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

Optimal. Leaf size=92 \[ \frac {(3 a d+b c) \tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{8 c^{5/2} d^{3/2}}+\frac {x (3 a d+b c)}{8 c^2 d \left (c+d x^2\right )}-\frac {x (b c-a d)}{4 c d \left (c+d x^2\right )^2} \]

[Out]

-1/4*(-a*d+b*c)*x/c/d/(d*x^2+c)^2+1/8*(3*a*d+b*c)*x/c^2/d/(d*x^2+c)+1/8*(3*a*d+b*c)*arctan(x*d^(1/2)/c^(1/2))/
c^(5/2)/d^(3/2)

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Rubi [A]  time = 0.03, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {385, 199, 205} \[ \frac {(3 a d+b c) \tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{8 c^{5/2} d^{3/2}}+\frac {x (3 a d+b c)}{8 c^2 d \left (c+d x^2\right )}-\frac {x (b c-a d)}{4 c d \left (c+d x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)/(c + d*x^2)^3,x]

[Out]

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

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 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 82, normalized size = 0.89 \[ \frac {(3 a d+b c) \tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{8 c^{5/2} d^{3/2}}+\frac {x \left (a d \left (5 c+3 d x^2\right )+b c \left (d x^2-c\right )\right )}{8 c^2 d \left (c+d x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.66, size = 300, normalized size = 3.26 \[ \left [\frac {2 \, {\left (b c^{2} d^{2} + 3 \, a c d^{3}\right )} x^{3} - {\left ({\left (b c d^{2} + 3 \, a d^{3}\right )} x^{4} + b c^{3} + 3 \, a c^{2} d + 2 \, {\left (b c^{2} d + 3 \, a c d^{2}\right )} x^{2}\right )} \sqrt {-c d} \log \left (\frac {d x^{2} - 2 \, \sqrt {-c d} x - c}{d x^{2} + c}\right ) - 2 \, {\left (b c^{3} d - 5 \, a c^{2} d^{2}\right )} x}{16 \, {\left (c^{3} d^{4} x^{4} + 2 \, c^{4} d^{3} x^{2} + c^{5} d^{2}\right )}}, \frac {{\left (b c^{2} d^{2} + 3 \, a c d^{3}\right )} x^{3} + {\left ({\left (b c d^{2} + 3 \, a d^{3}\right )} x^{4} + b c^{3} + 3 \, a c^{2} d + 2 \, {\left (b c^{2} d + 3 \, a c d^{2}\right )} x^{2}\right )} \sqrt {c d} \arctan \left (\frac {\sqrt {c d} x}{c}\right ) - {\left (b c^{3} d - 5 \, a c^{2} d^{2}\right )} x}{8 \, {\left (c^{3} d^{4} x^{4} + 2 \, c^{4} d^{3} x^{2} + c^{5} d^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(2*(b*c^2*d^2 + 3*a*c*d^3)*x^3 - ((b*c*d^2 + 3*a*d^3)*x^4 + b*c^3 + 3*a*c^2*d + 2*(b*c^2*d + 3*a*c*d^2)*
x^2)*sqrt(-c*d)*log((d*x^2 - 2*sqrt(-c*d)*x - c)/(d*x^2 + c)) - 2*(b*c^3*d - 5*a*c^2*d^2)*x)/(c^3*d^4*x^4 + 2*
c^4*d^3*x^2 + c^5*d^2), 1/8*((b*c^2*d^2 + 3*a*c*d^3)*x^3 + ((b*c*d^2 + 3*a*d^3)*x^4 + b*c^3 + 3*a*c^2*d + 2*(b
*c^2*d + 3*a*c*d^2)*x^2)*sqrt(c*d)*arctan(sqrt(c*d)*x/c) - (b*c^3*d - 5*a*c^2*d^2)*x)/(c^3*d^4*x^4 + 2*c^4*d^3
*x^2 + c^5*d^2)]

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giac [A]  time = 0.59, size = 78, normalized size = 0.85 \[ \frac {{\left (b c + 3 \, a d\right )} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{8 \, \sqrt {c d} c^{2} d} + \frac {b c d x^{3} + 3 \, a d^{2} x^{3} - b c^{2} x + 5 \, a c d x}{8 \, {\left (d x^{2} + c\right )}^{2} c^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 90, normalized size = 0.98 \[ \frac {3 a \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{8 \sqrt {c d}\, c^{2}}+\frac {b \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{8 \sqrt {c d}\, c d}+\frac {\frac {\left (3 a d +b c \right ) x^{3}}{8 c^{2}}+\frac {\left (5 a d -b c \right ) x}{8 c d}}{\left (d \,x^{2}+c \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(1/8*(3*a*d+b*c)/c^2*x^3+1/8*(5*a*d-b*c)/c/d*x)/(d*x^2+c)^2+3/8/c^2/(c*d)^(1/2)*arctan(1/(c*d)^(1/2)*d*x)*a+1/
8/c/d/(c*d)^(1/2)*arctan(1/(c*d)^(1/2)*d*x)*b

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maxima [A]  time = 2.96, size = 92, normalized size = 1.00 \[ \frac {{\left (b c d + 3 \, a d^{2}\right )} x^{3} - {\left (b c^{2} - 5 \, a c d\right )} x}{8 \, {\left (c^{2} d^{3} x^{4} + 2 \, c^{3} d^{2} x^{2} + c^{4} d\right )}} + \frac {{\left (b c + 3 \, a d\right )} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{8 \, \sqrt {c d} c^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*((b*c*d + 3*a*d^2)*x^3 - (b*c^2 - 5*a*c*d)*x)/(c^2*d^3*x^4 + 2*c^3*d^2*x^2 + c^4*d) + 1/8*(b*c + 3*a*d)*ar
ctan(d*x/sqrt(c*d))/(sqrt(c*d)*c^2*d)

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mupad [B]  time = 5.06, size = 82, normalized size = 0.89 \[ \frac {\frac {x^3\,\left (3\,a\,d+b\,c\right )}{8\,c^2}+\frac {x\,\left (5\,a\,d-b\,c\right )}{8\,c\,d}}{c^2+2\,c\,d\,x^2+d^2\,x^4}+\frac {\mathrm {atan}\left (\frac {\sqrt {d}\,x}{\sqrt {c}}\right )\,\left (3\,a\,d+b\,c\right )}{8\,c^{5/2}\,d^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)/(c + d*x^2)^3,x)

[Out]

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

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sympy [A]  time = 0.54, size = 150, normalized size = 1.63 \[ - \frac {\sqrt {- \frac {1}{c^{5} d^{3}}} \left (3 a d + b c\right ) \log {\left (- c^{3} d \sqrt {- \frac {1}{c^{5} d^{3}}} + x \right )}}{16} + \frac {\sqrt {- \frac {1}{c^{5} d^{3}}} \left (3 a d + b c\right ) \log {\left (c^{3} d \sqrt {- \frac {1}{c^{5} d^{3}}} + x \right )}}{16} + \frac {x^{3} \left (3 a d^{2} + b c d\right ) + x \left (5 a c d - b c^{2}\right )}{8 c^{4} d + 16 c^{3} d^{2} x^{2} + 8 c^{2} d^{3} x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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