\(\int \frac {1}{(a+b x^2)^2 (c+d x^2)^{5/2}} \, dx\) [780]

Optimal result

Integrand size = 21, antiderivative size = 201 \[ \int \frac {1}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}} \, dx=\frac {d (3 b c+2 a d) x}{6 a c (b c-a d)^2 \left (c+d x^2\right )^{3/2}}+\frac {b x}{2 a (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}}+\frac {d \left (3 b^2 c^2+16 a b c d-4 a^2 d^2\right ) x}{6 a c^2 (b c-a d)^3 \sqrt {c+d x^2}}+\frac {b^2 (b c-6 a d) \arctan \left (\frac {\sqrt {b c-a d} x}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 a^{3/2} (b c-a d)^{7/2}} \]

[Out]

1/6*d*(2*a*d+3*b*c)*x/a/c/(-a*d+b*c)^2/(d*x^2+c)^(3/2)+1/2*b*x/a/(-a*d+b*c)/(b*x^2+a)/(d*x^2+c)^(3/2)+1/2*b^2*
(-6*a*d+b*c)*arctan(x*(-a*d+b*c)^(1/2)/a^(1/2)/(d*x^2+c)^(1/2))/a^(3/2)/(-a*d+b*c)^(7/2)+1/6*d*(-4*a^2*d^2+16*
a*b*c*d+3*b^2*c^2)*x/a/c^2/(-a*d+b*c)^3/(d*x^2+c)^(1/2)

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {425, 541, 12, 385, 211} \[ \int \frac {1}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}} \, dx=\frac {b^2 (b c-6 a d) \arctan \left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 a^{3/2} (b c-a d)^{7/2}}+\frac {d x \left (-4 a^2 d^2+16 a b c d+3 b^2 c^2\right )}{6 a c^2 \sqrt {c+d x^2} (b c-a d)^3}+\frac {b x}{2 a \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2} (b c-a d)}+\frac {d x (2 a d+3 b c)}{6 a c \left (c+d x^2\right )^{3/2} (b c-a d)^2} \]

[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 12

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

Rule 211

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

Rule 385

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 425

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]) && IntBinomi
alQ[a, b, c, d, n, p, q, x]

Rule 541

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]

Rubi steps \begin{align*} \text {integral}& = \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 ) \text {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] (verified)

Time = 0.96 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.06 \[ \int \frac {1}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}} \, dx=\frac {x \left (3 b^3 c^2 \left (c+d x^2\right )^2-2 a^3 d^3 \left (3 c+2 d x^2\right )+2 a b^2 c d^2 x^2 \left (9 c+8 d x^2\right )+2 a^2 b d^2 \left (9 c^2+5 c d x^2-2 d^2 x^4\right )\right )}{6 a c^2 (b c-a d)^3 \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}}-\frac {b^2 (b c-6 a d) \arctan \left (\frac {a \sqrt {d}+b x \left (\sqrt {d} x-\sqrt {c+d x^2}\right )}{\sqrt {a} \sqrt {b c-a d}}\right )}{2 a^{3/2} (b c-a d)^{7/2}} \]

[In]

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

[Out]

(x*(3*b^3*c^2*(c + d*x^2)^2 - 2*a^3*d^3*(3*c + 2*d*x^2) + 2*a*b^2*c*d^2*x^2*(9*c + 8*d*x^2) + 2*a^2*b*d^2*(9*c
^2 + 5*c*d*x^2 - 2*d^2*x^4)))/(6*a*c^2*(b*c - a*d)^3*(a + b*x^2)*(c + d*x^2)^(3/2)) - (b^2*(b*c - 6*a*d)*ArcTa
n[(a*Sqrt[d] + b*x*(Sqrt[d]*x - Sqrt[c + d*x^2]))/(Sqrt[a]*Sqrt[b*c - a*d])])/(2*a^(3/2)*(b*c - a*d)^(7/2))

Maple [A] (verified)

Time = 3.10 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.79

method	result	size
pseudoelliptic	\(\frac {-\frac {b^{2} c^{2} \left (\frac {b \sqrt {d \,x^{2}+c}\, x}{b \,x^{2}+a}-\frac {\left (6 a d -b c \right ) \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}\, a}{x \sqrt {\left (a d -b c \right ) a}}\right )}{\sqrt {\left (a d -b c \right ) a}}\right )}{2 a \left (a d -b c \right )^{3}}+\frac {\left (a d -3 b c \right ) d^{2} x}{\left (a d -b c \right )^{3} \sqrt {d \,x^{2}+c}}-\frac {d^{3} x^{3}}{3 \left (a d -b c \right )^{2} \left (d \,x^{2}+c \right )^{\frac {3}{2}}}}{c^{2}}\)	\(158\)
default	\(\text {Expression too large to display}\)	\(3489\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 697 vs. \(2 (177) = 354\).

Time = 0.90 (sec) , antiderivative size = 1434, normalized size of antiderivative = 7.13 \[ \int \frac {1}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}} \, dx=\text {Too large to display} \]

[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)]

Sympy [F]

\[ \int \frac {1}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}} \, dx=\int \frac {1}{\left (a + b x^{2}\right )^{2} \left (c + d x^{2}\right )^{\frac {5}{2}}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {1}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{2} {\left (d x^{2} + c\right )}^{\frac {5}{2}}} \,d x } \]

[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)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 619 vs. \(2 (177) = 354\).

Time = 0.90 (sec) , antiderivative size = 619, normalized size of antiderivative = 3.08 \[ \int \frac {1}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}} \, dx=\frac {{\left (\frac {2 \, {\left (4 \, b^{4} c^{4} d^{4} - 13 \, a b^{3} c^{3} d^{5} + 15 \, a^{2} b^{2} c^{2} d^{6} - 7 \, a^{3} b c d^{7} + a^{4} d^{8}\right )} x^{2}}{b^{6} c^{8} d - 6 \, a b^{5} c^{7} d^{2} + 15 \, a^{2} b^{4} c^{6} d^{3} - 20 \, a^{3} b^{3} c^{5} d^{4} + 15 \, a^{4} b^{2} c^{4} d^{5} - 6 \, a^{5} b c^{3} d^{6} + a^{6} c^{2} d^{7}} + \frac {3 \, {\left (3 \, b^{4} c^{5} d^{3} - 10 \, a b^{3} c^{4} d^{4} + 12 \, a^{2} b^{2} c^{3} d^{5} - 6 \, a^{3} b c^{2} d^{6} + a^{4} c d^{7}\right )}}{b^{6} c^{8} d - 6 \, a b^{5} c^{7} d^{2} + 15 \, a^{2} b^{4} c^{6} d^{3} - 20 \, a^{3} b^{3} c^{5} d^{4} + 15 \, a^{4} b^{2} c^{4} d^{5} - 6 \, a^{5} b c^{3} d^{6} + a^{6} c^{2} d^{7}}\right )} x}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}}} + \frac {{\left (b^{3} c \sqrt {d} - 6 \, a b^{2} d^{\frac {3}{2}}\right )} \arctan \left (-\frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b - b c + 2 \, a d}{2 \, \sqrt {a b c d - a^{2} d^{2}}}\right )}{2 \, {\left (a b^{3} c^{3} - 3 \, a^{2} b^{2} c^{2} d + 3 \, a^{3} b c d^{2} - a^{4} d^{3}\right )} \sqrt {a b c d - a^{2} d^{2}}} - \frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b^{3} c \sqrt {d} - 2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a b^{2} d^{\frac {3}{2}} - b^{3} c^{2} \sqrt {d}}{{\left (a b^{3} c^{3} - 3 \, a^{2} b^{2} c^{2} d + 3 \, a^{3} b c d^{2} - a^{4} d^{3}\right )} {\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} b - 2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b c + 4 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a d + b c^{2}\right )}} \]

[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))

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}} \, dx=\int \frac {1}{{\left (b\,x^2+a\right )}^2\,{\left (d\,x^2+c\right )}^{5/2}} \,d x \]

[In]

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

[Out]

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