3.7.100 \(\int \frac {1}{x (a+b x^2) (c+d x^2)^{3/2}} \, dx\)

Optimal. Leaf size=107 \[ \frac {b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{a (b c-a d)^{3/2}}-\frac {d}{c \sqrt {c+d x^2} (b c-a d)}-\frac {\tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{a c^{3/2}} \]

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Rubi [A]  time = 0.11, antiderivative size = 107, 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, 85, 156, 63, 208} \begin {gather*} \frac {b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{a (b c-a d)^{3/2}}-\frac {d}{c \sqrt {c+d x^2} (b c-a d)}-\frac {\tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{a c^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x*(a + b*x^2)*(c + d*x^2)^(3/2)),x]

[Out]

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

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 85

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

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

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Mathematica [C]  time = 0.04, size = 87, normalized size = 0.81 \begin {gather*} \frac {(b c-a d) \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {d x^2}{c}+1\right )-b c \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {b \left (d x^2+c\right )}{b c-a d}\right )}{a c \sqrt {c+d x^2} (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(a + b*x^2)*(c + d*x^2)^(3/2)),x]

[Out]

(-(b*c*Hypergeometric2F1[-1/2, 1, 1/2, (b*(c + d*x^2))/(b*c - a*d)]) + (b*c - a*d)*Hypergeometric2F1[-1/2, 1,
1/2, 1 + (d*x^2)/c])/(a*c*(b*c - a*d)*Sqrt[c + d*x^2])

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IntegrateAlgebraic [A]  time = 0.20, size = 118, normalized size = 1.10 \begin {gather*} -\frac {b^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2} \sqrt {a d-b c}}{b c-a d}\right )}{a (a d-b c)^{3/2}}-\frac {d}{c \sqrt {c+d x^2} (b c-a d)}-\frac {\tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{a c^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(x*(a + b*x^2)*(c + d*x^2)^(3/2)),x]

[Out]

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

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fricas [B]  time = 2.02, size = 959, normalized size = 8.96 \begin {gather*} \left [-\frac {4 \, \sqrt {d x^{2} + c} a c d + {\left (b c^{2} d x^{2} + b c^{3}\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 (b c^{2} - a c d + {\left (b c d - a d^{2}\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 (a b c^{4} - a^{2} c^{3} d + {\left (a b c^{3} d - a^{2} c^{2} d^{2}\right )} x^{2}\right )}}, -\frac {4 \, \sqrt {d x^{2} + c} a c d - 4 \, {\left (b c^{2} - a c d + {\left (b c d - a d^{2}\right )} x^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) + {\left (b c^{2} d x^{2} + b c^{3}\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 (a b c^{4} - a^{2} c^{3} d + {\left (a b c^{3} d - a^{2} c^{2} d^{2}\right )} x^{2}\right )}}, -\frac {2 \, \sqrt {d x^{2} + c} a c d + {\left (b c^{2} d x^{2} + b c^{3}\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 (b c^{2} - a c d + {\left (b c d - a d^{2}\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 (a b c^{4} - a^{2} c^{3} d + {\left (a b c^{3} d - a^{2} c^{2} d^{2}\right )} x^{2}\right )}}, -\frac {2 \, \sqrt {d x^{2} + c} a c d + {\left (b c^{2} d x^{2} + b c^{3}\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 (b c^{2} - a c d + {\left (b c d - a d^{2}\right )} x^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right )}{2 \, {\left (a b c^{4} - a^{2} c^{3} d + {\left (a b c^{3} d - a^{2} c^{2} d^{2}\right )} x^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.36, size = 110, normalized size = 1.03 \begin {gather*} -\frac {b^{2} \arctan \left (\frac {\sqrt {d x^{2} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{{\left (a b c - a^{2} d\right )} \sqrt {-b^{2} c + a b d}} - \frac {d}{{\left (b c^{2} - a c d\right )} \sqrt {d x^{2} + c}} + \frac {\arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{a \sqrt {-c} c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.02, size = 681, normalized size = 6.36 \begin {gather*} -\frac {b \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 \left (a d -b c \right ) \sqrt {-\frac {a d -b c}{b}}\, a}-\frac {b \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 \left (a d -b c \right ) \sqrt {-\frac {a d -b c}{b}}\, a}+\frac {b}{2 \left (a d -b c \right ) \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}}\, a}+\frac {b}{2 \left (a d -b c \right ) \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}}\, a}+\frac {\sqrt {-a b}\, d x}{2 \left (a d -b c \right ) \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}}\, a c}-\frac {\sqrt {-a b}\, d x}{2 \left (a d -b c \right ) \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}}\, a c}-\frac {\ln \left (\frac {2 c +2 \sqrt {d \,x^{2}+c}\, \sqrt {c}}{x}\right )}{a \,c^{\frac {3}{2}}}+\frac {1}{\sqrt {d \,x^{2}+c}\, a c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (b x^{2} + a\right )} {\left (d x^{2} + c\right )}^{\frac {3}{2}} x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.42, size = 2296, normalized size = 21.46

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atan((((-b^3*(a*d - b*c)^3)^(1/2)*(((c + d*x^2)^(1/2)*(4*b^8*c^8*d^2 - 16*a*b^7*c^7*d^3 + 26*a^2*b^6*c^6*d^4
- 22*a^3*b^5*c^5*d^5 + 10*a^4*b^4*c^4*d^6 - 2*a^5*b^3*c^3*d^7))/2 - ((-b^3*(a*d - b*c)^3)^(1/2)*(18*a^3*b^6*c^
8*d^4 - 4*a^2*b^7*c^9*d^3 - 32*a^4*b^5*c^7*d^5 + 28*a^5*b^4*c^6*d^6 - 12*a^6*b^3*c^5*d^7 + 2*a^7*b^2*c^4*d^8 +
((-b^3*(a*d - b*c)^3)^(1/2)*(c + d*x^2)^(1/2)*(16*a^2*b^8*c^11*d^2 - 88*a^3*b^7*c^10*d^3 + 200*a^4*b^6*c^9*d^
4 - 240*a^5*b^5*c^8*d^5 + 160*a^6*b^4*c^7*d^6 - 56*a^7*b^3*c^6*d^7 + 8*a^8*b^2*c^5*d^8))/(4*a*(a*d - b*c)^3)))
/(2*a*(a*d - b*c)^3))*1i)/(a*(a*d - b*c)^3) + ((-b^3*(a*d - b*c)^3)^(1/2)*(((c + d*x^2)^(1/2)*(4*b^8*c^8*d^2 -
16*a*b^7*c^7*d^3 + 26*a^2*b^6*c^6*d^4 - 22*a^3*b^5*c^5*d^5 + 10*a^4*b^4*c^4*d^6 - 2*a^5*b^3*c^3*d^7))/2 - ((-
b^3*(a*d - b*c)^3)^(1/2)*(4*a^2*b^7*c^9*d^3 - 18*a^3*b^6*c^8*d^4 + 32*a^4*b^5*c^7*d^5 - 28*a^5*b^4*c^6*d^6 + 1
2*a^6*b^3*c^5*d^7 - 2*a^7*b^2*c^4*d^8 + ((-b^3*(a*d - b*c)^3)^(1/2)*(c + d*x^2)^(1/2)*(16*a^2*b^8*c^11*d^2 - 8
8*a^3*b^7*c^10*d^3 + 200*a^4*b^6*c^9*d^4 - 240*a^5*b^5*c^8*d^5 + 160*a^6*b^4*c^7*d^6 - 56*a^7*b^3*c^6*d^7 + 8*
a^8*b^2*c^5*d^8))/(4*a*(a*d - b*c)^3)))/(2*a*(a*d - b*c)^3))*1i)/(a*(a*d - b*c)^3))/(2*b^7*c^6*d^3 - 6*a*b^6*c
^5*d^4 + 6*a^2*b^5*c^4*d^5 - 2*a^3*b^4*c^3*d^6 + ((-b^3*(a*d - b*c)^3)^(1/2)*(((c + d*x^2)^(1/2)*(4*b^8*c^8*d^
2 - 16*a*b^7*c^7*d^3 + 26*a^2*b^6*c^6*d^4 - 22*a^3*b^5*c^5*d^5 + 10*a^4*b^4*c^4*d^6 - 2*a^5*b^3*c^3*d^7))/2 -
((-b^3*(a*d - b*c)^3)^(1/2)*(18*a^3*b^6*c^8*d^4 - 4*a^2*b^7*c^9*d^3 - 32*a^4*b^5*c^7*d^5 + 28*a^5*b^4*c^6*d^6
- 12*a^6*b^3*c^5*d^7 + 2*a^7*b^2*c^4*d^8 + ((-b^3*(a*d - b*c)^3)^(1/2)*(c + d*x^2)^(1/2)*(16*a^2*b^8*c^11*d^2
- 88*a^3*b^7*c^10*d^3 + 200*a^4*b^6*c^9*d^4 - 240*a^5*b^5*c^8*d^5 + 160*a^6*b^4*c^7*d^6 - 56*a^7*b^3*c^6*d^7 +
8*a^8*b^2*c^5*d^8))/(4*a*(a*d - b*c)^3)))/(2*a*(a*d - b*c)^3)))/(a*(a*d - b*c)^3) - ((-b^3*(a*d - b*c)^3)^(1/
2)*(((c + d*x^2)^(1/2)*(4*b^8*c^8*d^2 - 16*a*b^7*c^7*d^3 + 26*a^2*b^6*c^6*d^4 - 22*a^3*b^5*c^5*d^5 + 10*a^4*b^
4*c^4*d^6 - 2*a^5*b^3*c^3*d^7))/2 - ((-b^3*(a*d - b*c)^3)^(1/2)*(4*a^2*b^7*c^9*d^3 - 18*a^3*b^6*c^8*d^4 + 32*a
^4*b^5*c^7*d^5 - 28*a^5*b^4*c^6*d^6 + 12*a^6*b^3*c^5*d^7 - 2*a^7*b^2*c^4*d^8 + ((-b^3*(a*d - b*c)^3)^(1/2)*(c
+ d*x^2)^(1/2)*(16*a^2*b^8*c^11*d^2 - 88*a^3*b^7*c^10*d^3 + 200*a^4*b^6*c^9*d^4 - 240*a^5*b^5*c^8*d^5 + 160*a^
6*b^4*c^7*d^6 - 56*a^7*b^3*c^6*d^7 + 8*a^8*b^2*c^5*d^8))/(4*a*(a*d - b*c)^3)))/(2*a*(a*d - b*c)^3)))/(a*(a*d -
b*c)^3)))*(-b^3*(a*d - b*c)^3)^(1/2)*1i)/(a*(a*d - b*c)^3) - atanh((6*b^7*c^7*d^3*(c + d*x^2)^(1/2))/((c^3)^(
1/2)*(6*b^7*c^6*d^3 - 24*a*b^6*c^5*d^4 - 2*a^5*b^2*c*d^8 + 38*a^2*b^5*c^4*d^5 - 30*a^3*b^4*c^3*d^6 + 12*a^4*b^
3*c^2*d^7)) - (24*a*b^6*c^6*d^4*(c + d*x^2)^(1/2))/((c^3)^(1/2)*(6*b^7*c^6*d^3 - 24*a*b^6*c^5*d^4 - 2*a^5*b^2*
c*d^8 + 38*a^2*b^5*c^4*d^5 - 30*a^3*b^4*c^3*d^6 + 12*a^4*b^3*c^2*d^7)) + (38*a^2*b^5*c^5*d^5*(c + d*x^2)^(1/2)
)/((c^3)^(1/2)*(6*b^7*c^6*d^3 - 24*a*b^6*c^5*d^4 - 2*a^5*b^2*c*d^8 + 38*a^2*b^5*c^4*d^5 - 30*a^3*b^4*c^3*d^6 +
12*a^4*b^3*c^2*d^7)) - (30*a^3*b^4*c^4*d^6*(c + d*x^2)^(1/2))/((c^3)^(1/2)*(6*b^7*c^6*d^3 - 24*a*b^6*c^5*d^4
- 2*a^5*b^2*c*d^8 + 38*a^2*b^5*c^4*d^5 - 30*a^3*b^4*c^3*d^6 + 12*a^4*b^3*c^2*d^7)) + (12*a^4*b^3*c^3*d^7*(c +
d*x^2)^(1/2))/((c^3)^(1/2)*(6*b^7*c^6*d^3 - 24*a*b^6*c^5*d^4 - 2*a^5*b^2*c*d^8 + 38*a^2*b^5*c^4*d^5 - 30*a^3*b
^4*c^3*d^6 + 12*a^4*b^3*c^2*d^7)) - (2*a^5*b^2*c^2*d^8*(c + d*x^2)^(1/2))/((c^3)^(1/2)*(6*b^7*c^6*d^3 - 24*a*b
^6*c^5*d^4 - 2*a^5*b^2*c*d^8 + 38*a^2*b^5*c^4*d^5 - 30*a^3*b^4*c^3*d^6 + 12*a^4*b^3*c^2*d^7)))/(a*(c^3)^(1/2))
- d/((c + d*x^2)^(1/2)*(b*c^2 - a*c*d))

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sympy [A]  time = 21.00, size = 94, normalized size = 0.88 \begin {gather*} \frac {d}{c \sqrt {c + d x^{2}} \left (a d - b c\right )} + \frac {b \operatorname {atan}{\left (\frac {\sqrt {c + d x^{2}}}{\sqrt {\frac {a d - b c}{b}}} \right )}}{a \sqrt {\frac {a d - b c}{b}} \left (a d - b c\right )} + \frac {\operatorname {atan}{\left (\frac {\sqrt {c + d x^{2}}}{\sqrt {- c}} \right )}}{a c \sqrt {- c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d/(c*sqrt(c + d*x**2)*(a*d - b*c)) + b*atan(sqrt(c + d*x**2)/sqrt((a*d - b*c)/b))/(a*sqrt((a*d - b*c)/b)*(a*d
- b*c)) + atan(sqrt(c + d*x**2)/sqrt(-c))/(a*c*sqrt(-c))

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