3.8.43 \(\int \frac {\sqrt [3]{a+b x^2}}{(c x)^{17/3}} \, dx\) [743]

Optimal. Leaf size=57 \[ -\frac {3 \left (a+b x^2\right )^{4/3}}{8 a c (c x)^{14/3}}+\frac {9 \left (a+b x^2\right )^{7/3}}{56 a^2 c (c x)^{14/3}} \]

[Out]

-3/8*(b*x^2+a)^(4/3)/a/c/(c*x)^(14/3)+9/56*(b*x^2+a)^(7/3)/a^2/c/(c*x)^(14/3)

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Rubi [A]
time = 0.01, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {279, 270} \begin {gather*} \frac {9 \left (a+b x^2\right )^{7/3}}{56 a^2 c (c x)^{14/3}}-\frac {3 \left (a+b x^2\right )^{4/3}}{8 a c (c x)^{14/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(1/3)/(c*x)^(17/3),x]

[Out]

(-3*(a + b*x^2)^(4/3))/(8*a*c*(c*x)^(14/3)) + (9*(a + b*x^2)^(7/3))/(56*a^2*c*(c*x)^(14/3))

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*
c*(m + 1))), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rule 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-(c*x)^(m + 1))*((a + b*x^n)^(p + 1)/
(a*c*n*(p + 1))), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; Free
Q[{a, b, c, m, n, p}, x] && ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[p, -1]

Rubi steps

\begin {align*} \int \frac {\sqrt [3]{a+b x^2}}{(c x)^{17/3}} \, dx &=-\frac {3 \left (a+b x^2\right )^{4/3}}{8 a c (c x)^{14/3}}-\frac {3 \int \frac {\left (a+b x^2\right )^{4/3}}{(c x)^{17/3}} \, dx}{4 a}\\ &=-\frac {3 \left (a+b x^2\right )^{4/3}}{8 a c (c x)^{14/3}}+\frac {9 \left (a+b x^2\right )^{7/3}}{56 a^2 c (c x)^{14/3}}\\ \end {align*}

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Mathematica [A]
time = 1.73, size = 46, normalized size = 0.81 \begin {gather*} -\frac {3 x \sqrt [3]{a+b x^2} \left (4 a^2+a b x^2-3 b^2 x^4\right )}{56 a^2 (c x)^{17/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(1/3)/(c*x)^(17/3),x]

[Out]

(-3*x*(a + b*x^2)^(1/3)*(4*a^2 + a*b*x^2 - 3*b^2*x^4))/(56*a^2*(c*x)^(17/3))

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Maple [A]
time = 0.04, size = 31, normalized size = 0.54

method	result	size
gosper	\(-\frac {3 x \left (b \,x^{2}+a \right )^{\frac {4}{3}} \left (-3 b \,x^{2}+4 a \right )}{56 a^{2} \left (c x \right )^{\frac {17}{3}}}\)	\(31\)
risch	\(-\frac {3 \left (b \,x^{2}+a \right )^{\frac {1}{3}} \left (-3 b^{2} x^{4}+a b \,x^{2}+4 a^{2}\right )}{56 c^{5} \left (c x \right )^{\frac {2}{3}} x^{4} a^{2}}\)	\(46\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(1/3)/(c*x)^(17/3),x,method=_RETURNVERBOSE)

[Out]

-3/56*x*(b*x^2+a)^(4/3)*(-3*b*x^2+4*a)/a^2/(c*x)^(17/3)

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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((b*x^2+a)^(1/3)/(c*x)^(17/3),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^(1/3)/(c*x)^(17/3), x)

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Fricas [A]
time = 0.86, size = 46, normalized size = 0.81 \begin {gather*} \frac {3 \, {\left (3 \, b^{2} x^{4} - a b x^{2} - 4 \, a^{2}\right )} {\left (b x^{2} + a\right )}^{\frac {1}{3}} \left (c x\right )^{\frac {1}{3}}}{56 \, a^{2} c^{6} x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(1/3)/(c*x)^(17/3),x, algorithm="fricas")

[Out]

3/56*(3*b^2*x^4 - a*b*x^2 - 4*a^2)*(b*x^2 + a)^(1/3)*(c*x)^(1/3)/(a^2*c^6*x^5)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(1/3)/(c*x)**(17/3),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 5986 deep

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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((b*x^2+a)^(1/3)/(c*x)^(17/3),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^(1/3)/(c*x)^(17/3), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\left (b\,x^2+a\right )}^{1/3}}{{\left (c\,x\right )}^{17/3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(1/3)/(c*x)^(17/3),x)

[Out]

int((a + b*x^2)^(1/3)/(c*x)^(17/3), x)

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