3.6.51 \(\int \frac {1}{x^2 (a c+b c x^2+d \sqrt {a+b x^2})} \, dx\) [551]

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

[Out]

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

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Rubi [A]
time = 0.17, antiderivative size = 160, normalized size of antiderivative = 1.00, 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 = {2187, 331, 211, 491, 12, 385} \begin {gather*} \frac {\sqrt {b} c^2 \text {ArcTan}\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 \text {ArcTan}\left (\frac {\sqrt {b} c x}{\sqrt {a c^2-d^2}}\right )}{\left (a c^2-d^2\right )^{3/2}}+\frac {d \sqrt {a+b x^2}}{a x \left (a c^2-d^2\right )}-\frac {c}{x \left (a c^2-d^2\right )} \end {gather*}

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 12

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

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 331

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 385

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 491

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 2187

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]

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 ) \text {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*}

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Mathematica [A]
time = 0.35, size = 105, normalized size = 0.66 \begin {gather*} -\frac {a c-d \sqrt {a+b x^2}}{a^2 c^2 x-a d^2 x}+\frac {2 \sqrt {b} c^2 \tan ^{-1}\left (\frac {d+c \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{\sqrt {a c^2-d^2}}\right )}{\left (a c^2-d^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(2288\) vs. \(2(140)=280\).
time = 0.04, size = 2289, normalized size = 14.31

method	result	size
default	\(\text {Expression too large to display}\)	\(2289\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.42, size = 581, normalized size = 3.63 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{2} \left (a c + b c x^{2} + d \sqrt {a + b x^{2}}\right )}\, dx \end {gather*}

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 = 3.14, size = 211, normalized size = 1.32 \begin {gather*} -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 {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x^2\,\left (a\,c+d\,\sqrt {b\,x^2+a}+b\,c\,x^2\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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