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

Optimal. Leaf size=82 \[ -\frac {a^2 \sqrt {c+d x^2}}{c x}-\frac {b (b c-4 a d) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 d^{3/2}}+\frac {b^2 x \sqrt {c+d x^2}}{2 d} \]

[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

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Rubi [A]  time = 0.05, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {462, 388, 217, 206} \[ -\frac {a^2 \sqrt {c+d x^2}}{c x}-\frac {b (b c-4 a d) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 d^{3/2}}+\frac {b^2 x \sqrt {c+d x^2}}{2 d} \]

Antiderivative was successfully verified.

[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 206

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

Rule 217

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 388

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 462

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*} \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 {\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)) \operatorname {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*}

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Mathematica [A]  time = 0.08, size = 76, normalized size = 0.93 \[ \sqrt {c+d x^2} \left (\frac {b^2 x}{2 d}-\frac {a^2}{c x}\right )-\frac {b (b c-4 a d) \log \left (\sqrt {d} \sqrt {c+d x^2}+d x\right )}{2 d^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.79, size = 165, normalized size = 2.01 \[ \left [-\frac {{\left (b^{2} c^{2} - 4 \, a b c d\right )} \sqrt {d} x \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) - 2 \, {\left (b^{2} c d x^{2} - 2 \, a^{2} d^{2}\right )} \sqrt {d x^{2} + c}}{4 \, c d^{2} x}, \frac {{\left (b^{2} c^{2} - 4 \, a b c d\right )} \sqrt {-d} x \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) + {\left (b^{2} c d x^{2} - 2 \, a^{2} d^{2}\right )} \sqrt {d x^{2} + c}}{2 \, c d^{2} x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[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)]

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giac [A]  time = 0.46, size = 93, normalized size = 1.13 \[ \frac {\sqrt {d x^{2} + c} b^{2} x}{2 \, d} + \frac {2 \, a^{2} \sqrt {d}}{{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} - c} + \frac {{\left (b^{2} c \sqrt {d} - 4 \, a b d^{\frac {3}{2}}\right )} \log \left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2}\right )}{4 \, d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

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maple [A]  time = 0.01, size = 88, normalized size = 1.07 \[ \frac {2 a b \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )}{\sqrt {d}}-\frac {b^{2} c \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )}{2 d^{\frac {3}{2}}}+\frac {\sqrt {d \,x^{2}+c}\, b^{2} x}{2 d}-\frac {\sqrt {d \,x^{2}+c}\, a^{2}}{c x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.88, size = 73, normalized size = 0.89 \[ \frac {\sqrt {d x^{2} + c} b^{2} x}{2 \, d} - \frac {b^{2} c \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{2 \, d^{\frac {3}{2}}} + \frac {2 \, a b \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{\sqrt {d}} - \frac {\sqrt {d x^{2} + c} a^{2}}{c x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 1.46, size = 125, normalized size = 1.52 \[ \left \{\begin {array}{cl} \frac {-a^2+2\,a\,b\,x^2+\frac {b^2\,x^4}{3}}{\sqrt {c}\,x} & \text {\ if\ \ }d=0\\ \frac {2\,a\,b\,\ln \left (\sqrt {d}\,x+\sqrt {d\,x^2+c}\right )}{\sqrt {d}}+\frac {b^2\,x\,\sqrt {d\,x^2+c}}{2\,d}-\frac {a^2\,\sqrt {d\,x^2+c}}{c\,x}-\frac {b^2\,c\,\ln \left (2\,\sqrt {d}\,x+2\,\sqrt {d\,x^2+c}\right )}{2\,d^{3/2}} & \text {\ if\ \ }d\neq 0 \end {array}\right . \]

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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sympy [A]  time = 8.04, size = 155, normalized size = 1.89 \[ - \frac {a^{2} \sqrt {d} \sqrt {\frac {c}{d x^{2}} + 1}}{c} + 2 a b \left (\begin {cases} \frac {\sqrt {- \frac {c}{d}} \operatorname {asin}{\left (x \sqrt {- \frac {d}{c}} \right )}}{\sqrt {c}} & \text {for}\: c > 0 \wedge d < 0 \\\frac {\sqrt {\frac {c}{d}} \operatorname {asinh}{\left (x \sqrt {\frac {d}{c}} \right )}}{\sqrt {c}} & \text {for}\: c > 0 \wedge d > 0 \\\frac {\sqrt {- \frac {c}{d}} \operatorname {acosh}{\left (x \sqrt {- \frac {d}{c}} \right )}}{\sqrt {- c}} & \text {for}\: d > 0 \wedge c < 0 \end {cases}\right ) + \frac {b^{2} \sqrt {c} x \sqrt {1 + \frac {d x^{2}}{c}}}{2 d} - \frac {b^{2} c \operatorname {asinh}{\left (\frac {\sqrt {d} x}{\sqrt {c}} \right )}}{2 d^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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((sqrt(-c/d)*asin(x*sqrt(-d/c))/sqrt(c), (c > 0) & (d <
0)), (sqrt(c/d)*asinh(x*sqrt(d/c))/sqrt(c), (c > 0) & (d > 0)), (sqrt(-c/d)*acosh(x*sqrt(-d/c))/sqrt(-c), (d >
0) & (c < 0))) + b**2*sqrt(c)*x*sqrt(1 + d*x**2/c)/(2*d) - b**2*c*asinh(sqrt(d)*x/sqrt(c))/(2*d**(3/2))

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