3.521 \(\int \frac {(d+e x)^2}{(a+c x^2)^4} \, dx\)

Optimal. Leaf size=145 \[ \frac {\left (a e^2+5 c d^2\right ) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{16 a^{7/2} c^{3/2}}+\frac {x \left (a e^2+5 c d^2\right )}{16 a^3 c \left (a+c x^2\right )}-\frac {4 a d e-x \left (a e^2+5 c d^2\right )}{24 a^2 c \left (a+c x^2\right )^2}-\frac {(d+e x) (a e-c d x)}{6 a c \left (a+c x^2\right )^3} \]

[Out]

-1/6*(-c*d*x+a*e)*(e*x+d)/a/c/(c*x^2+a)^3+1/24*(-4*a*d*e+(a*e^2+5*c*d^2)*x)/a^2/c/(c*x^2+a)^2+1/16*(a*e^2+5*c*
d^2)*x/a^3/c/(c*x^2+a)+1/16*(a*e^2+5*c*d^2)*arctan(x*c^(1/2)/a^(1/2))/a^(7/2)/c^(3/2)

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Rubi [A]  time = 0.06, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {739, 639, 199, 205} \[ \frac {\left (a e^2+5 c d^2\right ) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{16 a^{7/2} c^{3/2}}+\frac {x \left (a e^2+5 c d^2\right )}{16 a^3 c \left (a+c x^2\right )}-\frac {4 a d e-x \left (a e^2+5 c d^2\right )}{24 a^2 c \left (a+c x^2\right )^2}-\frac {(d+e x) (a e-c d x)}{6 a c \left (a+c x^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^2/(a + c*x^2)^4,x]

[Out]

-((a*e - c*d*x)*(d + e*x))/(6*a*c*(a + c*x^2)^3) - (4*a*d*e - (5*c*d^2 + a*e^2)*x)/(24*a^2*c*(a + c*x^2)^2) +
((5*c*d^2 + a*e^2)*x)/(16*a^3*c*(a + c*x^2)) + ((5*c*d^2 + a*e^2)*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(16*a^(7/2)*c^(
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 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]

Rule 739

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m - 1)*(a*e - c*d*x)*(a
+ c*x^2)^(p + 1))/(2*a*c*(p + 1)), x] + Dist[1/((p + 1)*(-2*a*c)), Int[(d + e*x)^(m - 2)*Simp[a*e^2*(m - 1) -
c*d^2*(2*p + 3) - d*c*e*(m + 2*p + 2)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^
2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rubi steps

\begin {align*} \int \frac {(d+e x)^2}{\left (a+c x^2\right )^4} \, dx &=-\frac {(a e-c d x) (d+e x)}{6 a c \left (a+c x^2\right )^3}+\frac {\int \frac {5 c d^2+a e^2+4 c d e x}{\left (a+c x^2\right )^3} \, dx}{6 a c}\\ &=-\frac {(a e-c d x) (d+e x)}{6 a c \left (a+c x^2\right )^3}-\frac {4 a d e-\left (5 c d^2+a e^2\right ) x}{24 a^2 c \left (a+c x^2\right )^2}+\frac {\left (5 c d^2+a e^2\right ) \int \frac {1}{\left (a+c x^2\right )^2} \, dx}{8 a^2 c}\\ &=-\frac {(a e-c d x) (d+e x)}{6 a c \left (a+c x^2\right )^3}-\frac {4 a d e-\left (5 c d^2+a e^2\right ) x}{24 a^2 c \left (a+c x^2\right )^2}+\frac {\left (5 c d^2+a e^2\right ) x}{16 a^3 c \left (a+c x^2\right )}+\frac {\left (5 c d^2+a e^2\right ) \int \frac {1}{a+c x^2} \, dx}{16 a^3 c}\\ &=-\frac {(a e-c d x) (d+e x)}{6 a c \left (a+c x^2\right )^3}-\frac {4 a d e-\left (5 c d^2+a e^2\right ) x}{24 a^2 c \left (a+c x^2\right )^2}+\frac {\left (5 c d^2+a e^2\right ) x}{16 a^3 c \left (a+c x^2\right )}+\frac {\left (5 c d^2+a e^2\right ) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{16 a^{7/2} c^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 127, normalized size = 0.88 \[ \frac {\left (a e^2+5 c d^2\right ) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{16 a^{7/2} c^{3/2}}+\frac {-a^3 e (16 d+3 e x)+a^2 c x \left (33 d^2+8 e^2 x^2\right )+a c^2 x^3 \left (40 d^2+3 e^2 x^2\right )+15 c^3 d^2 x^5}{48 a^3 c \left (a+c x^2\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(15*c^3*d^2*x^5 - a^3*e*(16*d + 3*e*x) + a*c^2*x^3*(40*d^2 + 3*e^2*x^2) + a^2*c*x*(33*d^2 + 8*e^2*x^2))/(48*a^
3*c*(a + c*x^2)^3) + ((5*c*d^2 + a*e^2)*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(16*a^(7/2)*c^(3/2))

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fricas [A]  time = 0.84, size = 490, normalized size = 3.38 \[ \left [-\frac {32 \, a^{4} c d e - 6 \, {\left (5 \, a c^{4} d^{2} + a^{2} c^{3} e^{2}\right )} x^{5} - 16 \, {\left (5 \, a^{2} c^{3} d^{2} + a^{3} c^{2} e^{2}\right )} x^{3} + 3 \, {\left ({\left (5 \, c^{4} d^{2} + a c^{3} e^{2}\right )} x^{6} + 5 \, a^{3} c d^{2} + a^{4} e^{2} + 3 \, {\left (5 \, a c^{3} d^{2} + a^{2} c^{2} e^{2}\right )} x^{4} + 3 \, {\left (5 \, a^{2} c^{2} d^{2} + a^{3} c e^{2}\right )} x^{2}\right )} \sqrt {-a c} \log \left (\frac {c x^{2} - 2 \, \sqrt {-a c} x - a}{c x^{2} + a}\right ) - 6 \, {\left (11 \, a^{3} c^{2} d^{2} - a^{4} c e^{2}\right )} x}{96 \, {\left (a^{4} c^{5} x^{6} + 3 \, a^{5} c^{4} x^{4} + 3 \, a^{6} c^{3} x^{2} + a^{7} c^{2}\right )}}, -\frac {16 \, a^{4} c d e - 3 \, {\left (5 \, a c^{4} d^{2} + a^{2} c^{3} e^{2}\right )} x^{5} - 8 \, {\left (5 \, a^{2} c^{3} d^{2} + a^{3} c^{2} e^{2}\right )} x^{3} - 3 \, {\left ({\left (5 \, c^{4} d^{2} + a c^{3} e^{2}\right )} x^{6} + 5 \, a^{3} c d^{2} + a^{4} e^{2} + 3 \, {\left (5 \, a c^{3} d^{2} + a^{2} c^{2} e^{2}\right )} x^{4} + 3 \, {\left (5 \, a^{2} c^{2} d^{2} + a^{3} c e^{2}\right )} x^{2}\right )} \sqrt {a c} \arctan \left (\frac {\sqrt {a c} x}{a}\right ) - 3 \, {\left (11 \, a^{3} c^{2} d^{2} - a^{4} c e^{2}\right )} x}{48 \, {\left (a^{4} c^{5} x^{6} + 3 \, a^{5} c^{4} x^{4} + 3 \, a^{6} c^{3} x^{2} + a^{7} c^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/96*(32*a^4*c*d*e - 6*(5*a*c^4*d^2 + a^2*c^3*e^2)*x^5 - 16*(5*a^2*c^3*d^2 + a^3*c^2*e^2)*x^3 + 3*((5*c^4*d^
2 + a*c^3*e^2)*x^6 + 5*a^3*c*d^2 + a^4*e^2 + 3*(5*a*c^3*d^2 + a^2*c^2*e^2)*x^4 + 3*(5*a^2*c^2*d^2 + a^3*c*e^2)
*x^2)*sqrt(-a*c)*log((c*x^2 - 2*sqrt(-a*c)*x - a)/(c*x^2 + a)) - 6*(11*a^3*c^2*d^2 - a^4*c*e^2)*x)/(a^4*c^5*x^
6 + 3*a^5*c^4*x^4 + 3*a^6*c^3*x^2 + a^7*c^2), -1/48*(16*a^4*c*d*e - 3*(5*a*c^4*d^2 + a^2*c^3*e^2)*x^5 - 8*(5*a
^2*c^3*d^2 + a^3*c^2*e^2)*x^3 - 3*((5*c^4*d^2 + a*c^3*e^2)*x^6 + 5*a^3*c*d^2 + a^4*e^2 + 3*(5*a*c^3*d^2 + a^2*
c^2*e^2)*x^4 + 3*(5*a^2*c^2*d^2 + a^3*c*e^2)*x^2)*sqrt(a*c)*arctan(sqrt(a*c)*x/a) - 3*(11*a^3*c^2*d^2 - a^4*c*
e^2)*x)/(a^4*c^5*x^6 + 3*a^5*c^4*x^4 + 3*a^6*c^3*x^2 + a^7*c^2)]

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giac [A]  time = 0.17, size = 123, normalized size = 0.85 \[ \frac {{\left (5 \, c d^{2} + a e^{2}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{16 \, \sqrt {a c} a^{3} c} + \frac {15 \, c^{3} d^{2} x^{5} + 3 \, a c^{2} x^{5} e^{2} + 40 \, a c^{2} d^{2} x^{3} + 8 \, a^{2} c x^{3} e^{2} + 33 \, a^{2} c d^{2} x - 3 \, a^{3} x e^{2} - 16 \, a^{3} d e}{48 \, {\left (c x^{2} + a\right )}^{3} a^{3} c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(5*c*d^2 + a*e^2)*arctan(c*x/sqrt(a*c))/(sqrt(a*c)*a^3*c) + 1/48*(15*c^3*d^2*x^5 + 3*a*c^2*x^5*e^2 + 40*a
*c^2*d^2*x^3 + 8*a^2*c*x^3*e^2 + 33*a^2*c*d^2*x - 3*a^3*x*e^2 - 16*a^3*d*e)/((c*x^2 + a)^3*a^3*c)

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maple [A]  time = 0.05, size = 129, normalized size = 0.89 \[ \frac {e^{2} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{16 \sqrt {a c}\, a^{2} c}+\frac {5 d^{2} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{16 \sqrt {a c}\, a^{3}}+\frac {\frac {\left (a \,e^{2}+5 c \,d^{2}\right ) c \,x^{5}}{16 a^{3}}+\frac {\left (a \,e^{2}+5 c \,d^{2}\right ) x^{3}}{6 a^{2}}-\frac {d e}{3 c}-\frac {\left (a \,e^{2}-11 c \,d^{2}\right ) x}{16 a c}}{\left (c \,x^{2}+a \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(1/16*(a*e^2+5*c*d^2)/a^3*c*x^5+1/6/a^2*(a*e^2+5*c*d^2)*x^3-1/16*(a*e^2-11*c*d^2)/a/c*x-1/3/c*d*e)/(c*x^2+a)^3
+1/16/a^2/c/(a*c)^(1/2)*arctan(1/(a*c)^(1/2)*c*x)*e^2+5/16/a^3/(a*c)^(1/2)*arctan(1/(a*c)^(1/2)*c*x)*d^2

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maxima [A]  time = 3.07, size = 151, normalized size = 1.04 \[ \frac {3 \, {\left (5 \, c^{3} d^{2} + a c^{2} e^{2}\right )} x^{5} - 16 \, a^{3} d e + 8 \, {\left (5 \, a c^{2} d^{2} + a^{2} c e^{2}\right )} x^{3} + 3 \, {\left (11 \, a^{2} c d^{2} - a^{3} e^{2}\right )} x}{48 \, {\left (a^{3} c^{4} x^{6} + 3 \, a^{4} c^{3} x^{4} + 3 \, a^{5} c^{2} x^{2} + a^{6} c\right )}} + \frac {{\left (5 \, c d^{2} + a e^{2}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{16 \, \sqrt {a c} a^{3} c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(3*(5*c^3*d^2 + a*c^2*e^2)*x^5 - 16*a^3*d*e + 8*(5*a*c^2*d^2 + a^2*c*e^2)*x^3 + 3*(11*a^2*c*d^2 - a^3*e^2
)*x)/(a^3*c^4*x^6 + 3*a^4*c^3*x^4 + 3*a^5*c^2*x^2 + a^6*c) + 1/16*(5*c*d^2 + a*e^2)*arctan(c*x/sqrt(a*c))/(sqr
t(a*c)*a^3*c)

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mupad [B]  time = 0.36, size = 132, normalized size = 0.91 \[ \frac {\frac {x^3\,\left (5\,c\,d^2+a\,e^2\right )}{6\,a^2}-\frac {d\,e}{3\,c}-\frac {x\,\left (a\,e^2-11\,c\,d^2\right )}{16\,a\,c}+\frac {c\,x^5\,\left (5\,c\,d^2+a\,e^2\right )}{16\,a^3}}{a^3+3\,a^2\,c\,x^2+3\,a\,c^2\,x^4+c^3\,x^6}+\frac {\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {a}}\right )\,\left (5\,c\,d^2+a\,e^2\right )}{16\,a^{7/2}\,c^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^2/(a + c*x^2)^4,x)

[Out]

((x^3*(a*e^2 + 5*c*d^2))/(6*a^2) - (d*e)/(3*c) - (x*(a*e^2 - 11*c*d^2))/(16*a*c) + (c*x^5*(a*e^2 + 5*c*d^2))/(
16*a^3))/(a^3 + c^3*x^6 + 3*a^2*c*x^2 + 3*a*c^2*x^4) + (atan((c^(1/2)*x)/a^(1/2))*(a*e^2 + 5*c*d^2))/(16*a^(7/
2)*c^(3/2))

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sympy [A]  time = 1.04, size = 214, normalized size = 1.48 \[ - \frac {\sqrt {- \frac {1}{a^{7} c^{3}}} \left (a e^{2} + 5 c d^{2}\right ) \log {\left (- a^{4} c \sqrt {- \frac {1}{a^{7} c^{3}}} + x \right )}}{32} + \frac {\sqrt {- \frac {1}{a^{7} c^{3}}} \left (a e^{2} + 5 c d^{2}\right ) \log {\left (a^{4} c \sqrt {- \frac {1}{a^{7} c^{3}}} + x \right )}}{32} + \frac {- 16 a^{3} d e + x^{5} \left (3 a c^{2} e^{2} + 15 c^{3} d^{2}\right ) + x^{3} \left (8 a^{2} c e^{2} + 40 a c^{2} d^{2}\right ) + x \left (- 3 a^{3} e^{2} + 33 a^{2} c d^{2}\right )}{48 a^{6} c + 144 a^{5} c^{2} x^{2} + 144 a^{4} c^{3} x^{4} + 48 a^{3} c^{4} x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(-1/(a**7*c**3))*(a*e**2 + 5*c*d**2)*log(-a**4*c*sqrt(-1/(a**7*c**3)) + x)/32 + sqrt(-1/(a**7*c**3))*(a*e
**2 + 5*c*d**2)*log(a**4*c*sqrt(-1/(a**7*c**3)) + x)/32 + (-16*a**3*d*e + x**5*(3*a*c**2*e**2 + 15*c**3*d**2)
+ x**3*(8*a**2*c*e**2 + 40*a*c**2*d**2) + x*(-3*a**3*e**2 + 33*a**2*c*d**2))/(48*a**6*c + 144*a**5*c**2*x**2 +
144*a**4*c**3*x**4 + 48*a**3*c**4*x**6)

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