3.683 \(\int \frac {\sqrt {c+d x^2}}{x^3 (a+b x^2)} \, dx\)

Optimal. Leaf size=113 \[ \frac {(2 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{2 a^2 \sqrt {c}}-\frac {\sqrt {b} \sqrt {b c-a d} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{a^2}-\frac {\sqrt {c+d x^2}}{2 a x^2} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {446, 99, 156, 63, 208} \[ \frac {(2 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{2 a^2 \sqrt {c}}-\frac {\sqrt {b} \sqrt {b c-a d} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{a^2}-\frac {\sqrt {c+d x^2}}{2 a x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 99

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

Rule 156

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]

Rule 208

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

Rule 446

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

Rubi steps

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

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Mathematica [A]  time = 0.17, size = 107, normalized size = 0.95 \[ \frac {\frac {(2 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{\sqrt {c}}-2 \sqrt {b} \sqrt {b c-a d} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )-\frac {a \sqrt {c+d x^2}}{x^2}}{2 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.06, size = 708, normalized size = 6.27 \[ \left [\frac {\sqrt {b^{2} c - a b d} c x^{2} \log \left (\frac {b^{2} d^{2} x^{4} + 8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2} + 2 \, {\left (4 \, b^{2} c d - 3 \, a b d^{2}\right )} x^{2} - 4 \, {\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {b^{2} c - a b d} \sqrt {d x^{2} + c}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) - {\left (2 \, b c - a d\right )} \sqrt {c} x^{2} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) - 2 \, \sqrt {d x^{2} + c} a c}{4 \, a^{2} c x^{2}}, -\frac {2 \, {\left (2 \, b c - a d\right )} \sqrt {-c} x^{2} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) - \sqrt {b^{2} c - a b d} c x^{2} \log \left (\frac {b^{2} d^{2} x^{4} + 8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2} + 2 \, {\left (4 \, b^{2} c d - 3 \, a b d^{2}\right )} x^{2} - 4 \, {\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {b^{2} c - a b d} \sqrt {d x^{2} + c}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) + 2 \, \sqrt {d x^{2} + c} a c}{4 \, a^{2} c x^{2}}, -\frac {2 \, \sqrt {-b^{2} c + a b d} c x^{2} \arctan \left (-\frac {{\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {-b^{2} c + a b d} \sqrt {d x^{2} + c}}{2 \, {\left (b^{2} c^{2} - a b c d + {\left (b^{2} c d - a b d^{2}\right )} x^{2}\right )}}\right ) + {\left (2 \, b c - a d\right )} \sqrt {c} x^{2} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) + 2 \, \sqrt {d x^{2} + c} a c}{4 \, a^{2} c x^{2}}, -\frac {\sqrt {-b^{2} c + a b d} c x^{2} \arctan \left (-\frac {{\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {-b^{2} c + a b d} \sqrt {d x^{2} + c}}{2 \, {\left (b^{2} c^{2} - a b c d + {\left (b^{2} c d - a b d^{2}\right )} x^{2}\right )}}\right ) + {\left (2 \, b c - a d\right )} \sqrt {-c} x^{2} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) + \sqrt {d x^{2} + c} a c}{2 \, a^{2} c x^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(sqrt(b^2*c - a*b*d)*c*x^2*log((b^2*d^2*x^4 + 8*b^2*c^2 - 8*a*b*c*d + a^2*d^2 + 2*(4*b^2*c*d - 3*a*b*d^2)
*x^2 - 4*(b*d*x^2 + 2*b*c - a*d)*sqrt(b^2*c - a*b*d)*sqrt(d*x^2 + c))/(b^2*x^4 + 2*a*b*x^2 + a^2)) - (2*b*c -
a*d)*sqrt(c)*x^2*log(-(d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(c) + 2*c)/x^2) - 2*sqrt(d*x^2 + c)*a*c)/(a^2*c*x^2), -1/
4*(2*(2*b*c - a*d)*sqrt(-c)*x^2*arctan(sqrt(-c)/sqrt(d*x^2 + c)) - sqrt(b^2*c - a*b*d)*c*x^2*log((b^2*d^2*x^4
+ 8*b^2*c^2 - 8*a*b*c*d + a^2*d^2 + 2*(4*b^2*c*d - 3*a*b*d^2)*x^2 - 4*(b*d*x^2 + 2*b*c - a*d)*sqrt(b^2*c - a*b
*d)*sqrt(d*x^2 + c))/(b^2*x^4 + 2*a*b*x^2 + a^2)) + 2*sqrt(d*x^2 + c)*a*c)/(a^2*c*x^2), -1/4*(2*sqrt(-b^2*c +
a*b*d)*c*x^2*arctan(-1/2*(b*d*x^2 + 2*b*c - a*d)*sqrt(-b^2*c + a*b*d)*sqrt(d*x^2 + c)/(b^2*c^2 - a*b*c*d + (b^
2*c*d - a*b*d^2)*x^2)) + (2*b*c - a*d)*sqrt(c)*x^2*log(-(d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(c) + 2*c)/x^2) + 2*sqr
t(d*x^2 + c)*a*c)/(a^2*c*x^2), -1/2*(sqrt(-b^2*c + a*b*d)*c*x^2*arctan(-1/2*(b*d*x^2 + 2*b*c - a*d)*sqrt(-b^2*
c + a*b*d)*sqrt(d*x^2 + c)/(b^2*c^2 - a*b*c*d + (b^2*c*d - a*b*d^2)*x^2)) + (2*b*c - a*d)*sqrt(-c)*x^2*arctan(
sqrt(-c)/sqrt(d*x^2 + c)) + sqrt(d*x^2 + c)*a*c)/(a^2*c*x^2)]

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giac [A]  time = 0.47, size = 106, normalized size = 0.94 \[ \frac {{\left (b^{2} c - a b d\right )} \arctan \left (\frac {\sqrt {d x^{2} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d} a^{2}} - \frac {{\left (2 \, b c - a d\right )} \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{2 \, a^{2} \sqrt {-c}} - \frac {\sqrt {d x^{2} + c}}{2 \, a x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.01, size = 1054, normalized size = 9.33 \[ \frac {d \ln \left (\frac {\frac {2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {2 \left (a d -b c \right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} d +\frac {2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}}{x -\frac {\sqrt {-a b}}{b}}\right )}{2 \sqrt {-\frac {a d -b c}{b}}\, a}+\frac {d \ln \left (\frac {-\frac {2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {2 \left (a d -b c \right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} d -\frac {2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}}{x +\frac {\sqrt {-a b}}{b}}\right )}{2 \sqrt {-\frac {a d -b c}{b}}\, a}-\frac {b c \ln \left (\frac {\frac {2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {2 \left (a d -b c \right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} d +\frac {2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}}{x -\frac {\sqrt {-a b}}{b}}\right )}{2 \sqrt {-\frac {a d -b c}{b}}\, a^{2}}-\frac {b c \ln \left (\frac {-\frac {2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {2 \left (a d -b c \right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} d -\frac {2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}}{x +\frac {\sqrt {-a b}}{b}}\right )}{2 \sqrt {-\frac {a d -b c}{b}}\, a^{2}}-\frac {d \ln \left (\frac {2 c +2 \sqrt {d \,x^{2}+c}\, \sqrt {c}}{x}\right )}{2 a \sqrt {c}}+\frac {b \sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {d \,x^{2}+c}\, \sqrt {c}}{x}\right )}{a^{2}}-\frac {\sqrt {-a b}\, \sqrt {d}\, \ln \left (\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) d -\frac {\sqrt {-a b}\, d}{b}}{\sqrt {d}}+\sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} d -\frac {2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}\right )}{2 a^{2}}+\frac {\sqrt {-a b}\, \sqrt {d}\, \ln \left (\frac {\left (x -\frac {\sqrt {-a b}}{b}\right ) d +\frac {\sqrt {-a b}\, d}{b}}{\sqrt {d}}+\sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} d +\frac {2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}\right )}{2 a^{2}}+\frac {\sqrt {d \,x^{2}+c}\, d}{2 a c}+\frac {\sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} d -\frac {2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}\, b}{2 a^{2}}+\frac {\sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} d +\frac {2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}\, b}{2 a^{2}}-\frac {\sqrt {d \,x^{2}+c}\, b}{a^{2}}-\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}}}{2 a c \,x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {d x^{2} + c}}{{\left (b x^{2} + a\right )} x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(d*x^2 + c)/((b*x^2 + a)*x^3), x)

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mupad [B]  time = 0.99, size = 268, normalized size = 2.37 \[ \frac {\mathrm {atanh}\left (\frac {b^3\,d^4\,\sqrt {d\,x^2+c}\,\sqrt {b^2\,c-a\,b\,d}}{2\,\left (\frac {a\,b^3\,d^5}{2}-\frac {b^4\,c\,d^4}{2}\right )}\right )\,\sqrt {b^2\,c-a\,b\,d}}{a^2}-\frac {\sqrt {d\,x^2+c}}{2\,a\,x^2}-\frac {\mathrm {atanh}\left (\frac {b^4\,\sqrt {c}\,d^4\,\sqrt {d\,x^2+c}}{2\,\left (\frac {b^4\,c\,d^4}{2}-\frac {3\,a\,b^3\,d^5}{4}+\frac {a^2\,b^2\,d^6}{4\,c}\right )}-\frac {3\,b^3\,d^5\,\sqrt {d\,x^2+c}}{4\,\sqrt {c}\,\left (\frac {a\,b^2\,d^6}{4\,c}-\frac {3\,b^3\,d^5}{4}+\frac {b^4\,c\,d^4}{2\,a}\right )}+\frac {b^2\,d^6\,\sqrt {d\,x^2+c}}{4\,c^{3/2}\,\left (\frac {b^2\,d^6}{4\,c}-\frac {3\,b^3\,d^5}{4\,a}+\frac {b^4\,c\,d^4}{2\,a^2}\right )}\right )\,\left (a\,d-2\,b\,c\right )}{2\,a^2\,\sqrt {c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atanh((b^3*d^4*(c + d*x^2)^(1/2)*(b^2*c - a*b*d)^(1/2))/(2*((a*b^3*d^5)/2 - (b^4*c*d^4)/2)))*(b^2*c - a*b*d)^
(1/2))/a^2 - (c + d*x^2)^(1/2)/(2*a*x^2) - (atanh((b^4*c^(1/2)*d^4*(c + d*x^2)^(1/2))/(2*((b^4*c*d^4)/2 - (3*a
*b^3*d^5)/4 + (a^2*b^2*d^6)/(4*c))) - (3*b^3*d^5*(c + d*x^2)^(1/2))/(4*c^(1/2)*((a*b^2*d^6)/(4*c) - (3*b^3*d^5
)/4 + (b^4*c*d^4)/(2*a))) + (b^2*d^6*(c + d*x^2)^(1/2))/(4*c^(3/2)*((b^2*d^6)/(4*c) - (3*b^3*d^5)/(4*a) + (b^4
*c*d^4)/(2*a^2))))*(a*d - 2*b*c))/(2*a^2*c^(1/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c + d x^{2}}}{x^{3} \left (a + b x^{2}\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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