3.76 \(\int \frac{(a+b \sin (c+d x^3))^2}{x^5} \, dx\)

Optimal. Leaf size=283 \[ -\frac{3 i a b e^{i c} d^2 x^2 \text{Gamma}\left (\frac{2}{3},-i d x^3\right )}{4 \left (-i d x^3\right )^{2/3}}+\frac{3 i a b e^{-i c} d^2 x^2 \text{Gamma}\left (\frac{2}{3},i d x^3\right )}{4 \left (i d x^3\right )^{2/3}}-\frac{3 b^2 e^{2 i c} d^2 x^2 \text{Gamma}\left (\frac{2}{3},-2 i d x^3\right )}{4\ 2^{2/3} \left (-i d x^3\right )^{2/3}}-\frac{3 b^2 e^{-2 i c} d^2 x^2 \text{Gamma}\left (\frac{2}{3},2 i d x^3\right )}{4\ 2^{2/3} \left (i d x^3\right )^{2/3}}-\frac{2 a^2+b^2}{8 x^4}-\frac{a b \sin \left (c+d x^3\right )}{2 x^4}-\frac{3 a b d \cos \left (c+d x^3\right )}{2 x}-\frac{3 b^2 d \sin \left (2 c+2 d x^3\right )}{4 x}+\frac{b^2 \cos \left (2 c+2 d x^3\right )}{8 x^4} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.235946, antiderivative size = 283, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 7, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.389, Rules used = {3403, 6, 3388, 3387, 3390, 2218, 3389} \[ -\frac{3 i a b e^{i c} d^2 x^2 \text{Gamma}\left (\frac{2}{3},-i d x^3\right )}{4 \left (-i d x^3\right )^{2/3}}+\frac{3 i a b e^{-i c} d^2 x^2 \text{Gamma}\left (\frac{2}{3},i d x^3\right )}{4 \left (i d x^3\right )^{2/3}}-\frac{3 b^2 e^{2 i c} d^2 x^2 \text{Gamma}\left (\frac{2}{3},-2 i d x^3\right )}{4\ 2^{2/3} \left (-i d x^3\right )^{2/3}}-\frac{3 b^2 e^{-2 i c} d^2 x^2 \text{Gamma}\left (\frac{2}{3},2 i d x^3\right )}{4\ 2^{2/3} \left (i d x^3\right )^{2/3}}-\frac{2 a^2+b^2}{8 x^4}-\frac{a b \sin \left (c+d x^3\right )}{2 x^4}-\frac{3 a b d \cos \left (c+d x^3\right )}{2 x}-\frac{3 b^2 d \sin \left (2 c+2 d x^3\right )}{4 x}+\frac{b^2 \cos \left (2 c+2 d x^3\right )}{8 x^4} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[c + d*x^3])^2/x^5,x]

[Out]

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

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 3388

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

Rule 3387

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

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 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]

Rubi steps

\begin{align*} \int \frac{\left (a+b \sin \left (c+d x^3\right )\right )^2}{x^5} \, dx &=\int \left (\frac{a^2}{x^5}+\frac{b^2}{2 x^5}-\frac{b^2 \cos \left (2 c+2 d x^3\right )}{2 x^5}+\frac{2 a b \sin \left (c+d x^3\right )}{x^5}\right ) \, dx\\ &=\int \left (\frac{a^2+\frac{b^2}{2}}{x^5}-\frac{b^2 \cos \left (2 c+2 d x^3\right )}{2 x^5}+\frac{2 a b \sin \left (c+d x^3\right )}{x^5}\right ) \, dx\\ &=-\frac{2 a^2+b^2}{8 x^4}+(2 a b) \int \frac{\sin \left (c+d x^3\right )}{x^5} \, dx-\frac{1}{2} b^2 \int \frac{\cos \left (2 c+2 d x^3\right )}{x^5} \, dx\\ &=-\frac{2 a^2+b^2}{8 x^4}+\frac{b^2 \cos \left (2 c+2 d x^3\right )}{8 x^4}-\frac{a b \sin \left (c+d x^3\right )}{2 x^4}+\frac{1}{2} (3 a b d) \int \frac{\cos \left (c+d x^3\right )}{x^2} \, dx+\frac{1}{4} \left (3 b^2 d\right ) \int \frac{\sin \left (2 c+2 d x^3\right )}{x^2} \, dx\\ &=-\frac{2 a^2+b^2}{8 x^4}-\frac{3 a b d \cos \left (c+d x^3\right )}{2 x}+\frac{b^2 \cos \left (2 c+2 d x^3\right )}{8 x^4}-\frac{a b \sin \left (c+d x^3\right )}{2 x^4}-\frac{3 b^2 d \sin \left (2 c+2 d x^3\right )}{4 x}-\frac{1}{2} \left (9 a b d^2\right ) \int x \sin \left (c+d x^3\right ) \, dx+\frac{1}{2} \left (9 b^2 d^2\right ) \int x \cos \left (2 c+2 d x^3\right ) \, dx\\ &=-\frac{2 a^2+b^2}{8 x^4}-\frac{3 a b d \cos \left (c+d x^3\right )}{2 x}+\frac{b^2 \cos \left (2 c+2 d x^3\right )}{8 x^4}-\frac{a b \sin \left (c+d x^3\right )}{2 x^4}-\frac{3 b^2 d \sin \left (2 c+2 d x^3\right )}{4 x}-\frac{1}{4} \left (9 i a b d^2\right ) \int e^{-i c-i d x^3} x \, dx+\frac{1}{4} \left (9 i a b d^2\right ) \int e^{i c+i d x^3} x \, dx+\frac{1}{4} \left (9 b^2 d^2\right ) \int e^{-2 i c-2 i d x^3} x \, dx+\frac{1}{4} \left (9 b^2 d^2\right ) \int e^{2 i c+2 i d x^3} x \, dx\\ &=-\frac{2 a^2+b^2}{8 x^4}-\frac{3 a b d \cos \left (c+d x^3\right )}{2 x}+\frac{b^2 \cos \left (2 c+2 d x^3\right )}{8 x^4}-\frac{3 i a b d^2 e^{i c} x^2 \Gamma \left (\frac{2}{3},-i d x^3\right )}{4 \left (-i d x^3\right )^{2/3}}+\frac{3 i a b d^2 e^{-i c} x^2 \Gamma \left (\frac{2}{3},i d x^3\right )}{4 \left (i d x^3\right )^{2/3}}-\frac{3 b^2 d^2 e^{2 i c} x^2 \Gamma \left (\frac{2}{3},-2 i d x^3\right )}{4\ 2^{2/3} \left (-i d x^3\right )^{2/3}}-\frac{3 b^2 d^2 e^{-2 i c} x^2 \Gamma \left (\frac{2}{3},2 i d x^3\right )}{4\ 2^{2/3} \left (i d x^3\right )^{2/3}}-\frac{a b \sin \left (c+d x^3\right )}{2 x^4}-\frac{3 b^2 d \sin \left (2 c+2 d x^3\right )}{4 x}\\ \end{align*}

Mathematica [A]  time = 2.5053, size = 292, normalized size = 1.03 \[ -\frac{6 i a b \left (i d x^3\right )^{2/3} \sqrt [3]{d^2 x^6} (\cos (c)+i \sin (c)) \text{Gamma}\left (\frac{2}{3},-i d x^3\right )+6 i a b \left (i d x^3\right )^{4/3} (\cos (c)-i \sin (c)) \text{Gamma}\left (\frac{2}{3},i d x^3\right )-3 \sqrt [3]{2} b^2 \cos (2 c) \left (i d x^3\right )^{4/3} \text{Gamma}\left (\frac{2}{3},2 i d x^3\right )+3 i \sqrt [3]{2} b^2 \sin (2 c) \left (i d x^3\right )^{4/3} \text{Gamma}\left (\frac{2}{3},2 i d x^3\right )-3 \sqrt [3]{2} b^2 \left (-i d x^3\right )^{4/3} (\cos (2 c)+i \sin (2 c)) \text{Gamma}\left (\frac{2}{3},-2 i d x^3\right )+2 a^2+4 a b \sin \left (c+d x^3\right )+12 a b d x^3 \cos \left (c+d x^3\right )+6 b^2 d x^3 \sin \left (2 \left (c+d x^3\right )\right )-b^2 \cos \left (2 \left (c+d x^3\right )\right )+b^2}{8 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[c + d*x^3])^2/x^5,x]

[Out]

-(2*a^2 + b^2 + 12*a*b*d*x^3*Cos[c + d*x^3] - b^2*Cos[2*(c + d*x^3)] - 3*2^(1/3)*b^2*(I*d*x^3)^(4/3)*Cos[2*c]*
Gamma[2/3, (2*I)*d*x^3] + (6*I)*a*b*(I*d*x^3)^(4/3)*Gamma[2/3, I*d*x^3]*(Cos[c] - I*Sin[c]) + (6*I)*a*b*(I*d*x
^3)^(2/3)*(d^2*x^6)^(1/3)*Gamma[2/3, (-I)*d*x^3]*(Cos[c] + I*Sin[c]) - 3*2^(1/3)*b^2*((-I)*d*x^3)^(4/3)*Gamma[
2/3, (-2*I)*d*x^3]*(Cos[2*c] + I*Sin[2*c]) + (3*I)*2^(1/3)*b^2*(I*d*x^3)^(4/3)*Gamma[2/3, (2*I)*d*x^3]*Sin[2*c
] + 4*a*b*Sin[c + d*x^3] + 6*b^2*d*x^3*Sin[2*(c + d*x^3)])/(8*x^4)

________________________________________________________________________________________

Maple [F]  time = 0.243, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+b\sin \left ( d{x}^{3}+c \right ) \right ) ^{2}}{{x}^{5}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sin(d*x^3+c))^2/x^5,x)

[Out]

int((a+b*sin(d*x^3+c))^2/x^5,x)

________________________________________________________________________________________

Maxima [B]  time = 1.26353, size = 756, normalized size = 2.67 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x^3+c))^2/x^5,x, algorithm="maxima")

[Out]

-1/6*(x^3*abs(d))^(1/3)*(((I*gamma(-4/3, I*d*x^3) - I*gamma(-4/3, -I*d*x^3))*cos(2/3*pi + 4/3*arctan2(0, d)) +
(I*gamma(-4/3, I*d*x^3) - I*gamma(-4/3, -I*d*x^3))*cos(-2/3*pi + 4/3*arctan2(0, d)) - (gamma(-4/3, I*d*x^3) +
gamma(-4/3, -I*d*x^3))*sin(2/3*pi + 4/3*arctan2(0, d)) + (gamma(-4/3, I*d*x^3) + gamma(-4/3, -I*d*x^3))*sin(-
2/3*pi + 4/3*arctan2(0, d)))*cos(c) + ((gamma(-4/3, I*d*x^3) + gamma(-4/3, -I*d*x^3))*cos(2/3*pi + 4/3*arctan2
(0, d)) + (gamma(-4/3, I*d*x^3) + gamma(-4/3, -I*d*x^3))*cos(-2/3*pi + 4/3*arctan2(0, d)) + (I*gamma(-4/3, I*d
*x^3) - I*gamma(-4/3, -I*d*x^3))*sin(2/3*pi + 4/3*arctan2(0, d)) + (-I*gamma(-4/3, I*d*x^3) + I*gamma(-4/3, -I
*d*x^3))*sin(-2/3*pi + 4/3*arctan2(0, d)))*sin(c))*a*b*abs(d)/x + 1/24*(2^(1/3)*(x^3*abs(d))^(1/3)*((2*(gamma(
-4/3, 2*I*d*x^3) + gamma(-4/3, -2*I*d*x^3))*cos(2/3*pi + 4/3*arctan2(0, d)) + 2*(gamma(-4/3, 2*I*d*x^3) + gamm
a(-4/3, -2*I*d*x^3))*cos(-2/3*pi + 4/3*arctan2(0, d)) + (2*I*gamma(-4/3, 2*I*d*x^3) - 2*I*gamma(-4/3, -2*I*d*x
^3))*sin(2/3*pi + 4/3*arctan2(0, d)) + (-2*I*gamma(-4/3, 2*I*d*x^3) + 2*I*gamma(-4/3, -2*I*d*x^3))*sin(-2/3*pi
+ 4/3*arctan2(0, d)))*cos(2*c) + ((-2*I*gamma(-4/3, 2*I*d*x^3) + 2*I*gamma(-4/3, -2*I*d*x^3))*cos(2/3*pi + 4/
3*arctan2(0, d)) + (-2*I*gamma(-4/3, 2*I*d*x^3) + 2*I*gamma(-4/3, -2*I*d*x^3))*cos(-2/3*pi + 4/3*arctan2(0, d)
) + 2*(gamma(-4/3, 2*I*d*x^3) + gamma(-4/3, -2*I*d*x^3))*sin(2/3*pi + 4/3*arctan2(0, d)) - 2*(gamma(-4/3, 2*I*
d*x^3) + gamma(-4/3, -2*I*d*x^3))*sin(-2/3*pi + 4/3*arctan2(0, d)))*sin(2*c))*x^3*abs(d) - 3)*b^2/x^4 - 1/4*a^
2/x^4

________________________________________________________________________________________

Fricas [A]  time = 1.85463, size = 502, normalized size = 1.77 \begin{align*} \frac{3 i \, b^{2} \left (2 i \, d\right )^{\frac{1}{3}} d x^{4} e^{\left (-2 i \, c\right )} \Gamma \left (\frac{2}{3}, 2 i \, d x^{3}\right ) + 6 \, a b \left (i \, d\right )^{\frac{1}{3}} d x^{4} e^{\left (-i \, c\right )} \Gamma \left (\frac{2}{3}, i \, d x^{3}\right ) + 6 \, a b \left (-i \, d\right )^{\frac{1}{3}} d x^{4} e^{\left (i \, c\right )} \Gamma \left (\frac{2}{3}, -i \, d x^{3}\right ) - 3 i \, b^{2} \left (-2 i \, d\right )^{\frac{1}{3}} d x^{4} e^{\left (2 i \, c\right )} \Gamma \left (\frac{2}{3}, -2 i \, d x^{3}\right ) - 12 \, a b d x^{3} \cos \left (d x^{3} + c\right ) + 2 \, b^{2} \cos \left (d x^{3} + c\right )^{2} - 2 \, a^{2} - 2 \, b^{2} - 4 \,{\left (3 \, b^{2} d x^{3} \cos \left (d x^{3} + c\right ) + a b\right )} \sin \left (d x^{3} + c\right )}{8 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x^3+c))^2/x^5,x, algorithm="fricas")

[Out]

1/8*(3*I*b^2*(2*I*d)^(1/3)*d*x^4*e^(-2*I*c)*gamma(2/3, 2*I*d*x^3) + 6*a*b*(I*d)^(1/3)*d*x^4*e^(-I*c)*gamma(2/3
, I*d*x^3) + 6*a*b*(-I*d)^(1/3)*d*x^4*e^(I*c)*gamma(2/3, -I*d*x^3) - 3*I*b^2*(-2*I*d)^(1/3)*d*x^4*e^(2*I*c)*ga
mma(2/3, -2*I*d*x^3) - 12*a*b*d*x^3*cos(d*x^3 + c) + 2*b^2*cos(d*x^3 + c)^2 - 2*a^2 - 2*b^2 - 4*(3*b^2*d*x^3*c
os(d*x^3 + c) + a*b)*sin(d*x^3 + c))/x^4

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \sin{\left (c + d x^{3} \right )}\right )^{2}}{x^{5}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x**3+c))**2/x**5,x)

[Out]

Integral((a + b*sin(c + d*x**3))**2/x**5, x)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \sin \left (d x^{3} + c\right ) + a\right )}^{2}}{x^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x^3+c))^2/x^5,x, algorithm="giac")

[Out]

integrate((b*sin(d*x^3 + c) + a)^2/x^5, x)