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

Optimal. Leaf size=230 \[ \frac{d^{3/2} \left (3 a^2 d^2-14 a b c d+35 b^2 c^2\right ) \tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )}{8 c^{5/2} (b c-a d)^4}+\frac{b^{5/2} (b c-7 a d) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 a^{3/2} (b c-a d)^4}+\frac{d x (4 b c-a d) (3 a d+b c)}{8 a c^2 \left (c+d x^2\right ) (b c-a d)^3}+\frac{b x}{2 a \left (a+b x^2\right ) \left (c+d x^2\right )^2 (b c-a d)}+\frac{d x (a d+2 b c)}{4 a c \left (c+d x^2\right )^2 (b c-a d)^2} \]

[Out]

(d*(2*b*c + a*d)*x)/(4*a*c*(b*c - a*d)^2*(c + d*x^2)^2) + (b*x)/(2*a*(b*c - a*d)*(a + b*x^2)*(c + d*x^2)^2) +
(d*(4*b*c - a*d)*(b*c + 3*a*d)*x)/(8*a*c^2*(b*c - a*d)^3*(c + d*x^2)) + (b^(5/2)*(b*c - 7*a*d)*ArcTan[(Sqrt[b]
*x)/Sqrt[a]])/(2*a^(3/2)*(b*c - a*d)^4) + (d^(3/2)*(35*b^2*c^2 - 14*a*b*c*d + 3*a^2*d^2)*ArcTan[(Sqrt[d]*x)/Sq
rt[c]])/(8*c^(5/2)*(b*c - a*d)^4)

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Rubi [A]  time = 0.301425, antiderivative size = 230, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {414, 527, 522, 205} \[ \frac{d^{3/2} \left (3 a^2 d^2-14 a b c d+35 b^2 c^2\right ) \tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )}{8 c^{5/2} (b c-a d)^4}+\frac{b^{5/2} (b c-7 a d) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 a^{3/2} (b c-a d)^4}+\frac{d x (4 b c-a d) (3 a d+b c)}{8 a c^2 \left (c+d x^2\right ) (b c-a d)^3}+\frac{b x}{2 a \left (a+b x^2\right ) \left (c+d x^2\right )^2 (b c-a d)}+\frac{d x (a d+2 b c)}{4 a c \left (c+d x^2\right )^2 (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(2*b*c + a*d)*x)/(4*a*c*(b*c - a*d)^2*(c + d*x^2)^2) + (b*x)/(2*a*(b*c - a*d)*(a + b*x^2)*(c + d*x^2)^2) +
(d*(4*b*c - a*d)*(b*c + 3*a*d)*x)/(8*a*c^2*(b*c - a*d)^3*(c + d*x^2)) + (b^(5/2)*(b*c - 7*a*d)*ArcTan[(Sqrt[b]
*x)/Sqrt[a]])/(2*a^(3/2)*(b*c - a*d)^4) + (d^(3/2)*(35*b^2*c^2 - 14*a*b*c*d + 3*a^2*d^2)*ArcTan[(Sqrt[d]*x)/Sq
rt[c]])/(8*c^(5/2)*(b*c - a*d)^4)

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 522

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

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 )^3} \, dx &=\frac{b x}{2 a (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )^2}-\frac{\int \frac{-b c+2 a d-5 b d x^2}{\left (a+b x^2\right ) \left (c+d x^2\right )^3} \, dx}{2 a (b c-a d)}\\ &=\frac{d (2 b c+a d) x}{4 a c (b c-a d)^2 \left (c+d x^2\right )^2}+\frac{b x}{2 a (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )^2}-\frac{\int \frac{-2 \left (2 b^2 c^2-8 a b c d+3 a^2 d^2\right )-6 b d (2 b c+a d) x^2}{\left (a+b x^2\right ) \left (c+d x^2\right )^2} \, dx}{8 a c (b c-a d)^2}\\ &=\frac{d (2 b c+a d) x}{4 a c (b c-a d)^2 \left (c+d x^2\right )^2}+\frac{b x}{2 a (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )^2}+\frac{d (4 b c-a d) (b c+3 a d) x}{8 a c^2 (b c-a d)^3 \left (c+d x^2\right )}-\frac{\int \frac{-2 \left (4 b^3 c^3-24 a b^2 c^2 d+11 a^2 b c d^2-3 a^3 d^3\right )-2 b d (4 b c-a d) (b c+3 a d) x^2}{\left (a+b x^2\right ) \left (c+d x^2\right )} \, dx}{16 a c^2 (b c-a d)^3}\\ &=\frac{d (2 b c+a d) x}{4 a c (b c-a d)^2 \left (c+d x^2\right )^2}+\frac{b x}{2 a (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )^2}+\frac{d (4 b c-a d) (b c+3 a d) x}{8 a c^2 (b c-a d)^3 \left (c+d x^2\right )}+\frac{\left (b^3 (b c-7 a d)\right ) \int \frac{1}{a+b x^2} \, dx}{2 a (b c-a d)^4}+\frac{\left (d^2 \left (35 b^2 c^2-14 a b c d+3 a^2 d^2\right )\right ) \int \frac{1}{c+d x^2} \, dx}{8 c^2 (b c-a d)^4}\\ &=\frac{d (2 b c+a d) x}{4 a c (b c-a d)^2 \left (c+d x^2\right )^2}+\frac{b x}{2 a (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )^2}+\frac{d (4 b c-a d) (b c+3 a d) x}{8 a c^2 (b c-a d)^3 \left (c+d x^2\right )}+\frac{b^{5/2} (b c-7 a d) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 a^{3/2} (b c-a d)^4}+\frac{d^{3/2} \left (35 b^2 c^2-14 a b c d+3 a^2 d^2\right ) \tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )}{8 c^{5/2} (b c-a d)^4}\\ \end{align*}

Mathematica [A]  time = 0.406944, size = 197, normalized size = 0.86 \[ \frac{1}{8} \left (\frac{d^{3/2} \left (3 a^2 d^2-14 a b c d+35 b^2 c^2\right ) \tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )}{c^{5/2} (b c-a d)^4}+\frac{4 b^{5/2} (b c-7 a d) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{a^{3/2} (b c-a d)^4}-\frac{4 b^3 x}{a \left (a+b x^2\right ) (a d-b c)^3}+\frac{d^2 x (11 b c-3 a d)}{c^2 \left (c+d x^2\right ) (b c-a d)^3}+\frac{2 d^2 x}{c \left (c+d x^2\right )^2 (b c-a d)^2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-4*b^3*x)/(a*(-(b*c) + a*d)^3*(a + b*x^2)) + (2*d^2*x)/(c*(b*c - a*d)^2*(c + d*x^2)^2) + (d^2*(11*b*c - 3*a*
d)*x)/(c^2*(b*c - a*d)^3*(c + d*x^2)) + (4*b^(5/2)*(b*c - 7*a*d)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(a^(3/2)*(b*c -
a*d)^4) + (d^(3/2)*(35*b^2*c^2 - 14*a*b*c*d + 3*a^2*d^2)*ArcTan[(Sqrt[d]*x)/Sqrt[c]])/(c^(5/2)*(b*c - a*d)^4))
/8

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Maple [A]  time = 0.013, size = 403, normalized size = 1.8 \begin{align*}{\frac{3\,{d}^{5}{x}^{3}{a}^{2}}{8\, \left ( ad-bc \right ) ^{4} \left ( d{x}^{2}+c \right ) ^{2}{c}^{2}}}-{\frac{7\,{d}^{4}{x}^{3}ab}{4\, \left ( ad-bc \right ) ^{4} \left ( d{x}^{2}+c \right ) ^{2}c}}+{\frac{11\,{d}^{3}{b}^{2}{x}^{3}}{8\, \left ( ad-bc \right ) ^{4} \left ( d{x}^{2}+c \right ) ^{2}}}+{\frac{5\,{d}^{4}x{a}^{2}}{8\, \left ( ad-bc \right ) ^{4} \left ( d{x}^{2}+c \right ) ^{2}c}}-{\frac{9\,{d}^{3}abx}{4\, \left ( ad-bc \right ) ^{4} \left ( d{x}^{2}+c \right ) ^{2}}}+{\frac{13\,{d}^{2}{b}^{2}cx}{8\, \left ( ad-bc \right ) ^{4} \left ( d{x}^{2}+c \right ) ^{2}}}+{\frac{3\,{d}^{4}{a}^{2}}{8\, \left ( ad-bc \right ) ^{4}{c}^{2}}\arctan \left ({dx{\frac{1}{\sqrt{cd}}}} \right ){\frac{1}{\sqrt{cd}}}}-{\frac{7\,{d}^{3}ab}{4\, \left ( ad-bc \right ) ^{4}c}\arctan \left ({dx{\frac{1}{\sqrt{cd}}}} \right ){\frac{1}{\sqrt{cd}}}}+{\frac{35\,{d}^{2}{b}^{2}}{8\, \left ( ad-bc \right ) ^{4}}\arctan \left ({dx{\frac{1}{\sqrt{cd}}}} \right ){\frac{1}{\sqrt{cd}}}}-{\frac{{b}^{3}xd}{2\, \left ( ad-bc \right ) ^{4} \left ( b{x}^{2}+a \right ) }}+{\frac{{b}^{4}xc}{2\, \left ( ad-bc \right ) ^{4}a \left ( b{x}^{2}+a \right ) }}-{\frac{7\,{b}^{3}d}{2\, \left ( ad-bc \right ) ^{4}}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{{b}^{4}c}{2\, \left ( ad-bc \right ) ^{4}a}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/8*d^5/(a*d-b*c)^4/(d*x^2+c)^2/c^2*x^3*a^2-7/4*d^4/(a*d-b*c)^4/(d*x^2+c)^2/c*x^3*a*b+11/8*d^3/(a*d-b*c)^4/(d*
x^2+c)^2*b^2*x^3+5/8*d^4/(a*d-b*c)^4/(d*x^2+c)^2/c*x*a^2-9/4*d^3/(a*d-b*c)^4/(d*x^2+c)^2*a*b*x+13/8*d^2/(a*d-b
*c)^4/(d*x^2+c)^2*b^2*c*x+3/8*d^4/(a*d-b*c)^4/c^2/(c*d)^(1/2)*arctan(x*d/(c*d)^(1/2))*a^2-7/4*d^3/(a*d-b*c)^4/
c/(c*d)^(1/2)*arctan(x*d/(c*d)^(1/2))*a*b+35/8*d^2/(a*d-b*c)^4/(c*d)^(1/2)*arctan(x*d/(c*d)^(1/2))*b^2-1/2*b^3
/(a*d-b*c)^4*x/(b*x^2+a)*d+1/2*b^4/(a*d-b*c)^4*x/a/(b*x^2+a)*c-7/2*b^3/(a*d-b*c)^4/(a*b)^(1/2)*arctan(b*x/(a*b
)^(1/2))*d+1/2*b^4/(a*d-b*c)^4/a/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))*c

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Maxima [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)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 19.2844, size = 6472, normalized size = 28.14 \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)^3,x, algorithm="fricas")

[Out]

[1/16*(2*(4*b^4*c^3*d^2 + 7*a*b^3*c^2*d^3 - 14*a^2*b^2*c*d^4 + 3*a^3*b*d^5)*x^5 + 2*(8*b^4*c^4*d + 5*a*b^3*c^3
*d^2 - 7*a^2*b^2*c^2*d^3 - 9*a^3*b*c*d^4 + 3*a^4*d^5)*x^3 - 4*(a*b^3*c^5 - 7*a^2*b^2*c^4*d + (b^4*c^3*d^2 - 7*
a*b^3*c^2*d^3)*x^6 + (2*b^4*c^4*d - 13*a*b^3*c^3*d^2 - 7*a^2*b^2*c^2*d^3)*x^4 + (b^4*c^5 - 5*a*b^3*c^4*d - 14*
a^2*b^2*c^3*d^2)*x^2)*sqrt(-b/a)*log((b*x^2 - 2*a*x*sqrt(-b/a) - a)/(b*x^2 + a)) + (35*a^2*b^2*c^4*d - 14*a^3*
b*c^3*d^2 + 3*a^4*c^2*d^3 + (35*a*b^3*c^2*d^3 - 14*a^2*b^2*c*d^4 + 3*a^3*b*d^5)*x^6 + (70*a*b^3*c^3*d^2 + 7*a^
2*b^2*c^2*d^3 - 8*a^3*b*c*d^4 + 3*a^4*d^5)*x^4 + (35*a*b^3*c^4*d + 56*a^2*b^2*c^3*d^2 - 25*a^3*b*c^2*d^3 + 6*a
^4*c*d^4)*x^2)*sqrt(-d/c)*log((d*x^2 + 2*c*x*sqrt(-d/c) - c)/(d*x^2 + c)) + 2*(4*b^4*c^5 - 4*a*b^3*c^4*d + 13*
a^2*b^2*c^3*d^2 - 18*a^3*b*c^2*d^3 + 5*a^4*c*d^4)*x)/(a^2*b^4*c^8 - 4*a^3*b^3*c^7*d + 6*a^4*b^2*c^6*d^2 - 4*a^
5*b*c^5*d^3 + a^6*c^4*d^4 + (a*b^5*c^6*d^2 - 4*a^2*b^4*c^5*d^3 + 6*a^3*b^3*c^4*d^4 - 4*a^4*b^2*c^3*d^5 + a^5*b
*c^2*d^6)*x^6 + (2*a*b^5*c^7*d - 7*a^2*b^4*c^6*d^2 + 8*a^3*b^3*c^5*d^3 - 2*a^4*b^2*c^4*d^4 - 2*a^5*b*c^3*d^5 +
a^6*c^2*d^6)*x^4 + (a*b^5*c^8 - 2*a^2*b^4*c^7*d - 2*a^3*b^3*c^6*d^2 + 8*a^4*b^2*c^5*d^3 - 7*a^5*b*c^4*d^4 + 2
*a^6*c^3*d^5)*x^2), 1/8*((4*b^4*c^3*d^2 + 7*a*b^3*c^2*d^3 - 14*a^2*b^2*c*d^4 + 3*a^3*b*d^5)*x^5 + (8*b^4*c^4*d
+ 5*a*b^3*c^3*d^2 - 7*a^2*b^2*c^2*d^3 - 9*a^3*b*c*d^4 + 3*a^4*d^5)*x^3 + (35*a^2*b^2*c^4*d - 14*a^3*b*c^3*d^2
+ 3*a^4*c^2*d^3 + (35*a*b^3*c^2*d^3 - 14*a^2*b^2*c*d^4 + 3*a^3*b*d^5)*x^6 + (70*a*b^3*c^3*d^2 + 7*a^2*b^2*c^2
*d^3 - 8*a^3*b*c*d^4 + 3*a^4*d^5)*x^4 + (35*a*b^3*c^4*d + 56*a^2*b^2*c^3*d^2 - 25*a^3*b*c^2*d^3 + 6*a^4*c*d^4)
*x^2)*sqrt(d/c)*arctan(x*sqrt(d/c)) - 2*(a*b^3*c^5 - 7*a^2*b^2*c^4*d + (b^4*c^3*d^2 - 7*a*b^3*c^2*d^3)*x^6 + (
2*b^4*c^4*d - 13*a*b^3*c^3*d^2 - 7*a^2*b^2*c^2*d^3)*x^4 + (b^4*c^5 - 5*a*b^3*c^4*d - 14*a^2*b^2*c^3*d^2)*x^2)*
sqrt(-b/a)*log((b*x^2 - 2*a*x*sqrt(-b/a) - a)/(b*x^2 + a)) + (4*b^4*c^5 - 4*a*b^3*c^4*d + 13*a^2*b^2*c^3*d^2 -
18*a^3*b*c^2*d^3 + 5*a^4*c*d^4)*x)/(a^2*b^4*c^8 - 4*a^3*b^3*c^7*d + 6*a^4*b^2*c^6*d^2 - 4*a^5*b*c^5*d^3 + a^6
*c^4*d^4 + (a*b^5*c^6*d^2 - 4*a^2*b^4*c^5*d^3 + 6*a^3*b^3*c^4*d^4 - 4*a^4*b^2*c^3*d^5 + a^5*b*c^2*d^6)*x^6 + (
2*a*b^5*c^7*d - 7*a^2*b^4*c^6*d^2 + 8*a^3*b^3*c^5*d^3 - 2*a^4*b^2*c^4*d^4 - 2*a^5*b*c^3*d^5 + a^6*c^2*d^6)*x^4
+ (a*b^5*c^8 - 2*a^2*b^4*c^7*d - 2*a^3*b^3*c^6*d^2 + 8*a^4*b^2*c^5*d^3 - 7*a^5*b*c^4*d^4 + 2*a^6*c^3*d^5)*x^2
), 1/16*(2*(4*b^4*c^3*d^2 + 7*a*b^3*c^2*d^3 - 14*a^2*b^2*c*d^4 + 3*a^3*b*d^5)*x^5 + 2*(8*b^4*c^4*d + 5*a*b^3*c
^3*d^2 - 7*a^2*b^2*c^2*d^3 - 9*a^3*b*c*d^4 + 3*a^4*d^5)*x^3 + 8*(a*b^3*c^5 - 7*a^2*b^2*c^4*d + (b^4*c^3*d^2 -
7*a*b^3*c^2*d^3)*x^6 + (2*b^4*c^4*d - 13*a*b^3*c^3*d^2 - 7*a^2*b^2*c^2*d^3)*x^4 + (b^4*c^5 - 5*a*b^3*c^4*d - 1
4*a^2*b^2*c^3*d^2)*x^2)*sqrt(b/a)*arctan(x*sqrt(b/a)) + (35*a^2*b^2*c^4*d - 14*a^3*b*c^3*d^2 + 3*a^4*c^2*d^3 +
(35*a*b^3*c^2*d^3 - 14*a^2*b^2*c*d^4 + 3*a^3*b*d^5)*x^6 + (70*a*b^3*c^3*d^2 + 7*a^2*b^2*c^2*d^3 - 8*a^3*b*c*d
^4 + 3*a^4*d^5)*x^4 + (35*a*b^3*c^4*d + 56*a^2*b^2*c^3*d^2 - 25*a^3*b*c^2*d^3 + 6*a^4*c*d^4)*x^2)*sqrt(-d/c)*l
og((d*x^2 + 2*c*x*sqrt(-d/c) - c)/(d*x^2 + c)) + 2*(4*b^4*c^5 - 4*a*b^3*c^4*d + 13*a^2*b^2*c^3*d^2 - 18*a^3*b*
c^2*d^3 + 5*a^4*c*d^4)*x)/(a^2*b^4*c^8 - 4*a^3*b^3*c^7*d + 6*a^4*b^2*c^6*d^2 - 4*a^5*b*c^5*d^3 + a^6*c^4*d^4 +
(a*b^5*c^6*d^2 - 4*a^2*b^4*c^5*d^3 + 6*a^3*b^3*c^4*d^4 - 4*a^4*b^2*c^3*d^5 + a^5*b*c^2*d^6)*x^6 + (2*a*b^5*c^
7*d - 7*a^2*b^4*c^6*d^2 + 8*a^3*b^3*c^5*d^3 - 2*a^4*b^2*c^4*d^4 - 2*a^5*b*c^3*d^5 + a^6*c^2*d^6)*x^4 + (a*b^5*
c^8 - 2*a^2*b^4*c^7*d - 2*a^3*b^3*c^6*d^2 + 8*a^4*b^2*c^5*d^3 - 7*a^5*b*c^4*d^4 + 2*a^6*c^3*d^5)*x^2), 1/8*((4
*b^4*c^3*d^2 + 7*a*b^3*c^2*d^3 - 14*a^2*b^2*c*d^4 + 3*a^3*b*d^5)*x^5 + (8*b^4*c^4*d + 5*a*b^3*c^3*d^2 - 7*a^2*
b^2*c^2*d^3 - 9*a^3*b*c*d^4 + 3*a^4*d^5)*x^3 + 4*(a*b^3*c^5 - 7*a^2*b^2*c^4*d + (b^4*c^3*d^2 - 7*a*b^3*c^2*d^3
)*x^6 + (2*b^4*c^4*d - 13*a*b^3*c^3*d^2 - 7*a^2*b^2*c^2*d^3)*x^4 + (b^4*c^5 - 5*a*b^3*c^4*d - 14*a^2*b^2*c^3*d
^2)*x^2)*sqrt(b/a)*arctan(x*sqrt(b/a)) + (35*a^2*b^2*c^4*d - 14*a^3*b*c^3*d^2 + 3*a^4*c^2*d^3 + (35*a*b^3*c^2*
d^3 - 14*a^2*b^2*c*d^4 + 3*a^3*b*d^5)*x^6 + (70*a*b^3*c^3*d^2 + 7*a^2*b^2*c^2*d^3 - 8*a^3*b*c*d^4 + 3*a^4*d^5)
*x^4 + (35*a*b^3*c^4*d + 56*a^2*b^2*c^3*d^2 - 25*a^3*b*c^2*d^3 + 6*a^4*c*d^4)*x^2)*sqrt(d/c)*arctan(x*sqrt(d/c
)) + (4*b^4*c^5 - 4*a*b^3*c^4*d + 13*a^2*b^2*c^3*d^2 - 18*a^3*b*c^2*d^3 + 5*a^4*c*d^4)*x)/(a^2*b^4*c^8 - 4*a^3
*b^3*c^7*d + 6*a^4*b^2*c^6*d^2 - 4*a^5*b*c^5*d^3 + a^6*c^4*d^4 + (a*b^5*c^6*d^2 - 4*a^2*b^4*c^5*d^3 + 6*a^3*b^
3*c^4*d^4 - 4*a^4*b^2*c^3*d^5 + a^5*b*c^2*d^6)*x^6 + (2*a*b^5*c^7*d - 7*a^2*b^4*c^6*d^2 + 8*a^3*b^3*c^5*d^3 -
2*a^4*b^2*c^4*d^4 - 2*a^5*b*c^3*d^5 + a^6*c^2*d^6)*x^4 + (a*b^5*c^8 - 2*a^2*b^4*c^7*d - 2*a^3*b^3*c^6*d^2 + 8*
a^4*b^2*c^5*d^3 - 7*a^5*b*c^4*d^4 + 2*a^6*c^3*d^5)*x^2)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.17894, size = 448, normalized size = 1.95 \begin{align*} \frac{b^{3} x}{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 )}{\left (b x^{2} + a\right )}} + \frac{{\left (b^{4} c - 7 \, a b^{3} d\right )} \arctan \left (\frac{b x}{\sqrt{a b}}\right )}{2 \,{\left (a b^{4} c^{4} - 4 \, a^{2} b^{3} c^{3} d + 6 \, a^{3} b^{2} c^{2} d^{2} - 4 \, a^{4} b c d^{3} + a^{5} d^{4}\right )} \sqrt{a b}} + \frac{{\left (35 \, b^{2} c^{2} d^{2} - 14 \, a b c d^{3} + 3 \, a^{2} d^{4}\right )} \arctan \left (\frac{d x}{\sqrt{c d}}\right )}{8 \,{\left (b^{4} c^{6} - 4 \, a b^{3} c^{5} d + 6 \, a^{2} b^{2} c^{4} d^{2} - 4 \, a^{3} b c^{3} d^{3} + a^{4} c^{2} d^{4}\right )} \sqrt{c d}} + \frac{11 \, b c d^{3} x^{3} - 3 \, a d^{4} x^{3} + 13 \, b c^{2} d^{2} x - 5 \, a c d^{3} x}{8 \,{\left (b^{3} c^{5} - 3 \, a b^{2} c^{4} d + 3 \, a^{2} b c^{3} d^{2} - a^{3} c^{2} d^{3}\right )}{\left (d x^{2} + c\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b^3*x/((a*b^3*c^3 - 3*a^2*b^2*c^2*d + 3*a^3*b*c*d^2 - a^4*d^3)*(b*x^2 + a)) + 1/2*(b^4*c - 7*a*b^3*d)*arct
an(b*x/sqrt(a*b))/((a*b^4*c^4 - 4*a^2*b^3*c^3*d + 6*a^3*b^2*c^2*d^2 - 4*a^4*b*c*d^3 + a^5*d^4)*sqrt(a*b)) + 1/
8*(35*b^2*c^2*d^2 - 14*a*b*c*d^3 + 3*a^2*d^4)*arctan(d*x/sqrt(c*d))/((b^4*c^6 - 4*a*b^3*c^5*d + 6*a^2*b^2*c^4*
d^2 - 4*a^3*b*c^3*d^3 + a^4*c^2*d^4)*sqrt(c*d)) + 1/8*(11*b*c*d^3*x^3 - 3*a*d^4*x^3 + 13*b*c^2*d^2*x - 5*a*c*d
^3*x)/((b^3*c^5 - 3*a*b^2*c^4*d + 3*a^2*b*c^3*d^2 - a^3*c^2*d^3)*(d*x^2 + c)^2)