3.671 \(\int \frac {a^2+2 a b x^2+b^2 x^4}{(d x)^{7/2}} \, dx\)

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

[Out]

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

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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} \[ -\frac {2 a^2}{5 d (d x)^{5/2}}-\frac {4 a b}{d^3 \sqrt {d x}}+\frac {2 b^2 (d x)^{3/2}}{3 d^5} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x^2 + b^2*x^4)/(d*x)^(7/2),x]

[Out]

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

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Mathematica [A]  time = 0.01, size = 38, normalized size = 0.78 \[ \frac {2 \sqrt {d x} \left (-3 a^2-30 a b x^2+5 b^2 x^4\right )}{15 d^4 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)/(d*x)^(7/2),x]

[Out]

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

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fricas [A]  time = 0.84, size = 34, normalized size = 0.69 \[ \frac {2 \, {\left (5 \, b^{2} x^{4} - 30 \, a b x^{2} - 3 \, a^{2}\right )} \sqrt {d x}}{15 \, d^{4} x^{3}} \]

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)^(7/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.15, size = 48, normalized size = 0.98 \[ \frac {2 \, {\left (5 \, \sqrt {d x} b^{2} x - \frac {3 \, {\left (10 \, a b d^{3} x^{2} + a^{2} d^{3}\right )}}{\sqrt {d x} d^{2} x^{2}}\right )}}{15 \, d^{4}} \]

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)^(7/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 30, normalized size = 0.61 \[ -\frac {2 \left (-5 b^{2} x^{4}+30 a b \,x^{2}+3 a^{2}\right ) x}{15 \left (d x \right )^{\frac {7}{2}}} \]

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)^(7/2),x)

[Out]

-2/15*(-5*b^2*x^4+30*a*b*x^2+3*a^2)*x/(d*x)^(7/2)

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maxima [A]  time = 1.30, size = 47, normalized size = 0.96 \[ \frac {2 \, {\left (\frac {5 \, \left (d x\right )^{\frac {3}{2}} b^{2}}{d^{4}} - \frac {3 \, {\left (10 \, a b d^{2} x^{2} + a^{2} d^{2}\right )}}{\left (d x\right )^{\frac {5}{2}} d^{2}}\right )}}{15 \, d} \]

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)^(7/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.05, size = 34, normalized size = 0.69 \[ -\frac {6\,a^2+60\,a\,b\,x^2-10\,b^2\,x^4}{15\,d^3\,x^2\,\sqrt {d\,x}} \]

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)^(7/2),x)

[Out]

-(6*a^2 - 10*b^2*x^4 + 60*a*b*x^2)/(15*d^3*x^2*(d*x)^(1/2))

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sympy [A]  time = 1.97, size = 48, normalized size = 0.98 \[ - \frac {2 a^{2}}{5 d^{\frac {7}{2}} x^{\frac {5}{2}}} - \frac {4 a b}{d^{\frac {7}{2}} \sqrt {x}} + \frac {2 b^{2} x^{\frac {3}{2}}}{3 d^{\frac {7}{2}}} \]

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)**(7/2),x)

[Out]

-2*a**2/(5*d**(7/2)*x**(5/2)) - 4*a*b/(d**(7/2)*sqrt(x)) + 2*b**2*x**(3/2)/(3*d**(7/2))

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