3.282 \(\int \frac{(\frac{e (a+b x^2)}{c+d x^2})^{3/2}}{x^7} \, dx\)

Optimal. Leaf size=366 \[ -\frac{e^2 \left (-79 a^2 d^2+50 a b c d+5 b^2 c^2\right ) (b c-a d) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{48 a c^4 \left (a e-\frac{c e \left (a+b x^2\right )}{c+d x^2}\right )}+\frac{e^{3/2} \left (-35 a^2 d^2+10 a b c d+b^2 c^2\right ) (b c-a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt{a} \sqrt{e}}\right )}{16 a^{3/2} c^{9/2}}+\frac{d^2 e (b c-a d) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{c^4}+\frac{e^2 (b c-a d)^3 \left (\frac{e \left (a+b x^2\right )}{c+d x^2}\right )^{5/2}}{6 a c^2 \left (a e-\frac{c e \left (a+b x^2\right )}{c+d x^2}\right )^3}+\frac{e^3 (11 a d+b c) (b c-a d)^2 \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{24 c^4 \left (a e-\frac{c e \left (a+b x^2\right )}{c+d x^2}\right )^2} \]

[Out]

(d^2*(b*c - a*d)*e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/c^4 + ((b*c - a*d)^3*e^2*((e*(a + b*x^2))/(c + d*x^2))^(
5/2))/(6*a*c^2*(a*e - (c*e*(a + b*x^2))/(c + d*x^2))^3) + ((b*c - a*d)^2*(b*c + 11*a*d)*e^3*Sqrt[(e*(a + b*x^2
))/(c + d*x^2)])/(24*c^4*(a*e - (c*e*(a + b*x^2))/(c + d*x^2))^2) - ((b*c - a*d)*(5*b^2*c^2 + 50*a*b*c*d - 79*
a^2*d^2)*e^2*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/(48*a*c^4*(a*e - (c*e*(a + b*x^2))/(c + d*x^2))) + ((b*c - a*d
)*(b^2*c^2 + 10*a*b*c*d - 35*a^2*d^2)*e^(3/2)*ArcTanh[(Sqrt[c]*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/(Sqrt[a]*Sqr
t[e])])/(16*a^(3/2)*c^(9/2))

________________________________________________________________________________________

Rubi [A]  time = 0.368982, antiderivative size = 366, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {1960, 463, 455, 1157, 388, 208} \[ -\frac{e^2 \left (-79 a^2 d^2+50 a b c d+5 b^2 c^2\right ) (b c-a d) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{48 a c^4 \left (a e-\frac{c e \left (a+b x^2\right )}{c+d x^2}\right )}+\frac{e^{3/2} \left (-35 a^2 d^2+10 a b c d+b^2 c^2\right ) (b c-a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt{a} \sqrt{e}}\right )}{16 a^{3/2} c^{9/2}}+\frac{d^2 e (b c-a d) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{c^4}+\frac{e^2 (b c-a d)^3 \left (\frac{e \left (a+b x^2\right )}{c+d x^2}\right )^{5/2}}{6 a c^2 \left (a e-\frac{c e \left (a+b x^2\right )}{c+d x^2}\right )^3}+\frac{e^3 (11 a d+b c) (b c-a d)^2 \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{24 c^4 \left (a e-\frac{c e \left (a+b x^2\right )}{c+d x^2}\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[((e*(a + b*x^2))/(c + d*x^2))^(3/2)/x^7,x]

[Out]

(d^2*(b*c - a*d)*e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/c^4 + ((b*c - a*d)^3*e^2*((e*(a + b*x^2))/(c + d*x^2))^(
5/2))/(6*a*c^2*(a*e - (c*e*(a + b*x^2))/(c + d*x^2))^3) + ((b*c - a*d)^2*(b*c + 11*a*d)*e^3*Sqrt[(e*(a + b*x^2
))/(c + d*x^2)])/(24*c^4*(a*e - (c*e*(a + b*x^2))/(c + d*x^2))^2) - ((b*c - a*d)*(5*b^2*c^2 + 50*a*b*c*d - 79*
a^2*d^2)*e^2*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/(48*a*c^4*(a*e - (c*e*(a + b*x^2))/(c + d*x^2))) + ((b*c - a*d
)*(b^2*c^2 + 10*a*b*c*d - 35*a^2*d^2)*e^(3/2)*ArcTanh[(Sqrt[c]*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/(Sqrt[a]*Sqr
t[e])])/(16*a^(3/2)*c^(9/2))

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 1157

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ
uotient[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2,
x], x, 0]}, -Simp[(R*x*(d + e*x^2)^(q + 1))/(2*d*(q + 1)), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*
ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && N
eQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rule 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 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 \frac{\left (\frac{e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}}{x^7} \, dx &=((b c-a d) e) \operatorname{Subst}\left (\int \frac{x^4 \left (b e-d x^2\right )^2}{\left (-a e+c x^2\right )^4} \, dx,x,\sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}\right )\\ &=\frac{(b c-a d)^3 e^2 \left (\frac{e \left (a+b x^2\right )}{c+d x^2}\right )^{5/2}}{6 a c^2 \left (a e-\frac{c e \left (a+b x^2\right )}{c+d x^2}\right )^3}+\frac{(b c-a d) \operatorname{Subst}\left (\int \frac{x^4 \left (-6 b^2 c^2 e^2+5 (b c e-a d e)^2+6 a c d^2 e x^2\right )}{\left (-a e+c x^2\right )^3} \, dx,x,\sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}\right )}{6 a c^2}\\ &=\frac{(b c-a d)^3 e^2 \left (\frac{e \left (a+b x^2\right )}{c+d x^2}\right )^{5/2}}{6 a c^2 \left (a e-\frac{c e \left (a+b x^2\right )}{c+d x^2}\right )^3}+\frac{(b c-a d)^2 (b c+11 a d) e^3 \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{24 c^4 \left (a e-\frac{c e \left (a+b x^2\right )}{c+d x^2}\right )^2}-\frac{(b c-a d) \operatorname{Subst}\left (\int \frac{a c (b c-a d) (b c+11 a d) e^3+4 c^2 (b c-a d) (b c+11 a d) e^2 x^2-24 a c^3 d^2 e x^4}{\left (-a e+c x^2\right )^2} \, dx,x,\sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}\right )}{24 a c^5}\\ &=\frac{(b c-a d)^3 e^2 \left (\frac{e \left (a+b x^2\right )}{c+d x^2}\right )^{5/2}}{6 a c^2 \left (a e-\frac{c e \left (a+b x^2\right )}{c+d x^2}\right )^3}+\frac{(b c-a d)^2 (b c+11 a d) e^3 \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{24 c^4 \left (a e-\frac{c e \left (a+b x^2\right )}{c+d x^2}\right )^2}-\frac{(b c-a d) \left (5 b^2 c^2+50 a b c d-79 a^2 d^2\right ) e^2 \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{48 a c^4 \left (a e-\frac{c e \left (a+b x^2\right )}{c+d x^2}\right )}-\frac{(b c-a d) \operatorname{Subst}\left (\int \frac{3 a c \left (b^2 c^2+10 a b c d-19 a^2 d^2\right ) e^3-48 a^2 c^2 d^2 e^2 x^2}{-a e+c x^2} \, dx,x,\sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}\right )}{48 a^2 c^5 e}\\ &=\frac{d^2 (b c-a d) e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{c^4}+\frac{(b c-a d)^3 e^2 \left (\frac{e \left (a+b x^2\right )}{c+d x^2}\right )^{5/2}}{6 a c^2 \left (a e-\frac{c e \left (a+b x^2\right )}{c+d x^2}\right )^3}+\frac{(b c-a d)^2 (b c+11 a d) e^3 \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{24 c^4 \left (a e-\frac{c e \left (a+b x^2\right )}{c+d x^2}\right )^2}-\frac{(b c-a d) \left (5 b^2 c^2+50 a b c d-79 a^2 d^2\right ) e^2 \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{48 a c^4 \left (a e-\frac{c e \left (a+b x^2\right )}{c+d x^2}\right )}-\frac{\left ((b c-a d) \left (b^2 c^2+10 a b c d-35 a^2 d^2\right ) e^2\right ) \operatorname{Subst}\left (\int \frac{1}{-a e+c x^2} \, dx,x,\sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}\right )}{16 a c^4}\\ &=\frac{d^2 (b c-a d) e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{c^4}+\frac{(b c-a d)^3 e^2 \left (\frac{e \left (a+b x^2\right )}{c+d x^2}\right )^{5/2}}{6 a c^2 \left (a e-\frac{c e \left (a+b x^2\right )}{c+d x^2}\right )^3}+\frac{(b c-a d)^2 (b c+11 a d) e^3 \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{24 c^4 \left (a e-\frac{c e \left (a+b x^2\right )}{c+d x^2}\right )^2}-\frac{(b c-a d) \left (5 b^2 c^2+50 a b c d-79 a^2 d^2\right ) e^2 \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{48 a c^4 \left (a e-\frac{c e \left (a+b x^2\right )}{c+d x^2}\right )}+\frac{(b c-a d) \left (b^2 c^2+10 a b c d-35 a^2 d^2\right ) e^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt{a} \sqrt{e}}\right )}{16 a^{3/2} c^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.19232, size = 245, normalized size = 0.67 \[ \frac{e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \left (3 x^6 \sqrt{c+d x^2} \left (-45 a^2 b c d^2+35 a^3 d^3+9 a b^2 c^2 d+b^3 c^3\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x^2}}{\sqrt{a} \sqrt{c+d x^2}}\right )-\sqrt{a} \sqrt{c} \sqrt{a+b x^2} \left (a^2 \left (-14 c^2 d x^2+8 c^3+35 c d^2 x^4+105 d^3 x^6\right )+2 a b c x^2 \left (7 c^2-19 c d x^2-50 d^2 x^4\right )+3 b^2 c^2 x^4 \left (c+d x^2\right )\right )\right )}{48 a^{3/2} c^{9/2} x^6 \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((e*(a + b*x^2))/(c + d*x^2))^(3/2)/x^7,x]

[Out]

(e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(-(Sqrt[a]*Sqrt[c]*Sqrt[a + b*x^2]*(3*b^2*c^2*x^4*(c + d*x^2) + 2*a*b*c*x
^2*(7*c^2 - 19*c*d*x^2 - 50*d^2*x^4) + a^2*(8*c^3 - 14*c^2*d*x^2 + 35*c*d^2*x^4 + 105*d^3*x^6))) + 3*(b^3*c^3
+ 9*a*b^2*c^2*d - 45*a^2*b*c*d^2 + 35*a^3*d^3)*x^6*Sqrt[c + d*x^2]*ArcTanh[(Sqrt[c]*Sqrt[a + b*x^2])/(Sqrt[a]*
Sqrt[c + d*x^2])]))/(48*a^(3/2)*c^(9/2)*x^6*Sqrt[a + b*x^2])

________________________________________________________________________________________

Maple [B]  time = 0.018, size = 1498, normalized size = 4.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*(b*x^2+a)/(d*x^2+c))^(3/2)/x^7,x)

[Out]

-1/96*(-105*ln((a*d*x^2+b*c*x^2+2*(a*c)^(1/2)*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)+2*a*c)/x^2)*x^8*a^4*c*d^4-17
4*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(a*c)^(1/2)*x^8*a^3*d^4-105*ln((a*d*x^2+b*c*x^2+2*(a*c)^(1/2)*(b*d*x^4+a
*d*x^2+b*c*x^2+a*c)^(1/2)+2*a*c)/x^2)*x^6*a^4*c^2*d^3-3*ln((a*d*x^2+b*c*x^2+2*(a*c)^(1/2)*(b*d*x^4+a*d*x^2+b*c
*x^2+a*c)^(1/2)+2*a*c)/x^2)*x^6*a*b^3*c^5+174*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(3/2)*(a*c)^(1/2)*x^6*a^2*d^3+6*(b
*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(a*c)^(1/2)*x^6*b^3*c^4-6*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(3/2)*(a*c)^(1/2)*x^
4*b^2*c^3+6*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(a*c)^(1/2)*x^10*b^3*c^2*d^2+135*ln((a*d*x^2+b*c*x^2+2*(a*c)^(
1/2)*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)+2*a*c)/x^2)*x^8*a^3*b*c^2*d^3-27*ln((a*d*x^2+b*c*x^2+2*(a*c)^(1/2)*(b
*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)+2*a*c)/x^2)*x^8*a^2*b^2*c^3*d^2-3*ln((a*d*x^2+b*c*x^2+2*(a*c)^(1/2)*(b*d*x^4
+a*d*x^2+b*c*x^2+a*c)^(1/2)+2*a*c)/x^2)*x^8*a*b^3*c^4*d+12*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(a*c)^(1/2)*x^8
*b^3*c^3*d+135*ln((a*d*x^2+b*c*x^2+2*(a*c)^(1/2)*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)+2*a*c)/x^2)*x^6*a^3*b*c^3
*d^2-27*ln((a*d*x^2+b*c*x^2+2*(a*c)^(1/2)*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)+2*a*c)/x^2)*x^6*a^2*b^2*c^4*d+96
*((d*x^2+c)*(b*x^2+a))^(1/2)*(a*c)^(1/2)*x^6*a^3*c*d^3-6*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(3/2)*(a*c)^(1/2)*x^6*b
^2*c^2*d-174*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(a*c)^(1/2)*x^6*a^3*c*d^3+114*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(
3/2)*(a*c)^(1/2)*x^4*a^2*c*d^2-44*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(3/2)*(a*c)^(1/2)*x^2*a^2*c^2*d+12*(b*d*x^4+a*
d*x^2+b*c*x^2+a*c)^(3/2)*(a*c)^(1/2)*x^2*a*b*c^3+16*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(3/2)*(a*c)^(1/2)*a^2*c^3-17
4*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(a*c)^(1/2)*x^10*a^2*b*d^4-60*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(3/2)*(a*c)^
(1/2)*x^4*a*b*c^2*d+72*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(a*c)^(1/2)*x^10*a*b^2*c*d^3-216*(b*d*x^4+a*d*x^2+b
*c*x^2+a*c)^(1/2)*(a*c)^(1/2)*x^8*a^2*b*c*d^3+138*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(a*c)^(1/2)*x^8*a*b^2*c^
2*d^2-96*((d*x^2+c)*(b*x^2+a))^(1/2)*(a*c)^(1/2)*x^6*a^2*b*c^2*d^2-72*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(3/2)*(a*c
)^(1/2)*x^6*a*b*c*d^2-42*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(a*c)^(1/2)*x^6*a^2*b*c^2*d^2+66*(b*d*x^4+a*d*x^2
+b*c*x^2+a*c)^(1/2)*(a*c)^(1/2)*x^6*a*b^2*c^3*d)/a^2*(d*x^2+c)*(e*(b*x^2+a)/(d*x^2+c))^(3/2)/x^6/(a*c)^(1/2)/c
^5/(b*x^2+a)/((d*x^2+c)*(b*x^2+a))^(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((e*(b*x^2+a)/(d*x^2+c))^(3/2)/x^7,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 93.0066, size = 1211, normalized size = 3.31 \begin{align*} \left [\frac{3 \,{\left (b^{3} c^{3} + 9 \, a b^{2} c^{2} d - 45 \, a^{2} b c d^{2} + 35 \, a^{3} d^{3}\right )} e x^{6} \sqrt{\frac{e}{a c}} \log \left (\frac{{\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} e x^{4} + 8 \, a^{2} c^{2} e + 8 \,{\left (a b c^{2} + a^{2} c d\right )} e x^{2} + 4 \,{\left (2 \, a^{2} c^{3} +{\left (a b c^{2} d + a^{2} c d^{2}\right )} x^{4} +{\left (a b c^{3} + 3 \, a^{2} c^{2} d\right )} x^{2}\right )} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}} \sqrt{\frac{e}{a c}}}{x^{4}}\right ) - 4 \,{\left ({\left (3 \, b^{2} c^{2} d - 100 \, a b c d^{2} + 105 \, a^{2} d^{3}\right )} e x^{6} + 8 \, a^{2} c^{3} e +{\left (3 \, b^{2} c^{3} - 38 \, a b c^{2} d + 35 \, a^{2} c d^{2}\right )} e x^{4} + 14 \,{\left (a b c^{3} - a^{2} c^{2} d\right )} e x^{2}\right )} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}}}{192 \, a c^{4} x^{6}}, -\frac{3 \,{\left (b^{3} c^{3} + 9 \, a b^{2} c^{2} d - 45 \, a^{2} b c d^{2} + 35 \, a^{3} d^{3}\right )} e x^{6} \sqrt{-\frac{e}{a c}} \arctan \left (\frac{{\left ({\left (b c + a d\right )} x^{2} + 2 \, a c\right )} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}} \sqrt{-\frac{e}{a c}}}{2 \,{\left (b e x^{2} + a e\right )}}\right ) + 2 \,{\left ({\left (3 \, b^{2} c^{2} d - 100 \, a b c d^{2} + 105 \, a^{2} d^{3}\right )} e x^{6} + 8 \, a^{2} c^{3} e +{\left (3 \, b^{2} c^{3} - 38 \, a b c^{2} d + 35 \, a^{2} c d^{2}\right )} e x^{4} + 14 \,{\left (a b c^{3} - a^{2} c^{2} d\right )} e x^{2}\right )} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}}}{96 \, a c^{4} x^{6}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*(b*x^2+a)/(d*x^2+c))^(3/2)/x^7,x, algorithm="fricas")

[Out]

[1/192*(3*(b^3*c^3 + 9*a*b^2*c^2*d - 45*a^2*b*c*d^2 + 35*a^3*d^3)*e*x^6*sqrt(e/(a*c))*log(((b^2*c^2 + 6*a*b*c*
d + a^2*d^2)*e*x^4 + 8*a^2*c^2*e + 8*(a*b*c^2 + a^2*c*d)*e*x^2 + 4*(2*a^2*c^3 + (a*b*c^2*d + a^2*c*d^2)*x^4 +
(a*b*c^3 + 3*a^2*c^2*d)*x^2)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c))*sqrt(e/(a*c)))/x^4) - 4*((3*b^2*c^2*d - 100*a*b
*c*d^2 + 105*a^2*d^3)*e*x^6 + 8*a^2*c^3*e + (3*b^2*c^3 - 38*a*b*c^2*d + 35*a^2*c*d^2)*e*x^4 + 14*(a*b*c^3 - a^
2*c^2*d)*e*x^2)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c)))/(a*c^4*x^6), -1/96*(3*(b^3*c^3 + 9*a*b^2*c^2*d - 45*a^2*b*c
*d^2 + 35*a^3*d^3)*e*x^6*sqrt(-e/(a*c))*arctan(1/2*((b*c + a*d)*x^2 + 2*a*c)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c))
*sqrt(-e/(a*c))/(b*e*x^2 + a*e)) + 2*((3*b^2*c^2*d - 100*a*b*c*d^2 + 105*a^2*d^3)*e*x^6 + 8*a^2*c^3*e + (3*b^2
*c^3 - 38*a*b*c^2*d + 35*a^2*c*d^2)*e*x^4 + 14*(a*b*c^3 - a^2*c^2*d)*e*x^2)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c)))
/(a*c^4*x^6)]

________________________________________________________________________________________

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((e*(b*x**2+a)/(d*x**2+c))**(3/2)/x**7,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (\frac{{\left (b x^{2} + a\right )} e}{d x^{2} + c}\right )^{\frac{3}{2}}}{x^{7}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*(b*x^2+a)/(d*x^2+c))^(3/2)/x^7,x, algorithm="giac")

[Out]

integrate(((b*x^2 + a)*e/(d*x^2 + c))^(3/2)/x^7, x)