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

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

[Out]

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

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Rubi [A]  time = 0.240101, antiderivative size = 160, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.207, Rules used = {2156, 325, 205, 480, 12, 377} \[ \frac{d \sqrt{a+b x^2}}{a x \left (a c^2-d^2\right )}+\frac{\sqrt{b} c^2 \tan ^{-1}\left (\frac{\sqrt{b} d x}{\sqrt{a+b x^2} \sqrt{a c^2-d^2}}\right )}{\left (a c^2-d^2\right )^{3/2}}-\frac{\sqrt{b} c^2 \tan ^{-1}\left (\frac{\sqrt{b} c x}{\sqrt{a c^2-d^2}}\right )}{\left (a c^2-d^2\right )^{3/2}}-\frac{c}{x \left (a c^2-d^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2156

Int[(u_.)/((c_) + (d_.)*(x_)^(n_) + (e_.)*Sqrt[(a_) + (b_.)*(x_)^(n_)]), x_Symbol] :> Dist[c, Int[u/(c^2 - a*e
^2 + c*d*x^n), x], x] - Dist[a*e, Int[u/((c^2 - a*e^2 + c*d*x^n)*Sqrt[a + b*x^n]), x], x] /; FreeQ[{a, b, c, d
, e, n}, x] && EqQ[b*c - a*d, 0]

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, 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]

Rule 480

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[((e*x)^(m
+ 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*c*e*(m + 1)), x] - Dist[1/(a*c*e^n*(m + 1)), Int[(e*x)^(m +
n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[(b*c + a*d)*(m + n + 1) + n*(b*c*p + a*d*q) + b*d*(m + n*(p + q + 2) + 1)*
x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -1] && IntBino
mialQ[a, b, c, d, e, m, n, p, q, x]

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]

Rubi steps

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

Mathematica [A]  time = 0.276195, size = 139, normalized size = 0.87 \[ \frac{\sqrt{a c^2-d^2} \left (d \sqrt{a+b x^2}-a c\right )+a \sqrt{b} c^2 x \tan ^{-1}\left (\frac{\sqrt{b} d x}{\sqrt{a+b x^2} \sqrt{a c^2-d^2}}\right )-a \sqrt{b} c^2 x \tan ^{-1}\left (\frac{\sqrt{b} c x}{\sqrt{a c^2-d^2}}\right )}{a x \left (a c^2-d^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.027, size = 2289, normalized size = 14.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*c^2/d^2/(b*(a*c^2-d^2))^(1/2)*arctan(x*c*b/(b*(a*c^2-d^2))^(1/2))-a*c^4/(a*c^2-d^2)*b/d^2/(b*(a*c^2-d^2))^(1
/2)*arctan(x*c*b/(b*(a*c^2-d^2))^(1/2))-c/(a*c^2-d^2)/x-1/2*d*c^2*b^2/a/(-a*b)^(1/2)/((-a*b)^(1/2)*c^2+(-c^2*b
*(a*c^2-d^2))^(1/2))/((-a*b)^(1/2)*c^2-(-c^2*b*(a*c^2-d^2))^(1/2))*((x+(-a*b)^(1/2)/b)^2*b-2*(-a*b)^(1/2)*(x+(
-a*b)^(1/2)/b))^(1/2)+1/2*d*c^2*b^(3/2)/a/((-a*b)^(1/2)*c^2+(-c^2*b*(a*c^2-d^2))^(1/2))/((-a*b)^(1/2)*c^2-(-c^
2*b*(a*c^2-d^2))^(1/2))*ln((b*(x+(-a*b)^(1/2)/b)-(-a*b)^(1/2))/b^(1/2)+((x+(-a*b)^(1/2)/b)^2*b-2*(-a*b)^(1/2)*
(x+(-a*b)^(1/2)/b))^(1/2))-1/2*d*c^6*b^2/(a*c^2-d^2)/((-a*b)^(1/2)*c^2+(-c^2*b*(a*c^2-d^2))^(1/2))/((-a*b)^(1/
2)*c^2-(-c^2*b*(a*c^2-d^2))^(1/2))/(-c^2*b*(a*c^2-d^2))^(1/2)*((x-(-c^2*b*(a*c^2-d^2))^(1/2)/c^2/b)^2*b+2*(-c^
2*b*(a*c^2-d^2))^(1/2)/c^2*(x-(-c^2*b*(a*c^2-d^2))^(1/2)/c^2/b)+1/c^2*d^2)^(1/2)-1/2*d*c^4*b^(3/2)/(a*c^2-d^2)
/((-a*b)^(1/2)*c^2+(-c^2*b*(a*c^2-d^2))^(1/2))/((-a*b)^(1/2)*c^2-(-c^2*b*(a*c^2-d^2))^(1/2))*ln(((-c^2*b*(a*c^
2-d^2))^(1/2)/c^2+b*(x-(-c^2*b*(a*c^2-d^2))^(1/2)/c^2/b))/b^(1/2)+((x-(-c^2*b*(a*c^2-d^2))^(1/2)/c^2/b)^2*b+2*
(-c^2*b*(a*c^2-d^2))^(1/2)/c^2*(x-(-c^2*b*(a*c^2-d^2))^(1/2)/c^2/b)+1/c^2*d^2)^(1/2))+1/2*c^4*b^2/(a*c^2-d^2)/
((-a*b)^(1/2)*c^2+(-c^2*b*(a*c^2-d^2))^(1/2))/((-a*b)^(1/2)*c^2-(-c^2*b*(a*c^2-d^2))^(1/2))/(-c^2*b*(a*c^2-d^2
))^(1/2)*d^3/(1/c^2*d^2)^(1/2)*ln((2/c^2*d^2+2*(-c^2*b*(a*c^2-d^2))^(1/2)/c^2*(x-(-c^2*b*(a*c^2-d^2))^(1/2)/c^
2/b)+2*(1/c^2*d^2)^(1/2)*((x-(-c^2*b*(a*c^2-d^2))^(1/2)/c^2/b)^2*b+2*(-c^2*b*(a*c^2-d^2))^(1/2)/c^2*(x-(-c^2*b
*(a*c^2-d^2))^(1/2)/c^2/b)+1/c^2*d^2)^(1/2))/(x-(-c^2*b*(a*c^2-d^2))^(1/2)/c^2/b))+d/(a*c^2-d^2)/a^2/x*(b*x^2+
a)^(3/2)-d/(a*c^2-d^2)/a^2*b*x*(b*x^2+a)^(1/2)-d/(a*c^2-d^2)/a*b^(1/2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2))+1/2*d*c^6
*b^2/(a*c^2-d^2)/((-a*b)^(1/2)*c^2+(-c^2*b*(a*c^2-d^2))^(1/2))/((-a*b)^(1/2)*c^2-(-c^2*b*(a*c^2-d^2))^(1/2))/(
-c^2*b*(a*c^2-d^2))^(1/2)*((x+(-c^2*b*(a*c^2-d^2))^(1/2)/c^2/b)^2*b-2*(-c^2*b*(a*c^2-d^2))^(1/2)/c^2*(x+(-c^2*
b*(a*c^2-d^2))^(1/2)/c^2/b)+1/c^2*d^2)^(1/2)-1/2*d*c^4*b^(3/2)/(a*c^2-d^2)/((-a*b)^(1/2)*c^2+(-c^2*b*(a*c^2-d^
2))^(1/2))/((-a*b)^(1/2)*c^2-(-c^2*b*(a*c^2-d^2))^(1/2))*ln((-(-c^2*b*(a*c^2-d^2))^(1/2)/c^2+b*(x+(-c^2*b*(a*c
^2-d^2))^(1/2)/c^2/b))/b^(1/2)+((x+(-c^2*b*(a*c^2-d^2))^(1/2)/c^2/b)^2*b-2*(-c^2*b*(a*c^2-d^2))^(1/2)/c^2*(x+(
-c^2*b*(a*c^2-d^2))^(1/2)/c^2/b)+1/c^2*d^2)^(1/2))-1/2*c^4*b^2/(a*c^2-d^2)/((-a*b)^(1/2)*c^2+(-c^2*b*(a*c^2-d^
2))^(1/2))/((-a*b)^(1/2)*c^2-(-c^2*b*(a*c^2-d^2))^(1/2))/(-c^2*b*(a*c^2-d^2))^(1/2)*d^3/(1/c^2*d^2)^(1/2)*ln((
2/c^2*d^2-2*(-c^2*b*(a*c^2-d^2))^(1/2)/c^2*(x+(-c^2*b*(a*c^2-d^2))^(1/2)/c^2/b)+2*(1/c^2*d^2)^(1/2)*((x+(-c^2*
b*(a*c^2-d^2))^(1/2)/c^2/b)^2*b-2*(-c^2*b*(a*c^2-d^2))^(1/2)/c^2*(x+(-c^2*b*(a*c^2-d^2))^(1/2)/c^2/b)+1/c^2*d^
2)^(1/2))/(x+(-c^2*b*(a*c^2-d^2))^(1/2)/c^2/b))+1/2*d*c^2*b^2/a/(-a*b)^(1/2)/((-a*b)^(1/2)*c^2+(-c^2*b*(a*c^2-
d^2))^(1/2))/((-a*b)^(1/2)*c^2-(-c^2*b*(a*c^2-d^2))^(1/2))*((x-(-a*b)^(1/2)/b)^2*b+2*(-a*b)^(1/2)*(x-(-a*b)^(1
/2)/b))^(1/2)+1/2*d*c^2*b^(3/2)/a/((-a*b)^(1/2)*c^2+(-c^2*b*(a*c^2-d^2))^(1/2))/((-a*b)^(1/2)*c^2-(-c^2*b*(a*c
^2-d^2))^(1/2))*ln((b*(x-(-a*b)^(1/2)/b)+(-a*b)^(1/2))/b^(1/2)+((x-(-a*b)^(1/2)/b)^2*b+2*(-a*b)^(1/2)*(x-(-a*b
)^(1/2)/b))^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.84793, size = 1149, normalized size = 7.18 \begin{align*} \left [-\frac{a c^{2} x \sqrt{-\frac{b}{a c^{2} - d^{2}}} \log \left (\frac{a^{4} c^{4} - 2 \, a^{3} c^{2} d^{2} + a^{2} d^{4} +{\left (a^{2} b^{2} c^{4} - 8 \, a b^{2} c^{2} d^{2} + 8 \, b^{2} d^{4}\right )} x^{4} + 2 \,{\left (a^{3} b c^{4} - 5 \, a^{2} b c^{2} d^{2} + 4 \, a b d^{4}\right )} x^{2} + 4 \,{\left ({\left (a^{2} b c^{4} d - 3 \, a b c^{2} d^{3} + 2 \, b d^{5}\right )} x^{3} +{\left (a^{3} c^{4} d - 2 \, a^{2} c^{2} d^{3} + a d^{5}\right )} x\right )} \sqrt{b x^{2} + a} \sqrt{-\frac{b}{a c^{2} - d^{2}}}}{b^{2} c^{4} x^{4} + a^{2} c^{4} - 2 \, a c^{2} d^{2} + d^{4} + 2 \,{\left (a b c^{4} - b c^{2} d^{2}\right )} x^{2}}\right ) + 2 \, a c^{2} x \sqrt{-\frac{b}{a c^{2} - d^{2}}} \log \left (\frac{b c^{2} x^{2} - a c^{2} + 2 \,{\left (a c^{3} - c d^{2}\right )} x \sqrt{-\frac{b}{a c^{2} - d^{2}}} + d^{2}}{b c^{2} x^{2} + a c^{2} - d^{2}}\right ) + 4 \, a c - 4 \, \sqrt{b x^{2} + a} d}{4 \,{\left (a^{2} c^{2} - a d^{2}\right )} x}, -\frac{2 \, a c^{2} x \sqrt{\frac{b}{a c^{2} - d^{2}}} \arctan \left (c x \sqrt{\frac{b}{a c^{2} - d^{2}}}\right ) - a c^{2} x \sqrt{\frac{b}{a c^{2} - d^{2}}} \arctan \left (-\frac{{\left (a^{2} c^{2} - a d^{2} +{\left (a b c^{2} - 2 \, b d^{2}\right )} x^{2}\right )} \sqrt{b x^{2} + a} \sqrt{\frac{b}{a c^{2} - d^{2}}}}{2 \,{\left (b^{2} d x^{3} + a b d x\right )}}\right ) + 2 \, a c - 2 \, \sqrt{b x^{2} + a} d}{2 \,{\left (a^{2} c^{2} - a d^{2}\right )} x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.16282, size = 285, normalized size = 1.78 \begin{align*} -b^{\frac{3}{2}} d{\left (\frac{c^{2} \arctan \left (\frac{{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} c^{2} + a c^{2} - 2 \, d^{2}}{2 \, \sqrt{a c^{2} - d^{2}} d}\right )}{{\left (a b c^{2} - b d^{2}\right )} \sqrt{a c^{2} - d^{2}} d} + \frac{2}{{\left (a b c^{2} - b d^{2}\right )}{\left ({\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} - a\right )}}\right )} - \frac{b c^{2} \arctan \left (\frac{b c x}{\sqrt{a b c^{2} - b d^{2}}}\right )}{\sqrt{a b c^{2} - b d^{2}}{\left (a c^{2} - d^{2}\right )}} - \frac{c}{{\left (a c^{2} - d^{2}\right )} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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