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

Optimal. Leaf size=236 \[ \frac{b^{3/2} \left (35 a^2 d^2-14 a b c d+3 b^2 c^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{5/2} (b c-a d)^4}+\frac{d x (b c-4 a d) (a d+3 b c)}{8 a^2 c \left (c+d x^2\right ) (b c-a d)^3}+\frac{3 b x (b c-3 a d)}{8 a^2 \left (a+b x^2\right ) \left (c+d x^2\right ) (b c-a d)^2}-\frac{d^{5/2} (7 b c-a d) \tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )}{2 c^{3/2} (b c-a d)^4}+\frac{b x}{4 a \left (a+b x^2\right )^2 \left (c+d x^2\right ) (b c-a d)} \]

[Out]

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

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Rubi [A]  time = 0.310513, antiderivative size = 236, 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} \[ \frac{b^{3/2} \left (35 a^2 d^2-14 a b c d+3 b^2 c^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{5/2} (b c-a d)^4}+\frac{d x (b c-4 a d) (a d+3 b c)}{8 a^2 c \left (c+d x^2\right ) (b c-a d)^3}+\frac{3 b x (b c-3 a d)}{8 a^2 \left (a+b x^2\right ) \left (c+d x^2\right ) (b c-a d)^2}-\frac{d^{5/2} (7 b c-a d) \tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )}{2 c^{3/2} (b c-a d)^4}+\frac{b x}{4 a \left (a+b x^2\right )^2 \left (c+d x^2\right ) (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(b*c - 4*a*d)*(3*b*c + a*d)*x)/(8*a^2*c*(b*c - a*d)^3*(c + d*x^2)) + (b*x)/(4*a*(b*c - a*d)*(a + b*x^2)^2*(
c + d*x^2)) + (3*b*(b*c - 3*a*d)*x)/(8*a^2*(b*c - a*d)^2*(a + b*x^2)*(c + d*x^2)) + (b^(3/2)*(3*b^2*c^2 - 14*a
*b*c*d + 35*a^2*d^2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*a^(5/2)*(b*c - a*d)^4) - (d^(5/2)*(7*b*c - a*d)*ArcTan[(S
qrt[d]*x)/Sqrt[c]])/(2*c^(3/2)*(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 )^3 \left (c+d x^2\right )^2} \, dx &=\frac{b x}{4 a (b c-a d) \left (a+b x^2\right )^2 \left (c+d x^2\right )}-\frac{\int \frac{-3 b c+4 a d-5 b d x^2}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^2} \, dx}{4 a (b c-a d)}\\ &=\frac{b x}{4 a (b c-a d) \left (a+b x^2\right )^2 \left (c+d x^2\right )}+\frac{3 b (b c-3 a d) x}{8 a^2 (b c-a d)^2 \left (a+b x^2\right ) \left (c+d x^2\right )}+\frac{\int \frac{3 b^2 c^2-5 a b c d+8 a^2 d^2+9 b d (b c-3 a d) x^2}{\left (a+b x^2\right ) \left (c+d x^2\right )^2} \, dx}{8 a^2 (b c-a d)^2}\\ &=\frac{d (b c-4 a d) (3 b c+a d) x}{8 a^2 c (b c-a d)^3 \left (c+d x^2\right )}+\frac{b x}{4 a (b c-a d) \left (a+b x^2\right )^2 \left (c+d x^2\right )}+\frac{3 b (b c-3 a d) x}{8 a^2 (b c-a d)^2 \left (a+b x^2\right ) \left (c+d x^2\right )}+\frac{\int \frac{2 \left (3 b^3 c^3-11 a b^2 c^2 d+24 a^2 b c d^2-4 a^3 d^3\right )+2 b d (b c-4 a d) (3 b c+a d) x^2}{\left (a+b x^2\right ) \left (c+d x^2\right )} \, dx}{16 a^2 c (b c-a d)^3}\\ &=\frac{d (b c-4 a d) (3 b c+a d) x}{8 a^2 c (b c-a d)^3 \left (c+d x^2\right )}+\frac{b x}{4 a (b c-a d) \left (a+b x^2\right )^2 \left (c+d x^2\right )}+\frac{3 b (b c-3 a d) x}{8 a^2 (b c-a d)^2 \left (a+b x^2\right ) \left (c+d x^2\right )}-\frac{\left (d^3 (7 b c-a d)\right ) \int \frac{1}{c+d x^2} \, dx}{2 c (b c-a d)^4}+\frac{\left (b^2 \left (3 b^2 c^2-14 a b c d+35 a^2 d^2\right )\right ) \int \frac{1}{a+b x^2} \, dx}{8 a^2 (b c-a d)^4}\\ &=\frac{d (b c-4 a d) (3 b c+a d) x}{8 a^2 c (b c-a d)^3 \left (c+d x^2\right )}+\frac{b x}{4 a (b c-a d) \left (a+b x^2\right )^2 \left (c+d x^2\right )}+\frac{3 b (b c-3 a d) x}{8 a^2 (b c-a d)^2 \left (a+b x^2\right ) \left (c+d x^2\right )}+\frac{b^{3/2} \left (3 b^2 c^2-14 a b c d+35 a^2 d^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{5/2} (b c-a d)^4}-\frac{d^{5/2} (7 b c-a d) \tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )}{2 c^{3/2} (b c-a d)^4}\\ \end{align*}

Mathematica [A]  time = 0.41711, size = 197, normalized size = 0.83 \[ \frac{1}{8} \left (\frac{b^{3/2} \left (35 a^2 d^2-14 a b c d+3 b^2 c^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{a^{5/2} (b c-a d)^4}+\frac{b^2 x (11 a d-3 b c)}{a^2 \left (a+b x^2\right ) (a d-b c)^3}+\frac{2 b^2 x}{a \left (a+b x^2\right )^2 (b c-a d)^2}+\frac{4 d^{5/2} (a d-7 b c) \tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )}{c^{3/2} (b c-a d)^4}-\frac{4 d^3 x}{c \left (c+d x^2\right ) (b c-a d)^3}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.014, size = 403, normalized size = 1.7 \begin{align*}{\frac{{d}^{4}xa}{2\, \left ( ad-bc \right ) ^{4}c \left ( d{x}^{2}+c \right ) }}-{\frac{{d}^{3}xb}{2\, \left ( ad-bc \right ) ^{4} \left ( d{x}^{2}+c \right ) }}+{\frac{a{d}^{4}}{2\, \left ( ad-bc \right ) ^{4}c}\arctan \left ({dx{\frac{1}{\sqrt{cd}}}} \right ){\frac{1}{\sqrt{cd}}}}-{\frac{7\,{d}^{3}b}{2\, \left ( ad-bc \right ) ^{4}}\arctan \left ({dx{\frac{1}{\sqrt{cd}}}} \right ){\frac{1}{\sqrt{cd}}}}+{\frac{11\,{b}^{3}{x}^{3}{d}^{2}}{8\, \left ( ad-bc \right ) ^{4} \left ( b{x}^{2}+a \right ) ^{2}}}-{\frac{7\,{b}^{4}{x}^{3}cd}{4\, \left ( ad-bc \right ) ^{4} \left ( b{x}^{2}+a \right ) ^{2}a}}+{\frac{3\,{b}^{5}{x}^{3}{c}^{2}}{8\, \left ( ad-bc \right ) ^{4} \left ( b{x}^{2}+a \right ) ^{2}{a}^{2}}}+{\frac{13\,{b}^{2}xa{d}^{2}}{8\, \left ( ad-bc \right ) ^{4} \left ( b{x}^{2}+a \right ) ^{2}}}-{\frac{9\,{b}^{3}xcd}{4\, \left ( ad-bc \right ) ^{4} \left ( b{x}^{2}+a \right ) ^{2}}}+{\frac{5\,{b}^{4}x{c}^{2}}{8\, \left ( ad-bc \right ) ^{4} \left ( b{x}^{2}+a \right ) ^{2}a}}+{\frac{35\,{b}^{2}{d}^{2}}{8\, \left ( ad-bc \right ) ^{4}}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{7\,{b}^{3}cd}{4\, \left ( ad-bc \right ) ^{4}a}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{3\,{b}^{4}{c}^{2}}{8\, \left ( ad-bc \right ) ^{4}{a}^{2}}\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)^3/(d*x^2+c)^2,x)

[Out]

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

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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)^3/(d*x^2+c)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 23.1117, size = 6472, normalized size = 27.42 \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)^3/(d*x^2+c)^2,x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.13542, size = 450, normalized size = 1.91 \begin{align*} -\frac{d^{3} x}{2 \,{\left (b^{3} c^{4} - 3 \, a b^{2} c^{3} d + 3 \, a^{2} b c^{2} d^{2} - a^{3} c d^{3}\right )}{\left (d x^{2} + c\right )}} + \frac{{\left (3 \, b^{4} c^{2} - 14 \, a b^{3} c d + 35 \, a^{2} b^{2} d^{2}\right )} \arctan \left (\frac{b x}{\sqrt{a b}}\right )}{8 \,{\left (a^{2} b^{4} c^{4} - 4 \, a^{3} b^{3} c^{3} d + 6 \, a^{4} b^{2} c^{2} d^{2} - 4 \, a^{5} b c d^{3} + a^{6} d^{4}\right )} \sqrt{a b}} - \frac{{\left (7 \, b c d^{3} - a d^{4}\right )} \arctan \left (\frac{d x}{\sqrt{c d}}\right )}{2 \,{\left (b^{4} c^{5} - 4 \, a b^{3} c^{4} d + 6 \, a^{2} b^{2} c^{3} d^{2} - 4 \, a^{3} b c^{2} d^{3} + a^{4} c d^{4}\right )} \sqrt{c d}} + \frac{3 \, b^{4} c x^{3} - 11 \, a b^{3} d x^{3} + 5 \, a b^{3} c x - 13 \, a^{2} b^{2} d x}{8 \,{\left (a^{2} b^{3} c^{3} - 3 \, a^{3} b^{2} c^{2} d + 3 \, a^{4} b c d^{2} - a^{5} d^{3}\right )}{\left (b x^{2} + a\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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