3.670 \(\int \frac {a^2+2 a b x^2+b^2 x^4}{(d x)^{5/2}} \, dx\)

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

[Out]

-2/3*a^2/d/(d*x)^(3/2)+2/5*b^2*(d*x)^(5/2)/d^5+4*a*b*(d*x)^(1/2)/d^3

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Rubi [A]  time = 0.01, antiderivative size = 49, normalized size of antiderivative = 1.00, 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} \[ -\frac {2 a^2}{3 d (d x)^{3/2}}+\frac {4 a b \sqrt {d x}}{d^3}+\frac {2 b^2 (d x)^{5/2}}{5 d^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.01, size = 33, normalized size = 0.67 \[ \frac {x \left (-10 a^2+60 a b x^2+6 b^2 x^4\right )}{15 (d x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(-10*a^2 + 60*a*b*x^2 + 6*b^2*x^4))/(15*(d*x)^(5/2))

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fricas [A]  time = 0.73, size = 34, normalized size = 0.69 \[ \frac {2 \, {\left (3 \, b^{2} x^{4} + 30 \, a b x^{2} - 5 \, a^{2}\right )} \sqrt {d x}}{15 \, d^{3} x^{2}} \]

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)^(5/2),x, algorithm="fricas")

[Out]

2/15*(3*b^2*x^4 + 30*a*b*x^2 - 5*a^2)*sqrt(d*x)/(d^3*x^2)

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giac [A]  time = 0.15, size = 53, normalized size = 1.08 \[ -\frac {2 \, {\left (\frac {5 \, a^{2} d}{\sqrt {d x} x} - \frac {3 \, {\left (\sqrt {d x} b^{2} d^{10} x^{2} + 10 \, \sqrt {d x} a b d^{10}\right )}}{d^{10}}\right )}}{15 \, d^{3}} \]

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)^(5/2),x, algorithm="giac")

[Out]

-2/15*(5*a^2*d/(sqrt(d*x)*x) - 3*(sqrt(d*x)*b^2*d^10*x^2 + 10*sqrt(d*x)*a*b*d^10)/d^10)/d^3

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maple [A]  time = 0.01, size = 30, normalized size = 0.61 \[ -\frac {2 \left (-3 b^{2} x^{4}-30 a b \,x^{2}+5 a^{2}\right ) x}{15 \left (d x \right )^{\frac {5}{2}}} \]

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)^(5/2),x)

[Out]

-2/15*(-3*b^2*x^4-30*a*b*x^2+5*a^2)*x/(d*x)^(5/2)

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maxima [A]  time = 1.40, size = 43, normalized size = 0.88 \[ -\frac {2 \, {\left (\frac {5 \, a^{2}}{\left (d x\right )^{\frac {3}{2}}} - \frac {3 \, {\left (\left (d x\right )^{\frac {5}{2}} b^{2} + 10 \, \sqrt {d x} a b d^{2}\right )}}{d^{4}}\right )}}{15 \, d} \]

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)^(5/2),x, algorithm="maxima")

[Out]

-2/15*(5*a^2/(d*x)^(3/2) - 3*((d*x)^(5/2)*b^2 + 10*sqrt(d*x)*a*b*d^2)/d^4)/d

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mupad [B]  time = 4.23, size = 34, normalized size = 0.69 \[ \frac {-10\,a^2+60\,a\,b\,x^2+6\,b^2\,x^4}{15\,d^2\,x\,\sqrt {d\,x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2 + b^2*x^4 + 2*a*b*x^2)/(d*x)^(5/2),x)

[Out]

(6*b^2*x^4 - 10*a^2 + 60*a*b*x^2)/(15*d^2*x*(d*x)^(1/2))

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sympy [A]  time = 0.91, size = 48, normalized size = 0.98 \[ - \frac {2 a^{2}}{3 d^{\frac {5}{2}} x^{\frac {3}{2}}} + \frac {4 a b \sqrt {x}}{d^{\frac {5}{2}}} + \frac {2 b^{2} x^{\frac {5}{2}}}{5 d^{\frac {5}{2}}} \]

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)**(5/2),x)

[Out]

-2*a**2/(3*d**(5/2)*x**(3/2)) + 4*a*b*sqrt(x)/d**(5/2) + 2*b**2*x**(5/2)/(5*d**(5/2))

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