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

Optimal. Leaf size=160 \[ -\frac{\sqrt{d} \left (3 a^2 d^2-10 a b c d+15 b^2 c^2\right ) \tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )}{8 c^{5/2} (b c-a d)^3}+\frac{b^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{\sqrt{a} (b c-a d)^3}-\frac{d x (7 b c-3 a d)}{8 c^2 \left (c+d x^2\right ) (b c-a d)^2}-\frac{d x}{4 c \left (c+d x^2\right )^2 (b c-a d)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.188846, antiderivative size = 160, 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} \[ -\frac{\sqrt{d} \left (3 a^2 d^2-10 a b c d+15 b^2 c^2\right ) \tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )}{8 c^{5/2} (b c-a d)^3}+\frac{b^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{\sqrt{a} (b c-a d)^3}-\frac{d x (7 b c-3 a d)}{8 c^2 \left (c+d x^2\right ) (b c-a d)^2}-\frac{d x}{4 c \left (c+d x^2\right )^2 (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x^2)*(c + d*x^2)^3),x]

[Out]

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

Mathematica [A]  time = 0.230987, size = 158, normalized size = 0.99 \[ \frac{1}{8} \left (-\frac{\sqrt{d} \left (3 a^2 d^2-10 a b c d+15 b^2 c^2\right ) \tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )}{c^{5/2} (b c-a d)^3}-\frac{8 b^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{\sqrt{a} (a d-b c)^3}+\frac{d x (3 a d-7 b c)}{c^2 \left (c+d x^2\right ) (b c-a d)^2}-\frac{2 d x}{c \left (c+d x^2\right )^2 (b c-a d)}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0., size = 310, normalized size = 1.9 \begin{align*}{\frac{3\,{d}^{4}{x}^{3}{a}^{2}}{8\, \left ( ad-bc \right ) ^{3} \left ( d{x}^{2}+c \right ) ^{2}{c}^{2}}}-{\frac{5\,{d}^{3}{x}^{3}ab}{4\, \left ( ad-bc \right ) ^{3} \left ( d{x}^{2}+c \right ) ^{2}c}}+{\frac{7\,{d}^{2}{x}^{3}{b}^{2}}{8\, \left ( ad-bc \right ) ^{3} \left ( d{x}^{2}+c \right ) ^{2}}}+{\frac{5\,{d}^{3}x{a}^{2}}{8\, \left ( ad-bc \right ) ^{3} \left ( d{x}^{2}+c \right ) ^{2}c}}-{\frac{7\,{d}^{2}xab}{4\, \left ( ad-bc \right ) ^{3} \left ( d{x}^{2}+c \right ) ^{2}}}+{\frac{9\,cdx{b}^{2}}{8\, \left ( ad-bc \right ) ^{3} \left ( d{x}^{2}+c \right ) ^{2}}}+{\frac{3\,{a}^{2}{d}^{3}}{8\, \left ( ad-bc \right ) ^{3}{c}^{2}}\arctan \left ({dx{\frac{1}{\sqrt{cd}}}} \right ){\frac{1}{\sqrt{cd}}}}-{\frac{5\,a{d}^{2}b}{4\, \left ( ad-bc \right ) ^{3}c}\arctan \left ({dx{\frac{1}{\sqrt{cd}}}} \right ){\frac{1}{\sqrt{cd}}}}+{\frac{15\,{b}^{2}d}{8\, \left ( ad-bc \right ) ^{3}}\arctan \left ({dx{\frac{1}{\sqrt{cd}}}} \right ){\frac{1}{\sqrt{cd}}}}-{\frac{{b}^{3}}{ \left ( ad-bc \right ) ^{3}}\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)/(d*x^2+c)^3,x)

[Out]

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

________________________________________________________________________________________

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)/(d*x^2+c)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 6.45061, size = 3217, normalized size = 20.11 \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)/(d*x^2+c)^3,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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)/(d*x**2+c)**3,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.14651, size = 293, normalized size = 1.83 \begin{align*} \frac{b^{3} \arctan \left (\frac{b x}{\sqrt{a b}}\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{a b}} - \frac{{\left (15 \, b^{2} c^{2} d - 10 \, a b c d^{2} + 3 \, a^{2} d^{3}\right )} \arctan \left (\frac{d x}{\sqrt{c d}}\right )}{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 )} \sqrt{c d}} - \frac{7 \, b c d^{2} x^{3} - 3 \, a d^{3} x^{3} + 9 \, b c^{2} d x - 5 \, a c d^{2} x}{8 \,{\left (b^{2} c^{4} - 2 \, a b c^{3} d + a^{2} c^{2} d^{2}\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)/(d*x^2+c)^3,x, algorithm="giac")

[Out]

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