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

Optimal. Leaf size=49 \[ \frac{2 a^2 \sqrt{d x}}{d}+\frac{4 a b (d x)^{5/2}}{5 d^3}+\frac{2 b^2 (d x)^{9/2}}{9 d^5} \]

[Out]

(2*a^2*Sqrt[d*x])/d + (4*a*b*(d*x)^(5/2))/(5*d^3) + (2*b^2*(d*x)^(9/2))/(9*d^5)

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Rubi [A]  time = 0.0131858, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.038, Rules used = {14} \[ \frac{2 a^2 \sqrt{d x}}{d}+\frac{4 a b (d x)^{5/2}}{5 d^3}+\frac{2 b^2 (d x)^{9/2}}{9 d^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*Sqrt[d*x])/d + (4*a*b*(d*x)^(5/2))/(5*d^3) + (2*b^2*(d*x)^(9/2))/(9*d^5)

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]]

Rubi steps

\begin{align*} \int \frac{a^2+2 a b x^2+b^2 x^4}{\sqrt{d x}} \, dx &=\int \left (\frac{a^2}{\sqrt{d x}}+\frac{2 a b (d x)^{3/2}}{d^2}+\frac{b^2 (d x)^{7/2}}{d^4}\right ) \, dx\\ &=\frac{2 a^2 \sqrt{d x}}{d}+\frac{4 a b (d x)^{5/2}}{5 d^3}+\frac{2 b^2 (d x)^{9/2}}{9 d^5}\\ \end{align*}

Mathematica [A]  time = 0.0103703, size = 33, normalized size = 0.67 \[ \frac{2 \left (45 a^2 x+18 a b x^3+5 b^2 x^5\right )}{45 \sqrt{d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(45*a^2*x + 18*a*b*x^3 + 5*b^2*x^5))/(45*Sqrt[d*x])

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Maple [A]  time = 0.047, size = 30, normalized size = 0.6 \begin{align*}{\frac{ \left ( 10\,{b}^{2}{x}^{4}+36\,ab{x}^{2}+90\,{a}^{2} \right ) x}{45}{\frac{1}{\sqrt{dx}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/45*(5*b^2*x^4+18*a*b*x^2+45*a^2)*x/(d*x)^(1/2)

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Maxima [A]  time = 0.96012, size = 55, normalized size = 1.12 \begin{align*} \frac{2 \,{\left (45 \, \sqrt{d x} a^{2} + \frac{5 \, \left (d x\right )^{\frac{9}{2}} b^{2}}{d^{4}} + \frac{18 \, \left (d x\right )^{\frac{5}{2}} a b}{d^{2}}\right )}}{45 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/45*(45*sqrt(d*x)*a^2 + 5*(d*x)^(9/2)*b^2/d^4 + 18*(d*x)^(5/2)*a*b/d^2)/d

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Fricas [A]  time = 1.22918, size = 70, normalized size = 1.43 \begin{align*} \frac{2 \,{\left (5 \, b^{2} x^{4} + 18 \, a b x^{2} + 45 \, a^{2}\right )} \sqrt{d x}}{45 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/45*(5*b^2*x^4 + 18*a*b*x^2 + 45*a^2)*sqrt(d*x)/d

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Sympy [A]  time = 0.596318, size = 48, normalized size = 0.98 \begin{align*} \frac{2 a^{2} \sqrt{x}}{\sqrt{d}} + \frac{4 a b x^{\frac{5}{2}}}{5 \sqrt{d}} + \frac{2 b^{2} x^{\frac{9}{2}}}{9 \sqrt{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*a**2*sqrt(x)/sqrt(d) + 4*a*b*x**(5/2)/(5*sqrt(d)) + 2*b**2*x**(9/2)/(9*sqrt(d))

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Giac [A]  time = 1.11453, size = 55, normalized size = 1.12 \begin{align*} \frac{2 \,{\left (5 \, \sqrt{d x} b^{2} x^{4} + 18 \, \sqrt{d x} a b x^{2} + 45 \, \sqrt{d x} a^{2}\right )}}{45 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/45*(5*sqrt(d*x)*b^2*x^4 + 18*sqrt(d*x)*a*b*x^2 + 45*sqrt(d*x)*a^2)/d