3.6.53 \(\int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{\sqrt {d x}} \, dx\)

Optimal. Leaf size=91 \[ \frac {2 b (d x)^{5/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 d^3 \left (a+b x^2\right )}+\frac {2 a \sqrt {d x} \sqrt {a^2+2 a b x^2+b^2 x^4}}{d \left (a+b x^2\right )} \]

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Rubi [A]  time = 0.03, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {1112, 14} \begin {gather*} \frac {2 b (d x)^{5/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 d^3 \left (a+b x^2\right )}+\frac {2 a \sqrt {d x} \sqrt {a^2+2 a b x^2+b^2 x^4}}{d \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]/Sqrt[d*x],x]

[Out]

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 1112

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(a + b*x^2 + c*x^4)^FracPa
rt[p]/(c^IntPart[p]*(b/2 + c*x^2)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^2)^(2*p), x], x] /; FreeQ[{a, b, c,
d, m, p}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

Rubi steps

\begin {align*} \int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{\sqrt {d x}} \, dx &=\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {a b+b^2 x^2}{\sqrt {d x}} \, dx}{a b+b^2 x^2}\\ &=\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \left (\frac {a b}{\sqrt {d x}}+\frac {b^2 (d x)^{3/2}}{d^2}\right ) \, dx}{a b+b^2 x^2}\\ &=\frac {2 a \sqrt {d x} \sqrt {a^2+2 a b x^2+b^2 x^4}}{d \left (a+b x^2\right )}+\frac {2 b (d x)^{5/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 d^3 \left (a+b x^2\right )}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 43, normalized size = 0.47 \begin {gather*} \frac {2 \sqrt {\left (a+b x^2\right )^2} \left (5 a x+b x^3\right )}{5 \sqrt {d x} \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]/Sqrt[d*x],x]

[Out]

(2*Sqrt[(a + b*x^2)^2]*(5*a*x + b*x^3))/(5*Sqrt[d*x]*(a + b*x^2))

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IntegrateAlgebraic [A]  time = 23.99, size = 68, normalized size = 0.75 \begin {gather*} \frac {2 \left (a d^2+b d^2 x^2\right ) \left (5 a d^2 \sqrt {d x}+b (d x)^{5/2}\right )}{5 d^5 \sqrt {\frac {\left (a d^2+b d^2 x^2\right )^2}{d^4}}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]/Sqrt[d*x],x]

[Out]

(2*(a*d^2 + b*d^2*x^2)*(5*a*d^2*Sqrt[d*x] + b*(d*x)^(5/2)))/(5*d^5*Sqrt[(a*d^2 + b*d^2*x^2)^2/d^4])

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fricas [A]  time = 1.71, size = 19, normalized size = 0.21 \begin {gather*} \frac {2 \, {\left (b x^{2} + 5 \, a\right )} \sqrt {d x}}{5 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(b*x^2 + 5*a)*sqrt(d*x)/d

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giac [A]  time = 0.17, size = 40, normalized size = 0.44 \begin {gather*} \frac {2 \, {\left (\sqrt {d x} b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 5 \, \sqrt {d x} a \mathrm {sgn}\left (b x^{2} + a\right )\right )}}{5 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(sqrt(d*x)*b*x^2*sgn(b*x^2 + a) + 5*sqrt(d*x)*a*sgn(b*x^2 + a))/d

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maple [A]  time = 0.00, size = 38, normalized size = 0.42 \begin {gather*} \frac {2 \left (b \,x^{2}+5 a \right ) \sqrt {\left (b \,x^{2}+a \right )^{2}}\, x}{5 \left (b \,x^{2}+a \right ) \sqrt {d x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*x*(b*x^2+5*a)*((b*x^2+a)^2)^(1/2)/(b*x^2+a)/(d*x)^(1/2)

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maxima [A]  time = 1.37, size = 24, normalized size = 0.26 \begin {gather*} \frac {2 \, {\left (5 \, \sqrt {d x} a + \frac {\left (d x\right )^{\frac {5}{2}} b}{d^{2}}\right )}}{5 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(5*sqrt(d*x)*a + (d*x)^(5/2)*b/d^2)/d

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mupad [B]  time = 4.36, size = 47, normalized size = 0.52 \begin {gather*} \frac {\left (\frac {2\,x^3}{5}+\frac {2\,a\,x}{b}\right )\,\sqrt {{\left (b\,x^2+a\right )}^2}}{x^2\,\sqrt {d\,x}+\frac {a\,\sqrt {d\,x}}{b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(((2*x^3)/5 + (2*a*x)/b)*((a + b*x^2)^2)^(1/2))/(x^2*(d*x)^(1/2) + (a*(d*x)^(1/2))/b)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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