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

Optimal. Leaf size=21 \[ \frac {\tan ^{-1}\left (\frac {a+b x}{\sqrt {c}}\right )}{b \sqrt {c}} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {247, 203} \[ \frac {\tan ^{-1}\left (\frac {a+b x}{\sqrt {c}}\right )}{b \sqrt {c}} \]

Antiderivative was successfully verified.

[In]

Int[(c + (a + b*x)^2)^(-1),x]

[Out]

ArcTan[(a + b*x)/Sqrt[c]]/(b*Sqrt[c])

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 247

Int[((a_.) + (b_.)*(v_)^(n_))^(p_), x_Symbol] :> Dist[1/Coefficient[v, x, 1], Subst[Int[(a + b*x^n)^p, x], x,
v], x] /; FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && NeQ[v, x]

Rubi steps

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

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Mathematica [A]  time = 0.00, size = 21, normalized size = 1.00 \[ \frac {\tan ^{-1}\left (\frac {a+b x}{\sqrt {c}}\right )}{b \sqrt {c}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + (a + b*x)^2)^(-1),x]

[Out]

ArcTan[(a + b*x)/Sqrt[c]]/(b*Sqrt[c])

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fricas [A]  time = 0.66, size = 83, normalized size = 3.95 \[ \left [-\frac {\sqrt {-c} \log \left (\frac {b^{2} x^{2} + 2 \, a b x + a^{2} - 2 \, {\left (b x + a\right )} \sqrt {-c} - c}{b^{2} x^{2} + 2 \, a b x + a^{2} + c}\right )}{2 \, b c}, \frac {\arctan \left (\frac {b x + a}{\sqrt {c}}\right )}{b \sqrt {c}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.38, size = 17, normalized size = 0.81 \[ \frac {\arctan \left (\frac {b x + a}{\sqrt {c}}\right )}{b \sqrt {c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arctan((b*x + a)/sqrt(c))/(b*sqrt(c))

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maple [A]  time = 0.00, size = 28, normalized size = 1.33 \[ \frac {\arctan \left (\frac {2 b^{2} x +2 b a}{2 b \sqrt {c}}\right )}{b \sqrt {c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.52, size = 24, normalized size = 1.14 \[ \frac {\arctan \left (\frac {b^{2} x + a b}{b \sqrt {c}}\right )}{b \sqrt {c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.04, size = 17, normalized size = 0.81 \[ \frac {\mathrm {atan}\left (\frac {a+b\,x}{\sqrt {c}}\right )}{b\,\sqrt {c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c + (a + b*x)^2),x)

[Out]

atan((a + b*x)/c^(1/2))/(b*c^(1/2))

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sympy [B]  time = 0.19, size = 54, normalized size = 2.57 \[ \frac {- \frac {\sqrt {- \frac {1}{c}} \log {\left (x + \frac {a - c \sqrt {- \frac {1}{c}}}{b} \right )}}{2} + \frac {\sqrt {- \frac {1}{c}} \log {\left (x + \frac {a + c \sqrt {- \frac {1}{c}}}{b} \right )}}{2}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-sqrt(-1/c)*log(x + (a - c*sqrt(-1/c))/b)/2 + sqrt(-1/c)*log(x + (a + c*sqrt(-1/c))/b)/2)/b

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