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

Optimal. Leaf size=79 \[ \frac{b \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{\sqrt{a} (b c-a d)^{3/2}}-\frac{d x}{c \sqrt{c+d x^2} (b c-a d)} \]

[Out]

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

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Rubi [A]  time = 0.046164, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {382, 377, 205} \[ \frac{b \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{\sqrt{a} (b c-a d)^{3/2}}-\frac{d x}{c \sqrt{c+d x^2} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 382

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[(b*c + n*(p + 1)*(b*c - a*d))/(a*n*(p + 1)*(b*c - a*d
)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, q}, x] && NeQ[b*c - a*d, 0] && EqQ[
n*(p + q + 2) + 1, 0] && (LtQ[p, -1] ||  !LtQ[q, -1]) && NeQ[p, -1]

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

Mathematica [C]  time = 2.58988, size = 236, normalized size = 2.99 \[ -\frac{\frac{15 c \left (3 c+2 d x^2\right ) \left (c \left (a+b x^2\right ) \sqrt{\frac{a x^2 \left (c+d x^2\right ) (b c-a d)}{c^2 \left (a+b x^2\right )^2}}-a \left (c+d x^2\right ) \sin ^{-1}\left (\sqrt{\frac{x^2 (b c-a d)}{c \left (a+b x^2\right )}}\right )\right )}{\sqrt{\frac{a x^2 \left (c+d x^2\right ) (b c-a d)}{c^2 \left (a+b x^2\right )^2}}}+\frac{4 x^4 \left (c+d x^2\right ) (b c-a d)^2 \, _2F_1\left (2,2;\frac{7}{2};\frac{(b c-a d) x^2}{c \left (b x^2+a\right )}\right )}{a+b x^2}}{15 c^3 x \left (a+b x^2\right ) \sqrt{c+d x^2} (a d-b c)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-((15*c*(3*c + 2*d*x^2)*(c*(a + b*x^2)*Sqrt[(a*(b*c - a*d)*x^2*(c + d*x^2))/(c^2*(a + b*x^2)^2)] - a*(c + d*x^
2)*ArcSin[Sqrt[((b*c - a*d)*x^2)/(c*(a + b*x^2))]]))/Sqrt[(a*(b*c - a*d)*x^2*(c + d*x^2))/(c^2*(a + b*x^2)^2)]
+ (4*(b*c - a*d)^2*x^4*(c + d*x^2)*Hypergeometric2F1[2, 2, 7/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))])/(a + b*x^
2))/(15*c^3*(-(b*c) + a*d)*x*(a + b*x^2)*Sqrt[c + d*x^2])

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Maple [B]  time = 0.033, size = 628, normalized size = 8. \begin{align*}{\frac{b}{2\,ad-2\,bc}{\frac{1}{\sqrt{-ab}}}{\frac{1}{\sqrt{ \left ( x+{\frac{1}{b}\sqrt{-ab}} \right ) ^{2}d-2\,{\frac{d\sqrt{-ab}}{b} \left ( x+{\frac{\sqrt{-ab}}{b}} \right ) }-{\frac{ad-bc}{b}}}}}}+{\frac{dx}{ \left ( 2\,ad-2\,bc \right ) c}{\frac{1}{\sqrt{ \left ( x+{\frac{1}{b}\sqrt{-ab}} \right ) ^{2}d-2\,{\frac{d\sqrt{-ab}}{b} \left ( x+{\frac{\sqrt{-ab}}{b}} \right ) }-{\frac{ad-bc}{b}}}}}}-{\frac{b}{2\,ad-2\,bc}\ln \left ({ \left ( -2\,{\frac{ad-bc}{b}}-2\,{\frac{d\sqrt{-ab}}{b} \left ( x+{\frac{\sqrt{-ab}}{b}} \right ) }+2\,\sqrt{-{\frac{ad-bc}{b}}}\sqrt{ \left ( x+{\frac{\sqrt{-ab}}{b}} \right ) ^{2}d-2\,{\frac{d\sqrt{-ab}}{b} \left ( x+{\frac{\sqrt{-ab}}{b}} \right ) }-{\frac{ad-bc}{b}}} \right ) \left ( x+{\frac{1}{b}\sqrt{-ab}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{-ab}}}{\frac{1}{\sqrt{-{\frac{ad-bc}{b}}}}}}-{\frac{b}{2\,ad-2\,bc}{\frac{1}{\sqrt{-ab}}}{\frac{1}{\sqrt{ \left ( x-{\frac{1}{b}\sqrt{-ab}} \right ) ^{2}d+2\,{\frac{d\sqrt{-ab}}{b} \left ( x-{\frac{\sqrt{-ab}}{b}} \right ) }-{\frac{ad-bc}{b}}}}}}+{\frac{dx}{ \left ( 2\,ad-2\,bc \right ) c}{\frac{1}{\sqrt{ \left ( x-{\frac{1}{b}\sqrt{-ab}} \right ) ^{2}d+2\,{\frac{d\sqrt{-ab}}{b} \left ( x-{\frac{\sqrt{-ab}}{b}} \right ) }-{\frac{ad-bc}{b}}}}}}+{\frac{b}{2\,ad-2\,bc}\ln \left ({ \left ( -2\,{\frac{ad-bc}{b}}+2\,{\frac{d\sqrt{-ab}}{b} \left ( x-{\frac{\sqrt{-ab}}{b}} \right ) }+2\,\sqrt{-{\frac{ad-bc}{b}}}\sqrt{ \left ( x-{\frac{\sqrt{-ab}}{b}} \right ) ^{2}d+2\,{\frac{d\sqrt{-ab}}{b} \left ( x-{\frac{\sqrt{-ab}}{b}} \right ) }-{\frac{ad-bc}{b}}} \right ) \left ( x-{\frac{1}{b}\sqrt{-ab}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{-ab}}}{\frac{1}{\sqrt{-{\frac{ad-bc}{b}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/(-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)+1/2/(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
/2/(-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)))-1/2/(-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)+1/2/(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+1/2/(-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(-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)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.30654, size = 914, normalized size = 11.57 \begin{align*} \left [-\frac{4 \,{\left (a b c d - a^{2} d^{2}\right )} \sqrt{d x^{2} + c} x -{\left (b c d x^{2} + b c^{2}\right )} \sqrt{-a b c + a^{2} d} \log \left (\frac{{\left (b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{4} + a^{2} c^{2} - 2 \,{\left (3 \, a b c^{2} - 4 \, a^{2} c d\right )} x^{2} + 4 \,{\left ({\left (b c - 2 \, a d\right )} x^{3} - a c x\right )} \sqrt{-a b c + a^{2} d} \sqrt{d x^{2} + c}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right )}{4 \,{\left (a b^{2} c^{4} - 2 \, a^{2} b c^{3} d + a^{3} c^{2} d^{2} +{\left (a b^{2} c^{3} d - 2 \, a^{2} b c^{2} d^{2} + a^{3} c d^{3}\right )} x^{2}\right )}}, -\frac{2 \,{\left (a b c d - a^{2} d^{2}\right )} \sqrt{d x^{2} + c} x -{\left (b c d x^{2} + b c^{2}\right )} \sqrt{a b c - a^{2} d} \arctan \left (\frac{\sqrt{a b c - a^{2} d}{\left ({\left (b c - 2 \, a d\right )} x^{2} - a c\right )} \sqrt{d x^{2} + c}}{2 \,{\left ({\left (a b c d - a^{2} d^{2}\right )} x^{3} +{\left (a b c^{2} - a^{2} c d\right )} x\right )}}\right )}{2 \,{\left (a b^{2} c^{4} - 2 \, a^{2} b c^{3} d + a^{3} c^{2} d^{2} +{\left (a b^{2} c^{3} d - 2 \, a^{2} b c^{2} d^{2} + a^{3} c d^{3}\right )} x^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.1703, size = 144, normalized size = 1.82 \begin{align*} \frac{b \sqrt{d} \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 )}{\sqrt{a b c d - a^{2} d^{2}}{\left (b c - a d\right )}} - \frac{d x}{{\left (b c^{2} - a c d\right )} \sqrt{d x^{2} + c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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