3.8.57 \(\int \frac {(a+b x^2)^{4/3}}{(c x)^{5/3}} \, dx\) [757]

Optimal. Leaf size=153 \[ \frac {2 b (c x)^{4/3} \sqrt [3]{a+b x^2}}{c^3}-\frac {3 \left (a+b x^2\right )^{4/3}}{2 c (c x)^{2/3}}-\frac {2 a \sqrt [3]{b} \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{b} (c x)^{2/3}}{c^{2/3} \sqrt [3]{a+b x^2}}}{\sqrt {3}}\right )}{\sqrt {3} c^{5/3}}-\frac {a \sqrt [3]{b} \log \left (\sqrt [3]{b} (c x)^{2/3}-c^{2/3} \sqrt [3]{a+b x^2}\right )}{c^{5/3}} \]

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {283, 285, 335, 281, 337} \begin {gather*} -\frac {2 a \sqrt [3]{b} \text {ArcTan}\left (\frac {\frac {2 \sqrt [3]{b} (c x)^{2/3}}{c^{2/3} \sqrt [3]{a+b x^2}}+1}{\sqrt {3}}\right )}{\sqrt {3} c^{5/3}}-\frac {a \sqrt [3]{b} \log \left (\sqrt [3]{b} (c x)^{2/3}-c^{2/3} \sqrt [3]{a+b x^2}\right )}{c^{5/3}}+\frac {2 b (c x)^{4/3} \sqrt [3]{a+b x^2}}{c^3}-\frac {3 \left (a+b x^2\right )^{4/3}}{2 c (c x)^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(4/3)/(c*x)^(5/3),x]

[Out]

(2*b*(c*x)^(4/3)*(a + b*x^2)^(1/3))/c^3 - (3*(a + b*x^2)^(4/3))/(2*c*(c*x)^(2/3)) - (2*a*b^(1/3)*ArcTan[(1 + (
2*b^(1/3)*(c*x)^(2/3))/(c^(2/3)*(a + b*x^2)^(1/3)))/Sqrt[3]])/(Sqrt[3]*c^(5/3)) - (a*b^(1/3)*Log[b^(1/3)*(c*x)
^(2/3) - c^(2/3)*(a + b*x^2)^(1/3)])/c^(5/3)

Rule 281

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 283

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

Rule 285

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

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 337

Int[(x_)/((a_) + (b_.)*(x_)^3)^(2/3), x_Symbol] :> With[{q = Rt[b, 3]}, Simp[-ArcTan[(1 + 2*q*(x/(a + b*x^3)^(
1/3)))/Sqrt[3]]/(Sqrt[3]*q^2), x] - Simp[Log[q*x - (a + b*x^3)^(1/3)]/(2*q^2), x]] /; FreeQ[{a, b}, x]

Rubi steps

\begin {align*} \int \frac {\left (a+b x^2\right )^{4/3}}{(c x)^{5/3}} \, dx &=-\frac {3 \left (a+b x^2\right )^{4/3}}{2 c (c x)^{2/3}}+\frac {(4 b) \int \sqrt [3]{c x} \sqrt [3]{a+b x^2} \, dx}{c^2}\\ &=\frac {2 b (c x)^{4/3} \sqrt [3]{a+b x^2}}{c^3}-\frac {3 \left (a+b x^2\right )^{4/3}}{2 c (c x)^{2/3}}+\frac {(4 a b) \int \frac {\sqrt [3]{c x}}{\left (a+b x^2\right )^{2/3}} \, dx}{3 c^2}\\ &=\frac {2 b (c x)^{4/3} \sqrt [3]{a+b x^2}}{c^3}-\frac {3 \left (a+b x^2\right )^{4/3}}{2 c (c x)^{2/3}}+\frac {(4 a b) \text {Subst}\left (\int \frac {x^3}{\left (a+\frac {b x^6}{c^2}\right )^{2/3}} \, dx,x,\sqrt [3]{c x}\right )}{c^3}\\ &=\frac {2 b (c x)^{4/3} \sqrt [3]{a+b x^2}}{c^3}-\frac {3 \left (a+b x^2\right )^{4/3}}{2 c (c x)^{2/3}}+\frac {(2 a b) \text {Subst}\left (\int \frac {x}{\left (a+\frac {b x^3}{c^2}\right )^{2/3}} \, dx,x,(c x)^{2/3}\right )}{c^3}\\ &=\frac {2 b (c x)^{4/3} \sqrt [3]{a+b x^2}}{c^3}-\frac {3 \left (a+b x^2\right )^{4/3}}{2 c (c x)^{2/3}}+\frac {(2 a b) \text {Subst}\left (\int \frac {x}{1-\frac {b x^3}{c^2}} \, dx,x,\frac {(c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )}{c^3}\\ &=\frac {2 b (c x)^{4/3} \sqrt [3]{a+b x^2}}{c^3}-\frac {3 \left (a+b x^2\right )^{4/3}}{2 c (c x)^{2/3}}+\frac {\left (2 a b^{2/3}\right ) \text {Subst}\left (\int \frac {1}{1-\frac {\sqrt [3]{b} x}{c^{2/3}}} \, dx,x,\frac {(c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )}{3 c^{7/3}}-\frac {\left (2 a b^{2/3}\right ) \text {Subst}\left (\int \frac {1-\frac {\sqrt [3]{b} x}{c^{2/3}}}{1+\frac {\sqrt [3]{b} x}{c^{2/3}}+\frac {b^{2/3} x^2}{c^{4/3}}} \, dx,x,\frac {(c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )}{3 c^{7/3}}\\ &=\frac {2 b (c x)^{4/3} \sqrt [3]{a+b x^2}}{c^3}-\frac {3 \left (a+b x^2\right )^{4/3}}{2 c (c x)^{2/3}}-\frac {2 a \sqrt [3]{b} \log \left (c^{2/3}-\frac {\sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )}{3 c^{5/3}}-\frac {\left (a b^{2/3}\right ) \text {Subst}\left (\int \frac {1}{1+\frac {\sqrt [3]{b} x}{c^{2/3}}+\frac {b^{2/3} x^2}{c^{4/3}}} \, dx,x,\frac {(c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )}{c^{7/3}}+\frac {\left (a \sqrt [3]{b}\right ) \text {Subst}\left (\int \frac {\frac {\sqrt [3]{b}}{c^{2/3}}+\frac {2 b^{2/3} x}{c^{4/3}}}{1+\frac {\sqrt [3]{b} x}{c^{2/3}}+\frac {b^{2/3} x^2}{c^{4/3}}} \, dx,x,\frac {(c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )}{3 c^{5/3}}\\ &=\frac {2 b (c x)^{4/3} \sqrt [3]{a+b x^2}}{c^3}-\frac {3 \left (a+b x^2\right )^{4/3}}{2 c (c x)^{2/3}}-\frac {2 a \sqrt [3]{b} \log \left (c^{2/3}-\frac {\sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )}{3 c^{5/3}}+\frac {a \sqrt [3]{b} \log \left (c^{4/3}+\frac {b^{2/3} (c x)^{4/3}}{\left (a+b x^2\right )^{2/3}}+\frac {\sqrt [3]{b} c^{2/3} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )}{3 c^{5/3}}+\frac {\left (2 a \sqrt [3]{b}\right ) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 \sqrt [3]{b} (c x)^{2/3}}{c^{2/3} \sqrt [3]{a+b x^2}}\right )}{c^{5/3}}\\ &=\frac {2 b (c x)^{4/3} \sqrt [3]{a+b x^2}}{c^3}-\frac {3 \left (a+b x^2\right )^{4/3}}{2 c (c x)^{2/3}}-\frac {2 a \sqrt [3]{b} \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{b} (c x)^{2/3}}{c^{2/3} \sqrt [3]{a+b x^2}}}{\sqrt {3}}\right )}{\sqrt {3} c^{5/3}}-\frac {2 a \sqrt [3]{b} \log \left (c^{2/3}-\frac {\sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )}{3 c^{5/3}}+\frac {a \sqrt [3]{b} \log \left (c^{4/3}+\frac {b^{2/3} (c x)^{4/3}}{\left (a+b x^2\right )^{2/3}}+\frac {\sqrt [3]{b} c^{2/3} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )}{3 c^{5/3}}\\ \end {align*}

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Mathematica [A]
time = 0.74, size = 203, normalized size = 1.33 \begin {gather*} \frac {x \left (-9 a \sqrt [3]{a+b x^2}+3 b x^2 \sqrt [3]{a+b x^2}-4 \sqrt {3} a \sqrt [3]{b} x^{2/3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{b} x^{2/3}}{\sqrt [3]{b} x^{2/3}+2 \sqrt [3]{a+b x^2}}\right )-4 a \sqrt [3]{b} x^{2/3} \log \left (-\sqrt [3]{b} x^{2/3}+\sqrt [3]{a+b x^2}\right )+2 a \sqrt [3]{b} x^{2/3} \log \left (b^{2/3} x^{4/3}+\sqrt [3]{b} x^{2/3} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}\right )\right )}{6 (c x)^{5/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(4/3)/(c*x)^(5/3),x]

[Out]

(x*(-9*a*(a + b*x^2)^(1/3) + 3*b*x^2*(a + b*x^2)^(1/3) - 4*Sqrt[3]*a*b^(1/3)*x^(2/3)*ArcTan[(Sqrt[3]*b^(1/3)*x
^(2/3))/(b^(1/3)*x^(2/3) + 2*(a + b*x^2)^(1/3))] - 4*a*b^(1/3)*x^(2/3)*Log[-(b^(1/3)*x^(2/3)) + (a + b*x^2)^(1
/3)] + 2*a*b^(1/3)*x^(2/3)*Log[b^(2/3)*x^(4/3) + b^(1/3)*x^(2/3)*(a + b*x^2)^(1/3) + (a + b*x^2)^(2/3)]))/(6*(
c*x)^(5/3))

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (b \,x^{2}+a \right )^{\frac {4}{3}}}{\left (c x \right )^{\frac {5}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(4/3)/(c*x)^(5/3),x)

[Out]

int((b*x^2+a)^(4/3)/(c*x)^(5/3),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((b*x^2+a)^(4/3)/(c*x)^(5/3),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^(4/3)/(c*x)^(5/3), x)

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Fricas [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((b*x^2+a)^(4/3)/(c*x)^(5/3),x, algorithm="fricas")

[Out]

Timed out

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Sympy [C] Result contains complex when optimal does not.
time = 5.03, size = 49, normalized size = 0.32 \begin {gather*} \frac {a^{\frac {4}{3}} \Gamma \left (- \frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {4}{3}, - \frac {1}{3} \\ \frac {2}{3} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 c^{\frac {5}{3}} x^{\frac {2}{3}} \Gamma \left (\frac {2}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(4/3)/(c*x)**(5/3),x)

[Out]

a**(4/3)*gamma(-1/3)*hyper((-4/3, -1/3), (2/3,), b*x**2*exp_polar(I*pi)/a)/(2*c**(5/3)*x**(2/3)*gamma(2/3))

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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)^(5/3),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^(4/3)/(c*x)^(5/3), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (b\,x^2+a\right )}^{4/3}}{{\left (c\,x\right )}^{5/3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(4/3)/(c*x)^(5/3),x)

[Out]

int((a + b*x^2)^(4/3)/(c*x)^(5/3), x)

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