\(\int \frac {a^2+2 a b x^2+b^2 x^4}{(d x)^{3/2}} \, dx\) [669]

Optimal result

Integrand size = 26, antiderivative size = 49 \[ \int \frac {a^2+2 a b x^2+b^2 x^4}{(d x)^{3/2}} \, dx=-\frac {2 a^2}{d \sqrt {d x}}+\frac {4 a b (d x)^{3/2}}{3 d^3}+\frac {2 b^2 (d x)^{7/2}}{7 d^5} \]

[Out]

4/3*a*b*(d*x)^(3/2)/d^3+2/7*b^2*(d*x)^(7/2)/d^5-2*a^2/d/(d*x)^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {14} \[ \int \frac {a^2+2 a b x^2+b^2 x^4}{(d x)^{3/2}} \, dx=-\frac {2 a^2}{d \sqrt {d x}}+\frac {4 a b (d x)^{3/2}}{3 d^3}+\frac {2 b^2 (d x)^{7/2}}{7 d^5} \]

[In]

Int[(a^2 + 2*a*b*x^2 + b^2*x^4)/(d*x)^(3/2),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.67 \[ \int \frac {a^2+2 a b x^2+b^2 x^4}{(d x)^{3/2}} \, dx=-\frac {2 x \left (21 a^2-14 a b x^2-3 b^2 x^4\right )}{21 (d x)^{3/2}} \]

[In]

Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)/(d*x)^(3/2),x]

[Out]

(-2*x*(21*a^2 - 14*a*b*x^2 - 3*b^2*x^4))/(21*(d*x)^(3/2))

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.61

method	result	size
gosper	\(-\frac {2 \left (-3 b^{2} x^{4}-14 a b \,x^{2}+21 a^{2}\right ) x}{21 \left (d x \right )^{\frac {3}{2}}}\)	\(30\)
risch	\(-\frac {2 \left (-3 b^{2} x^{4}-14 a b \,x^{2}+21 a^{2}\right )}{21 d \sqrt {d x}}\)	\(32\)
pseudoelliptic	\(\frac {6 b^{2} x^{4}+28 a b \,x^{2}-42 a^{2}}{21 d \sqrt {d x}}\)	\(32\)
trager	\(-\frac {2 \left (-3 b^{2} x^{4}-14 a b \,x^{2}+21 a^{2}\right ) \sqrt {d x}}{21 d^{2} x}\)	\(35\)
derivativedivides	\(\frac {\frac {2 b^{2} \left (d x \right )^{\frac {7}{2}}}{7}+\frac {4 a b \,d^{2} \left (d x \right )^{\frac {3}{2}}}{3}-\frac {2 a^{2} d^{4}}{\sqrt {d x}}}{d^{5}}\)	\(42\)
default	\(\frac {\frac {2 b^{2} \left (d x \right )^{\frac {7}{2}}}{7}+\frac {4 a b \,d^{2} \left (d x \right )^{\frac {3}{2}}}{3}-\frac {2 a^{2} d^{4}}{\sqrt {d x}}}{d^{5}}\)	\(42\)

[In]

int((b^2*x^4+2*a*b*x^2+a^2)/(d*x)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-2/21*(-3*b^2*x^4-14*a*b*x^2+21*a^2)*x/(d*x)^(3/2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.69 \[ \int \frac {a^2+2 a b x^2+b^2 x^4}{(d x)^{3/2}} \, dx=\frac {2 \, {\left (3 \, b^{2} x^{4} + 14 \, a b x^{2} - 21 \, a^{2}\right )} \sqrt {d x}}{21 \, d^{2} x} \]

[In]

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

[Out]

2/21*(3*b^2*x^4 + 14*a*b*x^2 - 21*a^2)*sqrt(d*x)/(d^2*x)

Sympy [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.94 \[ \int \frac {a^2+2 a b x^2+b^2 x^4}{(d x)^{3/2}} \, dx=- \frac {2 a^{2} x}{\left (d x\right )^{\frac {3}{2}}} + \frac {4 a b x^{3}}{3 \left (d x\right )^{\frac {3}{2}}} + \frac {2 b^{2} x^{5}}{7 \left (d x\right )^{\frac {3}{2}}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.90 \[ \int \frac {a^2+2 a b x^2+b^2 x^4}{(d x)^{3/2}} \, dx=-\frac {2 \, {\left (\frac {21 \, a^{2}}{\sqrt {d x}} - \frac {3 \, \left (d x\right )^{\frac {7}{2}} b^{2} + 14 \, \left (d x\right )^{\frac {3}{2}} a b d^{2}}{d^{4}}\right )}}{21 \, d} \]

[In]

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

[Out]

-2/21*(21*a^2/sqrt(d*x) - (3*(d*x)^(7/2)*b^2 + 14*(d*x)^(3/2)*a*b*d^2)/d^4)/d

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.04 \[ \int \frac {a^2+2 a b x^2+b^2 x^4}{(d x)^{3/2}} \, dx=-\frac {2 \, {\left (\frac {21 \, a^{2}}{\sqrt {d x}} - \frac {3 \, \sqrt {d x} b^{2} d^{27} x^{3} + 14 \, \sqrt {d x} a b d^{27} x}{d^{28}}\right )}}{21 \, d} \]

[In]

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

[Out]

-2/21*(21*a^2/sqrt(d*x) - (3*sqrt(d*x)*b^2*d^27*x^3 + 14*sqrt(d*x)*a*b*d^27*x)/d^28)/d

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.63 \[ \int \frac {a^2+2 a b x^2+b^2 x^4}{(d x)^{3/2}} \, dx=\frac {-42\,a^2+28\,a\,b\,x^2+6\,b^2\,x^4}{21\,d\,\sqrt {d\,x}} \]

[In]

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

[Out]

(6*b^2*x^4 - 42*a^2 + 28*a*b*x^2)/(21*d*(d*x)^(1/2))