3.1.61 \(\int x^4 (a+b \sin (c+d x^3)) \, dx\) [61]

Optimal. Leaf size=112 \[ \frac {a x^5}{5}-\frac {b x^2 \cos \left (c+d x^3\right )}{3 d}-\frac {b e^{i c} x^2 \Gamma \left (\frac {2}{3},-i d x^3\right )}{9 d \left (-i d x^3\right )^{2/3}}-\frac {b e^{-i c} x^2 \Gamma \left (\frac {2}{3},i d x^3\right )}{9 d \left (i d x^3\right )^{2/3}} \]

[Out]

1/5*a*x^5-1/3*b*x^2*cos(d*x^3+c)/d-1/9*b*exp(I*c)*x^2*GAMMA(2/3,-I*d*x^3)/d/(-I*d*x^3)^(2/3)-1/9*b*x^2*GAMMA(2
/3,I*d*x^3)/d/exp(I*c)/(I*d*x^3)^(2/3)

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Rubi [A]
time = 0.06, antiderivative size = 112, 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, 3471, 2250} \begin {gather*} -\frac {b e^{i c} x^2 \text {Gamma}\left (\frac {2}{3},-i d x^3\right )}{9 d \left (-i d x^3\right )^{2/3}}-\frac {b e^{-i c} x^2 \text {Gamma}\left (\frac {2}{3},i d x^3\right )}{9 d \left (i d x^3\right )^{2/3}}+\frac {a x^5}{5}-\frac {b x^2 \cos \left (c+d x^3\right )}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^4*(a + b*Sin[c + d*x^3]),x]

[Out]

(a*x^5)/5 - (b*x^2*Cos[c + d*x^3])/(3*d) - (b*E^(I*c)*x^2*Gamma[2/3, (-I)*d*x^3])/(9*d*((-I)*d*x^3)^(2/3)) - (
b*x^2*Gamma[2/3, I*d*x^3])/(9*d*E^(I*c)*(I*d*x^3)^(2/3))

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 2250

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(-F^a)*((e +
f*x)^(m + 1)/(f*n*((-b)*(c + d*x)^n*Log[F])^((m + 1)/n)))*Gamma[(m + 1)/n, (-b)*(c + d*x)^n*Log[F]], x] /; Fre
eQ[{F, a, b, c, d, e, f, m, n}, x] && EqQ[d*e - c*f, 0]

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]

Rule 3471

Int[Cos[(c_.) + (d_.)*(x_)^(n_)]*((e_.)*(x_))^(m_.), x_Symbol] :> Dist[1/2, Int[(e*x)^m*E^((-c)*I - d*I*x^n),
x], x] + Dist[1/2, Int[(e*x)^m*E^(c*I + d*I*x^n), x], x] /; FreeQ[{c, d, e, m}, x] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int x^4 \left (a+b \sin \left (c+d x^3\right )\right ) \, dx &=\int \left (a x^4+b x^4 \sin \left (c+d x^3\right )\right ) \, dx\\ &=\frac {a x^5}{5}+b \int x^4 \sin \left (c+d x^3\right ) \, dx\\ &=\frac {a x^5}{5}-\frac {b x^2 \cos \left (c+d x^3\right )}{3 d}+\frac {(2 b) \int x \cos \left (c+d x^3\right ) \, dx}{3 d}\\ &=\frac {a x^5}{5}-\frac {b x^2 \cos \left (c+d x^3\right )}{3 d}+\frac {b \int e^{-i c-i d x^3} x \, dx}{3 d}+\frac {b \int e^{i c+i d x^3} x \, dx}{3 d}\\ &=\frac {a x^5}{5}-\frac {b x^2 \cos \left (c+d x^3\right )}{3 d}-\frac {b e^{i c} x^2 \Gamma \left (\frac {2}{3},-i d x^3\right )}{9 d \left (-i d x^3\right )^{2/3}}-\frac {b e^{-i c} x^2 \Gamma \left (\frac {2}{3},i d x^3\right )}{9 d \left (i d x^3\right )^{2/3}}\\ \end {align*}

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Mathematica [A]
time = 0.16, size = 124, normalized size = 1.11 \begin {gather*} \frac {d x^8 \left (3 \left (d^2 x^6\right )^{2/3} \left (3 a d x^3-5 b \cos \left (c+d x^3\right )\right )-5 b \left (-i d x^3\right )^{2/3} \Gamma \left (\frac {2}{3},i d x^3\right ) (\cos (c)-i \sin (c))-5 b \left (i d x^3\right )^{2/3} \Gamma \left (\frac {2}{3},-i d x^3\right ) (\cos (c)+i \sin (c))\right )}{45 \left (d^2 x^6\right )^{5/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^4*(a + b*Sin[c + d*x^3]),x]

[Out]

(d*x^8*(3*(d^2*x^6)^(2/3)*(3*a*d*x^3 - 5*b*Cos[c + d*x^3]) - 5*b*((-I)*d*x^3)^(2/3)*Gamma[2/3, I*d*x^3]*(Cos[c
] - I*Sin[c]) - 5*b*(I*d*x^3)^(2/3)*Gamma[2/3, (-I)*d*x^3]*(Cos[c] + I*Sin[c])))/(45*(d^2*x^6)^(5/3))

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int x^{4} \left (a +b \sin \left (d \,x^{3}+c \right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(a+b*sin(d*x^3+c)),x)

[Out]

int(x^4*(a+b*sin(d*x^3+c)),x)

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Maxima [A]
time = 0.31, size = 109, normalized size = 0.97 \begin {gather*} \frac {1}{5} \, a x^{5} - \frac {{\left (6 \, d x^{3} \cos \left (d x^{3} + c\right ) - \left (d x^{3}\right )^{\frac {1}{3}} {\left ({\left ({\left (i \, \sqrt {3} - 1\right )} \Gamma \left (\frac {2}{3}, i \, d x^{3}\right ) + {\left (-i \, \sqrt {3} - 1\right )} \Gamma \left (\frac {2}{3}, -i \, d x^{3}\right )\right )} \cos \left (c\right ) + {\left ({\left (\sqrt {3} + i\right )} \Gamma \left (\frac {2}{3}, i \, d x^{3}\right ) + {\left (\sqrt {3} - i\right )} \Gamma \left (\frac {2}{3}, -i \, d x^{3}\right )\right )} \sin \left (c\right )\right )}\right )} b}{18 \, d^{2} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*sin(d*x^3+c)),x, algorithm="maxima")

[Out]

1/5*a*x^5 - 1/18*(6*d*x^3*cos(d*x^3 + c) - (d*x^3)^(1/3)*(((I*sqrt(3) - 1)*gamma(2/3, I*d*x^3) + (-I*sqrt(3) -
1)*gamma(2/3, -I*d*x^3))*cos(c) + ((sqrt(3) + I)*gamma(2/3, I*d*x^3) + (sqrt(3) - I)*gamma(2/3, -I*d*x^3))*si
n(c)))*b/(d^2*x)

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Fricas [A]
time = 0.10, size = 70, normalized size = 0.62 \begin {gather*} \frac {9 \, a d^{2} x^{5} - 15 \, b d x^{2} \cos \left (d x^{3} + c\right ) + 5 i \, b \left (i \, d\right )^{\frac {1}{3}} e^{\left (-i \, c\right )} \Gamma \left (\frac {2}{3}, i \, d x^{3}\right ) - 5 i \, b \left (-i \, d\right )^{\frac {1}{3}} e^{\left (i \, c\right )} \Gamma \left (\frac {2}{3}, -i \, d x^{3}\right )}{45 \, d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*sin(d*x^3+c)),x, algorithm="fricas")

[Out]

1/45*(9*a*d^2*x^5 - 15*b*d*x^2*cos(d*x^3 + c) + 5*I*b*(I*d)^(1/3)*e^(-I*c)*gamma(2/3, I*d*x^3) - 5*I*b*(-I*d)^
(1/3)*e^(I*c)*gamma(2/3, -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^{4} \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**4*(a+b*sin(d*x**3+c)),x)

[Out]

Integral(x**4*(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^4*(a+b*sin(d*x^3+c)),x, algorithm="giac")

[Out]

integrate((b*sin(d*x^3 + c) + a)*x^4, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^4\,\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^4*(a + b*sin(c + d*x^3)),x)

[Out]

int(x^4*(a + b*sin(c + d*x^3)), x)

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