3.758 \(\int \frac{(a+b x^2)^{4/3}}{(c x)^{11/3}} \, dx\)

Optimal. Leaf size=157 \[ -\frac{3 b^{4/3} \log \left (\sqrt [3]{b} (c x)^{2/3}-c^{2/3} \sqrt [3]{a+b x^2}\right )}{4 c^{11/3}}-\frac{\sqrt{3} b^{4/3} \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} (c x)^{2/3}}{c^{2/3} \sqrt [3]{a+b x^2}}+1}{\sqrt{3}}\right )}{2 c^{11/3}}-\frac{3 b \sqrt [3]{a+b x^2}}{2 c^3 (c x)^{2/3}}-\frac{3 \left (a+b x^2\right )^{4/3}}{8 c (c x)^{8/3}} \]

[Out]

(-3*b*(a + b*x^2)^(1/3))/(2*c^3*(c*x)^(2/3)) - (3*(a + b*x^2)^(4/3))/(8*c*(c*x)^(8/3)) - (Sqrt[3]*b^(4/3)*ArcT
an[(1 + (2*b^(1/3)*(c*x)^(2/3))/(c^(2/3)*(a + b*x^2)^(1/3)))/Sqrt[3]])/(2*c^(11/3)) - (3*b^(4/3)*Log[b^(1/3)*(
c*x)^(2/3) - c^(2/3)*(a + b*x^2)^(1/3)])/(4*c^(11/3))

________________________________________________________________________________________

Rubi [A]  time = 0.290834, antiderivative size = 234, normalized size of antiderivative = 1.49, 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 = {277, 329, 275, 331, 292, 31, 634, 617, 204, 628} \[ -\frac{b^{4/3} \log \left (c^{2/3}-\frac{\sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )}{2 c^{11/3}}+\frac{b^{4/3} \log \left (\frac{b^{2/3} (c x)^{4/3}}{\left (a+b x^2\right )^{2/3}}+\frac{\sqrt [3]{b} c^{2/3} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}+c^{4/3}\right )}{4 c^{11/3}}-\frac{\sqrt{3} b^{4/3} \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}+c^{2/3}}{\sqrt{3} c^{2/3}}\right )}{2 c^{11/3}}-\frac{3 b \sqrt [3]{a+b x^2}}{2 c^3 (c x)^{2/3}}-\frac{3 \left (a+b x^2\right )^{4/3}}{8 c (c x)^{8/3}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(4/3)/(c*x)^(11/3),x]

[Out]

(-3*b*(a + b*x^2)^(1/3))/(2*c^3*(c*x)^(2/3)) - (3*(a + b*x^2)^(4/3))/(8*c*(c*x)^(8/3)) - (Sqrt[3]*b^(4/3)*ArcT
an[(c^(2/3) + (2*b^(1/3)*(c*x)^(2/3))/(a + b*x^2)^(1/3))/(Sqrt[3]*c^(2/3))])/(2*c^(11/3)) - (b^(4/3)*Log[c^(2/
3) - (b^(1/3)*(c*x)^(2/3))/(a + b*x^2)^(1/3)])/(2*c^(11/3)) + (b^(4/3)*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)])/(4*c^(11/3))

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 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 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 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 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 31

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

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]

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 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 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]

Rubi steps

\begin{align*} \int \frac{\left (a+b x^2\right )^{4/3}}{(c x)^{11/3}} \, dx &=-\frac{3 \left (a+b x^2\right )^{4/3}}{8 c (c x)^{8/3}}+\frac{b \int \frac{\sqrt [3]{a+b x^2}}{(c x)^{5/3}} \, dx}{c^2}\\ &=-\frac{3 b \sqrt [3]{a+b x^2}}{2 c^3 (c x)^{2/3}}-\frac{3 \left (a+b x^2\right )^{4/3}}{8 c (c x)^{8/3}}+\frac{b^2 \int \frac{\sqrt [3]{c x}}{\left (a+b x^2\right )^{2/3}} \, dx}{c^4}\\ &=-\frac{3 b \sqrt [3]{a+b x^2}}{2 c^3 (c x)^{2/3}}-\frac{3 \left (a+b x^2\right )^{4/3}}{8 c (c x)^{8/3}}+\frac{\left (3 b^2\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 )}{c^5}\\ &=-\frac{3 b \sqrt [3]{a+b x^2}}{2 c^3 (c x)^{2/3}}-\frac{3 \left (a+b x^2\right )^{4/3}}{8 c (c x)^{8/3}}+\frac{\left (3 b^2\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 )}{2 c^5}\\ &=-\frac{3 b \sqrt [3]{a+b x^2}}{2 c^3 (c x)^{2/3}}-\frac{3 \left (a+b x^2\right )^{4/3}}{8 c (c x)^{8/3}}+\frac{\left (3 b^2\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 )}{2 c^5}\\ &=-\frac{3 b \sqrt [3]{a+b x^2}}{2 c^3 (c x)^{2/3}}-\frac{3 \left (a+b x^2\right )^{4/3}}{8 c (c x)^{8/3}}+\frac{b^{5/3} \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 )}{2 c^{13/3}}-\frac{b^{5/3} \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 )}{2 c^{13/3}}\\ &=-\frac{3 b \sqrt [3]{a+b x^2}}{2 c^3 (c x)^{2/3}}-\frac{3 \left (a+b x^2\right )^{4/3}}{8 c (c x)^{8/3}}-\frac{b^{4/3} \log \left (c^{2/3}-\frac{\sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )}{2 c^{11/3}}-\frac{\left (3 b^{5/3}\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 )}{4 c^{13/3}}+\frac{b^{4/3} \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 )}{4 c^{11/3}}\\ &=-\frac{3 b \sqrt [3]{a+b x^2}}{2 c^3 (c x)^{2/3}}-\frac{3 \left (a+b x^2\right )^{4/3}}{8 c (c x)^{8/3}}-\frac{b^{4/3} \log \left (c^{2/3}-\frac{\sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )}{2 c^{11/3}}+\frac{b^{4/3} \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 )}{4 c^{11/3}}+\frac{\left (3 b^{4/3}\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 )}{2 c^{11/3}}\\ &=-\frac{3 b \sqrt [3]{a+b x^2}}{2 c^3 (c x)^{2/3}}-\frac{3 \left (a+b x^2\right )^{4/3}}{8 c (c x)^{8/3}}-\frac{\sqrt{3} b^{4/3} \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 )}{2 c^{11/3}}-\frac{b^{4/3} \log \left (c^{2/3}-\frac{\sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )}{2 c^{11/3}}+\frac{b^{4/3} \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 )}{4 c^{11/3}}\\ \end{align*}

Mathematica [C]  time = 0.0147145, size = 57, normalized size = 0.36 \[ -\frac{3 a x \sqrt [3]{a+b x^2} \, _2F_1\left (-\frac{4}{3},-\frac{4}{3};-\frac{1}{3};-\frac{b x^2}{a}\right )}{8 (c x)^{11/3} \sqrt [3]{\frac{b x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(4/3)/(c*x)^(11/3),x]

[Out]

(-3*a*x*(a + b*x^2)^(1/3)*Hypergeometric2F1[-4/3, -4/3, -1/3, -((b*x^2)/a)])/(8*(c*x)^(11/3)*(1 + (b*x^2)/a)^(
1/3))

________________________________________________________________________________________

Maple [F]  time = 0.023, size = 0, normalized size = 0. \begin{align*} \int{ \left ( b{x}^{2}+a \right ) ^{{\frac{4}{3}}} \left ( cx \right ) ^{-{\frac{11}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(4/3)/(c*x)^(11/3),x)

[Out]

int((b*x^2+a)^(4/3)/(c*x)^(11/3),x)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{2} + a\right )}^{\frac{4}{3}}}{\left (c x\right )^{\frac{11}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(4/3)/(c*x)^(11/3),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^(4/3)/(c*x)^(11/3), x)

________________________________________________________________________________________

Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(4/3)/(c*x)^(11/3),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(4/3)/(c*x)**(11/3),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{2} + a\right )}^{\frac{4}{3}}}{\left (c x\right )^{\frac{11}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(4/3)/(c*x)^(11/3),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^(4/3)/(c*x)^(11/3), x)