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

Optimal. Leaf size=285 \[ \frac {-2 a^2-b^2}{8 x^4}-\frac {3 i a b e^{i c} d^2 x^2 \Gamma \left (\frac {2}{3},-i d x^3\right )}{4 \left (-i d x^3\right )^{2/3}}+\frac {3 i a b e^{-i c} d^2 x^2 \Gamma \left (\frac {2}{3},i d x^3\right )}{4 \left (i d x^3\right )^{2/3}}-\frac {3 a b d \cos \left (c+d x^3\right )}{2 x}-\frac {a b \sin \left (c+d x^3\right )}{2 x^4}-\frac {3 b^2 e^{2 i c} d^2 x^2 \Gamma \left (\frac {2}{3},-2 i d x^3\right )}{4\ 2^{2/3} \left (-i d x^3\right )^{2/3}}-\frac {3 b^2 e^{-2 i c} d^2 x^2 \Gamma \left (\frac {2}{3},2 i d x^3\right )}{4\ 2^{2/3} \left (i d x^3\right )^{2/3}}-\frac {3 b^2 d \sin \left (2 c+2 d x^3\right )}{4 x}+\frac {b^2 \cos \left (2 c+2 d x^3\right )}{8 x^4} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.24, antiderivative size = 283, normalized size of antiderivative = 0.99, 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 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 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 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 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 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 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 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 \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 = 2.53, size = 292, normalized size = 1.02 \[ -\frac {2 a^2+6 i a b \left (i d x^3\right )^{2/3} \sqrt [3]{d^2 x^6} (\cos (c)+i \sin (c)) \Gamma \left (\frac {2}{3},-i d x^3\right )+4 a b \sin \left (c+d x^3\right )+12 a b d x^3 \cos \left (c+d x^3\right )+6 i a b \left (i d x^3\right )^{4/3} (\cos (c)-i \sin (c)) \Gamma \left (\frac {2}{3},i d x^3\right )+6 b^2 d x^3 \sin \left (2 \left (c+d x^3\right )\right )-b^2 \cos \left (2 \left (c+d x^3\right )\right )-3 \sqrt [3]{2} b^2 \cos (2 c) \left (i d x^3\right )^{4/3} \Gamma \left (\frac {2}{3},2 i d x^3\right )+3 i \sqrt [3]{2} b^2 \sin (2 c) \left (i d x^3\right )^{4/3} \Gamma \left (\frac {2}{3},2 i d x^3\right )-3 \sqrt [3]{2} b^2 \left (-i d x^3\right )^{4/3} (\cos (2 c)+i \sin (2 c)) \Gamma \left (\frac {2}{3},-2 i d x^3\right )+b^2}{8 x^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/8*(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)*Ga
mma[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)])/x^4

________________________________________________________________________________________

fricas [A]  time = 0.67, size = 180, normalized size = 0.63 \[ \frac {3 i \, b^{2} \left (2 i \, d\right )^{\frac {1}{3}} d x^{4} e^{\left (-2 i \, c\right )} \Gamma \left (\frac {2}{3}, 2 i \, d x^{3}\right ) + 6 \, a b \left (i \, d\right )^{\frac {1}{3}} d x^{4} e^{\left (-i \, c\right )} \Gamma \left (\frac {2}{3}, i \, d x^{3}\right ) + 6 \, a b \left (-i \, d\right )^{\frac {1}{3}} d x^{4} e^{\left (i \, c\right )} \Gamma \left (\frac {2}{3}, -i \, d x^{3}\right ) - 3 i \, b^{2} \left (-2 i \, d\right )^{\frac {1}{3}} d x^{4} e^{\left (2 i \, c\right )} \Gamma \left (\frac {2}{3}, -2 i \, d x^{3}\right ) - 12 \, a b d x^{3} \cos \left (d x^{3} + c\right ) + 2 \, b^{2} \cos \left (d x^{3} + c\right )^{2} - 2 \, a^{2} - 2 \, b^{2} - 4 \, {\left (3 \, b^{2} d x^{3} \cos \left (d x^{3} + c\right ) + a b\right )} \sin \left (d x^{3} + c\right )}{8 \, x^{4}} \]

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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \sin \left (d x^{3} + c\right ) + a\right )}^{2}}{x^{5}}\,{d x} \]

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)

________________________________________________________________________________________

maple [F]  time = 0.73, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +b \sin \left (d \,x^{3}+c \right )\right )^{2}}{x^{5}}\, dx \]

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 [A]  time = 0.60, size = 198, normalized size = 0.69 \[ \frac {\left (d x^{3}\right )^{\frac {1}{3}} {\left ({\left ({\left (\sqrt {3} + i\right )} \Gamma \left (-\frac {4}{3}, i \, d x^{3}\right ) + {\left (\sqrt {3} - i\right )} \Gamma \left (-\frac {4}{3}, -i \, d x^{3}\right )\right )} \cos \relax (c) - {\left ({\left (i \, \sqrt {3} - 1\right )} \Gamma \left (-\frac {4}{3}, i \, d x^{3}\right ) + {\left (-i \, \sqrt {3} - 1\right )} \Gamma \left (-\frac {4}{3}, -i \, d x^{3}\right )\right )} \sin \relax (c)\right )} a b d}{6 \, x} - \frac {{\left (2^{\frac {1}{3}} \left (d x^{3}\right )^{\frac {1}{3}} {\left (2 \, {\left ({\left (-i \, \sqrt {3} + 1\right )} \Gamma \left (-\frac {4}{3}, 2 i \, d x^{3}\right ) + {\left (i \, \sqrt {3} + 1\right )} \Gamma \left (-\frac {4}{3}, -2 i \, d x^{3}\right )\right )} \cos \left (2 \, c\right ) - {\left ({\left (2 \, \sqrt {3} + 2 i\right )} \Gamma \left (-\frac {4}{3}, 2 i \, d x^{3}\right ) + {\left (2 \, \sqrt {3} - 2 i\right )} \Gamma \left (-\frac {4}{3}, -2 i \, d x^{3}\right )\right )} \sin \left (2 \, c\right )\right )} d x^{3} + 3\right )} b^{2}}{24 \, x^{4}} - \frac {a^{2}}{4 \, x^{4}} \]

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*(d*x^3)^(1/3)*(((sqrt(3) + I)*gamma(-4/3, I*d*x^3) + (sqrt(3) - I)*gamma(-4/3, -I*d*x^3))*cos(c) - ((I*sqr
t(3) - 1)*gamma(-4/3, I*d*x^3) + (-I*sqrt(3) - 1)*gamma(-4/3, -I*d*x^3))*sin(c))*a*b*d/x - 1/24*(2^(1/3)*(d*x^
3)^(1/3)*(2*((-I*sqrt(3) + 1)*gamma(-4/3, 2*I*d*x^3) + (I*sqrt(3) + 1)*gamma(-4/3, -2*I*d*x^3))*cos(2*c) - ((2
*sqrt(3) + 2*I)*gamma(-4/3, 2*I*d*x^3) + (2*sqrt(3) - 2*I)*gamma(-4/3, -2*I*d*x^3))*sin(2*c))*d*x^3 + 3)*b^2/x
^4 - 1/4*a^2/x^4

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,\sin \left (d\,x^3+c\right )\right )}^2}{x^5} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sin(c + d*x^3))^2/x^5,x)

[Out]

int((a + b*sin(c + d*x^3))^2/x^5, x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \sin {\left (c + d x^{3} \right )}\right )^{2}}{x^{5}}\, dx \]

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)

________________________________________________________________________________________