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

Optimal. Leaf size=142 \[ \frac{b (b c-4 a d) \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{2 a^{3/2} (b c-a d)^{5/2}}+\frac{d x (2 a d+b c)}{2 a c \sqrt{c+d x^2} (b c-a d)^2}+\frac{b x}{2 a \left (a+b x^2\right ) \sqrt{c+d x^2} (b c-a d)} \]

[Out]

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

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Rubi [A]  time = 0.102182, antiderivative size = 142, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {414, 527, 12, 377, 205} \[ \frac{b (b c-4 a d) \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{2 a^{3/2} (b c-a d)^{5/2}}+\frac{d x (2 a d+b c)}{2 a c \sqrt{c+d x^2} (b c-a d)^2}+\frac{b x}{2 a \left (a+b x^2\right ) \sqrt{c+d x^2} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 12

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

Rule 377

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

Mathematica [C]  time = 4.4101, size = 210, normalized size = 1.48 \[ \frac{x \left (16 x^2 \left (c+d x^2\right )^2 (b c-a d) \text{HypergeometricPFQ}\left (\{2,2,3\},\left \{1,\frac{9}{2}\right \},\frac{x^2 (b c-a d)}{c \left (a+b x^2\right )}\right )+16 x^2 \left (4 c^2+7 c d x^2+3 d^2 x^4\right ) (b c-a d) \, _2F_1\left (2,3;\frac{9}{2};\frac{(b c-a d) x^2}{c \left (b x^2+a\right )}\right )+7 c \left (a+b x^2\right ) \left (15 c^2+20 c d x^2+8 d^2 x^4\right ) \, _2F_1\left (1,2;\frac{7}{2};\frac{(b c-a d) x^2}{c \left (b x^2+a\right )}\right )\right )}{105 c^4 \left (a+b x^2\right )^3 \sqrt{c+d x^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x*(7*c*(a + b*x^2)*(15*c^2 + 20*c*d*x^2 + 8*d^2*x^4)*Hypergeometric2F1[1, 2, 7/2, ((b*c - a*d)*x^2)/(c*(a + b
*x^2))] + 16*(b*c - a*d)*x^2*(4*c^2 + 7*c*d*x^2 + 3*d^2*x^4)*Hypergeometric2F1[2, 3, 9/2, ((b*c - a*d)*x^2)/(c
*(a + b*x^2))] + 16*(b*c - a*d)*x^2*(c + d*x^2)^2*HypergeometricPFQ[{2, 2, 3}, {1, 9/2}, ((b*c - a*d)*x^2)/(c*
(a + b*x^2))]))/(105*c^4*(a + b*x^2)^3*Sqrt[c + d*x^2])

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Maple [B]  time = 0.01, size = 1461, normalized size = 10.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/a/(a*d-b*c)/(x+1/b*(-a*b)^(1/2))/((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b
*c)/b)^(1/2)-3/4/a*d*(-a*b)^(1/2)/(a*d-b*c)^2/((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2)
)-(a*d-b*c)/b)^(1/2)+3/4*d^2/(a*d-b*c)^2/c/((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(
a*d-b*c)/b)^(1/2)*x+3/4/a*d*(-a*b)^(1/2)/(a*d-b*c)^2/(-(a*d-b*c)/b)^(1/2)*ln((-2*(a*d-b*c)/b-2*d*(-a*b)^(1/2)/
b*(x+1/b*(-a*b)^(1/2))+2*(-(a*d-b*c)/b)^(1/2)*((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2)
)-(a*d-b*c)/b)^(1/2))/(x+1/b*(-a*b)^(1/2)))-1/4/a/(a*d-b*c)/c/((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+
1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)*x*d-1/4/a/(a*d-b*c)/(x-1/b*(-a*b)^(1/2))/((x-1/b*(-a*b)^(1/2))^2*d+2*d*(-
a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)+3/4/a*d*(-a*b)^(1/2)/(a*d-b*c)^2/((x-1/b*(-a*b)^(1/2))^2*
d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)+3/4*d^2/(a*d-b*c)^2/c/((x-1/b*(-a*b)^(1/2))^2*d+2
*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)*x-3/4/a*d*(-a*b)^(1/2)/(a*d-b*c)^2/(-(a*d-b*c)/b)^(1
/2)*ln((-2*(a*d-b*c)/b+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))+2*(-(a*d-b*c)/b)^(1/2)*((x-1/b*(-a*b)^(1/2))^2*
d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))/(x-1/b*(-a*b)^(1/2)))-1/4/a/(a*d-b*c)/c/((x-1/b*
(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)*x*d+1/4/a/(-a*b)^(1/2)/(a*d-b*c)*
b/((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)-1/4/a/(-a*b)^(1/2)/(a*d
-b*c)*b/(-(a*d-b*c)/b)^(1/2)*ln((-2*(a*d-b*c)/b-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))+2*(-(a*d-b*c)/b)^(1/2)
*((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))/(x+1/b*(-a*b)^(1/2)))-1
/4/a/(-a*b)^(1/2)/(a*d-b*c)*b/((x-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(
1/2)+1/4/a/(-a*b)^(1/2)/(a*d-b*c)*b/(-(a*d-b*c)/b)^(1/2)*ln((-2*(a*d-b*c)/b+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(
1/2))+2*(-(a*d-b*c)/b)^(1/2)*((x-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1
/2))/(x-1/b*(-a*b)^(1/2)))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{2} + a\right )}^{2}{\left (d x^{2} + c\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 5.1823, size = 1705, normalized size = 12.01 \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)^(3/2),x, algorithm="fricas")

[Out]

[1/8*((a*b^2*c^3 - 4*a^2*b*c^2*d + (b^3*c^2*d - 4*a*b^2*c*d^2)*x^4 + (b^3*c^3 - 3*a*b^2*c^2*d - 4*a^2*b*c*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)*sqrt(-a*b*c + a^2*d)*sqrt(d*x^2 + c))/(b^2*x^4 + 2*a*b*x^2 + a^2)) + 4*((a*b
^3*c^2*d + a^2*b^2*c*d^2 - 2*a^3*b*d^3)*x^3 + (a*b^3*c^3 - a^2*b^2*c^2*d + 2*a^3*b*c*d^2 - 2*a^4*d^3)*x)*sqrt(
d*x^2 + c))/(a^3*b^3*c^5 - 3*a^4*b^2*c^4*d + 3*a^5*b*c^3*d^2 - a^6*c^2*d^3 + (a^2*b^4*c^4*d - 3*a^3*b^3*c^3*d^
2 + 3*a^4*b^2*c^2*d^3 - a^5*b*c*d^4)*x^4 + (a^2*b^4*c^5 - 2*a^3*b^3*c^4*d + 2*a^5*b*c^2*d^3 - a^6*c*d^4)*x^2),
1/4*((a*b^2*c^3 - 4*a^2*b*c^2*d + (b^3*c^2*d - 4*a*b^2*c*d^2)*x^4 + (b^3*c^3 - 3*a*b^2*c^2*d - 4*a^2*b*c*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*((a*b^3*c^2*d + a^2*b^2*c*d^2 - 2*a^3*b*d^3)*x^3 + (a*b^3*c^3 - a^
2*b^2*c^2*d + 2*a^3*b*c*d^2 - 2*a^4*d^3)*x)*sqrt(d*x^2 + c))/(a^3*b^3*c^5 - 3*a^4*b^2*c^4*d + 3*a^5*b*c^3*d^2
- a^6*c^2*d^3 + (a^2*b^4*c^4*d - 3*a^3*b^3*c^3*d^2 + 3*a^4*b^2*c^2*d^3 - a^5*b*c*d^4)*x^4 + (a^2*b^4*c^5 - 2*a
^3*b^3*c^4*d + 2*a^5*b*c^2*d^3 - a^6*c*d^4)*x^2)]

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Sympy [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)**(3/2),x)

[Out]

Exception raised: ValueError

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Giac [B]  time = 5.26109, size = 429, normalized size = 3.02 \begin{align*} \frac{d^{2} x}{{\left (b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2}\right )} \sqrt{d x^{2} + c}} - \frac{{\left (b^{2} c \sqrt{d} - 4 \, a b 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^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2}\right )} \sqrt{a b c d - a^{2} d^{2}}} - \frac{{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} b^{2} c \sqrt{d} - 2 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} a b d^{\frac{3}{2}} - b^{2} c^{2} \sqrt{d}}{{\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 )}{\left (a b^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d^2*x/((b^2*c^3 - 2*a*b*c^2*d + a^2*c*d^2)*sqrt(d*x^2 + c)) - 1/2*(b^2*c*sqrt(d) - 4*a*b*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^2*c^2 - 2*a^2*b*c*d + a^3*d^2)
*sqrt(a*b*c*d - a^2*d^2)) - ((sqrt(d)*x - sqrt(d*x^2 + c))^2*b^2*c*sqrt(d) - 2*(sqrt(d)*x - sqrt(d*x^2 + c))^2
*a*b*d^(3/2) - b^2*c^2*sqrt(d))/(((sqrt(d)*x - sqrt(d*x^2 + c))^4*b - 2*(sqrt(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)*(a*b^2*c^2 - 2*a^2*b*c*d + a^3*d^2))