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

Optimal. Leaf size=244 \[ \frac{\left (c+d x^2\right ) \left (-a^2 d^2-2 a b c d+11 b^2 c^2\right ) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{16 b^2 d^3}-\frac{\sqrt{e} (b c-a d) \left (a^2 d^2+2 a b c d+5 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt{b} \sqrt{e}}\right )}{16 b^{5/2} d^{7/2}}+\frac{\left (c+d x^2\right )^3 \left (\frac{e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}}{6 b d^2 e}-\frac{\left (c+d x^2\right )^2 (a d+3 b c) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{8 b d^3} \]

[Out]

((11*b^2*c^2 - 2*a*b*c*d - a^2*d^2)*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(c + d*x^2))/(16*b^2*d^3) - ((3*b*c + a*
d)*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(c + d*x^2)^2)/(8*b*d^3) + (((e*(a + b*x^2))/(c + d*x^2))^(3/2)*(c + d*x^
2)^3)/(6*b*d^2*e) - ((b*c - a*d)*(5*b^2*c^2 + 2*a*b*c*d + a^2*d^2)*Sqrt[e]*ArcTanh[(Sqrt[d]*Sqrt[(e*(a + b*x^2
))/(c + d*x^2)])/(Sqrt[b]*Sqrt[e])])/(16*b^(5/2)*d^(7/2))

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Rubi [A]  time = 0.332508, antiderivative size = 244, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {1960, 463, 455, 385, 208} \[ \frac{\left (c+d x^2\right ) \left (-a^2 d^2-2 a b c d+11 b^2 c^2\right ) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{16 b^2 d^3}-\frac{\sqrt{e} (b c-a d) \left (a^2 d^2+2 a b c d+5 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt{b} \sqrt{e}}\right )}{16 b^{5/2} d^{7/2}}+\frac{\left (c+d x^2\right )^3 \left (\frac{e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}}{6 b d^2 e}-\frac{\left (c+d x^2\right )^2 (a d+3 b c) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{8 b d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((11*b^2*c^2 - 2*a*b*c*d - a^2*d^2)*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(c + d*x^2))/(16*b^2*d^3) - ((3*b*c + a*
d)*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(c + d*x^2)^2)/(8*b*d^3) + (((e*(a + b*x^2))/(c + d*x^2))^(3/2)*(c + d*x^
2)^3)/(6*b*d^2*e) - ((b*c - a*d)*(5*b^2*c^2 + 2*a*b*c*d + a^2*d^2)*Sqrt[e]*ArcTanh[(Sqrt[d]*Sqrt[(e*(a + b*x^2
))/(c + d*x^2)])/(Sqrt[b]*Sqrt[e])])/(16*b^(5/2)*d^(7/2))

Rule 1960

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

Rule 463

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

Rule 455

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[((-a)^(m/2 - 1)*(b*c - a*d)*
x*(a + b*x^2)^(p + 1))/(2*b^(m/2 + 1)*(p + 1)), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[(a + b*x^2)^(p + 1)*E
xpandToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)]
- (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[
m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 385

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

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]

Rubi steps

\begin{align*} \int x^5 \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \, dx &=((b c-a d) e) \operatorname{Subst}\left (\int \frac{x^2 \left (-a e+c x^2\right )^2}{\left (b e-d x^2\right )^4} \, dx,x,\sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}\right )\\ &=\frac{\left (\frac{e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2} \left (c+d x^2\right )^3}{6 b d^2 e}-\frac{(b c-a d) \operatorname{Subst}\left (\int \frac{x^2 \left (-3 \left (2 a^2 d^2 e^2-(b c e-a d e)^2\right )+6 b c^2 d e x^2\right )}{\left (b e-d x^2\right )^3} \, dx,x,\sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}\right )}{6 b d^2}\\ &=-\frac{(3 b c+a d) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^2}{8 b d^3}+\frac{\left (\frac{e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2} \left (c+d x^2\right )^3}{6 b d^2 e}+\frac{(b c-a d) \operatorname{Subst}\left (\int \frac{3 d (b c-a d) (3 b c+a d) e^2+24 b c^2 d^2 e x^2}{\left (b e-d x^2\right )^2} \, dx,x,\sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}\right )}{24 b d^4}\\ &=\frac{\left (11 b^2 c^2-2 a b c d-a^2 d^2\right ) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{16 b^2 d^3}-\frac{(3 b c+a d) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^2}{8 b d^3}+\frac{\left (\frac{e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2} \left (c+d x^2\right )^3}{6 b d^2 e}-\frac{\left ((b c-a d) \left (5 b^2 c^2+2 a b c d+a^2 d^2\right ) e\right ) \operatorname{Subst}\left (\int \frac{1}{b e-d x^2} \, dx,x,\sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}\right )}{16 b^2 d^3}\\ &=\frac{\left (11 b^2 c^2-2 a b c d-a^2 d^2\right ) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{16 b^2 d^3}-\frac{(3 b c+a d) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^2}{8 b d^3}+\frac{\left (\frac{e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2} \left (c+d x^2\right )^3}{6 b d^2 e}-\frac{(b c-a d) \left (5 b^2 c^2+2 a b c d+a^2 d^2\right ) \sqrt{e} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt{b} \sqrt{e}}\right )}{16 b^{5/2} d^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.5243, size = 198, normalized size = 0.81 \[ \frac{\sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \left (-b \sqrt{d} \left (c+d x^2\right ) \left (3 a^2 d^2-2 a b d \left (d x^2-2 c\right )+b^2 \left (-15 c^2+10 c d x^2-8 d^2 x^4\right )\right )-\frac{3 \left (a^2 d^2+2 a b c d+5 b^2 c^2\right ) (b c-a d)^{3/2} \sqrt{\frac{b \left (c+d x^2\right )}{b c-a d}} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x^2}}{\sqrt{b c-a d}}\right )}{\sqrt{a+b x^2}}\right )}{48 b^3 d^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(-(b*Sqrt[d]*(c + d*x^2)*(3*a^2*d^2 - 2*a*b*d*(-2*c + d*x^2) + b^2*(-15*c^2
+ 10*c*d*x^2 - 8*d^2*x^4))) - (3*(b*c - a*d)^(3/2)*(5*b^2*c^2 + 2*a*b*c*d + a^2*d^2)*Sqrt[(b*(c + d*x^2))/(b*
c - a*d)]*ArcSinh[(Sqrt[d]*Sqrt[a + b*x^2])/Sqrt[b*c - a*d]])/Sqrt[a + b*x^2]))/(48*b^3*d^(7/2))

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Maple [B]  time = 0.056, size = 527, normalized size = 2.2 \begin{align*}{\frac{d{x}^{2}+c}{96\,{d}^{3}{b}^{2}}\sqrt{{\frac{e \left ( b{x}^{2}+a \right ) }{d{x}^{2}+c}}} \left ( -12\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}{x}^{2}ab{d}^{2}\sqrt{bd}-36\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}{x}^{2}c{b}^{2}d\sqrt{bd}+3\,{d}^{3}\ln \left ( 1/2\,{\frac{2\,bd{x}^{2}+2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){a}^{3}+3\,\ln \left ( 1/2\,{\frac{2\,bd{x}^{2}+2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){a}^{2}cb{d}^{2}+9\,\ln \left ( 1/2\,{\frac{2\,bd{x}^{2}+2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ) a{c}^{2}{b}^{2}d-15\,{b}^{3}\ln \left ( 1/2\,{\frac{2\,bd{x}^{2}+2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){c}^{3}+16\, \left ( bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac \right ) ^{3/2}bd\sqrt{bd}-6\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}{a}^{2}{d}^{2}\sqrt{bd}-24\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}acbd\sqrt{bd}+30\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}{c}^{2}{b}^{2}\sqrt{bd} \right ){\frac{1}{\sqrt{ \left ( d{x}^{2}+c \right ) \left ( b{x}^{2}+a \right ) }}}{\frac{1}{\sqrt{bd}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(e*(b*x^2+a)/(d*x^2+c))^(1/2)*(d*x^2+c)/d^3*(-12*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*x^2*a*b*d^2*(b*d)^(1
/2)-36*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*x^2*c*b^2*d*(b*d)^(1/2)+3*d^3*ln(1/2*(2*b*d*x^2+2*(b*d*x^4+a*d*x^2+
b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a^3+3*ln(1/2*(2*b*d*x^2+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(
1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a^2*c*b*d^2+9*ln(1/2*(2*b*d*x^2+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(
b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a*c^2*b^2*d-15*b^3*ln(1/2*(2*b*d*x^2+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b
*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*c^3+16*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(3/2)*b*d*(b*d)^(1/2)-6*(b*d*x^4+a*d*x^2+
b*c*x^2+a*c)^(1/2)*a^2*d^2*(b*d)^(1/2)-24*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*a*c*b*d*(b*d)^(1/2)+30*(b*d*x^4+
a*d*x^2+b*c*x^2+a*c)^(1/2)*c^2*b^2*(b*d)^(1/2))/((d*x^2+c)*(b*x^2+a))^(1/2)/b^2/(b*d)^(1/2)

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.5371, size = 1122, normalized size = 4.6 \begin{align*} \left [-\frac{3 \,{\left (5 \, b^{3} c^{3} - 3 \, a b^{2} c^{2} d - a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt{\frac{e}{b d}} \log \left (8 \, b^{2} d^{2} e x^{4} + 8 \,{\left (b^{2} c d + a b d^{2}\right )} e x^{2} +{\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} e + 4 \,{\left (2 \, b^{2} d^{3} x^{4} + b^{2} c^{2} d + a b c d^{2} +{\left (3 \, b^{2} c d^{2} + a b d^{3}\right )} x^{2}\right )} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}} \sqrt{\frac{e}{b d}}\right ) - 4 \,{\left (8 \, b^{2} d^{3} x^{6} + 15 \, b^{2} c^{3} - 4 \, a b c^{2} d - 3 \, a^{2} c d^{2} - 2 \,{\left (b^{2} c d^{2} - a b d^{3}\right )} x^{4} +{\left (5 \, b^{2} c^{2} d - 2 \, a b c d^{2} - 3 \, a^{2} d^{3}\right )} x^{2}\right )} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}}}{192 \, b^{2} d^{3}}, \frac{3 \,{\left (5 \, b^{3} c^{3} - 3 \, a b^{2} c^{2} d - a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt{-\frac{e}{b d}} \arctan \left (\frac{{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}} \sqrt{-\frac{e}{b d}}}{2 \,{\left (b e x^{2} + a e\right )}}\right ) + 2 \,{\left (8 \, b^{2} d^{3} x^{6} + 15 \, b^{2} c^{3} - 4 \, a b c^{2} d - 3 \, a^{2} c d^{2} - 2 \,{\left (b^{2} c d^{2} - a b d^{3}\right )} x^{4} +{\left (5 \, b^{2} c^{2} d - 2 \, a b c d^{2} - 3 \, a^{2} d^{3}\right )} x^{2}\right )} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}}}{96 \, b^{2} d^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/192*(3*(5*b^3*c^3 - 3*a*b^2*c^2*d - a^2*b*c*d^2 - a^3*d^3)*sqrt(e/(b*d))*log(8*b^2*d^2*e*x^4 + 8*(b^2*c*d
+ a*b*d^2)*e*x^2 + (b^2*c^2 + 6*a*b*c*d + a^2*d^2)*e + 4*(2*b^2*d^3*x^4 + b^2*c^2*d + a*b*c*d^2 + (3*b^2*c*d^2
+ a*b*d^3)*x^2)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c))*sqrt(e/(b*d))) - 4*(8*b^2*d^3*x^6 + 15*b^2*c^3 - 4*a*b*c^2*
d - 3*a^2*c*d^2 - 2*(b^2*c*d^2 - a*b*d^3)*x^4 + (5*b^2*c^2*d - 2*a*b*c*d^2 - 3*a^2*d^3)*x^2)*sqrt((b*e*x^2 + a
*e)/(d*x^2 + c)))/(b^2*d^3), 1/96*(3*(5*b^3*c^3 - 3*a*b^2*c^2*d - a^2*b*c*d^2 - a^3*d^3)*sqrt(-e/(b*d))*arctan
(1/2*(2*b*d*x^2 + b*c + a*d)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c))*sqrt(-e/(b*d))/(b*e*x^2 + a*e)) + 2*(8*b^2*d^3*
x^6 + 15*b^2*c^3 - 4*a*b*c^2*d - 3*a^2*c*d^2 - 2*(b^2*c*d^2 - a*b*d^3)*x^4 + (5*b^2*c^2*d - 2*a*b*c*d^2 - 3*a^
2*d^3)*x^2)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c)))/(b^2*d^3)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.49939, size = 451, normalized size = 1.85 \begin{align*} \frac{1}{48} \, \sqrt{b d x^{4} e + b c x^{2} e + a d x^{2} e + a c e}{\left (2 \,{\left (\frac{4 \, x^{2} \mathrm{sgn}\left (d x^{2} + c\right )}{d} - \frac{5 \, b^{3} c d^{2} \mathrm{sgn}\left (d x^{2} + c\right ) - a b^{2} d^{3} \mathrm{sgn}\left (d x^{2} + c\right )}{b^{3} d^{4}}\right )} x^{2} + \frac{15 \, b^{3} c^{2} d \mathrm{sgn}\left (d x^{2} + c\right ) - 4 \, a b^{2} c d^{2} \mathrm{sgn}\left (d x^{2} + c\right ) - 3 \, a^{2} b d^{3} \mathrm{sgn}\left (d x^{2} + c\right )}{b^{3} d^{4}}\right )} + \frac{{\left (5 \, b^{4} c^{3} d e \mathrm{sgn}\left (d x^{2} + c\right ) - 3 \, a b^{3} c^{2} d^{2} e \mathrm{sgn}\left (d x^{2} + c\right ) - a^{2} b^{2} c d^{3} e \mathrm{sgn}\left (d x^{2} + c\right ) - a^{3} b d^{4} e \mathrm{sgn}\left (d x^{2} + c\right )\right )} \sqrt{b d} e^{\left (-\frac{1}{2}\right )} \log \left ({\left | -\sqrt{b d} b c e^{\frac{1}{2}} - \sqrt{b d} a d e^{\frac{1}{2}} - 2 \,{\left (\sqrt{b d} x^{2} e^{\frac{1}{2}} - \sqrt{b d x^{4} e + b c x^{2} e + a d x^{2} e + a c e}\right )} b d \right |}\right )}{32 \, b^{4} d^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*sqrt(b*d*x^4*e + b*c*x^2*e + a*d*x^2*e + a*c*e)*(2*(4*x^2*sgn(d*x^2 + c)/d - (5*b^3*c*d^2*sgn(d*x^2 + c)
- a*b^2*d^3*sgn(d*x^2 + c))/(b^3*d^4))*x^2 + (15*b^3*c^2*d*sgn(d*x^2 + c) - 4*a*b^2*c*d^2*sgn(d*x^2 + c) - 3*a
^2*b*d^3*sgn(d*x^2 + c))/(b^3*d^4)) + 1/32*(5*b^4*c^3*d*e*sgn(d*x^2 + c) - 3*a*b^3*c^2*d^2*e*sgn(d*x^2 + c) -
a^2*b^2*c*d^3*e*sgn(d*x^2 + c) - a^3*b*d^4*e*sgn(d*x^2 + c))*sqrt(b*d)*e^(-1/2)*log(abs(-sqrt(b*d)*b*c*e^(1/2)
- sqrt(b*d)*a*d*e^(1/2) - 2*(sqrt(b*d)*x^2*e^(1/2) - sqrt(b*d*x^4*e + b*c*x^2*e + a*d*x^2*e + a*c*e))*b*d))/(
b^4*d^5)