3.73 \(\int x^4 (a+b \sin (c+d x^3))^2 \, dx\)

Optimal. Leaf size=249 \[ -\frac{2 a b e^{i c} x^2 \text{Gamma}\left (\frac{2}{3},-i d x^3\right )}{9 d \left (-i d x^3\right )^{2/3}}-\frac{2 a b e^{-i c} x^2 \text{Gamma}\left (\frac{2}{3},i d x^3\right )}{9 d \left (i d x^3\right )^{2/3}}+\frac{i b^2 e^{2 i c} x^2 \text{Gamma}\left (\frac{2}{3},-2 i d x^3\right )}{36\ 2^{2/3} d \left (-i d x^3\right )^{2/3}}-\frac{i b^2 e^{-2 i c} x^2 \text{Gamma}\left (\frac{2}{3},2 i d x^3\right )}{36\ 2^{2/3} d \left (i d x^3\right )^{2/3}}+\frac{1}{10} x^5 \left (2 a^2+b^2\right )-\frac{2 a b x^2 \cos \left (c+d x^3\right )}{3 d}-\frac{b^2 x^2 \sin \left (2 c+2 d x^3\right )}{12 d} \]

[Out]

((2*a^2 + b^2)*x^5)/10 - (2*a*b*x^2*Cos[c + d*x^3])/(3*d) - (2*a*b*E^(I*c)*x^2*Gamma[2/3, (-I)*d*x^3])/(9*d*((
-I)*d*x^3)^(2/3)) - (2*a*b*x^2*Gamma[2/3, I*d*x^3])/(9*d*E^(I*c)*(I*d*x^3)^(2/3)) + ((I/36)*b^2*E^((2*I)*c)*x^
2*Gamma[2/3, (-2*I)*d*x^3])/(2^(2/3)*d*((-I)*d*x^3)^(2/3)) - ((I/36)*b^2*x^2*Gamma[2/3, (2*I)*d*x^3])/(2^(2/3)
*d*E^((2*I)*c)*(I*d*x^3)^(2/3)) - (b^2*x^2*Sin[2*c + 2*d*x^3])/(12*d)

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Rubi [A]  time = 0.206465, antiderivative size = 249, normalized size of antiderivative = 1., 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, 3389, 2218, 3385, 3390} \[ -\frac{2 a b e^{i c} x^2 \text{Gamma}\left (\frac{2}{3},-i d x^3\right )}{9 d \left (-i d x^3\right )^{2/3}}-\frac{2 a b e^{-i c} x^2 \text{Gamma}\left (\frac{2}{3},i d x^3\right )}{9 d \left (i d x^3\right )^{2/3}}+\frac{i b^2 e^{2 i c} x^2 \text{Gamma}\left (\frac{2}{3},-2 i d x^3\right )}{36\ 2^{2/3} d \left (-i d x^3\right )^{2/3}}-\frac{i b^2 e^{-2 i c} x^2 \text{Gamma}\left (\frac{2}{3},2 i d x^3\right )}{36\ 2^{2/3} d \left (i d x^3\right )^{2/3}}+\frac{1}{10} x^5 \left (2 a^2+b^2\right )-\frac{2 a b x^2 \cos \left (c+d x^3\right )}{3 d}-\frac{b^2 x^2 \sin \left (2 c+2 d x^3\right )}{12 d} \]

Antiderivative was successfully verified.

[In]

Int[x^4*(a + b*Sin[c + d*x^3])^2,x]

[Out]

((2*a^2 + b^2)*x^5)/10 - (2*a*b*x^2*Cos[c + d*x^3])/(3*d) - (2*a*b*E^(I*c)*x^2*Gamma[2/3, (-I)*d*x^3])/(9*d*((
-I)*d*x^3)^(2/3)) - (2*a*b*x^2*Gamma[2/3, I*d*x^3])/(9*d*E^(I*c)*(I*d*x^3)^(2/3)) + ((I/36)*b^2*E^((2*I)*c)*x^
2*Gamma[2/3, (-2*I)*d*x^3])/(2^(2/3)*d*((-I)*d*x^3)^(2/3)) - ((I/36)*b^2*x^2*Gamma[2/3, (2*I)*d*x^3])/(2^(2/3)
*d*E^((2*I)*c)*(I*d*x^3)^(2/3)) - (b^2*x^2*Sin[2*c + 2*d*x^3])/(12*d)

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]

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 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 3389

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 2218

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

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 3390

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 )^2 \, dx &=\int \left (a^2 x^4+\frac{b^2 x^4}{2}-\frac{1}{2} b^2 x^4 \cos \left (2 c+2 d x^3\right )+2 a b x^4 \sin \left (c+d x^3\right )\right ) \, dx\\ &=\int \left (\left (a^2+\frac{b^2}{2}\right ) x^4-\frac{1}{2} b^2 x^4 \cos \left (2 c+2 d x^3\right )+2 a b x^4 \sin \left (c+d x^3\right )\right ) \, dx\\ &=\frac{1}{10} \left (2 a^2+b^2\right ) x^5+(2 a b) \int x^4 \sin \left (c+d x^3\right ) \, dx-\frac{1}{2} b^2 \int x^4 \cos \left (2 c+2 d x^3\right ) \, dx\\ &=\frac{1}{10} \left (2 a^2+b^2\right ) x^5-\frac{2 a b x^2 \cos \left (c+d x^3\right )}{3 d}-\frac{b^2 x^2 \sin \left (2 c+2 d x^3\right )}{12 d}+\frac{(4 a b) \int x \cos \left (c+d x^3\right ) \, dx}{3 d}+\frac{b^2 \int x \sin \left (2 c+2 d x^3\right ) \, dx}{6 d}\\ &=\frac{1}{10} \left (2 a^2+b^2\right ) x^5-\frac{2 a b x^2 \cos \left (c+d x^3\right )}{3 d}-\frac{b^2 x^2 \sin \left (2 c+2 d x^3\right )}{12 d}+\frac{(2 a b) \int e^{-i c-i d x^3} x \, dx}{3 d}+\frac{(2 a b) \int e^{i c+i d x^3} x \, dx}{3 d}+\frac{\left (i b^2\right ) \int e^{-2 i c-2 i d x^3} x \, dx}{12 d}-\frac{\left (i b^2\right ) \int e^{2 i c+2 i d x^3} x \, dx}{12 d}\\ &=\frac{1}{10} \left (2 a^2+b^2\right ) x^5-\frac{2 a b x^2 \cos \left (c+d x^3\right )}{3 d}-\frac{2 a 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{2 a 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{i b^2 e^{2 i c} x^2 \Gamma \left (\frac{2}{3},-2 i d x^3\right )}{36\ 2^{2/3} d \left (-i d x^3\right )^{2/3}}-\frac{i b^2 e^{-2 i c} x^2 \Gamma \left (\frac{2}{3},2 i d x^3\right )}{36\ 2^{2/3} d \left (i d x^3\right )^{2/3}}-\frac{b^2 x^2 \sin \left (2 c+2 d x^3\right )}{12 d}\\ \end{align*}

Mathematica [A]  time = 0.630199, size = 339, normalized size = 1.36 \[ \frac{d x^8 \left (-80 a b \left (-i d x^3\right )^{2/3} (\cos (c)-i \sin (c)) \text{Gamma}\left (\frac{2}{3},i d x^3\right )-80 a b \left (i d x^3\right )^{2/3} (\cos (c)+i \sin (c)) \text{Gamma}\left (\frac{2}{3},-i d x^3\right )+5 i \sqrt [3]{2} b^2 \cos (2 c) \left (i d x^3\right )^{2/3} \text{Gamma}\left (\frac{2}{3},-2 i d x^3\right )-5 i \sqrt [3]{2} b^2 \cos (2 c) \left (-i d x^3\right )^{2/3} \text{Gamma}\left (\frac{2}{3},2 i d x^3\right )-5 \sqrt [3]{2} b^2 \sin (2 c) \left (i d x^3\right )^{2/3} \text{Gamma}\left (\frac{2}{3},-2 i d x^3\right )-5 \sqrt [3]{2} b^2 \sin (2 c) \left (-i d x^3\right )^{2/3} \text{Gamma}\left (\frac{2}{3},2 i d x^3\right )+72 a^2 d x^3 \left (d^2 x^6\right )^{2/3}-240 a b \left (d^2 x^6\right )^{2/3} \cos \left (c+d x^3\right )-30 b^2 \left (d^2 x^6\right )^{2/3} \sin \left (2 \left (c+d x^3\right )\right )+36 b^2 d x^3 \left (d^2 x^6\right )^{2/3}\right )}{360 \left (d^2 x^6\right )^{5/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4*(a + b*Sin[c + d*x^3])^2,x]

[Out]

(d*x^8*(72*a^2*d*x^3*(d^2*x^6)^(2/3) + 36*b^2*d*x^3*(d^2*x^6)^(2/3) - 240*a*b*(d^2*x^6)^(2/3)*Cos[c + d*x^3] +
(5*I)*2^(1/3)*b^2*(I*d*x^3)^(2/3)*Cos[2*c]*Gamma[2/3, (-2*I)*d*x^3] - (5*I)*2^(1/3)*b^2*((-I)*d*x^3)^(2/3)*Co
s[2*c]*Gamma[2/3, (2*I)*d*x^3] - 80*a*b*((-I)*d*x^3)^(2/3)*Gamma[2/3, I*d*x^3]*(Cos[c] - I*Sin[c]) - 80*a*b*(I
*d*x^3)^(2/3)*Gamma[2/3, (-I)*d*x^3]*(Cos[c] + I*Sin[c]) - 5*2^(1/3)*b^2*(I*d*x^3)^(2/3)*Gamma[2/3, (-2*I)*d*x
^3]*Sin[2*c] - 5*2^(1/3)*b^2*((-I)*d*x^3)^(2/3)*Gamma[2/3, (2*I)*d*x^3]*Sin[2*c] - 30*b^2*(d^2*x^6)^(2/3)*Sin[
2*(c + d*x^3)]))/(360*(d^2*x^6)^(5/3))

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Maple [F]  time = 0.211, size = 0, normalized size = 0. \begin{align*} \int{x}^{4} \left ( a+b\sin \left ( d{x}^{3}+c \right ) \right ) ^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(a+b*sin(d*x^3+c))^2,x)

[Out]

int(x^4*(a+b*sin(d*x^3+c))^2,x)

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Maxima [B]  time = 1.28706, size = 830, normalized size = 3.33 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*sin(d*x^3+c))^2,x, algorithm="maxima")

[Out]

1/5*a^2*x^5 - 1/9*(6*x^3*abs(d)*cos(d*x^3 + c) + (x^3*abs(d))^(1/3)*(((gamma(2/3, I*d*x^3) + gamma(2/3, -I*d*x
^3))*cos(1/3*pi + 2/3*arctan2(0, d)) + (gamma(2/3, I*d*x^3) + gamma(2/3, -I*d*x^3))*cos(-1/3*pi + 2/3*arctan2(
0, d)) - (I*gamma(2/3, I*d*x^3) - I*gamma(2/3, -I*d*x^3))*sin(1/3*pi + 2/3*arctan2(0, d)) - (-I*gamma(2/3, I*d
*x^3) + I*gamma(2/3, -I*d*x^3))*sin(-1/3*pi + 2/3*arctan2(0, d)))*cos(c) - ((I*gamma(2/3, I*d*x^3) - I*gamma(2
/3, -I*d*x^3))*cos(1/3*pi + 2/3*arctan2(0, d)) + (I*gamma(2/3, I*d*x^3) - I*gamma(2/3, -I*d*x^3))*cos(-1/3*pi
+ 2/3*arctan2(0, d)) + (gamma(2/3, I*d*x^3) + gamma(2/3, -I*d*x^3))*sin(1/3*pi + 2/3*arctan2(0, d)) - (gamma(2
/3, I*d*x^3) + gamma(2/3, -I*d*x^3))*sin(-1/3*pi + 2/3*arctan2(0, d)))*sin(c)))*a*b/(d*x*abs(d)) + 1/720*(72*d
*x^6*abs(d) - 60*x^3*abs(d)*sin(2*d*x^3 + 2*c) + 2^(1/3)*(x^3*abs(d))^(1/3)*(((-5*I*gamma(2/3, 2*I*d*x^3) + 5*
I*gamma(2/3, -2*I*d*x^3))*cos(1/3*pi + 2/3*arctan2(0, d)) + (-5*I*gamma(2/3, 2*I*d*x^3) + 5*I*gamma(2/3, -2*I*
d*x^3))*cos(-1/3*pi + 2/3*arctan2(0, d)) - 5*(gamma(2/3, 2*I*d*x^3) + gamma(2/3, -2*I*d*x^3))*sin(1/3*pi + 2/3
*arctan2(0, d)) + 5*(gamma(2/3, 2*I*d*x^3) + gamma(2/3, -2*I*d*x^3))*sin(-1/3*pi + 2/3*arctan2(0, d)))*cos(2*c
) - (5*(gamma(2/3, 2*I*d*x^3) + gamma(2/3, -2*I*d*x^3))*cos(1/3*pi + 2/3*arctan2(0, d)) + 5*(gamma(2/3, 2*I*d*
x^3) + gamma(2/3, -2*I*d*x^3))*cos(-1/3*pi + 2/3*arctan2(0, d)) - (5*I*gamma(2/3, 2*I*d*x^3) - 5*I*gamma(2/3,
-2*I*d*x^3))*sin(1/3*pi + 2/3*arctan2(0, d)) - (-5*I*gamma(2/3, 2*I*d*x^3) + 5*I*gamma(2/3, -2*I*d*x^3))*sin(-
1/3*pi + 2/3*arctan2(0, d)))*sin(2*c)))*b^2/(d*x*abs(d))

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Fricas [A]  time = 1.81543, size = 446, normalized size = 1.79 \begin{align*} \frac{36 \,{\left (2 \, a^{2} + b^{2}\right )} d^{2} x^{5} - 60 \, b^{2} d x^{2} \cos \left (d x^{3} + c\right ) \sin \left (d x^{3} + c\right ) - 240 \, a b d x^{2} \cos \left (d x^{3} + c\right ) - 5 \, 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 ) + 80 i \, a b \left (i \, d\right )^{\frac{1}{3}} e^{\left (-i \, c\right )} \Gamma \left (\frac{2}{3}, i \, d x^{3}\right ) - 80 i \, a b \left (-i \, d\right )^{\frac{1}{3}} e^{\left (i \, c\right )} \Gamma \left (\frac{2}{3}, -i \, d x^{3}\right ) - 5 \, 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 )}{360 \, d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*sin(d*x^3+c))^2,x, algorithm="fricas")

[Out]

1/360*(36*(2*a^2 + b^2)*d^2*x^5 - 60*b^2*d*x^2*cos(d*x^3 + c)*sin(d*x^3 + c) - 240*a*b*d*x^2*cos(d*x^3 + c) -
5*b^2*(2*I*d)^(1/3)*e^(-2*I*c)*gamma(2/3, 2*I*d*x^3) + 80*I*a*b*(I*d)^(1/3)*e^(-I*c)*gamma(2/3, I*d*x^3) - 80*
I*a*b*(-I*d)^(1/3)*e^(I*c)*gamma(2/3, -I*d*x^3) - 5*b^2*(-2*I*d)^(1/3)*e^(2*I*c)*gamma(2/3, -2*I*d*x^3))/d^2

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{4} \left (a + b \sin{\left (c + d x^{3} \right )}\right )^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(a+b*sin(d*x**3+c))**2,x)

[Out]

Integral(x**4*(a + b*sin(c + d*x**3))**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sin \left (d x^{3} + c\right ) + a\right )}^{2} x^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*sin(d*x^3+c))^2,x, algorithm="giac")

[Out]

integrate((b*sin(d*x^3 + c) + a)^2*x^4, x)