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

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

[Out]

1/6*(-a*d+b*c)^3*e^2*(e*(b*x^2+a)/(d*x^2+c))^(5/2)/a/c^2/(a*e-c*e*(b*x^2+a)/(d*x^2+c))^3+1/16*(-a*d+b*c)*(-35*
a^2*d^2+10*a*b*c*d+b^2*c^2)*e^(3/2)*arctanh(c^(1/2)*(e*(b*x^2+a)/(d*x^2+c))^(1/2)/a^(1/2)/e^(1/2))/a^(3/2)/c^(
9/2)+d^2*(-a*d+b*c)*e*(e*(b*x^2+a)/(d*x^2+c))^(1/2)/c^4+1/24*(-a*d+b*c)^2*(11*a*d+b*c)*e^3*(e*(b*x^2+a)/(d*x^2
+c))^(1/2)/c^4/(a*e-c*e*(b*x^2+a)/(d*x^2+c))^2-1/48*(-a*d+b*c)*(-79*a^2*d^2+50*a*b*c*d+5*b^2*c^2)*e^2*(e*(b*x^
2+a)/(d*x^2+c))^(1/2)/a/c^4/(a*e-c*e*(b*x^2+a)/(d*x^2+c))

________________________________________________________________________________________

Rubi [A]  time = 0.37, antiderivative size = 366, normalized size of antiderivative = 1.00, 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 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]

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 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 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 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 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]]

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 = 0.19, size = 245, normalized size = 0.67 \[ \frac {e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (3 x^6 \sqrt {c+d x^2} \left (35 a^3 d^3-45 a^2 b c d^2+9 a b^2 c^2 d+b^3 c^3\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x^2}}{\sqrt {a} \sqrt {c+d x^2}}\right )-\sqrt {a} \sqrt {c} \sqrt {a+b x^2} \left (a^2 \left (8 c^3-14 c^2 d x^2+35 c d^2 x^4+105 d^3 x^6\right )+2 a b c x^2 \left (7 c^2-19 c d x^2-50 d^2 x^4\right )+3 b^2 c^2 x^4 \left (c+d x^2\right )\right )\right )}{48 a^{3/2} c^{9/2} x^6 \sqrt {a+b x^2}} \]

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])

________________________________________________________________________________________

fricas [A]  time = 14.42, size = 573, normalized size = 1.57 \[ \left [\frac {3 \, {\left (b^{3} c^{3} + 9 \, a b^{2} c^{2} d - 45 \, a^{2} b c d^{2} + 35 \, a^{3} d^{3}\right )} e x^{6} \sqrt {\frac {e}{a c}} \log \left (\frac {{\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} e x^{4} + 8 \, a^{2} c^{2} e + 8 \, {\left (a b c^{2} + a^{2} c d\right )} e x^{2} + 4 \, {\left (2 \, a^{2} c^{3} + {\left (a b c^{2} d + a^{2} c d^{2}\right )} x^{4} + {\left (a b c^{3} + 3 \, a^{2} c^{2} d\right )} x^{2}\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}} \sqrt {\frac {e}{a c}}}{x^{4}}\right ) - 4 \, {\left ({\left (3 \, b^{2} c^{2} d - 100 \, a b c d^{2} + 105 \, a^{2} d^{3}\right )} e x^{6} + 8 \, a^{2} c^{3} e + {\left (3 \, b^{2} c^{3} - 38 \, a b c^{2} d + 35 \, a^{2} c d^{2}\right )} e x^{4} + 14 \, {\left (a b c^{3} - a^{2} c^{2} d\right )} e x^{2}\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{192 \, a c^{4} x^{6}}, -\frac {3 \, {\left (b^{3} c^{3} + 9 \, a b^{2} c^{2} d - 45 \, a^{2} b c d^{2} + 35 \, a^{3} d^{3}\right )} e x^{6} \sqrt {-\frac {e}{a c}} \arctan \left (\frac {{\left ({\left (b c + a d\right )} x^{2} + 2 \, a c\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}} \sqrt {-\frac {e}{a c}}}{2 \, {\left (b e x^{2} + a e\right )}}\right ) + 2 \, {\left ({\left (3 \, b^{2} c^{2} d - 100 \, a b c d^{2} + 105 \, a^{2} d^{3}\right )} e x^{6} + 8 \, a^{2} c^{3} e + {\left (3 \, b^{2} c^{3} - 38 \, a b c^{2} d + 35 \, a^{2} c d^{2}\right )} e x^{4} + 14 \, {\left (a b c^{3} - a^{2} c^{2} d\right )} e x^{2}\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{96 \, a c^{4} x^{6}}\right ] \]

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)]

________________________________________________________________________________________

giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(x^2*d+c)]Evaluation time: 0.57Unable to divide, perhaps due to rounding error%%%{%%%{2,[5,1,5]%%%},[2,9,0]%%
%}+%%%{%%%{-10,[4,2,5]%%%},[2,8,1]%%%}+%%%{%%%{20,[3,3,5]%%%},[2,7,2]%%%}+%%%{%%%{-20,[2,4,5]%%%},[2,6,3]%%%}+
%%%{%%%{10,[1,5,5]%%%},[2,5,4]%%%}+%%%{%%%{-2,[0,6,5]%%%},[2,4,5]%%%}+%%%{%%{[%%%{-4,[5,0,5]%%%},0]:[1,0,%%%{-
1,[1,1,1]%%%}]%%},[1,10,0]%%%}+%%%{%%{[%%%{20,[4,1,5]%%%},0]:[1,0,%%%{-1,[1,1,1]%%%}]%%},[1,9,1]%%%}+%%%{%%{[%
%%{-40,[3,2,5]%%%},0]:[1,0,%%%{-1,[1,1,1]%%%}]%%},[1,8,2]%%%}+%%%{%%{[%%%{40,[2,3,5]%%%},0]:[1,0,%%%{-1,[1,1,1
]%%%}]%%},[1,7,3]%%%}+%%%{%%{[%%%{-20,[1,4,5]%%%},0]:[1,0,%%%{-1,[1,1,1]%%%}]%%},[1,6,4]%%%}+%%%{%%{[%%%{4,[0,
5,5]%%%},0]:[1,0,%%%{-1,[1,1,1]%%%}]%%},[1,5,5]%%%}+%%%{%%%{2,[6,0,6]%%%},[0,11,0]%%%}+%%%{%%%{-10,[5,1,6]%%%}
,[0,10,1]%%%}+%%%{%%%{20,[4,2,6]%%%},[0,9,2]%%%}+%%%{%%%{-20,[3,3,6]%%%},[0,8,3]%%%}+%%%{%%%{10,[2,4,6]%%%},[0
,7,4]%%%}+%%%{%%%{-2,[1,5,6]%%%},[0,6,5]%%%} / %%%{%%%{1,[0,2,0]%%%},[2,0,0]%%%}+%%%{%%{[%%%{-2,[0,1,0]%%%},0]
:[1,0,%%%{-1,[1,1,1]%%%}]%%},[1,1,0]%%%}+%%%{%%%{1,[1,1,1]%%%},[0,2,0]%%%} Error: Bad Argument Value

________________________________________________________________________________________

maple [B]  time = 0.10, size = 1498, normalized size = 4.09 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/96*(-174*(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+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^2-72*(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-96*(a*c)^(1/2)*((d*x^2+c)*(b*x^2+a))^(1/2)*x^6*a^2*b*c^2*d^2-60*(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-105*ln((a*d*x^2+b*c*x^2+2*a*c+2*(a*c)^(1/2)*(b*d*x^4
+a*d*x^2+b*c*x^2+a*c)^(1/2))/x^2)*x^6*a^4*c^2*d^3-3*ln((a*d*x^2+b*c*x^2+2*a*c+2*(a*c)^(1/2)*(b*d*x^4+a*d*x^2+b
*c*x^2+a*c)^(1/2))/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^3-105*ln((a*d*x^2+b*c*x^2+2*a*c+2*(a*c)^(1/2)*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2))/x^2)*x^8*a^4*c*d^4-174*
(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+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+2*(a*c)^(1/2)*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2))/x^2)*x^8*a^3
*b*c^2*d^3-27*ln((a*d*x^2+b*c*x^2+2*a*c+2*(a*c)^(1/2)*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2))/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+2*(a*c)^(1/2)*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2))/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+2*(a*c)^(1/2)*(b*d*
x^4+a*d*x^2+b*c*x^2+a*c)^(1/2))/x^2)*x^6*a^3*b*c^3*d^2-27*ln((a*d*x^2+b*c*x^2+2*a*c+2*(a*c)^(1/2)*(b*d*x^4+a*d
*x^2+b*c*x^2+a*c)^(1/2))/x^2)*x^6*a^2*b^2*c^4*d-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+96*(a*c)^(1/2)*((d*x^2+c)*(b*x^2+a))^(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)/a^2*(d*x^2+c)*((b*x^2+a)/(d*x^2+c)*e)^(3/2)/(a*c)^(1/2)/x^6/c
^5/(b*x^2+a)/((d*x^2+c)*(b*x^2+a))^(1/2)

________________________________________________________________________________________

maxima [A]  time = 3.04, size = 453, normalized size = 1.24 \[ \frac {1}{96} \, e {\left (\frac {2 \, {\left (3 \, {\left (b^{3} c^{5} - 23 \, a b^{2} c^{4} d + 51 \, a^{2} b c^{3} d^{2} - 29 \, a^{3} c^{2} d^{3}\right )} \left (\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}\right )^{\frac {5}{2}} e + 8 \, {\left (a b^{3} c^{4} + 9 \, a^{2} b^{2} c^{3} d - 27 \, a^{3} b c^{2} d^{2} + 17 \, a^{4} c d^{3}\right )} \left (\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}\right )^{\frac {3}{2}} e^{2} - 3 \, {\left (a^{2} b^{3} c^{3} + 9 \, a^{3} b^{2} c^{2} d - 29 \, a^{4} b c d^{2} + 19 \, a^{5} d^{3}\right )} \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}} e^{3}\right )}}{a^{4} c^{4} e^{3} - \frac {3 \, {\left (b x^{2} + a\right )} a^{3} c^{5} e^{3}}{d x^{2} + c} + \frac {3 \, {\left (b x^{2} + a\right )}^{2} a^{2} c^{6} e^{3}}{{\left (d x^{2} + c\right )}^{2}} - \frac {{\left (b x^{2} + a\right )}^{3} a c^{7} e^{3}}{{\left (d x^{2} + c\right )}^{3}}} + \frac {96 \, {\left (b c d^{2} - a d^{3}\right )} \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}}}{c^{4}} - \frac {3 \, {\left (b^{3} c^{3} + 9 \, a b^{2} c^{2} d - 45 \, a^{2} b c d^{2} + 35 \, a^{3} d^{3}\right )} e \log \left (\frac {c \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}} - \sqrt {a c e}}{c \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}} + \sqrt {a c e}}\right )}{\sqrt {a c e} a c^{4}}\right )} \]

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]

1/96*e*(2*(3*(b^3*c^5 - 23*a*b^2*c^4*d + 51*a^2*b*c^3*d^2 - 29*a^3*c^2*d^3)*((b*x^2 + a)*e/(d*x^2 + c))^(5/2)*
e + 8*(a*b^3*c^4 + 9*a^2*b^2*c^3*d - 27*a^3*b*c^2*d^2 + 17*a^4*c*d^3)*((b*x^2 + a)*e/(d*x^2 + c))^(3/2)*e^2 -
3*(a^2*b^3*c^3 + 9*a^3*b^2*c^2*d - 29*a^4*b*c*d^2 + 19*a^5*d^3)*sqrt((b*x^2 + a)*e/(d*x^2 + c))*e^3)/(a^4*c^4*
e^3 - 3*(b*x^2 + a)*a^3*c^5*e^3/(d*x^2 + c) + 3*(b*x^2 + a)^2*a^2*c^6*e^3/(d*x^2 + c)^2 - (b*x^2 + a)^3*a*c^7*
e^3/(d*x^2 + c)^3) + 96*(b*c*d^2 - a*d^3)*sqrt((b*x^2 + a)*e/(d*x^2 + c))/c^4 - 3*(b^3*c^3 + 9*a*b^2*c^2*d - 4
5*a^2*b*c*d^2 + 35*a^3*d^3)*e*log((c*sqrt((b*x^2 + a)*e/(d*x^2 + c)) - sqrt(a*c*e))/(c*sqrt((b*x^2 + a)*e/(d*x
^2 + c)) + sqrt(a*c*e)))/(sqrt(a*c*e)*a*c^4))

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (\frac {e\,\left (b\,x^2+a\right )}{d\,x^2+c}\right )}^{3/2}}{x^7} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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

________________________________________________________________________________________