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

Optimal. Leaf size=161 \[ \frac{\sqrt{b} \left (15 a^2 d^2-10 a b c d+3 b^2 c^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{5/2} (b c-a d)^3}+\frac{b x (3 b c-7 a d)}{8 a^2 \left (a+b x^2\right ) (b c-a d)^2}-\frac{d^{5/2} \tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )}{\sqrt{c} (b c-a d)^3}+\frac{b x}{4 a \left (a+b x^2\right )^2 (b c-a d)} \]

[Out]

(b*x)/(4*a*(b*c - a*d)*(a + b*x^2)^2) + (b*(3*b*c - 7*a*d)*x)/(8*a^2*(b*c - a*d)^2*(a + b*x^2)) + (Sqrt[b]*(3*
b^2*c^2 - 10*a*b*c*d + 15*a^2*d^2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*a^(5/2)*(b*c - a*d)^3) - (d^(5/2)*ArcTan[(S
qrt[d]*x)/Sqrt[c]])/(Sqrt[c]*(b*c - a*d)^3)

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Rubi [A]  time = 0.197347, antiderivative size = 161, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {414, 527, 522, 205} \[ \frac{\sqrt{b} \left (15 a^2 d^2-10 a b c d+3 b^2 c^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{5/2} (b c-a d)^3}+\frac{b x (3 b c-7 a d)}{8 a^2 \left (a+b x^2\right ) (b c-a d)^2}-\frac{d^{5/2} \tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )}{\sqrt{c} (b c-a d)^3}+\frac{b x}{4 a \left (a+b x^2\right )^2 (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x^2)^3*(c + d*x^2)),x]

[Out]

(b*x)/(4*a*(b*c - a*d)*(a + b*x^2)^2) + (b*(3*b*c - 7*a*d)*x)/(8*a^2*(b*c - a*d)^2*(a + b*x^2)) + (Sqrt[b]*(3*
b^2*c^2 - 10*a*b*c*d + 15*a^2*d^2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*a^(5/2)*(b*c - a*d)^3) - (d^(5/2)*ArcTan[(S
qrt[d]*x)/Sqrt[c]])/(Sqrt[c]*(b*c - a*d)^3)

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 )} \, dx &=\frac{b x}{4 a (b c-a d) \left (a+b x^2\right )^2}-\frac{\int \frac{-3 b c+4 a d-3 b d x^2}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx}{4 a (b c-a d)}\\ &=\frac{b x}{4 a (b c-a d) \left (a+b x^2\right )^2}+\frac{b (3 b c-7 a d) x}{8 a^2 (b c-a d)^2 \left (a+b x^2\right )}+\frac{\int \frac{3 b^2 c^2-7 a b c d+8 a^2 d^2+b d (3 b c-7 a d) x^2}{\left (a+b x^2\right ) \left (c+d x^2\right )} \, dx}{8 a^2 (b c-a d)^2}\\ &=\frac{b x}{4 a (b c-a d) \left (a+b x^2\right )^2}+\frac{b (3 b c-7 a d) x}{8 a^2 (b c-a d)^2 \left (a+b x^2\right )}-\frac{d^3 \int \frac{1}{c+d x^2} \, dx}{(b c-a d)^3}+\frac{\left (b \left (3 b^2 c^2-10 a b c d+15 a^2 d^2\right )\right ) \int \frac{1}{a+b x^2} \, dx}{8 a^2 (b c-a d)^3}\\ &=\frac{b x}{4 a (b c-a d) \left (a+b x^2\right )^2}+\frac{b (3 b c-7 a d) x}{8 a^2 (b c-a d)^2 \left (a+b x^2\right )}+\frac{\sqrt{b} \left (3 b^2 c^2-10 a b c d+15 a^2 d^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{5/2} (b c-a d)^3}-\frac{d^{5/2} \tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )}{\sqrt{c} (b c-a d)^3}\\ \end{align*}

Mathematica [A]  time = 0.264827, size = 158, normalized size = 0.98 \[ \frac{1}{8} \left (-\frac{\sqrt{b} \left (15 a^2 d^2-10 a b c d+3 b^2 c^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{a^{5/2} (a d-b c)^3}+\frac{b x (3 b c-7 a d)}{a^2 \left (a+b x^2\right ) (b c-a d)^2}-\frac{8 d^{5/2} \tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )}{\sqrt{c} (b c-a d)^3}-\frac{2 b x}{a \left (a+b x^2\right )^2 (a d-b c)}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*b*x)/(a*(-(b*c) + a*d)*(a + b*x^2)^2) + (b*(3*b*c - 7*a*d)*x)/(a^2*(b*c - a*d)^2*(a + b*x^2)) - (Sqrt[b]*
(3*b^2*c^2 - 10*a*b*c*d + 15*a^2*d^2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(a^(5/2)*(-(b*c) + a*d)^3) - (8*d^(5/2)*Arc
Tan[(Sqrt[d]*x)/Sqrt[c]])/(Sqrt[c]*(b*c - a*d)^3))/8

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Maple [B]  time = 0.01, size = 309, normalized size = 1.9 \begin{align*}{\frac{{d}^{3}}{ \left ( ad-bc \right ) ^{3}}\arctan \left ({dx{\frac{1}{\sqrt{cd}}}} \right ){\frac{1}{\sqrt{cd}}}}-{\frac{7\,{b}^{2}{x}^{3}{d}^{2}}{8\, \left ( ad-bc \right ) ^{3} \left ( b{x}^{2}+a \right ) ^{2}}}+{\frac{5\,{b}^{3}{x}^{3}cd}{4\, \left ( ad-bc \right ) ^{3} \left ( b{x}^{2}+a \right ) ^{2}a}}-{\frac{3\,{b}^{4}{x}^{3}{c}^{2}}{8\, \left ( ad-bc \right ) ^{3} \left ( b{x}^{2}+a \right ) ^{2}{a}^{2}}}-{\frac{9\,abx{d}^{2}}{8\, \left ( ad-bc \right ) ^{3} \left ( b{x}^{2}+a \right ) ^{2}}}+{\frac{7\,{b}^{2}xcd}{4\, \left ( ad-bc \right ) ^{3} \left ( b{x}^{2}+a \right ) ^{2}}}-{\frac{5\,{b}^{3}x{c}^{2}}{8\, \left ( ad-bc \right ) ^{3} \left ( b{x}^{2}+a \right ) ^{2}a}}-{\frac{15\,{d}^{2}b}{8\, \left ( ad-bc \right ) ^{3}}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{5\,{b}^{2}cd}{4\, \left ( ad-bc \right ) ^{3}a}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{3\,{b}^{3}{c}^{2}}{8\, \left ( ad-bc \right ) ^{3}{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),x)

[Out]

d^3/(a*d-b*c)^3/(c*d)^(1/2)*arctan(x*d/(c*d)^(1/2))-7/8*b^2/(a*d-b*c)^3/(b*x^2+a)^2*x^3*d^2+5/4*b^3/(a*d-b*c)^
3/(b*x^2+a)^2/a*x^3*c*d-3/8*b^4/(a*d-b*c)^3/(b*x^2+a)^2/a^2*x^3*c^2-9/8*b/(a*d-b*c)^3/(b*x^2+a)^2*x*a*d^2+7/4*
b^2/(a*d-b*c)^3/(b*x^2+a)^2*x*c*d-5/8*b^3/(a*d-b*c)^3/(b*x^2+a)^2*x/a*c^2-15/8*b/(a*d-b*c)^3/(a*b)^(1/2)*arcta
n(b*x/(a*b)^(1/2))*d^2+5/4*b^2/(a*d-b*c)^3/a/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))*c*d-3/8*b^3/(a*d-b*c)^3/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),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 6.13419, size = 3212, normalized size = 19.95 \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),x, algorithm="fricas")

[Out]

[1/16*(2*(3*b^4*c^2 - 10*a*b^3*c*d + 7*a^2*b^2*d^2)*x^3 - (3*a^2*b^2*c^2 - 10*a^3*b*c*d + 15*a^4*d^2 + (3*b^4*
c^2 - 10*a*b^3*c*d + 15*a^2*b^2*d^2)*x^4 + 2*(3*a*b^3*c^2 - 10*a^2*b^2*c*d + 15*a^3*b*d^2)*x^2)*sqrt(-b/a)*log
((b*x^2 - 2*a*x*sqrt(-b/a) - a)/(b*x^2 + a)) - 8*(a^2*b^2*d^2*x^4 + 2*a^3*b*d^2*x^2 + a^4*d^2)*sqrt(-d/c)*log(
(d*x^2 + 2*c*x*sqrt(-d/c) - c)/(d*x^2 + c)) + 2*(5*a*b^3*c^2 - 14*a^2*b^2*c*d + 9*a^3*b*d^2)*x)/(a^4*b^3*c^3 -
3*a^5*b^2*c^2*d + 3*a^6*b*c*d^2 - a^7*d^3 + (a^2*b^5*c^3 - 3*a^3*b^4*c^2*d + 3*a^4*b^3*c*d^2 - a^5*b^2*d^3)*x
^4 + 2*(a^3*b^4*c^3 - 3*a^4*b^3*c^2*d + 3*a^5*b^2*c*d^2 - a^6*b*d^3)*x^2), 1/16*(2*(3*b^4*c^2 - 10*a*b^3*c*d +
7*a^2*b^2*d^2)*x^3 - 16*(a^2*b^2*d^2*x^4 + 2*a^3*b*d^2*x^2 + a^4*d^2)*sqrt(d/c)*arctan(x*sqrt(d/c)) - (3*a^2*
b^2*c^2 - 10*a^3*b*c*d + 15*a^4*d^2 + (3*b^4*c^2 - 10*a*b^3*c*d + 15*a^2*b^2*d^2)*x^4 + 2*(3*a*b^3*c^2 - 10*a^
2*b^2*c*d + 15*a^3*b*d^2)*x^2)*sqrt(-b/a)*log((b*x^2 - 2*a*x*sqrt(-b/a) - a)/(b*x^2 + a)) + 2*(5*a*b^3*c^2 - 1
4*a^2*b^2*c*d + 9*a^3*b*d^2)*x)/(a^4*b^3*c^3 - 3*a^5*b^2*c^2*d + 3*a^6*b*c*d^2 - a^7*d^3 + (a^2*b^5*c^3 - 3*a^
3*b^4*c^2*d + 3*a^4*b^3*c*d^2 - a^5*b^2*d^3)*x^4 + 2*(a^3*b^4*c^3 - 3*a^4*b^3*c^2*d + 3*a^5*b^2*c*d^2 - a^6*b*
d^3)*x^2), 1/8*((3*b^4*c^2 - 10*a*b^3*c*d + 7*a^2*b^2*d^2)*x^3 + (3*a^2*b^2*c^2 - 10*a^3*b*c*d + 15*a^4*d^2 +
(3*b^4*c^2 - 10*a*b^3*c*d + 15*a^2*b^2*d^2)*x^4 + 2*(3*a*b^3*c^2 - 10*a^2*b^2*c*d + 15*a^3*b*d^2)*x^2)*sqrt(b/
a)*arctan(x*sqrt(b/a)) - 4*(a^2*b^2*d^2*x^4 + 2*a^3*b*d^2*x^2 + a^4*d^2)*sqrt(-d/c)*log((d*x^2 + 2*c*x*sqrt(-d
/c) - c)/(d*x^2 + c)) + (5*a*b^3*c^2 - 14*a^2*b^2*c*d + 9*a^3*b*d^2)*x)/(a^4*b^3*c^3 - 3*a^5*b^2*c^2*d + 3*a^6
*b*c*d^2 - a^7*d^3 + (a^2*b^5*c^3 - 3*a^3*b^4*c^2*d + 3*a^4*b^3*c*d^2 - a^5*b^2*d^3)*x^4 + 2*(a^3*b^4*c^3 - 3*
a^4*b^3*c^2*d + 3*a^5*b^2*c*d^2 - a^6*b*d^3)*x^2), 1/8*((3*b^4*c^2 - 10*a*b^3*c*d + 7*a^2*b^2*d^2)*x^3 + (3*a^
2*b^2*c^2 - 10*a^3*b*c*d + 15*a^4*d^2 + (3*b^4*c^2 - 10*a*b^3*c*d + 15*a^2*b^2*d^2)*x^4 + 2*(3*a*b^3*c^2 - 10*
a^2*b^2*c*d + 15*a^3*b*d^2)*x^2)*sqrt(b/a)*arctan(x*sqrt(b/a)) - 8*(a^2*b^2*d^2*x^4 + 2*a^3*b*d^2*x^2 + a^4*d^
2)*sqrt(d/c)*arctan(x*sqrt(d/c)) + (5*a*b^3*c^2 - 14*a^2*b^2*c*d + 9*a^3*b*d^2)*x)/(a^4*b^3*c^3 - 3*a^5*b^2*c^
2*d + 3*a^6*b*c*d^2 - a^7*d^3 + (a^2*b^5*c^3 - 3*a^3*b^4*c^2*d + 3*a^4*b^3*c*d^2 - a^5*b^2*d^3)*x^4 + 2*(a^3*b
^4*c^3 - 3*a^4*b^3*c^2*d + 3*a^5*b^2*c*d^2 - a^6*b*d^3)*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),x)

[Out]

Timed out

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Giac [A]  time = 1.14794, size = 294, normalized size = 1.83 \begin{align*} -\frac{d^{3} \arctan \left (\frac{d x}{\sqrt{c d}}\right )}{{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt{c d}} + \frac{{\left (3 \, b^{3} c^{2} - 10 \, a b^{2} c d + 15 \, a^{2} b d^{2}\right )} \arctan \left (\frac{b x}{\sqrt{a b}}\right )}{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 )} \sqrt{a b}} + \frac{3 \, b^{3} c x^{3} - 7 \, a b^{2} d x^{3} + 5 \, a b^{2} c x - 9 \, a^{2} b d x}{8 \,{\left (a^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} d^{2}\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),x, algorithm="giac")

[Out]

-d^3*arctan(d*x/sqrt(c*d))/((b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*sqrt(c*d)) + 1/8*(3*b^3*c^2 -
10*a*b^2*c*d + 15*a^2*b*d^2)*arctan(b*x/sqrt(a*b))/((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)*
sqrt(a*b)) + 1/8*(3*b^3*c*x^3 - 7*a*b^2*d*x^3 + 5*a*b^2*c*x - 9*a^2*b*d*x)/((a^2*b^2*c^2 - 2*a^3*b*c*d + a^4*d
^2)*(b*x^2 + a)^2)