3.307 \(\int \frac{x^5}{(\frac{e (a+b x^2)}{c+d x^2})^{3/2}} \, dx\)

Optimal. Leaf size=354 \[ \frac{\left (c+d x^2\right )^3 \left (7 a^2 d^2-2 a b c d+b^2 c^2\right ) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{6 b^2 d e^2 (b c-a d)^2}-\frac{a^2 \left (c+d x^2\right )^3}{b e (b c-a d)^2 \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}-\frac{(b c-a d) \left (5 a d (2 b c-7 a d)+b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt{b} \sqrt{e}}\right )}{16 b^{9/2} d^{3/2} e^{3/2}}-\frac{\left (c+d x^2\right )^2 \left (5 a d (2 b c-7 a d)+b^2 c^2\right ) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{24 b^3 d e^2 (b c-a d)}-\frac{\left (c+d x^2\right ) \left (5 a d (2 b c-7 a d)+b^2 c^2\right ) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{16 b^4 d e^2} \]

[Out]

-((b^2*c^2 + 5*a*d*(2*b*c - 7*a*d))*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(c + d*x^2))/(16*b^4*d*e^2) - ((b^2*c^2
+ 5*a*d*(2*b*c - 7*a*d))*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(c + d*x^2)^2)/(24*b^3*d*(b*c - a*d)*e^2) - (a^2*(c
+ d*x^2)^3)/(b*(b*c - a*d)^2*e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]) + ((b^2*c^2 - 2*a*b*c*d + 7*a^2*d^2)*Sqrt[(
e*(a + b*x^2))/(c + d*x^2)]*(c + d*x^2)^3)/(6*b^2*d*(b*c - a*d)^2*e^2) - ((b*c - a*d)*(b^2*c^2 + 5*a*d*(2*b*c
- 7*a*d))*ArcTanh[(Sqrt[d]*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/(Sqrt[b]*Sqrt[e])])/(16*b^(9/2)*d^(3/2)*e^(3/2))

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Rubi [A]  time = 0.378598, antiderivative size = 348, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {1960, 462, 385, 199, 208} \[ -\frac{a^2 \left (c+d x^2\right )^3}{b e (b c-a d)^2 \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}-\frac{(b c-a d) \left (5 a d (2 b c-7 a d)+b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt{b} \sqrt{e}}\right )}{16 b^{9/2} d^{3/2} e^{3/2}}+\frac{\left (c+d x^2\right )^3 \left (\frac{c^2}{d}-\frac{a (2 b c-7 a d)}{b^2}\right ) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{6 e^2 (b c-a d)^2}-\frac{\left (c+d x^2\right )^2 \left (\frac{5 a (2 b c-7 a d)}{b^2}+\frac{c^2}{d}\right ) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{24 b e^2 (b c-a d)}-\frac{\left (c+d x^2\right ) \left (5 a d (2 b c-7 a d)+b^2 c^2\right ) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{16 b^4 d e^2} \]

Antiderivative was successfully verified.

[In]

Int[x^5/((e*(a + b*x^2))/(c + d*x^2))^(3/2),x]

[Out]

-((b^2*c^2 + 5*a*d*(2*b*c - 7*a*d))*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(c + d*x^2))/(16*b^4*d*e^2) - ((c^2/d +
(5*a*(2*b*c - 7*a*d))/b^2)*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(c + d*x^2)^2)/(24*b*(b*c - a*d)*e^2) - (a^2*(c +
d*x^2)^3)/(b*(b*c - a*d)^2*e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]) + ((c^2/d - (a*(2*b*c - 7*a*d))/b^2)*Sqrt[(e*
(a + b*x^2))/(c + d*x^2)]*(c + d*x^2)^3)/(6*(b*c - a*d)^2*e^2) - ((b*c - a*d)*(b^2*c^2 + 5*a*d*(2*b*c - 7*a*d)
)*ArcTanh[(Sqrt[d]*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/(Sqrt[b]*Sqrt[e])])/(16*b^(9/2)*d^(3/2)*e^(3/2))

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 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 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}{\left (\frac{e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}} \, dx &=((b c-a d) e) \operatorname{Subst}\left (\int \frac{\left (-a e+c x^2\right )^2}{x^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{a^2 \left (c+d x^2\right )^3}{b (b c-a d)^2 e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}+\frac{(b c-a d) \operatorname{Subst}\left (\int \frac{-a (2 b c-7 a d) e^2+b c^2 e x^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 )}{b}\\ &=-\frac{a^2 \left (c+d x^2\right )^3}{b (b c-a d)^2 e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}+\frac{\left (b^2 c^2-2 a b c d+7 a^2 d^2\right ) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^3}{6 b^2 d (b c-a d)^2 e^2}-\frac{\left ((b c-a d) \left (b^2 c^2+5 a d (2 b c-7 a d)\right ) e\right ) \operatorname{Subst}\left (\int \frac{1}{\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^2 d}\\ &=-\frac{\left (b^2 c^2+5 a d (2 b c-7 a d)\right ) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^2}{24 b^3 d (b c-a d) e^2}-\frac{a^2 \left (c+d x^2\right )^3}{b (b c-a d)^2 e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}+\frac{\left (b^2 c^2-2 a b c d+7 a^2 d^2\right ) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^3}{6 b^2 d (b c-a d)^2 e^2}-\frac{\left ((b c-a d) \left (b^2 c^2+5 a d (2 b c-7 a d)\right )\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^3 d}\\ &=-\frac{\left (b^2 c^2+5 a d (2 b c-7 a d)\right ) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{16 b^4 d e^2}-\frac{\left (b^2 c^2+5 a d (2 b c-7 a d)\right ) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^2}{24 b^3 d (b c-a d) e^2}-\frac{a^2 \left (c+d x^2\right )^3}{b (b c-a d)^2 e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}+\frac{\left (b^2 c^2-2 a b c d+7 a^2 d^2\right ) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^3}{6 b^2 d (b c-a d)^2 e^2}-\frac{\left ((b c-a d) \left (b^2 c^2+5 a d (2 b c-7 a d)\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^4 d e}\\ &=-\frac{\left (b^2 c^2+5 a d (2 b c-7 a d)\right ) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{16 b^4 d e^2}-\frac{\left (b^2 c^2+5 a d (2 b c-7 a d)\right ) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^2}{24 b^3 d (b c-a d) e^2}-\frac{a^2 \left (c+d x^2\right )^3}{b (b c-a d)^2 e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}+\frac{\left (b^2 c^2-2 a b c d+7 a^2 d^2\right ) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^3}{6 b^2 d (b c-a d)^2 e^2}-\frac{(b c-a d) \left (b^2 c^2+5 a d (2 b c-7 a d)\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^{9/2} d^{3/2} e^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.501843, size = 247, normalized size = 0.7 \[ \frac{\sqrt{d} \sqrt{\frac{b \left (c+d x^2\right )}{b c-a d}} \left (5 a^2 b d \left (7 d x^2-20 c\right )+105 a^3 d^2+a b^2 \left (3 c^2-38 c d x^2-14 d^2 x^4\right )+b^3 x^2 \left (3 c^2+14 c d x^2+8 d^2 x^4\right )\right )-3 \sqrt{a+b x^2} \sqrt{b c-a d} \left (-35 a^2 d^2+10 a b c d+b^2 c^2\right ) \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x^2}}{\sqrt{b c-a d}}\right )}{48 b^4 d^{3/2} e \sqrt{\frac{b \left (c+d x^2\right )}{b c-a d}} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5/((e*(a + b*x^2))/(c + d*x^2))^(3/2),x]

[Out]

(Sqrt[d]*Sqrt[(b*(c + d*x^2))/(b*c - a*d)]*(105*a^3*d^2 + 5*a^2*b*d*(-20*c + 7*d*x^2) + a*b^2*(3*c^2 - 38*c*d*
x^2 - 14*d^2*x^4) + b^3*x^2*(3*c^2 + 14*c*d*x^2 + 8*d^2*x^4)) - 3*Sqrt[b*c - a*d]*(b^2*c^2 + 10*a*b*c*d - 35*a
^2*d^2)*Sqrt[a + b*x^2]*ArcSinh[(Sqrt[d]*Sqrt[a + b*x^2])/Sqrt[b*c - a*d]])/(48*b^4*d^(3/2)*e*Sqrt[(e*(a + b*x
^2))/(c + d*x^2)]*Sqrt[(b*(c + d*x^2))/(b*c - a*d)])

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Maple [B]  time = 0.031, size = 1027, normalized size = 2.9 \begin{align*} \text{result too large to display} \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))^(3/2),x)

[Out]

1/96*(-60*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*x^4*a*b^2*d^2+12*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)
*(b*d)^(1/2)*x^4*b^3*c*d-105*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))*x^2*a^3*b*d^3+135*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))*x^2*a^2*b^2*c*d^2-27*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))*x^2*a*b^3*c^2*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))*x^2*b^4*c^3+16*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(3/2)*(b*d)^(1/2)*x^2*b^2*d+54*(b*d*x^4+a*d*x^2+b*c*x^2+a*c
)^(1/2)*(b*d)^(1/2)*x^2*a^2*b*d^2-108*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*x^2*a*b^2*c*d+6*(b*d*x^4
+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*x^2*b^3*c^2-105*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^4*d^3+135*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*b*c*d^2-27*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*b^2*c^2*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*b^3*c^3+96*((d*x^2+c)*(b*x^2+a))^(1/2)*(b*d)^(1/2)*a^3*d^2-96*((d*x^2+c)*(b*x^2+a))^(1/2)
*(b*d)^(1/2)*a^2*b*c*d+16*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(3/2)*(b*d)^(1/2)*a*b*d+114*(b*d*x^4+a*d*x^2+b*c*x^2+a
*c)^(1/2)*(b*d)^(1/2)*a^3*d^2-120*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*a^2*b*c*d+6*(b*d*x^4+a*d*x^2
+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*a*b^2*c^2)/d/b^4*(b*x^2+a)/(b*d)^(1/2)/((d*x^2+c)*(b*x^2+a))^(1/2)/(d*x^2+c)/(
e*(b*x^2+a)/(d*x^2+c))^(3/2)

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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))^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 18.5063, size = 1652, normalized size = 4.67 \begin{align*} \text{result too large to display} \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))^(3/2),x, algorithm="fricas")

[Out]

[1/192*(3*(a*b^3*c^3 + 9*a^2*b^2*c^2*d - 45*a^3*b*c*d^2 + 35*a^4*d^3 + (b^4*c^3 + 9*a*b^3*c^2*d - 45*a^2*b^2*c
*d^2 + 35*a^3*b*d^3)*x^2)*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)*sqrt(b*d*e)*sqrt((b*e*x^2 + a*e)/(d*x^
2 + c))) + 4*(8*b^4*d^4*x^8 + 3*a*b^3*c^3*d - 100*a^2*b^2*c^2*d^2 + 105*a^3*b*c*d^3 + 2*(11*b^4*c*d^3 - 7*a*b^
3*d^4)*x^6 + (17*b^4*c^2*d^2 - 52*a*b^3*c*d^3 + 35*a^2*b^2*d^4)*x^4 + (3*b^4*c^3*d - 35*a*b^3*c^2*d^2 - 65*a^2
*b^2*c*d^3 + 105*a^3*b*d^4)*x^2)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c)))/(b^6*d^2*e^2*x^2 + a*b^5*d^2*e^2), 1/96*(3
*(a*b^3*c^3 + 9*a^2*b^2*c^2*d - 45*a^3*b*c*d^2 + 35*a^4*d^3 + (b^4*c^3 + 9*a*b^3*c^2*d - 45*a^2*b^2*c*d^2 + 35
*a^3*b*d^3)*x^2)*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^4*d^4*x^8 + 3*a*b^3*c^3*d - 100*a^2*b^2*c^2*d^2 + 105*a^3*b*c*d^3 + 2*(11*
b^4*c*d^3 - 7*a*b^3*d^4)*x^6 + (17*b^4*c^2*d^2 - 52*a*b^3*c*d^3 + 35*a^2*b^2*d^4)*x^4 + (3*b^4*c^3*d - 35*a*b^
3*c^2*d^2 - 65*a^2*b^2*c*d^3 + 105*a^3*b*d^4)*x^2)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c)))/(b^6*d^2*e^2*x^2 + a*b^5
*d^2*e^2)]

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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))**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{5}}{\left (\frac{{\left (b x^{2} + a\right )} e}{d x^{2} + c}\right )^{\frac{3}{2}}}\,{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))^(3/2),x, algorithm="giac")

[Out]

integrate(x^5/((b*x^2 + a)*e/(d*x^2 + c))^(3/2), x)