3.1.65 \(\int x^3 (a+b \sin (c+d x^3)) \, dx\) [65]

Optimal. Leaf size=106 \[ \frac {a x^4}{4}-\frac {b x \cos \left (c+d x^3\right )}{3 d}-\frac {b e^{i c} x \Gamma \left (\frac {1}{3},-i d x^3\right )}{18 d \sqrt [3]{-i d x^3}}-\frac {b e^{-i c} x \Gamma \left (\frac {1}{3},i d x^3\right )}{18 d \sqrt [3]{i d x^3}} \]

[Out]

1/4*a*x^4-1/3*b*x*cos(d*x^3+c)/d-1/18*b*exp(I*c)*x*GAMMA(1/3,-I*d*x^3)/d/(-I*d*x^3)^(1/3)-1/18*b*x*GAMMA(1/3,I
*d*x^3)/d/exp(I*c)/(I*d*x^3)^(1/3)

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Rubi [A]
time = 0.03, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {14, 3466, 3437, 2239} \begin {gather*} -\frac {b e^{i c} x \text {Gamma}\left (\frac {1}{3},-i d x^3\right )}{18 d \sqrt [3]{-i d x^3}}-\frac {b e^{-i c} x \text {Gamma}\left (\frac {1}{3},i d x^3\right )}{18 d \sqrt [3]{i d x^3}}+\frac {a x^4}{4}-\frac {b x \cos \left (c+d x^3\right )}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*Sin[c + d*x^3]),x]

[Out]

(a*x^4)/4 - (b*x*Cos[c + d*x^3])/(3*d) - (b*E^(I*c)*x*Gamma[1/3, (-I)*d*x^3])/(18*d*((-I)*d*x^3)^(1/3)) - (b*x
*Gamma[1/3, I*d*x^3])/(18*d*E^(I*c)*(I*d*x^3)^(1/3))

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2239

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_)), x_Symbol] :> Simp[(-F^a)*(c + d*x)*(Gamma[1/n, (-b)*(c + d
*x)^n*Log[F]]/(d*n*((-b)*(c + d*x)^n*Log[F])^(1/n))), x] /; FreeQ[{F, a, b, c, d, n}, x] &&  !IntegerQ[2/n]

Rule 3437

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

Rule 3466

Int[((e_.)*(x_))^(m_.)*Sin[(c_.) + (d_.)*(x_)^(n_)], x_Symbol] :> Simp[(-e^(n - 1))*(e*x)^(m - n + 1)*(Cos[c +
d*x^n]/(d*n)), x] + Dist[e^n*((m - n + 1)/(d*n)), Int[(e*x)^(m - n)*Cos[c + d*x^n], x], x] /; FreeQ[{c, d, e}
, x] && IGtQ[n, 0] && LtQ[n, m + 1]

Rubi steps

\begin {align*} \int x^3 \left (a+b \sin \left (c+d x^3\right )\right ) \, dx &=\int \left (a x^3+b x^3 \sin \left (c+d x^3\right )\right ) \, dx\\ &=\frac {a x^4}{4}+b \int x^3 \sin \left (c+d x^3\right ) \, dx\\ &=\frac {a x^4}{4}-\frac {b x \cos \left (c+d x^3\right )}{3 d}+\frac {b \int \cos \left (c+d x^3\right ) \, dx}{3 d}\\ &=\frac {a x^4}{4}-\frac {b x \cos \left (c+d x^3\right )}{3 d}+\frac {b \int e^{-i c-i d x^3} \, dx}{6 d}+\frac {b \int e^{i c+i d x^3} \, dx}{6 d}\\ &=\frac {a x^4}{4}-\frac {b x \cos \left (c+d x^3\right )}{3 d}-\frac {b e^{i c} x \Gamma \left (\frac {1}{3},-i d x^3\right )}{18 d \sqrt [3]{-i d x^3}}-\frac {b e^{-i c} x \Gamma \left (\frac {1}{3},i d x^3\right )}{18 d \sqrt [3]{i d x^3}}\\ \end {align*}

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Mathematica [A]
time = 0.13, size = 124, normalized size = 1.17 \begin {gather*} \frac {d x^7 \left (3 \sqrt [3]{d^2 x^6} \left (3 a d x^3-4 b \cos \left (c+d x^3\right )\right )-2 b \sqrt [3]{-i d x^3} \Gamma \left (\frac {1}{3},i d x^3\right ) (\cos (c)-i \sin (c))-2 b \sqrt [3]{i d x^3} \Gamma \left (\frac {1}{3},-i d x^3\right ) (\cos (c)+i \sin (c))\right )}{36 \left (d^2 x^6\right )^{4/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d*x^7*(3*(d^2*x^6)^(1/3)*(3*a*d*x^3 - 4*b*Cos[c + d*x^3]) - 2*b*((-I)*d*x^3)^(1/3)*Gamma[1/3, I*d*x^3]*(Cos[c
] - I*Sin[c]) - 2*b*(I*d*x^3)^(1/3)*Gamma[1/3, (-I)*d*x^3]*(Cos[c] + I*Sin[c])))/(36*(d^2*x^6)^(4/3))

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int x^{3} \left (a +b \sin \left (d \,x^{3}+c \right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.30, size = 110, normalized size = 1.04 \begin {gather*} \frac {1}{4} \, a x^{4} - \frac {{\left (12 \, \left (d x^{3}\right )^{\frac {1}{3}} x \cos \left (d x^{3} + c\right ) + {\left ({\left ({\left (\sqrt {3} - i\right )} \Gamma \left (\frac {1}{3}, i \, d x^{3}\right ) + {\left (\sqrt {3} + i\right )} \Gamma \left (\frac {1}{3}, -i \, d x^{3}\right )\right )} \cos \left (c\right ) + {\left ({\left (-i \, \sqrt {3} - 1\right )} \Gamma \left (\frac {1}{3}, i \, d x^{3}\right ) + {\left (i \, \sqrt {3} - 1\right )} \Gamma \left (\frac {1}{3}, -i \, d x^{3}\right )\right )} \sin \left (c\right )\right )} x\right )} b}{36 \, \left (d x^{3}\right )^{\frac {1}{3}} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a*x^4 - 1/36*(12*(d*x^3)^(1/3)*x*cos(d*x^3 + c) + (((sqrt(3) - I)*gamma(1/3, I*d*x^3) + (sqrt(3) + I)*gamm
a(1/3, -I*d*x^3))*cos(c) + ((-I*sqrt(3) - 1)*gamma(1/3, I*d*x^3) + (I*sqrt(3) - 1)*gamma(1/3, -I*d*x^3))*sin(c
))*x)*b/((d*x^3)^(1/3)*d)

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Fricas [A]
time = 0.11, size = 68, normalized size = 0.64 \begin {gather*} \frac {9 \, a d^{2} x^{4} - 12 \, b d x \cos \left (d x^{3} + c\right ) + 2 i \, b \left (i \, d\right )^{\frac {2}{3}} e^{\left (-i \, c\right )} \Gamma \left (\frac {1}{3}, i \, d x^{3}\right ) - 2 i \, b \left (-i \, d\right )^{\frac {2}{3}} e^{\left (i \, c\right )} \Gamma \left (\frac {1}{3}, -i \, d x^{3}\right )}{36 \, d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/36*(9*a*d^2*x^4 - 12*b*d*x*cos(d*x^3 + c) + 2*I*b*(I*d)^(2/3)*e^(-I*c)*gamma(1/3, I*d*x^3) - 2*I*b*(-I*d)^(2
/3)*e^(I*c)*gamma(1/3, -I*d*x^3))/d^2

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{3} \left (a + b \sin {\left (c + d x^{3} \right )}\right )\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**3*(a + b*sin(c + d*x**3)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^3\,\left (a+b\,\sin \left (d\,x^3+c\right )\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a + b*sin(c + d*x^3)),x)

[Out]

int(x^3*(a + b*sin(c + d*x^3)), x)

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