos \left (2 \left (a+b x^3\right )\right )-672 b x^3 \cos \left (4 \left (a+b x^3\right )\right )-192 b x^3 \cos \left (6 \left (a+b x^3\right )\right )-24 b x^3 \cos \left (8 \left (a+b x^3\right )\right )}{73728 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-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)

________________________________________________________________________________________

fricas [A]  time = 1.09, size = 85, normalized size = 0.66 \[ -\frac {384 \, b x^{3} \cos \left (b x^{3} + a\right )^{8} - 105 \, b x^{3} - {\left (48 \, \cos \left (b x^{3} + a\right )^{7} + 56 \, \cos \left (b x^{3} + a\right )^{5} + 70 \, \cos \left (b x^{3} + a\right )^{3} + 105 \, \cos \left (b x^{3} + a\right )\right )} \sin \left (b x^{3} + a\right )}{9216 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/9216*(384*b*x^3*cos(b*x^3 + a)^8 - 105*b*x^3 - (48*cos(b*x^3 + a)^7 + 56*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

________________________________________________________________________________________

giac [A]  time = 0.44, size = 126, normalized size = 0.98 \[ -\frac {24 \, b x^{3} \cos \left (8 \, b x^{3} + 8 \, a\right ) + 192 \, b x^{3} \cos \left (6 \, b x^{3} + 6 \, a\right ) + 672 \, b x^{3} \cos \left (4 \, b x^{3} + 4 \, a\right ) + 1344 \, b x^{3} \cos \left (2 \, b x^{3} + 2 \, a\right ) - 3 \, \sin \left (8 \, b x^{3} + 8 \, a\right ) - 32 \, \sin \left (6 \, b x^{3} + 6 \, a\right ) - 168 \, \sin \left (4 \, b x^{3} + 4 \, a\right ) - 672 \, \sin \left (2 \, b x^{3} + 2 \, a\right )}{73728 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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

________________________________________________________________________________________

maple [B]  time = 0.95, size = 403, normalized size = 3.12 \[ \frac {-\frac {4 x^{3}}{3 b}+\frac {4 \tan \left (b \,x^{3}+a \right )}{3 b^{2}}+\frac {4 x^{3} \left (\tan ^{2}\left (b \,x^{3}+a \right )\right )}{3 b}}{128+128 \left (\tan ^{2}\left (b \,x^{3}+a \right )\right )}+\frac {\frac {\tan \left (b \,x^{3}+a \right )}{b^{2}}-\frac {x^{3}}{b}-\frac {\tan ^{3}\left (b \,x^{3}+a \right )}{b^{2}}+\frac {6 x^{3} \left (\tan ^{2}\left (b \,x^{3}+a \right )\right )}{b}-\frac {x^{3} \left (\tan ^{4}\left (b \,x^{3}+a \right )\right )}{b}}{128 \left (1+\tan ^{2}\left (b \,x^{3}+a \right )\right )^{2}}+\frac {-6 x^{3} b \left (\cos ^{2}\left (3 b \,x^{3}+3 a \right )\right )-18 \left (\cos ^{2}\left (b \,x^{3}+a \right )\right ) b \,x^{3}+12 b \,x^{3}+\sin \left (3 b \,x^{3}+3 a \right ) \cos \left (3 b \,x^{3}+3 a \right )+9 \cos \left (b \,x^{3}+a \right ) \sin \left (b \,x^{3}+a \right )}{1152 b^{2}}+\frac {-\frac {x^{3}}{6 b}+\frac {\tan \left (2 b \,x^{3}+2 a \right )}{12 b^{2}}+\frac {x^{3} \left (\tan ^{2}\left (2 b \,x^{3}+2 a \right )\right )}{6 b}}{128+128 \left (\tan ^{2}\left (2 b \,x^{3}+2 a \right )\right )}+\frac {-\frac {x^{3}}{24 b}+\frac {\tan \left (2 b \,x^{3}+2 a \right )}{48 b^{2}}-\frac {\tan ^{3}\left (2 b \,x^{3}+2 a \right )}{48 b^{2}}+\frac {x^{3} \left (\tan ^{2}\left (2 b \,x^{3}+2 a \right )\right )}{4 b}-\frac {x^{3} \left (\tan ^{4}\left (2 b \,x^{3}+2 a \right )\right )}{24 b}}{128 \left (1+\tan ^{2}\left (2 b \,x^{3}+2 a \right )\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/128*(-4/3*x^3/b+4/3/b^2*tan(b*x^3+a)+4/3*x^3/b*tan(b*x^3+a)^2)/(1+tan(b*x^3+a)^2)+1/128*(1/b^2*tan(b*x^3+a)-
x^3/b-1/b^2*tan(b*x^3+a)^3+6*x^3/b*tan(b*x^3+a)^2-x^3/b*tan(b*x^3+a)^4)/(1+tan(b*x^3+a)^2)^2+1/1152*(-6*x^3*b*
cos(3*b*x^3+3*a)^2-18*cos(b*x^3+a)^2*b*x^3+12*b*x^3+sin(3*b*x^3+3*a)*cos(3*b*x^3+3*a)+9*cos(b*x^3+a)*sin(b*x^3
+a))/b^2+1/128*(-1/6*x^3/b+1/12/b^2*tan(2*b*x^3+2*a)+1/6*x^3/b*tan(2*b*x^3+2*a)^2)/(1+tan(2*b*x^3+2*a)^2)+1/12
8*(-1/24*x^3/b+1/48/b^2*tan(2*b*x^3+2*a)-1/48/b^2*tan(2*b*x^3+2*a)^3+1/4*x^3/b*tan(2*b*x^3+2*a)^2-1/24*x^3/b*t
an(2*b*x^3+2*a)^4)/(1+tan(2*b*x^3+2*a)^2)^2

________________________________________________________________________________________

maxima [A]  time = 0.34, size = 126, normalized size = 0.98 \[ -\frac {24 \, b x^{3} \cos \left (8 \, b x^{3} + 8 \, a\right ) + 192 \, b x^{3} \cos \left (6 \, b x^{3} + 6 \, a\right ) + 672 \, b x^{3} \cos \left (4 \, b x^{3} + 4 \, a\right ) + 1344 \, b x^{3} \cos \left (2 \, b x^{3} + 2 \, a\right ) - 3 \, \sin \left (8 \, b x^{3} + 8 \, a\right ) - 32 \, \sin \left (6 \, b x^{3} + 6 \, a\right ) - 168 \, \sin \left (4 \, b x^{3} + 4 \, a\right ) - 672 \, \sin \left (2 \, b x^{3} + 2 \, a\right )}{73728 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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

________________________________________________________________________________________

mupad [B]  time = 3.45, size = 147, normalized size = 1.14 \[ \frac {168\,\sin \left (2\,b\,x^3+2\,a\right )+42\,\sin \left (4\,b\,x^3+4\,a\right )+8\,\sin \left (6\,b\,x^3+6\,a\right )+\frac {3\,\sin \left (8\,b\,x^3+8\,a\right )}{4}+336\,b\,x^3\,\left (2\,{\sin \left (b\,x^3+a\right )}^2-1\right )+168\,b\,x^3\,\left (2\,{\sin \left (2\,b\,x^3+2\,a\right )}^2-1\right )+48\,b\,x^3\,\left (2\,{\sin \left (3\,b\,x^3+3\,a\right )}^2-1\right )+6\,b\,x^3\,\left (2\,{\sin \left (4\,b\,x^3+4\,a\right )}^2-1\right )}{18432\,b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(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)

________________________________________________________________________________________

sympy [A]  time = 74.97, size = 241, normalized size = 1.87 \[ \begin {cases} \frac {35 x^{3} \sin ^{8}{\left (a + b x^{3} \right )}}{3072 b} + \frac {35 x^{3} \sin ^{6}{\left (a + b x^{3} \right )} \cos ^{2}{\left (a + b x^{3} \right )}}{768 b} + \frac {35 x^{3} \sin ^{4}{\left (a + b x^{3} \right )} \cos ^{4}{\left (a + b x^{3} \right )}}{512 b} + \frac {35 x^{3} \sin ^{2}{\left (a + b x^{3} \right )} \cos ^{6}{\left (a + b x^{3} \right )}}{768 b} - \frac {31 x^{3} \cos ^{8}{\left (a + b x^{3} \right )}}{1024 b} + \frac {35 \sin ^{7}{\left (a + b x^{3} \right )} \cos {\left (a + b x^{3} \right )}}{3072 b^{2}} + \frac {385 \sin ^{5}{\left (a + b x^{3} \right )} \cos ^{3}{\left (a + b x^{3} \right )}}{9216 b^{2}} + \frac {511 \sin ^{3}{\left (a + b x^{3} \right )} \cos ^{5}{\left (a + b x^{3} \right )}}{9216 b^{2}} + \frac {31 \sin {\left (a + b x^{3} \right )} \cos ^{7}{\left (a + b x^{3} \right )}}{1024 b^{2}} & \text {for}\: b \neq 0 \\\frac {x^{6} \sin {\relax (a )} \cos ^{7}{\relax (a )}}{6} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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)/(3072*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)*co
s(a + b*x**3)**7/(1024*b**2), Ne(b, 0)), (x**6*sin(a)*cos(a)**7/6, True))

________________________________________________________________________________________