3.727 \(\int \frac{1}{x (a+b x^2) (c+d x^2)^{5/2}} \, dx\)

Optimal. Leaf size=145 \[ \frac{b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^2}}{\sqrt{b c-a d}}\right )}{a (b c-a d)^{5/2}}-\frac{d (2 b c-a d)}{c^2 \sqrt{c+d x^2} (b c-a d)^2}-\frac{d}{3 c \left (c+d x^2\right )^{3/2} (b c-a d)}-\frac{\tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{a c^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.184983, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {446, 85, 152, 156, 63, 208} \[ \frac{b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^2}}{\sqrt{b c-a d}}\right )}{a (b c-a d)^{5/2}}-\frac{d (2 b c-a d)}{c^2 \sqrt{c+d x^2} (b c-a d)^2}-\frac{d}{3 c \left (c+d x^2\right )^{3/2} (b c-a d)}-\frac{\tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{a c^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(a + b*x^2)*(c + d*x^2)^(5/2)),x]

[Out]

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

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 152

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] && IntegersQ[2*m, 2*n, 2*p]

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 )^{5/2}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{x (a+b x) (c+d x)^{5/2}} \, dx,x,x^2\right )\\ &=-\frac{d}{3 c (b c-a d) \left (c+d x^2\right )^{3/2}}+\frac{\operatorname{Subst}\left (\int \frac{b c-a d-b d x}{x (a+b x) (c+d x)^{3/2}} \, dx,x,x^2\right )}{2 c (b c-a d)}\\ &=-\frac{d}{3 c (b c-a d) \left (c+d x^2\right )^{3/2}}-\frac{d (2 b c-a d)}{c^2 (b c-a d)^2 \sqrt{c+d x^2}}-\frac{\operatorname{Subst}\left (\int \frac{-\frac{1}{2} (b c-a d)^2+\frac{1}{2} b d (2 b c-a d) x}{x (a+b x) \sqrt{c+d x}} \, dx,x,x^2\right )}{c^2 (b c-a d)^2}\\ &=-\frac{d}{3 c (b c-a d) \left (c+d x^2\right )^{3/2}}-\frac{d (2 b c-a d)}{c^2 (b c-a d)^2 \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^2}-\frac{b^3 \operatorname{Subst}\left (\int \frac{1}{(a+b x) \sqrt{c+d x}} \, dx,x,x^2\right )}{2 a (b c-a d)^2}\\ &=-\frac{d}{3 c (b c-a d) \left (c+d x^2\right )^{3/2}}-\frac{d (2 b c-a d)}{c^2 (b c-a d)^2 \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^2 d}-\frac{b^3 \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)^2}\\ &=-\frac{d}{3 c (b c-a d) \left (c+d x^2\right )^{3/2}}-\frac{d (2 b c-a d)}{c^2 (b c-a d)^2 \sqrt{c+d x^2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{a c^{5/2}}+\frac{b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^2}}{\sqrt{b c-a d}}\right )}{a (b c-a d)^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.0353496, size = 90, normalized size = 0.62 \[ \frac{(b c-a d) \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\frac{d x^2}{c}+1\right )-b c \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\frac{b \left (d x^2+c\right )}{b c-a d}\right )}{3 a c \left (c+d x^2\right )^{3/2} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(a + b*x^2)*(c + d*x^2)^(5/2)),x]

[Out]

(-(b*c*Hypergeometric2F1[-3/2, 1, -1/2, (b*(c + d*x^2))/(b*c - a*d)]) + (b*c - a*d)*Hypergeometric2F1[-3/2, 1,
-1/2, 1 + (d*x^2)/c])/(3*a*c*(b*c - a*d)*(c + d*x^2)^(3/2))

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Maple [B]  time = 0.011, size = 1186, normalized size = 8.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/(b*x^2+a)/(d*x^2+c)^(5/2),x)

[Out]

1/3/a/c/(d*x^2+c)^(3/2)+1/a/c^2/(d*x^2+c)^(1/2)-1/a/c^(5/2)*ln((2*c+2*c^(1/2)*(d*x^2+c)^(1/2))/x)+1/6/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)^(3/2)+1/6/a*d*(-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)^(3/2)*x+1/3/a*d*(
-a*b)^(1/2)/(a*d-b*c)/c^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-1/2/a*b^2/(a*d-b*c)^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)-
1/2/a*b/(a*d-b*c)^2*(-a*b)^(1/2)/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*b^2/(a*d-b*c)^2/(-(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)))+1/6/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)^(3/2)-1/6/a*d*(-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)^(3/2)*x-1/3/a*d*(-a*b)^(1/2)/(a*d-b*c)/c^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-1/2/a*b^2/(a*d-b*c)^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)+1/2/a*b/(a*d-b*c)^2*(-a*b)^(1/2)/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*b^2/(a*d-b*c)^2/(-(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{5}{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)^(5/2),x, algorithm="maxima")

[Out]

integrate(1/((b*x^2 + a)*(d*x^2 + c)^(5/2)*x), x)

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Fricas [B]  time = 9.38594, size = 3514, normalized size = 24.23 \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)^(5/2),x, algorithm="fricas")

[Out]

[1/12*(3*(b^2*c^3*d^2*x^4 + 2*b^2*c^4*d*x^2 + b^2*c^5)*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)) + 6*(b^2*c^4 - 2*a*b*c^3*d + a^2*c^2*d^2 +
(b^2*c^2*d^2 - 2*a*b*c*d^3 + a^2*d^4)*x^4 + 2*(b^2*c^3*d - 2*a*b*c^2*d^2 + a^2*c*d^3)*x^2)*sqrt(c)*log(-(d*x^
2 - 2*sqrt(d*x^2 + c)*sqrt(c) + 2*c)/x^2) - 4*(7*a*b*c^3*d - 4*a^2*c^2*d^2 + 3*(2*a*b*c^2*d^2 - a^2*c*d^3)*x^2
)*sqrt(d*x^2 + c))/(a*b^2*c^7 - 2*a^2*b*c^6*d + a^3*c^5*d^2 + (a*b^2*c^5*d^2 - 2*a^2*b*c^4*d^3 + a^3*c^3*d^4)*
x^4 + 2*(a*b^2*c^6*d - 2*a^2*b*c^5*d^2 + a^3*c^4*d^3)*x^2), 1/12*(12*(b^2*c^4 - 2*a*b*c^3*d + a^2*c^2*d^2 + (b
^2*c^2*d^2 - 2*a*b*c*d^3 + a^2*d^4)*x^4 + 2*(b^2*c^3*d - 2*a*b*c^2*d^2 + a^2*c*d^3)*x^2)*sqrt(-c)*arctan(sqrt(
-c)/sqrt(d*x^2 + c)) + 3*(b^2*c^3*d^2*x^4 + 2*b^2*c^4*d*x^2 + b^2*c^5)*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*(7*a*b*c^3*d - 4*a^2*c
^2*d^2 + 3*(2*a*b*c^2*d^2 - a^2*c*d^3)*x^2)*sqrt(d*x^2 + c))/(a*b^2*c^7 - 2*a^2*b*c^6*d + a^3*c^5*d^2 + (a*b^2
*c^5*d^2 - 2*a^2*b*c^4*d^3 + a^3*c^3*d^4)*x^4 + 2*(a*b^2*c^6*d - 2*a^2*b*c^5*d^2 + a^3*c^4*d^3)*x^2), -1/6*(3*
(b^2*c^3*d^2*x^4 + 2*b^2*c^4*d*x^2 + b^2*c^5)*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)) - 3*(b^2*c^4 - 2*a*b*c^3*d + a^2*c^2*d^2 + (b^2*c^2*d^2 - 2*a*b*
c*d^3 + a^2*d^4)*x^4 + 2*(b^2*c^3*d - 2*a*b*c^2*d^2 + a^2*c*d^3)*x^2)*sqrt(c)*log(-(d*x^2 - 2*sqrt(d*x^2 + c)*
sqrt(c) + 2*c)/x^2) + 2*(7*a*b*c^3*d - 4*a^2*c^2*d^2 + 3*(2*a*b*c^2*d^2 - a^2*c*d^3)*x^2)*sqrt(d*x^2 + c))/(a*
b^2*c^7 - 2*a^2*b*c^6*d + a^3*c^5*d^2 + (a*b^2*c^5*d^2 - 2*a^2*b*c^4*d^3 + a^3*c^3*d^4)*x^4 + 2*(a*b^2*c^6*d -
2*a^2*b*c^5*d^2 + a^3*c^4*d^3)*x^2), -1/6*(3*(b^2*c^3*d^2*x^4 + 2*b^2*c^4*d*x^2 + b^2*c^5)*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)) - 6*(b^2*c^4 - 2*a
*b*c^3*d + a^2*c^2*d^2 + (b^2*c^2*d^2 - 2*a*b*c*d^3 + a^2*d^4)*x^4 + 2*(b^2*c^3*d - 2*a*b*c^2*d^2 + a^2*c*d^3)
*x^2)*sqrt(-c)*arctan(sqrt(-c)/sqrt(d*x^2 + c)) + 2*(7*a*b*c^3*d - 4*a^2*c^2*d^2 + 3*(2*a*b*c^2*d^2 - a^2*c*d^
3)*x^2)*sqrt(d*x^2 + c))/(a*b^2*c^7 - 2*a^2*b*c^6*d + a^3*c^5*d^2 + (a*b^2*c^5*d^2 - 2*a^2*b*c^4*d^3 + a^3*c^3
*d^4)*x^4 + 2*(a*b^2*c^6*d - 2*a^2*b*c^5*d^2 + a^3*c^4*d^3)*x^2)]

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Sympy [A]  time = 17.0752, size = 133, normalized size = 0.92 \begin{align*} \frac{d}{3 c \left (c + d x^{2}\right )^{\frac{3}{2}} \left (a d - b c\right )} + \frac{d \left (a d - 2 b c\right )}{c^{2} \sqrt{c + d x^{2}} \left (a d - b c\right )^{2}} - \frac{b^{2} \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 )^{2}} + \frac{\operatorname{atan}{\left (\frac{\sqrt{c + d x^{2}}}{\sqrt{- c}} \right )}}{a c^{2} \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)**(5/2),x)

[Out]

d/(3*c*(c + d*x**2)**(3/2)*(a*d - b*c)) + d*(a*d - 2*b*c)/(c**2*sqrt(c + d*x**2)*(a*d - b*c)**2) - b**2*atan(s
qrt(c + d*x**2)/sqrt((a*d - b*c)/b))/(a*sqrt((a*d - b*c)/b)*(a*d - b*c)**2) + atan(sqrt(c + d*x**2)/sqrt(-c))/
(a*c**2*sqrt(-c))

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Giac [A]  time = 1.17202, size = 242, normalized size = 1.67 \begin{align*} -\frac{1}{3} \,{\left (\frac{3 \, b^{3} \arctan \left (\frac{\sqrt{d x^{2} + c} b}{\sqrt{-b^{2} c + a b d}}\right )}{{\left (a b^{2} c^{2} d - 2 \, a^{2} b c d^{2} + a^{3} d^{3}\right )} \sqrt{-b^{2} c + a b d}} + \frac{6 \,{\left (d x^{2} + c\right )} b c + b c^{2} - 3 \,{\left (d x^{2} + c\right )} a d - a c d}{{\left (b^{2} c^{4} - 2 \, a b c^{3} d + a^{2} c^{2} d^{2}\right )}{\left (d x^{2} + c\right )}^{\frac{3}{2}}} - \frac{3 \, \arctan \left (\frac{\sqrt{d x^{2} + c}}{\sqrt{-c}}\right )}{a \sqrt{-c} c^{2} 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)^(5/2),x, algorithm="giac")

[Out]

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