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

Optimal. Leaf size=184 \[ \frac {x \left (5 a^2-\frac {14 a c d^2}{e^2}-\frac {19 c^2 d^4}{e^4}\right )}{24 d^2 \left (d+e x^2\right )^2}-\frac {\left (-5 a^2 e^4-2 a c d^2 e^2+35 c^2 d^4\right ) \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{16 d^{7/2} e^{9/2}}+\frac {x \left (5 a^2+\frac {2 a c d^2}{e^2}+\frac {29 c^2 d^4}{e^4}\right )}{16 d^3 \left (d+e x^2\right )}+\frac {x \left (a e^2+c d^2\right )^2}{6 d e^4 \left (d+e x^2\right )^3}+\frac {c^2 x}{e^4} \]

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Rubi [A]  time = 0.30, antiderivative size = 184, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {1158, 1814, 1157, 388, 205} \begin {gather*} \frac {x \left (5 a^2+\frac {2 a c d^2}{e^2}+\frac {29 c^2 d^4}{e^4}\right )}{16 d^3 \left (d+e x^2\right )}+\frac {x \left (5 a^2-\frac {14 a c d^2}{e^2}-\frac {19 c^2 d^4}{e^4}\right )}{24 d^2 \left (d+e x^2\right )^2}-\frac {\left (-5 a^2 e^4-2 a c d^2 e^2+35 c^2 d^4\right ) \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{16 d^{7/2} e^{9/2}}+\frac {x \left (a e^2+c d^2\right )^2}{6 d e^4 \left (d+e x^2\right )^3}+\frac {c^2 x}{e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + c*x^4)^2/(d + e*x^2)^4,x]

[Out]

(c^2*x)/e^4 + ((c*d^2 + a*e^2)^2*x)/(6*d*e^4*(d + e*x^2)^3) + ((5*a^2 - (19*c^2*d^4)/e^4 - (14*a*c*d^2)/e^2)*x
)/(24*d^2*(d + e*x^2)^2) + ((5*a^2 + (29*c^2*d^4)/e^4 + (2*a*c*d^2)/e^2)*x)/(16*d^3*(d + e*x^2)) - ((35*c^2*d^
4 - 2*a*c*d^2*e^2 - 5*a^2*e^4)*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(16*d^(7/2)*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 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

Rule 1157

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ
uotient[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + 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, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && N
eQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

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 1814

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[P
olynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[((a
*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*(p + 1)), Int[(a + b*x^2)^(p + 1)*ExpandToS
um[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]

Rubi steps

\begin {align*} \int \frac {\left (a+c x^4\right )^2}{\left (d+e x^2\right )^4} \, dx &=\frac {\left (c d^2+a e^2\right )^2 x}{6 d e^4 \left (d+e x^2\right )^3}-\frac {\int \frac {-5 a^2+\frac {c^2 d^4}{e^4}+\frac {2 a c d^2}{e^2}-\frac {6 c d \left (c d^2+2 a e^2\right ) x^2}{e^3}+\frac {6 c^2 d^2 x^4}{e^2}-\frac {6 c^2 d x^6}{e}}{\left (d+e x^2\right )^3} \, dx}{6 d}\\ &=\frac {\left (c d^2+a e^2\right )^2 x}{6 d e^4 \left (d+e x^2\right )^3}+\frac {\left (5 a^2-\frac {19 c^2 d^4}{e^4}-\frac {14 a c d^2}{e^2}\right ) x}{24 d^2 \left (d+e x^2\right )^2}+\frac {\int \frac {3 \left (5 a^2+\frac {5 c^2 d^4}{e^4}+\frac {2 a c d^2}{e^2}\right )-\frac {48 c^2 d^3 x^2}{e^3}+\frac {24 c^2 d^2 x^4}{e^2}}{\left (d+e x^2\right )^2} \, dx}{24 d^2}\\ &=\frac {\left (c d^2+a e^2\right )^2 x}{6 d e^4 \left (d+e x^2\right )^3}+\frac {\left (5 a^2-\frac {19 c^2 d^4}{e^4}-\frac {14 a c d^2}{e^2}\right ) x}{24 d^2 \left (d+e x^2\right )^2}+\frac {\left (5 a^2+\frac {29 c^2 d^4}{e^4}+\frac {2 a c d^2}{e^2}\right ) x}{16 d^3 \left (d+e x^2\right )}-\frac {\int \frac {-3 \left (5 a^2-\frac {19 c^2 d^4}{e^4}+\frac {2 a c d^2}{e^2}\right )-\frac {48 c^2 d^3 x^2}{e^3}}{d+e x^2} \, dx}{48 d^3}\\ &=\frac {c^2 x}{e^4}+\frac {\left (c d^2+a e^2\right )^2 x}{6 d e^4 \left (d+e x^2\right )^3}+\frac {\left (5 a^2-\frac {19 c^2 d^4}{e^4}-\frac {14 a c d^2}{e^2}\right ) x}{24 d^2 \left (d+e x^2\right )^2}+\frac {\left (5 a^2+\frac {29 c^2 d^4}{e^4}+\frac {2 a c d^2}{e^2}\right ) x}{16 d^3 \left (d+e x^2\right )}-\frac {\left (35 c^2 d^4-2 a c d^2 e^2-5 a^2 e^4\right ) \int \frac {1}{d+e x^2} \, dx}{16 d^3 e^4}\\ &=\frac {c^2 x}{e^4}+\frac {\left (c d^2+a e^2\right )^2 x}{6 d e^4 \left (d+e x^2\right )^3}+\frac {\left (5 a^2-\frac {19 c^2 d^4}{e^4}-\frac {14 a c d^2}{e^2}\right ) x}{24 d^2 \left (d+e x^2\right )^2}+\frac {\left (5 a^2+\frac {29 c^2 d^4}{e^4}+\frac {2 a c d^2}{e^2}\right ) x}{16 d^3 \left (d+e x^2\right )}-\frac {\left (35 c^2 d^4-2 a c d^2 e^2-5 a^2 e^4\right ) \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{16 d^{7/2} e^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 174, normalized size = 0.95 \begin {gather*} \frac {x \left (a^2 e^4 \left (33 d^2+40 d e x^2+15 e^2 x^4\right )-2 a c d^2 e^2 \left (3 d^2+8 d e x^2-3 e^2 x^4\right )+c^2 d^3 \left (105 d^3+280 d^2 e x^2+231 d e^2 x^4+48 e^3 x^6\right )\right )}{48 d^3 e^4 \left (d+e x^2\right )^3}-\frac {\left (-5 a^2 e^4-2 a c d^2 e^2+35 c^2 d^4\right ) \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{16 d^{7/2} e^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(-2*a*c*d^2*e^2*(3*d^2 + 8*d*e*x^2 - 3*e^2*x^4) + a^2*e^4*(33*d^2 + 40*d*e*x^2 + 15*e^2*x^4) + c^2*d^3*(105
*d^3 + 280*d^2*e*x^2 + 231*d*e^2*x^4 + 48*e^3*x^6)))/(48*d^3*e^4*(d + e*x^2)^3) - ((35*c^2*d^4 - 2*a*c*d^2*e^2
- 5*a^2*e^4)*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(16*d^(7/2)*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 )^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.87, size = 662, normalized size = 3.60 \begin {gather*} \left [\frac {96 \, c^{2} d^{4} e^{4} x^{7} + 6 \, {\left (77 \, c^{2} d^{5} e^{3} + 2 \, a c d^{3} e^{5} + 5 \, a^{2} d e^{7}\right )} x^{5} + 16 \, {\left (35 \, c^{2} d^{6} e^{2} - 2 \, a c d^{4} e^{4} + 5 \, a^{2} d^{2} e^{6}\right )} x^{3} + 3 \, {\left (35 \, c^{2} d^{7} - 2 \, a c d^{5} e^{2} - 5 \, a^{2} d^{3} e^{4} + {\left (35 \, c^{2} d^{4} e^{3} - 2 \, a c d^{2} e^{5} - 5 \, a^{2} e^{7}\right )} x^{6} + 3 \, {\left (35 \, c^{2} d^{5} e^{2} - 2 \, a c d^{3} e^{4} - 5 \, a^{2} d e^{6}\right )} x^{4} + 3 \, {\left (35 \, c^{2} d^{6} e - 2 \, a c d^{4} e^{3} - 5 \, a^{2} d^{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 ) + 6 \, {\left (35 \, c^{2} d^{7} e - 2 \, a c d^{5} e^{3} + 11 \, a^{2} d^{3} e^{5}\right )} x}{96 \, {\left (d^{4} e^{8} x^{6} + 3 \, d^{5} e^{7} x^{4} + 3 \, d^{6} e^{6} x^{2} + d^{7} e^{5}\right )}}, \frac {48 \, c^{2} d^{4} e^{4} x^{7} + 3 \, {\left (77 \, c^{2} d^{5} e^{3} + 2 \, a c d^{3} e^{5} + 5 \, a^{2} d e^{7}\right )} x^{5} + 8 \, {\left (35 \, c^{2} d^{6} e^{2} - 2 \, a c d^{4} e^{4} + 5 \, a^{2} d^{2} e^{6}\right )} x^{3} - 3 \, {\left (35 \, c^{2} d^{7} - 2 \, a c d^{5} e^{2} - 5 \, a^{2} d^{3} e^{4} + {\left (35 \, c^{2} d^{4} e^{3} - 2 \, a c d^{2} e^{5} - 5 \, a^{2} e^{7}\right )} x^{6} + 3 \, {\left (35 \, c^{2} d^{5} e^{2} - 2 \, a c d^{3} e^{4} - 5 \, a^{2} d e^{6}\right )} x^{4} + 3 \, {\left (35 \, c^{2} d^{6} e - 2 \, a c d^{4} e^{3} - 5 \, a^{2} d^{2} e^{5}\right )} x^{2}\right )} \sqrt {d e} \arctan \left (\frac {\sqrt {d e} x}{d}\right ) + 3 \, {\left (35 \, c^{2} d^{7} e - 2 \, a c d^{5} e^{3} + 11 \, a^{2} d^{3} e^{5}\right )} x}{48 \, {\left (d^{4} e^{8} x^{6} + 3 \, d^{5} e^{7} x^{4} + 3 \, d^{6} e^{6} x^{2} + d^{7} 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)^4,x, algorithm="fricas")

[Out]

[1/96*(96*c^2*d^4*e^4*x^7 + 6*(77*c^2*d^5*e^3 + 2*a*c*d^3*e^5 + 5*a^2*d*e^7)*x^5 + 16*(35*c^2*d^6*e^2 - 2*a*c*
d^4*e^4 + 5*a^2*d^2*e^6)*x^3 + 3*(35*c^2*d^7 - 2*a*c*d^5*e^2 - 5*a^2*d^3*e^4 + (35*c^2*d^4*e^3 - 2*a*c*d^2*e^5
- 5*a^2*e^7)*x^6 + 3*(35*c^2*d^5*e^2 - 2*a*c*d^3*e^4 - 5*a^2*d*e^6)*x^4 + 3*(35*c^2*d^6*e - 2*a*c*d^4*e^3 - 5
*a^2*d^2*e^5)*x^2)*sqrt(-d*e)*log((e*x^2 - 2*sqrt(-d*e)*x - d)/(e*x^2 + d)) + 6*(35*c^2*d^7*e - 2*a*c*d^5*e^3
+ 11*a^2*d^3*e^5)*x)/(d^4*e^8*x^6 + 3*d^5*e^7*x^4 + 3*d^6*e^6*x^2 + d^7*e^5), 1/48*(48*c^2*d^4*e^4*x^7 + 3*(77
*c^2*d^5*e^3 + 2*a*c*d^3*e^5 + 5*a^2*d*e^7)*x^5 + 8*(35*c^2*d^6*e^2 - 2*a*c*d^4*e^4 + 5*a^2*d^2*e^6)*x^3 - 3*(
35*c^2*d^7 - 2*a*c*d^5*e^2 - 5*a^2*d^3*e^4 + (35*c^2*d^4*e^3 - 2*a*c*d^2*e^5 - 5*a^2*e^7)*x^6 + 3*(35*c^2*d^5*
e^2 - 2*a*c*d^3*e^4 - 5*a^2*d*e^6)*x^4 + 3*(35*c^2*d^6*e - 2*a*c*d^4*e^3 - 5*a^2*d^2*e^5)*x^2)*sqrt(d*e)*arcta
n(sqrt(d*e)*x/d) + 3*(35*c^2*d^7*e - 2*a*c*d^5*e^3 + 11*a^2*d^3*e^5)*x)/(d^4*e^8*x^6 + 3*d^5*e^7*x^4 + 3*d^6*e
^6*x^2 + d^7*e^5)]

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giac [A]  time = 0.16, size = 167, normalized size = 0.91 \begin {gather*} c^{2} x e^{\left (-4\right )} - \frac {{\left (35 \, c^{2} d^{4} - 2 \, a c d^{2} e^{2} - 5 \, a^{2} e^{4}\right )} \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\left (-\frac {9}{2}\right )}}{16 \, d^{\frac {7}{2}}} + \frac {{\left (87 \, c^{2} d^{4} x^{5} e^{2} + 136 \, c^{2} d^{5} x^{3} e + 6 \, a c d^{2} x^{5} e^{4} + 57 \, c^{2} d^{6} x - 16 \, a c d^{3} x^{3} e^{3} + 15 \, a^{2} x^{5} e^{6} - 6 \, a c d^{4} x e^{2} + 40 \, a^{2} d x^{3} e^{5} + 33 \, a^{2} d^{2} x e^{4}\right )} e^{\left (-4\right )}}{48 \, {\left (x^{2} e + d\right )}^{3} d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c^2*x*e^(-4) - 1/16*(35*c^2*d^4 - 2*a*c*d^2*e^2 - 5*a^2*e^4)*arctan(x*e^(1/2)/sqrt(d))*e^(-9/2)/d^(7/2) + 1/48
*(87*c^2*d^4*x^5*e^2 + 136*c^2*d^5*x^3*e + 6*a*c*d^2*x^5*e^4 + 57*c^2*d^6*x - 16*a*c*d^3*x^3*e^3 + 15*a^2*x^5*
e^6 - 6*a*c*d^4*x*e^2 + 40*a^2*d*x^3*e^5 + 33*a^2*d^2*x*e^4)*e^(-4)/((x^2*e + d)^3*d^3)

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maple [A]  time = 0.01, size = 262, normalized size = 1.42 \begin {gather*} \frac {5 a^{2} e^{2} x^{5}}{16 \left (e \,x^{2}+d \right )^{3} d^{3}}+\frac {a c \,x^{5}}{8 \left (e \,x^{2}+d \right )^{3} d}+\frac {29 c^{2} d \,x^{5}}{16 \left (e \,x^{2}+d \right )^{3} e^{2}}+\frac {5 a^{2} e \,x^{3}}{6 \left (e \,x^{2}+d \right )^{3} d^{2}}-\frac {a c \,x^{3}}{3 \left (e \,x^{2}+d \right )^{3} e}+\frac {17 c^{2} d^{2} x^{3}}{6 \left (e \,x^{2}+d \right )^{3} e^{3}}+\frac {11 a^{2} x}{16 \left (e \,x^{2}+d \right )^{3} d}-\frac {a c d x}{8 \left (e \,x^{2}+d \right )^{3} e^{2}}+\frac {19 c^{2} d^{3} x}{16 \left (e \,x^{2}+d \right )^{3} e^{4}}+\frac {5 a^{2} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{16 \sqrt {d e}\, d^{3}}+\frac {a c \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{8 \sqrt {d e}\, d \,e^{2}}-\frac {35 c^{2} d \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{16 \sqrt {d e}\, e^{4}}+\frac {c^{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)^4,x)

[Out]

c^2*x/e^4+5/16*e^2/(e*x^2+d)^3/d^3*x^5*a^2+1/8/(e*x^2+d)^3/d*x^5*a*c+29/16/e^2/(e*x^2+d)^3*d*x^5*c^2+5/6*e/(e*
x^2+d)^3/d^2*x^3*a^2-1/3/e/(e*x^2+d)^3*x^3*a*c+17/6/e^3/(e*x^2+d)^3*d^2*x^3*c^2+11/16/(e*x^2+d)^3/d*x*a^2-1/8/
e^2/(e*x^2+d)^3*d*x*a*c+19/16/e^4/(e*x^2+d)^3*d^3*x*c^2+5/16/d^3/(d*e)^(1/2)*arctan(1/(d*e)^(1/2)*e*x)*a^2+1/8
/e^2/d/(d*e)^(1/2)*arctan(1/(d*e)^(1/2)*e*x)*a*c-35/16/e^4*d/(d*e)^(1/2)*arctan(1/(d*e)^(1/2)*e*x)*c^2

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maxima [A]  time = 2.39, size = 205, normalized size = 1.11 \begin {gather*} \frac {3 \, {\left (29 \, c^{2} d^{4} e^{2} + 2 \, a c d^{2} e^{4} + 5 \, a^{2} e^{6}\right )} x^{5} + 8 \, {\left (17 \, c^{2} d^{5} e - 2 \, a c d^{3} e^{3} + 5 \, a^{2} d e^{5}\right )} x^{3} + 3 \, {\left (19 \, c^{2} d^{6} - 2 \, a c d^{4} e^{2} + 11 \, a^{2} d^{2} e^{4}\right )} x}{48 \, {\left (d^{3} e^{7} x^{6} + 3 \, d^{4} e^{6} x^{4} + 3 \, d^{5} e^{5} x^{2} + d^{6} e^{4}\right )}} + \frac {c^{2} x}{e^{4}} - \frac {{\left (35 \, c^{2} d^{4} - 2 \, a c d^{2} e^{2} - 5 \, a^{2} e^{4}\right )} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{16 \, \sqrt {d e} d^{3} 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)^4,x, algorithm="maxima")

[Out]

1/48*(3*(29*c^2*d^4*e^2 + 2*a*c*d^2*e^4 + 5*a^2*e^6)*x^5 + 8*(17*c^2*d^5*e - 2*a*c*d^3*e^3 + 5*a^2*d*e^5)*x^3
+ 3*(19*c^2*d^6 - 2*a*c*d^4*e^2 + 11*a^2*d^2*e^4)*x)/(d^3*e^7*x^6 + 3*d^4*e^6*x^4 + 3*d^5*e^5*x^2 + d^6*e^4) +
c^2*x/e^4 - 1/16*(35*c^2*d^4 - 2*a*c*d^2*e^2 - 5*a^2*e^4)*arctan(e*x/sqrt(d*e))/(sqrt(d*e)*d^3*e^4)

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mupad [B]  time = 4.49, size = 199, normalized size = 1.08 \begin {gather*} \frac {\frac {x^3\,\left (5\,a^2\,e^5-2\,a\,c\,d^2\,e^3+17\,c^2\,d^4\,e\right )}{6\,d^2}+\frac {x\,\left (11\,a^2\,e^4-2\,a\,c\,d^2\,e^2+19\,c^2\,d^4\right )}{16\,d}+\frac {x^5\,\left (5\,a^2\,e^6+2\,a\,c\,d^2\,e^4+29\,c^2\,d^4\,e^2\right )}{16\,d^3}}{d^3\,e^4+3\,d^2\,e^5\,x^2+3\,d\,e^6\,x^4+e^7\,x^6}+\frac {c^2\,x}{e^4}+\frac {\mathrm {atan}\left (\frac {\sqrt {e}\,x}{\sqrt {d}}\right )\,\left (5\,a^2\,e^4+2\,a\,c\,d^2\,e^2-35\,c^2\,d^4\right )}{16\,d^{7/2}\,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)^4,x)

[Out]

((x^3*(5*a^2*e^5 + 17*c^2*d^4*e - 2*a*c*d^2*e^3))/(6*d^2) + (x*(11*a^2*e^4 + 19*c^2*d^4 - 2*a*c*d^2*e^2))/(16*
d) + (x^5*(5*a^2*e^6 + 29*c^2*d^4*e^2 + 2*a*c*d^2*e^4))/(16*d^3))/(d^3*e^4 + e^7*x^6 + 3*d*e^6*x^4 + 3*d^2*e^5
*x^2) + (c^2*x)/e^4 + (atan((e^(1/2)*x)/d^(1/2))*(5*a^2*e^4 - 35*c^2*d^4 + 2*a*c*d^2*e^2))/(16*d^(7/2)*e^(9/2)
)

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sympy [A]  time = 2.61, size = 292, normalized size = 1.59 \begin {gather*} \frac {c^{2} x}{e^{4}} - \frac {\sqrt {- \frac {1}{d^{7} e^{9}}} \left (5 a^{2} e^{4} + 2 a c d^{2} e^{2} - 35 c^{2} d^{4}\right ) \log {\left (- d^{4} e^{4} \sqrt {- \frac {1}{d^{7} e^{9}}} + x \right )}}{32} + \frac {\sqrt {- \frac {1}{d^{7} e^{9}}} \left (5 a^{2} e^{4} + 2 a c d^{2} e^{2} - 35 c^{2} d^{4}\right ) \log {\left (d^{4} e^{4} \sqrt {- \frac {1}{d^{7} e^{9}}} + x \right )}}{32} + \frac {x^{5} \left (15 a^{2} e^{6} + 6 a c d^{2} e^{4} + 87 c^{2} d^{4} e^{2}\right ) + x^{3} \left (40 a^{2} d e^{5} - 16 a c d^{3} e^{3} + 136 c^{2} d^{5} e\right ) + x \left (33 a^{2} d^{2} e^{4} - 6 a c d^{4} e^{2} + 57 c^{2} d^{6}\right )}{48 d^{6} e^{4} + 144 d^{5} e^{5} x^{2} + 144 d^{4} e^{6} x^{4} + 48 d^{3} e^{7} x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c**2*x/e**4 - sqrt(-1/(d**7*e**9))*(5*a**2*e**4 + 2*a*c*d**2*e**2 - 35*c**2*d**4)*log(-d**4*e**4*sqrt(-1/(d**7
*e**9)) + x)/32 + sqrt(-1/(d**7*e**9))*(5*a**2*e**4 + 2*a*c*d**2*e**2 - 35*c**2*d**4)*log(d**4*e**4*sqrt(-1/(d
**7*e**9)) + x)/32 + (x**5*(15*a**2*e**6 + 6*a*c*d**2*e**4 + 87*c**2*d**4*e**2) + x**3*(40*a**2*d*e**5 - 16*a*
c*d**3*e**3 + 136*c**2*d**5*e) + x*(33*a**2*d**2*e**4 - 6*a*c*d**4*e**2 + 57*c**2*d**6))/(48*d**6*e**4 + 144*d
**5*e**5*x**2 + 144*d**4*e**6*x**4 + 48*d**3*e**7*x**6)

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