3.772 \(\int \frac {(c x)^{13/3}}{(a+b x^2)^{2/3}} \, dx\)

Optimal. Leaf size=167 \[ -\frac {5 a^2 c^{13/3} \log \left (\sqrt [3]{b} (c x)^{2/3}-c^{2/3} \sqrt [3]{a+b x^2}\right )}{12 b^{8/3}}-\frac {5 a^2 c^{13/3} \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b} (c x)^{2/3}}{c^{2/3} \sqrt [3]{a+b x^2}}+1}{\sqrt {3}}\right )}{6 \sqrt {3} b^{8/3}}-\frac {5 a c^3 (c x)^{4/3} \sqrt [3]{a+b x^2}}{12 b^2}+\frac {c (c x)^{10/3} \sqrt [3]{a+b x^2}}{4 b} \]

[Out]

-5/12*a*c^3*(c*x)^(4/3)*(b*x^2+a)^(1/3)/b^2+1/4*c*(c*x)^(10/3)*(b*x^2+a)^(1/3)/b-5/12*a^2*c^(13/3)*ln(b^(1/3)*
(c*x)^(2/3)-c^(2/3)*(b*x^2+a)^(1/3))/b^(8/3)-5/18*a^2*c^(13/3)*arctan(1/3*(1+2*b^(1/3)*(c*x)^(2/3)/c^(2/3)/(b*
x^2+a)^(1/3))*3^(1/2))/b^(8/3)*3^(1/2)

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Rubi [A]  time = 0.29, antiderivative size = 247, normalized size of antiderivative = 1.48, number of steps used = 11, number of rules used = 10, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.526, Rules used = {321, 329, 275, 331, 292, 31, 634, 617, 204, 628} \[ -\frac {5 a^2 c^{13/3} \log \left (c^{2/3}-\frac {\sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )}{18 b^{8/3}}+\frac {5 a^2 c^{13/3} \log \left (\frac {b^{2/3} (c x)^{4/3}}{\left (a+b x^2\right )^{2/3}}+\frac {\sqrt [3]{b} c^{2/3} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}+c^{4/3}\right )}{36 b^{8/3}}-\frac {5 a^2 c^{13/3} \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}+c^{2/3}}{\sqrt {3} c^{2/3}}\right )}{6 \sqrt {3} b^{8/3}}-\frac {5 a c^3 (c x)^{4/3} \sqrt [3]{a+b x^2}}{12 b^2}+\frac {c (c x)^{10/3} \sqrt [3]{a+b x^2}}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[(c*x)^(13/3)/(a + b*x^2)^(2/3),x]

[Out]

(-5*a*c^3*(c*x)^(4/3)*(a + b*x^2)^(1/3))/(12*b^2) + (c*(c*x)^(10/3)*(a + b*x^2)^(1/3))/(4*b) - (5*a^2*c^(13/3)
*ArcTan[(c^(2/3) + (2*b^(1/3)*(c*x)^(2/3))/(a + b*x^2)^(1/3))/(Sqrt[3]*c^(2/3))])/(6*Sqrt[3]*b^(8/3)) - (5*a^2
*c^(13/3)*Log[c^(2/3) - (b^(1/3)*(c*x)^(2/3))/(a + b*x^2)^(1/3)])/(18*b^(8/3)) + (5*a^2*c^(13/3)*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)])/(36*b^(8/3))

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 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 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, 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 {(c x)^{13/3}}{\left (a+b x^2\right )^{2/3}} \, dx &=\frac {c (c x)^{10/3} \sqrt [3]{a+b x^2}}{4 b}-\frac {\left (5 a c^2\right ) \int \frac {(c x)^{7/3}}{\left (a+b x^2\right )^{2/3}} \, dx}{6 b}\\ &=-\frac {5 a c^3 (c x)^{4/3} \sqrt [3]{a+b x^2}}{12 b^2}+\frac {c (c x)^{10/3} \sqrt [3]{a+b x^2}}{4 b}+\frac {\left (5 a^2 c^4\right ) \int \frac {\sqrt [3]{c x}}{\left (a+b x^2\right )^{2/3}} \, dx}{9 b^2}\\ &=-\frac {5 a c^3 (c x)^{4/3} \sqrt [3]{a+b x^2}}{12 b^2}+\frac {c (c x)^{10/3} \sqrt [3]{a+b x^2}}{4 b}+\frac {\left (5 a^2 c^3\right ) \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 )}{3 b^2}\\ &=-\frac {5 a c^3 (c x)^{4/3} \sqrt [3]{a+b x^2}}{12 b^2}+\frac {c (c x)^{10/3} \sqrt [3]{a+b x^2}}{4 b}+\frac {\left (5 a^2 c^3\right ) \operatorname {Subst}\left (\int \frac {x}{\left (a+\frac {b x^3}{c^2}\right )^{2/3}} \, dx,x,(c x)^{2/3}\right )}{6 b^2}\\ &=-\frac {5 a c^3 (c x)^{4/3} \sqrt [3]{a+b x^2}}{12 b^2}+\frac {c (c x)^{10/3} \sqrt [3]{a+b x^2}}{4 b}+\frac {\left (5 a^2 c^3\right ) \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 )}{6 b^2}\\ &=-\frac {5 a c^3 (c x)^{4/3} \sqrt [3]{a+b x^2}}{12 b^2}+\frac {c (c x)^{10/3} \sqrt [3]{a+b x^2}}{4 b}+\frac {\left (5 a^2 c^{11/3}\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 )}{18 b^{7/3}}-\frac {\left (5 a^2 c^{11/3}\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 )}{18 b^{7/3}}\\ &=-\frac {5 a c^3 (c x)^{4/3} \sqrt [3]{a+b x^2}}{12 b^2}+\frac {c (c x)^{10/3} \sqrt [3]{a+b x^2}}{4 b}-\frac {5 a^2 c^{13/3} \log \left (c^{2/3}-\frac {\sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )}{18 b^{8/3}}-\frac {\left (5 a^2 c^{11/3}\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 )}{12 b^{7/3}}+\frac {\left (5 a^2 c^{13/3}\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 )}{36 b^{8/3}}\\ &=-\frac {5 a c^3 (c x)^{4/3} \sqrt [3]{a+b x^2}}{12 b^2}+\frac {c (c x)^{10/3} \sqrt [3]{a+b x^2}}{4 b}-\frac {5 a^2 c^{13/3} \log \left (c^{2/3}-\frac {\sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )}{18 b^{8/3}}+\frac {5 a^2 c^{13/3} \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 )}{36 b^{8/3}}+\frac {\left (5 a^2 c^{13/3}\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 )}{6 b^{8/3}}\\ &=-\frac {5 a c^3 (c x)^{4/3} \sqrt [3]{a+b x^2}}{12 b^2}+\frac {c (c x)^{10/3} \sqrt [3]{a+b x^2}}{4 b}-\frac {5 a^2 c^{13/3} \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 )}{6 \sqrt {3} b^{8/3}}-\frac {5 a^2 c^{13/3} \log \left (c^{2/3}-\frac {\sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )}{18 b^{8/3}}+\frac {5 a^2 c^{13/3} \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 )}{36 b^{8/3}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 76, normalized size = 0.46 \[ \frac {c^3 (c x)^{4/3} \left (5 a^2 \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};\frac {b x^2}{b x^2+a}\right )-5 a^2-2 a b x^2+3 b^2 x^4\right )}{12 b^2 \left (a+b x^2\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*x)^(13/3)/(a + b*x^2)^(2/3),x]

[Out]

(c^3*(c*x)^(4/3)*(-5*a^2 - 2*a*b*x^2 + 3*b^2*x^4 + 5*a^2*Hypergeometric2F1[2/3, 1, 5/3, (b*x^2)/(a + b*x^2)]))
/(12*b^2*(a + b*x^2)^(2/3))

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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((c*x)^(13/3)/(b*x^2+a)^(2/3),x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (c x\right )^{\frac {13}{3}}}{{\left (b x^{2} + a\right )}^{\frac {2}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(13/3)/(b*x^2+a)^(2/3),x, algorithm="giac")

[Out]

integrate((c*x)^(13/3)/(b*x^2 + a)^(2/3), x)

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[ \int \frac {\left (c x \right )^{\frac {13}{3}}}{\left (b \,x^{2}+a \right )^{\frac {2}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^(13/3)/(b*x^2+a)^(2/3),x)

[Out]

int((c*x)^(13/3)/(b*x^2+a)^(2/3),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (c x\right )^{\frac {13}{3}}}{{\left (b x^{2} + a\right )}^{\frac {2}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(13/3)/(b*x^2+a)^(2/3),x, algorithm="maxima")

[Out]

integrate((c*x)^(13/3)/(b*x^2 + a)^(2/3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c\,x\right )}^{13/3}}{{\left (b\,x^2+a\right )}^{2/3}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^(13/3)/(a + b*x^2)^(2/3),x)

[Out]

int((c*x)^(13/3)/(a + b*x^2)^(2/3), x)

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sympy [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((c*x)**(13/3)/(b*x**2+a)**(2/3),x)

[Out]

Timed out

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