3.8.78 \(\int \frac {1}{(c x)^{23/3} (a+b x^2)^{2/3}} \, dx\) [778]

Optimal. Leaf size=113 \[ -\frac {3 \sqrt [3]{a+b x^2}}{2 a c (c x)^{20/3}}+\frac {27 \left (a+b x^2\right )^{4/3}}{8 a^2 c (c x)^{20/3}}-\frac {81 \left (a+b x^2\right )^{7/3}}{28 a^3 c (c x)^{20/3}}+\frac {243 \left (a+b x^2\right )^{10/3}}{280 a^4 c (c x)^{20/3}} \]

[Out]

-3/2*(b*x^2+a)^(1/3)/a/c/(c*x)^(20/3)+27/8*(b*x^2+a)^(4/3)/a^2/c/(c*x)^(20/3)-81/28*(b*x^2+a)^(7/3)/a^3/c/(c*x
)^(20/3)+243/280*(b*x^2+a)^(10/3)/a^4/c/(c*x)^(20/3)

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Rubi [A]
time = 0.03, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 4, 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 {243 \left (a+b x^2\right )^{10/3}}{280 a^4 c (c x)^{20/3}}-\frac {81 \left (a+b x^2\right )^{7/3}}{28 a^3 c (c x)^{20/3}}+\frac {27 \left (a+b x^2\right )^{4/3}}{8 a^2 c (c x)^{20/3}}-\frac {3 \sqrt [3]{a+b x^2}}{2 a c (c x)^{20/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((c*x)^(23/3)*(a + b*x^2)^(2/3)),x]

[Out]

(-3*(a + b*x^2)^(1/3))/(2*a*c*(c*x)^(20/3)) + (27*(a + b*x^2)^(4/3))/(8*a^2*c*(c*x)^(20/3)) - (81*(a + b*x^2)^
(7/3))/(28*a^3*c*(c*x)^(20/3)) + (243*(a + b*x^2)^(10/3))/(280*a^4*c*(c*x)^(20/3))

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 {1}{(c x)^{23/3} \left (a+b x^2\right )^{2/3}} \, dx &=-\frac {3 \sqrt [3]{a+b x^2}}{2 a c (c x)^{20/3}}-\frac {9 \int \frac {\sqrt [3]{a+b x^2}}{(c x)^{23/3}} \, dx}{a}\\ &=-\frac {3 \sqrt [3]{a+b x^2}}{2 a c (c x)^{20/3}}+\frac {27 \left (a+b x^2\right )^{4/3}}{8 a^2 c (c x)^{20/3}}+\frac {27 \int \frac {\left (a+b x^2\right )^{4/3}}{(c x)^{23/3}} \, dx}{2 a^2}\\ &=-\frac {3 \sqrt [3]{a+b x^2}}{2 a c (c x)^{20/3}}+\frac {27 \left (a+b x^2\right )^{4/3}}{8 a^2 c (c x)^{20/3}}-\frac {81 \left (a+b x^2\right )^{7/3}}{28 a^3 c (c x)^{20/3}}-\frac {81 \int \frac {\left (a+b x^2\right )^{7/3}}{(c x)^{23/3}} \, dx}{14 a^3}\\ &=-\frac {3 \sqrt [3]{a+b x^2}}{2 a c (c x)^{20/3}}+\frac {27 \left (a+b x^2\right )^{4/3}}{8 a^2 c (c x)^{20/3}}-\frac {81 \left (a+b x^2\right )^{7/3}}{28 a^3 c (c x)^{20/3}}+\frac {243 \left (a+b x^2\right )^{10/3}}{280 a^4 c (c x)^{20/3}}\\ \end {align*}

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Mathematica [A]
time = 4.95, size = 58, normalized size = 0.51 \begin {gather*} -\frac {3 x \sqrt [3]{a+b x^2} \left (14 a^3-18 a^2 b x^2+27 a b^2 x^4-81 b^3 x^6\right )}{280 a^4 (c x)^{23/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((c*x)^(23/3)*(a + b*x^2)^(2/3)),x]

[Out]

(-3*x*(a + b*x^2)^(1/3)*(14*a^3 - 18*a^2*b*x^2 + 27*a*b^2*x^4 - 81*b^3*x^6))/(280*a^4*(c*x)^(23/3))

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Maple [A]
time = 0.04, size = 53, normalized size = 0.47

method	result	size
gosper	\(-\frac {3 x \left (b \,x^{2}+a \right )^{\frac {1}{3}} \left (-81 b^{3} x^{6}+27 a \,b^{2} x^{4}-18 a^{2} b \,x^{2}+14 a^{3}\right )}{280 a^{4} \left (c x \right )^{\frac {23}{3}}}\)	\(53\)
risch	\(-\frac {3 \left (b \,x^{2}+a \right )^{\frac {1}{3}} \left (-81 b^{3} x^{6}+27 a \,b^{2} x^{4}-18 a^{2} b \,x^{2}+14 a^{3}\right )}{280 c^{7} \left (c x \right )^{\frac {2}{3}} a^{4} x^{6}}\)	\(58\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*x)^(23/3)/(b*x^2+a)^(2/3),x,method=_RETURNVERBOSE)

[Out]

-3/280*x*(b*x^2+a)^(1/3)*(-81*b^3*x^6+27*a*b^2*x^4-18*a^2*b*x^2+14*a^3)/a^4/(c*x)^(23/3)

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Maxima [A]
time = 0.29, size = 64, normalized size = 0.57 \begin {gather*} \frac {3 \, {\left (81 \, b^{4} x^{9} + 54 \, a b^{3} x^{7} - 9 \, a^{2} b^{2} x^{5} + 4 \, a^{3} b x^{3} - 14 \, a^{4} x\right )}}{280 \, {\left (b x^{2} + a\right )}^{\frac {2}{3}} a^{4} c^{\frac {23}{3}} x^{\frac {23}{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(23/3)/(b*x^2+a)^(2/3),x, algorithm="maxima")

[Out]

3/280*(81*b^4*x^9 + 54*a*b^3*x^7 - 9*a^2*b^2*x^5 + 4*a^3*b*x^3 - 14*a^4*x)/((b*x^2 + a)^(2/3)*a^4*c^(23/3)*x^(
23/3))

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Fricas [A]
time = 0.66, size = 57, normalized size = 0.50 \begin {gather*} \frac {3 \, {\left (81 \, b^{3} x^{6} - 27 \, a b^{2} x^{4} + 18 \, a^{2} b x^{2} - 14 \, a^{3}\right )} {\left (b x^{2} + a\right )}^{\frac {1}{3}} \left (c x\right )^{\frac {1}{3}}}{280 \, a^{4} c^{8} x^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(23/3)/(b*x^2+a)^(2/3),x, algorithm="fricas")

[Out]

3/280*(81*b^3*x^6 - 27*a*b^2*x^4 + 18*a^2*b*x^2 - 14*a^3)*(b*x^2 + a)^(1/3)*(c*x)^(1/3)/(a^4*c^8*x^7)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)**(23/3)/(b*x**2+a)**(2/3),x)

[Out]

Timed out

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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(1/(c*x)^(23/3)/(b*x^2+a)^(2/3),x, algorithm="giac")

[Out]

integrate(1/((b*x^2 + a)^(2/3)*(c*x)^(23/3)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (c\,x\right )}^{23/3}\,{\left (b\,x^2+a\right )}^{2/3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((c*x)^(23/3)*(a + b*x^2)^(2/3)),x)

[Out]

int(1/((c*x)^(23/3)*(a + b*x^2)^(2/3)), x)

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