3.2.15 \(\int \frac {(a+c x^4)^2}{(d+e x^2)^3} \, dx\)

Optimal. Leaf size=155 \[ \frac {x \left (3 a^2-\frac {10 a c d^2}{e^2}-\frac {13 c^2 d^4}{e^4}\right )}{8 d^2 \left (d+e x^2\right )}+\frac {\left (3 a^2 e^4+6 a c d^2 e^2+35 c^2 d^4\right ) \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{8 d^{5/2} e^{9/2}}+\frac {x \left (a e^2+c d^2\right )^2}{4 d e^4 \left (d+e x^2\right )^2}-\frac {3 c^2 d x}{e^4}+\frac {c^2 x^3}{3 e^3} \]

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Rubi [A]  time = 0.25, antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {1158, 1814, 1153, 205} \begin {gather*} \frac {x \left (3 a^2-\frac {10 a c d^2}{e^2}-\frac {13 c^2 d^4}{e^4}\right )}{8 d^2 \left (d+e x^2\right )}+\frac {\left (3 a^2 e^4+6 a c d^2 e^2+35 c^2 d^4\right ) \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{8 d^{5/2} e^{9/2}}+\frac {x \left (a e^2+c d^2\right )^2}{4 d e^4 \left (d+e x^2\right )^2}-\frac {3 c^2 d x}{e^4}+\frac {c^2 x^3}{3 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + c*x^4)^2/(d + e*x^2)^3,x]

[Out]

(-3*c^2*d*x)/e^4 + (c^2*x^3)/(3*e^3) + ((c*d^2 + a*e^2)^2*x)/(4*d*e^4*(d + e*x^2)^2) + ((3*a^2 - (13*c^2*d^4)/
e^4 - (10*a*c*d^2)/e^2)*x)/(8*d^2*(d + e*x^2)) + ((35*c^2*d^4 + 6*a*c*d^2*e^2 + 3*a^2*e^4)*ArcTan[(Sqrt[e]*x)/
Sqrt[d]])/(8*d^(5/2)*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 1153

Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 -
b*d*e + a*e^2, 0] && IGtQ[p, 0] && IGtQ[q, -2]

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 )^3} \, dx &=\frac {\left (c d^2+a e^2\right )^2 x}{4 d e^4 \left (d+e x^2\right )^2}-\frac {\int \frac {-3 a^2+\frac {c^2 d^4}{e^4}+\frac {2 a c d^2}{e^2}-\frac {4 c d \left (c d^2+2 a e^2\right ) x^2}{e^3}+\frac {4 c^2 d^2 x^4}{e^2}-\frac {4 c^2 d x^6}{e}}{\left (d+e x^2\right )^2} \, dx}{4 d}\\ &=\frac {\left (c d^2+a e^2\right )^2 x}{4 d e^4 \left (d+e x^2\right )^2}+\frac {\left (3 a^2-\frac {13 c^2 d^4}{e^4}-\frac {10 a c d^2}{e^2}\right ) x}{8 d^2 \left (d+e x^2\right )}+\frac {\int \frac {3 a^2+\frac {11 c^2 d^4}{e^4}+\frac {6 a c d^2}{e^2}-\frac {16 c^2 d^3 x^2}{e^3}+\frac {8 c^2 d^2 x^4}{e^2}}{d+e x^2} \, dx}{8 d^2}\\ &=\frac {\left (c d^2+a e^2\right )^2 x}{4 d e^4 \left (d+e x^2\right )^2}+\frac {\left (3 a^2-\frac {13 c^2 d^4}{e^4}-\frac {10 a c d^2}{e^2}\right ) x}{8 d^2 \left (d+e x^2\right )}+\frac {\int \left (-\frac {24 c^2 d^3}{e^4}+\frac {8 c^2 d^2 x^2}{e^3}+\frac {35 c^2 d^4+6 a c d^2 e^2+3 a^2 e^4}{e^4 \left (d+e x^2\right )}\right ) \, dx}{8 d^2}\\ &=-\frac {3 c^2 d x}{e^4}+\frac {c^2 x^3}{3 e^3}+\frac {\left (c d^2+a e^2\right )^2 x}{4 d e^4 \left (d+e x^2\right )^2}+\frac {\left (3 a^2-\frac {13 c^2 d^4}{e^4}-\frac {10 a c d^2}{e^2}\right ) x}{8 d^2 \left (d+e x^2\right )}+\frac {\left (35 c^2 d^4+6 a c d^2 e^2+3 a^2 e^4\right ) \int \frac {1}{d+e x^2} \, dx}{8 d^2 e^4}\\ &=-\frac {3 c^2 d x}{e^4}+\frac {c^2 x^3}{3 e^3}+\frac {\left (c d^2+a e^2\right )^2 x}{4 d e^4 \left (d+e x^2\right )^2}+\frac {\left (3 a^2-\frac {13 c^2 d^4}{e^4}-\frac {10 a c d^2}{e^2}\right ) x}{8 d^2 \left (d+e x^2\right )}+\frac {\left (35 c^2 d^4+6 a c d^2 e^2+3 a^2 e^4\right ) \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{8 d^{5/2} e^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 154, normalized size = 0.99 \begin {gather*} \frac {x \left (3 a^2 e^4 \left (5 d+3 e x^2\right )-6 a c d^2 e^2 \left (3 d+5 e x^2\right )-c^2 d^2 \left (105 d^3+175 d^2 e x^2+56 d e^2 x^4-8 e^3 x^6\right )\right )}{24 d^2 e^4 \left (d+e x^2\right )^2}+\frac {\left (3 a^2 e^4+6 a c d^2 e^2+35 c^2 d^4\right ) \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{8 d^{5/2} e^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + c*x^4)^2/(d + e*x^2)^3,x]

[Out]

(x*(3*a^2*e^4*(5*d + 3*e*x^2) - 6*a*c*d^2*e^2*(3*d + 5*e*x^2) - c^2*d^2*(105*d^3 + 175*d^2*e*x^2 + 56*d*e^2*x^
4 - 8*e^3*x^6)))/(24*d^2*e^4*(d + e*x^2)^2) + ((35*c^2*d^4 + 6*a*c*d^2*e^2 + 3*a^2*e^4)*ArcTan[(Sqrt[e]*x)/Sqr
t[d]])/(8*d^(5/2)*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 )^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(a + c*x^4)^2/(d + e*x^2)^3,x]

[Out]

IntegrateAlgebraic[(a + c*x^4)^2/(d + e*x^2)^3, x]

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fricas [A]  time = 0.88, size = 516, normalized size = 3.33 \begin {gather*} \left [\frac {16 \, c^{2} d^{3} e^{4} x^{7} - 112 \, c^{2} d^{4} e^{3} x^{5} - 2 \, {\left (175 \, c^{2} d^{5} e^{2} + 30 \, a c d^{3} e^{4} - 9 \, a^{2} d e^{6}\right )} x^{3} - 3 \, {\left (35 \, c^{2} d^{6} + 6 \, a c d^{4} e^{2} + 3 \, a^{2} d^{2} e^{4} + {\left (35 \, c^{2} d^{4} e^{2} + 6 \, a c d^{2} e^{4} + 3 \, a^{2} e^{6}\right )} x^{4} + 2 \, {\left (35 \, c^{2} d^{5} e + 6 \, a c d^{3} e^{3} + 3 \, a^{2} d 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^{6} e + 6 \, a c d^{4} e^{3} - 5 \, a^{2} d^{2} e^{5}\right )} x}{48 \, {\left (d^{3} e^{7} x^{4} + 2 \, d^{4} e^{6} x^{2} + d^{5} e^{5}\right )}}, \frac {8 \, c^{2} d^{3} e^{4} x^{7} - 56 \, c^{2} d^{4} e^{3} x^{5} - {\left (175 \, c^{2} d^{5} e^{2} + 30 \, a c d^{3} e^{4} - 9 \, a^{2} d e^{6}\right )} x^{3} + 3 \, {\left (35 \, c^{2} d^{6} + 6 \, a c d^{4} e^{2} + 3 \, a^{2} d^{2} e^{4} + {\left (35 \, c^{2} d^{4} e^{2} + 6 \, a c d^{2} e^{4} + 3 \, a^{2} e^{6}\right )} x^{4} + 2 \, {\left (35 \, c^{2} d^{5} e + 6 \, a c d^{3} e^{3} + 3 \, a^{2} d e^{5}\right )} x^{2}\right )} \sqrt {d e} \arctan \left (\frac {\sqrt {d e} x}{d}\right ) - 3 \, {\left (35 \, c^{2} d^{6} e + 6 \, a c d^{4} e^{3} - 5 \, a^{2} d^{2} e^{5}\right )} x}{24 \, {\left (d^{3} e^{7} x^{4} + 2 \, d^{4} e^{6} x^{2} + d^{5} 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)^3,x, algorithm="fricas")

[Out]

[1/48*(16*c^2*d^3*e^4*x^7 - 112*c^2*d^4*e^3*x^5 - 2*(175*c^2*d^5*e^2 + 30*a*c*d^3*e^4 - 9*a^2*d*e^6)*x^3 - 3*(
35*c^2*d^6 + 6*a*c*d^4*e^2 + 3*a^2*d^2*e^4 + (35*c^2*d^4*e^2 + 6*a*c*d^2*e^4 + 3*a^2*e^6)*x^4 + 2*(35*c^2*d^5*
e + 6*a*c*d^3*e^3 + 3*a^2*d*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^6
*e + 6*a*c*d^4*e^3 - 5*a^2*d^2*e^5)*x)/(d^3*e^7*x^4 + 2*d^4*e^6*x^2 + d^5*e^5), 1/24*(8*c^2*d^3*e^4*x^7 - 56*c
^2*d^4*e^3*x^5 - (175*c^2*d^5*e^2 + 30*a*c*d^3*e^4 - 9*a^2*d*e^6)*x^3 + 3*(35*c^2*d^6 + 6*a*c*d^4*e^2 + 3*a^2*
d^2*e^4 + (35*c^2*d^4*e^2 + 6*a*c*d^2*e^4 + 3*a^2*e^6)*x^4 + 2*(35*c^2*d^5*e + 6*a*c*d^3*e^3 + 3*a^2*d*e^5)*x^
2)*sqrt(d*e)*arctan(sqrt(d*e)*x/d) - 3*(35*c^2*d^6*e + 6*a*c*d^4*e^3 - 5*a^2*d^2*e^5)*x)/(d^3*e^7*x^4 + 2*d^4*
e^6*x^2 + d^5*e^5)]

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giac [A]  time = 0.17, size = 145, normalized size = 0.94 \begin {gather*} \frac {1}{3} \, {\left (c^{2} x^{3} e^{6} - 9 \, c^{2} d x e^{5}\right )} e^{\left (-9\right )} + \frac {{\left (35 \, c^{2} d^{4} + 6 \, a c d^{2} e^{2} + 3 \, a^{2} e^{4}\right )} \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\left (-\frac {9}{2}\right )}}{8 \, d^{\frac {5}{2}}} - \frac {{\left (13 \, c^{2} d^{4} x^{3} e + 11 \, c^{2} d^{5} x + 10 \, a c d^{2} x^{3} e^{3} + 6 \, a c d^{3} x e^{2} - 3 \, a^{2} x^{3} e^{5} - 5 \, a^{2} d x e^{4}\right )} e^{\left (-4\right )}}{8 \, {\left (x^{2} e + d\right )}^{2} d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+a)^2/(e*x^2+d)^3,x, algorithm="giac")

[Out]

1/3*(c^2*x^3*e^6 - 9*c^2*d*x*e^5)*e^(-9) + 1/8*(35*c^2*d^4 + 6*a*c*d^2*e^2 + 3*a^2*e^4)*arctan(x*e^(1/2)/sqrt(
d))*e^(-9/2)/d^(5/2) - 1/8*(13*c^2*d^4*x^3*e + 11*c^2*d^5*x + 10*a*c*d^2*x^3*e^3 + 6*a*c*d^3*x*e^2 - 3*a^2*x^3
*e^5 - 5*a^2*d*x*e^4)*e^(-4)/((x^2*e + d)^2*d^2)

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maple [A]  time = 0.01, size = 211, normalized size = 1.36 \begin {gather*} \frac {3 a^{2} e \,x^{3}}{8 \left (e \,x^{2}+d \right )^{2} d^{2}}-\frac {5 a c \,x^{3}}{4 \left (e \,x^{2}+d \right )^{2} e}-\frac {13 c^{2} d^{2} x^{3}}{8 \left (e \,x^{2}+d \right )^{2} e^{3}}+\frac {5 a^{2} x}{8 \left (e \,x^{2}+d \right )^{2} d}-\frac {3 a c d x}{4 \left (e \,x^{2}+d \right )^{2} e^{2}}-\frac {11 c^{2} d^{3} x}{8 \left (e \,x^{2}+d \right )^{2} e^{4}}+\frac {c^{2} x^{3}}{3 e^{3}}+\frac {3 a^{2} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{8 \sqrt {d e}\, d^{2}}+\frac {3 a c \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{4 \sqrt {d e}\, e^{2}}+\frac {35 c^{2} d^{2} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{8 \sqrt {d e}\, e^{4}}-\frac {3 c^{2} d 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)^3,x)

[Out]

1/3*c^2*x^3/e^3-3*c^2*d*x/e^4+3/8*e/(e*x^2+d)^2/d^2*x^3*a^2-5/4/e/(e*x^2+d)^2*x^3*a*c-13/8/e^3/(e*x^2+d)^2*d^2
*x^3*c^2+5/8/(e*x^2+d)^2/d*x*a^2-3/4/e^2/(e*x^2+d)^2*d*x*a*c-11/8/e^4/(e*x^2+d)^2*d^3*x*c^2+3/8/d^2/(d*e)^(1/2
)*arctan(1/(d*e)^(1/2)*e*x)*a^2+3/4/e^2/(d*e)^(1/2)*arctan(1/(d*e)^(1/2)*e*x)*a*c+35/8/e^4*d^2/(d*e)^(1/2)*arc
tan(1/(d*e)^(1/2)*e*x)*c^2

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maxima [A]  time = 2.31, size = 167, normalized size = 1.08 \begin {gather*} -\frac {{\left (13 \, c^{2} d^{4} e + 10 \, a c d^{2} e^{3} - 3 \, a^{2} e^{5}\right )} x^{3} + {\left (11 \, c^{2} d^{5} + 6 \, a c d^{3} e^{2} - 5 \, a^{2} d e^{4}\right )} x}{8 \, {\left (d^{2} e^{6} x^{4} + 2 \, d^{3} e^{5} x^{2} + d^{4} e^{4}\right )}} + \frac {c^{2} e x^{3} - 9 \, c^{2} d x}{3 \, e^{4}} + \frac {{\left (35 \, c^{2} d^{4} + 6 \, a c d^{2} e^{2} + 3 \, a^{2} e^{4}\right )} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{8 \, \sqrt {d e} d^{2} 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)^3,x, algorithm="maxima")

[Out]

-1/8*((13*c^2*d^4*e + 10*a*c*d^2*e^3 - 3*a^2*e^5)*x^3 + (11*c^2*d^5 + 6*a*c*d^3*e^2 - 5*a^2*d*e^4)*x)/(d^2*e^6
*x^4 + 2*d^3*e^5*x^2 + d^4*e^4) + 1/3*(c^2*e*x^3 - 9*c^2*d*x)/e^4 + 1/8*(35*c^2*d^4 + 6*a*c*d^2*e^2 + 3*a^2*e^
4)*arctan(e*x/sqrt(d*e))/(sqrt(d*e)*d^2*e^4)

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mupad [B]  time = 4.41, size = 164, normalized size = 1.06 \begin {gather*} \frac {c^2\,x^3}{3\,e^3}-\frac {\frac {x^3\,\left (-3\,a^2\,e^5+10\,a\,c\,d^2\,e^3+13\,c^2\,d^4\,e\right )}{8\,d^2}+\frac {x\,\left (-5\,a^2\,e^4+6\,a\,c\,d^2\,e^2+11\,c^2\,d^4\right )}{8\,d}}{d^2\,e^4+2\,d\,e^5\,x^2+e^6\,x^4}+\frac {\mathrm {atan}\left (\frac {\sqrt {e}\,x}{\sqrt {d}}\right )\,\left (3\,a^2\,e^4+6\,a\,c\,d^2\,e^2+35\,c^2\,d^4\right )}{8\,d^{5/2}\,e^{9/2}}-\frac {3\,c^2\,d\,x}{e^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + c*x^4)^2/(d + e*x^2)^3,x)

[Out]

(c^2*x^3)/(3*e^3) - ((x^3*(13*c^2*d^4*e - 3*a^2*e^5 + 10*a*c*d^2*e^3))/(8*d^2) + (x*(11*c^2*d^4 - 5*a^2*e^4 +
6*a*c*d^2*e^2))/(8*d))/(d^2*e^4 + e^6*x^4 + 2*d*e^5*x^2) + (atan((e^(1/2)*x)/d^(1/2))*(3*a^2*e^4 + 35*c^2*d^4
+ 6*a*c*d^2*e^2))/(8*d^(5/2)*e^(9/2)) - (3*c^2*d*x)/e^4

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sympy [A]  time = 1.71, size = 257, normalized size = 1.66 \begin {gather*} - \frac {3 c^{2} d x}{e^{4}} + \frac {c^{2} x^{3}}{3 e^{3}} - \frac {\sqrt {- \frac {1}{d^{5} e^{9}}} \left (3 a^{2} e^{4} + 6 a c d^{2} e^{2} + 35 c^{2} d^{4}\right ) \log {\left (- d^{3} e^{4} \sqrt {- \frac {1}{d^{5} e^{9}}} + x \right )}}{16} + \frac {\sqrt {- \frac {1}{d^{5} e^{9}}} \left (3 a^{2} e^{4} + 6 a c d^{2} e^{2} + 35 c^{2} d^{4}\right ) \log {\left (d^{3} e^{4} \sqrt {- \frac {1}{d^{5} e^{9}}} + x \right )}}{16} + \frac {x^{3} \left (3 a^{2} e^{5} - 10 a c d^{2} e^{3} - 13 c^{2} d^{4} e\right ) + x \left (5 a^{2} d e^{4} - 6 a c d^{3} e^{2} - 11 c^{2} d^{5}\right )}{8 d^{4} e^{4} + 16 d^{3} e^{5} x^{2} + 8 d^{2} e^{6} x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+a)**2/(e*x**2+d)**3,x)

[Out]

-3*c**2*d*x/e**4 + c**2*x**3/(3*e**3) - sqrt(-1/(d**5*e**9))*(3*a**2*e**4 + 6*a*c*d**2*e**2 + 35*c**2*d**4)*lo
g(-d**3*e**4*sqrt(-1/(d**5*e**9)) + x)/16 + sqrt(-1/(d**5*e**9))*(3*a**2*e**4 + 6*a*c*d**2*e**2 + 35*c**2*d**4
)*log(d**3*e**4*sqrt(-1/(d**5*e**9)) + x)/16 + (x**3*(3*a**2*e**5 - 10*a*c*d**2*e**3 - 13*c**2*d**4*e) + x*(5*
a**2*d*e**4 - 6*a*c*d**3*e**2 - 11*c**2*d**5))/(8*d**4*e**4 + 16*d**3*e**5*x**2 + 8*d**2*e**6*x**4)

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