3.1.75 \(\int \frac {(a+b \sin (c+d x^3))^2}{x^2} \, dx\) [75]

Optimal. Leaf size=231 \[ \frac {-2 a^2-b^2}{2 x}+\frac {b^2 \cos \left (2 c+2 d x^3\right )}{2 x}-\frac {a b d e^{i c} x^2 \Gamma \left (\frac {2}{3},-i d x^3\right )}{\left (-i d x^3\right )^{2/3}}-\frac {a b d e^{-i c} x^2 \Gamma \left (\frac {2}{3},i d x^3\right )}{\left (i d x^3\right )^{2/3}}+\frac {i b^2 d e^{2 i c} x^2 \Gamma \left (\frac {2}{3},-2 i d x^3\right )}{2\ 2^{2/3} \left (-i d x^3\right )^{2/3}}-\frac {i b^2 d e^{-2 i c} x^2 \Gamma \left (\frac {2}{3},2 i d x^3\right )}{2\ 2^{2/3} \left (i d x^3\right )^{2/3}}-\frac {2 a b \sin \left (c+d x^3\right )}{x} \]

[Out]

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

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Rubi [A]
time = 0.12, antiderivative size = 229, normalized size of antiderivative = 0.99, 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 = {3484, 6, 3469, 3470, 2250, 3468, 3471} \begin {gather*} -\frac {a b e^{i c} d x^2 \text {Gamma}\left (\frac {2}{3},-i d x^3\right )}{\left (-i d x^3\right )^{2/3}}-\frac {a b e^{-i c} d x^2 \text {Gamma}\left (\frac {2}{3},i d x^3\right )}{\left (i d x^3\right )^{2/3}}+\frac {i b^2 e^{2 i c} d x^2 \text {Gamma}\left (\frac {2}{3},-2 i d x^3\right )}{2\ 2^{2/3} \left (-i d x^3\right )^{2/3}}-\frac {i b^2 e^{-2 i c} d x^2 \text {Gamma}\left (\frac {2}{3},2 i d x^3\right )}{2\ 2^{2/3} \left (i d x^3\right )^{2/3}}-\frac {2 a^2+b^2}{2 x}-\frac {2 a b \sin \left (c+d x^3\right )}{x}+\frac {b^2 \cos \left (2 c+2 d x^3\right )}{2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[c + d*x^3])^2/x^2,x]

[Out]

-1/2*(2*a^2 + b^2)/x + (b^2*Cos[2*c + 2*d*x^3])/(2*x) - (a*b*d*E^(I*c)*x^2*Gamma[2/3, (-I)*d*x^3])/((-I)*d*x^3
)^(2/3) - (a*b*d*x^2*Gamma[2/3, I*d*x^3])/(E^(I*c)*(I*d*x^3)^(2/3)) + ((I/2)*b^2*d*E^((2*I)*c)*x^2*Gamma[2/3,
(-2*I)*d*x^3])/(2^(2/3)*((-I)*d*x^3)^(2/3)) - ((I/2)*b^2*d*x^2*Gamma[2/3, (2*I)*d*x^3])/(2^(2/3)*E^((2*I)*c)*(
I*d*x^3)^(2/3)) - (2*a*b*Sin[c + d*x^3])/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 2250

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

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 3470

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 3471

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

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Mathematica [A]
time = 0.37, size = 332, normalized size = 1.44 \begin {gather*} \frac {-4 a^2 \left (d^2 x^6\right )^{2/3}-2 b^2 \left (d^2 x^6\right )^{2/3}+2 b^2 \left (d^2 x^6\right )^{2/3} \cos \left (2 \left (c+d x^3\right )\right )+\sqrt [3]{2} b^2 \left (i d x^3\right )^{5/3} \cos (2 c) \Gamma \left (\frac {2}{3},-2 i d x^3\right )+\sqrt [3]{2} b^2 \left (-i d x^3\right )^{5/3} \cos (2 c) \Gamma \left (\frac {2}{3},2 i d x^3\right )-4 i a b \left (-i d x^3\right )^{5/3} \Gamma \left (\frac {2}{3},i d x^3\right ) (\cos (c)-i \sin (c))+4 i a b \left (i d x^3\right )^{5/3} \Gamma \left (\frac {2}{3},-i d x^3\right ) (\cos (c)+i \sin (c))+i \sqrt [3]{2} b^2 \left (i d x^3\right )^{5/3} \Gamma \left (\frac {2}{3},-2 i d x^3\right ) \sin (2 c)-i \sqrt [3]{2} b^2 \left (-i d x^3\right )^{5/3} \Gamma \left (\frac {2}{3},2 i d x^3\right ) \sin (2 c)-8 a b \left (d^2 x^6\right )^{2/3} \sin \left (c+d x^3\right )}{4 x \left (d^2 x^6\right )^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.17, size = 0, normalized size = 0.00 \[\int \frac {\left (a +b \sin \left (d \,x^{3}+c \right )\right )^{2}}{x^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.36, size = 187, normalized size = 0.81 \begin {gather*} -\frac {\left (d x^{3}\right )^{\frac {1}{3}} {\left ({\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 )} \cos \left (c\right ) + {\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 )} \sin \left (c\right )\right )} a b}{6 \, x} + \frac {{\left (2^{\frac {1}{3}} \left (d x^{3}\right )^{\frac {1}{3}} {\left ({\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 )} \cos \left (2 \, c\right ) - {\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 )} \sin \left (2 \, c\right )\right )} - 12\right )} b^{2}}{24 \, x} - \frac {a^{2}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(d*x^3)^(1/3)*(((I*sqrt(3) - 1)*gamma(-1/3, I*d*x^3) + (-I*sqrt(3) - 1)*gamma(-1/3, -I*d*x^3))*cos(c) + (
(sqrt(3) + I)*gamma(-1/3, I*d*x^3) + (sqrt(3) - I)*gamma(-1/3, -I*d*x^3))*sin(c))*a*b/x + 1/24*(2^(1/3)*(d*x^3
)^(1/3)*(((sqrt(3) + I)*gamma(-1/3, 2*I*d*x^3) + (sqrt(3) - I)*gamma(-1/3, -2*I*d*x^3))*cos(2*c) - ((I*sqrt(3)
- 1)*gamma(-1/3, 2*I*d*x^3) + (-I*sqrt(3) - 1)*gamma(-1/3, -2*I*d*x^3))*sin(2*c)) - 12)*b^2/x - a^2/x

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Fricas [A]
time = 0.13, size = 131, normalized size = 0.57 \begin {gather*} -\frac {b^{2} \left (2 i \, d\right )^{\frac {1}{3}} x e^{\left (-2 i \, c\right )} \Gamma \left (\frac {2}{3}, 2 i \, d x^{3}\right ) - 4 i \, a b \left (i \, d\right )^{\frac {1}{3}} x e^{\left (-i \, c\right )} \Gamma \left (\frac {2}{3}, i \, d x^{3}\right ) + 4 i \, a b \left (-i \, d\right )^{\frac {1}{3}} x e^{\left (i \, c\right )} \Gamma \left (\frac {2}{3}, -i \, d x^{3}\right ) + b^{2} \left (-2 i \, d\right )^{\frac {1}{3}} x e^{\left (2 i \, c\right )} \Gamma \left (\frac {2}{3}, -2 i \, d x^{3}\right ) - 4 \, b^{2} \cos \left (d x^{3} + c\right )^{2} + 8 \, a b \sin \left (d x^{3} + c\right ) + 4 \, a^{2} + 4 \, b^{2}}{4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

int((a + b*sin(c + d*x^3))^2/x^2, x)

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