3.179 \(\int \frac {a^2 c+a^2 d x+2 a b c x^2+2 a b d x^3+b^2 c x^4+b^2 d x^5}{(a+b x^2)^3} \, dx\)

Optimal. Leaf size=42 \[ \frac {c \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b}}+\frac {d \log \left (a+b x^2\right )}{2 b} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 54, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {1586, 635, 205, 260} \[ \frac {c \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b}}+\frac {d \log \left (a+b x^2\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[(a^2*c + a^2*d*x + 2*a*b*c*x^2 + 2*a*b*d*x^3 + b^2*c*x^4 + b^2*d*x^5)/(a + b*x^2)^3,x]

[Out]

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

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 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 1586

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

Rubi steps

\begin {align*} \int \frac {a^2 c+a^2 d x+2 a b c x^2+2 a b d x^3+b^2 c x^4+b^2 d x^5}{\left (a+b x^2\right )^3} \, dx &=\int \frac {a c+a d x+b c x^2+b d x^3}{\left (a+b x^2\right )^2} \, dx\\ &=\int \frac {c+d x}{a+b x^2} \, dx\\ &=c \int \frac {1}{a+b x^2} \, dx+d \int \frac {x}{a+b x^2} \, dx\\ &=\frac {c \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b}}+\frac {d \log \left (a+b x^2\right )}{2 b}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 42, normalized size = 1.00 \[ \frac {c \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b}}+\frac {d \log \left (a+b x^2\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2*c + a^2*d*x + 2*a*b*c*x^2 + 2*a*b*d*x^3 + b^2*c*x^4 + b^2*d*x^5)/(a + b*x^2)^3,x]

[Out]

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

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fricas [A]  time = 1.01, size = 98, normalized size = 2.33 \[ \left [\frac {a d \log \left (b x^{2} + a\right ) - \sqrt {-a b} c \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right )}{2 \, a b}, \frac {a d \log \left (b x^{2} + a\right ) + 2 \, \sqrt {a b} c \arctan \left (\frac {\sqrt {a b} x}{a}\right )}{2 \, a b}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*d*x^5+b^2*c*x^4+2*a*b*d*x^3+2*a*b*c*x^2+a^2*d*x+a^2*c)/(b*x^2+a)^3,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.40, size = 31, normalized size = 0.74 \[ \frac {c \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b}} + \frac {d \log \left (b x^{2} + a\right )}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*d*x^5+b^2*c*x^4+2*a*b*d*x^3+2*a*b*c*x^2+a^2*d*x+a^2*c)/(b*x^2+a)^3,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.00, size = 32, normalized size = 0.76 \[ \frac {c \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b}}+\frac {d \ln \left (b \,x^{2}+a \right )}{2 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b^2*d*x^5+b^2*c*x^4+2*a*b*d*x^3+2*a*b*c*x^2+a^2*d*x+a^2*c)/(b*x^2+a)^3,x)

[Out]

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

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maxima [A]  time = 1.49, size = 31, normalized size = 0.74 \[ \frac {c \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b}} + \frac {d \log \left (b x^{2} + a\right )}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*d*x^5+b^2*c*x^4+2*a*b*d*x^3+2*a*b*c*x^2+a^2*d*x+a^2*c)/(b*x^2+a)^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.13, size = 32, normalized size = 0.76 \[ \frac {d\,\ln \left (b\,x^2+a\right )}{2\,b}+\frac {c\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{\sqrt {a}\,\sqrt {b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2*c + b^2*c*x^4 + b^2*d*x^5 + a^2*d*x + 2*a*b*c*x^2 + 2*a*b*d*x^3)/(a + b*x^2)^3,x)

[Out]

(d*log(a + b*x^2))/(2*b) + (c*atan((b^(1/2)*x)/a^(1/2)))/(a^(1/2)*b^(1/2))

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sympy [B]  time = 0.30, size = 124, normalized size = 2.95 \[ \left (\frac {d}{2 b} - \frac {c \sqrt {- a b^{3}}}{2 a b^{2}}\right ) \log {\left (x + \frac {2 a b \left (\frac {d}{2 b} - \frac {c \sqrt {- a b^{3}}}{2 a b^{2}}\right ) - a d}{b c} \right )} + \left (\frac {d}{2 b} + \frac {c \sqrt {- a b^{3}}}{2 a b^{2}}\right ) \log {\left (x + \frac {2 a b \left (\frac {d}{2 b} + \frac {c \sqrt {- a b^{3}}}{2 a b^{2}}\right ) - a d}{b c} \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b**2*d*x**5+b**2*c*x**4+2*a*b*d*x**3+2*a*b*c*x**2+a**2*d*x+a**2*c)/(b*x**2+a)**3,x)

[Out]

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

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