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

Optimal. Leaf size=149 \[ \frac {\left (-3 a^2 e^4+6 a c d^2 e^2+c^2 d^4\right ) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{2 a^{3/2} c^{5/2}}-\frac {3 e^2 x \left (c d^2-a e^2\right )}{2 a c^2}+\frac {2 d e^3 \log \left (a+c x^2\right )}{c^2}-\frac {d e^3 x^2}{2 a c}-\frac {(d+e x)^3 (a e-c d x)}{2 a c \left (a+c x^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {739, 801, 635, 205, 260} \[ \frac {\left (-3 a^2 e^4+6 a c d^2 e^2+c^2 d^4\right ) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{2 a^{3/2} c^{5/2}}-\frac {3 e^2 x \left (c d^2-a e^2\right )}{2 a c^2}+\frac {2 d e^3 \log \left (a+c x^2\right )}{c^2}-\frac {d e^3 x^2}{2 a c}-\frac {(d+e x)^3 (a e-c d x)}{2 a c \left (a+c x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^4/(a + c*x^2)^2,x]

[Out]

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

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 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 635

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

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]

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rubi steps

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

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Mathematica [A]  time = 0.08, size = 137, normalized size = 0.92 \[ \frac {a^2 e^3 (4 d+e x)-2 a c d^2 e (2 d+3 e x)+c^2 d^4 x}{2 a c^2 \left (a+c x^2\right )}+\frac {\left (-3 a^2 e^4+6 a c d^2 e^2+c^2 d^4\right ) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{2 a^{3/2} c^{5/2}}+\frac {2 d e^3 \log \left (a+c x^2\right )}{c^2}+\frac {e^4 x}{c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.04, size = 433, normalized size = 2.91 \[ \left [\frac {4 \, a^{2} c^{2} e^{4} x^{3} - 8 \, a^{2} c^{2} d^{3} e + 8 \, a^{3} c d e^{3} + {\left (a c^{2} d^{4} + 6 \, a^{2} c d^{2} e^{2} - 3 \, a^{3} e^{4} + {\left (c^{3} d^{4} + 6 \, a c^{2} d^{2} e^{2} - 3 \, a^{2} c e^{4}\right )} x^{2}\right )} \sqrt {-a c} \log \left (\frac {c x^{2} + 2 \, \sqrt {-a c} x - a}{c x^{2} + a}\right ) + 2 \, {\left (a c^{3} d^{4} - 6 \, a^{2} c^{2} d^{2} e^{2} + 3 \, a^{3} c e^{4}\right )} x + 8 \, {\left (a^{2} c^{2} d e^{3} x^{2} + a^{3} c d e^{3}\right )} \log \left (c x^{2} + a\right )}{4 \, {\left (a^{2} c^{4} x^{2} + a^{3} c^{3}\right )}}, \frac {2 \, a^{2} c^{2} e^{4} x^{3} - 4 \, a^{2} c^{2} d^{3} e + 4 \, a^{3} c d e^{3} + {\left (a c^{2} d^{4} + 6 \, a^{2} c d^{2} e^{2} - 3 \, a^{3} e^{4} + {\left (c^{3} d^{4} + 6 \, a c^{2} d^{2} e^{2} - 3 \, a^{2} c e^{4}\right )} x^{2}\right )} \sqrt {a c} \arctan \left (\frac {\sqrt {a c} x}{a}\right ) + {\left (a c^{3} d^{4} - 6 \, a^{2} c^{2} d^{2} e^{2} + 3 \, a^{3} c e^{4}\right )} x + 4 \, {\left (a^{2} c^{2} d e^{3} x^{2} + a^{3} c d e^{3}\right )} \log \left (c x^{2} + a\right )}{2 \, {\left (a^{2} c^{4} x^{2} + a^{3} c^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(4*a^2*c^2*e^4*x^3 - 8*a^2*c^2*d^3*e + 8*a^3*c*d*e^3 + (a*c^2*d^4 + 6*a^2*c*d^2*e^2 - 3*a^3*e^4 + (c^3*d^
4 + 6*a*c^2*d^2*e^2 - 3*a^2*c*e^4)*x^2)*sqrt(-a*c)*log((c*x^2 + 2*sqrt(-a*c)*x - a)/(c*x^2 + a)) + 2*(a*c^3*d^
4 - 6*a^2*c^2*d^2*e^2 + 3*a^3*c*e^4)*x + 8*(a^2*c^2*d*e^3*x^2 + a^3*c*d*e^3)*log(c*x^2 + a))/(a^2*c^4*x^2 + a^
3*c^3), 1/2*(2*a^2*c^2*e^4*x^3 - 4*a^2*c^2*d^3*e + 4*a^3*c*d*e^3 + (a*c^2*d^4 + 6*a^2*c*d^2*e^2 - 3*a^3*e^4 +
(c^3*d^4 + 6*a*c^2*d^2*e^2 - 3*a^2*c*e^4)*x^2)*sqrt(a*c)*arctan(sqrt(a*c)*x/a) + (a*c^3*d^4 - 6*a^2*c^2*d^2*e^
2 + 3*a^3*c*e^4)*x + 4*(a^2*c^2*d*e^3*x^2 + a^3*c*d*e^3)*log(c*x^2 + a))/(a^2*c^4*x^2 + a^3*c^3)]

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giac [A]  time = 0.16, size = 131, normalized size = 0.88 \[ \frac {2 \, d e^{3} \log \left (c x^{2} + a\right )}{c^{2}} + \frac {x e^{4}}{c^{2}} + \frac {{\left (c^{2} d^{4} + 6 \, a c d^{2} e^{2} - 3 \, a^{2} e^{4}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \, \sqrt {a c} a c^{2}} - \frac {4 \, a c d^{3} e - 4 \, a^{2} d e^{3} - {\left (c^{2} d^{4} - 6 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} x}{2 \, {\left (c x^{2} + a\right )} a c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 192, normalized size = 1.29 \[ \frac {a \,e^{4} x}{2 \left (c \,x^{2}+a \right ) c^{2}}-\frac {3 a \,e^{4} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \sqrt {a c}\, c^{2}}+\frac {d^{4} x}{2 \left (c \,x^{2}+a \right ) a}+\frac {d^{4} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \sqrt {a c}\, a}-\frac {3 d^{2} e^{2} x}{\left (c \,x^{2}+a \right ) c}+\frac {3 d^{2} e^{2} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c}\, c}+\frac {2 a d \,e^{3}}{\left (c \,x^{2}+a \right ) c^{2}}-\frac {2 d^{3} e}{\left (c \,x^{2}+a \right ) c}+\frac {2 d \,e^{3} \ln \left (c \,x^{2}+a \right )}{c^{2}}+\frac {e^{4} x}{c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^2*e^4*x+1/2/c^2/(c*x^2+a)*a*x*e^4-3/c/(c*x^2+a)*x*d^2*e^2+1/2/(c*x^2+a)/a*x*d^4+2/c^2/(c*x^2+a)*d*e^3*a-2/
c/(c*x^2+a)*d^3*e+2*d*e^3*ln(c*x^2+a)/c^2-3/2/c^2*a/(a*c)^(1/2)*arctan(1/(a*c)^(1/2)*c*x)*e^4+3/c/(a*c)^(1/2)*
arctan(1/(a*c)^(1/2)*c*x)*d^2*e^2+1/2/a/(a*c)^(1/2)*arctan(1/(a*c)^(1/2)*c*x)*d^4

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maxima [A]  time = 3.01, size = 140, normalized size = 0.94 \[ \frac {e^{4} x}{c^{2}} + \frac {2 \, d e^{3} \log \left (c x^{2} + a\right )}{c^{2}} - \frac {4 \, a c d^{3} e - 4 \, a^{2} d e^{3} - {\left (c^{2} d^{4} - 6 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} x}{2 \, {\left (a c^{3} x^{2} + a^{2} c^{2}\right )}} + \frac {{\left (c^{2} d^{4} + 6 \, a c d^{2} e^{2} - 3 \, a^{2} e^{4}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \, \sqrt {a c} a c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.34, size = 131, normalized size = 0.88 \[ \frac {\frac {x\,\left (a^2\,e^4-6\,a\,c\,d^2\,e^2+c^2\,d^4\right )}{2\,a}+2\,a\,d\,e^3-2\,c\,d^3\,e}{c^3\,x^2+a\,c^2}+\frac {e^4\,x}{c^2}+\frac {2\,d\,e^3\,\ln \left (c\,x^2+a\right )}{c^2}+\frac {\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {a}}\right )\,\left (-3\,a^2\,e^4+6\,a\,c\,d^2\,e^2+c^2\,d^4\right )}{2\,a^{3/2}\,c^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^4/(a + c*x^2)^2,x)

[Out]

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

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sympy [B]  time = 1.40, size = 403, normalized size = 2.70 \[ \left (\frac {2 d e^{3}}{c^{2}} - \frac {\sqrt {- a^{3} c^{5}} \left (3 a^{2} e^{4} - 6 a c d^{2} e^{2} - c^{2} d^{4}\right )}{4 a^{3} c^{5}}\right ) \log {\left (x + \frac {- 4 a^{2} c^{2} \left (\frac {2 d e^{3}}{c^{2}} - \frac {\sqrt {- a^{3} c^{5}} \left (3 a^{2} e^{4} - 6 a c d^{2} e^{2} - c^{2} d^{4}\right )}{4 a^{3} c^{5}}\right ) + 8 a^{2} d e^{3}}{3 a^{2} e^{4} - 6 a c d^{2} e^{2} - c^{2} d^{4}} \right )} + \left (\frac {2 d e^{3}}{c^{2}} + \frac {\sqrt {- a^{3} c^{5}} \left (3 a^{2} e^{4} - 6 a c d^{2} e^{2} - c^{2} d^{4}\right )}{4 a^{3} c^{5}}\right ) \log {\left (x + \frac {- 4 a^{2} c^{2} \left (\frac {2 d e^{3}}{c^{2}} + \frac {\sqrt {- a^{3} c^{5}} \left (3 a^{2} e^{4} - 6 a c d^{2} e^{2} - c^{2} d^{4}\right )}{4 a^{3} c^{5}}\right ) + 8 a^{2} d e^{3}}{3 a^{2} e^{4} - 6 a c d^{2} e^{2} - c^{2} d^{4}} \right )} + \frac {4 a^{2} d e^{3} - 4 a c d^{3} e + x \left (a^{2} e^{4} - 6 a c d^{2} e^{2} + c^{2} d^{4}\right )}{2 a^{2} c^{2} + 2 a c^{3} x^{2}} + \frac {e^{4} x}{c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*d*e**3/c**2 - sqrt(-a**3*c**5)*(3*a**2*e**4 - 6*a*c*d**2*e**2 - c**2*d**4)/(4*a**3*c**5))*log(x + (-4*a**2*
c**2*(2*d*e**3/c**2 - sqrt(-a**3*c**5)*(3*a**2*e**4 - 6*a*c*d**2*e**2 - c**2*d**4)/(4*a**3*c**5)) + 8*a**2*d*e
**3)/(3*a**2*e**4 - 6*a*c*d**2*e**2 - c**2*d**4)) + (2*d*e**3/c**2 + sqrt(-a**3*c**5)*(3*a**2*e**4 - 6*a*c*d**
2*e**2 - c**2*d**4)/(4*a**3*c**5))*log(x + (-4*a**2*c**2*(2*d*e**3/c**2 + sqrt(-a**3*c**5)*(3*a**2*e**4 - 6*a*
c*d**2*e**2 - c**2*d**4)/(4*a**3*c**5)) + 8*a**2*d*e**3)/(3*a**2*e**4 - 6*a*c*d**2*e**2 - c**2*d**4)) + (4*a**
2*d*e**3 - 4*a*c*d**3*e + x*(a**2*e**4 - 6*a*c*d**2*e**2 + c**2*d**4))/(2*a**2*c**2 + 2*a*c**3*x**2) + e**4*x/
c**2

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