3.2.27 \(\int \frac {a+c x^4}{(d+e x^2)^4} \, dx\) [127]

Optimal. Leaf size=123 \[ \frac {\left (a+\frac {c d^2}{e^2}\right ) x}{6 d \left (d+e x^2\right )^3}+\frac {\left (\frac {5 a}{d^2}-\frac {7 c}{e^2}\right ) x}{24 \left (d+e x^2\right )^2}+\frac {\left (\frac {5 a}{d^2}+\frac {c}{e^2}\right ) x}{16 d \left (d+e x^2\right )}+\frac {\left (c d^2+5 a e^2\right ) \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{16 d^{7/2} e^{5/2}} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 123, 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 = {1172, 393, 205, 211} \begin {gather*} \frac {\left (5 a e^2+c d^2\right ) \text {ArcTan}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{16 d^{7/2} e^{5/2}}+\frac {x \left (\frac {5 a}{d^2}+\frac {c}{e^2}\right )}{16 d \left (d+e x^2\right )}+\frac {x \left (\frac {5 a}{d^2}-\frac {7 c}{e^2}\right )}{24 \left (d+e x^2\right )^2}+\frac {x \left (a+\frac {c d^2}{e^2}\right )}{6 d \left (d+e x^2\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + c*x^4)/(d + e*x^2)^4,x]

[Out]

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

Rule 205

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] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

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 393

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

Rule 1172

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]

Rubi steps

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

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Mathematica [A]
time = 0.05, size = 113, normalized size = 0.92 \begin {gather*} \frac {x \left (c d^2 \left (-3 d^2-8 d e x^2+3 e^2 x^4\right )+a e^2 \left (33 d^2+40 d e x^2+15 e^2 x^4\right )\right )}{48 d^3 e^2 \left (d+e x^2\right )^3}+\frac {\left (c d^2+5 a e^2\right ) \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{16 d^{7/2} e^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(c*d^2*(-3*d^2 - 8*d*e*x^2 + 3*e^2*x^4) + a*e^2*(33*d^2 + 40*d*e*x^2 + 15*e^2*x^4)))/(48*d^3*e^2*(d + e*x^2
)^3) + ((c*d^2 + 5*a*e^2)*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(16*d^(7/2)*e^(5/2))

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Maple [A]
time = 0.15, size = 113, normalized size = 0.92

method	result	size
default	\(\frac {\frac {\left (5 a \,e^{2}+c \,d^{2}\right ) x^{5}}{16 d^{3}}+\frac {\left (5 a \,e^{2}-c \,d^{2}\right ) x^{3}}{6 d^{2} e}+\frac {\left (11 a \,e^{2}-c \,d^{2}\right ) x}{16 d \,e^{2}}}{\left (e \,x^{2}+d \right )^{3}}+\frac {\left (5 a \,e^{2}+c \,d^{2}\right ) \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{16 d^{3} e^{2} \sqrt {d e}}\)	\(113\)
risch	\(\frac {\frac {\left (5 a \,e^{2}+c \,d^{2}\right ) x^{5}}{16 d^{3}}+\frac {\left (5 a \,e^{2}-c \,d^{2}\right ) x^{3}}{6 d^{2} e}+\frac {\left (11 a \,e^{2}-c \,d^{2}\right ) x}{16 d \,e^{2}}}{\left (e \,x^{2}+d \right )^{3}}-\frac {5 \ln \left (e x +\sqrt {-d e}\right ) a}{32 \sqrt {-d e}\, d^{3}}-\frac {\ln \left (e x +\sqrt {-d e}\right ) c}{32 \sqrt {-d e}\, e^{2} d}+\frac {5 \ln \left (-e x +\sqrt {-d e}\right ) a}{32 \sqrt {-d e}\, d^{3}}+\frac {\ln \left (-e x +\sqrt {-d e}\right ) c}{32 \sqrt {-d e}\, e^{2} d}\)	\(179\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.52, size = 122, normalized size = 0.99 \begin {gather*} \frac {3 \, {\left (c d^{2} e^{2} + 5 \, a e^{4}\right )} x^{5} - 8 \, {\left (c d^{3} e - 5 \, a d e^{3}\right )} x^{3} - 3 \, {\left (c d^{4} - 11 \, a d^{2} e^{2}\right )} x}{48 \, {\left (d^{3} x^{6} e^{5} + 3 \, d^{4} x^{4} e^{4} + 3 \, d^{5} x^{2} e^{3} + d^{6} e^{2}\right )}} + \frac {{\left (c d^{2} + 5 \, a e^{2}\right )} \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\left (-\frac {5}{2}\right )}}{16 \, d^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.36, size = 407, normalized size = 3.31 \begin {gather*} \left [-\frac {16 \, c d^{4} x^{3} e^{2} - 30 \, a d x^{5} e^{5} + 6 \, c d^{5} x e - 80 \, a d^{2} x^{3} e^{4} + 3 \, {\left (5 \, a x^{6} e^{5} + 3 \, c d^{4} x^{2} e + 15 \, a d x^{4} e^{4} + c d^{5} + {\left (c d^{2} x^{6} + 15 \, a d^{2} x^{2}\right )} e^{3} + {\left (3 \, c d^{3} x^{4} + 5 \, a d^{3}\right )} e^{2}\right )} \sqrt {-d e} \log \left (\frac {x^{2} e - 2 \, \sqrt {-d e} x - d}{x^{2} e + d}\right ) - 6 \, {\left (c d^{3} x^{5} + 11 \, a d^{3} x\right )} e^{3}}{96 \, {\left (d^{4} x^{6} e^{6} + 3 \, d^{5} x^{4} e^{5} + 3 \, d^{6} x^{2} e^{4} + d^{7} e^{3}\right )}}, -\frac {8 \, c d^{4} x^{3} e^{2} - 15 \, a d x^{5} e^{5} + 3 \, c d^{5} x e - 40 \, a d^{2} x^{3} e^{4} - 3 \, {\left (5 \, a x^{6} e^{5} + 3 \, c d^{4} x^{2} e + 15 \, a d x^{4} e^{4} + c d^{5} + {\left (c d^{2} x^{6} + 15 \, a d^{2} x^{2}\right )} e^{3} + {\left (3 \, c d^{3} x^{4} + 5 \, a d^{3}\right )} e^{2}\right )} \sqrt {d} \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\frac {1}{2}} - 3 \, {\left (c d^{3} x^{5} + 11 \, a d^{3} x\right )} e^{3}}{48 \, {\left (d^{4} x^{6} e^{6} + 3 \, d^{5} x^{4} e^{5} + 3 \, d^{6} x^{2} e^{4} + d^{7} e^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/96*(16*c*d^4*x^3*e^2 - 30*a*d*x^5*e^5 + 6*c*d^5*x*e - 80*a*d^2*x^3*e^4 + 3*(5*a*x^6*e^5 + 3*c*d^4*x^2*e +
15*a*d*x^4*e^4 + c*d^5 + (c*d^2*x^6 + 15*a*d^2*x^2)*e^3 + (3*c*d^3*x^4 + 5*a*d^3)*e^2)*sqrt(-d*e)*log((x^2*e -
2*sqrt(-d*e)*x - d)/(x^2*e + d)) - 6*(c*d^3*x^5 + 11*a*d^3*x)*e^3)/(d^4*x^6*e^6 + 3*d^5*x^4*e^5 + 3*d^6*x^2*e
^4 + d^7*e^3), -1/48*(8*c*d^4*x^3*e^2 - 15*a*d*x^5*e^5 + 3*c*d^5*x*e - 40*a*d^2*x^3*e^4 - 3*(5*a*x^6*e^5 + 3*c
*d^4*x^2*e + 15*a*d*x^4*e^4 + c*d^5 + (c*d^2*x^6 + 15*a*d^2*x^2)*e^3 + (3*c*d^3*x^4 + 5*a*d^3)*e^2)*sqrt(d)*ar
ctan(x*e^(1/2)/sqrt(d))*e^(1/2) - 3*(c*d^3*x^5 + 11*a*d^3*x)*e^3)/(d^4*x^6*e^6 + 3*d^5*x^4*e^5 + 3*d^6*x^2*e^4
+ d^7*e^3)]

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Sympy [A]
time = 0.47, size = 204, normalized size = 1.66 \begin {gather*} - \frac {\sqrt {- \frac {1}{d^{7} e^{5}}} \cdot \left (5 a e^{2} + c d^{2}\right ) \log {\left (- d^{4} e^{2} \sqrt {- \frac {1}{d^{7} e^{5}}} + x \right )}}{32} + \frac {\sqrt {- \frac {1}{d^{7} e^{5}}} \cdot \left (5 a e^{2} + c d^{2}\right ) \log {\left (d^{4} e^{2} \sqrt {- \frac {1}{d^{7} e^{5}}} + x \right )}}{32} + \frac {x^{5} \cdot \left (15 a e^{4} + 3 c d^{2} e^{2}\right ) + x^{3} \cdot \left (40 a d e^{3} - 8 c d^{3} e\right ) + x \left (33 a d^{2} e^{2} - 3 c d^{4}\right )}{48 d^{6} e^{2} + 144 d^{5} e^{3} x^{2} + 144 d^{4} e^{4} x^{4} + 48 d^{3} e^{5} x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 3.16, size = 100, normalized size = 0.81 \begin {gather*} \frac {{\left (c d^{2} + 5 \, a e^{2}\right )} \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\left (-\frac {5}{2}\right )}}{16 \, d^{\frac {7}{2}}} + \frac {{\left (3 \, c d^{2} x^{5} e^{2} - 8 \, c d^{3} x^{3} e + 15 \, a x^{5} e^{4} - 3 \, c d^{4} x + 40 \, a d x^{3} e^{3} + 33 \, a d^{2} x e^{2}\right )} e^{\left (-2\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)/(e*x^2+d)^4,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 4.48, size = 129, normalized size = 1.05 \begin {gather*} \frac {\frac {x^5\,\left (c\,d^2+5\,a\,e^2\right )}{16\,d^3}+\frac {x^3\,\left (5\,a\,e^2-c\,d^2\right )}{6\,d^2\,e}+\frac {x\,\left (11\,a\,e^2-c\,d^2\right )}{16\,d\,e^2}}{d^3+3\,d^2\,e\,x^2+3\,d\,e^2\,x^4+e^3\,x^6}+\frac {\mathrm {atan}\left (\frac {\sqrt {e}\,x}{\sqrt {d}}\right )\,\left (c\,d^2+5\,a\,e^2\right )}{16\,d^{7/2}\,e^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + c*x^4)/(d + e*x^2)^4,x)

[Out]

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

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