\(\int \frac {(a+b x^2)^2}{x^2 \sqrt {c+d x^2}} \, dx\) [642]

Optimal result

Integrand size = 24, antiderivative size = 82 \[ \int \frac {\left (a+b x^2\right )^2}{x^2 \sqrt {c+d x^2}} \, dx=-\frac {a^2 \sqrt {c+d x^2}}{c x}+\frac {b^2 x \sqrt {c+d x^2}}{2 d}-\frac {b (b c-4 a d) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 d^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {473, 396, 223, 212} \[ \int \frac {\left (a+b x^2\right )^2}{x^2 \sqrt {c+d x^2}} \, dx=-\frac {a^2 \sqrt {c+d x^2}}{c x}-\frac {b (b c-4 a d) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 d^{3/2}}+\frac {b^2 x \sqrt {c+d x^2}}{2 d} \]

[In]

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

[Out]

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

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 396

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[d*x*((a + b*x^n)^(p + 1)/(b*(n*(
p + 1) + 1))), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

Rule 473

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> Simp[c^2*(e*x)^(m
+ 1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] - Dist[1/(a*e^n*(m + 1)), Int[(e*x)^(m + n)*(a + b*x^n)^p*Simp[b
*c^2*n*(p + 1) + c*(b*c - 2*a*d)*(m + 1) - a*(m + 1)*d^2*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && Ne
Q[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -1] && GtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {a^2 \sqrt {c+d x^2}}{c x}+\frac {\int \frac {2 a b c+b^2 c x^2}{\sqrt {c+d x^2}} \, dx}{c} \\ & = -\frac {a^2 \sqrt {c+d x^2}}{c x}+\frac {b^2 x \sqrt {c+d x^2}}{2 d}-\frac {(b (b c-4 a d)) \int \frac {1}{\sqrt {c+d x^2}} \, dx}{2 d} \\ & = -\frac {a^2 \sqrt {c+d x^2}}{c x}+\frac {b^2 x \sqrt {c+d x^2}}{2 d}-\frac {(b (b c-4 a d)) \text {Subst}\left (\int \frac {1}{1-d x^2} \, dx,x,\frac {x}{\sqrt {c+d x^2}}\right )}{2 d} \\ & = -\frac {a^2 \sqrt {c+d x^2}}{c x}+\frac {b^2 x \sqrt {c+d x^2}}{2 d}-\frac {b (b c-4 a d) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 d^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.96 \[ \int \frac {\left (a+b x^2\right )^2}{x^2 \sqrt {c+d x^2}} \, dx=\frac {\left (-2 a^2 d+b^2 c x^2\right ) \sqrt {c+d x^2}}{2 c d x}+\frac {b (b c-4 a d) \log \left (-\sqrt {d} x+\sqrt {c+d x^2}\right )}{2 d^{3/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.97 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.84

method	result	size
risch	\(-\frac {\sqrt {d \,x^{2}+c}\, \left (-b^{2} c \,x^{2}+2 a^{2} d \right )}{2 d c x}+\frac {\left (4 a d -b c \right ) b \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{2 d^{\frac {3}{2}}}\)	\(69\)
default	\(b^{2} \left (\frac {x \sqrt {d \,x^{2}+c}}{2 d}-\frac {c \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{2 d^{\frac {3}{2}}}\right )-\frac {a^{2} \sqrt {d \,x^{2}+c}}{c x}+\frac {2 a b \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{\sqrt {d}}\)	\(87\)
pseudoelliptic	\(\frac {b^{2} c \,x^{2} \sqrt {d \,x^{2}+c}\, \sqrt {d}+4 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}}{x \sqrt {d}}\right ) a b c d x -\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}}{x \sqrt {d}}\right ) b^{2} c^{2} x -2 \sqrt {d \,x^{2}+c}\, a^{2} d^{\frac {3}{2}}}{2 d^{\frac {3}{2}} x c}\)	\(100\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 165, normalized size of antiderivative = 2.01 \[ \int \frac {\left (a+b x^2\right )^2}{x^2 \sqrt {c+d x^2}} \, dx=\left [-\frac {{\left (b^{2} c^{2} - 4 \, a b c d\right )} \sqrt {d} x \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) - 2 \, {\left (b^{2} c d x^{2} - 2 \, a^{2} d^{2}\right )} \sqrt {d x^{2} + c}}{4 \, c d^{2} x}, \frac {{\left (b^{2} c^{2} - 4 \, a b c d\right )} \sqrt {-d} x \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) + {\left (b^{2} c d x^{2} - 2 \, a^{2} d^{2}\right )} \sqrt {d x^{2} + c}}{2 \, c d^{2} x}\right ] \]

[In]

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

[Out]

[-1/4*((b^2*c^2 - 4*a*b*c*d)*sqrt(d)*x*log(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) - 2*(b^2*c*d*x^2 - 2*a^
2*d^2)*sqrt(d*x^2 + c))/(c*d^2*x), 1/2*((b^2*c^2 - 4*a*b*c*d)*sqrt(-d)*x*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) +
(b^2*c*d*x^2 - 2*a^2*d^2)*sqrt(d*x^2 + c))/(c*d^2*x)]

Sympy [A] (verification not implemented)

Time = 0.83 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.84 \[ \int \frac {\left (a+b x^2\right )^2}{x^2 \sqrt {c+d x^2}} \, dx=- \frac {a^{2} \sqrt {d} \sqrt {\frac {c}{d x^{2}} + 1}}{c} + 2 a b \left (\begin {cases} \frac {\log {\left (2 \sqrt {d} \sqrt {c + d x^{2}} + 2 d x \right )}}{\sqrt {d}} & \text {for}\: c \neq 0 \wedge d \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {d x^{2}}} & \text {for}\: d \neq 0 \\\frac {x}{\sqrt {c}} & \text {otherwise} \end {cases}\right ) + b^{2} \left (\begin {cases} - \frac {c \left (\begin {cases} \frac {\log {\left (2 \sqrt {d} \sqrt {c + d x^{2}} + 2 d x \right )}}{\sqrt {d}} & \text {for}\: c \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {d x^{2}}} & \text {otherwise} \end {cases}\right )}{2 d} + \frac {x \sqrt {c + d x^{2}}}{2 d} & \text {for}\: d \neq 0 \\\frac {x^{3}}{3 \sqrt {c}} & \text {otherwise} \end {cases}\right ) \]

[In]

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

[Out]

-a**2*sqrt(d)*sqrt(c/(d*x**2) + 1)/c + 2*a*b*Piecewise((log(2*sqrt(d)*sqrt(c + d*x**2) + 2*d*x)/sqrt(d), Ne(c,
0) & Ne(d, 0)), (x*log(x)/sqrt(d*x**2), Ne(d, 0)), (x/sqrt(c), True)) + b**2*Piecewise((-c*Piecewise((log(2*s
qrt(d)*sqrt(c + d*x**2) + 2*d*x)/sqrt(d), Ne(c, 0)), (x*log(x)/sqrt(d*x**2), True))/(2*d) + x*sqrt(c + d*x**2)
/(2*d), Ne(d, 0)), (x**3/(3*sqrt(c)), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.89 \[ \int \frac {\left (a+b x^2\right )^2}{x^2 \sqrt {c+d x^2}} \, dx=\frac {\sqrt {d x^{2} + c} b^{2} x}{2 \, d} - \frac {b^{2} c \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{2 \, d^{\frac {3}{2}}} + \frac {2 \, a b \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{\sqrt {d}} - \frac {\sqrt {d x^{2} + c} a^{2}}{c x} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.07 \[ \int \frac {\left (a+b x^2\right )^2}{x^2 \sqrt {c+d x^2}} \, dx=\frac {\sqrt {d x^{2} + c} b^{2} x}{2 \, d} + \frac {2 \, a^{2} \sqrt {d}}{{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} - c} + \frac {{\left (b^{2} c - 4 \, a b d\right )} \log \left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2}\right )}{4 \, d^{\frac {3}{2}}} \]

[In]

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

[Out]

1/2*sqrt(d*x^2 + c)*b^2*x/d + 2*a^2*sqrt(d)/((sqrt(d)*x - sqrt(d*x^2 + c))^2 - c) + 1/4*(b^2*c - 4*a*b*d)*log(
(sqrt(d)*x - sqrt(d*x^2 + c))^2)/d^(3/2)

Mupad [B] (verification not implemented)

Time = 6.21 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.52 \[ \int \frac {\left (a+b x^2\right )^2}{x^2 \sqrt {c+d x^2}} \, dx=\left \{\begin {array}{cl} \frac {-a^2+2\,a\,b\,x^2+\frac {b^2\,x^4}{3}}{\sqrt {c}\,x} & \text {\ if\ \ }d=0\\ \frac {2\,a\,b\,\ln \left (\sqrt {d}\,x+\sqrt {d\,x^2+c}\right )}{\sqrt {d}}+\frac {b^2\,x\,\sqrt {d\,x^2+c}}{2\,d}-\frac {a^2\,\sqrt {d\,x^2+c}}{c\,x}-\frac {b^2\,c\,\ln \left (2\,\sqrt {d}\,x+2\,\sqrt {d\,x^2+c}\right )}{2\,d^{3/2}} & \text {\ if\ \ }d\neq 0 \end {array}\right . \]

[In]

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

[Out]

piecewise(d == 0, (- a^2 + (b^2*x^4)/3 + 2*a*b*x^2)/(c^(1/2)*x), d ~= 0, (2*a*b*log(d^(1/2)*x + (c + d*x^2)^(1
/2)))/d^(1/2) + (b^2*x*(c + d*x^2)^(1/2))/(2*d) - (a^2*(c + d*x^2)^(1/2))/(c*x) - (b^2*c*log(2*d^(1/2)*x + 2*(
c + d*x^2)^(1/2)))/(2*d^(3/2)))