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

Optimal. Leaf size=131 \[ \frac {x \left (a e^2+c d^2\right )^2}{2 d e^4 \left (d+e x^2\right )}+\frac {c x \left (2 a e^2+3 c d^2\right )}{e^4}-\frac {\left (7 c d^2-a e^2\right ) \left (a e^2+c d^2\right ) \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 d^{3/2} e^{9/2}}-\frac {2 c^2 d x^3}{3 e^3}+\frac {c^2 x^5}{5 e^2} \]

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Rubi [A]  time = 0.19, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {1158, 1810, 205} \begin {gather*} \frac {x \left (a e^2+c d^2\right )^2}{2 d e^4 \left (d+e x^2\right )}+\frac {c x \left (2 a e^2+3 c d^2\right )}{e^4}-\frac {\left (7 c d^2-a e^2\right ) \left (a e^2+c d^2\right ) \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 d^{3/2} e^{9/2}}-\frac {2 c^2 d x^3}{3 e^3}+\frac {c^2 x^5}{5 e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + c*x^4)^2/(d + e*x^2)^2,x]

[Out]

(c*(3*c*d^2 + 2*a*e^2)*x)/e^4 - (2*c^2*d*x^3)/(3*e^3) + (c^2*x^5)/(5*e^2) + ((c*d^2 + a*e^2)^2*x)/(2*d*e^4*(d
+ e*x^2)) - ((7*c*d^2 - a*e^2)*(c*d^2 + a*e^2)*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(2*d^(3/2)*e^(9/2))

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 1158

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQuotient[(a + c*
x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + c*x^4)^p, d + e*x^2, x], x, 0]}, -Simp[(R*x*(d + e*x
^2)^(q + 1))/(2*d*(q + 1)), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*ExpandToSum[2*d*(q + 1)*Qx + R*
(2*q + 3), x], x], x]] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rule 1810

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a,
b}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps

\begin {align*} \int \frac {\left (a+c x^4\right )^2}{\left (d+e x^2\right )^2} \, dx &=\frac {\left (c d^2+a e^2\right )^2 x}{2 d e^4 \left (d+e x^2\right )}-\frac {\int \frac {-a^2+\frac {c^2 d^4}{e^4}+\frac {2 a c d^2}{e^2}-\frac {2 c d \left (c d^2+2 a e^2\right ) x^2}{e^3}+\frac {2 c^2 d^2 x^4}{e^2}-\frac {2 c^2 d x^6}{e}}{d+e x^2} \, dx}{2 d}\\ &=\frac {\left (c d^2+a e^2\right )^2 x}{2 d e^4 \left (d+e x^2\right )}-\frac {\int \left (-\frac {2 c d \left (3 c d^2+2 a e^2\right )}{e^4}+\frac {4 c^2 d^2 x^2}{e^3}-\frac {2 c^2 d x^4}{e^2}+\frac {7 c^2 d^4+6 a c d^2 e^2-a^2 e^4}{e^4 \left (d+e x^2\right )}\right ) \, dx}{2 d}\\ &=\frac {c \left (3 c d^2+2 a e^2\right ) x}{e^4}-\frac {2 c^2 d x^3}{3 e^3}+\frac {c^2 x^5}{5 e^2}+\frac {\left (c d^2+a e^2\right )^2 x}{2 d e^4 \left (d+e x^2\right )}-\frac {\left (\left (7 c d^2-a e^2\right ) \left (c d^2+a e^2\right )\right ) \int \frac {1}{d+e x^2} \, dx}{2 d e^4}\\ &=\frac {c \left (3 c d^2+2 a e^2\right ) x}{e^4}-\frac {2 c^2 d x^3}{3 e^3}+\frac {c^2 x^5}{5 e^2}+\frac {\left (c d^2+a e^2\right )^2 x}{2 d e^4 \left (d+e x^2\right )}-\frac {\left (7 c d^2-a e^2\right ) \left (c d^2+a e^2\right ) \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 d^{3/2} e^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 134, normalized size = 1.02 \begin {gather*} -\frac {\left (-a^2 e^4+6 a c d^2 e^2+7 c^2 d^4\right ) \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 d^{3/2} e^{9/2}}+\frac {x \left (a e^2+c d^2\right )^2}{2 d e^4 \left (d+e x^2\right )}+\frac {c x \left (2 a e^2+3 c d^2\right )}{e^4}-\frac {2 c^2 d x^3}{3 e^3}+\frac {c^2 x^5}{5 e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(c*(3*c*d^2 + 2*a*e^2)*x)/e^4 - (2*c^2*d*x^3)/(3*e^3) + (c^2*x^5)/(5*e^2) + ((c*d^2 + a*e^2)^2*x)/(2*d*e^4*(d
+ e*x^2)) - ((7*c^2*d^4 + 6*a*c*d^2*e^2 - a^2*e^4)*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(2*d^(3/2)*e^(9/2))

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a+c x^4\right )^2}{\left (d+e x^2\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.60, size = 394, normalized size = 3.01 \begin {gather*} \left [\frac {12 \, c^{2} d^{2} e^{4} x^{7} - 28 \, c^{2} d^{3} e^{3} x^{5} + 20 \, {\left (7 \, c^{2} d^{4} e^{2} + 6 \, a c d^{2} e^{4}\right )} x^{3} + 15 \, {\left (7 \, c^{2} d^{5} + 6 \, a c d^{3} e^{2} - a^{2} d e^{4} + {\left (7 \, c^{2} d^{4} e + 6 \, a c d^{2} e^{3} - a^{2} e^{5}\right )} x^{2}\right )} \sqrt {-d e} \log \left (\frac {e x^{2} - 2 \, \sqrt {-d e} x - d}{e x^{2} + d}\right ) + 30 \, {\left (7 \, c^{2} d^{5} e + 6 \, a c d^{3} e^{3} + a^{2} d e^{5}\right )} x}{60 \, {\left (d^{2} e^{6} x^{2} + d^{3} e^{5}\right )}}, \frac {6 \, c^{2} d^{2} e^{4} x^{7} - 14 \, c^{2} d^{3} e^{3} x^{5} + 10 \, {\left (7 \, c^{2} d^{4} e^{2} + 6 \, a c d^{2} e^{4}\right )} x^{3} - 15 \, {\left (7 \, c^{2} d^{5} + 6 \, a c d^{3} e^{2} - a^{2} d e^{4} + {\left (7 \, c^{2} d^{4} e + 6 \, a c d^{2} e^{3} - a^{2} e^{5}\right )} x^{2}\right )} \sqrt {d e} \arctan \left (\frac {\sqrt {d e} x}{d}\right ) + 15 \, {\left (7 \, c^{2} d^{5} e + 6 \, a c d^{3} e^{3} + a^{2} d e^{5}\right )} x}{30 \, {\left (d^{2} e^{6} x^{2} + d^{3} e^{5}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/60*(12*c^2*d^2*e^4*x^7 - 28*c^2*d^3*e^3*x^5 + 20*(7*c^2*d^4*e^2 + 6*a*c*d^2*e^4)*x^3 + 15*(7*c^2*d^5 + 6*a*
c*d^3*e^2 - a^2*d*e^4 + (7*c^2*d^4*e + 6*a*c*d^2*e^3 - a^2*e^5)*x^2)*sqrt(-d*e)*log((e*x^2 - 2*sqrt(-d*e)*x -
d)/(e*x^2 + d)) + 30*(7*c^2*d^5*e + 6*a*c*d^3*e^3 + a^2*d*e^5)*x)/(d^2*e^6*x^2 + d^3*e^5), 1/30*(6*c^2*d^2*e^4
*x^7 - 14*c^2*d^3*e^3*x^5 + 10*(7*c^2*d^4*e^2 + 6*a*c*d^2*e^4)*x^3 - 15*(7*c^2*d^5 + 6*a*c*d^3*e^2 - a^2*d*e^4
+ (7*c^2*d^4*e + 6*a*c*d^2*e^3 - a^2*e^5)*x^2)*sqrt(d*e)*arctan(sqrt(d*e)*x/d) + 15*(7*c^2*d^5*e + 6*a*c*d^3*
e^3 + a^2*d*e^5)*x)/(d^2*e^6*x^2 + d^3*e^5)]

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giac [A]  time = 0.17, size = 128, normalized size = 0.98 \begin {gather*} \frac {1}{15} \, {\left (3 \, c^{2} x^{5} e^{8} - 10 \, c^{2} d x^{3} e^{7} + 45 \, c^{2} d^{2} x e^{6} + 30 \, a c x e^{8}\right )} e^{\left (-10\right )} - \frac {{\left (7 \, c^{2} d^{4} + 6 \, a c d^{2} e^{2} - a^{2} e^{4}\right )} \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\left (-\frac {9}{2}\right )}}{2 \, d^{\frac {3}{2}}} + \frac {{\left (c^{2} d^{4} x + 2 \, a c d^{2} x e^{2} + a^{2} x e^{4}\right )} e^{\left (-4\right )}}{2 \, {\left (x^{2} e + d\right )} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(3*c^2*x^5*e^8 - 10*c^2*d*x^3*e^7 + 45*c^2*d^2*x*e^6 + 30*a*c*x*e^8)*e^(-10) - 1/2*(7*c^2*d^4 + 6*a*c*d^2
*e^2 - a^2*e^4)*arctan(x*e^(1/2)/sqrt(d))*e^(-9/2)/d^(3/2) + 1/2*(c^2*d^4*x + 2*a*c*d^2*x*e^2 + a^2*x*e^4)*e^(
-4)/((x^2*e + d)*d)

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maple [A]  time = 0.01, size = 170, normalized size = 1.30 \begin {gather*} \frac {c^{2} x^{5}}{5 e^{2}}-\frac {2 c^{2} d \,x^{3}}{3 e^{3}}+\frac {a^{2} x}{2 \left (e \,x^{2}+d \right ) d}+\frac {a^{2} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{2 \sqrt {d e}\, d}+\frac {a c d x}{\left (e \,x^{2}+d \right ) e^{2}}-\frac {3 a c d \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{\sqrt {d e}\, e^{2}}+\frac {c^{2} d^{3} x}{2 \left (e \,x^{2}+d \right ) e^{4}}-\frac {7 c^{2} d^{3} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{2 \sqrt {d e}\, e^{4}}+\frac {2 a c x}{e^{2}}+\frac {3 c^{2} d^{2} x}{e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.28, size = 142, normalized size = 1.08 \begin {gather*} \frac {{\left (c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} x}{2 \, {\left (d e^{5} x^{2} + d^{2} e^{4}\right )}} + \frac {3 \, c^{2} e^{2} x^{5} - 10 \, c^{2} d e x^{3} + 15 \, {\left (3 \, c^{2} d^{2} + 2 \, a c e^{2}\right )} x}{15 \, e^{4}} - \frac {{\left (7 \, c^{2} d^{4} + 6 \, a c d^{2} e^{2} - a^{2} e^{4}\right )} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{2 \, \sqrt {d e} d e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(c^2*d^4 + 2*a*c*d^2*e^2 + a^2*e^4)*x/(d*e^5*x^2 + d^2*e^4) + 1/15*(3*c^2*e^2*x^5 - 10*c^2*d*e*x^3 + 15*(3
*c^2*d^2 + 2*a*c*e^2)*x)/e^4 - 1/2*(7*c^2*d^4 + 6*a*c*d^2*e^2 - a^2*e^4)*arctan(e*x/sqrt(d*e))/(sqrt(d*e)*d*e^
4)

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mupad [B]  time = 4.40, size = 183, normalized size = 1.40 \begin {gather*} x\,\left (\frac {3\,c^2\,d^2}{e^4}+\frac {2\,a\,c}{e^2}\right )+\frac {c^2\,x^5}{5\,e^2}-\frac {2\,c^2\,d\,x^3}{3\,e^3}+\frac {x\,\left (a^2\,e^4+2\,a\,c\,d^2\,e^2+c^2\,d^4\right )}{2\,d\,\left (e^5\,x^2+d\,e^4\right )}-\frac {\mathrm {atan}\left (\frac {\sqrt {e}\,x\,\left (c\,d^2+a\,e^2\right )\,\left (a\,e^2-7\,c\,d^2\right )}{\sqrt {d}\,\left (-a^2\,e^4+6\,a\,c\,d^2\,e^2+7\,c^2\,d^4\right )}\right )\,\left (c\,d^2+a\,e^2\right )\,\left (a\,e^2-7\,c\,d^2\right )}{2\,d^{3/2}\,e^{9/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + c*x^4)^2/(d + e*x^2)^2,x)

[Out]

x*((3*c^2*d^2)/e^4 + (2*a*c)/e^2) + (c^2*x^5)/(5*e^2) - (2*c^2*d*x^3)/(3*e^3) + (x*(a^2*e^4 + c^2*d^4 + 2*a*c*
d^2*e^2))/(2*d*(d*e^4 + e^5*x^2)) - (atan((e^(1/2)*x*(a*e^2 + c*d^2)*(a*e^2 - 7*c*d^2))/(d^(1/2)*(7*c^2*d^4 -
a^2*e^4 + 6*a*c*d^2*e^2)))*(a*e^2 + c*d^2)*(a*e^2 - 7*c*d^2))/(2*d^(3/2)*e^(9/2))

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sympy [B]  time = 0.93, size = 314, normalized size = 2.40 \begin {gather*} - \frac {2 c^{2} d x^{3}}{3 e^{3}} + \frac {c^{2} x^{5}}{5 e^{2}} + x \left (\frac {2 a c}{e^{2}} + \frac {3 c^{2} d^{2}}{e^{4}}\right ) + \frac {x \left (a^{2} e^{4} + 2 a c d^{2} e^{2} + c^{2} d^{4}\right )}{2 d^{2} e^{4} + 2 d e^{5} x^{2}} - \frac {\sqrt {- \frac {1}{d^{3} e^{9}}} \left (a e^{2} - 7 c d^{2}\right ) \left (a e^{2} + c d^{2}\right ) \log {\left (- \frac {d^{2} e^{4} \sqrt {- \frac {1}{d^{3} e^{9}}} \left (a e^{2} - 7 c d^{2}\right ) \left (a e^{2} + c d^{2}\right )}{a^{2} e^{4} - 6 a c d^{2} e^{2} - 7 c^{2} d^{4}} + x \right )}}{4} + \frac {\sqrt {- \frac {1}{d^{3} e^{9}}} \left (a e^{2} - 7 c d^{2}\right ) \left (a e^{2} + c d^{2}\right ) \log {\left (\frac {d^{2} e^{4} \sqrt {- \frac {1}{d^{3} e^{9}}} \left (a e^{2} - 7 c d^{2}\right ) \left (a e^{2} + c d^{2}\right )}{a^{2} e^{4} - 6 a c d^{2} e^{2} - 7 c^{2} d^{4}} + x \right )}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*c**2*d*x**3/(3*e**3) + c**2*x**5/(5*e**2) + x*(2*a*c/e**2 + 3*c**2*d**2/e**4) + x*(a**2*e**4 + 2*a*c*d**2*e
**2 + c**2*d**4)/(2*d**2*e**4 + 2*d*e**5*x**2) - sqrt(-1/(d**3*e**9))*(a*e**2 - 7*c*d**2)*(a*e**2 + c*d**2)*lo
g(-d**2*e**4*sqrt(-1/(d**3*e**9))*(a*e**2 - 7*c*d**2)*(a*e**2 + c*d**2)/(a**2*e**4 - 6*a*c*d**2*e**2 - 7*c**2*
d**4) + x)/4 + sqrt(-1/(d**3*e**9))*(a*e**2 - 7*c*d**2)*(a*e**2 + c*d**2)*log(d**2*e**4*sqrt(-1/(d**3*e**9))*(
a*e**2 - 7*c*d**2)*(a*e**2 + c*d**2)/(a**2*e**4 - 6*a*c*d**2*e**2 - 7*c**2*d**4) + x)/4

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