Integrand size = 16, antiderivative size = 106 \[ \int x^3 \left (a+b \sin \left (c+d x^3\right )\right ) \, dx=\frac {a x^4}{4}-\frac {b x \cos \left (c+d x^3\right )}{3 d}-\frac {b e^{i c} x \Gamma \left (\frac {1}{3},-i d x^3\right )}{18 d \sqrt [3]{-i d x^3}}-\frac {b e^{-i c} x \Gamma \left (\frac {1}{3},i d x^3\right )}{18 d \sqrt [3]{i d x^3}} \] Output:
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)
Time = 0.24 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.17 \[ \int x^3 \left (a+b \sin \left (c+d x^3\right )\right ) \, dx=\frac {d x^7 \left (3 \sqrt [3]{d^2 x^6} \left (3 a d x^3-4 b \cos \left (c+d x^3\right )\right )-2 b \sqrt [3]{-i d x^3} \Gamma \left (\frac {1}{3},i d x^3\right ) (\cos (c)-i \sin (c))-2 b \sqrt [3]{i d x^3} \Gamma \left (\frac {1}{3},-i d x^3\right ) (\cos (c)+i \sin (c))\right )}{36 \left (d^2 x^6\right )^{4/3}} \] Input:
Integrate[x^3*(a + b*Sin[c + d*x^3]),x]
Output:
(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)*Ga mma[1/3, (-I)*d*x^3]*(Cos[c] + I*Sin[c])))/(36*(d^2*x^6)^(4/3))
Time = 0.24 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2010, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x^3 \left (a+b \sin \left (c+d x^3\right )\right ) \, dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \int \left (a x^3+b x^3 \sin \left (c+d x^3\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a x^4}{4}-\frac {b x \cos \left (c+d x^3\right )}{3 d}-\frac {b e^{i c} x \Gamma \left (\frac {1}{3},-i d x^3\right )}{18 d \sqrt [3]{-i d x^3}}-\frac {b e^{-i c} x \Gamma \left (\frac {1}{3},i d x^3\right )}{18 d \sqrt [3]{i d x^3}}\) |
Input:
Int[x^3*(a + b*Sin[c + d*x^3]),x]
Output:
(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))
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]]
\[\int x^{3} \left (a +b \sin \left (d \,x^{3}+c \right )\right )d x\]
Input:
int(x^3*(a+b*sin(d*x^3+c)),x)
Output:
int(x^3*(a+b*sin(d*x^3+c)),x)
Time = 0.08 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.75 \[ \int x^3 \left (a+b \sin \left (c+d x^3\right )\right ) \, dx=\frac {9 \, a d^{2} x^{4} - 12 \, b d x \cos \left (d x^{3} + c\right ) - 2 \, {\left (-i \, b \cos \left (c\right ) - b \sin \left (c\right )\right )} \left (i \, d\right )^{\frac {2}{3}} \Gamma \left (\frac {1}{3}, i \, d x^{3}\right ) - 2 \, {\left (i \, b \cos \left (c\right ) - b \sin \left (c\right )\right )} \left (-i \, d\right )^{\frac {2}{3}} \Gamma \left (\frac {1}{3}, -i \, d x^{3}\right )}{36 \, d^{2}} \] Input:
integrate(x^3*(a+b*sin(d*x^3+c)),x, algorithm="fricas")
Output:
1/36*(9*a*d^2*x^4 - 12*b*d*x*cos(d*x^3 + c) - 2*(-I*b*cos(c) - b*sin(c))*( I*d)^(2/3)*gamma(1/3, I*d*x^3) - 2*(I*b*cos(c) - b*sin(c))*(-I*d)^(2/3)*ga mma(1/3, -I*d*x^3))/d^2
\[ \int x^3 \left (a+b \sin \left (c+d x^3\right )\right ) \, dx=\int x^{3} \left (a + b \sin {\left (c + d x^{3} \right )}\right )\, dx \] Input:
integrate(x**3*(a+b*sin(d*x**3+c)),x)
Output:
Integral(x**3*(a + b*sin(c + d*x**3)), x)
Time = 0.18 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.04 \[ \int x^3 \left (a+b \sin \left (c+d x^3\right )\right ) \, dx=\frac {1}{4} \, a x^{4} - \frac {{\left (12 \, \left (d x^{3}\right )^{\frac {1}{3}} x \cos \left (d x^{3} + c\right ) + {\left ({\left ({\left (\sqrt {3} - i\right )} \Gamma \left (\frac {1}{3}, i \, d x^{3}\right ) + {\left (\sqrt {3} + i\right )} \Gamma \left (\frac {1}{3}, -i \, d x^{3}\right )\right )} \cos \left (c\right ) + {\left ({\left (-i \, \sqrt {3} - 1\right )} \Gamma \left (\frac {1}{3}, i \, d x^{3}\right ) + {\left (i \, \sqrt {3} - 1\right )} \Gamma \left (\frac {1}{3}, -i \, d x^{3}\right )\right )} \sin \left (c\right )\right )} x\right )} b}{36 \, \left (d x^{3}\right )^{\frac {1}{3}} d} \] Input:
integrate(x^3*(a+b*sin(d*x^3+c)),x, algorithm="maxima")
Output:
1/4*a*x^4 - 1/36*(12*(d*x^3)^(1/3)*x*cos(d*x^3 + c) + (((sqrt(3) - I)*gamm a(1/3, I*d*x^3) + (sqrt(3) + I)*gamma(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)
\[ \int x^3 \left (a+b \sin \left (c+d x^3\right )\right ) \, dx=\int { {\left (b \sin \left (d x^{3} + c\right ) + a\right )} x^{3} \,d x } \] Input:
integrate(x^3*(a+b*sin(d*x^3+c)),x, algorithm="giac")
Output:
integrate((b*sin(d*x^3 + c) + a)*x^3, x)
Timed out. \[ \int x^3 \left (a+b \sin \left (c+d x^3\right )\right ) \, dx=\int x^3\,\left (a+b\,\sin \left (d\,x^3+c\right )\right ) \,d x \] Input:
int(x^3*(a + b*sin(c + d*x^3)),x)
Output:
int(x^3*(a + b*sin(c + d*x^3)), x)
\[ \int x^3 \left (a+b \sin \left (c+d x^3\right )\right ) \, dx=\left (\int \sin \left (d \,x^{3}+c \right ) x^{3}d x \right ) b +\frac {a \,x^{4}}{4} \] Input:
int(x^3*(a+b*sin(d*x^3+c)),x)
Output:
(4*int(sin(c + d*x**3)*x**3,x)*b + a*x**4)/4