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

Optimal. Leaf size=155 \[ \frac {\left (a e^2+c d^2\right ) \left (a e^2+5 c d^2\right ) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{16 a^{7/2} c^{5/2}}-\frac {(d+e x) \left (a e^2+5 c d^2\right ) (a e-c d x)}{16 a^3 c^2 \left (a+c x^2\right )}-\frac {(d+e x)^3 (a e-5 c d x)}{24 a^2 c \left (a+c x^2\right )^2}+\frac {x (d+e x)^4}{6 a \left (a+c x^2\right )^3} \]

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Rubi [A]  time = 0.08, antiderivative size = 155, 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 = {737, 805, 723, 205} \begin {gather*} -\frac {(d+e x) \left (a e^2+5 c d^2\right ) (a e-c d x)}{16 a^3 c^2 \left (a+c x^2\right )}+\frac {\left (a e^2+c d^2\right ) \left (a e^2+5 c d^2\right ) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{16 a^{7/2} c^{5/2}}-\frac {(d+e x)^3 (a e-5 c d x)}{24 a^2 c \left (a+c x^2\right )^2}+\frac {x (d+e x)^4}{6 a \left (a+c x^2\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^4/(a + c*x^2)^4,x]

[Out]

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

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 723

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[((2*p + 3)*(c*d^2 + a*e^2))/(2*a*c*(p + 1)), Int[(d + e*x)^(m -
2)*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ[m + 2*p + 2, 0] && Lt
Q[p, -1]

Rule 737

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(x*(d + e*x)^m*(a + c*x^2)^(p + 1)
)/(2*a*(p + 1)), x] + Dist[1/(2*a*(p + 1)), Int[(d + e*x)^(m - 1)*(d*(2*p + 3) + e*(m + 2*p + 3)*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, 0] && (LtQ[m, 1]
|| (ILtQ[m + 2*p + 3, 0] && NeQ[m, 2])) && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 805

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^m*
(a + c*x^2)^(p + 1)*(a*g - c*f*x))/(2*a*c*(p + 1)), x] - Dist[(m*(c*d*f + a*e*g))/(2*a*c*(p + 1)), Int[(d + e*
x)^(m - 1)*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ[Simplif
y[m + 2*p + 3], 0] && LtQ[p, -1]

Rubi steps

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

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Mathematica [A]  time = 0.15, size = 197, normalized size = 1.27 \begin {gather*} \frac {\left (a^2 e^4+6 a c d^2 e^2+5 c^2 d^4\right ) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{16 a^{7/2} c^{5/2}}+\frac {-a^4 e^3 (16 d+3 e x)-2 a^3 c e \left (16 d^3+9 d^2 e x+24 d e^2 x^2+4 e^3 x^3\right )+3 a^2 c^2 x \left (11 d^4+16 d^2 e^2 x^2+e^4 x^4\right )+2 a c^3 d^2 x^3 \left (20 d^2+9 e^2 x^2\right )+15 c^4 d^4 x^5}{48 a^3 c^2 \left (a+c x^2\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(15*c^4*d^4*x^5 - a^4*e^3*(16*d + 3*e*x) + 2*a*c^3*d^2*x^3*(20*d^2 + 9*e^2*x^2) - 2*a^3*c*e*(16*d^3 + 9*d^2*e*
x + 24*d*e^2*x^2 + 4*e^3*x^3) + 3*a^2*c^2*x*(11*d^4 + 16*d^2*e^2*x^2 + e^4*x^4))/(48*a^3*c^2*(a + c*x^2)^3) +
((5*c^2*d^4 + 6*a*c*d^2*e^2 + a^2*e^4)*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(16*a^(7/2)*c^(5/2))

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [B]  time = 0.42, size = 746, normalized size = 4.81 \begin {gather*} \left [-\frac {96 \, a^{4} c^{2} d e^{3} x^{2} + 64 \, a^{4} c^{2} d^{3} e + 32 \, a^{5} c d e^{3} - 6 \, {\left (5 \, a c^{5} d^{4} + 6 \, a^{2} c^{4} d^{2} e^{2} + a^{3} c^{3} e^{4}\right )} x^{5} - 16 \, {\left (5 \, a^{2} c^{4} d^{4} + 6 \, a^{3} c^{3} d^{2} e^{2} - a^{4} c^{2} e^{4}\right )} x^{3} + 3 \, {\left (5 \, a^{3} c^{2} d^{4} + 6 \, a^{4} c d^{2} e^{2} + a^{5} e^{4} + {\left (5 \, c^{5} d^{4} + 6 \, a c^{4} d^{2} e^{2} + a^{2} c^{3} e^{4}\right )} x^{6} + 3 \, {\left (5 \, a c^{4} d^{4} + 6 \, a^{2} c^{3} d^{2} e^{2} + a^{3} c^{2} e^{4}\right )} x^{4} + 3 \, {\left (5 \, a^{2} c^{3} d^{4} + 6 \, a^{3} c^{2} d^{2} e^{2} + a^{4} 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 ) - 6 \, {\left (11 \, a^{3} c^{3} d^{4} - 6 \, a^{4} c^{2} d^{2} e^{2} - a^{5} c e^{4}\right )} x}{96 \, {\left (a^{4} c^{6} x^{6} + 3 \, a^{5} c^{5} x^{4} + 3 \, a^{6} c^{4} x^{2} + a^{7} c^{3}\right )}}, -\frac {48 \, a^{4} c^{2} d e^{3} x^{2} + 32 \, a^{4} c^{2} d^{3} e + 16 \, a^{5} c d e^{3} - 3 \, {\left (5 \, a c^{5} d^{4} + 6 \, a^{2} c^{4} d^{2} e^{2} + a^{3} c^{3} e^{4}\right )} x^{5} - 8 \, {\left (5 \, a^{2} c^{4} d^{4} + 6 \, a^{3} c^{3} d^{2} e^{2} - a^{4} c^{2} e^{4}\right )} x^{3} - 3 \, {\left (5 \, a^{3} c^{2} d^{4} + 6 \, a^{4} c d^{2} e^{2} + a^{5} e^{4} + {\left (5 \, c^{5} d^{4} + 6 \, a c^{4} d^{2} e^{2} + a^{2} c^{3} e^{4}\right )} x^{6} + 3 \, {\left (5 \, a c^{4} d^{4} + 6 \, a^{2} c^{3} d^{2} e^{2} + a^{3} c^{2} e^{4}\right )} x^{4} + 3 \, {\left (5 \, a^{2} c^{3} d^{4} + 6 \, a^{3} c^{2} d^{2} e^{2} + a^{4} c e^{4}\right )} x^{2}\right )} \sqrt {a c} \arctan \left (\frac {\sqrt {a c} x}{a}\right ) - 3 \, {\left (11 \, a^{3} c^{3} d^{4} - 6 \, a^{4} c^{2} d^{2} e^{2} - a^{5} c e^{4}\right )} x}{48 \, {\left (a^{4} c^{6} x^{6} + 3 \, a^{5} c^{5} x^{4} + 3 \, a^{6} c^{4} x^{2} + a^{7} c^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/96*(96*a^4*c^2*d*e^3*x^2 + 64*a^4*c^2*d^3*e + 32*a^5*c*d*e^3 - 6*(5*a*c^5*d^4 + 6*a^2*c^4*d^2*e^2 + a^3*c^
3*e^4)*x^5 - 16*(5*a^2*c^4*d^4 + 6*a^3*c^3*d^2*e^2 - a^4*c^2*e^4)*x^3 + 3*(5*a^3*c^2*d^4 + 6*a^4*c*d^2*e^2 + a
^5*e^4 + (5*c^5*d^4 + 6*a*c^4*d^2*e^2 + a^2*c^3*e^4)*x^6 + 3*(5*a*c^4*d^4 + 6*a^2*c^3*d^2*e^2 + a^3*c^2*e^4)*x
^4 + 3*(5*a^2*c^3*d^4 + 6*a^3*c^2*d^2*e^2 + a^4*c*e^4)*x^2)*sqrt(-a*c)*log((c*x^2 - 2*sqrt(-a*c)*x - a)/(c*x^2
+ a)) - 6*(11*a^3*c^3*d^4 - 6*a^4*c^2*d^2*e^2 - a^5*c*e^4)*x)/(a^4*c^6*x^6 + 3*a^5*c^5*x^4 + 3*a^6*c^4*x^2 +
a^7*c^3), -1/48*(48*a^4*c^2*d*e^3*x^2 + 32*a^4*c^2*d^3*e + 16*a^5*c*d*e^3 - 3*(5*a*c^5*d^4 + 6*a^2*c^4*d^2*e^2
+ a^3*c^3*e^4)*x^5 - 8*(5*a^2*c^4*d^4 + 6*a^3*c^3*d^2*e^2 - a^4*c^2*e^4)*x^3 - 3*(5*a^3*c^2*d^4 + 6*a^4*c*d^2
*e^2 + a^5*e^4 + (5*c^5*d^4 + 6*a*c^4*d^2*e^2 + a^2*c^3*e^4)*x^6 + 3*(5*a*c^4*d^4 + 6*a^2*c^3*d^2*e^2 + a^3*c^
2*e^4)*x^4 + 3*(5*a^2*c^3*d^4 + 6*a^3*c^2*d^2*e^2 + a^4*c*e^4)*x^2)*sqrt(a*c)*arctan(sqrt(a*c)*x/a) - 3*(11*a^
3*c^3*d^4 - 6*a^4*c^2*d^2*e^2 - a^5*c*e^4)*x)/(a^4*c^6*x^6 + 3*a^5*c^5*x^4 + 3*a^6*c^4*x^2 + a^7*c^3)]

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giac [A]  time = 0.20, size = 205, normalized size = 1.32 \begin {gather*} \frac {{\left (5 \, c^{2} d^{4} + 6 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{16 \, \sqrt {a c} a^{3} c^{2}} + \frac {15 \, c^{4} d^{4} x^{5} + 18 \, a c^{3} d^{2} x^{5} e^{2} + 40 \, a c^{3} d^{4} x^{3} + 3 \, a^{2} c^{2} x^{5} e^{4} + 48 \, a^{2} c^{2} d^{2} x^{3} e^{2} + 33 \, a^{2} c^{2} d^{4} x - 8 \, a^{3} c x^{3} e^{4} - 48 \, a^{3} c d x^{2} e^{3} - 18 \, a^{3} c d^{2} x e^{2} - 32 \, a^{3} c d^{3} e - 3 \, a^{4} x e^{4} - 16 \, a^{4} d e^{3}}{48 \, {\left (c x^{2} + a\right )}^{3} a^{3} c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(5*c^2*d^4 + 6*a*c*d^2*e^2 + a^2*e^4)*arctan(c*x/sqrt(a*c))/(sqrt(a*c)*a^3*c^2) + 1/48*(15*c^4*d^4*x^5 +
18*a*c^3*d^2*x^5*e^2 + 40*a*c^3*d^4*x^3 + 3*a^2*c^2*x^5*e^4 + 48*a^2*c^2*d^2*x^3*e^2 + 33*a^2*c^2*d^4*x - 8*a^
3*c*x^3*e^4 - 48*a^3*c*d*x^2*e^3 - 18*a^3*c*d^2*x*e^2 - 32*a^3*c*d^3*e - 3*a^4*x*e^4 - 16*a^4*d*e^3)/((c*x^2 +
a)^3*a^3*c^2)

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maple [A]  time = 0.05, size = 225, normalized size = 1.45 \begin {gather*} \frac {e^{4} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{16 \sqrt {a c}\, a \,c^{2}}+\frac {3 d^{2} e^{2} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{8 \sqrt {a c}\, a^{2} c}+\frac {5 d^{4} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{16 \sqrt {a c}\, a^{3}}+\frac {-\frac {d \,e^{3} x^{2}}{c}+\frac {\left (a^{2} e^{4}+6 a c \,d^{2} e^{2}+5 c^{2} d^{4}\right ) x^{5}}{16 a^{3}}-\frac {\left (a^{2} e^{4}-6 a c \,d^{2} e^{2}-5 c^{2} d^{4}\right ) x^{3}}{6 a^{2} c}-\frac {\left (a \,e^{2}+2 c \,d^{2}\right ) d e}{3 c^{2}}-\frac {\left (a^{2} e^{4}+6 a c \,d^{2} e^{2}-11 c^{2} d^{4}\right ) x}{16 a \,c^{2}}}{\left (c \,x^{2}+a \right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 3.04, size = 235, normalized size = 1.52 \begin {gather*} -\frac {48 \, a^{3} c d e^{3} x^{2} + 32 \, a^{3} c d^{3} e + 16 \, a^{4} d e^{3} - 3 \, {\left (5 \, c^{4} d^{4} + 6 \, a c^{3} d^{2} e^{2} + a^{2} c^{2} e^{4}\right )} x^{5} - 8 \, {\left (5 \, a c^{3} d^{4} + 6 \, a^{2} c^{2} d^{2} e^{2} - a^{3} c e^{4}\right )} x^{3} - 3 \, {\left (11 \, a^{2} c^{2} d^{4} - 6 \, a^{3} c d^{2} e^{2} - a^{4} e^{4}\right )} x}{48 \, {\left (a^{3} c^{5} x^{6} + 3 \, a^{4} c^{4} x^{4} + 3 \, a^{5} c^{3} x^{2} + a^{6} c^{2}\right )}} + \frac {{\left (5 \, c^{2} d^{4} + 6 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{16 \, \sqrt {a c} a^{3} c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(48*a^3*c*d*e^3*x^2 + 32*a^3*c*d^3*e + 16*a^4*d*e^3 - 3*(5*c^4*d^4 + 6*a*c^3*d^2*e^2 + a^2*c^2*e^4)*x^5
- 8*(5*a*c^3*d^4 + 6*a^2*c^2*d^2*e^2 - a^3*c*e^4)*x^3 - 3*(11*a^2*c^2*d^4 - 6*a^3*c*d^2*e^2 - a^4*e^4)*x)/(a^3
*c^5*x^6 + 3*a^4*c^4*x^4 + 3*a^5*c^3*x^2 + a^6*c^2) + 1/16*(5*c^2*d^4 + 6*a*c*d^2*e^2 + a^2*e^4)*arctan(c*x/sq
rt(a*c))/(sqrt(a*c)*a^3*c^2)

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mupad [B]  time = 0.44, size = 263, normalized size = 1.70 \begin {gather*} \frac {\mathrm {atan}\left (\frac {\sqrt {c}\,x\,\left (c\,d^2+a\,e^2\right )\,\left (5\,c\,d^2+a\,e^2\right )}{\sqrt {a}\,\left (a^2\,e^4+6\,a\,c\,d^2\,e^2+5\,c^2\,d^4\right )}\right )\,\left (c\,d^2+a\,e^2\right )\,\left (5\,c\,d^2+a\,e^2\right )}{16\,a^{7/2}\,c^{5/2}}-\frac {\frac {d\,e^3\,x^2}{c}-\frac {x^5\,\left (a^2\,e^4+6\,a\,c\,d^2\,e^2+5\,c^2\,d^4\right )}{16\,a^3}+\frac {x\,\left (a^2\,e^4+6\,a\,c\,d^2\,e^2-11\,c^2\,d^4\right )}{16\,a\,c^2}+\frac {d\,e\,\left (2\,c\,d^2+a\,e^2\right )}{3\,c^2}-\frac {x^3\,\left (-a^2\,e^4+6\,a\,c\,d^2\,e^2+5\,c^2\,d^4\right )}{6\,a^2\,c}}{a^3+3\,a^2\,c\,x^2+3\,a\,c^2\,x^4+c^3\,x^6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^4/(a + c*x^2)^4,x)

[Out]

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

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sympy [B]  time = 3.15, size = 413, normalized size = 2.66 \begin {gather*} - \frac {\sqrt {- \frac {1}{a^{7} c^{5}}} \left (a e^{2} + c d^{2}\right ) \left (a e^{2} + 5 c d^{2}\right ) \log {\left (- \frac {a^{4} c^{2} \sqrt {- \frac {1}{a^{7} c^{5}}} \left (a e^{2} + c d^{2}\right ) \left (a e^{2} + 5 c d^{2}\right )}{a^{2} e^{4} + 6 a c d^{2} e^{2} + 5 c^{2} d^{4}} + x \right )}}{32} + \frac {\sqrt {- \frac {1}{a^{7} c^{5}}} \left (a e^{2} + c d^{2}\right ) \left (a e^{2} + 5 c d^{2}\right ) \log {\left (\frac {a^{4} c^{2} \sqrt {- \frac {1}{a^{7} c^{5}}} \left (a e^{2} + c d^{2}\right ) \left (a e^{2} + 5 c d^{2}\right )}{a^{2} e^{4} + 6 a c d^{2} e^{2} + 5 c^{2} d^{4}} + x \right )}}{32} + \frac {- 16 a^{4} d e^{3} - 32 a^{3} c d^{3} e - 48 a^{3} c d e^{3} x^{2} + x^{5} \left (3 a^{2} c^{2} e^{4} + 18 a c^{3} d^{2} e^{2} + 15 c^{4} d^{4}\right ) + x^{3} \left (- 8 a^{3} c e^{4} + 48 a^{2} c^{2} d^{2} e^{2} + 40 a c^{3} d^{4}\right ) + x \left (- 3 a^{4} e^{4} - 18 a^{3} c d^{2} e^{2} + 33 a^{2} c^{2} d^{4}\right )}{48 a^{6} c^{2} + 144 a^{5} c^{3} x^{2} + 144 a^{4} c^{4} x^{4} + 48 a^{3} c^{5} x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(-1/(a**7*c**5))*(a*e**2 + c*d**2)*(a*e**2 + 5*c*d**2)*log(-a**4*c**2*sqrt(-1/(a**7*c**5))*(a*e**2 + c*d*
*2)*(a*e**2 + 5*c*d**2)/(a**2*e**4 + 6*a*c*d**2*e**2 + 5*c**2*d**4) + x)/32 + sqrt(-1/(a**7*c**5))*(a*e**2 + c
*d**2)*(a*e**2 + 5*c*d**2)*log(a**4*c**2*sqrt(-1/(a**7*c**5))*(a*e**2 + c*d**2)*(a*e**2 + 5*c*d**2)/(a**2*e**4
+ 6*a*c*d**2*e**2 + 5*c**2*d**4) + x)/32 + (-16*a**4*d*e**3 - 32*a**3*c*d**3*e - 48*a**3*c*d*e**3*x**2 + x**5
*(3*a**2*c**2*e**4 + 18*a*c**3*d**2*e**2 + 15*c**4*d**4) + x**3*(-8*a**3*c*e**4 + 48*a**2*c**2*d**2*e**2 + 40*
a*c**3*d**4) + x*(-3*a**4*e**4 - 18*a**3*c*d**2*e**2 + 33*a**2*c**2*d**4))/(48*a**6*c**2 + 144*a**5*c**3*x**2
+ 144*a**4*c**4*x**4 + 48*a**3*c**5*x**6)

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