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

Optimal. Leaf size=159 \[ -\frac {\sqrt {b} (4 b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{2 a^3 \sqrt {b c-a d}}+\frac {(4 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{2 a^3 \sqrt {c}}-\frac {b \sqrt {c+d x^2}}{a^2 \left (a+b x^2\right )}-\frac {\sqrt {c+d x^2}}{2 a x^2 \left (a+b x^2\right )} \]

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Rubi [A]  time = 0.21, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {446, 99, 151, 156, 63, 208} \begin {gather*} -\frac {b \sqrt {c+d x^2}}{a^2 \left (a+b x^2\right )}-\frac {\sqrt {b} (4 b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{2 a^3 \sqrt {b c-a d}}+\frac {(4 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{2 a^3 \sqrt {c}}-\frac {\sqrt {c+d x^2}}{2 a x^2 \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 151

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*
f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegerQ[m]

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

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Mathematica [A]  time = 0.23, size = 190, normalized size = 1.19 \begin {gather*} \frac {\sqrt {c} \left (a \left (a+2 b x^2\right ) \sqrt {c+d x^2} (b c-a d)+\sqrt {b} x^2 \left (a+b x^2\right ) (4 b c-3 a d) \sqrt {b c-a d} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )\right )-x^2 \left (a+b x^2\right ) \left (a^2 d^2-5 a b c d+4 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{2 a^3 \sqrt {c} x^2 \left (a+b x^2\right ) (a d-b c)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-((4*b^2*c^2 - 5*a*b*c*d + a^2*d^2)*x^2*(a + b*x^2)*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]]) + Sqrt[c]*(a*(b*c - a*d
)*(a + 2*b*x^2)*Sqrt[c + d*x^2] + Sqrt[b]*(4*b*c - 3*a*d)*Sqrt[b*c - a*d]*x^2*(a + b*x^2)*ArcTanh[(Sqrt[b]*Sqr
t[c + d*x^2])/Sqrt[b*c - a*d]]))/(2*a^3*Sqrt[c]*(-(b*c) + a*d)*x^2*(a + b*x^2))

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IntegrateAlgebraic [A]  time = 0.56, size = 157, normalized size = 0.99 \begin {gather*} \frac {\left (3 a \sqrt {b} d-4 b^{3/2} c\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2} \sqrt {a d-b c}}{b c-a d}\right )}{2 a^3 \sqrt {a d-b c}}+\frac {(4 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{2 a^3 \sqrt {c}}+\frac {\left (-a-2 b x^2\right ) \sqrt {c+d x^2}}{2 a^2 x^2 \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.64, size = 1043, normalized size = 6.56 \begin {gather*} \left [-\frac {{\left ({\left (4 \, b^{2} c^{2} - 3 \, a b c d\right )} x^{4} + {\left (4 \, a b c^{2} - 3 \, a^{2} c d\right )} x^{2}\right )} \sqrt {\frac {b}{b c - a d}} \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 (2 \, b^{2} c^{2} - 3 \, a b c d + a^{2} d^{2} + {\left (b^{2} c d - a b d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c} \sqrt {\frac {b}{b c - a d}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) + 2 \, {\left ({\left (4 \, b^{2} c - a b d\right )} x^{4} + {\left (4 \, a b c - a^{2} d\right )} x^{2}\right )} \sqrt {c} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) + 4 \, {\left (2 \, a b c x^{2} + a^{2} c\right )} \sqrt {d x^{2} + c}}{8 \, {\left (a^{3} b c x^{4} + a^{4} c x^{2}\right )}}, -\frac {4 \, {\left ({\left (4 \, b^{2} c - a b d\right )} x^{4} + {\left (4 \, a b c - a^{2} d\right )} x^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) + {\left ({\left (4 \, b^{2} c^{2} - 3 \, a b c d\right )} x^{4} + {\left (4 \, a b c^{2} - 3 \, a^{2} c d\right )} x^{2}\right )} \sqrt {\frac {b}{b c - a d}} \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 (2 \, b^{2} c^{2} - 3 \, a b c d + a^{2} d^{2} + {\left (b^{2} c d - a b d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c} \sqrt {\frac {b}{b c - a d}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) + 4 \, {\left (2 \, a b c x^{2} + a^{2} c\right )} \sqrt {d x^{2} + c}}{8 \, {\left (a^{3} b c x^{4} + a^{4} c x^{2}\right )}}, \frac {{\left ({\left (4 \, b^{2} c^{2} - 3 \, a b c d\right )} x^{4} + {\left (4 \, a b c^{2} - 3 \, a^{2} c d\right )} x^{2}\right )} \sqrt {-\frac {b}{b c - a d}} \arctan \left (\frac {{\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {b}{b c - a d}}}{2 \, {\left (b d x^{2} + b c\right )}}\right ) - {\left ({\left (4 \, b^{2} c - a b d\right )} x^{4} + {\left (4 \, a b c - a^{2} d\right )} x^{2}\right )} \sqrt {c} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) - 2 \, {\left (2 \, a b c x^{2} + a^{2} c\right )} \sqrt {d x^{2} + c}}{4 \, {\left (a^{3} b c x^{4} + a^{4} c x^{2}\right )}}, \frac {{\left ({\left (4 \, b^{2} c^{2} - 3 \, a b c d\right )} x^{4} + {\left (4 \, a b c^{2} - 3 \, a^{2} c d\right )} x^{2}\right )} \sqrt {-\frac {b}{b c - a d}} \arctan \left (\frac {{\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {b}{b c - a d}}}{2 \, {\left (b d x^{2} + b c\right )}}\right ) - 2 \, {\left ({\left (4 \, b^{2} c - a b d\right )} x^{4} + {\left (4 \, a b c - a^{2} d\right )} x^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) - 2 \, {\left (2 \, a b c x^{2} + a^{2} c\right )} \sqrt {d x^{2} + c}}{4 \, {\left (a^{3} b c x^{4} + a^{4} c x^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/8*(((4*b^2*c^2 - 3*a*b*c*d)*x^4 + (4*a*b*c^2 - 3*a^2*c*d)*x^2)*sqrt(b/(b*c - a*d))*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*(2*b^2*c^2 - 3*a*b*c*d + a^2*d^2 + (b^2*c*d -
a*b*d^2)*x^2)*sqrt(d*x^2 + c)*sqrt(b/(b*c - a*d)))/(b^2*x^4 + 2*a*b*x^2 + a^2)) + 2*((4*b^2*c - a*b*d)*x^4 + (
4*a*b*c - a^2*d)*x^2)*sqrt(c)*log(-(d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(c) + 2*c)/x^2) + 4*(2*a*b*c*x^2 + a^2*c)*sq
rt(d*x^2 + c))/(a^3*b*c*x^4 + a^4*c*x^2), -1/8*(4*((4*b^2*c - a*b*d)*x^4 + (4*a*b*c - a^2*d)*x^2)*sqrt(-c)*arc
tan(sqrt(-c)/sqrt(d*x^2 + c)) + ((4*b^2*c^2 - 3*a*b*c*d)*x^4 + (4*a*b*c^2 - 3*a^2*c*d)*x^2)*sqrt(b/(b*c - a*d)
)*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*(2*b^2*c^2 - 3*a*b*c*
d + a^2*d^2 + (b^2*c*d - a*b*d^2)*x^2)*sqrt(d*x^2 + c)*sqrt(b/(b*c - a*d)))/(b^2*x^4 + 2*a*b*x^2 + a^2)) + 4*(
2*a*b*c*x^2 + a^2*c)*sqrt(d*x^2 + c))/(a^3*b*c*x^4 + a^4*c*x^2), 1/4*(((4*b^2*c^2 - 3*a*b*c*d)*x^4 + (4*a*b*c^
2 - 3*a^2*c*d)*x^2)*sqrt(-b/(b*c - a*d))*arctan(1/2*(b*d*x^2 + 2*b*c - a*d)*sqrt(d*x^2 + c)*sqrt(-b/(b*c - a*d
))/(b*d*x^2 + b*c)) - ((4*b^2*c - a*b*d)*x^4 + (4*a*b*c - a^2*d)*x^2)*sqrt(c)*log(-(d*x^2 - 2*sqrt(d*x^2 + c)*
sqrt(c) + 2*c)/x^2) - 2*(2*a*b*c*x^2 + a^2*c)*sqrt(d*x^2 + c))/(a^3*b*c*x^4 + a^4*c*x^2), 1/4*(((4*b^2*c^2 - 3
*a*b*c*d)*x^4 + (4*a*b*c^2 - 3*a^2*c*d)*x^2)*sqrt(-b/(b*c - a*d))*arctan(1/2*(b*d*x^2 + 2*b*c - a*d)*sqrt(d*x^
2 + c)*sqrt(-b/(b*c - a*d))/(b*d*x^2 + b*c)) - 2*((4*b^2*c - a*b*d)*x^4 + (4*a*b*c - a^2*d)*x^2)*sqrt(-c)*arct
an(sqrt(-c)/sqrt(d*x^2 + c)) - 2*(2*a*b*c*x^2 + a^2*c)*sqrt(d*x^2 + c))/(a^3*b*c*x^4 + a^4*c*x^2)]

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giac [A]  time = 0.36, size = 183, normalized size = 1.15 \begin {gather*} \frac {{\left (4 \, b^{2} c - 3 \, a b d\right )} \arctan \left (\frac {\sqrt {d x^{2} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{2 \, \sqrt {-b^{2} c + a b d} a^{3}} - \frac {{\left (4 \, b c - a d\right )} \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{2 \, a^{3} \sqrt {-c}} - \frac {2 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b d - 2 \, \sqrt {d x^{2} + c} b c d + \sqrt {d x^{2} + c} a d^{2}}{2 \, {\left ({\left (d x^{2} + c\right )}^{2} b - 2 \, {\left (d x^{2} + c\right )} b c + b c^{2} + {\left (d x^{2} + c\right )} a d - a c d\right )} a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.02, size = 2669, normalized size = 16.79

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.69, size = 1193, normalized size = 7.50 \begin {gather*} \frac {\mathrm {atanh}\left (\frac {b^2\,d^6\,\sqrt {d\,x^2+c}}{4\,c^{3/2}\,\left (\frac {b^3\,d^5}{a}-\frac {b^2\,d^6}{4\,c}\right )}-\frac {b^3\,d^5\,\sqrt {d\,x^2+c}}{\sqrt {c}\,\left (b^3\,d^5-\frac {a\,b^2\,d^6}{4\,c}\right )}\right )\,\left (a\,d-4\,b\,c\right )}{2\,a^3\,\sqrt {c}}-\frac {\frac {b\,d\,{\left (d\,x^2+c\right )}^{3/2}}{a^2}+\frac {d\,\sqrt {d\,x^2+c}\,\left (a\,d-2\,b\,c\right )}{2\,a^2}}{\left (d\,x^2+c\right )\,\left (a\,d-2\,b\,c\right )+b\,{\left (d\,x^2+c\right )}^2+b\,c^2-a\,c\,d}+\frac {\mathrm {atan}\left (\frac {\frac {\sqrt {-b\,\left (a\,d-b\,c\right )}\,\left (\frac {\sqrt {d\,x^2+c}\,\left (5\,a^2\,b^3\,d^4-16\,a\,b^4\,c\,d^3+16\,b^5\,c^2\,d^2\right )}{a^4}-\frac {\left (\frac {2\,a^7\,b^2\,d^4-4\,a^6\,b^3\,c\,d^3}{a^6}-\frac {\left (8\,a^7\,b^2\,d^3-16\,a^6\,b^3\,c\,d^2\right )\,\sqrt {d\,x^2+c}\,\sqrt {-b\,\left (a\,d-b\,c\right )}\,\left (3\,a\,d-4\,b\,c\right )}{4\,a^4\,\left (a^4\,d-a^3\,b\,c\right )}\right )\,\sqrt {-b\,\left (a\,d-b\,c\right )}\,\left (3\,a\,d-4\,b\,c\right )}{4\,\left (a^4\,d-a^3\,b\,c\right )}\right )\,\left (3\,a\,d-4\,b\,c\right )\,1{}\mathrm {i}}{4\,\left (a^4\,d-a^3\,b\,c\right )}+\frac {\sqrt {-b\,\left (a\,d-b\,c\right )}\,\left (\frac {\sqrt {d\,x^2+c}\,\left (5\,a^2\,b^3\,d^4-16\,a\,b^4\,c\,d^3+16\,b^5\,c^2\,d^2\right )}{a^4}+\frac {\left (\frac {2\,a^7\,b^2\,d^4-4\,a^6\,b^3\,c\,d^3}{a^6}+\frac {\left (8\,a^7\,b^2\,d^3-16\,a^6\,b^3\,c\,d^2\right )\,\sqrt {d\,x^2+c}\,\sqrt {-b\,\left (a\,d-b\,c\right )}\,\left (3\,a\,d-4\,b\,c\right )}{4\,a^4\,\left (a^4\,d-a^3\,b\,c\right )}\right )\,\sqrt {-b\,\left (a\,d-b\,c\right )}\,\left (3\,a\,d-4\,b\,c\right )}{4\,\left (a^4\,d-a^3\,b\,c\right )}\right )\,\left (3\,a\,d-4\,b\,c\right )\,1{}\mathrm {i}}{4\,\left (a^4\,d-a^3\,b\,c\right )}}{\frac {\frac {3\,a^2\,b^3\,d^5}{2}-8\,a\,b^4\,c\,d^4+8\,b^5\,c^2\,d^3}{a^6}-\frac {\sqrt {-b\,\left (a\,d-b\,c\right )}\,\left (\frac {\sqrt {d\,x^2+c}\,\left (5\,a^2\,b^3\,d^4-16\,a\,b^4\,c\,d^3+16\,b^5\,c^2\,d^2\right )}{a^4}-\frac {\left (\frac {2\,a^7\,b^2\,d^4-4\,a^6\,b^3\,c\,d^3}{a^6}-\frac {\left (8\,a^7\,b^2\,d^3-16\,a^6\,b^3\,c\,d^2\right )\,\sqrt {d\,x^2+c}\,\sqrt {-b\,\left (a\,d-b\,c\right )}\,\left (3\,a\,d-4\,b\,c\right )}{4\,a^4\,\left (a^4\,d-a^3\,b\,c\right )}\right )\,\sqrt {-b\,\left (a\,d-b\,c\right )}\,\left (3\,a\,d-4\,b\,c\right )}{4\,\left (a^4\,d-a^3\,b\,c\right )}\right )\,\left (3\,a\,d-4\,b\,c\right )}{4\,\left (a^4\,d-a^3\,b\,c\right )}+\frac {\sqrt {-b\,\left (a\,d-b\,c\right )}\,\left (\frac {\sqrt {d\,x^2+c}\,\left (5\,a^2\,b^3\,d^4-16\,a\,b^4\,c\,d^3+16\,b^5\,c^2\,d^2\right )}{a^4}+\frac {\left (\frac {2\,a^7\,b^2\,d^4-4\,a^6\,b^3\,c\,d^3}{a^6}+\frac {\left (8\,a^7\,b^2\,d^3-16\,a^6\,b^3\,c\,d^2\right )\,\sqrt {d\,x^2+c}\,\sqrt {-b\,\left (a\,d-b\,c\right )}\,\left (3\,a\,d-4\,b\,c\right )}{4\,a^4\,\left (a^4\,d-a^3\,b\,c\right )}\right )\,\sqrt {-b\,\left (a\,d-b\,c\right )}\,\left (3\,a\,d-4\,b\,c\right )}{4\,\left (a^4\,d-a^3\,b\,c\right )}\right )\,\left (3\,a\,d-4\,b\,c\right )}{4\,\left (a^4\,d-a^3\,b\,c\right )}}\right )\,\sqrt {-b\,\left (a\,d-b\,c\right )}\,\left (3\,a\,d-4\,b\,c\right )\,1{}\mathrm {i}}{2\,\left (a^4\,d-a^3\,b\,c\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atan((((-b*(a*d - b*c))^(1/2)*(((c + d*x^2)^(1/2)*(5*a^2*b^3*d^4 + 16*b^5*c^2*d^2 - 16*a*b^4*c*d^3))/a^4 - ((
(2*a^7*b^2*d^4 - 4*a^6*b^3*c*d^3)/a^6 - ((8*a^7*b^2*d^3 - 16*a^6*b^3*c*d^2)*(c + d*x^2)^(1/2)*(-b*(a*d - b*c))
^(1/2)*(3*a*d - 4*b*c))/(4*a^4*(a^4*d - a^3*b*c)))*(-b*(a*d - b*c))^(1/2)*(3*a*d - 4*b*c))/(4*(a^4*d - a^3*b*c
)))*(3*a*d - 4*b*c)*1i)/(4*(a^4*d - a^3*b*c)) + ((-b*(a*d - b*c))^(1/2)*(((c + d*x^2)^(1/2)*(5*a^2*b^3*d^4 + 1
6*b^5*c^2*d^2 - 16*a*b^4*c*d^3))/a^4 + (((2*a^7*b^2*d^4 - 4*a^6*b^3*c*d^3)/a^6 + ((8*a^7*b^2*d^3 - 16*a^6*b^3*
c*d^2)*(c + d*x^2)^(1/2)*(-b*(a*d - b*c))^(1/2)*(3*a*d - 4*b*c))/(4*a^4*(a^4*d - a^3*b*c)))*(-b*(a*d - b*c))^(
1/2)*(3*a*d - 4*b*c))/(4*(a^4*d - a^3*b*c)))*(3*a*d - 4*b*c)*1i)/(4*(a^4*d - a^3*b*c)))/(((3*a^2*b^3*d^5)/2 +
8*b^5*c^2*d^3 - 8*a*b^4*c*d^4)/a^6 - ((-b*(a*d - b*c))^(1/2)*(((c + d*x^2)^(1/2)*(5*a^2*b^3*d^4 + 16*b^5*c^2*d
^2 - 16*a*b^4*c*d^3))/a^4 - (((2*a^7*b^2*d^4 - 4*a^6*b^3*c*d^3)/a^6 - ((8*a^7*b^2*d^3 - 16*a^6*b^3*c*d^2)*(c +
d*x^2)^(1/2)*(-b*(a*d - b*c))^(1/2)*(3*a*d - 4*b*c))/(4*a^4*(a^4*d - a^3*b*c)))*(-b*(a*d - b*c))^(1/2)*(3*a*d
- 4*b*c))/(4*(a^4*d - a^3*b*c)))*(3*a*d - 4*b*c))/(4*(a^4*d - a^3*b*c)) + ((-b*(a*d - b*c))^(1/2)*(((c + d*x^
2)^(1/2)*(5*a^2*b^3*d^4 + 16*b^5*c^2*d^2 - 16*a*b^4*c*d^3))/a^4 + (((2*a^7*b^2*d^4 - 4*a^6*b^3*c*d^3)/a^6 + ((
8*a^7*b^2*d^3 - 16*a^6*b^3*c*d^2)*(c + d*x^2)^(1/2)*(-b*(a*d - b*c))^(1/2)*(3*a*d - 4*b*c))/(4*a^4*(a^4*d - a^
3*b*c)))*(-b*(a*d - b*c))^(1/2)*(3*a*d - 4*b*c))/(4*(a^4*d - a^3*b*c)))*(3*a*d - 4*b*c))/(4*(a^4*d - a^3*b*c))
))*(-b*(a*d - b*c))^(1/2)*(3*a*d - 4*b*c)*1i)/(2*(a^4*d - a^3*b*c)) - ((b*d*(c + d*x^2)^(3/2))/a^2 + (d*(c + d
*x^2)^(1/2)*(a*d - 2*b*c))/(2*a^2))/((c + d*x^2)*(a*d - 2*b*c) + b*(c + d*x^2)^2 + b*c^2 - a*c*d) + (atanh((b^
2*d^6*(c + d*x^2)^(1/2))/(4*c^(3/2)*((b^3*d^5)/a - (b^2*d^6)/(4*c))) - (b^3*d^5*(c + d*x^2)^(1/2))/(c^(1/2)*(b
^3*d^5 - (a*b^2*d^6)/(4*c))))*(a*d - 4*b*c))/(2*a^3*c^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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