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

Optimal. Leaf size=103 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {b} c x}{\sqrt {a c^2-d^2}}\right )}{\sqrt {b} \sqrt {a c^2-d^2}}-\frac {\tan ^{-1}\left (\frac {\sqrt {b} d x}{\sqrt {a+b x^2} \sqrt {a c^2-d^2}}\right )}{\sqrt {b} \sqrt {a c^2-d^2}} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2156, 205, 377} \[ \frac {\tan ^{-1}\left (\frac {\sqrt {b} c x}{\sqrt {a c^2-d^2}}\right )}{\sqrt {b} \sqrt {a c^2-d^2}}-\frac {\tan ^{-1}\left (\frac {\sqrt {b} d x}{\sqrt {a+b x^2} \sqrt {a c^2-d^2}}\right )}{\sqrt {b} \sqrt {a c^2-d^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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]

Rubi steps

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

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Mathematica [A]  time = 0.13, size = 83, normalized size = 0.81 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {b} c x}{\sqrt {a c^2-d^2}}\right )-\tan ^{-1}\left (\frac {\sqrt {b} d x}{\sqrt {a+b x^2} \sqrt {a c^2-d^2}}\right )}{\sqrt {b} \sqrt {a c^2-d^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.48, size = 510, normalized size = 4.95 \[ \left [-\frac {\sqrt {-a b c^{2} + b d^{2}} \log \left (\frac {a^{4} c^{4} - 2 \, a^{3} c^{2} d^{2} + a^{2} d^{4} + {\left (a^{2} b^{2} c^{4} - 8 \, a b^{2} c^{2} d^{2} + 8 \, b^{2} d^{4}\right )} x^{4} + 2 \, {\left (a^{3} b c^{4} - 5 \, a^{2} b c^{2} d^{2} + 4 \, a b d^{4}\right )} x^{2} - 4 \, \sqrt {-a b c^{2} + b d^{2}} {\left ({\left (a b c^{2} d - 2 \, b d^{3}\right )} x^{3} + {\left (a^{2} c^{2} d - a d^{3}\right )} x\right )} \sqrt {b x^{2} + a}}{b^{2} c^{4} x^{4} + a^{2} c^{4} - 2 \, a c^{2} d^{2} + d^{4} + 2 \, {\left (a b c^{4} - b c^{2} d^{2}\right )} x^{2}}\right ) + 2 \, \sqrt {-a b c^{2} + b d^{2}} \log \left (\frac {b c^{2} x^{2} - a c^{2} - 2 \, \sqrt {-a b c^{2} + b d^{2}} c x + d^{2}}{b c^{2} x^{2} + a c^{2} - d^{2}}\right )}{4 \, {\left (a b c^{2} - b d^{2}\right )}}, -\frac {2 \, \sqrt {a b c^{2} - b d^{2}} \arctan \left (-\frac {\sqrt {a b c^{2} - b d^{2}} c x}{a c^{2} - d^{2}}\right ) - \sqrt {a b c^{2} - b d^{2}} \arctan \left (\frac {{\left (a^{2} c^{2} - a d^{2} + {\left (a b c^{2} - 2 \, b d^{2}\right )} x^{2}\right )} \sqrt {a b c^{2} - b d^{2}} \sqrt {b x^{2} + a}}{2 \, {\left ({\left (a b^{2} c^{2} d - b^{2} d^{3}\right )} x^{3} + {\left (a^{2} b c^{2} d - a b d^{3}\right )} x\right )}}\right )}{2 \, {\left (a b c^{2} - b d^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.36, size = 107, normalized size = 1.04 \[ \frac {\arctan \left (\frac {b c x}{\sqrt {a b c^{2} - b d^{2}}}\right )}{\sqrt {a b c^{2} - b d^{2}}} + \frac {\arctan \left (\frac {{\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} c^{2} + a c^{2} - 2 \, d^{2}}{2 \, \sqrt {a c^{2} - d^{2}} d}\right )}{\sqrt {a c^{2} - d^{2}} \sqrt {b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.04, size = 1995, normalized size = 19.37 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{b c x^{2} + a c + \sqrt {b x^{2} + a} d}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \left \{\begin {array}{cl} \frac {\mathrm {atan}\left (\frac {b\,c\,x}{\sqrt {a\,b\,c^2-b\,d^2}}\right )}{\sqrt {a\,b\,c^2-b\,d^2}}-\frac {d\,x}{\sqrt {a}\,\left (a\,c^2-d^2\right )} & \text {\ if\ \ }b=0\vee d=0\\ \frac {\mathrm {atan}\left (\frac {b\,c\,x}{\sqrt {a\,b\,c^2-b\,d^2}}\right )}{\sqrt {a\,b\,c^2-b\,d^2}}-\frac {d\,\mathrm {atan}\left (\frac {x\,\sqrt {a\,b\,c^2-b\,\left (a\,c^2-d^2\right )}}{\sqrt {a\,c^2-d^2}\,\sqrt {b\,x^2+a}}\right )}{\sqrt {-\left (a\,c^2-d^2\right )\,\left (b\,\left (a\,c^2-d^2\right )-a\,b\,c^2\right )}} & \text {\ if\ \ }0<b\,d^2\\ \frac {\mathrm {atan}\left (\frac {b\,c\,x}{\sqrt {a\,b\,c^2-b\,d^2}}\right )}{\sqrt {a\,b\,c^2-b\,d^2}}-\frac {d\,\ln \left (\frac {\sqrt {\left (a\,c^2-d^2\right )\,\left (b\,x^2+a\right )}+x\,\sqrt {b\,\left (a\,c^2-d^2\right )-a\,b\,c^2}}{\sqrt {\left (a\,c^2-d^2\right )\,\left (b\,x^2+a\right )}-x\,\sqrt {b\,\left (a\,c^2-d^2\right )-a\,b\,c^2}}\right )}{2\,\sqrt {\left (a\,c^2-d^2\right )\,\left (b\,\left (a\,c^2-d^2\right )-a\,b\,c^2\right )}} & \text {\ if\ \ }b\,d^2<0\\ \int \frac {1}{a\,c+d\,\sqrt {b\,x^2+a}+b\,c\,x^2} \,d x & \text {\ if\ \ }b\,d^2\notin \mathbb {R} \end {array}\right . \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

piecewise(b == 0 | d == 0, atan((b*c*x)/(- b*d^2 + a*b*c^2)^(1/2))/(- b*d^2 + a*b*c^2)^(1/2) - (d*x)/(a^(1/2)*
(a*c^2 - d^2)), 0 < b*d^2, atan((b*c*x)/(- b*d^2 + a*b*c^2)^(1/2))/(- b*d^2 + a*b*c^2)^(1/2) - (d*atan((x*(- b
*(a*c^2 - d^2) + a*b*c^2)^(1/2))/((a*c^2 - d^2)^(1/2)*(a + b*x^2)^(1/2))))/(-(a*c^2 - d^2)*(b*(a*c^2 - d^2) -
a*b*c^2))^(1/2), b*d^2 < 0, atan((b*c*x)/(- b*d^2 + a*b*c^2)^(1/2))/(- b*d^2 + a*b*c^2)^(1/2) - (d*log((((a*c^
2 - d^2)*(a + b*x^2))^(1/2) + x*(b*(a*c^2 - d^2) - a*b*c^2)^(1/2))/(((a*c^2 - d^2)*(a + b*x^2))^(1/2) - x*(b*(
a*c^2 - d^2) - a*b*c^2)^(1/2))))/(2*((a*c^2 - d^2)*(b*(a*c^2 - d^2) - a*b*c^2))^(1/2)), ~in(b*d^2, 'real'), in
t(1/(a*c + d*(a + b*x^2)^(1/2) + b*c*x^2), x))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{a c + b c x^{2} + d \sqrt {a + b x^{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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