3.1.80 \(\int \frac {(a+b \sin (c+d x^3))^2}{x^6} \, dx\) [80]

Optimal. Leaf size=277 \[ \frac {-2 a^2-b^2}{10 x^5}-\frac {3 a b d \cos \left (c+d x^3\right )}{5 x^2}+\frac {b^2 \cos \left (2 c+2 d x^3\right )}{10 x^5}-\frac {3 i a b d^2 e^{i c} x \Gamma \left (\frac {1}{3},-i d x^3\right )}{10 \sqrt [3]{-i d x^3}}+\frac {3 i a b d^2 e^{-i c} x \Gamma \left (\frac {1}{3},i d x^3\right )}{10 \sqrt [3]{i d x^3}}-\frac {3 b^2 d^2 e^{2 i c} x \Gamma \left (\frac {1}{3},-2 i d x^3\right )}{10 \sqrt [3]{2} \sqrt [3]{-i d x^3}}-\frac {3 b^2 d^2 e^{-2 i c} x \Gamma \left (\frac {1}{3},2 i d x^3\right )}{10 \sqrt [3]{2} \sqrt [3]{i d x^3}}-\frac {2 a b \sin \left (c+d x^3\right )}{5 x^5}-\frac {3 b^2 d \sin \left (2 c+2 d x^3\right )}{10 x^2} \]

[Out]

1/10*(-2*a^2-b^2)/x^5-3/5*a*b*d*cos(d*x^3+c)/x^2+1/10*b^2*cos(2*d*x^3+2*c)/x^5-3/10*I*a*b*d^2*exp(I*c)*x*GAMMA
(1/3,-I*d*x^3)/(-I*d*x^3)^(1/3)+3/10*I*a*b*d^2*x*GAMMA(1/3,I*d*x^3)/exp(I*c)/(I*d*x^3)^(1/3)-3/20*b^2*d^2*exp(
2*I*c)*x*GAMMA(1/3,-2*I*d*x^3)*2^(2/3)/(-I*d*x^3)^(1/3)-3/20*b^2*d^2*x*GAMMA(1/3,2*I*d*x^3)*2^(2/3)/exp(2*I*c)
/(I*d*x^3)^(1/3)-2/5*a*b*sin(d*x^3+c)/x^5-3/10*b^2*d*sin(2*d*x^3+2*c)/x^2

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Rubi [A]
time = 0.11, antiderivative size = 275, 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 = {3484, 6, 3469, 3468, 3437, 2239, 3436} \begin {gather*} -\frac {3 i a b e^{i c} d^2 x \text {Gamma}\left (\frac {1}{3},-i d x^3\right )}{10 \sqrt [3]{-i d x^3}}+\frac {3 i a b e^{-i c} d^2 x \text {Gamma}\left (\frac {1}{3},i d x^3\right )}{10 \sqrt [3]{i d x^3}}-\frac {3 b^2 e^{2 i c} d^2 x \text {Gamma}\left (\frac {1}{3},-2 i d x^3\right )}{10 \sqrt [3]{2} \sqrt [3]{-i d x^3}}-\frac {3 b^2 e^{-2 i c} d^2 x \text {Gamma}\left (\frac {1}{3},2 i d x^3\right )}{10 \sqrt [3]{2} \sqrt [3]{i d x^3}}-\frac {2 a^2+b^2}{10 x^5}-\frac {2 a b \sin \left (c+d x^3\right )}{5 x^5}-\frac {3 a b d \cos \left (c+d x^3\right )}{5 x^2}+\frac {b^2 \cos \left (2 c+2 d x^3\right )}{10 x^5}-\frac {3 b^2 d \sin \left (2 c+2 d x^3\right )}{10 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[c + d*x^3])^2/x^6,x]

[Out]

-1/10*(2*a^2 + b^2)/x^5 - (3*a*b*d*Cos[c + d*x^3])/(5*x^2) + (b^2*Cos[2*c + 2*d*x^3])/(10*x^5) - (((3*I)/10)*a
*b*d^2*E^(I*c)*x*Gamma[1/3, (-I)*d*x^3])/((-I)*d*x^3)^(1/3) + (((3*I)/10)*a*b*d^2*x*Gamma[1/3, I*d*x^3])/(E^(I
*c)*(I*d*x^3)^(1/3)) - (3*b^2*d^2*E^((2*I)*c)*x*Gamma[1/3, (-2*I)*d*x^3])/(10*2^(1/3)*((-I)*d*x^3)^(1/3)) - (3
*b^2*d^2*x*Gamma[1/3, (2*I)*d*x^3])/(10*2^(1/3)*E^((2*I)*c)*(I*d*x^3)^(1/3)) - (2*a*b*Sin[c + d*x^3])/(5*x^5)
- (3*b^2*d*Sin[2*c + 2*d*x^3])/(10*x^2)

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 2239

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 3436

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 3437

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 3468

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 3469

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 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 \frac {\left (a+b \sin \left (c+d x^3\right )\right )^2}{x^6} \, dx &=\int \left (\frac {a^2}{x^6}+\frac {b^2}{2 x^6}-\frac {b^2 \cos \left (2 c+2 d x^3\right )}{2 x^6}+\frac {2 a b \sin \left (c+d x^3\right )}{x^6}\right ) \, dx\\ &=\int \left (\frac {a^2+\frac {b^2}{2}}{x^6}-\frac {b^2 \cos \left (2 c+2 d x^3\right )}{2 x^6}+\frac {2 a b \sin \left (c+d x^3\right )}{x^6}\right ) \, dx\\ &=-\frac {2 a^2+b^2}{10 x^5}+(2 a b) \int \frac {\sin \left (c+d x^3\right )}{x^6} \, dx-\frac {1}{2} b^2 \int \frac {\cos \left (2 c+2 d x^3\right )}{x^6} \, dx\\ &=-\frac {2 a^2+b^2}{10 x^5}+\frac {b^2 \cos \left (2 c+2 d x^3\right )}{10 x^5}-\frac {2 a b \sin \left (c+d x^3\right )}{5 x^5}+\frac {1}{5} (6 a b d) \int \frac {\cos \left (c+d x^3\right )}{x^3} \, dx+\frac {1}{5} \left (3 b^2 d\right ) \int \frac {\sin \left (2 c+2 d x^3\right )}{x^3} \, dx\\ &=-\frac {2 a^2+b^2}{10 x^5}-\frac {3 a b d \cos \left (c+d x^3\right )}{5 x^2}+\frac {b^2 \cos \left (2 c+2 d x^3\right )}{10 x^5}-\frac {2 a b \sin \left (c+d x^3\right )}{5 x^5}-\frac {3 b^2 d \sin \left (2 c+2 d x^3\right )}{10 x^2}-\frac {1}{5} \left (9 a b d^2\right ) \int \sin \left (c+d x^3\right ) \, dx+\frac {1}{5} \left (9 b^2 d^2\right ) \int \cos \left (2 c+2 d x^3\right ) \, dx\\ &=-\frac {2 a^2+b^2}{10 x^5}-\frac {3 a b d \cos \left (c+d x^3\right )}{5 x^2}+\frac {b^2 \cos \left (2 c+2 d x^3\right )}{10 x^5}-\frac {2 a b \sin \left (c+d x^3\right )}{5 x^5}-\frac {3 b^2 d \sin \left (2 c+2 d x^3\right )}{10 x^2}-\frac {1}{10} \left (9 i a b d^2\right ) \int e^{-i c-i d x^3} \, dx+\frac {1}{10} \left (9 i a b d^2\right ) \int e^{i c+i d x^3} \, dx+\frac {1}{10} \left (9 b^2 d^2\right ) \int e^{-2 i c-2 i d x^3} \, dx+\frac {1}{10} \left (9 b^2 d^2\right ) \int e^{2 i c+2 i d x^3} \, dx\\ &=-\frac {2 a^2+b^2}{10 x^5}-\frac {3 a b d \cos \left (c+d x^3\right )}{5 x^2}+\frac {b^2 \cos \left (2 c+2 d x^3\right )}{10 x^5}-\frac {3 i a b d^2 e^{i c} x \Gamma \left (\frac {1}{3},-i d x^3\right )}{10 \sqrt [3]{-i d x^3}}+\frac {3 i a b d^2 e^{-i c} x \Gamma \left (\frac {1}{3},i d x^3\right )}{10 \sqrt [3]{i d x^3}}-\frac {3 b^2 d^2 e^{2 i c} x \Gamma \left (\frac {1}{3},-2 i d x^3\right )}{10 \sqrt [3]{2} \sqrt [3]{-i d x^3}}-\frac {3 b^2 d^2 e^{-2 i c} x \Gamma \left (\frac {1}{3},2 i d x^3\right )}{10 \sqrt [3]{2} \sqrt [3]{i d x^3}}-\frac {2 a b \sin \left (c+d x^3\right )}{5 x^5}-\frac {3 b^2 d \sin \left (2 c+2 d x^3\right )}{10 x^2}\\ \end {align*}

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Mathematica [A]
time = 1.58, size = 294, normalized size = 1.06 \begin {gather*} -\frac {4 a^2+2 b^2+12 a b d x^3 \cos \left (c+d x^3\right )-2 b^2 \cos \left (2 \left (c+d x^3\right )\right )-3\ 2^{2/3} b^2 \left (i d x^3\right )^{5/3} \cos (2 c) \Gamma \left (\frac {1}{3},2 i d x^3\right )+6 i a b \left (i d x^3\right )^{5/3} \Gamma \left (\frac {1}{3},i d x^3\right ) (\cos (c)-i \sin (c))+6 i a b \sqrt [3]{i d x^3} \left (d^2 x^6\right )^{2/3} \Gamma \left (\frac {1}{3},-i d x^3\right ) (\cos (c)+i \sin (c))-3\ 2^{2/3} b^2 \left (-i d x^3\right )^{5/3} \Gamma \left (\frac {1}{3},-2 i d x^3\right ) (\cos (2 c)+i \sin (2 c))+3 i 2^{2/3} b^2 \left (i d x^3\right )^{5/3} \Gamma \left (\frac {1}{3},2 i d x^3\right ) \sin (2 c)+8 a b \sin \left (c+d x^3\right )+6 b^2 d x^3 \sin \left (2 \left (c+d x^3\right )\right )}{20 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[c + d*x^3])^2/x^6,x]

[Out]

-1/20*(4*a^2 + 2*b^2 + 12*a*b*d*x^3*Cos[c + d*x^3] - 2*b^2*Cos[2*(c + d*x^3)] - 3*2^(2/3)*b^2*(I*d*x^3)^(5/3)*
Cos[2*c]*Gamma[1/3, (2*I)*d*x^3] + (6*I)*a*b*(I*d*x^3)^(5/3)*Gamma[1/3, I*d*x^3]*(Cos[c] - I*Sin[c]) + (6*I)*a
*b*(I*d*x^3)^(1/3)*(d^2*x^6)^(2/3)*Gamma[1/3, (-I)*d*x^3]*(Cos[c] + I*Sin[c]) - 3*2^(2/3)*b^2*((-I)*d*x^3)^(5/
3)*Gamma[1/3, (-2*I)*d*x^3]*(Cos[2*c] + I*Sin[2*c]) + (3*I)*2^(2/3)*b^2*(I*d*x^3)^(5/3)*Gamma[1/3, (2*I)*d*x^3
]*Sin[2*c] + 8*a*b*Sin[c + d*x^3] + 6*b^2*d*x^3*Sin[2*(c + d*x^3)])/x^5

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Maple [F]
time = 0.16, size = 0, normalized size = 0.00 \[\int \frac {\left (a +b \sin \left (d \,x^{3}+c \right )\right )^{2}}{x^{6}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sin(d*x^3+c))^2/x^6,x)

[Out]

int((a+b*sin(d*x^3+c))^2/x^6,x)

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Maxima [A]
time = 0.36, size = 193, normalized size = 0.70 \begin {gather*} -\frac {\left (d x^{3}\right )^{\frac {2}{3}} {\left ({\left ({\left (-i \, \sqrt {3} - 1\right )} \Gamma \left (-\frac {5}{3}, i \, d x^{3}\right ) + {\left (i \, \sqrt {3} - 1\right )} \Gamma \left (-\frac {5}{3}, -i \, d x^{3}\right )\right )} \cos \left (c\right ) - {\left ({\left (\sqrt {3} - i\right )} \Gamma \left (-\frac {5}{3}, i \, d x^{3}\right ) + {\left (\sqrt {3} + i\right )} \Gamma \left (-\frac {5}{3}, -i \, d x^{3}\right )\right )} \sin \left (c\right )\right )} a b d}{6 \, x^{2}} - \frac {{\left (5 \cdot 2^{\frac {2}{3}} \left (d x^{3}\right )^{\frac {2}{3}} {\left ({\left ({\left (\sqrt {3} - i\right )} \Gamma \left (-\frac {5}{3}, 2 i \, d x^{3}\right ) + {\left (\sqrt {3} + i\right )} \Gamma \left (-\frac {5}{3}, -2 i \, d x^{3}\right )\right )} \cos \left (2 \, c\right ) + {\left ({\left (-i \, \sqrt {3} - 1\right )} \Gamma \left (-\frac {5}{3}, 2 i \, d x^{3}\right ) + {\left (i \, \sqrt {3} - 1\right )} \Gamma \left (-\frac {5}{3}, -2 i \, d x^{3}\right )\right )} \sin \left (2 \, c\right )\right )} d x^{3} + 6\right )} b^{2}}{60 \, x^{5}} - \frac {a^{2}}{5 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x^3+c))^2/x^6,x, algorithm="maxima")

[Out]

-1/6*(d*x^3)^(2/3)*(((-I*sqrt(3) - 1)*gamma(-5/3, I*d*x^3) + (I*sqrt(3) - 1)*gamma(-5/3, -I*d*x^3))*cos(c) - (
(sqrt(3) - I)*gamma(-5/3, I*d*x^3) + (sqrt(3) + I)*gamma(-5/3, -I*d*x^3))*sin(c))*a*b*d/x^2 - 1/60*(5*2^(2/3)*
(d*x^3)^(2/3)*(((sqrt(3) - I)*gamma(-5/3, 2*I*d*x^3) + (sqrt(3) + I)*gamma(-5/3, -2*I*d*x^3))*cos(2*c) + ((-I*
sqrt(3) - 1)*gamma(-5/3, 2*I*d*x^3) + (I*sqrt(3) - 1)*gamma(-5/3, -2*I*d*x^3))*sin(2*c))*d*x^3 + 6)*b^2/x^5 -
1/5*a^2/x^5

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Fricas [A]
time = 0.13, size = 181, normalized size = 0.65 \begin {gather*} \frac {3 i \, b^{2} \left (2 i \, d\right )^{\frac {2}{3}} d x^{5} e^{\left (-2 i \, c\right )} \Gamma \left (\frac {1}{3}, 2 i \, d x^{3}\right ) + 6 \, a b \left (i \, d\right )^{\frac {2}{3}} d x^{5} e^{\left (-i \, c\right )} \Gamma \left (\frac {1}{3}, i \, d x^{3}\right ) + 6 \, a b \left (-i \, d\right )^{\frac {2}{3}} d x^{5} e^{\left (i \, c\right )} \Gamma \left (\frac {1}{3}, -i \, d x^{3}\right ) - 3 i \, b^{2} \left (-2 i \, d\right )^{\frac {2}{3}} d x^{5} e^{\left (2 i \, c\right )} \Gamma \left (\frac {1}{3}, -2 i \, d x^{3}\right ) - 12 \, a b d x^{3} \cos \left (d x^{3} + c\right ) + 4 \, b^{2} \cos \left (d x^{3} + c\right )^{2} - 4 \, a^{2} - 4 \, b^{2} - 4 \, {\left (3 \, b^{2} d x^{3} \cos \left (d x^{3} + c\right ) + 2 \, a b\right )} \sin \left (d x^{3} + c\right )}{20 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x^3+c))^2/x^6,x, algorithm="fricas")

[Out]

1/20*(3*I*b^2*(2*I*d)^(2/3)*d*x^5*e^(-2*I*c)*gamma(1/3, 2*I*d*x^3) + 6*a*b*(I*d)^(2/3)*d*x^5*e^(-I*c)*gamma(1/
3, I*d*x^3) + 6*a*b*(-I*d)^(2/3)*d*x^5*e^(I*c)*gamma(1/3, -I*d*x^3) - 3*I*b^2*(-2*I*d)^(2/3)*d*x^5*e^(2*I*c)*g
amma(1/3, -2*I*d*x^3) - 12*a*b*d*x^3*cos(d*x^3 + c) + 4*b^2*cos(d*x^3 + c)^2 - 4*a^2 - 4*b^2 - 4*(3*b^2*d*x^3*
cos(d*x^3 + c) + 2*a*b)*sin(d*x^3 + c))/x^5

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b \sin {\left (c + d x^{3} \right )}\right )^{2}}{x^{6}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x**3+c))**2/x**6,x)

[Out]

Integral((a + b*sin(c + d*x**3))**2/x**6, x)

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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((a+b*sin(d*x^3+c))^2/x^6,x, algorithm="giac")

[Out]

integrate((b*sin(d*x^3 + c) + a)^2/x^6, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+b\,\sin \left (d\,x^3+c\right )\right )}^2}{x^6} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sin(c + d*x^3))^2/x^6,x)

[Out]

int((a + b*sin(c + d*x^3))^2/x^6, x)

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