3.12.77 \(\int \frac {A+B x}{(a+c x^2)^3} \, dx\)

Optimal. Leaf size=75 \[ \frac {3 A \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{8 a^{5/2} \sqrt {c}}+\frac {3 A x}{8 a^2 \left (a+c x^2\right )}+\frac {A c x-a B}{4 a c \left (a+c x^2\right )^2} \]

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Rubi [A]  time = 0.02, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {639, 199, 205} \begin {gather*} \frac {3 A x}{8 a^2 \left (a+c x^2\right )}+\frac {3 A \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{8 a^{5/2} \sqrt {c}}-\frac {a B-A c x}{4 a c \left (a+c x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(a + c*x^2)^3,x]

[Out]

-(a*B - A*c*x)/(4*a*c*(a + c*x^2)^2) + (3*A*x)/(8*a^2*(a + c*x^2)) + (3*A*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(8*a^(5
/2)*Sqrt[c])

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

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 639

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((a*e - c*d*x)*(a + c*x^2)^(p + 1))/(2*a
*c*(p + 1)), x] + Dist[(d*(2*p + 3))/(2*a*(p + 1)), Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x]
&& LtQ[p, -1] && NeQ[p, -3/2]

Rubi steps

\begin {align*} \int \frac {A+B x}{\left (a+c x^2\right )^3} \, dx &=-\frac {a B-A c x}{4 a c \left (a+c x^2\right )^2}+\frac {(3 A) \int \frac {1}{\left (a+c x^2\right )^2} \, dx}{4 a}\\ &=-\frac {a B-A c x}{4 a c \left (a+c x^2\right )^2}+\frac {3 A x}{8 a^2 \left (a+c x^2\right )}+\frac {(3 A) \int \frac {1}{a+c x^2} \, dx}{8 a^2}\\ &=-\frac {a B-A c x}{4 a c \left (a+c x^2\right )^2}+\frac {3 A x}{8 a^2 \left (a+c x^2\right )}+\frac {3 A \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{8 a^{5/2} \sqrt {c}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 71, normalized size = 0.95 \begin {gather*} \frac {\frac {\sqrt {a} \left (-2 a^2 B+5 a A c x+3 A c^2 x^3\right )}{\left (a+c x^2\right )^2}+3 A \sqrt {c} \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{8 a^{5/2} c} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(a + c*x^2)^3,x]

[Out]

((Sqrt[a]*(-2*a^2*B + 5*a*A*c*x + 3*A*c^2*x^3))/(a + c*x^2)^2 + 3*A*Sqrt[c]*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(8*a^
(5/2)*c)

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A+B x}{\left (a+c x^2\right )^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(A + B*x)/(a + c*x^2)^3,x]

[Out]

IntegrateAlgebraic[(A + B*x)/(a + c*x^2)^3, x]

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fricas [A]  time = 0.43, size = 212, normalized size = 2.83 \begin {gather*} \left [\frac {6 \, A a c^{2} x^{3} + 10 \, A a^{2} c x - 4 \, B a^{3} - 3 \, {\left (A c^{2} x^{4} + 2 \, A a c x^{2} + A a^{2}\right )} \sqrt {-a c} \log \left (\frac {c x^{2} - 2 \, \sqrt {-a c} x - a}{c x^{2} + a}\right )}{16 \, {\left (a^{3} c^{3} x^{4} + 2 \, a^{4} c^{2} x^{2} + a^{5} c\right )}}, \frac {3 \, A a c^{2} x^{3} + 5 \, A a^{2} c x - 2 \, B a^{3} + 3 \, {\left (A c^{2} x^{4} + 2 \, A a c x^{2} + A a^{2}\right )} \sqrt {a c} \arctan \left (\frac {\sqrt {a c} x}{a}\right )}{8 \, {\left (a^{3} c^{3} x^{4} + 2 \, a^{4} c^{2} x^{2} + a^{5} c\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(c*x^2+a)^3,x, algorithm="fricas")

[Out]

[1/16*(6*A*a*c^2*x^3 + 10*A*a^2*c*x - 4*B*a^3 - 3*(A*c^2*x^4 + 2*A*a*c*x^2 + A*a^2)*sqrt(-a*c)*log((c*x^2 - 2*
sqrt(-a*c)*x - a)/(c*x^2 + a)))/(a^3*c^3*x^4 + 2*a^4*c^2*x^2 + a^5*c), 1/8*(3*A*a*c^2*x^3 + 5*A*a^2*c*x - 2*B*
a^3 + 3*(A*c^2*x^4 + 2*A*a*c*x^2 + A*a^2)*sqrt(a*c)*arctan(sqrt(a*c)*x/a))/(a^3*c^3*x^4 + 2*a^4*c^2*x^2 + a^5*
c)]

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giac [A]  time = 0.19, size = 60, normalized size = 0.80 \begin {gather*} \frac {3 \, A \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{8 \, \sqrt {a c} a^{2}} + \frac {3 \, A c^{2} x^{3} + 5 \, A a c x - 2 \, B a^{2}}{8 \, {\left (c x^{2} + a\right )}^{2} a^{2} c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(c*x^2+a)^3,x, algorithm="giac")

[Out]

3/8*A*arctan(c*x/sqrt(a*c))/(sqrt(a*c)*a^2) + 1/8*(3*A*c^2*x^3 + 5*A*a*c*x - 2*B*a^2)/((c*x^2 + a)^2*a^2*c)

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maple [A]  time = 0.05, size = 65, normalized size = 0.87 \begin {gather*} \frac {3 A x}{8 \left (c \,x^{2}+a \right ) a^{2}}+\frac {3 A \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{8 \sqrt {a c}\, a^{2}}+\frac {2 A c x -2 B a}{8 \left (c \,x^{2}+a \right )^{2} a c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/(c*x^2+a)^3,x)

[Out]

1/8*(2*A*c*x-2*B*a)/a/c/(c*x^2+a)^2+3/8*A*x/a^2/(c*x^2+a)+3/8*A/a^2/(a*c)^(1/2)*arctan(1/(a*c)^(1/2)*c*x)

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maxima [A]  time = 1.30, size = 74, normalized size = 0.99 \begin {gather*} \frac {3 \, A c^{2} x^{3} + 5 \, A a c x - 2 \, B a^{2}}{8 \, {\left (a^{2} c^{3} x^{4} + 2 \, a^{3} c^{2} x^{2} + a^{4} c\right )}} + \frac {3 \, A \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{8 \, \sqrt {a c} a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(c*x^2+a)^3,x, algorithm="maxima")

[Out]

1/8*(3*A*c^2*x^3 + 5*A*a*c*x - 2*B*a^2)/(a^2*c^3*x^4 + 2*a^3*c^2*x^2 + a^4*c) + 3/8*A*arctan(c*x/sqrt(a*c))/(s
qrt(a*c)*a^2)

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mupad [B]  time = 0.08, size = 64, normalized size = 0.85 \begin {gather*} \frac {\frac {5\,A\,x}{8\,a}-\frac {B}{4\,c}+\frac {3\,A\,c\,x^3}{8\,a^2}}{a^2+2\,a\,c\,x^2+c^2\,x^4}+\frac {3\,A\,\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {a}}\right )}{8\,a^{5/2}\,\sqrt {c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/(a + c*x^2)^3,x)

[Out]

((5*A*x)/(8*a) - B/(4*c) + (3*A*c*x^3)/(8*a^2))/(a^2 + c^2*x^4 + 2*a*c*x^2) + (3*A*atan((c^(1/2)*x)/a^(1/2)))/
(8*a^(5/2)*c^(1/2))

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sympy [A]  time = 0.45, size = 124, normalized size = 1.65 \begin {gather*} A \left (- \frac {3 \sqrt {- \frac {1}{a^{5} c}} \log {\left (- a^{3} \sqrt {- \frac {1}{a^{5} c}} + x \right )}}{16} + \frac {3 \sqrt {- \frac {1}{a^{5} c}} \log {\left (a^{3} \sqrt {- \frac {1}{a^{5} c}} + x \right )}}{16}\right ) + \frac {5 A a c x + 3 A c^{2} x^{3} - 2 B a^{2}}{8 a^{4} c + 16 a^{3} c^{2} x^{2} + 8 a^{2} c^{3} x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(c*x**2+a)**3,x)

[Out]

A*(-3*sqrt(-1/(a**5*c))*log(-a**3*sqrt(-1/(a**5*c)) + x)/16 + 3*sqrt(-1/(a**5*c))*log(a**3*sqrt(-1/(a**5*c)) +
x)/16) + (5*A*a*c*x + 3*A*c**2*x**3 - 2*B*a**2)/(8*a**4*c + 16*a**3*c**2*x**2 + 8*a**2*c**3*x**4)

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