3.417 \(\int \frac {x^{3/2} (a+b x^2)^2}{c+d x^2} \, dx\)

Optimal. Leaf size=288 \[ \frac {\sqrt [4]{c} (b c-a d)^2 \log \left (-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{2 \sqrt {2} d^{13/4}}-\frac {\sqrt [4]{c} (b c-a d)^2 \log \left (\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{2 \sqrt {2} d^{13/4}}+\frac {\sqrt [4]{c} (b c-a d)^2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{\sqrt {2} d^{13/4}}-\frac {\sqrt [4]{c} (b c-a d)^2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}+1\right )}{\sqrt {2} d^{13/4}}+\frac {2 \sqrt {x} (b c-a d)^2}{d^3}-\frac {2 b x^{5/2} (b c-2 a d)}{5 d^2}+\frac {2 b^2 x^{9/2}}{9 d} \]

[Out]

-2/5*b*(-2*a*d+b*c)*x^(5/2)/d^2+2/9*b^2*x^(9/2)/d+1/2*c^(1/4)*(-a*d+b*c)^2*arctan(1-d^(1/4)*2^(1/2)*x^(1/2)/c^
(1/4))/d^(13/4)*2^(1/2)-1/2*c^(1/4)*(-a*d+b*c)^2*arctan(1+d^(1/4)*2^(1/2)*x^(1/2)/c^(1/4))/d^(13/4)*2^(1/2)+1/
4*c^(1/4)*(-a*d+b*c)^2*ln(c^(1/2)+x*d^(1/2)-c^(1/4)*d^(1/4)*2^(1/2)*x^(1/2))/d^(13/4)*2^(1/2)-1/4*c^(1/4)*(-a*
d+b*c)^2*ln(c^(1/2)+x*d^(1/2)+c^(1/4)*d^(1/4)*2^(1/2)*x^(1/2))/d^(13/4)*2^(1/2)+2*(-a*d+b*c)^2*x^(1/2)/d^3

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Rubi [A]  time = 0.24, antiderivative size = 288, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {461, 321, 329, 211, 1165, 628, 1162, 617, 204} \[ -\frac {2 b x^{5/2} (b c-2 a d)}{5 d^2}+\frac {2 \sqrt {x} (b c-a d)^2}{d^3}+\frac {\sqrt [4]{c} (b c-a d)^2 \log \left (-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{2 \sqrt {2} d^{13/4}}-\frac {\sqrt [4]{c} (b c-a d)^2 \log \left (\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{2 \sqrt {2} d^{13/4}}+\frac {\sqrt [4]{c} (b c-a d)^2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{\sqrt {2} d^{13/4}}-\frac {\sqrt [4]{c} (b c-a d)^2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}+1\right )}{\sqrt {2} d^{13/4}}+\frac {2 b^2 x^{9/2}}{9 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(b*c - a*d)^2*Sqrt[x])/d^3 - (2*b*(b*c - 2*a*d)*x^(5/2))/(5*d^2) + (2*b^2*x^(9/2))/(9*d) + (c^(1/4)*(b*c -
a*d)^2*ArcTan[1 - (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/(Sqrt[2]*d^(13/4)) - (c^(1/4)*(b*c - a*d)^2*ArcTan[1 + (
Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/(Sqrt[2]*d^(13/4)) + (c^(1/4)*(b*c - a*d)^2*Log[Sqrt[c] - Sqrt[2]*c^(1/4)*d
^(1/4)*Sqrt[x] + Sqrt[d]*x])/(2*Sqrt[2]*d^(13/4)) - (c^(1/4)*(b*c - a*d)^2*Log[Sqrt[c] + Sqrt[2]*c^(1/4)*d^(1/
4)*Sqrt[x] + Sqrt[d]*x])/(2*Sqrt[2]*d^(13/4))

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 211

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

Rule 321

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

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 461

Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegr
and[((e*x)^m*(a + b*x^n)^p)/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n
, 0] && IGtQ[p, 0] && (IntegerQ[m] || IGtQ[2*(m + 1), 0] ||  !RationalQ[m])

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rubi steps

\begin {align*} \int \frac {x^{3/2} \left (a+b x^2\right )^2}{c+d x^2} \, dx &=\int \left (-\frac {b (b c-2 a d) x^{3/2}}{d^2}+\frac {b^2 x^{7/2}}{d}+\frac {\left (b^2 c^2-2 a b c d+a^2 d^2\right ) x^{3/2}}{d^2 \left (c+d x^2\right )}\right ) \, dx\\ &=-\frac {2 b (b c-2 a d) x^{5/2}}{5 d^2}+\frac {2 b^2 x^{9/2}}{9 d}+\frac {(b c-a d)^2 \int \frac {x^{3/2}}{c+d x^2} \, dx}{d^2}\\ &=\frac {2 (b c-a d)^2 \sqrt {x}}{d^3}-\frac {2 b (b c-2 a d) x^{5/2}}{5 d^2}+\frac {2 b^2 x^{9/2}}{9 d}-\frac {\left (c (b c-a d)^2\right ) \int \frac {1}{\sqrt {x} \left (c+d x^2\right )} \, dx}{d^3}\\ &=\frac {2 (b c-a d)^2 \sqrt {x}}{d^3}-\frac {2 b (b c-2 a d) x^{5/2}}{5 d^2}+\frac {2 b^2 x^{9/2}}{9 d}-\frac {\left (2 c (b c-a d)^2\right ) \operatorname {Subst}\left (\int \frac {1}{c+d x^4} \, dx,x,\sqrt {x}\right )}{d^3}\\ &=\frac {2 (b c-a d)^2 \sqrt {x}}{d^3}-\frac {2 b (b c-2 a d) x^{5/2}}{5 d^2}+\frac {2 b^2 x^{9/2}}{9 d}-\frac {\left (\sqrt {c} (b c-a d)^2\right ) \operatorname {Subst}\left (\int \frac {\sqrt {c}-\sqrt {d} x^2}{c+d x^4} \, dx,x,\sqrt {x}\right )}{d^3}-\frac {\left (\sqrt {c} (b c-a d)^2\right ) \operatorname {Subst}\left (\int \frac {\sqrt {c}+\sqrt {d} x^2}{c+d x^4} \, dx,x,\sqrt {x}\right )}{d^3}\\ &=\frac {2 (b c-a d)^2 \sqrt {x}}{d^3}-\frac {2 b (b c-2 a d) x^{5/2}}{5 d^2}+\frac {2 b^2 x^{9/2}}{9 d}-\frac {\left (\sqrt {c} (b c-a d)^2\right ) \operatorname {Subst}\left (\int \frac {1}{\frac {\sqrt {c}}{\sqrt {d}}-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{d}}+x^2} \, dx,x,\sqrt {x}\right )}{2 d^{7/2}}-\frac {\left (\sqrt {c} (b c-a d)^2\right ) \operatorname {Subst}\left (\int \frac {1}{\frac {\sqrt {c}}{\sqrt {d}}+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{d}}+x^2} \, dx,x,\sqrt {x}\right )}{2 d^{7/2}}+\frac {\left (\sqrt [4]{c} (b c-a d)^2\right ) \operatorname {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt [4]{c}}{\sqrt [4]{d}}+2 x}{-\frac {\sqrt {c}}{\sqrt {d}}-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{d}}-x^2} \, dx,x,\sqrt {x}\right )}{2 \sqrt {2} d^{13/4}}+\frac {\left (\sqrt [4]{c} (b c-a d)^2\right ) \operatorname {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt [4]{c}}{\sqrt [4]{d}}-2 x}{-\frac {\sqrt {c}}{\sqrt {d}}+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{d}}-x^2} \, dx,x,\sqrt {x}\right )}{2 \sqrt {2} d^{13/4}}\\ &=\frac {2 (b c-a d)^2 \sqrt {x}}{d^3}-\frac {2 b (b c-2 a d) x^{5/2}}{5 d^2}+\frac {2 b^2 x^{9/2}}{9 d}+\frac {\sqrt [4]{c} (b c-a d)^2 \log \left (\sqrt {c}-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {d} x\right )}{2 \sqrt {2} d^{13/4}}-\frac {\sqrt [4]{c} (b c-a d)^2 \log \left (\sqrt {c}+\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {d} x\right )}{2 \sqrt {2} d^{13/4}}-\frac {\left (\sqrt [4]{c} (b c-a d)^2\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{\sqrt {2} d^{13/4}}+\frac {\left (\sqrt [4]{c} (b c-a d)^2\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{\sqrt {2} d^{13/4}}\\ &=\frac {2 (b c-a d)^2 \sqrt {x}}{d^3}-\frac {2 b (b c-2 a d) x^{5/2}}{5 d^2}+\frac {2 b^2 x^{9/2}}{9 d}+\frac {\sqrt [4]{c} (b c-a d)^2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{\sqrt {2} d^{13/4}}-\frac {\sqrt [4]{c} (b c-a d)^2 \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{\sqrt {2} d^{13/4}}+\frac {\sqrt [4]{c} (b c-a d)^2 \log \left (\sqrt {c}-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {d} x\right )}{2 \sqrt {2} d^{13/4}}-\frac {\sqrt [4]{c} (b c-a d)^2 \log \left (\sqrt {c}+\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {d} x\right )}{2 \sqrt {2} d^{13/4}}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 276, normalized size = 0.96 \[ \frac {-72 b d^{5/4} x^{5/2} (b c-2 a d)+360 \sqrt [4]{d} \sqrt {x} (b c-a d)^2+45 \sqrt {2} \sqrt [4]{c} (b c-a d)^2 \log \left (-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )-45 \sqrt {2} \sqrt [4]{c} (b c-a d)^2 \log \left (\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )+90 \sqrt {2} \sqrt [4]{c} (b c-a d)^2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )-90 \sqrt {2} \sqrt [4]{c} (b c-a d)^2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}+1\right )+40 b^2 d^{9/4} x^{9/2}}{180 d^{13/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(360*d^(1/4)*(b*c - a*d)^2*Sqrt[x] - 72*b*d^(5/4)*(b*c - 2*a*d)*x^(5/2) + 40*b^2*d^(9/4)*x^(9/2) + 90*Sqrt[2]*
c^(1/4)*(b*c - a*d)^2*ArcTan[1 - (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)] - 90*Sqrt[2]*c^(1/4)*(b*c - a*d)^2*ArcTan[
1 + (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)] + 45*Sqrt[2]*c^(1/4)*(b*c - a*d)^2*Log[Sqrt[c] - Sqrt[2]*c^(1/4)*d^(1/4
)*Sqrt[x] + Sqrt[d]*x] - 45*Sqrt[2]*c^(1/4)*(b*c - a*d)^2*Log[Sqrt[c] + Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt
[d]*x])/(180*d^(13/4))

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fricas [B]  time = 0.54, size = 1268, normalized size = 4.40 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/90*(180*d^3*(-(b^8*c^9 - 8*a*b^7*c^8*d + 28*a^2*b^6*c^7*d^2 - 56*a^3*b^5*c^6*d^3 + 70*a^4*b^4*c^5*d^4 - 56*
a^5*b^3*c^4*d^5 + 28*a^6*b^2*c^3*d^6 - 8*a^7*b*c^2*d^7 + a^8*c*d^8)/d^13)^(1/4)*arctan((sqrt(d^6*sqrt(-(b^8*c^
9 - 8*a*b^7*c^8*d + 28*a^2*b^6*c^7*d^2 - 56*a^3*b^5*c^6*d^3 + 70*a^4*b^4*c^5*d^4 - 56*a^5*b^3*c^4*d^5 + 28*a^6
*b^2*c^3*d^6 - 8*a^7*b*c^2*d^7 + a^8*c*d^8)/d^13) + (b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d
^3 + a^4*d^4)*x)*d^10*(-(b^8*c^9 - 8*a*b^7*c^8*d + 28*a^2*b^6*c^7*d^2 - 56*a^3*b^5*c^6*d^3 + 70*a^4*b^4*c^5*d^
4 - 56*a^5*b^3*c^4*d^5 + 28*a^6*b^2*c^3*d^6 - 8*a^7*b*c^2*d^7 + a^8*c*d^8)/d^13)^(3/4) - (b^2*c^2*d^10 - 2*a*b
*c*d^11 + a^2*d^12)*sqrt(x)*(-(b^8*c^9 - 8*a*b^7*c^8*d + 28*a^2*b^6*c^7*d^2 - 56*a^3*b^5*c^6*d^3 + 70*a^4*b^4*
c^5*d^4 - 56*a^5*b^3*c^4*d^5 + 28*a^6*b^2*c^3*d^6 - 8*a^7*b*c^2*d^7 + a^8*c*d^8)/d^13)^(3/4))/(b^8*c^9 - 8*a*b
^7*c^8*d + 28*a^2*b^6*c^7*d^2 - 56*a^3*b^5*c^6*d^3 + 70*a^4*b^4*c^5*d^4 - 56*a^5*b^3*c^4*d^5 + 28*a^6*b^2*c^3*
d^6 - 8*a^7*b*c^2*d^7 + a^8*c*d^8)) + 45*d^3*(-(b^8*c^9 - 8*a*b^7*c^8*d + 28*a^2*b^6*c^7*d^2 - 56*a^3*b^5*c^6*
d^3 + 70*a^4*b^4*c^5*d^4 - 56*a^5*b^3*c^4*d^5 + 28*a^6*b^2*c^3*d^6 - 8*a^7*b*c^2*d^7 + a^8*c*d^8)/d^13)^(1/4)*
log(d^3*(-(b^8*c^9 - 8*a*b^7*c^8*d + 28*a^2*b^6*c^7*d^2 - 56*a^3*b^5*c^6*d^3 + 70*a^4*b^4*c^5*d^4 - 56*a^5*b^3
*c^4*d^5 + 28*a^6*b^2*c^3*d^6 - 8*a^7*b*c^2*d^7 + a^8*c*d^8)/d^13)^(1/4) + (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqr
t(x)) - 45*d^3*(-(b^8*c^9 - 8*a*b^7*c^8*d + 28*a^2*b^6*c^7*d^2 - 56*a^3*b^5*c^6*d^3 + 70*a^4*b^4*c^5*d^4 - 56*
a^5*b^3*c^4*d^5 + 28*a^6*b^2*c^3*d^6 - 8*a^7*b*c^2*d^7 + a^8*c*d^8)/d^13)^(1/4)*log(-d^3*(-(b^8*c^9 - 8*a*b^7*
c^8*d + 28*a^2*b^6*c^7*d^2 - 56*a^3*b^5*c^6*d^3 + 70*a^4*b^4*c^5*d^4 - 56*a^5*b^3*c^4*d^5 + 28*a^6*b^2*c^3*d^6
- 8*a^7*b*c^2*d^7 + a^8*c*d^8)/d^13)^(1/4) + (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqrt(x)) - 4*(5*b^2*d^2*x^4 + 45
*b^2*c^2 - 90*a*b*c*d + 45*a^2*d^2 - 9*(b^2*c*d - 2*a*b*d^2)*x^2)*sqrt(x))/d^3

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giac [A]  time = 0.46, size = 385, normalized size = 1.34 \[ -\frac {\sqrt {2} {\left (\left (c d^{3}\right )^{\frac {1}{4}} b^{2} c^{2} - 2 \, \left (c d^{3}\right )^{\frac {1}{4}} a b c d + \left (c d^{3}\right )^{\frac {1}{4}} a^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {c}{d}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {c}{d}\right )^{\frac {1}{4}}}\right )}{2 \, d^{4}} - \frac {\sqrt {2} {\left (\left (c d^{3}\right )^{\frac {1}{4}} b^{2} c^{2} - 2 \, \left (c d^{3}\right )^{\frac {1}{4}} a b c d + \left (c d^{3}\right )^{\frac {1}{4}} a^{2} d^{2}\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {c}{d}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {c}{d}\right )^{\frac {1}{4}}}\right )}{2 \, d^{4}} - \frac {\sqrt {2} {\left (\left (c d^{3}\right )^{\frac {1}{4}} b^{2} c^{2} - 2 \, \left (c d^{3}\right )^{\frac {1}{4}} a b c d + \left (c d^{3}\right )^{\frac {1}{4}} a^{2} d^{2}\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {c}{d}\right )^{\frac {1}{4}} + x + \sqrt {\frac {c}{d}}\right )}{4 \, d^{4}} + \frac {\sqrt {2} {\left (\left (c d^{3}\right )^{\frac {1}{4}} b^{2} c^{2} - 2 \, \left (c d^{3}\right )^{\frac {1}{4}} a b c d + \left (c d^{3}\right )^{\frac {1}{4}} a^{2} d^{2}\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {c}{d}\right )^{\frac {1}{4}} + x + \sqrt {\frac {c}{d}}\right )}{4 \, d^{4}} + \frac {2 \, {\left (5 \, b^{2} d^{8} x^{\frac {9}{2}} - 9 \, b^{2} c d^{7} x^{\frac {5}{2}} + 18 \, a b d^{8} x^{\frac {5}{2}} + 45 \, b^{2} c^{2} d^{6} \sqrt {x} - 90 \, a b c d^{7} \sqrt {x} + 45 \, a^{2} d^{8} \sqrt {x}\right )}}{45 \, d^{9}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(2)*((c*d^3)^(1/4)*b^2*c^2 - 2*(c*d^3)^(1/4)*a*b*c*d + (c*d^3)^(1/4)*a^2*d^2)*arctan(1/2*sqrt(2)*(sqr
t(2)*(c/d)^(1/4) + 2*sqrt(x))/(c/d)^(1/4))/d^4 - 1/2*sqrt(2)*((c*d^3)^(1/4)*b^2*c^2 - 2*(c*d^3)^(1/4)*a*b*c*d
+ (c*d^3)^(1/4)*a^2*d^2)*arctan(-1/2*sqrt(2)*(sqrt(2)*(c/d)^(1/4) - 2*sqrt(x))/(c/d)^(1/4))/d^4 - 1/4*sqrt(2)*
((c*d^3)^(1/4)*b^2*c^2 - 2*(c*d^3)^(1/4)*a*b*c*d + (c*d^3)^(1/4)*a^2*d^2)*log(sqrt(2)*sqrt(x)*(c/d)^(1/4) + x
+ sqrt(c/d))/d^4 + 1/4*sqrt(2)*((c*d^3)^(1/4)*b^2*c^2 - 2*(c*d^3)^(1/4)*a*b*c*d + (c*d^3)^(1/4)*a^2*d^2)*log(-
sqrt(2)*sqrt(x)*(c/d)^(1/4) + x + sqrt(c/d))/d^4 + 2/45*(5*b^2*d^8*x^(9/2) - 9*b^2*c*d^7*x^(5/2) + 18*a*b*d^8*
x^(5/2) + 45*b^2*c^2*d^6*sqrt(x) - 90*a*b*c*d^7*sqrt(x) + 45*a^2*d^8*sqrt(x))/d^9

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maple [B]  time = 0.01, size = 495, normalized size = 1.72 \[ \frac {2 b^{2} x^{\frac {9}{2}}}{9 d}+\frac {4 a b \,x^{\frac {5}{2}}}{5 d}-\frac {2 b^{2} c \,x^{\frac {5}{2}}}{5 d^{2}}-\frac {\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, a^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )}{2 d}-\frac {\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, a^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )}{2 d}-\frac {\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, a^{2} \ln \left (\frac {x +\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {c}{d}}}{x -\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {c}{d}}}\right )}{4 d}+\frac {\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, a b c \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )}{d^{2}}+\frac {\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, a b c \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )}{d^{2}}+\frac {\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, a b c \ln \left (\frac {x +\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {c}{d}}}{x -\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {c}{d}}}\right )}{2 d^{2}}-\frac {\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, b^{2} c^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )}{2 d^{3}}-\frac {\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, b^{2} c^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )}{2 d^{3}}-\frac {\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, b^{2} c^{2} \ln \left (\frac {x +\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {c}{d}}}{x -\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {c}{d}}}\right )}{4 d^{3}}+\frac {2 a^{2} \sqrt {x}}{d}-\frac {4 a b c \sqrt {x}}{d^{2}}+\frac {2 b^{2} c^{2} \sqrt {x}}{d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)*(b*x^2+a)^2/(d*x^2+c),x)

[Out]

2/9*b^2*x^(9/2)/d+4/5/d*x^(5/2)*a*b-2/5/d^2*x^(5/2)*b^2*c+2/d*a^2*x^(1/2)-4/d^2*a*b*c*x^(1/2)+2/d^3*b^2*c^2*x^
(1/2)-1/4/d*(c/d)^(1/4)*2^(1/2)*ln((x+(c/d)^(1/4)*2^(1/2)*x^(1/2)+(c/d)^(1/2))/(x-(c/d)^(1/4)*2^(1/2)*x^(1/2)+
(c/d)^(1/2)))*a^2+1/2/d^2*(c/d)^(1/4)*2^(1/2)*ln((x+(c/d)^(1/4)*2^(1/2)*x^(1/2)+(c/d)^(1/2))/(x-(c/d)^(1/4)*2^
(1/2)*x^(1/2)+(c/d)^(1/2)))*a*b*c-1/4/d^3*(c/d)^(1/4)*2^(1/2)*ln((x+(c/d)^(1/4)*2^(1/2)*x^(1/2)+(c/d)^(1/2))/(
x-(c/d)^(1/4)*2^(1/2)*x^(1/2)+(c/d)^(1/2)))*b^2*c^2-1/2/d*(c/d)^(1/4)*2^(1/2)*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/
2)+1)*a^2+1/d^2*(c/d)^(1/4)*2^(1/2)*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)+1)*a*b*c-1/2/d^3*(c/d)^(1/4)*2^(1/2)*ar
ctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)+1)*b^2*c^2-1/2/d*(c/d)^(1/4)*2^(1/2)*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)-1)*a^
2+1/d^2*(c/d)^(1/4)*2^(1/2)*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)-1)*a*b*c-1/2/d^3*(c/d)^(1/4)*2^(1/2)*arctan(2^(
1/2)/(c/d)^(1/4)*x^(1/2)-1)*b^2*c^2

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maxima [A]  time = 2.43, size = 324, normalized size = 1.12 \[ -\frac {{\left (\frac {2 \, \sqrt {2} {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} + 2 \, \sqrt {d} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {c} \sqrt {d}}}\right )}{\sqrt {c} \sqrt {\sqrt {c} \sqrt {d}}} + \frac {2 \, \sqrt {2} {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} - 2 \, \sqrt {d} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {c} \sqrt {d}}}\right )}{\sqrt {c} \sqrt {\sqrt {c} \sqrt {d}}} + \frac {\sqrt {2} {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left (\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} \sqrt {x} + \sqrt {d} x + \sqrt {c}\right )}{c^{\frac {3}{4}} d^{\frac {1}{4}}} - \frac {\sqrt {2} {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left (-\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} \sqrt {x} + \sqrt {d} x + \sqrt {c}\right )}{c^{\frac {3}{4}} d^{\frac {1}{4}}}\right )} c}{4 \, d^{3}} + \frac {2 \, {\left (5 \, b^{2} d^{2} x^{\frac {9}{2}} - 9 \, {\left (b^{2} c d - 2 \, a b d^{2}\right )} x^{\frac {5}{2}} + 45 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {x}\right )}}{45 \, d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*sqrt(2)*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*arctan(1/2*sqrt(2)*(sqrt(2)*c^(1/4)*d^(1/4) + 2*sqrt(d)*sqrt(x
))/sqrt(sqrt(c)*sqrt(d)))/(sqrt(c)*sqrt(sqrt(c)*sqrt(d))) + 2*sqrt(2)*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*arctan(-
1/2*sqrt(2)*(sqrt(2)*c^(1/4)*d^(1/4) - 2*sqrt(d)*sqrt(x))/sqrt(sqrt(c)*sqrt(d)))/(sqrt(c)*sqrt(sqrt(c)*sqrt(d)
)) + sqrt(2)*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*log(sqrt(2)*c^(1/4)*d^(1/4)*sqrt(x) + sqrt(d)*x + sqrt(c))/(c^(3/
4)*d^(1/4)) - sqrt(2)*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*log(-sqrt(2)*c^(1/4)*d^(1/4)*sqrt(x) + sqrt(d)*x + sqrt(
c))/(c^(3/4)*d^(1/4)))*c/d^3 + 2/45*(5*b^2*d^2*x^(9/2) - 9*(b^2*c*d - 2*a*b*d^2)*x^(5/2) + 45*(b^2*c^2 - 2*a*b
*c*d + a^2*d^2)*sqrt(x))/d^3

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mupad [B]  time = 0.35, size = 1175, normalized size = 4.08 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^(3/2)*(a + b*x^2)^2)/(c + d*x^2),x)

[Out]

x^(1/2)*((2*a^2)/d + (c*((2*b^2*c)/d^2 - (4*a*b)/d))/d) - x^(5/2)*((2*b^2*c)/(5*d^2) - (4*a*b)/(5*d)) + (2*b^2
*x^(9/2))/(9*d) - ((-c)^(1/4)*atan((((-c)^(1/4)*(a*d - b*c)^2*((8*x^(1/2)*(b^4*c^6 + a^4*c^2*d^4 - 4*a^3*b*c^3
*d^3 + 6*a^2*b^2*c^4*d^2 - 4*a*b^3*c^5*d))/d^3 - ((-c)^(1/4)*(a*d - b*c)^2*(16*b^2*c^4 + 16*a^2*c^2*d^2 - 32*a
*b*c^3*d))/(2*d^(13/4)))*1i)/d^(13/4) + ((-c)^(1/4)*(a*d - b*c)^2*((8*x^(1/2)*(b^4*c^6 + a^4*c^2*d^4 - 4*a^3*b
*c^3*d^3 + 6*a^2*b^2*c^4*d^2 - 4*a*b^3*c^5*d))/d^3 + ((-c)^(1/4)*(a*d - b*c)^2*(16*b^2*c^4 + 16*a^2*c^2*d^2 -
32*a*b*c^3*d))/(2*d^(13/4)))*1i)/d^(13/4))/(((-c)^(1/4)*(a*d - b*c)^2*((8*x^(1/2)*(b^4*c^6 + a^4*c^2*d^4 - 4*a
^3*b*c^3*d^3 + 6*a^2*b^2*c^4*d^2 - 4*a*b^3*c^5*d))/d^3 - ((-c)^(1/4)*(a*d - b*c)^2*(16*b^2*c^4 + 16*a^2*c^2*d^
2 - 32*a*b*c^3*d))/(2*d^(13/4))))/d^(13/4) - ((-c)^(1/4)*(a*d - b*c)^2*((8*x^(1/2)*(b^4*c^6 + a^4*c^2*d^4 - 4*
a^3*b*c^3*d^3 + 6*a^2*b^2*c^4*d^2 - 4*a*b^3*c^5*d))/d^3 + ((-c)^(1/4)*(a*d - b*c)^2*(16*b^2*c^4 + 16*a^2*c^2*d
^2 - 32*a*b*c^3*d))/(2*d^(13/4))))/d^(13/4)))*(a*d - b*c)^2*1i)/d^(13/4) - ((-c)^(1/4)*atan((((-c)^(1/4)*(a*d
- b*c)^2*((8*x^(1/2)*(b^4*c^6 + a^4*c^2*d^4 - 4*a^3*b*c^3*d^3 + 6*a^2*b^2*c^4*d^2 - 4*a*b^3*c^5*d))/d^3 - ((-c
)^(1/4)*(a*d - b*c)^2*(16*b^2*c^4 + 16*a^2*c^2*d^2 - 32*a*b*c^3*d)*1i)/(2*d^(13/4))))/d^(13/4) + ((-c)^(1/4)*(
a*d - b*c)^2*((8*x^(1/2)*(b^4*c^6 + a^4*c^2*d^4 - 4*a^3*b*c^3*d^3 + 6*a^2*b^2*c^4*d^2 - 4*a*b^3*c^5*d))/d^3 +
((-c)^(1/4)*(a*d - b*c)^2*(16*b^2*c^4 + 16*a^2*c^2*d^2 - 32*a*b*c^3*d)*1i)/(2*d^(13/4))))/d^(13/4))/(((-c)^(1/
4)*(a*d - b*c)^2*((8*x^(1/2)*(b^4*c^6 + a^4*c^2*d^4 - 4*a^3*b*c^3*d^3 + 6*a^2*b^2*c^4*d^2 - 4*a*b^3*c^5*d))/d^
3 - ((-c)^(1/4)*(a*d - b*c)^2*(16*b^2*c^4 + 16*a^2*c^2*d^2 - 32*a*b*c^3*d)*1i)/(2*d^(13/4)))*1i)/d^(13/4) - ((
-c)^(1/4)*(a*d - b*c)^2*((8*x^(1/2)*(b^4*c^6 + a^4*c^2*d^4 - 4*a^3*b*c^3*d^3 + 6*a^2*b^2*c^4*d^2 - 4*a*b^3*c^5
*d))/d^3 + ((-c)^(1/4)*(a*d - b*c)^2*(16*b^2*c^4 + 16*a^2*c^2*d^2 - 32*a*b*c^3*d)*1i)/(2*d^(13/4)))*1i)/d^(13/
4)))*(a*d - b*c)^2)/d^(13/4)

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sympy [A]  time = 53.53, size = 656, normalized size = 2.28 \[ \begin {cases} \tilde {\infty } \left (2 a^{2} \sqrt {x} + \frac {4 a b x^{\frac {5}{2}}}{5} + \frac {2 b^{2} x^{\frac {9}{2}}}{9}\right ) & \text {for}\: c = 0 \wedge d = 0 \\\frac {\frac {2 a^{2} x^{\frac {5}{2}}}{5} + \frac {4 a b x^{\frac {9}{2}}}{9} + \frac {2 b^{2} x^{\frac {13}{2}}}{13}}{c} & \text {for}\: d = 0 \\\frac {2 a^{2} \sqrt {x} + \frac {4 a b x^{\frac {5}{2}}}{5} + \frac {2 b^{2} x^{\frac {9}{2}}}{9}}{d} & \text {for}\: c = 0 \\\frac {\sqrt [4]{-1} a^{2} \sqrt [4]{c} \sqrt [4]{\frac {1}{d}} \log {\left (- \sqrt [4]{-1} \sqrt [4]{c} \sqrt [4]{\frac {1}{d}} + \sqrt {x} \right )}}{2 d} - \frac {\sqrt [4]{-1} a^{2} \sqrt [4]{c} \sqrt [4]{\frac {1}{d}} \log {\left (\sqrt [4]{-1} \sqrt [4]{c} \sqrt [4]{\frac {1}{d}} + \sqrt {x} \right )}}{2 d} + \frac {\sqrt [4]{-1} a^{2} \sqrt [4]{c} \sqrt [4]{\frac {1}{d}} \operatorname {atan}{\left (\frac {\left (-1\right )^{\frac {3}{4}} \sqrt {x}}{\sqrt [4]{c} \sqrt [4]{\frac {1}{d}}} \right )}}{d} + \frac {2 a^{2} \sqrt {x}}{d} - \frac {\sqrt [4]{-1} a b c^{\frac {5}{4}} \sqrt [4]{\frac {1}{d}} \log {\left (- \sqrt [4]{-1} \sqrt [4]{c} \sqrt [4]{\frac {1}{d}} + \sqrt {x} \right )}}{d^{2}} + \frac {\sqrt [4]{-1} a b c^{\frac {5}{4}} \sqrt [4]{\frac {1}{d}} \log {\left (\sqrt [4]{-1} \sqrt [4]{c} \sqrt [4]{\frac {1}{d}} + \sqrt {x} \right )}}{d^{2}} - \frac {2 \sqrt [4]{-1} a b c^{\frac {5}{4}} \sqrt [4]{\frac {1}{d}} \operatorname {atan}{\left (\frac {\left (-1\right )^{\frac {3}{4}} \sqrt {x}}{\sqrt [4]{c} \sqrt [4]{\frac {1}{d}}} \right )}}{d^{2}} - \frac {4 a b c \sqrt {x}}{d^{2}} + \frac {4 a b x^{\frac {5}{2}}}{5 d} + \frac {\sqrt [4]{-1} b^{2} c^{\frac {9}{4}} \sqrt [4]{\frac {1}{d}} \log {\left (- \sqrt [4]{-1} \sqrt [4]{c} \sqrt [4]{\frac {1}{d}} + \sqrt {x} \right )}}{2 d^{3}} - \frac {\sqrt [4]{-1} b^{2} c^{\frac {9}{4}} \sqrt [4]{\frac {1}{d}} \log {\left (\sqrt [4]{-1} \sqrt [4]{c} \sqrt [4]{\frac {1}{d}} + \sqrt {x} \right )}}{2 d^{3}} + \frac {\sqrt [4]{-1} b^{2} c^{\frac {9}{4}} \sqrt [4]{\frac {1}{d}} \operatorname {atan}{\left (\frac {\left (-1\right )^{\frac {3}{4}} \sqrt {x}}{\sqrt [4]{c} \sqrt [4]{\frac {1}{d}}} \right )}}{d^{3}} + \frac {2 b^{2} c^{2} \sqrt {x}}{d^{3}} - \frac {2 b^{2} c x^{\frac {5}{2}}}{5 d^{2}} + \frac {2 b^{2} x^{\frac {9}{2}}}{9 d} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*(2*a**2*sqrt(x) + 4*a*b*x**(5/2)/5 + 2*b**2*x**(9/2)/9), Eq(c, 0) & Eq(d, 0)), ((2*a**2*x**(5/2
)/5 + 4*a*b*x**(9/2)/9 + 2*b**2*x**(13/2)/13)/c, Eq(d, 0)), ((2*a**2*sqrt(x) + 4*a*b*x**(5/2)/5 + 2*b**2*x**(9
/2)/9)/d, Eq(c, 0)), ((-1)**(1/4)*a**2*c**(1/4)*(1/d)**(1/4)*log(-(-1)**(1/4)*c**(1/4)*(1/d)**(1/4) + sqrt(x))
/(2*d) - (-1)**(1/4)*a**2*c**(1/4)*(1/d)**(1/4)*log((-1)**(1/4)*c**(1/4)*(1/d)**(1/4) + sqrt(x))/(2*d) + (-1)*
*(1/4)*a**2*c**(1/4)*(1/d)**(1/4)*atan((-1)**(3/4)*sqrt(x)/(c**(1/4)*(1/d)**(1/4)))/d + 2*a**2*sqrt(x)/d - (-1
)**(1/4)*a*b*c**(5/4)*(1/d)**(1/4)*log(-(-1)**(1/4)*c**(1/4)*(1/d)**(1/4) + sqrt(x))/d**2 + (-1)**(1/4)*a*b*c*
*(5/4)*(1/d)**(1/4)*log((-1)**(1/4)*c**(1/4)*(1/d)**(1/4) + sqrt(x))/d**2 - 2*(-1)**(1/4)*a*b*c**(5/4)*(1/d)**
(1/4)*atan((-1)**(3/4)*sqrt(x)/(c**(1/4)*(1/d)**(1/4)))/d**2 - 4*a*b*c*sqrt(x)/d**2 + 4*a*b*x**(5/2)/(5*d) + (
-1)**(1/4)*b**2*c**(9/4)*(1/d)**(1/4)*log(-(-1)**(1/4)*c**(1/4)*(1/d)**(1/4) + sqrt(x))/(2*d**3) - (-1)**(1/4)
*b**2*c**(9/4)*(1/d)**(1/4)*log((-1)**(1/4)*c**(1/4)*(1/d)**(1/4) + sqrt(x))/(2*d**3) + (-1)**(1/4)*b**2*c**(9
/4)*(1/d)**(1/4)*atan((-1)**(3/4)*sqrt(x)/(c**(1/4)*(1/d)**(1/4)))/d**3 + 2*b**2*c**2*sqrt(x)/d**3 - 2*b**2*c*
x**(5/2)/(5*d**2) + 2*b**2*x**(9/2)/(9*d), True))

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