3.4.32 \(\int \frac {\sqrt [3]{c \sin ^3(a+b x^n)}}{x^2} \, dx\) [332]

Optimal. Leaf size=139 \[ \frac {i e^{i a} \left (-i b x^n\right )^{\frac {1}{n}} \csc \left (a+b x^n\right ) \Gamma \left (-\frac {1}{n},-i b x^n\right ) \sqrt [3]{c \sin ^3\left (a+b x^n\right )}}{2 n x}-\frac {i e^{-i a} \left (i b x^n\right )^{\frac {1}{n}} \csc \left (a+b x^n\right ) \Gamma \left (-\frac {1}{n},i b x^n\right ) \sqrt [3]{c \sin ^3\left (a+b x^n\right )}}{2 n x} \]

[Out]

1/2*I*exp(I*a)*(-I*b*x^n)^(1/n)*csc(a+b*x^n)*GAMMA(-1/n,-I*b*x^n)*(c*sin(a+b*x^n)^3)^(1/3)/n/x-1/2*I*(I*b*x^n)
^(1/n)*csc(a+b*x^n)*GAMMA(-1/n,I*b*x^n)*(c*sin(a+b*x^n)^3)^(1/3)/exp(I*a)/n/x

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Rubi [A]
time = 0.14, antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {6852, 3504, 2250} \begin {gather*} \frac {i e^{i a} \left (-i b x^n\right )^{\frac {1}{n}} \text {Gamma}\left (-\frac {1}{n},-i b x^n\right ) \csc \left (a+b x^n\right ) \sqrt [3]{c \sin ^3\left (a+b x^n\right )}}{2 n x}-\frac {i e^{-i a} \left (i b x^n\right )^{\frac {1}{n}} \text {Gamma}\left (-\frac {1}{n},i b x^n\right ) \csc \left (a+b x^n\right ) \sqrt [3]{c \sin ^3\left (a+b x^n\right )}}{2 n x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c*Sin[a + b*x^n]^3)^(1/3)/x^2,x]

[Out]

((I/2)*E^(I*a)*((-I)*b*x^n)^n^(-1)*Csc[a + b*x^n]*Gamma[-n^(-1), (-I)*b*x^n]*(c*Sin[a + b*x^n]^3)^(1/3))/(n*x)
- ((I/2)*(I*b*x^n)^n^(-1)*Csc[a + b*x^n]*Gamma[-n^(-1), I*b*x^n]*(c*Sin[a + b*x^n]^3)^(1/3))/(E^(I*a)*n*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 3504

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, n}, x]

Rule 6852

Int[(u_.)*((a_.)*(v_)^(m_.))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a*v^m)^FracPart[p]/v^(m*FracPart[p])), Int
[u*v^(m*p), x], x] /; FreeQ[{a, m, p}, x] &&  !IntegerQ[p] &&  !FreeQ[v, x] &&  !(EqQ[a, 1] && EqQ[m, 1]) &&
!(EqQ[v, x] && EqQ[m, 1])

Rubi steps

\begin {align*} \int \frac {\sqrt [3]{c \sin ^3\left (a+b x^n\right )}}{x^2} \, dx &=\left (\csc \left (a+b x^n\right ) \sqrt [3]{c \sin ^3\left (a+b x^n\right )}\right ) \int \frac {\sin \left (a+b x^n\right )}{x^2} \, dx\\ &=\frac {1}{2} \left (i \csc \left (a+b x^n\right ) \sqrt [3]{c \sin ^3\left (a+b x^n\right )}\right ) \int \frac {e^{-i a-i b x^n}}{x^2} \, dx-\frac {1}{2} \left (i \csc \left (a+b x^n\right ) \sqrt [3]{c \sin ^3\left (a+b x^n\right )}\right ) \int \frac {e^{i a+i b x^n}}{x^2} \, dx\\ &=\frac {i e^{i a} \left (-i b x^n\right )^{\frac {1}{n}} \csc \left (a+b x^n\right ) \Gamma \left (-\frac {1}{n},-i b x^n\right ) \sqrt [3]{c \sin ^3\left (a+b x^n\right )}}{2 n x}-\frac {i e^{-i a} \left (i b x^n\right )^{\frac {1}{n}} \csc \left (a+b x^n\right ) \Gamma \left (-\frac {1}{n},i b x^n\right ) \sqrt [3]{c \sin ^3\left (a+b x^n\right )}}{2 n x}\\ \end {align*}

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Mathematica [A]
time = 0.11, size = 110, normalized size = 0.79 \begin {gather*} \frac {i \csc \left (a+b x^n\right ) \left (-\left (i b x^n\right )^{\frac {1}{n}} \Gamma \left (-\frac {1}{n},i b x^n\right ) (\cos (a)-i \sin (a))+\left (-i b x^n\right )^{\frac {1}{n}} \Gamma \left (-\frac {1}{n},-i b x^n\right ) (\cos (a)+i \sin (a))\right ) \sqrt [3]{c \sin ^3\left (a+b x^n\right )}}{2 n x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c*Sin[a + b*x^n]^3)^(1/3)/x^2,x]

[Out]

((I/2)*Csc[a + b*x^n]*(-((I*b*x^n)^n^(-1)*Gamma[-n^(-1), I*b*x^n]*(Cos[a] - I*Sin[a])) + ((-I)*b*x^n)^n^(-1)*G
amma[-n^(-1), (-I)*b*x^n]*(Cos[a] + I*Sin[a]))*(c*Sin[a + b*x^n]^3)^(1/3))/(n*x)

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Maple [F]
time = 0.16, size = 0, normalized size = 0.00 \[\int \frac {\left (c \left (\sin ^{3}\left (a +b \,x^{n}\right )\right )\right )^{\frac {1}{3}}}{x^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*sin(a+b*x^n)^3)^(1/3)/x^2,x)

[Out]

int((c*sin(a+b*x^n)^3)^(1/3)/x^2,x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*sin(a+b*x^n)^3)^(1/3)/x^2,x, algorithm="maxima")

[Out]

integrate((c*sin(b*x^n + a)^3)^(1/3)/x^2, x)

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Fricas [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((c*sin(a+b*x^n)^3)^(1/3)/x^2,x, algorithm="fricas")

[Out]

integral((-(c*cos(b*x^n + a)^2 - c)*sin(b*x^n + a))^(1/3)/x^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*sin(a+b*x**n)**3)**(1/3)/x**2,x)

[Out]

Integral((c*sin(a + b*x**n)**3)**(1/3)/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((c*sin(a+b*x^n)^3)^(1/3)/x^2,x, algorithm="giac")

[Out]

integrate((c*sin(b*x^n + a)^3)^(1/3)/x^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (c\,{\sin \left (a+b\,x^n\right )}^3\right )}^{1/3}}{x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*sin(a + b*x^n)^3)^(1/3)/x^2,x)

[Out]

int((c*sin(a + b*x^n)^3)^(1/3)/x^2, x)

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