3.729 \(\int \frac{1}{x^3 (a+b x^2) (c+d x^2)^{5/2}} \, dx\)

Optimal. Leaf size=211 \[ -\frac{d \left (5 a^2 d^2-8 a b c d+b^2 c^2\right )}{2 a c^3 \sqrt{c+d x^2} (b c-a d)^2}-\frac{b^{7/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^2}}{\sqrt{b c-a d}}\right )}{a^2 (b c-a d)^{5/2}}+\frac{(5 a d+2 b c) \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{2 a^2 c^{7/2}}-\frac{d (3 b c-5 a d)}{6 a c^2 \left (c+d x^2\right )^{3/2} (b c-a d)}-\frac{1}{2 a c x^2 \left (c+d x^2\right )^{3/2}} \]

[Out]

-(d*(3*b*c - 5*a*d))/(6*a*c^2*(b*c - a*d)*(c + d*x^2)^(3/2)) - 1/(2*a*c*x^2*(c + d*x^2)^(3/2)) - (d*(b^2*c^2 -
8*a*b*c*d + 5*a^2*d^2))/(2*a*c^3*(b*c - a*d)^2*Sqrt[c + d*x^2]) + ((2*b*c + 5*a*d)*ArcTanh[Sqrt[c + d*x^2]/Sq
rt[c]])/(2*a^2*c^(7/2)) - (b^(7/2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^2])/Sqrt[b*c - a*d]])/(a^2*(b*c - a*d)^(5/2))

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Rubi [A]  time = 0.32226, antiderivative size = 211, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {446, 103, 152, 156, 63, 208} \[ -\frac{d \left (5 a^2 d^2-8 a b c d+b^2 c^2\right )}{2 a c^3 \sqrt{c+d x^2} (b c-a d)^2}-\frac{b^{7/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^2}}{\sqrt{b c-a d}}\right )}{a^2 (b c-a d)^{5/2}}+\frac{(5 a d+2 b c) \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{2 a^2 c^{7/2}}-\frac{d (3 b c-5 a d)}{6 a c^2 \left (c+d x^2\right )^{3/2} (b c-a d)}-\frac{1}{2 a c x^2 \left (c+d x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a + b*x^2)*(c + d*x^2)^(5/2)),x]

[Out]

-(d*(3*b*c - 5*a*d))/(6*a*c^2*(b*c - a*d)*(c + d*x^2)^(3/2)) - 1/(2*a*c*x^2*(c + d*x^2)^(3/2)) - (d*(b^2*c^2 -
8*a*b*c*d + 5*a^2*d^2))/(2*a*c^3*(b*c - a*d)^2*Sqrt[c + d*x^2]) + ((2*b*c + 5*a*d)*ArcTanh[Sqrt[c + d*x^2]/Sq
rt[c]])/(2*a^2*c^(7/2)) - (b^(7/2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^2])/Sqrt[b*c - a*d]])/(a^2*(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 103

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(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*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] &&
IntegerQ[m] && (IntegerQ[n] || IntegersQ[2*n, 2*p])

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^3 \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^2 (a+b x) (c+d x)^{5/2}} \, dx,x,x^2\right )\\ &=-\frac{1}{2 a c x^2 \left (c+d x^2\right )^{3/2}}-\frac{\operatorname{Subst}\left (\int \frac{\frac{1}{2} (2 b c+5 a d)+\frac{5 b d x}{2}}{x (a+b x) (c+d x)^{5/2}} \, dx,x,x^2\right )}{2 a c}\\ &=-\frac{d (3 b c-5 a d)}{6 a c^2 (b c-a d) \left (c+d x^2\right )^{3/2}}-\frac{1}{2 a c x^2 \left (c+d x^2\right )^{3/2}}+\frac{\operatorname{Subst}\left (\int \frac{-\frac{3}{4} (b c-a d) (2 b c+5 a d)-\frac{3}{4} b d (3 b c-5 a d) x}{x (a+b x) (c+d x)^{3/2}} \, dx,x,x^2\right )}{3 a c^2 (b c-a d)}\\ &=-\frac{d (3 b c-5 a d)}{6 a c^2 (b c-a d) \left (c+d x^2\right )^{3/2}}-\frac{1}{2 a c x^2 \left (c+d x^2\right )^{3/2}}-\frac{d \left (b^2 c^2-8 a b c d+5 a^2 d^2\right )}{2 a c^3 (b c-a d)^2 \sqrt{c+d x^2}}-\frac{2 \operatorname{Subst}\left (\int \frac{\frac{3}{8} (b c-a d)^2 (2 b c+5 a d)+\frac{3}{8} b d \left (b^2 c^2-8 a b c d+5 a^2 d^2\right ) x}{x (a+b x) \sqrt{c+d x}} \, dx,x,x^2\right )}{3 a c^3 (b c-a d)^2}\\ &=-\frac{d (3 b c-5 a d)}{6 a c^2 (b c-a d) \left (c+d x^2\right )^{3/2}}-\frac{1}{2 a c x^2 \left (c+d x^2\right )^{3/2}}-\frac{d \left (b^2 c^2-8 a b c d+5 a^2 d^2\right )}{2 a c^3 (b c-a d)^2 \sqrt{c+d x^2}}+\frac{b^4 \operatorname{Subst}\left (\int \frac{1}{(a+b x) \sqrt{c+d x}} \, dx,x,x^2\right )}{2 a^2 (b c-a d)^2}-\frac{(2 b c+5 a d) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{c+d x}} \, dx,x,x^2\right )}{4 a^2 c^3}\\ &=-\frac{d (3 b c-5 a d)}{6 a c^2 (b c-a d) \left (c+d x^2\right )^{3/2}}-\frac{1}{2 a c x^2 \left (c+d x^2\right )^{3/2}}-\frac{d \left (b^2 c^2-8 a b c d+5 a^2 d^2\right )}{2 a c^3 (b c-a d)^2 \sqrt{c+d x^2}}+\frac{b^4 \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 (b c-a d)^2}-\frac{(2 b c+5 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 c^3 d}\\ &=-\frac{d (3 b c-5 a d)}{6 a c^2 (b c-a d) \left (c+d x^2\right )^{3/2}}-\frac{1}{2 a c x^2 \left (c+d x^2\right )^{3/2}}-\frac{d \left (b^2 c^2-8 a b c d+5 a^2 d^2\right )}{2 a c^3 (b c-a d)^2 \sqrt{c+d x^2}}+\frac{(2 b c+5 a d) \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{2 a^2 c^{7/2}}-\frac{b^{7/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^2}}{\sqrt{b c-a d}}\right )}{a^2 (b c-a d)^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.0529899, size = 118, normalized size = 0.56 \[ \frac{2 b^2 c^2 x^2 \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\frac{b \left (d x^2+c\right )}{b c-a d}\right )+(a d-b c) \left (x^2 (5 a d+2 b c) \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\frac{d x^2}{c}+1\right )+3 a c\right )}{6 a^2 c^2 x^2 \left (c+d x^2\right )^{3/2} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(a + b*x^2)*(c + d*x^2)^(5/2)),x]

[Out]

(2*b^2*c^2*x^2*Hypergeometric2F1[-3/2, 1, -1/2, (b*(c + d*x^2))/(b*c - a*d)] + (-(b*c) + a*d)*(3*a*c + (2*b*c
+ 5*a*d)*x^2*Hypergeometric2F1[-3/2, 1, -1/2, 1 + (d*x^2)/c]))/(6*a^2*c^2*(b*c - a*d)*x^2*(c + d*x^2)^(3/2))

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Maple [B]  time = 0.013, size = 1289, normalized size = 6.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(b*x^2+a)/(d*x^2+c)^(5/2),x)

[Out]

-1/3*b/a^2/c/(d*x^2+c)^(3/2)-b/a^2/c^2/(d*x^2+c)^(1/2)+b/a^2/c^(5/2)*ln((2*c+2*c^(1/2)*(d*x^2+c)^(1/2))/x)-1/6
*b^2/a^2/(a*d-b*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)-1/6*b/
a^2*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*b/a^2*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*b^3/a^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*b^2/a^2/(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*b^3/a^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*b^2/a^2/(a*d-b*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)+1/6*b/a^2*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*b/a^2*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*b^3/a^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*b^2/a^2/(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*b^3/a^
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)))-1/2/a/c/x^2/(d*x^2+c)^(3/2)-5/6/a*d/c^2/(d*x^2+c)^(3/2)-5/2/a*d/c^3/(d*x^2+c)^(1/2)+5/2/a*d/c^(7/2)*ln((2
*c+2*c^(1/2)*(d*x^2+c)^(1/2))/x)

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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^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(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^3), x)

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Fricas [B]  time = 19.0385, size = 4504, normalized size = 21.35 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b*x^2+a)/(d*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

[1/12*(3*(b^3*c^4*d^2*x^6 + 2*b^3*c^5*d*x^4 + b^3*c^6*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)) + 3*((2*b^3*c^3*d^2 + a*b^2*c^2*d^3 -
8*a^2*b*c*d^4 + 5*a^3*d^5)*x^6 + 2*(2*b^3*c^4*d + a*b^2*c^3*d^2 - 8*a^2*b*c^2*d^3 + 5*a^3*c*d^4)*x^4 + (2*b^3*
c^5 + a*b^2*c^4*d - 8*a^2*b*c^3*d^2 + 5*a^3*c^2*d^3)*x^2)*sqrt(c)*log(-(d*x^2 + 2*sqrt(d*x^2 + c)*sqrt(c) + 2*
c)/x^2) - 2*(3*a*b^2*c^5 - 6*a^2*b*c^4*d + 3*a^3*c^3*d^2 + 3*(a*b^2*c^3*d^2 - 8*a^2*b*c^2*d^3 + 5*a^3*c*d^4)*x
^4 + 2*(3*a*b^2*c^4*d - 16*a^2*b*c^3*d^2 + 10*a^3*c^2*d^3)*x^2)*sqrt(d*x^2 + c))/((a^2*b^2*c^6*d^2 - 2*a^3*b*c
^5*d^3 + a^4*c^4*d^4)*x^6 + 2*(a^2*b^2*c^7*d - 2*a^3*b*c^6*d^2 + a^4*c^5*d^3)*x^4 + (a^2*b^2*c^8 - 2*a^3*b*c^7
*d + a^4*c^6*d^2)*x^2), -1/12*(6*((2*b^3*c^3*d^2 + a*b^2*c^2*d^3 - 8*a^2*b*c*d^4 + 5*a^3*d^5)*x^6 + 2*(2*b^3*c
^4*d + a*b^2*c^3*d^2 - 8*a^2*b*c^2*d^3 + 5*a^3*c*d^4)*x^4 + (2*b^3*c^5 + a*b^2*c^4*d - 8*a^2*b*c^3*d^2 + 5*a^3
*c^2*d^3)*x^2)*sqrt(-c)*arctan(sqrt(-c)/sqrt(d*x^2 + c)) - 3*(b^3*c^4*d^2*x^6 + 2*b^3*c^5*d*x^4 + b^3*c^6*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*(3*a*b^2*c^5 - 6*a^2*b*c^4*d + 3*a^3*c^3*d^2 + 3*(a*b^2*c^3*d^2 - 8*a^2*b*c^2*d^3 + 5*a^3*c*
d^4)*x^4 + 2*(3*a*b^2*c^4*d - 16*a^2*b*c^3*d^2 + 10*a^3*c^2*d^3)*x^2)*sqrt(d*x^2 + c))/((a^2*b^2*c^6*d^2 - 2*a
^3*b*c^5*d^3 + a^4*c^4*d^4)*x^6 + 2*(a^2*b^2*c^7*d - 2*a^3*b*c^6*d^2 + a^4*c^5*d^3)*x^4 + (a^2*b^2*c^8 - 2*a^3
*b*c^7*d + a^4*c^6*d^2)*x^2), 1/12*(6*(b^3*c^4*d^2*x^6 + 2*b^3*c^5*d*x^4 + b^3*c^6*x^2)*sqrt(-b/(b*c - a*d))*a
rctan(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*((2*b^3*c^3*d^2 +
a*b^2*c^2*d^3 - 8*a^2*b*c*d^4 + 5*a^3*d^5)*x^6 + 2*(2*b^3*c^4*d + a*b^2*c^3*d^2 - 8*a^2*b*c^2*d^3 + 5*a^3*c*d^
4)*x^4 + (2*b^3*c^5 + a*b^2*c^4*d - 8*a^2*b*c^3*d^2 + 5*a^3*c^2*d^3)*x^2)*sqrt(c)*log(-(d*x^2 + 2*sqrt(d*x^2 +
c)*sqrt(c) + 2*c)/x^2) - 2*(3*a*b^2*c^5 - 6*a^2*b*c^4*d + 3*a^3*c^3*d^2 + 3*(a*b^2*c^3*d^2 - 8*a^2*b*c^2*d^3
+ 5*a^3*c*d^4)*x^4 + 2*(3*a*b^2*c^4*d - 16*a^2*b*c^3*d^2 + 10*a^3*c^2*d^3)*x^2)*sqrt(d*x^2 + c))/((a^2*b^2*c^6
*d^2 - 2*a^3*b*c^5*d^3 + a^4*c^4*d^4)*x^6 + 2*(a^2*b^2*c^7*d - 2*a^3*b*c^6*d^2 + a^4*c^5*d^3)*x^4 + (a^2*b^2*c
^8 - 2*a^3*b*c^7*d + a^4*c^6*d^2)*x^2), 1/6*(3*(b^3*c^4*d^2*x^6 + 2*b^3*c^5*d*x^4 + b^3*c^6*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)) - 3*((2*b^3*c
^3*d^2 + a*b^2*c^2*d^3 - 8*a^2*b*c*d^4 + 5*a^3*d^5)*x^6 + 2*(2*b^3*c^4*d + a*b^2*c^3*d^2 - 8*a^2*b*c^2*d^3 + 5
*a^3*c*d^4)*x^4 + (2*b^3*c^5 + a*b^2*c^4*d - 8*a^2*b*c^3*d^2 + 5*a^3*c^2*d^3)*x^2)*sqrt(-c)*arctan(sqrt(-c)/sq
rt(d*x^2 + c)) - (3*a*b^2*c^5 - 6*a^2*b*c^4*d + 3*a^3*c^3*d^2 + 3*(a*b^2*c^3*d^2 - 8*a^2*b*c^2*d^3 + 5*a^3*c*d
^4)*x^4 + 2*(3*a*b^2*c^4*d - 16*a^2*b*c^3*d^2 + 10*a^3*c^2*d^3)*x^2)*sqrt(d*x^2 + c))/((a^2*b^2*c^6*d^2 - 2*a^
3*b*c^5*d^3 + a^4*c^4*d^4)*x^6 + 2*(a^2*b^2*c^7*d - 2*a^3*b*c^6*d^2 + a^4*c^5*d^3)*x^4 + (a^2*b^2*c^8 - 2*a^3*
b*c^7*d + a^4*c^6*d^2)*x^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{3} \left (a + b x^{2}\right ) \left (c + d x^{2}\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**3/(b*x**2+a)/(d*x**2+c)**(5/2),x)

[Out]

Integral(1/(x**3*(a + b*x**2)*(c + d*x**2)**(5/2)), x)

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Giac [A]  time = 1.15693, size = 294, normalized size = 1.39 \begin{align*} \frac{1}{6} \,{\left (\frac{6 \, b^{4} \arctan \left (\frac{\sqrt{d x^{2} + c} b}{\sqrt{-b^{2} c + a b d}}\right )}{{\left (a^{2} b^{2} c^{2} d^{2} - 2 \, a^{3} b c d^{3} + a^{4} d^{4}\right )} \sqrt{-b^{2} c + a b d}} + \frac{2 \,{\left (9 \,{\left (d x^{2} + c\right )} b c + b c^{2} - 6 \,{\left (d x^{2} + c\right )} a d - a c d\right )}}{{\left (b^{2} c^{5} - 2 \, a b c^{4} d + a^{2} c^{3} d^{2}\right )}{\left (d x^{2} + c\right )}^{\frac{3}{2}}} - \frac{3 \,{\left (2 \, b c + 5 \, a d\right )} \arctan \left (\frac{\sqrt{d x^{2} + c}}{\sqrt{-c}}\right )}{a^{2} \sqrt{-c} c^{3} d^{2}} - \frac{3 \, \sqrt{d x^{2} + c}}{a c^{3} d^{2} x^{2}}\right )} d^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b*x^2+a)/(d*x^2+c)^(5/2),x, algorithm="giac")

[Out]

1/6*(6*b^4*arctan(sqrt(d*x^2 + c)*b/sqrt(-b^2*c + a*b*d))/((a^2*b^2*c^2*d^2 - 2*a^3*b*c*d^3 + a^4*d^4)*sqrt(-b
^2*c + a*b*d)) + 2*(9*(d*x^2 + c)*b*c + b*c^2 - 6*(d*x^2 + c)*a*d - a*c*d)/((b^2*c^5 - 2*a*b*c^4*d + a^2*c^3*d
^2)*(d*x^2 + c)^(3/2)) - 3*(2*b*c + 5*a*d)*arctan(sqrt(d*x^2 + c)/sqrt(-c))/(a^2*sqrt(-c)*c^3*d^2) - 3*sqrt(d*
x^2 + c)/(a*c^3*d^2*x^2))*d^2