3.360 \(\int \frac {x^2 (a+b x^2)}{\sqrt {-c+d x} \sqrt {c+d x}} \, dx\)

Optimal. Leaf size=118 \[ \frac {c^2 \left (4 a d^2+3 b c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d x-c}}{\sqrt {c+d x}}\right )}{4 d^5}+\frac {x \sqrt {d x-c} \sqrt {c+d x} \left (4 a d^2+3 b c^2\right )}{8 d^4}+\frac {b x^3 \sqrt {d x-c} \sqrt {c+d x}}{4 d^2} \]

[Out]

1/4*c^2*(4*a*d^2+3*b*c^2)*arctanh((d*x-c)^(1/2)/(d*x+c)^(1/2))/d^5+1/8*(4*a*d^2+3*b*c^2)*x*(d*x-c)^(1/2)*(d*x+
c)^(1/2)/d^4+1/4*b*x^3*(d*x-c)^(1/2)*(d*x+c)^(1/2)/d^2

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Rubi [A]  time = 0.10, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {460, 90, 12, 63, 217, 206} \[ \frac {x \sqrt {d x-c} \sqrt {c+d x} \left (4 a d^2+3 b c^2\right )}{8 d^4}+\frac {c^2 \left (4 a d^2+3 b c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d x-c}}{\sqrt {c+d x}}\right )}{4 d^5}+\frac {b x^3 \sqrt {d x-c} \sqrt {c+d x}}{4 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*b*c^2 + 4*a*d^2)*x*Sqrt[-c + d*x]*Sqrt[c + d*x])/(8*d^4) + (b*x^3*Sqrt[-c + d*x]*Sqrt[c + d*x])/(4*d^2) +
(c^2*(3*b*c^2 + 4*a*d^2)*ArcTanh[Sqrt[-c + d*x]/Sqrt[c + d*x]])/(4*d^5)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 90

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a + b*
x)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 3)), x] + Dist[1/(d*f*(n + p + 3)), Int[(c + d*x)^n*(e +
f*x)^p*Simp[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(n + p + 4) - b*(d*e*(
n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]

Rule 206

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

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 460

Int[((e_.)*(x_))^(m_.)*((a1_) + (b1_.)*(x_)^(non2_.))^(p_.)*((a2_) + (b2_.)*(x_)^(non2_.))^(p_.)*((c_) + (d_.)
*(x_)^(n_)), x_Symbol] :> Simp[(d*(e*x)^(m + 1)*(a1 + b1*x^(n/2))^(p + 1)*(a2 + b2*x^(n/2))^(p + 1))/(b1*b2*e*
(m + n*(p + 1) + 1)), x] - Dist[(a1*a2*d*(m + 1) - b1*b2*c*(m + n*(p + 1) + 1))/(b1*b2*(m + n*(p + 1) + 1)), I
nt[(e*x)^m*(a1 + b1*x^(n/2))^p*(a2 + b2*x^(n/2))^p, x], x] /; FreeQ[{a1, b1, a2, b2, c, d, e, m, n, p}, x] &&
EqQ[non2, n/2] && EqQ[a2*b1 + a1*b2, 0] && NeQ[m + n*(p + 1) + 1, 0]

Rubi steps

\begin {align*} \int \frac {x^2 \left (a+b x^2\right )}{\sqrt {-c+d x} \sqrt {c+d x}} \, dx &=\frac {b x^3 \sqrt {-c+d x} \sqrt {c+d x}}{4 d^2}-\frac {1}{4} \left (-4 a-\frac {3 b c^2}{d^2}\right ) \int \frac {x^2}{\sqrt {-c+d x} \sqrt {c+d x}} \, dx\\ &=\frac {\left (3 b c^2+4 a d^2\right ) x \sqrt {-c+d x} \sqrt {c+d x}}{8 d^4}+\frac {b x^3 \sqrt {-c+d x} \sqrt {c+d x}}{4 d^2}+\frac {\left (3 b c^2+4 a d^2\right ) \int \frac {c^2}{\sqrt {-c+d x} \sqrt {c+d x}} \, dx}{8 d^4}\\ &=\frac {\left (3 b c^2+4 a d^2\right ) x \sqrt {-c+d x} \sqrt {c+d x}}{8 d^4}+\frac {b x^3 \sqrt {-c+d x} \sqrt {c+d x}}{4 d^2}+\frac {\left (c^2 \left (3 b c^2+4 a d^2\right )\right ) \int \frac {1}{\sqrt {-c+d x} \sqrt {c+d x}} \, dx}{8 d^4}\\ &=\frac {\left (3 b c^2+4 a d^2\right ) x \sqrt {-c+d x} \sqrt {c+d x}}{8 d^4}+\frac {b x^3 \sqrt {-c+d x} \sqrt {c+d x}}{4 d^2}+\frac {\left (c^2 \left (3 b c^2+4 a d^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {2 c+x^2}} \, dx,x,\sqrt {-c+d x}\right )}{4 d^5}\\ &=\frac {\left (3 b c^2+4 a d^2\right ) x \sqrt {-c+d x} \sqrt {c+d x}}{8 d^4}+\frac {b x^3 \sqrt {-c+d x} \sqrt {c+d x}}{4 d^2}+\frac {\left (c^2 \left (3 b c^2+4 a d^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {\sqrt {-c+d x}}{\sqrt {c+d x}}\right )}{4 d^5}\\ &=\frac {\left (3 b c^2+4 a d^2\right ) x \sqrt {-c+d x} \sqrt {c+d x}}{8 d^4}+\frac {b x^3 \sqrt {-c+d x} \sqrt {c+d x}}{4 d^2}+\frac {c^2 \left (3 b c^2+4 a d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {-c+d x}}{\sqrt {c+d x}}\right )}{4 d^5}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 121, normalized size = 1.03 \[ \frac {d x \left (d^2 x^2-c^2\right ) \left (4 a d^2+3 b c^2+2 b d^2 x^2\right )+c^2 \sqrt {d^2 x^2-c^2} \left (4 a d^2+3 b c^2\right ) \tanh ^{-1}\left (\frac {d x}{\sqrt {d^2 x^2-c^2}}\right )}{8 d^5 \sqrt {d x-c} \sqrt {c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*x*(-c^2 + d^2*x^2)*(3*b*c^2 + 4*a*d^2 + 2*b*d^2*x^2) + c^2*(3*b*c^2 + 4*a*d^2)*Sqrt[-c^2 + d^2*x^2]*ArcTanh
[(d*x)/Sqrt[-c^2 + d^2*x^2]])/(8*d^5*Sqrt[-c + d*x]*Sqrt[c + d*x])

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fricas [A]  time = 0.91, size = 90, normalized size = 0.76 \[ \frac {{\left (2 \, b d^{3} x^{3} + {\left (3 \, b c^{2} d + 4 \, a d^{3}\right )} x\right )} \sqrt {d x + c} \sqrt {d x - c} - {\left (3 \, b c^{4} + 4 \, a c^{2} d^{2}\right )} \log \left (-d x + \sqrt {d x + c} \sqrt {d x - c}\right )}{8 \, d^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*((2*b*d^3*x^3 + (3*b*c^2*d + 4*a*d^3)*x)*sqrt(d*x + c)*sqrt(d*x - c) - (3*b*c^4 + 4*a*c^2*d^2)*log(-d*x +
sqrt(d*x + c)*sqrt(d*x - c)))/d^5

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giac [A]  time = 0.23, size = 140, normalized size = 1.19 \[ \frac {{\left ({\left (d x + c\right )} {\left (2 \, {\left (d x + c\right )} {\left (\frac {{\left (d x + c\right )} b}{d^{4}} - \frac {3 \, b c}{d^{4}}\right )} + \frac {9 \, b c^{2} d^{16} + 4 \, a d^{18}}{d^{20}}\right )} - \frac {5 \, b c^{3} d^{16} + 4 \, a c d^{18}}{d^{20}}\right )} \sqrt {d x + c} \sqrt {d x - c} - \frac {2 \, {\left (3 \, b c^{4} + 4 \, a c^{2} d^{2}\right )} \log \left ({\left | -\sqrt {d x + c} + \sqrt {d x - c} \right |}\right )}{d^{4}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(((d*x + c)*(2*(d*x + c)*((d*x + c)*b/d^4 - 3*b*c/d^4) + (9*b*c^2*d^16 + 4*a*d^18)/d^20) - (5*b*c^3*d^16 +
4*a*c*d^18)/d^20)*sqrt(d*x + c)*sqrt(d*x - c) - 2*(3*b*c^4 + 4*a*c^2*d^2)*log(abs(-sqrt(d*x + c) + sqrt(d*x -
c)))/d^4)/d

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maple [C]  time = 0.08, size = 182, normalized size = 1.54 \[ \frac {\sqrt {d x -c}\, \sqrt {d x +c}\, \left (2 \sqrt {d^{2} x^{2}-c^{2}}\, b \,d^{3} x^{3} \mathrm {csgn}\relax (d )+4 a \,c^{2} d^{2} \ln \left (\left (d x +\sqrt {d^{2} x^{2}-c^{2}}\, \mathrm {csgn}\relax (d )\right ) \mathrm {csgn}\relax (d )\right )+4 \sqrt {d^{2} x^{2}-c^{2}}\, a \,d^{3} x \,\mathrm {csgn}\relax (d )+3 b \,c^{4} \ln \left (\left (d x +\sqrt {d^{2} x^{2}-c^{2}}\, \mathrm {csgn}\relax (d )\right ) \mathrm {csgn}\relax (d )\right )+3 \sqrt {d^{2} x^{2}-c^{2}}\, b \,c^{2} d x \,\mathrm {csgn}\relax (d )\right ) \mathrm {csgn}\relax (d )}{8 \sqrt {d^{2} x^{2}-c^{2}}\, d^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(d*x-c)^(1/2)*(d*x+c)^(1/2)*(2*(d^2*x^2-c^2)^(1/2)*b*d^3*x^3*csgn(d)+4*(d^2*x^2-c^2)^(1/2)*a*d^3*x*csgn(d)
+3*(d^2*x^2-c^2)^(1/2)*b*c^2*d*x*csgn(d)+4*a*c^2*d^2*ln((d*x+(d^2*x^2-c^2)^(1/2)*csgn(d))*csgn(d))+3*b*c^4*ln(
(d*x+(d^2*x^2-c^2)^(1/2)*csgn(d))*csgn(d)))*csgn(d)/d^5/(d^2*x^2-c^2)^(1/2)

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maxima [A]  time = 0.64, size = 142, normalized size = 1.20 \[ \frac {\sqrt {d^{2} x^{2} - c^{2}} b x^{3}}{4 \, d^{2}} + \frac {3 \, b c^{4} \log \left (2 \, d^{2} x + 2 \, \sqrt {d^{2} x^{2} - c^{2}} d\right )}{8 \, d^{5}} + \frac {a c^{2} \log \left (2 \, d^{2} x + 2 \, \sqrt {d^{2} x^{2} - c^{2}} d\right )}{2 \, d^{3}} + \frac {3 \, \sqrt {d^{2} x^{2} - c^{2}} b c^{2} x}{8 \, d^{4}} + \frac {\sqrt {d^{2} x^{2} - c^{2}} a x}{2 \, d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(d^2*x^2 - c^2)*b*x^3/d^2 + 3/8*b*c^4*log(2*d^2*x + 2*sqrt(d^2*x^2 - c^2)*d)/d^5 + 1/2*a*c^2*log(2*d^2
*x + 2*sqrt(d^2*x^2 - c^2)*d)/d^3 + 3/8*sqrt(d^2*x^2 - c^2)*b*c^2*x/d^4 + 1/2*sqrt(d^2*x^2 - c^2)*a*x/d^2

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mupad [B]  time = 25.51, size = 1048, normalized size = 8.88 \[ \frac {\frac {2\,a\,c^2\,\left (\sqrt {c+d\,x}-\sqrt {c}\right )}{\sqrt {-c}-\sqrt {d\,x-c}}+\frac {14\,a\,c^2\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^3}{{\left (\sqrt {-c}-\sqrt {d\,x-c}\right )}^3}+\frac {14\,a\,c^2\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^5}{{\left (\sqrt {-c}-\sqrt {d\,x-c}\right )}^5}+\frac {2\,a\,c^2\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^7}{{\left (\sqrt {-c}-\sqrt {d\,x-c}\right )}^7}}{d^3-\frac {4\,d^3\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^2}{{\left (\sqrt {-c}-\sqrt {d\,x-c}\right )}^2}+\frac {6\,d^3\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^4}{{\left (\sqrt {-c}-\sqrt {d\,x-c}\right )}^4}-\frac {4\,d^3\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^6}{{\left (\sqrt {-c}-\sqrt {d\,x-c}\right )}^6}+\frac {d^3\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^8}{{\left (\sqrt {-c}-\sqrt {d\,x-c}\right )}^8}}-\frac {\frac {23\,b\,c^4\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^3}{2\,{\left (\sqrt {-c}-\sqrt {d\,x-c}\right )}^3}-\frac {3\,b\,c^4\,\left (\sqrt {c+d\,x}-\sqrt {c}\right )}{2\,\left (\sqrt {-c}-\sqrt {d\,x-c}\right )}+\frac {333\,b\,c^4\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^5}{2\,{\left (\sqrt {-c}-\sqrt {d\,x-c}\right )}^5}+\frac {671\,b\,c^4\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^7}{2\,{\left (\sqrt {-c}-\sqrt {d\,x-c}\right )}^7}+\frac {671\,b\,c^4\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^9}{2\,{\left (\sqrt {-c}-\sqrt {d\,x-c}\right )}^9}+\frac {333\,b\,c^4\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^{11}}{2\,{\left (\sqrt {-c}-\sqrt {d\,x-c}\right )}^{11}}+\frac {23\,b\,c^4\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^{13}}{2\,{\left (\sqrt {-c}-\sqrt {d\,x-c}\right )}^{13}}-\frac {3\,b\,c^4\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^{15}}{2\,{\left (\sqrt {-c}-\sqrt {d\,x-c}\right )}^{15}}}{d^5-\frac {8\,d^5\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^2}{{\left (\sqrt {-c}-\sqrt {d\,x-c}\right )}^2}+\frac {28\,d^5\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^4}{{\left (\sqrt {-c}-\sqrt {d\,x-c}\right )}^4}-\frac {56\,d^5\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^6}{{\left (\sqrt {-c}-\sqrt {d\,x-c}\right )}^6}+\frac {70\,d^5\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^8}{{\left (\sqrt {-c}-\sqrt {d\,x-c}\right )}^8}-\frac {56\,d^5\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^{10}}{{\left (\sqrt {-c}-\sqrt {d\,x-c}\right )}^{10}}+\frac {28\,d^5\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^{12}}{{\left (\sqrt {-c}-\sqrt {d\,x-c}\right )}^{12}}-\frac {8\,d^5\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^{14}}{{\left (\sqrt {-c}-\sqrt {d\,x-c}\right )}^{14}}+\frac {d^5\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^{16}}{{\left (\sqrt {-c}-\sqrt {d\,x-c}\right )}^{16}}}-\frac {2\,a\,c^2\,\mathrm {atanh}\left (\frac {\sqrt {c+d\,x}-\sqrt {c}}{\sqrt {-c}-\sqrt {d\,x-c}}\right )}{d^3}-\frac {3\,b\,c^4\,\mathrm {atanh}\left (\frac {\sqrt {c+d\,x}-\sqrt {c}}{\sqrt {-c}-\sqrt {d\,x-c}}\right )}{2\,d^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((2*a*c^2*((c + d*x)^(1/2) - c^(1/2)))/((-c)^(1/2) - (d*x - c)^(1/2)) + (14*a*c^2*((c + d*x)^(1/2) - c^(1/2))^
3)/((-c)^(1/2) - (d*x - c)^(1/2))^3 + (14*a*c^2*((c + d*x)^(1/2) - c^(1/2))^5)/((-c)^(1/2) - (d*x - c)^(1/2))^
5 + (2*a*c^2*((c + d*x)^(1/2) - c^(1/2))^7)/((-c)^(1/2) - (d*x - c)^(1/2))^7)/(d^3 - (4*d^3*((c + d*x)^(1/2) -
c^(1/2))^2)/((-c)^(1/2) - (d*x - c)^(1/2))^2 + (6*d^3*((c + d*x)^(1/2) - c^(1/2))^4)/((-c)^(1/2) - (d*x - c)^
(1/2))^4 - (4*d^3*((c + d*x)^(1/2) - c^(1/2))^6)/((-c)^(1/2) - (d*x - c)^(1/2))^6 + (d^3*((c + d*x)^(1/2) - c^
(1/2))^8)/((-c)^(1/2) - (d*x - c)^(1/2))^8) - ((23*b*c^4*((c + d*x)^(1/2) - c^(1/2))^3)/(2*((-c)^(1/2) - (d*x
- c)^(1/2))^3) - (3*b*c^4*((c + d*x)^(1/2) - c^(1/2)))/(2*((-c)^(1/2) - (d*x - c)^(1/2))) + (333*b*c^4*((c + d
*x)^(1/2) - c^(1/2))^5)/(2*((-c)^(1/2) - (d*x - c)^(1/2))^5) + (671*b*c^4*((c + d*x)^(1/2) - c^(1/2))^7)/(2*((
-c)^(1/2) - (d*x - c)^(1/2))^7) + (671*b*c^4*((c + d*x)^(1/2) - c^(1/2))^9)/(2*((-c)^(1/2) - (d*x - c)^(1/2))^
9) + (333*b*c^4*((c + d*x)^(1/2) - c^(1/2))^11)/(2*((-c)^(1/2) - (d*x - c)^(1/2))^11) + (23*b*c^4*((c + d*x)^(
1/2) - c^(1/2))^13)/(2*((-c)^(1/2) - (d*x - c)^(1/2))^13) - (3*b*c^4*((c + d*x)^(1/2) - c^(1/2))^15)/(2*((-c)^
(1/2) - (d*x - c)^(1/2))^15))/(d^5 - (8*d^5*((c + d*x)^(1/2) - c^(1/2))^2)/((-c)^(1/2) - (d*x - c)^(1/2))^2 +
(28*d^5*((c + d*x)^(1/2) - c^(1/2))^4)/((-c)^(1/2) - (d*x - c)^(1/2))^4 - (56*d^5*((c + d*x)^(1/2) - c^(1/2))^
6)/((-c)^(1/2) - (d*x - c)^(1/2))^6 + (70*d^5*((c + d*x)^(1/2) - c^(1/2))^8)/((-c)^(1/2) - (d*x - c)^(1/2))^8
- (56*d^5*((c + d*x)^(1/2) - c^(1/2))^10)/((-c)^(1/2) - (d*x - c)^(1/2))^10 + (28*d^5*((c + d*x)^(1/2) - c^(1/
2))^12)/((-c)^(1/2) - (d*x - c)^(1/2))^12 - (8*d^5*((c + d*x)^(1/2) - c^(1/2))^14)/((-c)^(1/2) - (d*x - c)^(1/
2))^14 + (d^5*((c + d*x)^(1/2) - c^(1/2))^16)/((-c)^(1/2) - (d*x - c)^(1/2))^16) - (2*a*c^2*atanh(((c + d*x)^(
1/2) - c^(1/2))/((-c)^(1/2) - (d*x - c)^(1/2))))/d^3 - (3*b*c^4*atanh(((c + d*x)^(1/2) - c^(1/2))/((-c)^(1/2)
- (d*x - c)^(1/2))))/(2*d^5)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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