3.45 \(\int \frac {\sqrt {a x+b x^3}}{x^4} \, dx\)

Optimal. Leaf size=283 \[ \frac {2 b^{5/4} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 a^{3/4} \sqrt {a x+b x^3}}-\frac {4 b^{5/4} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 a^{3/4} \sqrt {a x+b x^3}}+\frac {4 b^{3/2} x \left (a+b x^2\right )}{5 a \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {a x+b x^3}}-\frac {4 b \sqrt {a x+b x^3}}{5 a x}-\frac {2 \sqrt {a x+b x^3}}{5 x^3} \]

[Out]

4/5*b^(3/2)*x*(b*x^2+a)/a/(a^(1/2)+x*b^(1/2))/(b*x^3+a*x)^(1/2)-2/5*(b*x^3+a*x)^(1/2)/x^3-4/5*b*(b*x^3+a*x)^(1
/2)/a/x-4/5*b^(5/4)*(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)))*El
lipticE(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)/a^(3/4)/(b*x^3+a*x)^(1/2)+2/5*b^(5/4)*(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)))*EllipticF(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)/a^(3/4)/(b*x^3+a*x)^(1/2)

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Rubi [A]  time = 0.25, antiderivative size = 283, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {2020, 2025, 2032, 329, 305, 220, 1196} \[ \frac {2 b^{5/4} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 a^{3/4} \sqrt {a x+b x^3}}-\frac {4 b^{5/4} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 a^{3/4} \sqrt {a x+b x^3}}+\frac {4 b^{3/2} x \left (a+b x^2\right )}{5 a \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {a x+b x^3}}-\frac {4 b \sqrt {a x+b x^3}}{5 a x}-\frac {2 \sqrt {a x+b x^3}}{5 x^3} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a*x + b*x^3]/x^4,x]

[Out]

(4*b^(3/2)*x*(a + b*x^2))/(5*a*(Sqrt[a] + Sqrt[b]*x)*Sqrt[a*x + b*x^3]) - (2*Sqrt[a*x + b*x^3])/(5*x^3) - (4*b
*Sqrt[a*x + b*x^3])/(5*a*x) - (4*b^(5/4)*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])/(5*a^(3/4)*Sqrt[a*x + b*x^3]) + (2*b^(5/4)*Sqrt[x]*(Sq
rt[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])/(5*a^(3/4)*Sqrt[a*x + b*x^3])

Rule 220

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)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 305

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 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 1196

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)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rule 2020

Int[((c_.)*(x_))^(m_)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a*x^j + b*
x^n)^p)/(c*(m + j*p + 1)), x] - Dist[(b*p*(n - j))/(c^n*(m + j*p + 1)), Int[(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] && (IntegersQ[j, n] || GtQ[c, 0]) && GtQ[p
, 0] && LtQ[m + j*p + 1, 0]

Rule 2025

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

Rule 2032

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(c^IntPart[m]*(c*x)^FracP
art[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 {\sqrt {a x+b x^3}}{x^4} \, dx &=-\frac {2 \sqrt {a x+b x^3}}{5 x^3}+\frac {1}{5} (2 b) \int \frac {1}{x \sqrt {a x+b x^3}} \, dx\\ &=-\frac {2 \sqrt {a x+b x^3}}{5 x^3}-\frac {4 b \sqrt {a x+b x^3}}{5 a x}+\frac {\left (2 b^2\right ) \int \frac {x}{\sqrt {a x+b x^3}} \, dx}{5 a}\\ &=-\frac {2 \sqrt {a x+b x^3}}{5 x^3}-\frac {4 b \sqrt {a x+b x^3}}{5 a x}+\frac {\left (2 b^2 \sqrt {x} \sqrt {a+b x^2}\right ) \int \frac {\sqrt {x}}{\sqrt {a+b x^2}} \, dx}{5 a \sqrt {a x+b x^3}}\\ &=-\frac {2 \sqrt {a x+b x^3}}{5 x^3}-\frac {4 b \sqrt {a x+b x^3}}{5 a x}+\frac {\left (4 b^2 \sqrt {x} \sqrt {a+b x^2}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {a+b x^4}} \, dx,x,\sqrt {x}\right )}{5 a \sqrt {a x+b x^3}}\\ &=-\frac {2 \sqrt {a x+b x^3}}{5 x^3}-\frac {4 b \sqrt {a x+b x^3}}{5 a x}+\frac {\left (4 b^{3/2} \sqrt {x} \sqrt {a+b x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^4}} \, dx,x,\sqrt {x}\right )}{5 \sqrt {a} \sqrt {a x+b x^3}}-\frac {\left (4 b^{3/2} \sqrt {x} \sqrt {a+b x^2}\right ) \operatorname {Subst}\left (\int \frac {1-\frac {\sqrt {b} x^2}{\sqrt {a}}}{\sqrt {a+b x^4}} \, dx,x,\sqrt {x}\right )}{5 \sqrt {a} \sqrt {a x+b x^3}}\\ &=\frac {4 b^{3/2} x \left (a+b x^2\right )}{5 a \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {a x+b x^3}}-\frac {2 \sqrt {a x+b x^3}}{5 x^3}-\frac {4 b \sqrt {a x+b x^3}}{5 a x}-\frac {4 b^{5/4} \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 )}{5 a^{3/4} \sqrt {a x+b x^3}}+\frac {2 b^{5/4} \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 )}{5 a^{3/4} \sqrt {a x+b x^3}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 53, normalized size = 0.19 \[ -\frac {2 \sqrt {x \left (a+b x^2\right )} \, _2F_1\left (-\frac {5}{4},-\frac {1}{2};-\frac {1}{4};-\frac {b x^2}{a}\right )}{5 x^3 \sqrt {\frac {b x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a*x + b*x^3]/x^4,x]

[Out]

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

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fricas [F]  time = 0.73, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b x^{3} + a x}}{x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a*x)^(1/2)/x^4,x, algorithm="fricas")

[Out]

integral(sqrt(b*x^3 + a*x)/x^4, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {b x^{3} + a x}}{x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a*x)^(1/2)/x^4,x, algorithm="giac")

[Out]

integrate(sqrt(b*x^3 + a*x)/x^4, x)

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maple [A]  time = 0.04, size = 201, normalized size = 0.71 \[ -\frac {4 \left (b \,x^{2}+a \right ) b}{5 \sqrt {\left (b \,x^{2}+a \right ) x}\, a}+\frac {2 \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, \EllipticE \left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}+\frac {\sqrt {-a b}\, \EllipticF \left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}\right ) b}{5 \sqrt {b \,x^{3}+a x}\, a}-\frac {2 \sqrt {b \,x^{3}+a x}}{5 x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a*x)^(1/2)/x^4,x)

[Out]

-2/5*(b*x^3+a*x)^(1/2)/x^3-4/5*(b*x^2+a)*b/a/((b*x^2+a)*x)^(1/2)+2/5*b/a*(-a*b)^(1/2)*((x+(-a*b)^(1/2)/b)/(-a*
b)^(1/2)*b)^(1/2)*(-2*(x-(-a*b)^(1/2)/b)/(-a*b)^(1/2)*b)^(1/2)*(-1/(-a*b)^(1/2)*b*x)^(1/2)/(b*x^3+a*x)^(1/2)*(
-2*(-a*b)^(1/2)/b*EllipticE(((x+(-a*b)^(1/2)/b)/(-a*b)^(1/2)*b)^(1/2),1/2*2^(1/2))+(-a*b)^(1/2)/b*EllipticF(((
x+(-a*b)^(1/2)/b)/(-a*b)^(1/2)*b)^(1/2),1/2*2^(1/2)))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {b x^{3} + a x}}{x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a*x)^(1/2)/x^4,x, algorithm="maxima")

[Out]

integrate(sqrt(b*x^3 + a*x)/x^4, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\sqrt {b\,x^3+a\,x}}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x + b*x^3)^(1/2)/x^4,x)

[Out]

int((a*x + b*x^3)^(1/2)/x^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {x \left (a + b x^{2}\right )}}{x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a*x)**(1/2)/x**4,x)

[Out]

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

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