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

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

[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 = 0.29, antiderivative size = 233, normalized size of antiderivative = 1.52, number of steps used = 11, number of rules used = 11, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.579, Rules used = {277, 279, 329, 275, 331, 292, 31, 634, 617, 204, 628} \[ \frac {a \sqrt [3]{b} \log \left (\frac {b^{2/3} (c x)^{4/3}}{\left (a+b x^2\right )^{2/3}}+\frac {\sqrt [3]{b} c^{2/3} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}+c^{4/3}\right )}{3 c^{5/3}}+\frac {2 b (c x)^{4/3} \sqrt [3]{a+b x^2}}{c^3}-\frac {2 a \sqrt [3]{b} \log \left (c^{2/3}-\frac {\sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )}{3 c^{5/3}}-\frac {2 a \sqrt [3]{b} \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}+c^{2/3}}{\sqrt {3} c^{2/3}}\right )}{\sqrt {3} c^{5/3}}-\frac {3 \left (a+b x^2\right )^{4/3}}{2 c (c x)^{2/3}} \]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 275

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 277

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 279

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 292

Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> -Dist[(3*Rt[a, 3]*Rt[b, 3])^(-1), Int[1/(Rt[a, 3] + Rt[b, 3]*x),
x], x] + Dist[1/(3*Rt[a, 3]*Rt[b, 3]), Int[(Rt[a, 3] + Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3
]^2*x^2), x], x] /; FreeQ[{a, b}, x]

Rule 329

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 331

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 1)/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[
p, -2^(-1)] && IntegersQ[m, p + (m + 1)/n]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

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) \operatorname {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) \operatorname {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) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 [C]  time = 0.01, size = 57, normalized size = 0.37 \[ -\frac {3 a x \sqrt [3]{a+b x^2} \, _2F_1\left (-\frac {4}{3},-\frac {1}{3};\frac {2}{3};-\frac {b x^2}{a}\right )}{2 (c x)^{5/3} \sqrt [3]{\frac {b x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a*x*(a + b*x^2)^(1/3)*Hypergeometric2F1[-4/3, -1/3, 2/3, -((b*x^2)/a)])/(2*(c*x)^(5/3)*(1 + (b*x^2)/a)^(1/
3))

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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

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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maple [F]  time = 0.05, 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 \[ \int \frac {{\left (b x^{2} + a\right )}^{\frac {4}{3}}}{\left (c x\right )^{\frac {5}{3}}}\,{d x} \]

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

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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sympy [C]  time = 7.23, size = 49, normalized size = 0.32 \[ \frac {a^{\frac {4}{3}} \Gamma \left (- \frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {4}{3}, - \frac {1}{3} \\ \frac {2}{3} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 c^{\frac {5}{3}} x^{\frac {2}{3}} \Gamma \left (\frac {2}{3}\right )} \]

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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