3.318 \(\int x^5 \sqrt{a+\frac{b}{c+d x^2}} \, dx\)

Optimal. Leaf size=216 \[ -\frac{\left (-8 a^2 c^2+4 a b c+b^2\right ) \left (c+d x^2\right ) \sqrt{\frac{a c+a d x^2+b}{c+d x^2}}}{16 a^2 d^3}+\frac{b \left (8 a^2 c^2+4 a b c+b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{\frac{a c+a d x^2+b}{c+d x^2}}}{\sqrt{a}}\right )}{16 a^{5/2} d^3}+\frac{\left (c+d x^2\right )^3 \left (\frac{a c+a d x^2+b}{c+d x^2}\right )^{3/2}}{6 a d^3}-\frac{(4 a c+b) \left (c+d x^2\right )^2 \sqrt{\frac{a c+a d x^2+b}{c+d x^2}}}{8 a d^3} \]

[Out]

-((b^2 + 4*a*b*c - 8*a^2*c^2)*(c + d*x^2)*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)])/(16*a^2*d^3) - ((b + 4*a*c)*(
c + d*x^2)^2*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)])/(8*a*d^3) + ((c + d*x^2)^3*((b + a*c + a*d*x^2)/(c + d*x^2
))^(3/2))/(6*a*d^3) + (b*(b^2 + 4*a*b*c + 8*a^2*c^2)*ArcTanh[Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)]/Sqrt[a]])/(
16*a^(5/2)*d^3)

________________________________________________________________________________________

Rubi [A]  time = 0.620321, antiderivative size = 259, normalized size of antiderivative = 1.2, number of steps used = 9, number of rules used = 9, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {6722, 1975, 446, 90, 80, 50, 63, 217, 206} \[ \frac{\left (8 a^2 c^2+4 a b c+b^2\right ) \left (c+d x^2\right ) \sqrt{a+\frac{b}{c+d x^2}}}{16 a^2 d^3}+\frac{b \left (8 a^2 c^2+4 a b c+b^2\right ) \sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{c+d x^2}}{\sqrt{a \left (c+d x^2\right )+b}}\right )}{16 a^{5/2} d^3 \sqrt{a \left (c+d x^2\right )+b}}-\frac{(8 a c+3 b) \left (c+d x^2\right ) \sqrt{a+\frac{b}{c+d x^2}} \left (a \left (c+d x^2\right )+b\right )}{24 a^2 d^3}+\frac{x^2 \left (c+d x^2\right ) \sqrt{a+\frac{b}{c+d x^2}} \left (a \left (c+d x^2\right )+b\right )}{6 a d^2} \]

Antiderivative was successfully verified.

[In]

Int[x^5*Sqrt[a + b/(c + d*x^2)],x]

[Out]

((b^2 + 4*a*b*c + 8*a^2*c^2)*(c + d*x^2)*Sqrt[a + b/(c + d*x^2)])/(16*a^2*d^3) - ((3*b + 8*a*c)*(c + d*x^2)*Sq
rt[a + b/(c + d*x^2)]*(b + a*(c + d*x^2)))/(24*a^2*d^3) + (x^2*(c + d*x^2)*Sqrt[a + b/(c + d*x^2)]*(b + a*(c +
d*x^2)))/(6*a*d^2) + (b*(b^2 + 4*a*b*c + 8*a^2*c^2)*Sqrt[c + d*x^2]*Sqrt[a + b/(c + d*x^2)]*ArcTanh[(Sqrt[a]*
Sqrt[c + d*x^2])/Sqrt[b + a*(c + d*x^2)]])/(16*a^(5/2)*d^3*Sqrt[b + a*(c + d*x^2)])

Rule 6722

Int[(u_.)*((a_.) + (b_.)*(v_)^(n_))^(p_), x_Symbol] :> Dist[(a + b*v^n)^FracPart[p]/(v^(n*FracPart[p])*(b + a/
v^n)^FracPart[p]), Int[u*v^(n*p)*(b + a/v^n)^p, x], x] /; FreeQ[{a, b, p}, x] &&  !IntegerQ[p] && ILtQ[n, 0] &
& BinomialQ[v, x] &&  !LinearQ[v, x]

Rule 1975

Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*ExpandToSum[u, x]^p*ExpandToSum[v, x]^q
, x] /; FreeQ[{e, m, p, q}, x] && BinomialQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0]
&&  !BinomialMatchQ[{u, v}, x]

Rule 446

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

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 80

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, 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 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 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])

Rubi steps

\begin{align*} \int x^5 \sqrt{a+\frac{b}{c+d x^2}} \, dx &=\frac{\left (\sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}}\right ) \int \frac{x^5 \sqrt{b+a \left (c+d x^2\right )}}{\sqrt{c+d x^2}} \, dx}{\sqrt{b+a \left (c+d x^2\right )}}\\ &=\frac{\left (\sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}}\right ) \int \frac{x^5 \sqrt{b+a c+a d x^2}}{\sqrt{c+d x^2}} \, dx}{\sqrt{b+a \left (c+d x^2\right )}}\\ &=\frac{\left (\sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}}\right ) \operatorname{Subst}\left (\int \frac{x^2 \sqrt{b+a c+a d x}}{\sqrt{c+d x}} \, dx,x,x^2\right )}{2 \sqrt{b+a \left (c+d x^2\right )}}\\ &=\frac{x^2 \left (c+d x^2\right ) \sqrt{a+\frac{b}{c+d x^2}} \left (b+a \left (c+d x^2\right )\right )}{6 a d^2}+\frac{\left (\sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{b+a c+a d x} \left (-c (b+a c)-\frac{1}{2} (3 b+8 a c) d x\right )}{\sqrt{c+d x}} \, dx,x,x^2\right )}{6 a d^2 \sqrt{b+a \left (c+d x^2\right )}}\\ &=-\frac{(3 b+8 a c) \left (c+d x^2\right ) \sqrt{a+\frac{b}{c+d x^2}} \left (b+a \left (c+d x^2\right )\right )}{24 a^2 d^3}+\frac{x^2 \left (c+d x^2\right ) \sqrt{a+\frac{b}{c+d x^2}} \left (b+a \left (c+d x^2\right )\right )}{6 a d^2}+\frac{\left (\left (-2 a c (b+a c) d^2+\frac{1}{2} (3 b+8 a c) d \left (\frac{3 a c d}{2}+\frac{1}{2} (b+a c) d\right )\right ) \sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{b+a c+a d x}}{\sqrt{c+d x}} \, dx,x,x^2\right )}{12 a^2 d^4 \sqrt{b+a \left (c+d x^2\right )}}\\ &=\frac{\left (b^2+4 a b c+8 a^2 c^2\right ) \left (c+d x^2\right ) \sqrt{a+\frac{b}{c+d x^2}}}{16 a^2 d^3}-\frac{(3 b+8 a c) \left (c+d x^2\right ) \sqrt{a+\frac{b}{c+d x^2}} \left (b+a \left (c+d x^2\right )\right )}{24 a^2 d^3}+\frac{x^2 \left (c+d x^2\right ) \sqrt{a+\frac{b}{c+d x^2}} \left (b+a \left (c+d x^2\right )\right )}{6 a d^2}+\frac{\left (b \left (-2 a c (b+a c) d^2+\frac{1}{2} (3 b+8 a c) d \left (\frac{3 a c d}{2}+\frac{1}{2} (b+a c) d\right )\right ) \sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c+d x} \sqrt{b+a c+a d x}} \, dx,x,x^2\right )}{24 a^2 d^4 \sqrt{b+a \left (c+d x^2\right )}}\\ &=\frac{\left (b^2+4 a b c+8 a^2 c^2\right ) \left (c+d x^2\right ) \sqrt{a+\frac{b}{c+d x^2}}}{16 a^2 d^3}-\frac{(3 b+8 a c) \left (c+d x^2\right ) \sqrt{a+\frac{b}{c+d x^2}} \left (b+a \left (c+d x^2\right )\right )}{24 a^2 d^3}+\frac{x^2 \left (c+d x^2\right ) \sqrt{a+\frac{b}{c+d x^2}} \left (b+a \left (c+d x^2\right )\right )}{6 a d^2}+\frac{\left (b \left (-2 a c (b+a c) d^2+\frac{1}{2} (3 b+8 a c) d \left (\frac{3 a c d}{2}+\frac{1}{2} (b+a c) d\right )\right ) \sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{b+a x^2}} \, dx,x,\sqrt{c+d x^2}\right )}{12 a^2 d^5 \sqrt{b+a \left (c+d x^2\right )}}\\ &=\frac{\left (b^2+4 a b c+8 a^2 c^2\right ) \left (c+d x^2\right ) \sqrt{a+\frac{b}{c+d x^2}}}{16 a^2 d^3}-\frac{(3 b+8 a c) \left (c+d x^2\right ) \sqrt{a+\frac{b}{c+d x^2}} \left (b+a \left (c+d x^2\right )\right )}{24 a^2 d^3}+\frac{x^2 \left (c+d x^2\right ) \sqrt{a+\frac{b}{c+d x^2}} \left (b+a \left (c+d x^2\right )\right )}{6 a d^2}+\frac{\left (b \left (-2 a c (b+a c) d^2+\frac{1}{2} (3 b+8 a c) d \left (\frac{3 a c d}{2}+\frac{1}{2} (b+a c) d\right )\right ) \sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}}\right ) \operatorname{Subst}\left (\int \frac{1}{1-a x^2} \, dx,x,\frac{\sqrt{c+d x^2}}{\sqrt{b+a \left (c+d x^2\right )}}\right )}{12 a^2 d^5 \sqrt{b+a \left (c+d x^2\right )}}\\ &=\frac{\left (b^2+4 a b c+8 a^2 c^2\right ) \left (c+d x^2\right ) \sqrt{a+\frac{b}{c+d x^2}}}{16 a^2 d^3}-\frac{(3 b+8 a c) \left (c+d x^2\right ) \sqrt{a+\frac{b}{c+d x^2}} \left (b+a \left (c+d x^2\right )\right )}{24 a^2 d^3}+\frac{x^2 \left (c+d x^2\right ) \sqrt{a+\frac{b}{c+d x^2}} \left (b+a \left (c+d x^2\right )\right )}{6 a d^2}+\frac{b \left (b^2+4 a b c+8 a^2 c^2\right ) \sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{c+d x^2}}{\sqrt{b+a \left (c+d x^2\right )}}\right )}{16 a^{5/2} d^3 \sqrt{b+a \left (c+d x^2\right )}}\\ \end{align*}

Mathematica [A]  time = 0.309146, size = 137, normalized size = 0.63 \[ \frac{\sqrt{a} \left (c+d x^2\right ) \sqrt{\frac{a c+a d x^2+b}{c+d x^2}} \left (8 a^2 \left (c^2-c d x^2+d^2 x^4\right )+2 a b \left (d x^2-5 c\right )-3 b^2\right )+3 b \left (8 a^2 c^2+4 a b c+b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{c+d x^2}}}{\sqrt{a}}\right )}{48 a^{5/2} d^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5*Sqrt[a + b/(c + d*x^2)],x]

[Out]

(Sqrt[a]*(c + d*x^2)*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)]*(-3*b^2 + 2*a*b*(-5*c + d*x^2) + 8*a^2*(c^2 - c*d*x
^2 + d^2*x^4)) + 3*b*(b^2 + 4*a*b*c + 8*a^2*c^2)*ArcTanh[Sqrt[a + b/(c + d*x^2)]/Sqrt[a]])/(48*a^(5/2)*d^3)

________________________________________________________________________________________

Maple [B]  time = 0.05, size = 533, normalized size = 2.5 \begin{align*}{\frac{d{x}^{2}+c}{96\,{a}^{2}{d}^{3}}\sqrt{{\frac{ad{x}^{2}+ac+b}{d{x}^{2}+c}}} \left ( -48\,\sqrt{a{d}^{2}{x}^{4}+2\,acd{x}^{2}+bd{x}^{2}+{c}^{2}a+bc}{x}^{2}c{a}^{2}d\sqrt{a{d}^{2}}-12\,\sqrt{a{d}^{2}{x}^{4}+2\,acd{x}^{2}+bd{x}^{2}+{c}^{2}a+bc}{x}^{2}bad\sqrt{a{d}^{2}}+24\,\ln \left ( 1/2\,{\frac{2\,a{d}^{2}{x}^{2}+2\,acd+2\,\sqrt{a{d}^{2}{x}^{4}+2\,acd{x}^{2}+bd{x}^{2}+{c}^{2}a+bc}\sqrt{a{d}^{2}}+bd}{\sqrt{a{d}^{2}}}} \right ){a}^{2}b{c}^{2}d+12\,\ln \left ( 1/2\,{\frac{2\,a{d}^{2}{x}^{2}+2\,acd+2\,\sqrt{a{d}^{2}{x}^{4}+2\,acd{x}^{2}+bd{x}^{2}+{c}^{2}a+bc}\sqrt{a{d}^{2}}+bd}{\sqrt{a{d}^{2}}}} \right ){b}^{2}cad+16\, \left ( a{d}^{2}{x}^{4}+2\,acd{x}^{2}+bd{x}^{2}+{c}^{2}a+bc \right ) ^{3/2}a\sqrt{a{d}^{2}}-36\,\sqrt{a{d}^{2}{x}^{4}+2\,acd{x}^{2}+bd{x}^{2}+{c}^{2}a+bc}cba\sqrt{a{d}^{2}}+3\,\ln \left ( 1/2\,{\frac{2\,a{d}^{2}{x}^{2}+2\,acd+2\,\sqrt{a{d}^{2}{x}^{4}+2\,acd{x}^{2}+bd{x}^{2}+{c}^{2}a+bc}\sqrt{a{d}^{2}}+bd}{\sqrt{a{d}^{2}}}} \right ){b}^{3}d-6\,\sqrt{a{d}^{2}{x}^{4}+2\,acd{x}^{2}+bd{x}^{2}+{c}^{2}a+bc}{b}^{2}\sqrt{a{d}^{2}} \right ){\frac{1}{\sqrt{ \left ( d{x}^{2}+c \right ) \left ( ad{x}^{2}+ac+b \right ) }}}{\frac{1}{\sqrt{a{d}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(a+b/(d*x^2+c))^(1/2),x)

[Out]

1/96*((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)*(d*x^2+c)/d^3*(-48*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*x^2*
c*a^2*d*(a*d^2)^(1/2)-12*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*x^2*b*a*d*(a*d^2)^(1/2)+24*ln(1/2*(2*
a*d^2*x^2+2*a*c*d+2*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*(a*d^2)^(1/2)+b*d)/(a*d^2)^(1/2))*a^2*b*c^
2*d+12*ln(1/2*(2*a*d^2*x^2+2*a*c*d+2*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*(a*d^2)^(1/2)+b*d)/(a*d^2
)^(1/2))*b^2*c*a*d+16*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(3/2)*a*(a*d^2)^(1/2)-36*(a*d^2*x^4+2*a*c*d*x^
2+b*d*x^2+a*c^2+b*c)^(1/2)*c*b*a*(a*d^2)^(1/2)+3*ln(1/2*(2*a*d^2*x^2+2*a*c*d+2*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+
a*c^2+b*c)^(1/2)*(a*d^2)^(1/2)+b*d)/(a*d^2)^(1/2))*b^3*d-6*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*b^2
*(a*d^2)^(1/2))/((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)/a^2/(a*d^2)^(1/2)

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b/(d*x^2+c))^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.37123, size = 944, normalized size = 4.37 \begin{align*} \left [\frac{3 \,{\left (8 \, a^{2} b c^{2} + 4 \, a b^{2} c + b^{3}\right )} \sqrt{a} \log \left (8 \, a^{2} d^{2} x^{4} + 8 \, a^{2} c^{2} + 8 \,{\left (2 \, a^{2} c + a b\right )} d x^{2} + 8 \, a b c + b^{2} + 4 \,{\left (2 \, a d^{2} x^{4} +{\left (4 \, a c + b\right )} d x^{2} + 2 \, a c^{2} + b c\right )} \sqrt{a} \sqrt{\frac{a d x^{2} + a c + b}{d x^{2} + c}}\right ) + 4 \,{\left (8 \, a^{3} d^{3} x^{6} + 2 \, a^{2} b d^{2} x^{4} + 8 \, a^{3} c^{3} - 10 \, a^{2} b c^{2} - 3 \, a b^{2} c -{\left (8 \, a^{2} b c + 3 \, a b^{2}\right )} d x^{2}\right )} \sqrt{\frac{a d x^{2} + a c + b}{d x^{2} + c}}}{192 \, a^{3} d^{3}}, -\frac{3 \,{\left (8 \, a^{2} b c^{2} + 4 \, a b^{2} c + b^{3}\right )} \sqrt{-a} \arctan \left (\frac{{\left (2 \, a d x^{2} + 2 \, a c + b\right )} \sqrt{-a} \sqrt{\frac{a d x^{2} + a c + b}{d x^{2} + c}}}{2 \,{\left (a^{2} d x^{2} + a^{2} c + a b\right )}}\right ) - 2 \,{\left (8 \, a^{3} d^{3} x^{6} + 2 \, a^{2} b d^{2} x^{4} + 8 \, a^{3} c^{3} - 10 \, a^{2} b c^{2} - 3 \, a b^{2} c -{\left (8 \, a^{2} b c + 3 \, a b^{2}\right )} d x^{2}\right )} \sqrt{\frac{a d x^{2} + a c + b}{d x^{2} + c}}}{96 \, a^{3} d^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b/(d*x^2+c))^(1/2),x, algorithm="fricas")

[Out]

[1/192*(3*(8*a^2*b*c^2 + 4*a*b^2*c + b^3)*sqrt(a)*log(8*a^2*d^2*x^4 + 8*a^2*c^2 + 8*(2*a^2*c + a*b)*d*x^2 + 8*
a*b*c + b^2 + 4*(2*a*d^2*x^4 + (4*a*c + b)*d*x^2 + 2*a*c^2 + b*c)*sqrt(a)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)
)) + 4*(8*a^3*d^3*x^6 + 2*a^2*b*d^2*x^4 + 8*a^3*c^3 - 10*a^2*b*c^2 - 3*a*b^2*c - (8*a^2*b*c + 3*a*b^2)*d*x^2)*
sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)))/(a^3*d^3), -1/96*(3*(8*a^2*b*c^2 + 4*a*b^2*c + b^3)*sqrt(-a)*arctan(1/2
*(2*a*d*x^2 + 2*a*c + b)*sqrt(-a)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c))/(a^2*d*x^2 + a^2*c + a*b)) - 2*(8*a^3*
d^3*x^6 + 2*a^2*b*d^2*x^4 + 8*a^3*c^3 - 10*a^2*b*c^2 - 3*a*b^2*c - (8*a^2*b*c + 3*a*b^2)*d*x^2)*sqrt((a*d*x^2
+ a*c + b)/(d*x^2 + c)))/(a^3*d^3)]

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{5} \sqrt{\frac{a c + a d x^{2} + b}{c + d x^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(a+b/(d*x**2+c))**(1/2),x)

[Out]

Integral(x**5*sqrt((a*c + a*d*x**2 + b)/(c + d*x**2)), x)

________________________________________________________________________________________

Giac [A]  time = 1.68737, size = 408, normalized size = 1.89 \begin{align*} \frac{1}{48} \, \sqrt{a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c}{\left (2 \,{\left (\frac{4 \, x^{2} \mathrm{sgn}\left (d x^{2} + c\right )}{d} - \frac{4 \, a^{3} c d^{4} \mathrm{sgn}\left (d x^{2} + c\right ) - a^{2} b d^{4} \mathrm{sgn}\left (d x^{2} + c\right )}{a^{3} d^{6}}\right )} x^{2} + \frac{8 \, a^{3} c^{2} d^{3} \mathrm{sgn}\left (d x^{2} + c\right ) - 10 \, a^{2} b c d^{3} \mathrm{sgn}\left (d x^{2} + c\right ) - 3 \, a b^{2} d^{3} \mathrm{sgn}\left (d x^{2} + c\right )}{a^{3} d^{6}}\right )} - \frac{{\left (8 \, a^{3} b c^{2} d^{4} \mathrm{sgn}\left (d x^{2} + c\right ) + 4 \, a^{2} b^{2} c d^{4} \mathrm{sgn}\left (d x^{2} + c\right ) + a b^{3} d^{4} \mathrm{sgn}\left (d x^{2} + c\right )\right )} \log \left ({\left | -2 \, a^{\frac{3}{2}} c d - 2 \,{\left (\sqrt{a d^{2}} x^{2} - \sqrt{a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c}\right )} a{\left | d \right |} - \sqrt{a} b d \right |}\right )}{32 \, a^{\frac{7}{2}} d^{6}{\left | d \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b/(d*x^2+c))^(1/2),x, algorithm="giac")

[Out]

1/48*sqrt(a*d^2*x^4 + 2*a*c*d*x^2 + b*d*x^2 + a*c^2 + b*c)*(2*(4*x^2*sgn(d*x^2 + c)/d - (4*a^3*c*d^4*sgn(d*x^2
+ c) - a^2*b*d^4*sgn(d*x^2 + c))/(a^3*d^6))*x^2 + (8*a^3*c^2*d^3*sgn(d*x^2 + c) - 10*a^2*b*c*d^3*sgn(d*x^2 +
c) - 3*a*b^2*d^3*sgn(d*x^2 + c))/(a^3*d^6)) - 1/32*(8*a^3*b*c^2*d^4*sgn(d*x^2 + c) + 4*a^2*b^2*c*d^4*sgn(d*x^2
+ c) + a*b^3*d^4*sgn(d*x^2 + c))*log(abs(-2*a^(3/2)*c*d - 2*(sqrt(a*d^2)*x^2 - sqrt(a*d^2*x^4 + 2*a*c*d*x^2 +
b*d*x^2 + a*c^2 + b*c))*a*abs(d) - sqrt(a)*b*d))/(a^(7/2)*d^6*abs(d))