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

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

[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))

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Rubi [A]  time = 0.112166, antiderivative size = 107, normalized size of antiderivative = 1., 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} \[ \frac{b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^2}}{\sqrt{b c-a d}}\right )}{a (b c-a d)^{3/2}}-\frac{d}{c \sqrt{c+d x^2} (b c-a d)}-\frac{\tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{a c^{3/2}} \]

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 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]]

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 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 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]

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*}

Mathematica [C]  time = 0.0293199, size = 87, normalized size = 0.81 \[ \frac{(b c-a d) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{d x^2}{c}+1\right )-b c \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{b \left (d x^2+c\right )}{b c-a d}\right )}{a c \sqrt{c+d x^2} (b c-a d)} \]

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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Maple [B]  time = 0.012, size = 681, normalized size = 6.4 \begin{align*}{\frac{1}{ac}{\frac{1}{\sqrt{d{x}^{2}+c}}}}-{\frac{1}{a}\ln \left ({\frac{1}{x} \left ( 2\,c+2\,\sqrt{c}\sqrt{d{x}^{2}+c} \right ) } \right ){c}^{-{\frac{3}{2}}}}+{\frac{b}{2\,a \left ( ad-bc \right ) }{\frac{1}{\sqrt{ \left ( x+{\frac{1}{b}\sqrt{-ab}} \right ) ^{2}d-2\,{\frac{d\sqrt{-ab}}{b} \left ( x+{\frac{\sqrt{-ab}}{b}} \right ) }-{\frac{ad-bc}{b}}}}}}+{\frac{dx}{2\,a \left ( ad-bc \right ) c}\sqrt{-ab}{\frac{1}{\sqrt{ \left ( x+{\frac{1}{b}\sqrt{-ab}} \right ) ^{2}d-2\,{\frac{d\sqrt{-ab}}{b} \left ( x+{\frac{\sqrt{-ab}}{b}} \right ) }-{\frac{ad-bc}{b}}}}}}-{\frac{b}{2\,a \left ( ad-bc \right ) }\ln \left ({ \left ( -2\,{\frac{ad-bc}{b}}-2\,{\frac{d\sqrt{-ab}}{b} \left ( x+{\frac{\sqrt{-ab}}{b}} \right ) }+2\,\sqrt{-{\frac{ad-bc}{b}}}\sqrt{ \left ( x+{\frac{\sqrt{-ab}}{b}} \right ) ^{2}d-2\,{\frac{d\sqrt{-ab}}{b} \left ( x+{\frac{\sqrt{-ab}}{b}} \right ) }-{\frac{ad-bc}{b}}} \right ) \left ( x+{\frac{1}{b}\sqrt{-ab}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{-{\frac{ad-bc}{b}}}}}}+{\frac{b}{2\,a \left ( ad-bc \right ) }{\frac{1}{\sqrt{ \left ( x-{\frac{1}{b}\sqrt{-ab}} \right ) ^{2}d+2\,{\frac{d\sqrt{-ab}}{b} \left ( x-{\frac{\sqrt{-ab}}{b}} \right ) }-{\frac{ad-bc}{b}}}}}}-{\frac{dx}{2\,a \left ( ad-bc \right ) c}\sqrt{-ab}{\frac{1}{\sqrt{ \left ( x-{\frac{1}{b}\sqrt{-ab}} \right ) ^{2}d+2\,{\frac{d\sqrt{-ab}}{b} \left ( x-{\frac{\sqrt{-ab}}{b}} \right ) }-{\frac{ad-bc}{b}}}}}}-{\frac{b}{2\,a \left ( ad-bc \right ) }\ln \left ({ \left ( -2\,{\frac{ad-bc}{b}}+2\,{\frac{d\sqrt{-ab}}{b} \left ( x-{\frac{\sqrt{-ab}}{b}} \right ) }+2\,\sqrt{-{\frac{ad-bc}{b}}}\sqrt{ \left ( x-{\frac{\sqrt{-ab}}{b}} \right ) ^{2}d+2\,{\frac{d\sqrt{-ab}}{b} \left ( x-{\frac{\sqrt{-ab}}{b}} \right ) }-{\frac{ad-bc}{b}}} \right ) \left ( x-{\frac{1}{b}\sqrt{-ab}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{-{\frac{ad-bc}{b}}}}}} \end{align*}

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

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

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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Fricas [B]  time = 3.32774, size = 2030, normalized size = 18.97 \begin{align*} \text{result too large to display} \end{align*}

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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Sympy [A]  time = 10.612, size = 94, normalized size = 0.88 \begin{align*} \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{align*}

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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Giac [A]  time = 1.12866, size = 158, normalized size = 1.48 \begin{align*} -{\left (\frac{b^{2} \arctan \left (\frac{\sqrt{d x^{2} + c} b}{\sqrt{-b^{2} c + a b d}}\right )}{{\left (a b c d - a^{2} d^{2}\right )} \sqrt{-b^{2} c + a b d}} + \frac{1}{{\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 d}\right )} d \end{align*}

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