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

Optimal. Leaf size=100 \[ \frac{(b c-2 a d) \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{2 a^{3/2} (b c-a d)^{3/2}}+\frac{b x \sqrt{c+d x^2}}{2 a \left (a+b x^2\right ) (b c-a d)} \]

[Out]

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

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Rubi [A]  time = 0.0529042, antiderivative size = 100, 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} \[ \frac{(b c-2 a d) \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{2 a^{3/2} (b c-a d)^{3/2}}+\frac{b x \sqrt{c+d x^2}}{2 a \left (a+b x^2\right ) (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [C]  time = 0.766238, size = 405, normalized size = 4.05 \[ \frac{x \sqrt{c+d x^2} \left (-30 d x^2 \sqrt{\frac{a x^2 \left (c+d x^2\right ) (b c-a d)}{c^2 \left (a+b x^2\right )^2}}-45 c \sqrt{\frac{a x^2 \left (c+d x^2\right ) (b c-a d)}{c^2 \left (a+b x^2\right )^2}}+16 d x^2 \sqrt{\frac{a \left (c+d x^2\right )}{c \left (a+b x^2\right )}} \left (\frac{x^2 (b c-a d)}{c \left (a+b x^2\right )}\right )^{5/2} \, _2F_1\left (2,3;\frac{7}{2};\frac{(b c-a d) x^2}{c \left (b x^2+a\right )}\right )+16 c \sqrt{\frac{a \left (c+d x^2\right )}{c \left (a+b x^2\right )}} \left (\frac{x^2 (b c-a d)}{c \left (a+b x^2\right )}\right )^{5/2} \, _2F_1\left (2,3;\frac{7}{2};\frac{(b c-a d) x^2}{c \left (b x^2+a\right )}\right )+30 d x^2 \sin ^{-1}\left (\sqrt{\frac{x^2 (b c-a d)}{c \left (a+b x^2\right )}}\right )+45 c \sin ^{-1}\left (\sqrt{\frac{x^2 (b c-a d)}{c \left (a+b x^2\right )}}\right )\right )}{30 c^2 \left (a+b x^2\right )^2 \left (\frac{x^2 (b c-a d)}{c \left (a+b x^2\right )}\right )^{3/2} \sqrt{\frac{a \left (c+d x^2\right )}{c \left (a+b x^2\right )}}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x*Sqrt[c + d*x^2]*(-45*c*Sqrt[(a*(b*c - a*d)*x^2*(c + d*x^2))/(c^2*(a + b*x^2)^2)] - 30*d*x^2*Sqrt[(a*(b*c -
a*d)*x^2*(c + d*x^2))/(c^2*(a + b*x^2)^2)] + 45*c*ArcSin[Sqrt[((b*c - a*d)*x^2)/(c*(a + b*x^2))]] + 30*d*x^2*A
rcSin[Sqrt[((b*c - a*d)*x^2)/(c*(a + b*x^2))]] + 16*c*(((b*c - a*d)*x^2)/(c*(a + b*x^2)))^(5/2)*Sqrt[(a*(c + d
*x^2))/(c*(a + b*x^2))]*Hypergeometric2F1[2, 3, 7/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] + 16*d*x^2*(((b*c - a*
d)*x^2)/(c*(a + b*x^2)))^(5/2)*Sqrt[(a*(c + d*x^2))/(c*(a + b*x^2))]*Hypergeometric2F1[2, 3, 7/2, ((b*c - a*d)
*x^2)/(c*(a + b*x^2))]))/(30*c^2*(((b*c - a*d)*x^2)/(c*(a + b*x^2)))^(3/2)*(a + b*x^2)^2*Sqrt[(a*(c + d*x^2))/
(c*(a + b*x^2))])

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Maple [B]  time = 0.01, size = 823, normalized size = 8.2 \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)^(1/2),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)-1/4/b/a*d*(-a*b)^(1/2)/(a*d-b*c)/(-(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)/(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)+1/4/b/a*d*(-a*b)^(1/2)/(a*d-b*c)/(-(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)^(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)^(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} \sqrt{d x^{2} + c}}\,{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)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 3.06731, size = 954, normalized size = 9.54 \begin{align*} \left [\frac{4 \,{\left (a b^{2} c - a^{2} b d\right )} \sqrt{d x^{2} + c} x -{\left (a b c - 2 \, a^{2} d +{\left (b^{2} c - 2 \, a b d\right )} x^{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 )}{8 \,{\left (a^{3} b^{2} c^{2} - 2 \, a^{4} b c d + a^{5} d^{2} +{\left (a^{2} b^{3} c^{2} - 2 \, a^{3} b^{2} c d + a^{4} b d^{2}\right )} x^{2}\right )}}, \frac{2 \,{\left (a b^{2} c - a^{2} b d\right )} \sqrt{d x^{2} + c} x + \sqrt{a b c - a^{2} d}{\left (a b c - 2 \, a^{2} d +{\left (b^{2} c - 2 \, a b d\right )} x^{2}\right )} \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 )}{4 \,{\left (a^{3} b^{2} c^{2} - 2 \, a^{4} b c d + a^{5} d^{2} +{\left (a^{2} b^{3} c^{2} - 2 \, a^{3} b^{2} c d + a^{4} b d^{2}\right )} x^{2}\right )}}\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)^(1/2),x, algorithm="fricas")

[Out]

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

[Out]

Exception raised: ValueError

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Giac [B]  time = 1.14046, size = 304, normalized size = 3.04 \begin{align*} -\frac{1}{2} \, d^{\frac{3}{2}}{\left (\frac{{\left (b c - 2 \, a d\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 )}{{\left (a b c d - a^{2} d^{2}\right )}^{\frac{3}{2}}} + \frac{2 \,{\left ({\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} b c - 2 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} a d - b c^{2}\right )}}{{\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 c d - a^{2} d^{2}\right )}}\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)^(1/2),x, algorithm="giac")

[Out]

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