3.1.74 \(\int x (a+b \sin (c+d x^3))^2 \, dx\) [74]

Optimal. Leaf size=193 \[ \frac {1}{4} \left (2 a^2+b^2\right ) x^2+\frac {i a b e^{i c} x^2 \Gamma \left (\frac {2}{3},-i d x^3\right )}{3 \left (-i d x^3\right )^{2/3}}-\frac {i a b e^{-i c} x^2 \Gamma \left (\frac {2}{3},i d x^3\right )}{3 \left (i d x^3\right )^{2/3}}+\frac {b^2 e^{2 i c} x^2 \Gamma \left (\frac {2}{3},-2 i d x^3\right )}{12\ 2^{2/3} \left (-i d x^3\right )^{2/3}}+\frac {b^2 e^{-2 i c} x^2 \Gamma \left (\frac {2}{3},2 i d x^3\right )}{12\ 2^{2/3} \left (i d x^3\right )^{2/3}} \]

[Out]

1/4*(2*a^2+b^2)*x^2+1/3*I*a*b*exp(I*c)*x^2*GAMMA(2/3,-I*d*x^3)/(-I*d*x^3)^(2/3)-1/3*I*a*b*x^2*GAMMA(2/3,I*d*x^
3)/exp(I*c)/(I*d*x^3)^(2/3)+1/24*b^2*exp(2*I*c)*x^2*GAMMA(2/3,-2*I*d*x^3)*2^(1/3)/(-I*d*x^3)^(2/3)+1/24*b^2*x^
2*GAMMA(2/3,2*I*d*x^3)*2^(1/3)/exp(2*I*c)/(I*d*x^3)^(2/3)

________________________________________________________________________________________

Rubi [A]
time = 0.09, antiderivative size = 193, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {3484, 6, 3471, 2250, 3470} \begin {gather*} \frac {i a b e^{i c} x^2 \text {Gamma}\left (\frac {2}{3},-i d x^3\right )}{3 \left (-i d x^3\right )^{2/3}}-\frac {i a b e^{-i c} x^2 \text {Gamma}\left (\frac {2}{3},i d x^3\right )}{3 \left (i d x^3\right )^{2/3}}+\frac {b^2 e^{2 i c} x^2 \text {Gamma}\left (\frac {2}{3},-2 i d x^3\right )}{12\ 2^{2/3} \left (-i d x^3\right )^{2/3}}+\frac {b^2 e^{-2 i c} x^2 \text {Gamma}\left (\frac {2}{3},2 i d x^3\right )}{12\ 2^{2/3} \left (i d x^3\right )^{2/3}}+\frac {1}{4} x^2 \left (2 a^2+b^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*(a + b*Sin[c + d*x^3])^2,x]

[Out]

((2*a^2 + b^2)*x^2)/4 + ((I/3)*a*b*E^(I*c)*x^2*Gamma[2/3, (-I)*d*x^3])/((-I)*d*x^3)^(2/3) - ((I/3)*a*b*x^2*Gam
ma[2/3, I*d*x^3])/(E^(I*c)*(I*d*x^3)^(2/3)) + (b^2*E^((2*I)*c)*x^2*Gamma[2/3, (-2*I)*d*x^3])/(12*2^(2/3)*((-I)
*d*x^3)^(2/3)) + (b^2*x^2*Gamma[2/3, (2*I)*d*x^3])/(12*2^(2/3)*E^((2*I)*c)*(I*d*x^3)^(2/3))

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

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 3470

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

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]

Rule 3484

Int[((e_.)*(x_))^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_), x_Symbol] :> Int[ExpandTrigReduce[(e
*x)^m, (a + b*Sin[c + d*x^n])^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[p, 1] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int x \left (a+b \sin \left (c+d x^3\right )\right )^2 \, dx &=\int \left (a^2 x+\frac {b^2 x}{2}-\frac {1}{2} b^2 x \cos \left (2 c+2 d x^3\right )+2 a b x \sin \left (c+d x^3\right )\right ) \, dx\\ &=\int \left (\left (a^2+\frac {b^2}{2}\right ) x-\frac {1}{2} b^2 x \cos \left (2 c+2 d x^3\right )+2 a b x \sin \left (c+d x^3\right )\right ) \, dx\\ &=\frac {1}{4} \left (2 a^2+b^2\right ) x^2+(2 a b) \int x \sin \left (c+d x^3\right ) \, dx-\frac {1}{2} b^2 \int x \cos \left (2 c+2 d x^3\right ) \, dx\\ &=\frac {1}{4} \left (2 a^2+b^2\right ) x^2+(i a b) \int e^{-i c-i d x^3} x \, dx-(i a b) \int e^{i c+i d x^3} x \, dx-\frac {1}{4} b^2 \int e^{-2 i c-2 i d x^3} x \, dx-\frac {1}{4} b^2 \int e^{2 i c+2 i d x^3} x \, dx\\ &=\frac {1}{4} \left (2 a^2+b^2\right ) x^2+\frac {i a b e^{i c} x^2 \Gamma \left (\frac {2}{3},-i d x^3\right )}{3 \left (-i d x^3\right )^{2/3}}-\frac {i a b e^{-i c} x^2 \Gamma \left (\frac {2}{3},i d x^3\right )}{3 \left (i d x^3\right )^{2/3}}+\frac {b^2 e^{2 i c} x^2 \Gamma \left (\frac {2}{3},-2 i d x^3\right )}{12\ 2^{2/3} \left (-i d x^3\right )^{2/3}}+\frac {b^2 e^{-2 i c} x^2 \Gamma \left (\frac {2}{3},2 i d x^3\right )}{12\ 2^{2/3} \left (i d x^3\right )^{2/3}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.22, size = 283, normalized size = 1.47 \begin {gather*} \frac {x^2 \left (12 a^2 \left (d^2 x^6\right )^{2/3}+6 b^2 \left (d^2 x^6\right )^{2/3}+\sqrt [3]{2} b^2 \left (i d x^3\right )^{2/3} \cos (2 c) \Gamma \left (\frac {2}{3},-2 i d x^3\right )+\sqrt [3]{2} b^2 \left (-i d x^3\right )^{2/3} \cos (2 c) \Gamma \left (\frac {2}{3},2 i d x^3\right )-8 i a b \left (-i d x^3\right )^{2/3} \Gamma \left (\frac {2}{3},i d x^3\right ) (\cos (c)-i \sin (c))+8 i a b \left (i d x^3\right )^{2/3} \Gamma \left (\frac {2}{3},-i d x^3\right ) (\cos (c)+i \sin (c))+i \sqrt [3]{2} b^2 \left (i d x^3\right )^{2/3} \Gamma \left (\frac {2}{3},-2 i d x^3\right ) \sin (2 c)-i \sqrt [3]{2} b^2 \left (-i d x^3\right )^{2/3} \Gamma \left (\frac {2}{3},2 i d x^3\right ) \sin (2 c)\right )}{24 \left (d^2 x^6\right )^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^2*(12*a^2*(d^2*x^6)^(2/3) + 6*b^2*(d^2*x^6)^(2/3) + 2^(1/3)*b^2*(I*d*x^3)^(2/3)*Cos[2*c]*Gamma[2/3, (-2*I)*
d*x^3] + 2^(1/3)*b^2*((-I)*d*x^3)^(2/3)*Cos[2*c]*Gamma[2/3, (2*I)*d*x^3] - (8*I)*a*b*((-I)*d*x^3)^(2/3)*Gamma[
2/3, I*d*x^3]*(Cos[c] - I*Sin[c]) + (8*I)*a*b*(I*d*x^3)^(2/3)*Gamma[2/3, (-I)*d*x^3]*(Cos[c] + I*Sin[c]) + I*2
^(1/3)*b^2*(I*d*x^3)^(2/3)*Gamma[2/3, (-2*I)*d*x^3]*Sin[2*c] - I*2^(1/3)*b^2*((-I)*d*x^3)^(2/3)*Gamma[2/3, (2*
I)*d*x^3]*Sin[2*c]))/(24*(d^2*x^6)^(2/3))

________________________________________________________________________________________

Maple [F]
time = 0.18, size = 0, normalized size = 0.00 \[\int x \left (a +b \sin \left (d \,x^{3}+c \right )\right )^{2}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.35, size = 199, normalized size = 1.03 \begin {gather*} \frac {1}{2} \, a^{2} x^{2} - \frac {\left (d x^{3}\right )^{\frac {1}{3}} {\left ({\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 )} \cos \left (c\right ) - {\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 )} \sin \left (c\right )\right )} a b}{6 \, d x} + \frac {{\left (12 \, d x^{3} - 2^{\frac {1}{3}} \left (d x^{3}\right )^{\frac {1}{3}} {\left ({\left ({\left (i \, \sqrt {3} - 1\right )} \Gamma \left (\frac {2}{3}, 2 i \, d x^{3}\right ) + {\left (-i \, \sqrt {3} - 1\right )} \Gamma \left (\frac {2}{3}, -2 i \, d x^{3}\right )\right )} \cos \left (2 \, c\right ) + {\left ({\left (\sqrt {3} + i\right )} \Gamma \left (\frac {2}{3}, 2 i \, d x^{3}\right ) + {\left (\sqrt {3} - i\right )} \Gamma \left (\frac {2}{3}, -2 i \, d x^{3}\right )\right )} \sin \left (2 \, c\right )\right )}\right )} b^{2}}{48 \, d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a^2*x^2 - 1/6*(d*x^3)^(1/3)*(((sqrt(3) + I)*gamma(2/3, I*d*x^3) + (sqrt(3) - I)*gamma(2/3, -I*d*x^3))*cos(
c) - ((I*sqrt(3) - 1)*gamma(2/3, I*d*x^3) + (-I*sqrt(3) - 1)*gamma(2/3, -I*d*x^3))*sin(c))*a*b/(d*x) + 1/48*(1
2*d*x^3 - 2^(1/3)*(d*x^3)^(1/3)*(((I*sqrt(3) - 1)*gamma(2/3, 2*I*d*x^3) + (-I*sqrt(3) - 1)*gamma(2/3, -2*I*d*x
^3))*cos(2*c) + ((sqrt(3) + I)*gamma(2/3, 2*I*d*x^3) + (sqrt(3) - I)*gamma(2/3, -2*I*d*x^3))*sin(2*c)))*b^2/(d
*x)

________________________________________________________________________________________

Fricas [A]
time = 0.14, size = 107, normalized size = 0.55 \begin {gather*} \frac {-i \, b^{2} \left (2 i \, d\right )^{\frac {1}{3}} e^{\left (-2 i \, c\right )} \Gamma \left (\frac {2}{3}, 2 i \, d x^{3}\right ) - 8 \, a b \left (i \, d\right )^{\frac {1}{3}} e^{\left (-i \, c\right )} \Gamma \left (\frac {2}{3}, i \, d x^{3}\right ) - 8 \, a b \left (-i \, d\right )^{\frac {1}{3}} e^{\left (i \, c\right )} \Gamma \left (\frac {2}{3}, -i \, d x^{3}\right ) + i \, b^{2} \left (-2 i \, d\right )^{\frac {1}{3}} e^{\left (2 i \, c\right )} \Gamma \left (\frac {2}{3}, -2 i \, d x^{3}\right ) + 6 \, {\left (2 \, a^{2} + b^{2}\right )} d x^{2}}{24 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(-I*b^2*(2*I*d)^(1/3)*e^(-2*I*c)*gamma(2/3, 2*I*d*x^3) - 8*a*b*(I*d)^(1/3)*e^(-I*c)*gamma(2/3, I*d*x^3) -
8*a*b*(-I*d)^(1/3)*e^(I*c)*gamma(2/3, -I*d*x^3) + I*b^2*(-2*I*d)^(1/3)*e^(2*I*c)*gamma(2/3, -2*I*d*x^3) + 6*(
2*a^2 + b^2)*d*x^2)/d

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x \left (a + b \sin {\left (c + d x^{3} \right )}\right )^{2}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x\,{\left (a+b\,\sin \left (d\,x^3+c\right )\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + b*sin(c + d*x^3))^2,x)

[Out]

int(x*(a + b*sin(c + d*x^3))^2, x)

________________________________________________________________________________________