3.69 \(\int x^5 (a+b \sin (c+d x^3))^2 \, dx\)

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

[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)

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Rubi [A]  time = 0.132765, antiderivative size = 107, normalized size of antiderivative = 1., 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 = {3379, 3317, 3296, 2637, 3310, 30} \[ \frac{a^2 x^6}{6}+\frac{2 a b \sin \left (c+d x^3\right )}{3 d^2}-\frac{2 a b x^3 \cos \left (c+d x^3\right )}{3 d}+\frac{b^2 \sin ^2\left (c+d x^3\right )}{12 d^2}-\frac{b^2 x^3 \sin \left (c+d x^3\right ) \cos \left (c+d x^3\right )}{6 d}+\frac{b^2 x^6}{12} \]

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 3379

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]))

Rule 3317

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 3296

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 2637

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

Rule 3310

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 30

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

Rubi steps

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

Mathematica [A]  time = 0.281155, size = 92, normalized size = 0.86 \[ \frac{4 a^2 d^2 x^6+16 a b \sin \left (c+d x^3\right )-16 a b d x^3 \cos \left (c+d x^3\right )-2 b^2 d x^3 \sin \left (2 \left (c+d x^3\right )\right )-b^2 \cos \left (2 \left (c+d x^3\right )\right )+2 b^2 d^2 x^6}{24 d^2} \]

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 = 0.095, size = 137, normalized size = 1.3 \begin{align*}{\frac{{a}^{2}{x}^{6}}{6}}+{\frac{{b}^{2}{x}^{6}}{6}}-{\frac{{b}^{2}}{2} \left ({\frac{{x}^{6}}{6}}+{\frac{1}{1+ \left ( \tan \left ( d{x}^{3}+c \right ) \right ) ^{2}} \left ({\frac{1}{6\,{d}^{2}}}+{\frac{{x}^{3}\tan \left ( d{x}^{3}+c \right ) }{3\,d}} \right ) } \right ) }+{\frac{1}{2} \left ({\frac{8\,ab}{3\,{d}^{2}}\tan \left ({\frac{d{x}^{3}}{2}}+{\frac{c}{2}} \right ) }-{\frac{4\,ab{x}^{3}}{3\,d}}+{\frac{4\,ab{x}^{3}}{3\,d} \left ( \tan \left ({\frac{d{x}^{3}}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}} \right ) \left ( 1+ \left ( \tan \left ({\frac{d{x}^{3}}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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*a*b/d^2
*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 = 1.01646, size = 117, normalized size = 1.09 \begin{align*} \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{align*}

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 = 1.76182, size = 189, normalized size = 1.77 \begin{align*} \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{align*}

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 = 8.30871, size = 143, normalized size = 1.34 \begin{align*} \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} \cos ^{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{align*}

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*cos(
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 = 1.10524, size = 166, normalized size = 1.55 \begin{align*} \frac{\frac{4 \,{\left ({\left (d x^{3} + c\right )}^{2} - 2 \,{\left (d x^{3} + c\right )} c\right )} a^{2}}{d} - \frac{16 \,{\left (d x^{3} \cos \left (d x^{3} + c\right ) - \sin \left (d x^{3} + c\right )\right )} a b}{d} - \frac{{\left (2 \, d x^{3} \sin \left (2 \, d x^{3} + 2 \, c\right ) - 2 \,{\left (d x^{3} + c\right )}^{2} + 4 \,{\left (d x^{3} + c\right )} c + \cos \left (2 \, d x^{3} + 2 \, c\right )\right )} b^{2}}{d}}{24 \, d} \end{align*}

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