3.1.68 \(\int \frac {x^4}{(a x+b x^3)^{3/2}} \, dx\) [68]

Optimal. Leaf size=253 \[ -\frac {x^2}{b \sqrt {a x+b x^3}}+\frac {3 x \left (a+b x^2\right )}{b^{3/2} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {a x+b x^3}}-\frac {3 \sqrt [4]{a} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{b^{7/4} \sqrt {a x+b x^3}}+\frac {3 \sqrt [4]{a} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 b^{7/4} \sqrt {a x+b x^3}} \]

[Out]

-x^2/b/(b*x^3+a*x)^(1/2)+3*x*(b*x^2+a)/b^(3/2)/(a^(1/2)+x*b^(1/2))/(b*x^3+a*x)^(1/2)-3*a^(1/4)*(cos(2*arctan(b
^(1/4)*x^(1/2)/a^(1/4)))^2)^(1/2)/cos(2*arctan(b^(1/4)*x^(1/2)/a^(1/4)))*EllipticE(sin(2*arctan(b^(1/4)*x^(1/2
)/a^(1/4))),1/2*2^(1/2))*(a^(1/2)+x*b^(1/2))*x^(1/2)*((b*x^2+a)/(a^(1/2)+x*b^(1/2))^2)^(1/2)/b^(7/4)/(b*x^3+a*
x)^(1/2)+3/2*a^(1/4)*(cos(2*arctan(b^(1/4)*x^(1/2)/a^(1/4)))^2)^(1/2)/cos(2*arctan(b^(1/4)*x^(1/2)/a^(1/4)))*E
llipticF(sin(2*arctan(b^(1/4)*x^(1/2)/a^(1/4))),1/2*2^(1/2))*(a^(1/2)+x*b^(1/2))*x^(1/2)*((b*x^2+a)/(a^(1/2)+x
*b^(1/2))^2)^(1/2)/b^(7/4)/(b*x^3+a*x)^(1/2)

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Rubi [A]
time = 0.15, antiderivative size = 253, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {2047, 2057, 335, 311, 226, 1210} \begin {gather*} \frac {3 \sqrt [4]{a} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} F\left (2 \text {ArcTan}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 b^{7/4} \sqrt {a x+b x^3}}-\frac {3 \sqrt [4]{a} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} E\left (2 \text {ArcTan}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{b^{7/4} \sqrt {a x+b x^3}}+\frac {3 x \left (a+b x^2\right )}{b^{3/2} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {a x+b x^3}}-\frac {x^2}{b \sqrt {a x+b x^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^4/(a*x + b*x^3)^(3/2),x]

[Out]

-(x^2/(b*Sqrt[a*x + b*x^3])) + (3*x*(a + b*x^2))/(b^(3/2)*(Sqrt[a] + Sqrt[b]*x)*Sqrt[a*x + b*x^3]) - (3*a^(1/4
)*Sqrt[x]*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*EllipticE[2*ArcTan[(b^(1/4)*Sqrt[x])
/a^(1/4)], 1/2])/(b^(7/4)*Sqrt[a*x + b*x^3]) + (3*a^(1/4)*Sqrt[x]*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt
[a] + Sqrt[b]*x)^2]*EllipticF[2*ArcTan[(b^(1/4)*Sqrt[x])/a^(1/4)], 1/2])/(2*b^(7/4)*Sqrt[a*x + b*x^3])

Rule 226

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[(1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))*EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 311

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, Dist[1/q, Int[1/Sqrt[a + b*x^4], x],
x] - Dist[1/q, Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

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 1210

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a +
c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d*(1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4
]))*EllipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rule 2047

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

Rule 2057

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[c^IntPart[m]*(c*x)^FracPa
rt[m]*((a*x^j + b*x^n)^FracPart[p]/(x^(FracPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p])), Int[x^(m
+ j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && PosQ[n
- j]

Rubi steps

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

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.04, size = 57, normalized size = 0.23 \begin {gather*} -\frac {2 x^2 \left (-1+\sqrt {1+\frac {b x^2}{a}} \, _2F_1\left (\frac {3}{4},\frac {3}{2};\frac {7}{4};-\frac {b x^2}{a}\right )\right )}{b \sqrt {x \left (a+b x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^4/(a*x + b*x^3)^(3/2),x]

[Out]

(-2*x^2*(-1 + Sqrt[1 + (b*x^2)/a]*Hypergeometric2F1[3/4, 3/2, 7/4, -((b*x^2)/a)]))/(b*Sqrt[x*(a + b*x^2)])

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Maple [A]
time = 0.35, size = 182, normalized size = 0.72

method	result	size
default	\(-\frac {x^{2}}{b \sqrt {\left (x^{2}+\frac {a}{b}\right ) b x}}+\frac {3 \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, \EllipticE \left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}+\frac {\sqrt {-a b}\, \EllipticF \left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}\right )}{2 b^{2} \sqrt {b \,x^{3}+a x}}\)	\(182\)
elliptic	\(-\frac {x^{2}}{b \sqrt {\left (x^{2}+\frac {a}{b}\right ) b x}}+\frac {3 \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, \EllipticE \left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}+\frac {\sqrt {-a b}\, \EllipticF \left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}\right )}{2 b^{2} \sqrt {b \,x^{3}+a x}}\)	\(182\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(b*x^3+a*x)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/b*x^2/((x^2+a/b)*b*x)^(1/2)+3/2/b^2*(-a*b)^(1/2)*((x+1/b*(-a*b)^(1/2))*b/(-a*b)^(1/2))^(1/2)*(-2*(x-1/b*(-a
*b)^(1/2))*b/(-a*b)^(1/2))^(1/2)*(-x*b/(-a*b)^(1/2))^(1/2)/(b*x^3+a*x)^(1/2)*(-2/b*(-a*b)^(1/2)*EllipticE(((x+
1/b*(-a*b)^(1/2))*b/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))+1/b*(-a*b)^(1/2)*EllipticF(((x+1/b*(-a*b)^(1/2))*b/(-a*b)
^(1/2))^(1/2),1/2*2^(1/2)))

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

[Out]

integrate(x^4/(b*x^3 + a*x)^(3/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.51, size = 61, normalized size = 0.24 \begin {gather*} -\frac {\sqrt {b x^{3} + a x} b x + 3 \, {\left (b x^{2} + a\right )} \sqrt {b} {\rm weierstrassZeta}\left (-\frac {4 \, a}{b}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, a}{b}, 0, x\right )\right )}{b^{3} x^{2} + a b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(sqrt(b*x^3 + a*x)*b*x + 3*(b*x^2 + a)*sqrt(b)*weierstrassZeta(-4*a/b, 0, weierstrassPInverse(-4*a/b, 0, x)))
/(b^3*x^2 + a*b^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{4}}{\left (x \left (a + b x^{2}\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**4/(x*(a + b*x**2))**(3/2), x)

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

[Out]

integrate(x^4/(b*x^3 + a*x)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(a*x + b*x^3)^(3/2),x)

[Out]

int(x^4/(a*x + b*x^3)^(3/2), x)

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