3.29.63 \(\int \frac {\sqrt {-b+x^2} (c+x^4) \sqrt {x+\sqrt {-b+x^2}}}{x} \, dx\)

Optimal. Leaf size=301 \[ \sqrt {2} b^{3/4} c \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {\sqrt {x^2-b}+x}}{\sqrt {x^2-b}-\sqrt {b}+x}\right )+\sqrt {2} b^{3/4} c \tanh ^{-1}\left (\frac {\frac {\sqrt {x^2-b}}{\sqrt {2} \sqrt [4]{b}}+\frac {x}{\sqrt {2} \sqrt [4]{b}}+\frac {\sqrt [4]{b}}{\sqrt {2}}}{\sqrt {\sqrt {x^2-b}+x}}\right )+\frac {2 \sqrt {x^2-b} \left (-1368 b^4 x+3705 b^3 x^3+10395 b^2 c x+1335 b^2 x^5-32340 b c x^3-8100 b x^7+18480 c x^5+5040 x^9\right )+2 \left (304 b^5-3078 b^4 x^2-2310 b^3 c+3735 b^3 x^4+24255 b^2 c x^2+4755 b^2 x^6-41580 b c x^4-10620 b x^8+18480 c x^6+5040 x^{10}\right )}{3465 \left (\sqrt {x^2-b}+x\right )^{9/2}} \]

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Rubi [A]  time = 1.25, antiderivative size = 391, normalized size of antiderivative = 1.30, number of steps used = 18, number of rules used = 12, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.324, Rules used = {6742, 2120, 462, 459, 329, 297, 1162, 617, 204, 1165, 628, 448} \begin {gather*} -\frac {b^{3/4} c \log \left (\sqrt {x^2-b}-\sqrt {2} \sqrt [4]{b} \sqrt {\sqrt {x^2-b}+x}+\sqrt {b}+x\right )}{\sqrt {2}}+\frac {b^{3/4} c \log \left (\sqrt {x^2-b}+\sqrt {2} \sqrt [4]{b} \sqrt {\sqrt {x^2-b}+x}+\sqrt {b}+x\right )}{\sqrt {2}}+\sqrt {2} b^{3/4} c \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {\sqrt {x^2-b}+x}}{\sqrt [4]{b}}\right )-\sqrt {2} b^{3/4} c \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {\sqrt {x^2-b}+x}}{\sqrt [4]{b}}+1\right )-\frac {b^5}{144 \left (\sqrt {x^2-b}+x\right )^{9/2}}-\frac {b^4}{80 \left (\sqrt {x^2-b}+x\right )^{5/2}}+\frac {b^3}{8 \sqrt {\sqrt {x^2-b}+x}}-\frac {1}{24} b^2 \left (\sqrt {x^2-b}+x\right )^{3/2}+\frac {1}{3} c \left (\sqrt {x^2-b}+x\right )^{3/2}-\frac {b c}{\sqrt {\sqrt {x^2-b}+x}}+\frac {1}{176} \left (\sqrt {x^2-b}+x\right )^{11/2}+\frac {1}{112} b \left (\sqrt {x^2-b}+x\right )^{7/2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/144*b^5/(x + Sqrt[-b + x^2])^(9/2) - b^4/(80*(x + Sqrt[-b + x^2])^(5/2)) + b^3/(8*Sqrt[x + Sqrt[-b + x^2]])
- (b*c)/Sqrt[x + Sqrt[-b + x^2]] - (b^2*(x + Sqrt[-b + x^2])^(3/2))/24 + (c*(x + Sqrt[-b + x^2])^(3/2))/3 + (
b*(x + Sqrt[-b + x^2])^(7/2))/112 + (x + Sqrt[-b + x^2])^(11/2)/176 + Sqrt[2]*b^(3/4)*c*ArcTan[1 - (Sqrt[2]*Sq
rt[x + Sqrt[-b + x^2]])/b^(1/4)] - Sqrt[2]*b^(3/4)*c*ArcTan[1 + (Sqrt[2]*Sqrt[x + Sqrt[-b + x^2]])/b^(1/4)] -
(b^(3/4)*c*Log[Sqrt[b] + x + Sqrt[-b + x^2] - Sqrt[2]*b^(1/4)*Sqrt[x + Sqrt[-b + x^2]]])/Sqrt[2] + (b^(3/4)*c*
Log[Sqrt[b] + x + Sqrt[-b + x^2] + Sqrt[2]*b^(1/4)*Sqrt[x + Sqrt[-b + x^2]]])/Sqrt[2]

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 297

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

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 448

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

Rule 459

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

Rule 462

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> Simp[(c^2*(e*x)^(
m + 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] - Dist[1/(a*e^n*(m + 1)), Int[(e*x)^(m + n)*(a + b*x^n)^p*Simp[b
*c^2*n*(p + 1) + c*(b*c - 2*a*d)*(m + 1) - a*(m + 1)*d^2*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && Ne
Q[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -1] && GtQ[n, 0]

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]

Rule 2120

Int[(x_)^(p_.)*((g_) + (i_.)*(x_)^2)^(m_.)*((e_.)*(x_) + (f_.)*Sqrt[(a_) + (c_.)*(x_)^2])^(n_.), x_Symbol] :>
Dist[(1*(i/c)^m)/(2^(2*m + p + 1)*e^(p + 1)*f^(2*m)), Subst[Int[x^(n - 2*m - p - 2)*(-(a*f^2) + x^2)^p*(a*f^2
+ x^2)^(2*m + 1), x], x, e*x + f*Sqrt[a + c*x^2]], x] /; FreeQ[{a, c, e, f, g, i, n}, x] && EqQ[e^2 - c*f^2, 0
] && EqQ[c*g - a*i, 0] && IntegersQ[p, 2*m] && (IntegerQ[m] || GtQ[i/c, 0])

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {align*} \int \frac {\sqrt {-b+x^2} \left (c+x^4\right ) \sqrt {x+\sqrt {-b+x^2}}}{x} \, dx &=\int \left (\frac {c \sqrt {-b+x^2} \sqrt {x+\sqrt {-b+x^2}}}{x}+x^3 \sqrt {-b+x^2} \sqrt {x+\sqrt {-b+x^2}}\right ) \, dx\\ &=c \int \frac {\sqrt {-b+x^2} \sqrt {x+\sqrt {-b+x^2}}}{x} \, dx+\int x^3 \sqrt {-b+x^2} \sqrt {x+\sqrt {-b+x^2}} \, dx\\ &=\frac {1}{32} \operatorname {Subst}\left (\int \frac {\left (-b+x^2\right )^2 \left (b+x^2\right )^3}{x^{11/2}} \, dx,x,x+\sqrt {-b+x^2}\right )+\frac {1}{2} c \operatorname {Subst}\left (\int \frac {\left (-b+x^2\right )^2}{x^{3/2} \left (b+x^2\right )} \, dx,x,x+\sqrt {-b+x^2}\right )\\ &=-\frac {b c}{\sqrt {x+\sqrt {-b+x^2}}}+\frac {1}{32} \operatorname {Subst}\left (\int \left (\frac {b^5}{x^{11/2}}+\frac {b^4}{x^{7/2}}-\frac {2 b^3}{x^{3/2}}-2 b^2 \sqrt {x}+b x^{5/2}+x^{9/2}\right ) \, dx,x,x+\sqrt {-b+x^2}\right )+\frac {c \operatorname {Subst}\left (\int \frac {\sqrt {x} \left (-\frac {3 b^2}{2}+\frac {b x^2}{2}\right )}{b+x^2} \, dx,x,x+\sqrt {-b+x^2}\right )}{b}\\ &=-\frac {b^5}{144 \left (x+\sqrt {-b+x^2}\right )^{9/2}}-\frac {b^4}{80 \left (x+\sqrt {-b+x^2}\right )^{5/2}}+\frac {b^3}{8 \sqrt {x+\sqrt {-b+x^2}}}-\frac {b c}{\sqrt {x+\sqrt {-b+x^2}}}-\frac {1}{24} b^2 \left (x+\sqrt {-b+x^2}\right )^{3/2}+\frac {1}{3} c \left (x+\sqrt {-b+x^2}\right )^{3/2}+\frac {1}{112} b \left (x+\sqrt {-b+x^2}\right )^{7/2}+\frac {1}{176} \left (x+\sqrt {-b+x^2}\right )^{11/2}-(2 b c) \operatorname {Subst}\left (\int \frac {\sqrt {x}}{b+x^2} \, dx,x,x+\sqrt {-b+x^2}\right )\\ &=-\frac {b^5}{144 \left (x+\sqrt {-b+x^2}\right )^{9/2}}-\frac {b^4}{80 \left (x+\sqrt {-b+x^2}\right )^{5/2}}+\frac {b^3}{8 \sqrt {x+\sqrt {-b+x^2}}}-\frac {b c}{\sqrt {x+\sqrt {-b+x^2}}}-\frac {1}{24} b^2 \left (x+\sqrt {-b+x^2}\right )^{3/2}+\frac {1}{3} c \left (x+\sqrt {-b+x^2}\right )^{3/2}+\frac {1}{112} b \left (x+\sqrt {-b+x^2}\right )^{7/2}+\frac {1}{176} \left (x+\sqrt {-b+x^2}\right )^{11/2}-(4 b c) \operatorname {Subst}\left (\int \frac {x^2}{b+x^4} \, dx,x,\sqrt {x+\sqrt {-b+x^2}}\right )\\ &=-\frac {b^5}{144 \left (x+\sqrt {-b+x^2}\right )^{9/2}}-\frac {b^4}{80 \left (x+\sqrt {-b+x^2}\right )^{5/2}}+\frac {b^3}{8 \sqrt {x+\sqrt {-b+x^2}}}-\frac {b c}{\sqrt {x+\sqrt {-b+x^2}}}-\frac {1}{24} b^2 \left (x+\sqrt {-b+x^2}\right )^{3/2}+\frac {1}{3} c \left (x+\sqrt {-b+x^2}\right )^{3/2}+\frac {1}{112} b \left (x+\sqrt {-b+x^2}\right )^{7/2}+\frac {1}{176} \left (x+\sqrt {-b+x^2}\right )^{11/2}+(2 b c) \operatorname {Subst}\left (\int \frac {\sqrt {b}-x^2}{b+x^4} \, dx,x,\sqrt {x+\sqrt {-b+x^2}}\right )-(2 b c) \operatorname {Subst}\left (\int \frac {\sqrt {b}+x^2}{b+x^4} \, dx,x,\sqrt {x+\sqrt {-b+x^2}}\right )\\ &=-\frac {b^5}{144 \left (x+\sqrt {-b+x^2}\right )^{9/2}}-\frac {b^4}{80 \left (x+\sqrt {-b+x^2}\right )^{5/2}}+\frac {b^3}{8 \sqrt {x+\sqrt {-b+x^2}}}-\frac {b c}{\sqrt {x+\sqrt {-b+x^2}}}-\frac {1}{24} b^2 \left (x+\sqrt {-b+x^2}\right )^{3/2}+\frac {1}{3} c \left (x+\sqrt {-b+x^2}\right )^{3/2}+\frac {1}{112} b \left (x+\sqrt {-b+x^2}\right )^{7/2}+\frac {1}{176} \left (x+\sqrt {-b+x^2}\right )^{11/2}-\frac {\left (b^{3/4} c\right ) \operatorname {Subst}\left (\int \frac {\sqrt {2} \sqrt [4]{b}+2 x}{-\sqrt {b}-\sqrt {2} \sqrt [4]{b} x-x^2} \, dx,x,\sqrt {x+\sqrt {-b+x^2}}\right )}{\sqrt {2}}-\frac {\left (b^{3/4} c\right ) \operatorname {Subst}\left (\int \frac {\sqrt {2} \sqrt [4]{b}-2 x}{-\sqrt {b}+\sqrt {2} \sqrt [4]{b} x-x^2} \, dx,x,\sqrt {x+\sqrt {-b+x^2}}\right )}{\sqrt {2}}-(b c) \operatorname {Subst}\left (\int \frac {1}{\sqrt {b}-\sqrt {2} \sqrt [4]{b} x+x^2} \, dx,x,\sqrt {x+\sqrt {-b+x^2}}\right )-(b c) \operatorname {Subst}\left (\int \frac {1}{\sqrt {b}+\sqrt {2} \sqrt [4]{b} x+x^2} \, dx,x,\sqrt {x+\sqrt {-b+x^2}}\right )\\ &=-\frac {b^5}{144 \left (x+\sqrt {-b+x^2}\right )^{9/2}}-\frac {b^4}{80 \left (x+\sqrt {-b+x^2}\right )^{5/2}}+\frac {b^3}{8 \sqrt {x+\sqrt {-b+x^2}}}-\frac {b c}{\sqrt {x+\sqrt {-b+x^2}}}-\frac {1}{24} b^2 \left (x+\sqrt {-b+x^2}\right )^{3/2}+\frac {1}{3} c \left (x+\sqrt {-b+x^2}\right )^{3/2}+\frac {1}{112} b \left (x+\sqrt {-b+x^2}\right )^{7/2}+\frac {1}{176} \left (x+\sqrt {-b+x^2}\right )^{11/2}-\frac {b^{3/4} c \log \left (\sqrt {b}+x+\sqrt {-b+x^2}-\sqrt {2} \sqrt [4]{b} \sqrt {x+\sqrt {-b+x^2}}\right )}{\sqrt {2}}+\frac {b^{3/4} c \log \left (\sqrt {b}+x+\sqrt {-b+x^2}+\sqrt {2} \sqrt [4]{b} \sqrt {x+\sqrt {-b+x^2}}\right )}{\sqrt {2}}-\left (\sqrt {2} b^{3/4} c\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt {x+\sqrt {-b+x^2}}}{\sqrt [4]{b}}\right )+\left (\sqrt {2} b^{3/4} c\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt {x+\sqrt {-b+x^2}}}{\sqrt [4]{b}}\right )\\ &=-\frac {b^5}{144 \left (x+\sqrt {-b+x^2}\right )^{9/2}}-\frac {b^4}{80 \left (x+\sqrt {-b+x^2}\right )^{5/2}}+\frac {b^3}{8 \sqrt {x+\sqrt {-b+x^2}}}-\frac {b c}{\sqrt {x+\sqrt {-b+x^2}}}-\frac {1}{24} b^2 \left (x+\sqrt {-b+x^2}\right )^{3/2}+\frac {1}{3} c \left (x+\sqrt {-b+x^2}\right )^{3/2}+\frac {1}{112} b \left (x+\sqrt {-b+x^2}\right )^{7/2}+\frac {1}{176} \left (x+\sqrt {-b+x^2}\right )^{11/2}+\sqrt {2} b^{3/4} c \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {x+\sqrt {-b+x^2}}}{\sqrt [4]{b}}\right )-\sqrt {2} b^{3/4} c \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {x+\sqrt {-b+x^2}}}{\sqrt [4]{b}}\right )-\frac {b^{3/4} c \log \left (\sqrt {b}+x+\sqrt {-b+x^2}-\sqrt {2} \sqrt [4]{b} \sqrt {x+\sqrt {-b+x^2}}\right )}{\sqrt {2}}+\frac {b^{3/4} c \log \left (\sqrt {b}+x+\sqrt {-b+x^2}+\sqrt {2} \sqrt [4]{b} \sqrt {x+\sqrt {-b+x^2}}\right )}{\sqrt {2}}\\ \end {align*}

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Mathematica [A]  time = 8.63, size = 524, normalized size = 1.74 \begin {gather*} \frac {1}{6} c \left (-3 \sqrt {2} b^{3/4} \log \left (\sqrt {x^2-b}-\sqrt {2} \sqrt [4]{b} \sqrt {\sqrt {x^2-b}+x}+\sqrt {b}+x\right )+3 \sqrt {2} b^{3/4} \log \left (\sqrt {x^2-b}+\sqrt {2} \sqrt [4]{b} \sqrt {\sqrt {x^2-b}+x}+\sqrt {b}+x\right )+6 \sqrt {2} b^{3/4} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {\sqrt {x^2-b}+x}}{\sqrt [4]{b}}\right )-6 \sqrt {2} b^{3/4} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {\sqrt {x^2-b}+x}}{\sqrt [4]{b}}+1\right )+\frac {4 x \left (\sqrt {x^2-b}+x\right )-8 b}{\sqrt {\sqrt {x^2-b}+x}}\right )+\frac {2 \sqrt {x^2-b} \left (304 b^5-342 b^4 x \left (4 \sqrt {x^2-b}+9 x\right )+15 b^3 x^3 \left (247 \sqrt {x^2-b}+249 x\right )+15 b^2 x^5 \left (89 \sqrt {x^2-b}+317 x\right )+5040 x^9 \left (\sqrt {x^2-b}+x\right )-180 b x^7 \left (45 \sqrt {x^2-b}+59 x\right )\right ) \left (\sqrt {x^2-b}+x\right )^{9/2}}{3465 \left (-b^5+b^4 x \left (9 \sqrt {x^2-b}+41 x\right )-40 b^3 x^3 \left (3 \sqrt {x^2-b}+7 x\right )+16 b^2 x^5 \left (27 \sqrt {x^2-b}+43 x\right )+256 x^9 \left (\sqrt {x^2-b}+x\right )-64 b x^7 \left (9 \sqrt {x^2-b}+11 x\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[-b + x^2]*(x + Sqrt[-b + x^2])^(9/2)*(304*b^5 + 5040*x^9*(x + Sqrt[-b + x^2]) - 342*b^4*x*(9*x + 4*Sqr
t[-b + x^2]) - 180*b*x^7*(59*x + 45*Sqrt[-b + x^2]) + 15*b^2*x^5*(317*x + 89*Sqrt[-b + x^2]) + 15*b^3*x^3*(249
*x + 247*Sqrt[-b + x^2])))/(3465*(-b^5 + 256*x^9*(x + Sqrt[-b + x^2]) - 40*b^3*x^3*(7*x + 3*Sqrt[-b + x^2]) -
64*b*x^7*(11*x + 9*Sqrt[-b + x^2]) + b^4*x*(41*x + 9*Sqrt[-b + x^2]) + 16*b^2*x^5*(43*x + 27*Sqrt[-b + x^2])))
+ (c*((-8*b + 4*x*(x + Sqrt[-b + x^2]))/Sqrt[x + Sqrt[-b + x^2]] + 6*Sqrt[2]*b^(3/4)*ArcTan[1 - (Sqrt[2]*Sqrt
[x + Sqrt[-b + x^2]])/b^(1/4)] - 6*Sqrt[2]*b^(3/4)*ArcTan[1 + (Sqrt[2]*Sqrt[x + Sqrt[-b + x^2]])/b^(1/4)] - 3*
Sqrt[2]*b^(3/4)*Log[Sqrt[b] + x + Sqrt[-b + x^2] - Sqrt[2]*b^(1/4)*Sqrt[x + Sqrt[-b + x^2]]] + 3*Sqrt[2]*b^(3/
4)*Log[Sqrt[b] + x + Sqrt[-b + x^2] + Sqrt[2]*b^(1/4)*Sqrt[x + Sqrt[-b + x^2]]]))/6

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IntegrateAlgebraic [A]  time = 0.64, size = 301, normalized size = 1.00 \begin {gather*} \frac {2 \sqrt {-b+x^2} \left (-1368 b^4 x+10395 b^2 c x+3705 b^3 x^3-32340 b c x^3+1335 b^2 x^5+18480 c x^5-8100 b x^7+5040 x^9\right )+2 \left (304 b^5-2310 b^3 c-3078 b^4 x^2+24255 b^2 c x^2+3735 b^3 x^4-41580 b c x^4+4755 b^2 x^6+18480 c x^6-10620 b x^8+5040 x^{10}\right )}{3465 \left (x+\sqrt {-b+x^2}\right )^{9/2}}+\sqrt {2} b^{3/4} c \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x+\sqrt {-b+x^2}}}{-\sqrt {b}+x+\sqrt {-b+x^2}}\right )+\sqrt {2} b^{3/4} c \tanh ^{-1}\left (\frac {\frac {\sqrt [4]{b}}{\sqrt {2}}+\frac {x}{\sqrt {2} \sqrt [4]{b}}+\frac {\sqrt {-b+x^2}}{\sqrt {2} \sqrt [4]{b}}}{\sqrt {x+\sqrt {-b+x^2}}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(Sqrt[-b + x^2]*(c + x^4)*Sqrt[x + Sqrt[-b + x^2]])/x,x]

[Out]

(2*Sqrt[-b + x^2]*(-1368*b^4*x + 10395*b^2*c*x + 3705*b^3*x^3 - 32340*b*c*x^3 + 1335*b^2*x^5 + 18480*c*x^5 - 8
100*b*x^7 + 5040*x^9) + 2*(304*b^5 - 2310*b^3*c - 3078*b^4*x^2 + 24255*b^2*c*x^2 + 3735*b^3*x^4 - 41580*b*c*x^
4 + 4755*b^2*x^6 + 18480*c*x^6 - 10620*b*x^8 + 5040*x^10))/(3465*(x + Sqrt[-b + x^2])^(9/2)) + Sqrt[2]*b^(3/4)
*c*ArcTan[(Sqrt[2]*b^(1/4)*Sqrt[x + Sqrt[-b + x^2]])/(-Sqrt[b] + x + Sqrt[-b + x^2])] + Sqrt[2]*b^(3/4)*c*ArcT
anh[(b^(1/4)/Sqrt[2] + x/(Sqrt[2]*b^(1/4)) + Sqrt[-b + x^2]/(Sqrt[2]*b^(1/4)))/Sqrt[x + Sqrt[-b + x^2]]]

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fricas [A]  time = 0.48, size = 269, normalized size = 0.89 \begin {gather*} -\frac {2}{3465} \, {\left (35 \, x^{5} - 19 \, b x^{3} - {\left (152 \, b^{2} - 1155 \, c\right )} x - 2 \, {\left (175 \, x^{4} - 57 \, b x^{2} - 152 \, b^{2} + 1155 \, c\right )} \sqrt {x^{2} - b}\right )} \sqrt {x + \sqrt {x^{2} - b}} + 4 \, \left (-b^{3} c^{4}\right )^{\frac {1}{4}} \arctan \left (-\frac {\left (-b^{3} c^{4}\right )^{\frac {1}{4}} b^{2} c^{3} \sqrt {x + \sqrt {x^{2} - b}} - \sqrt {b^{4} c^{6} x + \sqrt {x^{2} - b} b^{4} c^{6} - \sqrt {-b^{3} c^{4}} b^{3} c^{4}} \left (-b^{3} c^{4}\right )^{\frac {1}{4}}}{b^{3} c^{4}}\right ) - \left (-b^{3} c^{4}\right )^{\frac {1}{4}} \log \left (b^{2} c^{3} \sqrt {x + \sqrt {x^{2} - b}} + \left (-b^{3} c^{4}\right )^{\frac {3}{4}}\right ) + \left (-b^{3} c^{4}\right )^{\frac {1}{4}} \log \left (b^{2} c^{3} \sqrt {x + \sqrt {x^{2} - b}} - \left (-b^{3} c^{4}\right )^{\frac {3}{4}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3465*(35*x^5 - 19*b*x^3 - (152*b^2 - 1155*c)*x - 2*(175*x^4 - 57*b*x^2 - 152*b^2 + 1155*c)*sqrt(x^2 - b))*s
qrt(x + sqrt(x^2 - b)) + 4*(-b^3*c^4)^(1/4)*arctan(-((-b^3*c^4)^(1/4)*b^2*c^3*sqrt(x + sqrt(x^2 - b)) - sqrt(b
^4*c^6*x + sqrt(x^2 - b)*b^4*c^6 - sqrt(-b^3*c^4)*b^3*c^4)*(-b^3*c^4)^(1/4))/(b^3*c^4)) - (-b^3*c^4)^(1/4)*log
(b^2*c^3*sqrt(x + sqrt(x^2 - b)) + (-b^3*c^4)^(3/4)) + (-b^3*c^4)^(1/4)*log(b^2*c^3*sqrt(x + sqrt(x^2 - b)) -
(-b^3*c^4)^(3/4))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{4} + c\right )} \sqrt {x^{2} - b} \sqrt {x + \sqrt {x^{2} - b}}}{x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((x^4 + c)*sqrt(x^2 - b)*sqrt(x + sqrt(x^2 - b))/x, x)

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maple [F]  time = 0.05, size = 0, normalized size = 0.00 \[\int \frac {\sqrt {x^{2}-b}\, \left (x^{4}+c \right ) \sqrt {x +\sqrt {x^{2}-b}}}{x}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2-b)^(1/2)*(x^4+c)*(x+(x^2-b)^(1/2))^(1/2)/x,x)

[Out]

int((x^2-b)^(1/2)*(x^4+c)*(x+(x^2-b)^(1/2))^(1/2)/x,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{4} + c\right )} \sqrt {x^{2} - b} \sqrt {x + \sqrt {x^{2} - b}}}{x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((x^4 + c)*sqrt(x^2 - b)*sqrt(x + sqrt(x^2 - b))/x, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\sqrt {x+\sqrt {x^2-b}}\,\left (x^4+c\right )\,\sqrt {x^2-b}}{x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x + (x^2 - b)^(1/2))^(1/2)*(c + x^4)*(x^2 - b)^(1/2))/x,x)

[Out]

int(((x + (x^2 - b)^(1/2))^(1/2)*(c + x^4)*(x^2 - b)^(1/2))/x, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {- b + x^{2}} \left (c + x^{4}\right ) \sqrt {x + \sqrt {- b + x^{2}}}}{x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**2-b)**(1/2)*(x**4+c)*(x+(x**2-b)**(1/2))**(1/2)/x,x)

[Out]

Integral(sqrt(-b + x**2)*(c + x**4)*sqrt(x + sqrt(-b + x**2))/x, x)

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