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

Optimal. Leaf size=229 \[ -\frac{a b e^{i c} d x^2 \text{Gamma}\left (\frac{2}{3},-i d x^3\right )}{\left (-i d x^3\right )^{2/3}}-\frac{a b e^{-i c} d x^2 \text{Gamma}\left (\frac{2}{3},i d x^3\right )}{\left (i d x^3\right )^{2/3}}+\frac{i b^2 e^{2 i c} d x^2 \text{Gamma}\left (\frac{2}{3},-2 i d x^3\right )}{2\ 2^{2/3} \left (-i d x^3\right )^{2/3}}-\frac{i b^2 e^{-2 i c} d x^2 \text{Gamma}\left (\frac{2}{3},2 i d x^3\right )}{2\ 2^{2/3} \left (i d x^3\right )^{2/3}}-\frac{2 a^2+b^2}{2 x}-\frac{2 a b \sin \left (c+d x^3\right )}{x}+\frac{b^2 \cos \left (2 c+2 d x^3\right )}{2 x} \]

[Out]

-(2*a^2 + b^2)/(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

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Rubi [A]  time = 0.189251, antiderivative size = 229, 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, 3388, 3389, 2218, 3387, 3390} \[ -\frac{a b e^{i c} d x^2 \text{Gamma}\left (\frac{2}{3},-i d x^3\right )}{\left (-i d x^3\right )^{2/3}}-\frac{a b e^{-i c} d x^2 \text{Gamma}\left (\frac{2}{3},i d x^3\right )}{\left (i d x^3\right )^{2/3}}+\frac{i b^2 e^{2 i c} d x^2 \text{Gamma}\left (\frac{2}{3},-2 i d x^3\right )}{2\ 2^{2/3} \left (-i d x^3\right )^{2/3}}-\frac{i b^2 e^{-2 i c} d x^2 \text{Gamma}\left (\frac{2}{3},2 i d x^3\right )}{2\ 2^{2/3} \left (i d x^3\right )^{2/3}}-\frac{2 a^2+b^2}{2 x}-\frac{2 a b \sin \left (c+d x^3\right )}{x}+\frac{b^2 \cos \left (2 c+2 d x^3\right )}{2 x} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[c + d*x^3])^2/x^2,x]

[Out]

-(2*a^2 + b^2)/(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 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 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 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]

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*}

Mathematica [A]  time = 0.584463, size = 332, normalized size = 1.45 \[ \frac{-4 i a b \left (-i d x^3\right )^{5/3} (\cos (c)-i \sin (c)) \text{Gamma}\left (\frac{2}{3},i d x^3\right )+4 i a b \left (i d x^3\right )^{5/3} (\cos (c)+i \sin (c)) \text{Gamma}\left (\frac{2}{3},-i d x^3\right )+\sqrt [3]{2} b^2 \cos (2 c) \left (i d x^3\right )^{5/3} \text{Gamma}\left (\frac{2}{3},-2 i d x^3\right )+\sqrt [3]{2} b^2 \cos (2 c) \left (-i d x^3\right )^{5/3} \text{Gamma}\left (\frac{2}{3},2 i d x^3\right )+i \sqrt [3]{2} b^2 \sin (2 c) \left (i d x^3\right )^{5/3} \text{Gamma}\left (\frac{2}{3},-2 i d x^3\right )-i \sqrt [3]{2} b^2 \sin (2 c) \left (-i d x^3\right )^{5/3} \text{Gamma}\left (\frac{2}{3},2 i d x^3\right )-4 a^2 \left (d^2 x^6\right )^{2/3}-8 a b \left (d^2 x^6\right )^{2/3} \sin \left (c+d x^3\right )+2 b^2 \left (d^2 x^6\right )^{2/3} \cos \left (2 \left (c+d x^3\right )\right )-2 b^2 \left (d^2 x^6\right )^{2/3}}{4 x \left (d^2 x^6\right )^{2/3}} \]

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 = 0.197, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+b\sin \left ( d{x}^{3}+c \right ) \right ) ^{2}}{{x}^{2}}}\, dx \end{align*}

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 [B]  time = 1.27987, size = 743, normalized size = 3.24 \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^2,x, algorithm="maxima")

[Out]

-1/6*(x^3*abs(d))^(1/3)*(((I*gamma(-1/3, I*d*x^3) - I*gamma(-1/3, -I*d*x^3))*cos(1/6*pi + 1/3*arctan2(0, d)) +
(I*gamma(-1/3, I*d*x^3) - I*gamma(-1/3, -I*d*x^3))*cos(-1/6*pi + 1/3*arctan2(0, d)) - (gamma(-1/3, I*d*x^3) +
gamma(-1/3, -I*d*x^3))*sin(1/6*pi + 1/3*arctan2(0, d)) + (gamma(-1/3, I*d*x^3) + gamma(-1/3, -I*d*x^3))*sin(-
1/6*pi + 1/3*arctan2(0, d)))*cos(c) + ((gamma(-1/3, I*d*x^3) + gamma(-1/3, -I*d*x^3))*cos(1/6*pi + 1/3*arctan2
(0, d)) + (gamma(-1/3, I*d*x^3) + gamma(-1/3, -I*d*x^3))*cos(-1/6*pi + 1/3*arctan2(0, d)) + (I*gamma(-1/3, I*d
*x^3) - I*gamma(-1/3, -I*d*x^3))*sin(1/6*pi + 1/3*arctan2(0, d)) + (-I*gamma(-1/3, I*d*x^3) + I*gamma(-1/3, -I
*d*x^3))*sin(-1/6*pi + 1/3*arctan2(0, d)))*sin(c))*a*b/x + 1/24*(2^(1/3)*(x^3*abs(d))^(1/3)*(((gamma(-1/3, 2*I
*d*x^3) + gamma(-1/3, -2*I*d*x^3))*cos(1/6*pi + 1/3*arctan2(0, d)) + (gamma(-1/3, 2*I*d*x^3) + gamma(-1/3, -2*
I*d*x^3))*cos(-1/6*pi + 1/3*arctan2(0, d)) + (I*gamma(-1/3, 2*I*d*x^3) - I*gamma(-1/3, -2*I*d*x^3))*sin(1/6*pi
+ 1/3*arctan2(0, d)) + (-I*gamma(-1/3, 2*I*d*x^3) + I*gamma(-1/3, -2*I*d*x^3))*sin(-1/6*pi + 1/3*arctan2(0, d
)))*cos(2*c) + ((-I*gamma(-1/3, 2*I*d*x^3) + I*gamma(-1/3, -2*I*d*x^3))*cos(1/6*pi + 1/3*arctan2(0, d)) + (-I*
gamma(-1/3, 2*I*d*x^3) + I*gamma(-1/3, -2*I*d*x^3))*cos(-1/6*pi + 1/3*arctan2(0, d)) + (gamma(-1/3, 2*I*d*x^3)
+ gamma(-1/3, -2*I*d*x^3))*sin(1/6*pi + 1/3*arctan2(0, d)) - (gamma(-1/3, 2*I*d*x^3) + gamma(-1/3, -2*I*d*x^3
))*sin(-1/6*pi + 1/3*arctan2(0, d)))*sin(2*c)) - 12)*b^2/x - a^2/x

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Fricas [A]  time = 1.8324, size = 392, normalized size = 1.71 \begin{align*} -\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{align*}

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

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., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \sin \left (d x^{3} + c\right ) + a\right )}^{2}}{x^{2}}\,{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^2,x, algorithm="giac")

[Out]

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