3.1.69 \(\int x^5 (a+b \sin (c+d x^3))^2 \, dx\) [69]

Optimal. Leaf size=107 \[ \frac {a^2 x^6}{6}+\frac {b^2 x^6}{12}-\frac {2 a b x^3 \cos \left (c+d x^3\right )}{3 d}+\frac {2 a b \sin \left (c+d x^3\right )}{3 d^2}-\frac {b^2 x^3 \cos \left (c+d x^3\right ) \sin \left (c+d x^3\right )}{6 d}+\frac {b^2 \sin ^2\left (c+d x^3\right )}{12 d^2} \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3460, 3398, 3377, 2717, 3391, 30} \begin {gather*} \frac {a^2 x^6}{6}+\frac {2 a b \sin \left (c+d x^3\right )}{3 d^2}-\frac {2 a b x^3 \cos \left (c+d x^3\right )}{3 d}+\frac {b^2 \sin ^2\left (c+d x^3\right )}{12 d^2}-\frac {b^2 x^3 \sin \left (c+d x^3\right ) \cos \left (c+d x^3\right )}{6 d}+\frac {b^2 x^6}{12} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^5*(a + b*Sin[c + d*x^3])^2,x]

[Out]

(a^2*x^6)/6 + (b^2*x^6)/12 - (2*a*b*x^3*Cos[c + d*x^3])/(3*d) + (2*a*b*Sin[c + d*x^3])/(3*d^2) - (b^2*x^3*Cos[
c + d*x^3]*Sin[c + d*x^3])/(6*d) + (b^2*Sin[c + d*x^3]^2)/(12*d^2)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3377

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(-(c + d*x)^m)*(Cos[e + f*x]/f), x]
+ Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 3391

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

Rule 3398

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 3460

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

Rubi steps

\begin {align*} \int x^5 \left (a+b \sin \left (c+d x^3\right )\right )^2 \, dx &=\frac {1}{3} \text {Subst}\left (\int x (a+b \sin (c+d x))^2 \, dx,x,x^3\right )\\ &=\frac {1}{3} \text {Subst}\left (\int \left (a^2 x+2 a b x \sin (c+d x)+b^2 x \sin ^2(c+d x)\right ) \, dx,x,x^3\right )\\ &=\frac {a^2 x^6}{6}+\frac {1}{3} (2 a b) \text {Subst}\left (\int x \sin (c+d x) \, dx,x,x^3\right )+\frac {1}{3} b^2 \text {Subst}\left (\int x \sin ^2(c+d x) \, dx,x,x^3\right )\\ &=\frac {a^2 x^6}{6}-\frac {2 a b x^3 \cos \left (c+d x^3\right )}{3 d}-\frac {b^2 x^3 \cos \left (c+d x^3\right ) \sin \left (c+d x^3\right )}{6 d}+\frac {b^2 \sin ^2\left (c+d x^3\right )}{12 d^2}+\frac {1}{6} b^2 \text {Subst}\left (\int x \, dx,x,x^3\right )+\frac {(2 a b) \text {Subst}\left (\int \cos (c+d x) \, dx,x,x^3\right )}{3 d}\\ &=\frac {a^2 x^6}{6}+\frac {b^2 x^6}{12}-\frac {2 a b x^3 \cos \left (c+d x^3\right )}{3 d}+\frac {2 a b \sin \left (c+d x^3\right )}{3 d^2}-\frac {b^2 x^3 \cos \left (c+d x^3\right ) \sin \left (c+d x^3\right )}{6 d}+\frac {b^2 \sin ^2\left (c+d x^3\right )}{12 d^2}\\ \end {align*}

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Mathematica [A]
time = 0.16, size = 92, normalized size = 0.86 \begin {gather*} \frac {4 a^2 d^2 x^6+2 b^2 d^2 x^6-16 a b d x^3 \cos \left (c+d x^3\right )-b^2 \cos \left (2 \left (c+d x^3\right )\right )+16 a b \sin \left (c+d x^3\right )-2 b^2 d x^3 \sin \left (2 \left (c+d x^3\right )\right )}{24 d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^5*(a + b*Sin[c + d*x^3])^2,x]

[Out]

(4*a^2*d^2*x^6 + 2*b^2*d^2*x^6 - 16*a*b*d*x^3*Cos[c + d*x^3] - b^2*Cos[2*(c + d*x^3)] + 16*a*b*Sin[c + d*x^3]
- 2*b^2*d*x^3*Sin[2*(c + d*x^3)])/(24*d^2)

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Maple [A]
time = 0.22, size = 137, normalized size = 1.28

method	result	size
risch	\(\frac {a^{2} x^{6}}{6}+\frac {b^{2} x^{6}}{12}-\frac {2 a b \,x^{3} \cos \left (d \,x^{3}+c \right )}{3 d}+\frac {2 a b \sin \left (d \,x^{3}+c \right )}{3 d^{2}}-\frac {b^{2} \cos \left (2 d \,x^{3}+2 c \right )}{24 d^{2}}-\frac {b^{2} x^{3} \sin \left (2 d \,x^{3}+2 c \right )}{12 d}\)	\(92\)
default	\(\frac {a^{2} x^{6}}{6}+\frac {b^{2} x^{6}}{6}-\frac {b^{2} \left (\frac {x^{6}}{6}+\frac {\frac {1}{6 d^{2}}+\frac {x^{3} \tan \left (d \,x^{3}+c \right )}{3 d}}{1+\tan ^{2}\left (d \,x^{3}+c \right )}\right )}{2}-\frac {-\frac {8 a b \tan \left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right )}{3 d^{2}}+\frac {4 a b \,x^{3}}{3 d}-\frac {4 a b \,x^{3} \left (\tan ^{2}\left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right )\right )}{3 d}}{2 \left (1+\tan ^{2}\left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right )\right )}\)	\(137\)
norman	\(\frac {\left (\frac {a^{2}}{6}+\frac {b^{2}}{12}\right ) x^{6}+\left (\frac {a^{2}}{3}+\frac {b^{2}}{6}\right ) x^{6} \left (\tan ^{2}\left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right )\right )+\left (\frac {a^{2}}{6}+\frac {b^{2}}{12}\right ) x^{6} \left (\tan ^{4}\left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right )\right )+\frac {b^{2} \left (\tan ^{2}\left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right )\right )}{3 d^{2}}-\frac {b^{2} x^{3} \tan \left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right )}{3 d}+\frac {b^{2} x^{3} \left (\tan ^{3}\left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {4 a b \tan \left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right )}{3 d^{2}}+\frac {4 a b \left (\tan ^{3}\left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right )\right )}{3 d^{2}}-\frac {2 a b \,x^{3}}{3 d}+\frac {2 a b \,x^{3} \left (\tan ^{4}\left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (1+\tan ^{2}\left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right )\right )^{2}}\)	\(229\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(a+b*sin(d*x^3+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/6*a^2*x^6+1/6*b^2*x^6-1/2*b^2*(1/6*x^6+(1/6/d^2+1/3*x^3/d*tan(d*x^3+c))/(1+tan(d*x^3+c)^2))-1/2*(-8/3/d^2*a*
b*tan(1/2*d*x^3+1/2*c)+4/3/d*a*b*x^3-4/3/d*a*b*x^3*tan(1/2*d*x^3+1/2*c)^2)/(1+tan(1/2*d*x^3+1/2*c)^2)

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Maxima [A]
time = 0.29, size = 87, normalized size = 0.81 \begin {gather*} \frac {1}{6} \, a^{2} x^{6} - \frac {2 \, {\left (d x^{3} \cos \left (d x^{3} + c\right ) - \sin \left (d x^{3} + c\right )\right )} a b}{3 \, d^{2}} + \frac {{\left (2 \, d^{2} x^{6} - 2 \, d x^{3} \sin \left (2 \, d x^{3} + 2 \, c\right ) - \cos \left (2 \, d x^{3} + 2 \, c\right )\right )} b^{2}}{24 \, d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*sin(d*x^3+c))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.36, size = 84, normalized size = 0.79 \begin {gather*} \frac {{\left (2 \, a^{2} + b^{2}\right )} d^{2} x^{6} - 8 \, a b d x^{3} \cos \left (d x^{3} + c\right ) - b^{2} \cos \left (d x^{3} + c\right )^{2} - 2 \, {\left (b^{2} d x^{3} \cos \left (d x^{3} + c\right ) - 4 \, a b\right )} \sin \left (d x^{3} + c\right )}{12 \, d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*sin(d*x^3+c))^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.61, size = 143, normalized size = 1.34 \begin {gather*} \begin {cases} \frac {a^{2} x^{6}}{6} - \frac {2 a b x^{3} \cos {\left (c + d x^{3} \right )}}{3 d} + \frac {2 a b \sin {\left (c + d x^{3} \right )}}{3 d^{2}} + \frac {b^{2} x^{6} \sin ^{2}{\left (c + d x^{3} \right )}}{12} + \frac {b^{2} x^{6} \cos ^{2}{\left (c + d x^{3} \right )}}{12} - \frac {b^{2} x^{3} \sin {\left (c + d x^{3} \right )} \cos {\left (c + d x^{3} \right )}}{6 d} + \frac {b^{2} \sin ^{2}{\left (c + d x^{3} \right )}}{12 d^{2}} & \text {for}\: d \neq 0 \\\frac {x^{6} \left (a + b \sin {\left (c \right )}\right )^{2}}{6} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(a+b*sin(d*x**3+c))**2,x)

[Out]

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

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Giac [A]
time = 7.24, size = 165, normalized size = 1.54 \begin {gather*} \frac {4 \, {\left (d x^{3} + c\right )}^{2} a^{2} + 2 \, {\left (d x^{3} + c\right )}^{2} b^{2} - 16 \, {\left (d x^{3} + c\right )} a b \cos \left (d x^{3} + c\right ) - 2 \, {\left (d x^{3} + c\right )} b^{2} \sin \left (2 \, d x^{3} + 2 \, c\right ) - b^{2} \cos \left (2 \, d x^{3} + 2 \, c\right ) + 16 \, a b \sin \left (d x^{3} + c\right )}{24 \, d^{2}} - \frac {4 \, {\left (d x^{3} + c\right )} a^{2} c + {\left (2 \, d x^{3} + 2 \, c - \sin \left (2 \, d x^{3} + 2 \, c\right )\right )} b^{2} c - 8 \, a b c \cos \left (d x^{3} + c\right )}{12 \, d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*sin(d*x^3+c))^2,x, algorithm="giac")

[Out]

1/24*(4*(d*x^3 + c)^2*a^2 + 2*(d*x^3 + c)^2*b^2 - 16*(d*x^3 + c)*a*b*cos(d*x^3 + c) - 2*(d*x^3 + c)*b^2*sin(2*
d*x^3 + 2*c) - b^2*cos(2*d*x^3 + 2*c) + 16*a*b*sin(d*x^3 + c))/d^2 - 1/12*(4*(d*x^3 + c)*a^2*c + (2*d*x^3 + 2*
c - sin(2*d*x^3 + 2*c))*b^2*c - 8*a*b*c*cos(d*x^3 + c))/d^2

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Mupad [B]
time = 0.27, size = 95, normalized size = 0.89 \begin {gather*} -\frac {b^2\,{\cos \left (d\,x^3+c\right )}^2-2\,a^2\,d^2\,x^6-b^2\,d^2\,x^6-8\,a\,b\,\sin \left (d\,x^3+c\right )+8\,a\,b\,d\,x^3\,\cos \left (d\,x^3+c\right )+2\,b^2\,d\,x^3\,\cos \left (d\,x^3+c\right )\,\sin \left (d\,x^3+c\right )}{12\,d^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(a + b*sin(c + d*x^3))^2,x)

[Out]

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

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