3.3.70 \(\int \frac {c+d x^2}{x^4 (a+b x^2)^2} \, dx\) [270]

Optimal. Leaf size=90 \[ -\frac {c}{3 a^2 x^3}+\frac {2 b c-a d}{a^3 x}+\frac {b (b c-a d) x}{2 a^3 \left (a+b x^2\right )}+\frac {\sqrt {b} (5 b c-3 a d) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{7/2}} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {467, 1275, 211} \begin {gather*} \frac {\sqrt {b} \text {ArcTan}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) (5 b c-3 a d)}{2 a^{7/2}}+\frac {b x (b c-a d)}{2 a^3 \left (a+b x^2\right )}+\frac {2 b c-a d}{a^3 x}-\frac {c}{3 a^2 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*c/(a^2*x^3) + (2*b*c - a*d)/(a^3*x) + (b*(b*c - a*d)*x)/(2*a^3*(a + b*x^2)) + (Sqrt[b]*(5*b*c - 3*a*d)*Ar
cTan[(Sqrt[b]*x)/Sqrt[a]])/(2*a^(7/2))

Rule 211

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

Rule 467

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x
*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p + 1))), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[x^m*(a + b*x^2)^(p +
1)*ExpandToSum[2*b*(p + 1)*Together[(b^(m/2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d)*x^(-m + 2))/(a + b*x^2)]
- ((-a)^(m/2 - 1)*(b*c - a*d))/x^m, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &
& ILtQ[m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 1275

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] &&
NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]

Rubi steps

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

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Mathematica [A]
time = 0.05, size = 90, normalized size = 1.00 \begin {gather*} -\frac {c}{3 a^2 x^3}+\frac {2 b c-a d}{a^3 x}-\frac {b (-b c+a d) x}{2 a^3 \left (a+b x^2\right )}-\frac {\sqrt {b} (-5 b c+3 a d) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*c/(a^2*x^3) + (2*b*c - a*d)/(a^3*x) - (b*(-(b*c) + a*d)*x)/(2*a^3*(a + b*x^2)) - (Sqrt[b]*(-5*b*c + 3*a*d
)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(2*a^(7/2))

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Maple [A]
time = 0.08, size = 79, normalized size = 0.88

method	result	size
default	\(-\frac {b \left (\frac {\left (\frac {a d}{2}-\frac {b c}{2}\right ) x}{b \,x^{2}+a}+\frac {\left (3 a d -5 b c \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{a^{3}}-\frac {c}{3 a^{2} x^{3}}-\frac {a d -2 b c}{a^{3} x}\)	\(79\)
risch	\(\frac {-\frac {b \left (3 a d -5 b c \right ) x^{4}}{2 a^{3}}-\frac {\left (3 a d -5 b c \right ) x^{2}}{3 a^{2}}-\frac {c}{3 a}}{x^{3} \left (b \,x^{2}+a \right )}+\frac {\left (\munderset {\textit {\_R} =\RootOf \left (a^{7} \textit {\_Z}^{2}+9 a^{2} b \,d^{2}-30 a \,b^{2} c d +25 b^{3} c^{2}\right )}{\sum }\textit {\_R} \ln \left (\left (3 \textit {\_R}^{2} a^{7}+18 a^{2} b \,d^{2}-60 a \,b^{2} c d +50 b^{3} c^{2}\right ) x +\left (3 a^{5} d -5 a^{4} b c \right ) \textit {\_R} \right )\right )}{4}\)	\(152\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x^2+c)/x^4/(b*x^2+a)^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.50, size = 93, normalized size = 1.03 \begin {gather*} \frac {3 \, {\left (5 \, b^{2} c - 3 \, a b d\right )} x^{4} - 2 \, a^{2} c + 2 \, {\left (5 \, a b c - 3 \, a^{2} d\right )} x^{2}}{6 \, {\left (a^{3} b x^{5} + a^{4} x^{3}\right )}} + \frac {{\left (5 \, b^{2} c - 3 \, a b d\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.91, size = 250, normalized size = 2.78 \begin {gather*} \left [\frac {6 \, {\left (5 \, b^{2} c - 3 \, a b d\right )} x^{4} - 4 \, a^{2} c + 4 \, {\left (5 \, a b c - 3 \, a^{2} d\right )} x^{2} - 3 \, {\left ({\left (5 \, b^{2} c - 3 \, a b d\right )} x^{5} + {\left (5 \, a b c - 3 \, a^{2} d\right )} x^{3}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x^{2} - 2 \, a x \sqrt {-\frac {b}{a}} - a}{b x^{2} + a}\right )}{12 \, {\left (a^{3} b x^{5} + a^{4} x^{3}\right )}}, \frac {3 \, {\left (5 \, b^{2} c - 3 \, a b d\right )} x^{4} - 2 \, a^{2} c + 2 \, {\left (5 \, a b c - 3 \, a^{2} d\right )} x^{2} + 3 \, {\left ({\left (5 \, b^{2} c - 3 \, a b d\right )} x^{5} + {\left (5 \, a b c - 3 \, a^{2} d\right )} x^{3}\right )} \sqrt {\frac {b}{a}} \arctan \left (x \sqrt {\frac {b}{a}}\right )}{6 \, {\left (a^{3} b x^{5} + a^{4} x^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/12*(6*(5*b^2*c - 3*a*b*d)*x^4 - 4*a^2*c + 4*(5*a*b*c - 3*a^2*d)*x^2 - 3*((5*b^2*c - 3*a*b*d)*x^5 + (5*a*b*c
- 3*a^2*d)*x^3)*sqrt(-b/a)*log((b*x^2 - 2*a*x*sqrt(-b/a) - a)/(b*x^2 + a)))/(a^3*b*x^5 + a^4*x^3), 1/6*(3*(5*
b^2*c - 3*a*b*d)*x^4 - 2*a^2*c + 2*(5*a*b*c - 3*a^2*d)*x^2 + 3*((5*b^2*c - 3*a*b*d)*x^5 + (5*a*b*c - 3*a^2*d)*
x^3)*sqrt(b/a)*arctan(x*sqrt(b/a)))/(a^3*b*x^5 + a^4*x^3)]

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 184 vs. \(2 (82) = 164\).
time = 0.30, size = 184, normalized size = 2.04 \begin {gather*} \frac {\sqrt {- \frac {b}{a^{7}}} \cdot \left (3 a d - 5 b c\right ) \log {\left (- \frac {a^{4} \sqrt {- \frac {b}{a^{7}}} \cdot \left (3 a d - 5 b c\right )}{3 a b d - 5 b^{2} c} + x \right )}}{4} - \frac {\sqrt {- \frac {b}{a^{7}}} \cdot \left (3 a d - 5 b c\right ) \log {\left (\frac {a^{4} \sqrt {- \frac {b}{a^{7}}} \cdot \left (3 a d - 5 b c\right )}{3 a b d - 5 b^{2} c} + x \right )}}{4} + \frac {- 2 a^{2} c + x^{4} \left (- 9 a b d + 15 b^{2} c\right ) + x^{2} \left (- 6 a^{2} d + 10 a b c\right )}{6 a^{4} x^{3} + 6 a^{3} b x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(-b/a**7)*(3*a*d - 5*b*c)*log(-a**4*sqrt(-b/a**7)*(3*a*d - 5*b*c)/(3*a*b*d - 5*b**2*c) + x)/4 - sqrt(-b/a*
*7)*(3*a*d - 5*b*c)*log(a**4*sqrt(-b/a**7)*(3*a*d - 5*b*c)/(3*a*b*d - 5*b**2*c) + x)/4 + (-2*a**2*c + x**4*(-9
*a*b*d + 15*b**2*c) + x**2*(-6*a**2*d + 10*a*b*c))/(6*a**4*x**3 + 6*a**3*b*x**5)

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Giac [A]
time = 1.46, size = 86, normalized size = 0.96 \begin {gather*} \frac {{\left (5 \, b^{2} c - 3 \, a b d\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a^{3}} + \frac {b^{2} c x - a b d x}{2 \, {\left (b x^{2} + a\right )} a^{3}} + \frac {6 \, b c x^{2} - 3 \, a d x^{2} - a c}{3 \, a^{3} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(5*b^2*c - 3*a*b*d)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^3) + 1/2*(b^2*c*x - a*b*d*x)/((b*x^2 + a)*a^3) + 1/
3*(6*b*c*x^2 - 3*a*d*x^2 - a*c)/(a^3*x^3)

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Mupad [B]
time = 0.11, size = 84, normalized size = 0.93 \begin {gather*} -\frac {\frac {c}{3\,a}+\frac {x^2\,\left (3\,a\,d-5\,b\,c\right )}{3\,a^2}+\frac {b\,x^4\,\left (3\,a\,d-5\,b\,c\right )}{2\,a^3}}{b\,x^5+a\,x^3}-\frac {\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )\,\left (3\,a\,d-5\,b\,c\right )}{2\,a^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (c/(3*a) + (x^2*(3*a*d - 5*b*c))/(3*a^2) + (b*x^4*(3*a*d - 5*b*c))/(2*a^3))/(a*x^3 + b*x^5) - (b^(1/2)*atan(
(b^(1/2)*x)/a^(1/2))*(3*a*d - 5*b*c))/(2*a^(7/2))

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