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

Optimal. Leaf size=237 \[ \frac {1}{8} x^4 \left (2 a^2+b^2\right )-\frac {2 a b x \cos \left (c+d x^3\right )}{3 d}-\frac {a b e^{i c} x \Gamma \left (\frac {1}{3},-i d x^3\right )}{9 d \sqrt [3]{-i d x^3}}-\frac {a b e^{-i c} x \Gamma \left (\frac {1}{3},i d x^3\right )}{9 d \sqrt [3]{i d x^3}}-\frac {b^2 x \sin \left (2 c+2 d x^3\right )}{12 d}+\frac {i b^2 e^{2 i c} x \Gamma \left (\frac {1}{3},-2 i d x^3\right )}{72 \sqrt [3]{2} d \sqrt [3]{-i d x^3}}-\frac {i b^2 e^{-2 i c} x \Gamma \left (\frac {1}{3},2 i d x^3\right )}{72 \sqrt [3]{2} d \sqrt [3]{i d x^3}} \]

[Out]

1/8*(2*a^2+b^2)*x^4-2/3*a*b*x*cos(d*x^3+c)/d-1/9*a*b*exp(I*c)*x*GAMMA(1/3,-I*d*x^3)/d/(-I*d*x^3)^(1/3)-1/9*a*b
*x*GAMMA(1/3,I*d*x^3)/d/exp(I*c)/(I*d*x^3)^(1/3)+1/144*I*b^2*exp(2*I*c)*x*GAMMA(1/3,-2*I*d*x^3)*2^(2/3)/d/(-I*
d*x^3)^(1/3)-1/144*I*b^2*x*GAMMA(1/3,2*I*d*x^3)*2^(2/3)/d/exp(2*I*c)/(I*d*x^3)^(1/3)-1/12*b^2*x*sin(2*d*x^3+2*
c)/d

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Rubi [A]  time = 0.15, antiderivative size = 237, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {3403, 6, 3386, 3355, 2208, 3385, 3356} \[ -\frac {a b e^{i c} x \text {Gamma}\left (\frac {1}{3},-i d x^3\right )}{9 d \sqrt [3]{-i d x^3}}-\frac {a b e^{-i c} x \text {Gamma}\left (\frac {1}{3},i d x^3\right )}{9 d \sqrt [3]{i d x^3}}+\frac {i b^2 e^{2 i c} x \text {Gamma}\left (\frac {1}{3},-2 i d x^3\right )}{72 \sqrt [3]{2} d \sqrt [3]{-i d x^3}}-\frac {i b^2 e^{-2 i c} x \text {Gamma}\left (\frac {1}{3},2 i d x^3\right )}{72 \sqrt [3]{2} d \sqrt [3]{i d x^3}}+\frac {1}{8} x^4 \left (2 a^2+b^2\right )-\frac {2 a b x \cos \left (c+d x^3\right )}{3 d}-\frac {b^2 x \sin \left (2 c+2 d x^3\right )}{12 d} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*Sin[c + d*x^3])^2,x]

[Out]

((2*a^2 + b^2)*x^4)/8 - (2*a*b*x*Cos[c + d*x^3])/(3*d) - (a*b*E^(I*c)*x*Gamma[1/3, (-I)*d*x^3])/(9*d*((-I)*d*x
^3)^(1/3)) - (a*b*x*Gamma[1/3, I*d*x^3])/(9*d*E^(I*c)*(I*d*x^3)^(1/3)) + ((I/72)*b^2*E^((2*I)*c)*x*Gamma[1/3,
(-2*I)*d*x^3])/(2^(1/3)*d*((-I)*d*x^3)^(1/3)) - ((I/72)*b^2*x*Gamma[1/3, (2*I)*d*x^3])/(2^(1/3)*d*E^((2*I)*c)*
(I*d*x^3)^(1/3)) - (b^2*x*Sin[2*c + 2*d*x^3])/(12*d)

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 2208

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 3355

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

Rule 3356

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 3385

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 3386

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

Rule 3403

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^3 \left (a+b \sin \left (c+d x^3\right )\right )^2 \, dx &=\int \left (a^2 x^3+\frac {b^2 x^3}{2}-\frac {1}{2} b^2 x^3 \cos \left (2 c+2 d x^3\right )+2 a b x^3 \sin \left (c+d x^3\right )\right ) \, dx\\ &=\int \left (\left (a^2+\frac {b^2}{2}\right ) x^3-\frac {1}{2} b^2 x^3 \cos \left (2 c+2 d x^3\right )+2 a b x^3 \sin \left (c+d x^3\right )\right ) \, dx\\ &=\frac {1}{8} \left (2 a^2+b^2\right ) x^4+(2 a b) \int x^3 \sin \left (c+d x^3\right ) \, dx-\frac {1}{2} b^2 \int x^3 \cos \left (2 c+2 d x^3\right ) \, dx\\ &=\frac {1}{8} \left (2 a^2+b^2\right ) x^4-\frac {2 a b x \cos \left (c+d x^3\right )}{3 d}-\frac {b^2 x \sin \left (2 c+2 d x^3\right )}{12 d}+\frac {(2 a b) \int \cos \left (c+d x^3\right ) \, dx}{3 d}+\frac {b^2 \int \sin \left (2 c+2 d x^3\right ) \, dx}{12 d}\\ &=\frac {1}{8} \left (2 a^2+b^2\right ) x^4-\frac {2 a b x \cos \left (c+d x^3\right )}{3 d}-\frac {b^2 x \sin \left (2 c+2 d x^3\right )}{12 d}+\frac {(a b) \int e^{-i c-i d x^3} \, dx}{3 d}+\frac {(a b) \int e^{i c+i d x^3} \, dx}{3 d}+\frac {\left (i b^2\right ) \int e^{-2 i c-2 i d x^3} \, dx}{24 d}-\frac {\left (i b^2\right ) \int e^{2 i c+2 i d x^3} \, dx}{24 d}\\ &=\frac {1}{8} \left (2 a^2+b^2\right ) x^4-\frac {2 a b x \cos \left (c+d x^3\right )}{3 d}-\frac {a b e^{i c} x \Gamma \left (\frac {1}{3},-i d x^3\right )}{9 d \sqrt [3]{-i d x^3}}-\frac {a b e^{-i c} x \Gamma \left (\frac {1}{3},i d x^3\right )}{9 d \sqrt [3]{i d x^3}}+\frac {i b^2 e^{2 i c} x \Gamma \left (\frac {1}{3},-2 i d x^3\right )}{72 \sqrt [3]{2} d \sqrt [3]{-i d x^3}}-\frac {i b^2 e^{-2 i c} x \Gamma \left (\frac {1}{3},2 i d x^3\right )}{72 \sqrt [3]{2} d \sqrt [3]{i d x^3}}-\frac {b^2 x \sin \left (2 c+2 d x^3\right )}{12 d}\\ \end {align*}

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Mathematica [A]  time = 0.58, size = 339, normalized size = 1.43 \[ \frac {d x^7 \left (36 a^2 d x^3 \sqrt [3]{d^2 x^6}-96 a b \sqrt [3]{d^2 x^6} \cos \left (c+d x^3\right )-16 a b \sqrt [3]{-i d x^3} (\cos (c)-i \sin (c)) \Gamma \left (\frac {1}{3},i d x^3\right )-16 a b \sqrt [3]{i d x^3} (\cos (c)+i \sin (c)) \Gamma \left (\frac {1}{3},-i d x^3\right )-12 b^2 \sqrt [3]{d^2 x^6} \sin \left (2 \left (c+d x^3\right )\right )+i 2^{2/3} b^2 \cos (2 c) \sqrt [3]{i d x^3} \Gamma \left (\frac {1}{3},-2 i d x^3\right )-i 2^{2/3} b^2 \cos (2 c) \sqrt [3]{-i d x^3} \Gamma \left (\frac {1}{3},2 i d x^3\right )-2^{2/3} b^2 \sin (2 c) \sqrt [3]{i d x^3} \Gamma \left (\frac {1}{3},-2 i d x^3\right )-2^{2/3} b^2 \sin (2 c) \sqrt [3]{-i d x^3} \Gamma \left (\frac {1}{3},2 i d x^3\right )+18 b^2 d x^3 \sqrt [3]{d^2 x^6}\right )}{144 \left (d^2 x^6\right )^{4/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*x^7*(36*a^2*d*x^3*(d^2*x^6)^(1/3) + 18*b^2*d*x^3*(d^2*x^6)^(1/3) - 96*a*b*(d^2*x^6)^(1/3)*Cos[c + d*x^3] +
I*2^(2/3)*b^2*(I*d*x^3)^(1/3)*Cos[2*c]*Gamma[1/3, (-2*I)*d*x^3] - I*2^(2/3)*b^2*((-I)*d*x^3)^(1/3)*Cos[2*c]*Ga
mma[1/3, (2*I)*d*x^3] - 16*a*b*((-I)*d*x^3)^(1/3)*Gamma[1/3, I*d*x^3]*(Cos[c] - I*Sin[c]) - 16*a*b*(I*d*x^3)^(
1/3)*Gamma[1/3, (-I)*d*x^3]*(Cos[c] + I*Sin[c]) - 2^(2/3)*b^2*(I*d*x^3)^(1/3)*Gamma[1/3, (-2*I)*d*x^3]*Sin[2*c
] - 2^(2/3)*b^2*((-I)*d*x^3)^(1/3)*Gamma[1/3, (2*I)*d*x^3]*Sin[2*c] - 12*b^2*(d^2*x^6)^(1/3)*Sin[2*(c + d*x^3)
]))/(144*(d^2*x^6)^(4/3))

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fricas [A]  time = 0.93, size = 146, normalized size = 0.62 \[ \frac {18 \, {\left (2 \, a^{2} + b^{2}\right )} d^{2} x^{4} - 24 \, b^{2} d x \cos \left (d x^{3} + c\right ) \sin \left (d x^{3} + c\right ) - 96 \, a b d x \cos \left (d x^{3} + c\right ) - b^{2} \left (2 i \, d\right )^{\frac {2}{3}} e^{\left (-2 i \, c\right )} \Gamma \left (\frac {1}{3}, 2 i \, d x^{3}\right ) + 16 i \, a b \left (i \, d\right )^{\frac {2}{3}} e^{\left (-i \, c\right )} \Gamma \left (\frac {1}{3}, i \, d x^{3}\right ) - 16 i \, a b \left (-i \, d\right )^{\frac {2}{3}} e^{\left (i \, c\right )} \Gamma \left (\frac {1}{3}, -i \, d x^{3}\right ) - b^{2} \left (-2 i \, d\right )^{\frac {2}{3}} e^{\left (2 i \, c\right )} \Gamma \left (\frac {1}{3}, -2 i \, d x^{3}\right )}{144 \, d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/144*(18*(2*a^2 + b^2)*d^2*x^4 - 24*b^2*d*x*cos(d*x^3 + c)*sin(d*x^3 + c) - 96*a*b*d*x*cos(d*x^3 + c) - b^2*(
2*I*d)^(2/3)*e^(-2*I*c)*gamma(1/3, 2*I*d*x^3) + 16*I*a*b*(I*d)^(2/3)*e^(-I*c)*gamma(1/3, I*d*x^3) - 16*I*a*b*(
-I*d)^(2/3)*e^(I*c)*gamma(1/3, -I*d*x^3) - b^2*(-2*I*d)^(2/3)*e^(2*I*c)*gamma(1/3, -2*I*d*x^3))/d^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.49, size = 0, normalized size = 0.00 \[ \int x^{3} \left (a +b \sin \left (d \,x^{3}+c \right )\right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.58, size = 240, normalized size = 1.01 \[ \frac {1}{4} \, a^{2} x^{4} - \frac {2^{\frac {2}{3}} {\left ({\left ({\left ({\left (i \, \sqrt {3} + 1\right )} \Gamma \left (\frac {1}{3}, 2 i \, d x^{3}\right ) + {\left (-i \, \sqrt {3} + 1\right )} \Gamma \left (\frac {1}{3}, -2 i \, d x^{3}\right )\right )} \cos \left (2 \, c\right ) + {\left ({\left (\sqrt {3} - i\right )} \Gamma \left (\frac {1}{3}, 2 i \, d x^{3}\right ) + {\left (\sqrt {3} + i\right )} \Gamma \left (\frac {1}{3}, -2 i \, d x^{3}\right )\right )} \sin \left (2 \, c\right )\right )} x - 6 \cdot 2^{\frac {1}{3}} {\left (3 \, d x^{4} - 2 \, x \sin \left (2 \, d x^{3} + 2 \, c\right )\right )} \left (d x^{3}\right )^{\frac {1}{3}}\right )} b^{2}}{288 \, \left (d x^{3}\right )^{\frac {1}{3}} d} - \frac {{\left (12 \, \left (d x^{3}\right )^{\frac {1}{3}} x \cos \left (d x^{3} + c\right ) + {\left ({\left ({\left (\sqrt {3} - i\right )} \Gamma \left (\frac {1}{3}, i \, d x^{3}\right ) + {\left (\sqrt {3} + i\right )} \Gamma \left (\frac {1}{3}, -i \, d x^{3}\right )\right )} \cos \relax (c) + {\left ({\left (-i \, \sqrt {3} - 1\right )} \Gamma \left (\frac {1}{3}, i \, d x^{3}\right ) + {\left (i \, \sqrt {3} - 1\right )} \Gamma \left (\frac {1}{3}, -i \, d x^{3}\right )\right )} \sin \relax (c)\right )} x\right )} a b}{18 \, \left (d x^{3}\right )^{\frac {1}{3}} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a^2*x^4 - 1/288*2^(2/3)*((((I*sqrt(3) + 1)*gamma(1/3, 2*I*d*x^3) + (-I*sqrt(3) + 1)*gamma(1/3, -2*I*d*x^3)
)*cos(2*c) + ((sqrt(3) - I)*gamma(1/3, 2*I*d*x^3) + (sqrt(3) + I)*gamma(1/3, -2*I*d*x^3))*sin(2*c))*x - 6*2^(1
/3)*(3*d*x^4 - 2*x*sin(2*d*x^3 + 2*c))*(d*x^3)^(1/3))*b^2/((d*x^3)^(1/3)*d) - 1/18*(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)*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)*a*b/((d*x^3)^(1/3)*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a + b*sin(c + d*x^3))^2,x)

[Out]

int(x^3*(a + b*sin(c + d*x^3))^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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