3.136 \(\int \frac {a+b x^2+c x^4}{x^3 \sqrt {d-e x} \sqrt {d+e x}} \, dx\)

Optimal. Leaf size=99 \[ -\frac {\left (a e^2+2 b d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d-e x} \sqrt {d+e x}}{d}\right )}{2 d^3}-\frac {a \sqrt {d-e x} \sqrt {d+e x}}{2 d^2 x^2}-\frac {c \sqrt {d-e x} \sqrt {d+e x}}{e^2} \]

[Out]

-1/2*(a*e^2+2*b*d^2)*arctanh((-e*x+d)^(1/2)*(e*x+d)^(1/2)/d)/d^3-c*(-e*x+d)^(1/2)*(e*x+d)^(1/2)/e^2-1/2*a*(-e*
x+d)^(1/2)*(e*x+d)^(1/2)/d^2/x^2

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Rubi [A]  time = 0.25, antiderivative size = 155, normalized size of antiderivative = 1.57, number of steps used = 6, number of rules used = 6, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {520, 1251, 897, 1157, 388, 208} \[ -\frac {\sqrt {d^2-e^2 x^2} \left (a e^2+2 b d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d^3 \sqrt {d-e x} \sqrt {d+e x}}-\frac {a \left (d^2-e^2 x^2\right )}{2 d^2 x^2 \sqrt {d-e x} \sqrt {d+e x}}-\frac {c \left (d^2-e^2 x^2\right )}{e^2 \sqrt {d-e x} \sqrt {d+e x}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2 + c*x^4)/(x^3*Sqrt[d - e*x]*Sqrt[d + e*x]),x]

[Out]

-((c*(d^2 - e^2*x^2))/(e^2*Sqrt[d - e*x]*Sqrt[d + e*x])) - (a*(d^2 - e^2*x^2))/(2*d^2*x^2*Sqrt[d - e*x]*Sqrt[d
+ e*x]) - ((2*b*d^2 + a*e^2)*Sqrt[d^2 - e^2*x^2]*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/(2*d^3*Sqrt[d - e*x]*Sqrt[d
+ e*x])

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[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 520

Int[(u_.)*((c_) + (d_.)*(x_)^(n_.) + (e_.)*(x_)^(n2_.))^(q_.)*((a1_) + (b1_.)*(x_)^(non2_.))^(p_.)*((a2_) + (b
2_.)*(x_)^(non2_.))^(p_.), x_Symbol] :> Dist[((a1 + b1*x^(n/2))^FracPart[p]*(a2 + b2*x^(n/2))^FracPart[p])/(a1
*a2 + b1*b2*x^n)^FracPart[p], Int[u*(a1*a2 + b1*b2*x^n)^p*(c + d*x^n + e*x^(2*n))^q, x], x] /; FreeQ[{a1, b1,
a2, b2, c, d, e, n, p, q}, x] && EqQ[non2, n/2] && EqQ[n2, 2*n] && EqQ[a2*b1 + a1*b2, 0]

Rule 897

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> With[{q = Denominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + (g*x^q)/e)^n*((c*d^2 - b*d
*e + a*e^2)/e^2 - ((2*c*d - b*e)*x^q)/e^2 + (c*x^(2*q))/e^2)^p, x], x, (d + e*x)^(1/q)], x]] /; FreeQ[{a, b, c
, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegersQ[n,
p] && FractionQ[m]

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 1251

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2,
Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] &&
IntegerQ[(m - 1)/2]

Rubi steps

\begin {align*} \int \frac {a+b x^2+c x^4}{x^3 \sqrt {d-e x} \sqrt {d+e x}} \, dx &=\frac {\sqrt {d^2-e^2 x^2} \int \frac {a+b x^2+c x^4}{x^3 \sqrt {d^2-e^2 x^2}} \, dx}{\sqrt {d-e x} \sqrt {d+e x}}\\ &=\frac {\sqrt {d^2-e^2 x^2} \operatorname {Subst}\left (\int \frac {a+b x+c x^2}{x^2 \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )}{2 \sqrt {d-e x} \sqrt {d+e x}}\\ &=-\frac {\sqrt {d^2-e^2 x^2} \operatorname {Subst}\left (\int \frac {\frac {c d^4+b d^2 e^2+a e^4}{e^4}-\frac {\left (2 c d^2+b e^2\right ) x^2}{e^4}+\frac {c x^4}{e^4}}{\left (\frac {d^2}{e^2}-\frac {x^2}{e^2}\right )^2} \, dx,x,\sqrt {d^2-e^2 x^2}\right )}{e^2 \sqrt {d-e x} \sqrt {d+e x}}\\ &=-\frac {a \left (d^2-e^2 x^2\right )}{2 d^2 x^2 \sqrt {d-e x} \sqrt {d+e x}}+\frac {\sqrt {d^2-e^2 x^2} \operatorname {Subst}\left (\int \frac {-a-\frac {2 \left (c d^4+b d^2 e^2\right )}{e^4}+\frac {2 c d^2 x^2}{e^4}}{\frac {d^2}{e^2}-\frac {x^2}{e^2}} \, dx,x,\sqrt {d^2-e^2 x^2}\right )}{2 d^2 \sqrt {d-e x} \sqrt {d+e x}}\\ &=-\frac {c \left (d^2-e^2 x^2\right )}{e^2 \sqrt {d-e x} \sqrt {d+e x}}-\frac {a \left (d^2-e^2 x^2\right )}{2 d^2 x^2 \sqrt {d-e x} \sqrt {d+e x}}+\frac {\left (e^2 \left (\frac {2 c d^4}{e^6}+\frac {-a-\frac {2 \left (c d^4+b d^2 e^2\right )}{e^4}}{e^2}\right ) \sqrt {d^2-e^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\frac {d^2}{e^2}-\frac {x^2}{e^2}} \, dx,x,\sqrt {d^2-e^2 x^2}\right )}{2 d^2 \sqrt {d-e x} \sqrt {d+e x}}\\ &=-\frac {c \left (d^2-e^2 x^2\right )}{e^2 \sqrt {d-e x} \sqrt {d+e x}}-\frac {a \left (d^2-e^2 x^2\right )}{2 d^2 x^2 \sqrt {d-e x} \sqrt {d+e x}}-\frac {\left (2 b d^2+a e^2\right ) \sqrt {d^2-e^2 x^2} \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d^3 \sqrt {d-e x} \sqrt {d+e x}}\\ \end {align*}

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Mathematica [B]  time = 0.21, size = 233, normalized size = 2.35 \[ \frac {-e^2 x^2 \sqrt {d^2-e^2 x^2} \left (a e^2+2 b d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )-a d^3 e^2+a d e^4 x^2-4 c d^{9/2} x^2 \sqrt {d-e x} \sqrt {\frac {e x}{d}+1} \sin ^{-1}\left (\frac {\sqrt {d-e x}}{\sqrt {2} \sqrt {d}}\right )-2 c d^5 x^2+4 c d^4 x^2 \sqrt {d-e x} \sqrt {d+e x} \tan ^{-1}\left (\frac {\sqrt {d-e x}}{\sqrt {d+e x}}\right )+2 c d^3 e^2 x^4}{2 d^3 e^2 x^2 \sqrt {d-e x} \sqrt {d+e x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(a*d^3*e^2) - 2*c*d^5*x^2 + a*d*e^4*x^2 + 2*c*d^3*e^2*x^4 - 4*c*d^(9/2)*x^2*Sqrt[d - e*x]*Sqrt[1 + (e*x)/d]*
ArcSin[Sqrt[d - e*x]/(Sqrt[2]*Sqrt[d])] + 4*c*d^4*x^2*Sqrt[d - e*x]*Sqrt[d + e*x]*ArcTan[Sqrt[d - e*x]/Sqrt[d
+ e*x]] - e^2*(2*b*d^2 + a*e^2)*x^2*Sqrt[d^2 - e^2*x^2]*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/(2*d^3*e^2*x^2*Sqrt[d
- e*x]*Sqrt[d + e*x])

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fricas [A]  time = 0.80, size = 98, normalized size = 0.99 \[ -\frac {2 \, c d^{4} x^{2} - {\left (2 \, b d^{2} e^{2} + a e^{4}\right )} x^{2} \log \left (\frac {\sqrt {e x + d} \sqrt {-e x + d} - d}{x}\right ) + {\left (2 \, c d^{3} x^{2} + a d e^{2}\right )} \sqrt {e x + d} \sqrt {-e x + d}}{2 \, d^{3} e^{2} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)/x^3/(-e*x+d)^(1/2)/(e*x+d)^(1/2),x, algorithm="fricas")

[Out]

-1/2*(2*c*d^4*x^2 - (2*b*d^2*e^2 + a*e^4)*x^2*log((sqrt(e*x + d)*sqrt(-e*x + d) - d)/x) + (2*c*d^3*x^2 + a*d*e
^2)*sqrt(e*x + d)*sqrt(-e*x + d))/(d^3*e^2*x^2)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)/x^3/(-e*x+d)^(1/2)/(e*x+d)^(1/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: schur row 1 1.55494e-10Francis algorithm
not precise enough for[1.0,-220.862474643,10162.5484803,-174574.213802,1032773.91614]schur row 1 3.66198e-10F
rancis algorithm not precise enough for[1.0,-467.909596927,45612.3731035,-1659969.6644,20804885.8013]Bad condi
tionned root j= 2 value 38.9905751966 ratio 0.000133135092941 mindist 0.002415226181251/exp(1)*(-(2*a*(2*sqrt(
d+x*exp(1))/(2*sqrt(2)*sqrt(d)-2*sqrt(-d-x*exp(1)+2*d))-1/2*(2*sqrt(2)*sqrt(d)-2*sqrt(-d-x*exp(1)+2*d))/sqrt(d
+x*exp(1)))^3*exp(1)^3+8*a*(2*sqrt(d+x*exp(1))/(2*sqrt(2)*sqrt(d)-2*sqrt(-d-x*exp(1)+2*d))-1/2*(2*sqrt(2)*sqrt
(d)-2*sqrt(-d-x*exp(1)+2*d))/sqrt(d+x*exp(1)))*exp(1)^3)/d^3/((2*sqrt(d+x*exp(1))/(2*sqrt(2)*sqrt(d)-2*sqrt(-d
-x*exp(1)+2*d))-1/2*(2*sqrt(2)*sqrt(d)-2*sqrt(-d-x*exp(1)+2*d))/sqrt(d+x*exp(1)))^2-4)^2-1/2*(a*exp(1)^3+2*b*d
^2*exp(1))*ln(abs(2*sqrt(d+x*exp(1))/(2*sqrt(2)*sqrt(d)-2*sqrt(-d-x*exp(1)+2*d))-1/2*(2*sqrt(2)*sqrt(d)-2*sqrt
(-d-x*exp(1)+2*d))/sqrt(d+x*exp(1))+2))/d^3+1/2*(a*exp(1)^3+2*b*d^2*exp(1))*ln(abs(2*sqrt(d+x*exp(1))/(2*sqrt(
2)*sqrt(d)-2*sqrt(-d-x*exp(1)+2*d))-1/2*(2*sqrt(2)*sqrt(d)-2*sqrt(-d-x*exp(1)+2*d))/sqrt(d+x*exp(1))-2))/d^3-2
*c*exp(1)/2/exp(1)^2*sqrt(d+x*exp(1))*sqrt(-d-x*exp(1)+2*d))

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maple [C]  time = 0.02, size = 163, normalized size = 1.65 \[ -\frac {\sqrt {-e x +d}\, \sqrt {e x +d}\, \left (a \,e^{4} x^{2} \ln \left (\frac {2 \left (d +\sqrt {-e^{2} x^{2}+d^{2}}\, \mathrm {csgn}\relax (d )\right ) d}{x}\right )+2 b \,d^{2} e^{2} x^{2} \ln \left (\frac {2 \left (d +\sqrt {-e^{2} x^{2}+d^{2}}\, \mathrm {csgn}\relax (d )\right ) d}{x}\right )+2 \sqrt {-e^{2} x^{2}+d^{2}}\, c \,d^{3} x^{2} \mathrm {csgn}\relax (d )+\sqrt {-e^{2} x^{2}+d^{2}}\, a d \,e^{2} \mathrm {csgn}\relax (d )\right ) \mathrm {csgn}\relax (d )}{2 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{3} e^{2} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+b*x^2+a)/x^3/(-e*x+d)^(1/2)/(e*x+d)^(1/2),x)

[Out]

-1/2*(-e*x+d)^(1/2)*(e*x+d)^(1/2)/d^3*(2*csgn(d)*x^2*c*d^3*(-e^2*x^2+d^2)^(1/2)+ln(2*(d+(-e^2*x^2+d^2)^(1/2)*c
sgn(d))*d/x)*x^2*a*e^4+2*ln(2*(d+(-e^2*x^2+d^2)^(1/2)*csgn(d))*d/x)*x^2*b*d^2*e^2+csgn(d)*a*d*e^2*(-e^2*x^2+d^
2)^(1/2))*csgn(d)/(-e^2*x^2+d^2)^(1/2)/e^2/x^2

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maxima [A]  time = 1.02, size = 123, normalized size = 1.24 \[ -\frac {b \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right )}{d} - \frac {a e^{2} \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right )}{2 \, d^{3}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} c}{e^{2}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} a}{2 \, d^{2} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)/x^3/(-e*x+d)^(1/2)/(e*x+d)^(1/2),x, algorithm="maxima")

[Out]

-b*log(2*d^2/abs(x) + 2*sqrt(-e^2*x^2 + d^2)*d/abs(x))/d - 1/2*a*e^2*log(2*d^2/abs(x) + 2*sqrt(-e^2*x^2 + d^2)
*d/abs(x))/d^3 - sqrt(-e^2*x^2 + d^2)*c/e^2 - 1/2*sqrt(-e^2*x^2 + d^2)*a/(d^2*x^2)

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mupad [B]  time = 5.15, size = 422, normalized size = 4.26 \[ \frac {b\,\left (\ln \left (\frac {{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^2}{{\left (\sqrt {d-e\,x}-\sqrt {d}\right )}^2}-1\right )-\ln \left (\frac {\sqrt {d+e\,x}-\sqrt {d}}{\sqrt {d-e\,x}-\sqrt {d}}\right )\right )}{d}-\frac {\left (\frac {c\,d}{e^2}+\frac {c\,x}{e}\right )\,\sqrt {d-e\,x}}{\sqrt {d+e\,x}}-\frac {\frac {a\,e^2\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^2}{{\left (\sqrt {d-e\,x}-\sqrt {d}\right )}^2}-\frac {a\,e^2}{2}+\frac {15\,a\,e^2\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^4}{2\,{\left (\sqrt {d-e\,x}-\sqrt {d}\right )}^4}}{\frac {16\,d^3\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^2}{{\left (\sqrt {d-e\,x}-\sqrt {d}\right )}^2}-\frac {32\,d^3\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^4}{{\left (\sqrt {d-e\,x}-\sqrt {d}\right )}^4}+\frac {16\,d^3\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^6}{{\left (\sqrt {d-e\,x}-\sqrt {d}\right )}^6}}-\frac {a\,e^2\,\ln \left (\frac {\sqrt {d+e\,x}-\sqrt {d}}{\sqrt {d-e\,x}-\sqrt {d}}\right )}{2\,d^3}+\frac {a\,e^2\,\ln \left (\frac {{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^2}{{\left (\sqrt {d-e\,x}-\sqrt {d}\right )}^2}-1\right )}{2\,d^3}+\frac {a\,e^2\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^2}{32\,d^3\,{\left (\sqrt {d-e\,x}-\sqrt {d}\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2 + c*x^4)/(x^3*(d + e*x)^(1/2)*(d - e*x)^(1/2)),x)

[Out]

(b*(log(((d + e*x)^(1/2) - d^(1/2))^2/((d - e*x)^(1/2) - d^(1/2))^2 - 1) - log(((d + e*x)^(1/2) - d^(1/2))/((d
- e*x)^(1/2) - d^(1/2)))))/d - (((c*d)/e^2 + (c*x)/e)*(d - e*x)^(1/2))/(d + e*x)^(1/2) - ((a*e^2*((d + e*x)^(
1/2) - d^(1/2))^2)/((d - e*x)^(1/2) - d^(1/2))^2 - (a*e^2)/2 + (15*a*e^2*((d + e*x)^(1/2) - d^(1/2))^4)/(2*((d
- e*x)^(1/2) - d^(1/2))^4))/((16*d^3*((d + e*x)^(1/2) - d^(1/2))^2)/((d - e*x)^(1/2) - d^(1/2))^2 - (32*d^3*(
(d + e*x)^(1/2) - d^(1/2))^4)/((d - e*x)^(1/2) - d^(1/2))^4 + (16*d^3*((d + e*x)^(1/2) - d^(1/2))^6)/((d - e*x
)^(1/2) - d^(1/2))^6) - (a*e^2*log(((d + e*x)^(1/2) - d^(1/2))/((d - e*x)^(1/2) - d^(1/2))))/(2*d^3) + (a*e^2*
log(((d + e*x)^(1/2) - d^(1/2))^2/((d - e*x)^(1/2) - d^(1/2))^2 - 1))/(2*d^3) + (a*e^2*((d + e*x)^(1/2) - d^(1
/2))^2)/(32*d^3*((d - e*x)^(1/2) - d^(1/2))^2)

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sympy [C]  time = 133.79, size = 270, normalized size = 2.73 \[ \frac {i a e^{2} {G_{6, 6}^{5, 3}\left (\begin {matrix} \frac {7}{4}, \frac {9}{4}, 1 & 2, 2, \frac {5}{2} \\\frac {3}{2}, \frac {7}{4}, 2, \frac {9}{4}, \frac {5}{2} & 0 \end {matrix} \middle | {\frac {d^{2}}{e^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d^{3}} - \frac {a e^{2} {G_{6, 6}^{2, 6}\left (\begin {matrix} 1, \frac {5}{4}, \frac {3}{2}, \frac {7}{4}, 2, 1 & \\\frac {5}{4}, \frac {7}{4} & 1, \frac {3}{2}, \frac {3}{2}, 0 \end {matrix} \middle | {\frac {d^{2} e^{- 2 i \pi }}{e^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d^{3}} + \frac {i b {G_{6, 6}^{5, 3}\left (\begin {matrix} \frac {3}{4}, \frac {5}{4}, 1 & 1, 1, \frac {3}{2} \\\frac {1}{2}, \frac {3}{4}, 1, \frac {5}{4}, \frac {3}{2} & 0 \end {matrix} \middle | {\frac {d^{2}}{e^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d} - \frac {b {G_{6, 6}^{2, 6}\left (\begin {matrix} 0, \frac {1}{4}, \frac {1}{2}, \frac {3}{4}, 1, 1 & \\\frac {1}{4}, \frac {3}{4} & 0, \frac {1}{2}, \frac {1}{2}, 0 \end {matrix} \middle | {\frac {d^{2} e^{- 2 i \pi }}{e^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d} - \frac {i c d {G_{6, 6}^{6, 2}\left (\begin {matrix} - \frac {1}{4}, \frac {1}{4} & 0, 0, \frac {1}{2}, 1 \\- \frac {1}{2}, - \frac {1}{4}, 0, \frac {1}{4}, \frac {1}{2}, 0 & \end {matrix} \middle | {\frac {d^{2}}{e^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} e^{2}} - \frac {c d {G_{6, 6}^{2, 6}\left (\begin {matrix} -1, - \frac {3}{4}, - \frac {1}{2}, - \frac {1}{4}, 0, 1 & \\- \frac {3}{4}, - \frac {1}{4} & -1, - \frac {1}{2}, - \frac {1}{2}, 0 \end {matrix} \middle | {\frac {d^{2} e^{- 2 i \pi }}{e^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**2+a)/x**3/(-e*x+d)**(1/2)/(e*x+d)**(1/2),x)

[Out]

I*a*e**2*meijerg(((7/4, 9/4, 1), (2, 2, 5/2)), ((3/2, 7/4, 2, 9/4, 5/2), (0,)), d**2/(e**2*x**2))/(4*pi**(3/2)
*d**3) - a*e**2*meijerg(((1, 5/4, 3/2, 7/4, 2, 1), ()), ((5/4, 7/4), (1, 3/2, 3/2, 0)), d**2*exp_polar(-2*I*pi
)/(e**2*x**2))/(4*pi**(3/2)*d**3) + I*b*meijerg(((3/4, 5/4, 1), (1, 1, 3/2)), ((1/2, 3/4, 1, 5/4, 3/2), (0,)),
d**2/(e**2*x**2))/(4*pi**(3/2)*d) - b*meijerg(((0, 1/4, 1/2, 3/4, 1, 1), ()), ((1/4, 3/4), (0, 1/2, 1/2, 0)),
d**2*exp_polar(-2*I*pi)/(e**2*x**2))/(4*pi**(3/2)*d) - I*c*d*meijerg(((-1/4, 1/4), (0, 0, 1/2, 1)), ((-1/2, -
1/4, 0, 1/4, 1/2, 0), ()), d**2/(e**2*x**2))/(4*pi**(3/2)*e**2) - c*d*meijerg(((-1, -3/4, -1/2, -1/4, 0, 1), (
)), ((-3/4, -1/4), (-1, -1/2, -1/2, 0)), d**2*exp_polar(-2*I*pi)/(e**2*x**2))/(4*pi**(3/2)*e**2)

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