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

Optimal. Leaf size=152 \[ -\frac {x \left (5 a^2 d^2-2 a b c d+b^2 c^2\right )}{4 c^2 d \left (c+d x^2\right )^2}-\frac {a^2}{c x \left (c+d x^2\right )^2}+\frac {\left (3 a d (2 b c-5 a d)+b^2 c^2\right ) \tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{8 c^{7/2} d^{3/2}}+\frac {x \left (3 a d (2 b c-5 a d)+b^2 c^2\right )}{8 c^3 d \left (c+d x^2\right )} \]

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Rubi [A]  time = 0.11, antiderivative size = 151, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {462, 385, 199, 205} \begin {gather*} -\frac {x \left (5 a^2 d^2-2 a b c d+b^2 c^2\right )}{4 c^2 d \left (c+d x^2\right )^2}-\frac {a^2}{c x \left (c+d x^2\right )^2}+\frac {\left (3 a d (2 b c-5 a d)+b^2 c^2\right ) \tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{8 c^{7/2} d^{3/2}}+\frac {x \left (\frac {3 a (2 b c-5 a d)}{c^2}+\frac {b^2}{d}\right )}{8 c \left (c+d x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2/(c*x*(c + d*x^2)^2)) - ((b^2*c^2 - 2*a*b*c*d + 5*a^2*d^2)*x)/(4*c^2*d*(c + d*x^2)^2) + ((b^2/d + (3*a*(2
*b*c - 5*a*d))/c^2)*x)/(8*c*(c + d*x^2)) + ((b^2*c^2 + 3*a*d*(2*b*c - 5*a*d))*ArcTan[(Sqrt[d]*x)/Sqrt[c]])/(8*
c^(7/2)*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])

Rule 462

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> Simp[(c^2*(e*x)^(
m + 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] - Dist[1/(a*e^n*(m + 1)), Int[(e*x)^(m + n)*(a + b*x^n)^p*Simp[b
*c^2*n*(p + 1) + c*(b*c - 2*a*d)*(m + 1) - a*(m + 1)*d^2*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && Ne
Q[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -1] && GtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {\left (a+b x^2\right )^2}{x^2 \left (c+d x^2\right )^3} \, dx &=-\frac {a^2}{c x \left (c+d x^2\right )^2}+\frac {\int \frac {a (2 b c-5 a d)+b^2 c x^2}{\left (c+d x^2\right )^3} \, dx}{c}\\ &=-\frac {a^2}{c x \left (c+d x^2\right )^2}+\frac {\left (2 a b-\frac {b^2 c}{d}-\frac {5 a^2 d}{c}\right ) x}{4 c \left (c+d x^2\right )^2}+\frac {1}{4} \left (\frac {b^2}{d}+\frac {3 a (2 b c-5 a d)}{c^2}\right ) \int \frac {1}{\left (c+d x^2\right )^2} \, dx\\ &=-\frac {a^2}{c x \left (c+d x^2\right )^2}+\frac {\left (2 a b-\frac {b^2 c}{d}-\frac {5 a^2 d}{c}\right ) x}{4 c \left (c+d x^2\right )^2}+\frac {\left (\frac {b^2}{d}+\frac {3 a (2 b c-5 a d)}{c^2}\right ) x}{8 c \left (c+d x^2\right )}+\frac {\left (\frac {b^2}{d}+\frac {3 a (2 b c-5 a d)}{c^2}\right ) \int \frac {1}{c+d x^2} \, dx}{8 c}\\ &=-\frac {a^2}{c x \left (c+d x^2\right )^2}+\frac {\left (2 a b-\frac {b^2 c}{d}-\frac {5 a^2 d}{c}\right ) x}{4 c \left (c+d x^2\right )^2}+\frac {\left (\frac {b^2}{d}+\frac {3 a (2 b c-5 a d)}{c^2}\right ) x}{8 c \left (c+d x^2\right )}+\frac {\left (b^2 c^2+6 a b c d-15 a^2 d^2\right ) \tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{8 c^{7/2} d^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 133, normalized size = 0.88 \begin {gather*} \frac {\left (-15 a^2 d^2+6 a b c d+b^2 c^2\right ) \tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{8 c^{7/2} d^{3/2}}+\frac {x \left (-7 a^2 d^2+6 a b c d+b^2 c^2\right )}{8 c^3 d \left (c+d x^2\right )}-\frac {a^2}{c^3 x}-\frac {x (b c-a d)^2}{4 c^2 d \left (c+d x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2/(c^3*x)) - ((b*c - a*d)^2*x)/(4*c^2*d*(c + d*x^2)^2) + ((b^2*c^2 + 6*a*b*c*d - 7*a^2*d^2)*x)/(8*c^3*d*(c
+ d*x^2)) + ((b^2*c^2 + 6*a*b*c*d - 15*a^2*d^2)*ArcTan[(Sqrt[d]*x)/Sqrt[c]])/(8*c^(7/2)*d^(3/2))

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.85, size = 475, normalized size = 3.12 \begin {gather*} \left [-\frac {16 \, a^{2} c^{3} d^{2} - 2 \, {\left (b^{2} c^{3} d^{2} + 6 \, a b c^{2} d^{3} - 15 \, a^{2} c d^{4}\right )} x^{4} + 2 \, {\left (b^{2} c^{4} d - 10 \, a b c^{3} d^{2} + 25 \, a^{2} c^{2} d^{3}\right )} x^{2} - {\left ({\left (b^{2} c^{2} d^{2} + 6 \, a b c d^{3} - 15 \, a^{2} d^{4}\right )} x^{5} + 2 \, {\left (b^{2} c^{3} d + 6 \, a b c^{2} d^{2} - 15 \, a^{2} c d^{3}\right )} x^{3} + {\left (b^{2} c^{4} + 6 \, a b c^{3} d - 15 \, a^{2} c^{2} d^{2}\right )} x\right )} \sqrt {-c d} \log \left (\frac {d x^{2} + 2 \, \sqrt {-c d} x - c}{d x^{2} + c}\right )}{16 \, {\left (c^{4} d^{4} x^{5} + 2 \, c^{5} d^{3} x^{3} + c^{6} d^{2} x\right )}}, -\frac {8 \, a^{2} c^{3} d^{2} - {\left (b^{2} c^{3} d^{2} + 6 \, a b c^{2} d^{3} - 15 \, a^{2} c d^{4}\right )} x^{4} + {\left (b^{2} c^{4} d - 10 \, a b c^{3} d^{2} + 25 \, a^{2} c^{2} d^{3}\right )} x^{2} - {\left ({\left (b^{2} c^{2} d^{2} + 6 \, a b c d^{3} - 15 \, a^{2} d^{4}\right )} x^{5} + 2 \, {\left (b^{2} c^{3} d + 6 \, a b c^{2} d^{2} - 15 \, a^{2} c d^{3}\right )} x^{3} + {\left (b^{2} c^{4} + 6 \, a b c^{3} d - 15 \, a^{2} c^{2} d^{2}\right )} x\right )} \sqrt {c d} \arctan \left (\frac {\sqrt {c d} x}{c}\right )}{8 \, {\left (c^{4} d^{4} x^{5} + 2 \, c^{5} d^{3} x^{3} + c^{6} d^{2} x\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/16*(16*a^2*c^3*d^2 - 2*(b^2*c^3*d^2 + 6*a*b*c^2*d^3 - 15*a^2*c*d^4)*x^4 + 2*(b^2*c^4*d - 10*a*b*c^3*d^2 +
25*a^2*c^2*d^3)*x^2 - ((b^2*c^2*d^2 + 6*a*b*c*d^3 - 15*a^2*d^4)*x^5 + 2*(b^2*c^3*d + 6*a*b*c^2*d^2 - 15*a^2*c*
d^3)*x^3 + (b^2*c^4 + 6*a*b*c^3*d - 15*a^2*c^2*d^2)*x)*sqrt(-c*d)*log((d*x^2 + 2*sqrt(-c*d)*x - c)/(d*x^2 + c)
))/(c^4*d^4*x^5 + 2*c^5*d^3*x^3 + c^6*d^2*x), -1/8*(8*a^2*c^3*d^2 - (b^2*c^3*d^2 + 6*a*b*c^2*d^3 - 15*a^2*c*d^
4)*x^4 + (b^2*c^4*d - 10*a*b*c^3*d^2 + 25*a^2*c^2*d^3)*x^2 - ((b^2*c^2*d^2 + 6*a*b*c*d^3 - 15*a^2*d^4)*x^5 + 2
*(b^2*c^3*d + 6*a*b*c^2*d^2 - 15*a^2*c*d^3)*x^3 + (b^2*c^4 + 6*a*b*c^3*d - 15*a^2*c^2*d^2)*x)*sqrt(c*d)*arctan
(sqrt(c*d)*x/c))/(c^4*d^4*x^5 + 2*c^5*d^3*x^3 + c^6*d^2*x)]

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giac [A]  time = 0.31, size = 135, normalized size = 0.89 \begin {gather*} -\frac {a^{2}}{c^{3} x} + \frac {{\left (b^{2} c^{2} + 6 \, a b c d - 15 \, a^{2} d^{2}\right )} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{8 \, \sqrt {c d} c^{3} d} + \frac {b^{2} c^{2} d x^{3} + 6 \, a b c d^{2} x^{3} - 7 \, a^{2} d^{3} x^{3} - b^{2} c^{3} x + 10 \, a b c^{2} d x - 9 \, a^{2} c d^{2} x}{8 \, {\left (d x^{2} + c\right )}^{2} c^{3} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a^2/(c^3*x) + 1/8*(b^2*c^2 + 6*a*b*c*d - 15*a^2*d^2)*arctan(d*x/sqrt(c*d))/(sqrt(c*d)*c^3*d) + 1/8*(b^2*c^2*d
*x^3 + 6*a*b*c*d^2*x^3 - 7*a^2*d^3*x^3 - b^2*c^3*x + 10*a*b*c^2*d*x - 9*a^2*c*d^2*x)/((d*x^2 + c)^2*c^3*d)

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maple [A]  time = 0.02, size = 199, normalized size = 1.31 \begin {gather*} -\frac {7 a^{2} d^{2} x^{3}}{8 \left (d \,x^{2}+c \right )^{2} c^{3}}+\frac {3 a b d \,x^{3}}{4 \left (d \,x^{2}+c \right )^{2} c^{2}}+\frac {b^{2} x^{3}}{8 \left (d \,x^{2}+c \right )^{2} c}-\frac {9 a^{2} d x}{8 \left (d \,x^{2}+c \right )^{2} c^{2}}+\frac {5 a b x}{4 \left (d \,x^{2}+c \right )^{2} c}-\frac {b^{2} x}{8 \left (d \,x^{2}+c \right )^{2} d}-\frac {15 a^{2} d \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{8 \sqrt {c d}\, c^{3}}+\frac {3 a b \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{4 \sqrt {c d}\, c^{2}}+\frac {b^{2} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{8 \sqrt {c d}\, c d}-\frac {a^{2}}{c^{3} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-7/8/c^3/(d*x^2+c)^2*x^3*a^2*d^2+3/4/c^2/(d*x^2+c)^2*x^3*a*b*d+1/8/c/(d*x^2+c)^2*x^3*b^2-9/8/c^2/(d*x^2+c)^2*a
^2*d*x+5/4/c/(d*x^2+c)^2*a*b*x-1/8/(d*x^2+c)^2/d*x*b^2-15/8/c^3*d/(c*d)^(1/2)*arctan(1/(c*d)^(1/2)*d*x)*a^2+3/
4/c^2/(c*d)^(1/2)*arctan(1/(c*d)^(1/2)*d*x)*a*b+1/8/c/d/(c*d)^(1/2)*arctan(1/(c*d)^(1/2)*d*x)*b^2-a^2/c^3/x

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maxima [A]  time = 2.41, size = 146, normalized size = 0.96 \begin {gather*} -\frac {8 \, a^{2} c^{2} d - {\left (b^{2} c^{2} d + 6 \, a b c d^{2} - 15 \, a^{2} d^{3}\right )} x^{4} + {\left (b^{2} c^{3} - 10 \, a b c^{2} d + 25 \, a^{2} c d^{2}\right )} x^{2}}{8 \, {\left (c^{3} d^{3} x^{5} + 2 \, c^{4} d^{2} x^{3} + c^{5} d x\right )}} + \frac {{\left (b^{2} c^{2} + 6 \, a b c d - 15 \, a^{2} d^{2}\right )} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{8 \, \sqrt {c d} c^{3} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(8*a^2*c^2*d - (b^2*c^2*d + 6*a*b*c*d^2 - 15*a^2*d^3)*x^4 + (b^2*c^3 - 10*a*b*c^2*d + 25*a^2*c*d^2)*x^2)/
(c^3*d^3*x^5 + 2*c^4*d^2*x^3 + c^5*d*x) + 1/8*(b^2*c^2 + 6*a*b*c*d - 15*a^2*d^2)*arctan(d*x/sqrt(c*d))/(sqrt(c
*d)*c^3*d)

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mupad [B]  time = 0.18, size = 135, normalized size = 0.89 \begin {gather*} \frac {\mathrm {atan}\left (\frac {\sqrt {d}\,x}{\sqrt {c}}\right )\,\left (-15\,a^2\,d^2+6\,a\,b\,c\,d+b^2\,c^2\right )}{8\,c^{7/2}\,d^{3/2}}-\frac {\frac {a^2}{c}-\frac {x^4\,\left (-15\,a^2\,d^2+6\,a\,b\,c\,d+b^2\,c^2\right )}{8\,c^3}+\frac {x^2\,\left (25\,a^2\,d^2-10\,a\,b\,c\,d+b^2\,c^2\right )}{8\,c^2\,d}}{c^2\,x+2\,c\,d\,x^3+d^2\,x^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atan((d^(1/2)*x)/c^(1/2))*(b^2*c^2 - 15*a^2*d^2 + 6*a*b*c*d))/(8*c^(7/2)*d^(3/2)) - (a^2/c - (x^4*(b^2*c^2 -
15*a^2*d^2 + 6*a*b*c*d))/(8*c^3) + (x^2*(25*a^2*d^2 + b^2*c^2 - 10*a*b*c*d))/(8*c^2*d))/(c^2*x + d^2*x^5 + 2*c
*d*x^3)

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sympy [A]  time = 1.20, size = 224, normalized size = 1.47 \begin {gather*} \frac {\sqrt {- \frac {1}{c^{7} d^{3}}} \left (15 a^{2} d^{2} - 6 a b c d - b^{2} c^{2}\right ) \log {\left (- c^{4} d \sqrt {- \frac {1}{c^{7} d^{3}}} + x \right )}}{16} - \frac {\sqrt {- \frac {1}{c^{7} d^{3}}} \left (15 a^{2} d^{2} - 6 a b c d - b^{2} c^{2}\right ) \log {\left (c^{4} d \sqrt {- \frac {1}{c^{7} d^{3}}} + x \right )}}{16} + \frac {- 8 a^{2} c^{2} d + x^{4} \left (- 15 a^{2} d^{3} + 6 a b c d^{2} + b^{2} c^{2} d\right ) + x^{2} \left (- 25 a^{2} c d^{2} + 10 a b c^{2} d - b^{2} c^{3}\right )}{8 c^{5} d x + 16 c^{4} d^{2} x^{3} + 8 c^{3} d^{3} x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(-1/(c**7*d**3))*(15*a**2*d**2 - 6*a*b*c*d - b**2*c**2)*log(-c**4*d*sqrt(-1/(c**7*d**3)) + x)/16 - sqrt(-1
/(c**7*d**3))*(15*a**2*d**2 - 6*a*b*c*d - b**2*c**2)*log(c**4*d*sqrt(-1/(c**7*d**3)) + x)/16 + (-8*a**2*c**2*d
+ x**4*(-15*a**2*d**3 + 6*a*b*c*d**2 + b**2*c**2*d) + x**2*(-25*a**2*c*d**2 + 10*a*b*c**2*d - b**2*c**3))/(8*
c**5*d*x + 16*c**4*d**2*x**3 + 8*c**3*d**3*x**5)

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