\(\int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{(d x)^{7/2}} \, dx\) [734]

Optimal result

Integrand size = 30, antiderivative size = 91 \[ \int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{(d x)^{7/2}} \, dx=-\frac {2 a \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 d (d x)^{5/2} \left (a+b x^2\right )}-\frac {2 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{d^3 \sqrt {d x} \left (a+b x^2\right )} \]

[Out]

-2/5*a*((b*x^2+a)^2)^(1/2)/d/(d*x)^(5/2)/(b*x^2+a)-2*b*((b*x^2+a)^2)^(1/2)/d^3/(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} \[ \int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{(d x)^{7/2}} \, dx=-\frac {2 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{d^3 \sqrt {d x} \left (a+b x^2\right )}-\frac {2 a \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 d (d x)^{5/2} \left (a+b x^2\right )} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.46 \[ \int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{(d x)^{7/2}} \, dx=-\frac {2 x \sqrt {\left (a+b x^2\right )^2} \left (a+5 b x^2\right )}{5 (d x)^{7/2} \left (a+b x^2\right )} \]

[In]

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

[Out]

(-2*x*Sqrt[(a + b*x^2)^2]*(a + 5*b*x^2))/(5*(d*x)^(7/2)*(a + b*x^2))

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.41

method	result	size
gosper	\(-\frac {2 x \left (5 b \,x^{2}+a \right ) \sqrt {\left (b \,x^{2}+a \right )^{2}}}{5 \left (b \,x^{2}+a \right ) \left (d x \right )^{\frac {7}{2}}}\)	\(37\)
default	\(-\frac {2 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (5 b \,x^{2}+a \right )}{5 d \left (b \,x^{2}+a \right ) \left (d x \right )^{\frac {5}{2}}}\)	\(39\)
risch	\(-\frac {2 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (5 b \,x^{2}+a \right )}{5 d^{3} \left (b \,x^{2}+a \right ) x^{2} \sqrt {d x}}\)	\(42\)

[In]

int(((b*x^2+a)^2)^(1/2)/(d*x)^(7/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.23 \[ \int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{(d x)^{7/2}} \, dx=-\frac {2 \, {\left (5 \, b x^{2} + a\right )} \sqrt {d x}}{5 \, d^{4} x^{3}} \]

[In]

integrate(((b*x^2+a)^2)^(1/2)/(d*x)^(7/2),x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{(d x)^{7/2}} \, dx=\text {Timed out} \]

[In]

integrate(((b*x**2+a)**2)**(1/2)/(d*x)**(7/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.27 \[ \int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{(d x)^{7/2}} \, dx=-\frac {2 \, {\left (5 \, b d^{2} x^{2} + a d^{2}\right )}}{5 \, \left (d x\right )^{\frac {5}{2}} d^{3}} \]

[In]

integrate(((b*x^2+a)^2)^(1/2)/(d*x)^(7/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.48 \[ \int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{(d x)^{7/2}} \, dx=-\frac {2 \, {\left (5 \, b d^{3} x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + a d^{3} \mathrm {sgn}\left (b x^{2} + a\right )\right )}}{5 \, \sqrt {d x} d^{6} x^{2}} \]

[In]

integrate(((b*x^2+a)^2)^(1/2)/(d*x)^(7/2),x, algorithm="giac")

[Out]

-2/5*(5*b*d^3*x^2*sgn(b*x^2 + a) + a*d^3*sgn(b*x^2 + a))/(sqrt(d*x)*d^6*x^2)

Mupad [B] (verification not implemented)

Time = 13.42 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.62 \[ \int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{(d x)^{7/2}} \, dx=-\frac {\left (\frac {2\,x^2}{d^3}+\frac {2\,a}{5\,b\,d^3}\right )\,\sqrt {{\left (b\,x^2+a\right )}^2}}{x^4\,\sqrt {d\,x}+\frac {a\,x^2\,\sqrt {d\,x}}{b}} \]

[In]

int(((a + b*x^2)^2)^(1/2)/(d*x)^(7/2),x)

[Out]

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