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

Optimal. Leaf size=201 \[ \frac{d x \left (-4 a^2 d^2+16 a b c d+3 b^2 c^2\right )}{6 a c^2 \sqrt{c+d x^2} (b c-a d)^3}+\frac{b^2 (b c-6 a d) \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{2 a^{3/2} (b c-a d)^{7/2}}+\frac{b x}{2 a \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2} (b c-a d)}+\frac{d x (2 a d+3 b c)}{6 a c \left (c+d x^2\right )^{3/2} (b c-a d)^2} \]

[Out]

(d*(3*b*c + 2*a*d)*x)/(6*a*c*(b*c - a*d)^2*(c + d*x^2)^(3/2)) + (b*x)/(2*a*(b*c - a*d)*(a + b*x^2)*(c + d*x^2)
^(3/2)) + (d*(3*b^2*c^2 + 16*a*b*c*d - 4*a^2*d^2)*x)/(6*a*c^2*(b*c - a*d)^3*Sqrt[c + d*x^2]) + (b^2*(b*c - 6*a
*d)*ArcTan[(Sqrt[b*c - a*d]*x)/(Sqrt[a]*Sqrt[c + d*x^2])])/(2*a^(3/2)*(b*c - a*d)^(7/2))

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Rubi [A]  time = 0.221752, antiderivative size = 201, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {414, 527, 12, 377, 205} \[ \frac{d x \left (-4 a^2 d^2+16 a b c d+3 b^2 c^2\right )}{6 a c^2 \sqrt{c+d x^2} (b c-a d)^3}+\frac{b^2 (b c-6 a d) \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{2 a^{3/2} (b c-a d)^{7/2}}+\frac{b x}{2 a \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2} (b c-a d)}+\frac{d x (2 a d+3 b c)}{6 a c \left (c+d x^2\right )^{3/2} (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(3*b*c + 2*a*d)*x)/(6*a*c*(b*c - a*d)^2*(c + d*x^2)^(3/2)) + (b*x)/(2*a*(b*c - a*d)*(a + b*x^2)*(c + d*x^2)
^(3/2)) + (d*(3*b^2*c^2 + 16*a*b*c*d - 4*a^2*d^2)*x)/(6*a*c^2*(b*c - a*d)^3*Sqrt[c + d*x^2]) + (b^2*(b*c - 6*a
*d)*ArcTan[(Sqrt[b*c - a*d]*x)/(Sqrt[a]*Sqrt[c + d*x^2])])/(2*a^(3/2)*(b*c - a*d)^(7/2))

Rule 414

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1)*
(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,
n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomial
Q[a, b, c, d, n, p, q, x]

Rule 527

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{1}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}} \, dx &=\frac{b x}{2 a (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}}-\frac{\int \frac{-b c+2 a d-4 b d x^2}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx}{2 a (b c-a d)}\\ &=\frac{d (3 b c+2 a d) x}{6 a c (b c-a d)^2 \left (c+d x^2\right )^{3/2}}+\frac{b x}{2 a (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}}-\frac{\int \frac{-3 b^2 c^2+12 a b c d-4 a^2 d^2-2 b d (3 b c+2 a d) x^2}{\left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}} \, dx}{6 a c (b c-a d)^2}\\ &=\frac{d (3 b c+2 a d) x}{6 a c (b c-a d)^2 \left (c+d x^2\right )^{3/2}}+\frac{b x}{2 a (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}}+\frac{d \left (3 b^2 c^2+16 a b c d-4 a^2 d^2\right ) x}{6 a c^2 (b c-a d)^3 \sqrt{c+d x^2}}-\frac{\int -\frac{3 b^2 c^2 (b c-6 a d)}{\left (a+b x^2\right ) \sqrt{c+d x^2}} \, dx}{6 a c^2 (b c-a d)^3}\\ &=\frac{d (3 b c+2 a d) x}{6 a c (b c-a d)^2 \left (c+d x^2\right )^{3/2}}+\frac{b x}{2 a (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}}+\frac{d \left (3 b^2 c^2+16 a b c d-4 a^2 d^2\right ) x}{6 a c^2 (b c-a d)^3 \sqrt{c+d x^2}}+\frac{\left (b^2 (b c-6 a d)\right ) \int \frac{1}{\left (a+b x^2\right ) \sqrt{c+d x^2}} \, dx}{2 a (b c-a d)^3}\\ &=\frac{d (3 b c+2 a d) x}{6 a c (b c-a d)^2 \left (c+d x^2\right )^{3/2}}+\frac{b x}{2 a (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}}+\frac{d \left (3 b^2 c^2+16 a b c d-4 a^2 d^2\right ) x}{6 a c^2 (b c-a d)^3 \sqrt{c+d x^2}}+\frac{\left (b^2 (b c-6 a d)\right ) \operatorname{Subst}\left (\int \frac{1}{a-(-b c+a d) x^2} \, dx,x,\frac{x}{\sqrt{c+d x^2}}\right )}{2 a (b c-a d)^3}\\ &=\frac{d (3 b c+2 a d) x}{6 a c (b c-a d)^2 \left (c+d x^2\right )^{3/2}}+\frac{b x}{2 a (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}}+\frac{d \left (3 b^2 c^2+16 a b c d-4 a^2 d^2\right ) x}{6 a c^2 (b c-a d)^3 \sqrt{c+d x^2}}+\frac{b^2 (b c-6 a d) \tan ^{-1}\left (\frac{\sqrt{b c-a d} x}{\sqrt{a} \sqrt{c+d x^2}}\right )}{2 a^{3/2} (b c-a d)^{7/2}}\\ \end{align*}

Mathematica [A]  time = 5.37815, size = 170, normalized size = 0.85 \[ \frac{1}{6} \left (\frac{3 b^2 (b c-6 a d) \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{a^{3/2} (b c-a d)^{7/2}}+x \sqrt{c+d x^2} \left (-\frac{3 b^3}{a \left (a+b x^2\right ) (a d-b c)^3}+\frac{4 d^2 (4 b c-a d)}{c^2 \left (c+d x^2\right ) (b c-a d)^3}+\frac{2 d^2}{c \left (c+d x^2\right )^2 (b c-a d)^2}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sqrt[c + d*x^2]*((-3*b^3)/(a*(-(b*c) + a*d)^3*(a + b*x^2)) + (2*d^2)/(c*(b*c - a*d)^2*(c + d*x^2)^2) + (4*d
^2*(4*b*c - a*d))/(c^2*(b*c - a*d)^3*(c + d*x^2))) + (3*b^2*(b*c - 6*a*d)*ArcTan[(Sqrt[b*c - a*d]*x)/(Sqrt[a]*
Sqrt[c + d*x^2])])/(a^(3/2)*(b*c - a*d)^(7/2)))/6

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Maple [B]  time = 0.013, size = 2405, normalized size = 12. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/(-a*b)^(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)-1/4/a*b/(a*d-b*c)^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-1/4/a*b/(a*d-b*c)^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-1/12/(-a*b)^(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)^(3/2)+1/4/(-a*b)^(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)+1/12/(-a*b)^(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)^(3/2)+5/12*d^2/(a*d-b*c)^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)^(3/2)*x+1/4/(-a*b)^(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)))-5/4*b*d^2/(a*d-b*c)^3/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-1/4/a*d/(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-1/2/a*d/(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-5/4*b/a*d*(-a*b)^(1/2)/(a*d-b*c)^3/((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)-1/4/(-a*b)^(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)))-1/4/a*d/(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-1/2/a*d/(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+5/4*b/a*d*(-a*b)^(
1/2)/(a*d-b*c)^3/((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)-5/4*b*d^
2/(a*d-b*c)^3/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-1/4/a/(
a*d-b*c)/(x-1/b*(-a*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)^(
3/2)-1/4/a/(a*d-b*c)/(x+1/b*(-a*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)^(3/2)+5/6*d^2/(a*d-b*c)^2/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+5/12/a*d*(-a*b)^(1/2)/(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)^(3/2)+5/6*d^2/(a*d-b*c)^2/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+5/12*d^2/(a*d-b*c)^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)^(3/2)*x-5/12/a*d*(-a*b)^(1/2)/(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)^(3/2)+5/4*b/a*d*(-a*b)^(1/2)/(a*d-b*c)^3/(-(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)))-5/4*b/a*d*(-a*b)^(1/2)/(a*d-b*c)^3/(-(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 )}^{2}{\left (d x^{2} + c\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 11.2082, size = 2849, normalized size = 14.17 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/24*(3*(a*b^3*c^5 - 6*a^2*b^2*c^4*d + (b^4*c^3*d^2 - 6*a*b^3*c^2*d^3)*x^6 + (2*b^4*c^4*d - 11*a*b^3*c^3*d^2
- 6*a^2*b^2*c^2*d^3)*x^4 + (b^4*c^5 - 4*a*b^3*c^4*d - 12*a^2*b^2*c^3*d^2)*x^2)*sqrt(-a*b*c + a^2*d)*log(((b^2
*c^2 - 8*a*b*c*d + 8*a^2*d^2)*x^4 + a^2*c^2 - 2*(3*a*b*c^2 - 4*a^2*c*d)*x^2 - 4*((b*c - 2*a*d)*x^3 - a*c*x)*sq
rt(-a*b*c + a^2*d)*sqrt(d*x^2 + c))/(b^2*x^4 + 2*a*b*x^2 + a^2)) - 4*((3*a*b^4*c^3*d^2 + 13*a^2*b^3*c^2*d^3 -
20*a^3*b^2*c*d^4 + 4*a^4*b*d^5)*x^5 + 2*(3*a*b^4*c^4*d + 6*a^2*b^3*c^3*d^2 - 4*a^3*b^2*c^2*d^3 - 7*a^4*b*c*d^4
+ 2*a^5*d^5)*x^3 + 3*(a*b^4*c^5 - a^2*b^3*c^4*d + 6*a^3*b^2*c^3*d^2 - 8*a^4*b*c^2*d^3 + 2*a^5*c*d^4)*x)*sqrt(
d*x^2 + c))/(a^3*b^4*c^8 - 4*a^4*b^3*c^7*d + 6*a^5*b^2*c^6*d^2 - 4*a^6*b*c^5*d^3 + a^7*c^4*d^4 + (a^2*b^5*c^6*
d^2 - 4*a^3*b^4*c^5*d^3 + 6*a^4*b^3*c^4*d^4 - 4*a^5*b^2*c^3*d^5 + a^6*b*c^2*d^6)*x^6 + (2*a^2*b^5*c^7*d - 7*a^
3*b^4*c^6*d^2 + 8*a^4*b^3*c^5*d^3 - 2*a^5*b^2*c^4*d^4 - 2*a^6*b*c^3*d^5 + a^7*c^2*d^6)*x^4 + (a^2*b^5*c^8 - 2*
a^3*b^4*c^7*d - 2*a^4*b^3*c^6*d^2 + 8*a^5*b^2*c^5*d^3 - 7*a^6*b*c^4*d^4 + 2*a^7*c^3*d^5)*x^2), 1/12*(3*(a*b^3*
c^5 - 6*a^2*b^2*c^4*d + (b^4*c^3*d^2 - 6*a*b^3*c^2*d^3)*x^6 + (2*b^4*c^4*d - 11*a*b^3*c^3*d^2 - 6*a^2*b^2*c^2*
d^3)*x^4 + (b^4*c^5 - 4*a*b^3*c^4*d - 12*a^2*b^2*c^3*d^2)*x^2)*sqrt(a*b*c - a^2*d)*arctan(1/2*sqrt(a*b*c - a^2
*d)*((b*c - 2*a*d)*x^2 - a*c)*sqrt(d*x^2 + c)/((a*b*c*d - a^2*d^2)*x^3 + (a*b*c^2 - a^2*c*d)*x)) + 2*((3*a*b^4
*c^3*d^2 + 13*a^2*b^3*c^2*d^3 - 20*a^3*b^2*c*d^4 + 4*a^4*b*d^5)*x^5 + 2*(3*a*b^4*c^4*d + 6*a^2*b^3*c^3*d^2 - 4
*a^3*b^2*c^2*d^3 - 7*a^4*b*c*d^4 + 2*a^5*d^5)*x^3 + 3*(a*b^4*c^5 - a^2*b^3*c^4*d + 6*a^3*b^2*c^3*d^2 - 8*a^4*b
*c^2*d^3 + 2*a^5*c*d^4)*x)*sqrt(d*x^2 + c))/(a^3*b^4*c^8 - 4*a^4*b^3*c^7*d + 6*a^5*b^2*c^6*d^2 - 4*a^6*b*c^5*d
^3 + a^7*c^4*d^4 + (a^2*b^5*c^6*d^2 - 4*a^3*b^4*c^5*d^3 + 6*a^4*b^3*c^4*d^4 - 4*a^5*b^2*c^3*d^5 + a^6*b*c^2*d^
6)*x^6 + (2*a^2*b^5*c^7*d - 7*a^3*b^4*c^6*d^2 + 8*a^4*b^3*c^5*d^3 - 2*a^5*b^2*c^4*d^4 - 2*a^6*b*c^3*d^5 + a^7*
c^2*d^6)*x^4 + (a^2*b^5*c^8 - 2*a^3*b^4*c^7*d - 2*a^4*b^3*c^6*d^2 + 8*a^5*b^2*c^5*d^3 - 7*a^6*b*c^4*d^4 + 2*a^
7*c^3*d^5)*x^2)]

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Giac [B]  time = 6.51488, size = 836, normalized size = 4.16 \begin{align*} \frac{{\left (\frac{2 \,{\left (4 \, b^{4} c^{4} d^{4} - 13 \, a b^{3} c^{3} d^{5} + 15 \, a^{2} b^{2} c^{2} d^{6} - 7 \, a^{3} b c d^{7} + a^{4} d^{8}\right )} x^{2}}{b^{6} c^{8} d - 6 \, a b^{5} c^{7} d^{2} + 15 \, a^{2} b^{4} c^{6} d^{3} - 20 \, a^{3} b^{3} c^{5} d^{4} + 15 \, a^{4} b^{2} c^{4} d^{5} - 6 \, a^{5} b c^{3} d^{6} + a^{6} c^{2} d^{7}} + \frac{3 \,{\left (3 \, b^{4} c^{5} d^{3} - 10 \, a b^{3} c^{4} d^{4} + 12 \, a^{2} b^{2} c^{3} d^{5} - 6 \, a^{3} b c^{2} d^{6} + a^{4} c d^{7}\right )}}{b^{6} c^{8} d - 6 \, a b^{5} c^{7} d^{2} + 15 \, a^{2} b^{4} c^{6} d^{3} - 20 \, a^{3} b^{3} c^{5} d^{4} + 15 \, a^{4} b^{2} c^{4} d^{5} - 6 \, a^{5} b c^{3} d^{6} + a^{6} c^{2} d^{7}}\right )} x}{3 \,{\left (d x^{2} + c\right )}^{\frac{3}{2}}} + \frac{{\left (b^{3} c \sqrt{d} - 6 \, a b^{2} d^{\frac{3}{2}}\right )} \arctan \left (-\frac{{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} b - b c + 2 \, a d}{2 \, \sqrt{a b c d - a^{2} d^{2}}}\right )}{2 \,{\left (a b^{3} c^{3} - 3 \, a^{2} b^{2} c^{2} d + 3 \, a^{3} b c d^{2} - a^{4} d^{3}\right )} \sqrt{a b c d - a^{2} d^{2}}} - \frac{{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} b^{3} c \sqrt{d} - 2 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} a b^{2} d^{\frac{3}{2}} - b^{3} c^{2} \sqrt{d}}{{\left (a b^{3} c^{3} - 3 \, a^{2} b^{2} c^{2} d + 3 \, a^{3} b c d^{2} - a^{4} d^{3}\right )}{\left ({\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{4} b - 2 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} b c + 4 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} a d + b c^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(2*(4*b^4*c^4*d^4 - 13*a*b^3*c^3*d^5 + 15*a^2*b^2*c^2*d^6 - 7*a^3*b*c*d^7 + a^4*d^8)*x^2/(b^6*c^8*d - 6*a*
b^5*c^7*d^2 + 15*a^2*b^4*c^6*d^3 - 20*a^3*b^3*c^5*d^4 + 15*a^4*b^2*c^4*d^5 - 6*a^5*b*c^3*d^6 + a^6*c^2*d^7) +
3*(3*b^4*c^5*d^3 - 10*a*b^3*c^4*d^4 + 12*a^2*b^2*c^3*d^5 - 6*a^3*b*c^2*d^6 + a^4*c*d^7)/(b^6*c^8*d - 6*a*b^5*c
^7*d^2 + 15*a^2*b^4*c^6*d^3 - 20*a^3*b^3*c^5*d^4 + 15*a^4*b^2*c^4*d^5 - 6*a^5*b*c^3*d^6 + a^6*c^2*d^7))*x/(d*x
^2 + c)^(3/2) + 1/2*(b^3*c*sqrt(d) - 6*a*b^2*d^(3/2))*arctan(-1/2*((sqrt(d)*x - sqrt(d*x^2 + c))^2*b - b*c + 2
*a*d)/sqrt(a*b*c*d - a^2*d^2))/((a*b^3*c^3 - 3*a^2*b^2*c^2*d + 3*a^3*b*c*d^2 - a^4*d^3)*sqrt(a*b*c*d - a^2*d^2
)) - ((sqrt(d)*x - sqrt(d*x^2 + c))^2*b^3*c*sqrt(d) - 2*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a*b^2*d^(3/2) - b^3*c^
2*sqrt(d))/((a*b^3*c^3 - 3*a^2*b^2*c^2*d + 3*a^3*b*c*d^2 - a^4*d^3)*((sqrt(d)*x - sqrt(d*x^2 + c))^4*b - 2*(sq
rt(d)*x - sqrt(d*x^2 + c))^2*b*c + 4*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a*d + b*c^2))