3.8.59 \(\int \frac {(a+b x^2)^{4/3}}{(c x)^{17/3}} \, dx\) [759]

Optimal. Leaf size=28 \[ -\frac {3 \left (a+b x^2\right )^{7/3}}{14 a c (c x)^{14/3}} \]

[Out]

-3/14*(b*x^2+a)^(7/3)/a/c/(c*x)^(14/3)

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Rubi [A]
time = 0.00, antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {270} \begin {gather*} -\frac {3 \left (a+b x^2\right )^{7/3}}{14 a c (c x)^{14/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(4/3)/(c*x)^(17/3),x]

[Out]

(-3*(a + b*x^2)^(7/3))/(14*a*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]

Rubi steps

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

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Mathematica [A]
time = 1.97, size = 26, normalized size = 0.93 \begin {gather*} -\frac {3 x \left (a+b x^2\right )^{7/3}}{14 a (c x)^{17/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(4/3)/(c*x)^(17/3),x]

[Out]

(-3*x*(a + b*x^2)^(7/3))/(14*a*(c*x)^(17/3))

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Maple [A]
time = 0.04, size = 21, normalized size = 0.75

method	result	size
gosper	\(-\frac {3 x \left (b \,x^{2}+a \right )^{\frac {7}{3}}}{14 a \left (c x \right )^{\frac {17}{3}}}\)	\(21\)
risch	\(-\frac {3 \left (b \,x^{2}+a \right )^{\frac {1}{3}} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )}{14 c^{5} \left (c x \right )^{\frac {2}{3}} x^{4} a}\)	\(44\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(4/3)/(c*x)^(17/3),x,method=_RETURNVERBOSE)

[Out]

-3/14*x*(b*x^2+a)^(7/3)/a/(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)^(4/3)/(c*x)^(17/3),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^(4/3)/(c*x)^(17/3), x)

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Fricas [A]
time = 1.38, size = 43, normalized size = 1.54 \begin {gather*} -\frac {3 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )} {\left (b x^{2} + a\right )}^{\frac {1}{3}} \left (c x\right )^{\frac {1}{3}}}{14 \, a c^{6} x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(4/3)/(c*x)^(17/3),x, algorithm="fricas")

[Out]

-3/14*(b^2*x^4 + 2*a*b*x^2 + a^2)*(b*x^2 + a)^(1/3)*(c*x)^(1/3)/(a*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)**(4/3)/(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)^(4/3)/(c*x)^(17/3),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^(4/3)/(c*x)^(17/3), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {{\left (b\,x^2+a\right )}^{4/3}}{{\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)^(4/3)/(c*x)^(17/3),x)

[Out]

int((a + b*x^2)^(4/3)/(c*x)^(17/3), x)

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