\(\int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{(d x)^{3/2}} \, dx\) [732]

Optimal result

Integrand size = 30, antiderivative size = 91 \[ \int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{(d x)^{3/2}} \, dx=-\frac {2 a \sqrt {a^2+2 a b x^2+b^2 x^4}}{d \sqrt {d x} \left (a+b x^2\right )}+\frac {2 b (d x)^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 d^3 \left (a+b x^2\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {1126, 14} \[ \int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{(d x)^{3/2}} \, dx=\frac {2 b (d x)^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 d^3 \left (a+b x^2\right )}-\frac {2 a \sqrt {a^2+2 a b x^2+b^2 x^4}}{d \sqrt {d x} \left (a+b x^2\right )} \]

[In]

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

[Out]

(-2*a*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(d*Sqrt[d*x]*(a + b*x^2)) + (2*b*(d*x)^(3/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2
*x^4])/(3*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 1126

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.48 \[ \int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{(d x)^{3/2}} \, dx=-\frac {2 x \left (3 a-b x^2\right ) \sqrt {\left (a+b x^2\right )^2}}{3 (d x)^{3/2} \left (a+b x^2\right )} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.43

method	result	size
gosper	\(-\frac {2 x \left (-b \,x^{2}+3 a \right ) \sqrt {\left (b \,x^{2}+a \right )^{2}}}{3 \left (b \,x^{2}+a \right ) \left (d x \right )^{\frac {3}{2}}}\)	\(39\)
default	\(-\frac {2 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-b \,x^{2}+3 a \right )}{3 d \left (b \,x^{2}+a \right ) \sqrt {d x}}\)	\(41\)
risch	\(-\frac {2 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-b \,x^{2}+3 a \right )}{3 d \left (b \,x^{2}+a \right ) \sqrt {d x}}\)	\(41\)

[In]

int(((b*x^2+a)^2)^(1/2)/(d*x)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.24 \[ \int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{(d x)^{3/2}} \, dx=\frac {2 \, {\left (b x^{2} - 3 \, a\right )} \sqrt {d x}}{3 \, d^{2} x} \]

[In]

integrate(((b*x^2+a)^2)^(1/2)/(d*x)^(3/2),x, algorithm="fricas")

[Out]

2/3*(b*x^2 - 3*a)*sqrt(d*x)/(d^2*x)

Sympy [F]

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

[In]

integrate(((b*x**2+a)**2)**(1/2)/(d*x)**(3/2),x)

[Out]

Integral(sqrt((a + b*x**2)**2)/(d*x)**(3/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.27 \[ \int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{(d x)^{3/2}} \, dx=-\frac {2 \, {\left (\frac {3 \, a}{\sqrt {d x}} - \frac {\left (d x\right )^{\frac {3}{2}} b}{d^{2}}\right )}}{3 \, d} \]

[In]

integrate(((b*x^2+a)^2)^(1/2)/(d*x)^(3/2),x, algorithm="maxima")

[Out]

-2/3*(3*a/sqrt(d*x) - (d*x)^(3/2)*b/d^2)/d

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.45 \[ \int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{(d x)^{3/2}} \, dx=\frac {2 \, {\left (\frac {\sqrt {d x} b x \mathrm {sgn}\left (b x^{2} + a\right )}{d} - \frac {3 \, a \mathrm {sgn}\left (b x^{2} + a\right )}{\sqrt {d x}}\right )}}{3 \, d} \]

[In]

integrate(((b*x^2+a)^2)^(1/2)/(d*x)^(3/2),x, algorithm="giac")

[Out]

2/3*(sqrt(d*x)*b*x*sgn(b*x^2 + a)/d - 3*a*sgn(b*x^2 + a)/sqrt(d*x))/d

Mupad [B] (verification not implemented)

Time = 13.60 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.57 \[ \int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{(d x)^{3/2}} \, dx=\frac {\left (\frac {2\,x^2}{3\,d}-\frac {2\,a}{b\,d}\right )\,\sqrt {{\left (b\,x^2+a\right )}^2}}{x^2\,\sqrt {d\,x}+\frac {a\,\sqrt {d\,x}}{b}} \]

[In]

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

[Out]

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