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

Optimal. Leaf size=281 \[ \frac{\left (c+d x^2\right ) \left (5 a^2 d^2+2 a b c d+b^2 c^2\right ) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{16 b^3 d^2 e}+\frac{(b c-a d) \left (5 a^2 d^2+2 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt{b} \sqrt{e}}\right )}{16 b^{7/2} d^{5/2} \sqrt{e}}-\frac{\left (c+d x^2\right )^2 (5 a d+3 b c) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{24 b^2 d^2 e}-\frac{\left (c+d x^2\right )^3 \left (a-\frac{c \left (a+b x^2\right )}{c+d x^2}\right ) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{6 b d e (b c-a d)} \]

[Out]

((b^2*c^2 + 2*a*b*c*d + 5*a^2*d^2)*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(c + d*x^2))/(16*b^3*d^2*e) - ((3*b*c + 5
*a*d)*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(c + d*x^2)^2)/(24*b^2*d^2*e) - (Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(c
+ d*x^2)^3*(a - (c*(a + b*x^2))/(c + d*x^2)))/(6*b*d*(b*c - a*d)*e) + ((b*c - a*d)*(b^2*c^2 + 2*a*b*c*d + 5*a^
2*d^2)*ArcTanh[(Sqrt[d]*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/(Sqrt[b]*Sqrt[e])])/(16*b^(7/2)*d^(5/2)*Sqrt[e])

________________________________________________________________________________________

Rubi [A]  time = 0.290242, antiderivative size = 281, 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, 413, 385, 199, 208} \[ \frac{\left (c+d x^2\right ) \left (5 a^2 d^2+2 a b c d+b^2 c^2\right ) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{16 b^3 d^2 e}+\frac{(b c-a d) \left (5 a^2 d^2+2 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt{b} \sqrt{e}}\right )}{16 b^{7/2} d^{5/2} \sqrt{e}}-\frac{\left (c+d x^2\right )^2 (5 a d+3 b c) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{24 b^2 d^2 e}-\frac{\left (c+d x^2\right )^3 \left (a-\frac{c \left (a+b x^2\right )}{c+d x^2}\right ) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{6 b d e (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b^2*c^2 + 2*a*b*c*d + 5*a^2*d^2)*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(c + d*x^2))/(16*b^3*d^2*e) - ((3*b*c + 5
*a*d)*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(c + d*x^2)^2)/(24*b^2*d^2*e) - (Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(c
+ d*x^2)^3*(a - (c*(a + b*x^2))/(c + d*x^2)))/(6*b*d*(b*c - a*d)*e) + ((b*c - a*d)*(b^2*c^2 + 2*a*b*c*d + 5*a^
2*d^2)*ArcTanh[(Sqrt[d]*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/(Sqrt[b]*Sqrt[e])])/(16*b^(7/2)*d^(5/2)*Sqrt[e])

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 413

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

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 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

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

Mathematica [A]  time = 0.388769, size = 224, normalized size = 0.8 \[ \frac{\sqrt{a+b x^2} \left (\sqrt{d} \sqrt{a+b x^2} \sqrt{\frac{b \left (c+d x^2\right )}{b c-a d}} \left (15 a^2 d^2-2 a b d \left (2 c+5 d x^2\right )+b^2 \left (-3 c^2+2 c d x^2+8 d^2 x^4\right )\right )+3 \sqrt{b c-a d} \left (5 a^2 d^2+2 a b c d+b^2 c^2\right ) \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x^2}}{\sqrt{b c-a d}}\right )\right )}{48 b^3 d^{5/2} \sqrt{\frac{b \left (c+d x^2\right )}{b c-a d}} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.023, size = 527, normalized size = 1.9 \begin{align*}{\frac{b{x}^{2}+a}{96\,{b}^{3}{d}^{2}} \left ( -36\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}{x}^{2}ab{d}^{2}\sqrt{bd}-12\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}{x}^{2}c{b}^{2}d\sqrt{bd}-15\,{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}+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}^{2}cb{d}^{2}+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{c}^{2}{b}^{2}d+3\,{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}+30\,\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}-6\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}{c}^{2}{b}^{2}\sqrt{bd} \right ){\frac{1}{\sqrt{{\frac{e \left ( b{x}^{2}+a \right ) }{d{x}^{2}+c}}}}}{\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*(b*x^2+a)/b^3*(-36*(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)-12*(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)-15*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+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^2*c*b*d^2+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*c^2*b^2*d+3*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)+30*(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)-6*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*c^2
*b^2*(b*d)^(1/2))/(e*(b*x^2+a)/(d*x^2+c))^(1/2)/((d*x^2+c)*(b*x^2+a))^(1/2)/d^2/(b*d)^(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/(e*(b*x^2+a)/(d*x^2+c))^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.82024, size = 1153, normalized size = 4.1 \begin{align*} \left [-\frac{3 \,{\left (b^{3} c^{3} + a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - 5 \, a^{3} d^{3}\right )} \sqrt{b d e} \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 d^{2} x^{4} + b c^{2} + a c d +{\left (3 \, b c d + a d^{2}\right )} x^{2}\right )} \sqrt{b d e} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}}\right ) - 4 \,{\left (8 \, b^{3} d^{4} x^{6} - 3 \, b^{3} c^{3} d - 4 \, a b^{2} c^{2} d^{2} + 15 \, a^{2} b c d^{3} + 10 \,{\left (b^{3} c d^{3} - a b^{2} d^{4}\right )} x^{4} -{\left (b^{3} c^{2} d^{2} + 14 \, a b^{2} c d^{3} - 15 \, a^{2} b d^{4}\right )} x^{2}\right )} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}}}{192 \, b^{4} d^{3} e}, -\frac{3 \,{\left (b^{3} c^{3} + a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - 5 \, a^{3} d^{3}\right )} \sqrt{-b d e} \arctan \left (\frac{{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt{-b d e} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}}}{2 \,{\left (b^{2} d e x^{2} + a b d e\right )}}\right ) - 2 \,{\left (8 \, b^{3} d^{4} x^{6} - 3 \, b^{3} c^{3} d - 4 \, a b^{2} c^{2} d^{2} + 15 \, a^{2} b c d^{3} + 10 \,{\left (b^{3} c d^{3} - a b^{2} d^{4}\right )} x^{4} -{\left (b^{3} c^{2} d^{2} + 14 \, a b^{2} c d^{3} - 15 \, a^{2} b d^{4}\right )} x^{2}\right )} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}}}{96 \, b^{4} d^{3} e}\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*(b^3*c^3 + a*b^2*c^2*d + 3*a^2*b*c*d^2 - 5*a^3*d^3)*sqrt(b*d*e)*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*d^2*x^4 + b*c^2 + a*c*d + (3*b*c*d + a*d^2)*x^2)*s
qrt(b*d*e)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c))) - 4*(8*b^3*d^4*x^6 - 3*b^3*c^3*d - 4*a*b^2*c^2*d^2 + 15*a^2*b*c*
d^3 + 10*(b^3*c*d^3 - a*b^2*d^4)*x^4 - (b^3*c^2*d^2 + 14*a*b^2*c*d^3 - 15*a^2*b*d^4)*x^2)*sqrt((b*e*x^2 + a*e)
/(d*x^2 + c)))/(b^4*d^3*e), -1/96*(3*(b^3*c^3 + a*b^2*c^2*d + 3*a^2*b*c*d^2 - 5*a^3*d^3)*sqrt(-b*d*e)*arctan(1
/2*(2*b*d*x^2 + b*c + a*d)*sqrt(-b*d*e)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c))/(b^2*d*e*x^2 + a*b*d*e)) - 2*(8*b^3*
d^4*x^6 - 3*b^3*c^3*d - 4*a*b^2*c^2*d^2 + 15*a^2*b*c*d^3 + 10*(b^3*c*d^3 - a*b^2*d^4)*x^4 - (b^3*c^2*d^2 + 14*
a*b^2*c*d^3 - 15*a^2*b*d^4)*x^2)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c)))/(b^4*d^3*e)]

________________________________________________________________________________________

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

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{5}}{\sqrt{\frac{{\left (b x^{2} + a\right )} e}{d x^{2} + c}}}\,{d x} \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]

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