\(\int \frac {x^{7/2} (a+b x^2)^2}{(c+d x^2)^2} \, dx\) [425]

Optimal result

Integrand size = 24, antiderivative size = 375 \[ \int \frac {x^{7/2} \left (a+b x^2\right )^2}{\left (c+d x^2\right )^2} \, dx=\frac {(13 b c-5 a d) (b c-a d) \sqrt {x}}{2 d^4}-\frac {(13 b c-5 a d) (b c-a d) x^{5/2}}{10 c d^3}+\frac {2 b^2 x^{9/2}}{9 d^2}+\frac {(b c-a d)^2 x^{9/2}}{2 c d^2 \left (c+d x^2\right )}+\frac {\sqrt [4]{c} (13 b c-5 a d) (b c-a d) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{4 \sqrt {2} d^{17/4}}-\frac {\sqrt [4]{c} (13 b c-5 a d) (b c-a d) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{4 \sqrt {2} d^{17/4}}+\frac {\sqrt [4]{c} (13 b c-5 a d) (b c-a d) \log \left (\sqrt {c}-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {d} x\right )}{8 \sqrt {2} d^{17/4}}-\frac {\sqrt [4]{c} (13 b c-5 a d) (b c-a d) \log \left (\sqrt {c}+\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {d} x\right )}{8 \sqrt {2} d^{17/4}} \]

[Out]

-1/10*(-5*a*d+13*b*c)*(-a*d+b*c)*x^(5/2)/c/d^3+2/9*b^2*x^(9/2)/d^2+1/2*(-a*d+b*c)^2*x^(9/2)/c/d^2/(d*x^2+c)+1/
8*c^(1/4)*(-5*a*d+13*b*c)*(-a*d+b*c)*arctan(1-d^(1/4)*2^(1/2)*x^(1/2)/c^(1/4))/d^(17/4)*2^(1/2)-1/8*c^(1/4)*(-
5*a*d+13*b*c)*(-a*d+b*c)*arctan(1+d^(1/4)*2^(1/2)*x^(1/2)/c^(1/4))/d^(17/4)*2^(1/2)+1/16*c^(1/4)*(-5*a*d+13*b*
c)*(-a*d+b*c)*ln(c^(1/2)+x*d^(1/2)-c^(1/4)*d^(1/4)*2^(1/2)*x^(1/2))/d^(17/4)*2^(1/2)-1/16*c^(1/4)*(-5*a*d+13*b
*c)*(-a*d+b*c)*ln(c^(1/2)+x*d^(1/2)+c^(1/4)*d^(1/4)*2^(1/2)*x^(1/2))/d^(17/4)*2^(1/2)+1/2*(-5*a*d+13*b*c)*(-a*
d+b*c)*x^(1/2)/d^4

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 375, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {474, 470, 327, 335, 217, 1179, 642, 1176, 631, 210} \[ \int \frac {x^{7/2} \left (a+b x^2\right )^2}{\left (c+d x^2\right )^2} \, dx=\frac {\sqrt [4]{c} (13 b c-5 a d) (b c-a d) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{4 \sqrt {2} d^{17/4}}-\frac {\sqrt [4]{c} (13 b c-5 a d) (b c-a d) \arctan \left (\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}+1\right )}{4 \sqrt {2} d^{17/4}}+\frac {\sqrt [4]{c} (13 b c-5 a d) (b c-a d) \log \left (-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{8 \sqrt {2} d^{17/4}}-\frac {\sqrt [4]{c} (13 b c-5 a d) (b c-a d) \log \left (\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{8 \sqrt {2} d^{17/4}}+\frac {\sqrt {x} (13 b c-5 a d) (b c-a d)}{2 d^4}-\frac {x^{5/2} (13 b c-5 a d) (b c-a d)}{10 c d^3}+\frac {x^{9/2} (b c-a d)^2}{2 c d^2 \left (c+d x^2\right )}+\frac {2 b^2 x^{9/2}}{9 d^2} \]

[In]

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

[Out]

((13*b*c - 5*a*d)*(b*c - a*d)*Sqrt[x])/(2*d^4) - ((13*b*c - 5*a*d)*(b*c - a*d)*x^(5/2))/(10*c*d^3) + (2*b^2*x^
(9/2))/(9*d^2) + ((b*c - a*d)^2*x^(9/2))/(2*c*d^2*(c + d*x^2)) + (c^(1/4)*(13*b*c - 5*a*d)*(b*c - a*d)*ArcTan[
1 - (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/(4*Sqrt[2]*d^(17/4)) - (c^(1/4)*(13*b*c - 5*a*d)*(b*c - a*d)*ArcTan[1
+ (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/(4*Sqrt[2]*d^(17/4)) + (c^(1/4)*(13*b*c - 5*a*d)*(b*c - a*d)*Log[Sqrt[c]
- Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x])/(8*Sqrt[2]*d^(17/4)) - (c^(1/4)*(13*b*c - 5*a*d)*(b*c - a*d)*
Log[Sqrt[c] + Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x])/(8*Sqrt[2]*d^(17/4))

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

Rule 327

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

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 470

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

Rule 474

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 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rubi steps \begin{align*} \text {integral}& = \frac {(b c-a d)^2 x^{9/2}}{2 c d^2 \left (c+d x^2\right )}-\frac {\int \frac {x^{7/2} \left (\frac {1}{2} (3 b c-5 a d) (3 b c-a d)-2 b^2 c d x^2\right )}{c+d x^2} \, dx}{2 c d^2} \\ & = \frac {2 b^2 x^{9/2}}{9 d^2}+\frac {(b c-a d)^2 x^{9/2}}{2 c d^2 \left (c+d x^2\right )}-\frac {((13 b c-5 a d) (b c-a d)) \int \frac {x^{7/2}}{c+d x^2} \, dx}{4 c d^2} \\ & = -\frac {(13 b c-5 a d) (b c-a d) x^{5/2}}{10 c d^3}+\frac {2 b^2 x^{9/2}}{9 d^2}+\frac {(b c-a d)^2 x^{9/2}}{2 c d^2 \left (c+d x^2\right )}+\frac {((13 b c-5 a d) (b c-a d)) \int \frac {x^{3/2}}{c+d x^2} \, dx}{4 d^3} \\ & = \frac {(13 b c-5 a d) (b c-a d) \sqrt {x}}{2 d^4}-\frac {(13 b c-5 a d) (b c-a d) x^{5/2}}{10 c d^3}+\frac {2 b^2 x^{9/2}}{9 d^2}+\frac {(b c-a d)^2 x^{9/2}}{2 c d^2 \left (c+d x^2\right )}-\frac {(c (13 b c-5 a d) (b c-a d)) \int \frac {1}{\sqrt {x} \left (c+d x^2\right )} \, dx}{4 d^4} \\ & = \frac {(13 b c-5 a d) (b c-a d) \sqrt {x}}{2 d^4}-\frac {(13 b c-5 a d) (b c-a d) x^{5/2}}{10 c d^3}+\frac {2 b^2 x^{9/2}}{9 d^2}+\frac {(b c-a d)^2 x^{9/2}}{2 c d^2 \left (c+d x^2\right )}-\frac {(c (13 b c-5 a d) (b c-a d)) \text {Subst}\left (\int \frac {1}{c+d x^4} \, dx,x,\sqrt {x}\right )}{2 d^4} \\ & = \frac {(13 b c-5 a d) (b c-a d) \sqrt {x}}{2 d^4}-\frac {(13 b c-5 a d) (b c-a d) x^{5/2}}{10 c d^3}+\frac {2 b^2 x^{9/2}}{9 d^2}+\frac {(b c-a d)^2 x^{9/2}}{2 c d^2 \left (c+d x^2\right )}-\frac {\left (\sqrt {c} (13 b c-5 a d) (b c-a d)\right ) \text {Subst}\left (\int \frac {\sqrt {c}-\sqrt {d} x^2}{c+d x^4} \, dx,x,\sqrt {x}\right )}{4 d^4}-\frac {\left (\sqrt {c} (13 b c-5 a d) (b c-a d)\right ) \text {Subst}\left (\int \frac {\sqrt {c}+\sqrt {d} x^2}{c+d x^4} \, dx,x,\sqrt {x}\right )}{4 d^4} \\ & = \frac {(13 b c-5 a d) (b c-a d) \sqrt {x}}{2 d^4}-\frac {(13 b c-5 a d) (b c-a d) x^{5/2}}{10 c d^3}+\frac {2 b^2 x^{9/2}}{9 d^2}+\frac {(b c-a d)^2 x^{9/2}}{2 c d^2 \left (c+d x^2\right )}-\frac {\left (\sqrt {c} (13 b c-5 a d) (b c-a d)\right ) \text {Subst}\left (\int \frac {1}{\frac {\sqrt {c}}{\sqrt {d}}-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{d}}+x^2} \, dx,x,\sqrt {x}\right )}{8 d^{9/2}}-\frac {\left (\sqrt {c} (13 b c-5 a d) (b c-a d)\right ) \text {Subst}\left (\int \frac {1}{\frac {\sqrt {c}}{\sqrt {d}}+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{d}}+x^2} \, dx,x,\sqrt {x}\right )}{8 d^{9/2}}+\frac {\left (\sqrt [4]{c} (13 b c-5 a d) (b c-a d)\right ) \text {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt [4]{c}}{\sqrt [4]{d}}+2 x}{-\frac {\sqrt {c}}{\sqrt {d}}-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{d}}-x^2} \, dx,x,\sqrt {x}\right )}{8 \sqrt {2} d^{17/4}}+\frac {\left (\sqrt [4]{c} (13 b c-5 a d) (b c-a d)\right ) \text {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt [4]{c}}{\sqrt [4]{d}}-2 x}{-\frac {\sqrt {c}}{\sqrt {d}}+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{d}}-x^2} \, dx,x,\sqrt {x}\right )}{8 \sqrt {2} d^{17/4}} \\ & = \frac {(13 b c-5 a d) (b c-a d) \sqrt {x}}{2 d^4}-\frac {(13 b c-5 a d) (b c-a d) x^{5/2}}{10 c d^3}+\frac {2 b^2 x^{9/2}}{9 d^2}+\frac {(b c-a d)^2 x^{9/2}}{2 c d^2 \left (c+d x^2\right )}+\frac {\sqrt [4]{c} (13 b c-5 a d) (b c-a d) \log \left (\sqrt {c}-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {d} x\right )}{8 \sqrt {2} d^{17/4}}-\frac {\sqrt [4]{c} (13 b c-5 a d) (b c-a d) \log \left (\sqrt {c}+\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {d} x\right )}{8 \sqrt {2} d^{17/4}}-\frac {\left (\sqrt [4]{c} (13 b c-5 a d) (b c-a d)\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{4 \sqrt {2} d^{17/4}}+\frac {\left (\sqrt [4]{c} (13 b c-5 a d) (b c-a d)\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{4 \sqrt {2} d^{17/4}} \\ & = \frac {(13 b c-5 a d) (b c-a d) \sqrt {x}}{2 d^4}-\frac {(13 b c-5 a d) (b c-a d) x^{5/2}}{10 c d^3}+\frac {2 b^2 x^{9/2}}{9 d^2}+\frac {(b c-a d)^2 x^{9/2}}{2 c d^2 \left (c+d x^2\right )}+\frac {\sqrt [4]{c} (13 b c-5 a d) (b c-a d) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{4 \sqrt {2} d^{17/4}}-\frac {\sqrt [4]{c} (13 b c-5 a d) (b c-a d) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{4 \sqrt {2} d^{17/4}}+\frac {\sqrt [4]{c} (13 b c-5 a d) (b c-a d) \log \left (\sqrt {c}-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {d} x\right )}{8 \sqrt {2} d^{17/4}}-\frac {\sqrt [4]{c} (13 b c-5 a d) (b c-a d) \log \left (\sqrt {c}+\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {d} x\right )}{8 \sqrt {2} d^{17/4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.75 (sec) , antiderivative size = 255, normalized size of antiderivative = 0.68 \[ \int \frac {x^{7/2} \left (a+b x^2\right )^2}{\left (c+d x^2\right )^2} \, dx=\frac {\frac {4 \sqrt [4]{d} \sqrt {x} \left (45 a^2 d^2 \left (5 c+4 d x^2\right )+18 a b d \left (-45 c^2-36 c d x^2+4 d^2 x^4\right )+b^2 \left (585 c^3+468 c^2 d x^2-52 c d^2 x^4+20 d^3 x^6\right )\right )}{c+d x^2}+45 \sqrt {2} \sqrt [4]{c} \left (13 b^2 c^2-18 a b c d+5 a^2 d^2\right ) \arctan \left (\frac {\sqrt {c}-\sqrt {d} x}{\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}}\right )-45 \sqrt {2} \sqrt [4]{c} \left (13 b^2 c^2-18 a b c d+5 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}}{\sqrt {c}+\sqrt {d} x}\right )}{360 d^{17/4}} \]

[In]

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

[Out]

((4*d^(1/4)*Sqrt[x]*(45*a^2*d^2*(5*c + 4*d*x^2) + 18*a*b*d*(-45*c^2 - 36*c*d*x^2 + 4*d^2*x^4) + b^2*(585*c^3 +
468*c^2*d*x^2 - 52*c*d^2*x^4 + 20*d^3*x^6)))/(c + d*x^2) + 45*Sqrt[2]*c^(1/4)*(13*b^2*c^2 - 18*a*b*c*d + 5*a^
2*d^2)*ArcTan[(Sqrt[c] - Sqrt[d]*x)/(Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x])] - 45*Sqrt[2]*c^(1/4)*(13*b^2*c^2 - 18*a
*b*c*d + 5*a^2*d^2)*ArcTanh[(Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x])/(Sqrt[c] + Sqrt[d]*x)])/(360*d^(17/4))

Maple [A] (verified)

Time = 2.81 (sec) , antiderivative size = 216, normalized size of antiderivative = 0.58

method	result	size
risch	\(\frac {2 \left (5 b^{2} d^{2} x^{4}+18 x^{2} a b \,d^{2}-18 x^{2} b^{2} c d +45 a^{2} d^{2}-180 a b c d +135 b^{2} c^{2}\right ) \sqrt {x}}{45 d^{4}}-\frac {c \left (2 a d -2 b c \right ) \left (\frac {\left (-\frac {a d}{4}+\frac {b c}{4}\right ) \sqrt {x}}{d \,x^{2}+c}+\frac {\left (5 a d -13 b c \right ) \left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {c}{d}}}{x -\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {c}{d}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )\right )}{32 c}\right )}{d^{4}}\)	\(216\)
derivativedivides	\(\frac {\frac {2 b^{2} d^{2} x^{\frac {9}{2}}}{9}+\frac {4 a b \,d^{2} x^{\frac {5}{2}}}{5}-\frac {4 b^{2} c d \,x^{\frac {5}{2}}}{5}+2 a^{2} d^{2} \sqrt {x}-8 a b c d \sqrt {x}+6 b^{2} c^{2} \sqrt {x}}{d^{4}}-\frac {2 c \left (\frac {\left (-\frac {1}{4} a^{2} d^{2}+\frac {1}{2} a b c d -\frac {1}{4} b^{2} c^{2}\right ) \sqrt {x}}{d \,x^{2}+c}+\frac {\left (5 a^{2} d^{2}-18 a b c d +13 b^{2} c^{2}\right ) \left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {c}{d}}}{x -\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {c}{d}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )\right )}{32 c}\right )}{d^{4}}\)	\(240\)
default	\(\frac {\frac {2 b^{2} d^{2} x^{\frac {9}{2}}}{9}+\frac {4 a b \,d^{2} x^{\frac {5}{2}}}{5}-\frac {4 b^{2} c d \,x^{\frac {5}{2}}}{5}+2 a^{2} d^{2} \sqrt {x}-8 a b c d \sqrt {x}+6 b^{2} c^{2} \sqrt {x}}{d^{4}}-\frac {2 c \left (\frac {\left (-\frac {1}{4} a^{2} d^{2}+\frac {1}{2} a b c d -\frac {1}{4} b^{2} c^{2}\right ) \sqrt {x}}{d \,x^{2}+c}+\frac {\left (5 a^{2} d^{2}-18 a b c d +13 b^{2} c^{2}\right ) \left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {c}{d}}}{x -\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {c}{d}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )\right )}{32 c}\right )}{d^{4}}\)	\(240\)

[In]

int(x^(7/2)*(b*x^2+a)^2/(d*x^2+c)^2,x,method=_RETURNVERBOSE)

[Out]

2/45*(5*b^2*d^2*x^4+18*a*b*d^2*x^2-18*b^2*c*d*x^2+45*a^2*d^2-180*a*b*c*d+135*b^2*c^2)*x^(1/2)/d^4-c/d^4*(2*a*d
-2*b*c)*((-1/4*a*d+1/4*b*c)*x^(1/2)/(d*x^2+c)+1/32*(5*a*d-13*b*c)*(c/d)^(1/4)/c*2^(1/2)*(ln((x+(c/d)^(1/4)*x^(
1/2)*2^(1/2)+(c/d)^(1/2))/(x-(c/d)^(1/4)*x^(1/2)*2^(1/2)+(c/d)^(1/2)))+2*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)+1)
+2*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)-1)))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.30 (sec) , antiderivative size = 1248, normalized size of antiderivative = 3.33 \[ \int \frac {x^{7/2} \left (a+b x^2\right )^2}{\left (c+d x^2\right )^2} \, dx=\text {Too large to display} \]

[In]

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

[Out]

-1/360*(45*(d^5*x^2 + c*d^4)*(-(28561*b^8*c^9 - 158184*a*b^7*c^8*d + 372476*a^2*b^6*c^7*d^2 - 485784*a^3*b^5*c
^6*d^3 + 383046*a^4*b^4*c^5*d^4 - 186840*a^5*b^3*c^4*d^5 + 55100*a^6*b^2*c^3*d^6 - 9000*a^7*b*c^2*d^7 + 625*a^
8*c*d^8)/d^17)^(1/4)*log(d^4*(-(28561*b^8*c^9 - 158184*a*b^7*c^8*d + 372476*a^2*b^6*c^7*d^2 - 485784*a^3*b^5*c
^6*d^3 + 383046*a^4*b^4*c^5*d^4 - 186840*a^5*b^3*c^4*d^5 + 55100*a^6*b^2*c^3*d^6 - 9000*a^7*b*c^2*d^7 + 625*a^
8*c*d^8)/d^17)^(1/4) + (13*b^2*c^2 - 18*a*b*c*d + 5*a^2*d^2)*sqrt(x)) + 45*(I*d^5*x^2 + I*c*d^4)*(-(28561*b^8*
c^9 - 158184*a*b^7*c^8*d + 372476*a^2*b^6*c^7*d^2 - 485784*a^3*b^5*c^6*d^3 + 383046*a^4*b^4*c^5*d^4 - 186840*a
^5*b^3*c^4*d^5 + 55100*a^6*b^2*c^3*d^6 - 9000*a^7*b*c^2*d^7 + 625*a^8*c*d^8)/d^17)^(1/4)*log(I*d^4*(-(28561*b^
8*c^9 - 158184*a*b^7*c^8*d + 372476*a^2*b^6*c^7*d^2 - 485784*a^3*b^5*c^6*d^3 + 383046*a^4*b^4*c^5*d^4 - 186840
*a^5*b^3*c^4*d^5 + 55100*a^6*b^2*c^3*d^6 - 9000*a^7*b*c^2*d^7 + 625*a^8*c*d^8)/d^17)^(1/4) + (13*b^2*c^2 - 18*
a*b*c*d + 5*a^2*d^2)*sqrt(x)) + 45*(-I*d^5*x^2 - I*c*d^4)*(-(28561*b^8*c^9 - 158184*a*b^7*c^8*d + 372476*a^2*b
^6*c^7*d^2 - 485784*a^3*b^5*c^6*d^3 + 383046*a^4*b^4*c^5*d^4 - 186840*a^5*b^3*c^4*d^5 + 55100*a^6*b^2*c^3*d^6
- 9000*a^7*b*c^2*d^7 + 625*a^8*c*d^8)/d^17)^(1/4)*log(-I*d^4*(-(28561*b^8*c^9 - 158184*a*b^7*c^8*d + 372476*a^
2*b^6*c^7*d^2 - 485784*a^3*b^5*c^6*d^3 + 383046*a^4*b^4*c^5*d^4 - 186840*a^5*b^3*c^4*d^5 + 55100*a^6*b^2*c^3*d
^6 - 9000*a^7*b*c^2*d^7 + 625*a^8*c*d^8)/d^17)^(1/4) + (13*b^2*c^2 - 18*a*b*c*d + 5*a^2*d^2)*sqrt(x)) - 45*(d^
5*x^2 + c*d^4)*(-(28561*b^8*c^9 - 158184*a*b^7*c^8*d + 372476*a^2*b^6*c^7*d^2 - 485784*a^3*b^5*c^6*d^3 + 38304
6*a^4*b^4*c^5*d^4 - 186840*a^5*b^3*c^4*d^5 + 55100*a^6*b^2*c^3*d^6 - 9000*a^7*b*c^2*d^7 + 625*a^8*c*d^8)/d^17)
^(1/4)*log(-d^4*(-(28561*b^8*c^9 - 158184*a*b^7*c^8*d + 372476*a^2*b^6*c^7*d^2 - 485784*a^3*b^5*c^6*d^3 + 3830
46*a^4*b^4*c^5*d^4 - 186840*a^5*b^3*c^4*d^5 + 55100*a^6*b^2*c^3*d^6 - 9000*a^7*b*c^2*d^7 + 625*a^8*c*d^8)/d^17
)^(1/4) + (13*b^2*c^2 - 18*a*b*c*d + 5*a^2*d^2)*sqrt(x)) - 4*(20*b^2*d^3*x^6 + 585*b^2*c^3 - 810*a*b*c^2*d + 2
25*a^2*c*d^2 - 4*(13*b^2*c*d^2 - 18*a*b*d^3)*x^4 + 36*(13*b^2*c^2*d - 18*a*b*c*d^2 + 5*a^2*d^3)*x^2)*sqrt(x))/
(d^5*x^2 + c*d^4)

Sympy [F(-1)]

Timed out. \[ \int \frac {x^{7/2} \left (a+b x^2\right )^2}{\left (c+d x^2\right )^2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 377, normalized size of antiderivative = 1.01 \[ \int \frac {x^{7/2} \left (a+b x^2\right )^2}{\left (c+d x^2\right )^2} \, dx=\frac {{\left (b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2}\right )} \sqrt {x}}{2 \, {\left (d^{5} x^{2} + c d^{4}\right )}} - \frac {{\left (\frac {2 \, \sqrt {2} {\left (13 \, b^{2} c^{2} - 18 \, a b c d + 5 \, a^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} + 2 \, \sqrt {d} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {c} \sqrt {d}}}\right )}{\sqrt {c} \sqrt {\sqrt {c} \sqrt {d}}} + \frac {2 \, \sqrt {2} {\left (13 \, b^{2} c^{2} - 18 \, a b c d + 5 \, a^{2} d^{2}\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} - 2 \, \sqrt {d} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {c} \sqrt {d}}}\right )}{\sqrt {c} \sqrt {\sqrt {c} \sqrt {d}}} + \frac {\sqrt {2} {\left (13 \, b^{2} c^{2} - 18 \, a b c d + 5 \, a^{2} d^{2}\right )} \log \left (\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} \sqrt {x} + \sqrt {d} x + \sqrt {c}\right )}{c^{\frac {3}{4}} d^{\frac {1}{4}}} - \frac {\sqrt {2} {\left (13 \, b^{2} c^{2} - 18 \, a b c d + 5 \, a^{2} d^{2}\right )} \log \left (-\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} \sqrt {x} + \sqrt {d} x + \sqrt {c}\right )}{c^{\frac {3}{4}} d^{\frac {1}{4}}}\right )} c}{16 \, d^{4}} + \frac {2 \, {\left (5 \, b^{2} d^{2} x^{\frac {9}{2}} - 18 \, {\left (b^{2} c d - a b d^{2}\right )} x^{\frac {5}{2}} + 45 \, {\left (3 \, b^{2} c^{2} - 4 \, a b c d + a^{2} d^{2}\right )} \sqrt {x}\right )}}{45 \, d^{4}} \]

[In]

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

[Out]

1/2*(b^2*c^3 - 2*a*b*c^2*d + a^2*c*d^2)*sqrt(x)/(d^5*x^2 + c*d^4) - 1/16*(2*sqrt(2)*(13*b^2*c^2 - 18*a*b*c*d +
5*a^2*d^2)*arctan(1/2*sqrt(2)*(sqrt(2)*c^(1/4)*d^(1/4) + 2*sqrt(d)*sqrt(x))/sqrt(sqrt(c)*sqrt(d)))/(sqrt(c)*s
qrt(sqrt(c)*sqrt(d))) + 2*sqrt(2)*(13*b^2*c^2 - 18*a*b*c*d + 5*a^2*d^2)*arctan(-1/2*sqrt(2)*(sqrt(2)*c^(1/4)*d
^(1/4) - 2*sqrt(d)*sqrt(x))/sqrt(sqrt(c)*sqrt(d)))/(sqrt(c)*sqrt(sqrt(c)*sqrt(d))) + sqrt(2)*(13*b^2*c^2 - 18*
a*b*c*d + 5*a^2*d^2)*log(sqrt(2)*c^(1/4)*d^(1/4)*sqrt(x) + sqrt(d)*x + sqrt(c))/(c^(3/4)*d^(1/4)) - sqrt(2)*(1
3*b^2*c^2 - 18*a*b*c*d + 5*a^2*d^2)*log(-sqrt(2)*c^(1/4)*d^(1/4)*sqrt(x) + sqrt(d)*x + sqrt(c))/(c^(3/4)*d^(1/
4)))*c/d^4 + 2/45*(5*b^2*d^2*x^(9/2) - 18*(b^2*c*d - a*b*d^2)*x^(5/2) + 45*(3*b^2*c^2 - 4*a*b*c*d + a^2*d^2)*s
qrt(x))/d^4

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 440, normalized size of antiderivative = 1.17 \[ \int \frac {x^{7/2} \left (a+b x^2\right )^2}{\left (c+d x^2\right )^2} \, dx=-\frac {\sqrt {2} {\left (13 \, \left (c d^{3}\right )^{\frac {1}{4}} b^{2} c^{2} - 18 \, \left (c d^{3}\right )^{\frac {1}{4}} a b c d + 5 \, \left (c d^{3}\right )^{\frac {1}{4}} a^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {c}{d}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {c}{d}\right )^{\frac {1}{4}}}\right )}{8 \, d^{5}} - \frac {\sqrt {2} {\left (13 \, \left (c d^{3}\right )^{\frac {1}{4}} b^{2} c^{2} - 18 \, \left (c d^{3}\right )^{\frac {1}{4}} a b c d + 5 \, \left (c d^{3}\right )^{\frac {1}{4}} a^{2} d^{2}\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {c}{d}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {c}{d}\right )^{\frac {1}{4}}}\right )}{8 \, d^{5}} - \frac {\sqrt {2} {\left (13 \, \left (c d^{3}\right )^{\frac {1}{4}} b^{2} c^{2} - 18 \, \left (c d^{3}\right )^{\frac {1}{4}} a b c d + 5 \, \left (c d^{3}\right )^{\frac {1}{4}} a^{2} d^{2}\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {c}{d}\right )^{\frac {1}{4}} + x + \sqrt {\frac {c}{d}}\right )}{16 \, d^{5}} + \frac {\sqrt {2} {\left (13 \, \left (c d^{3}\right )^{\frac {1}{4}} b^{2} c^{2} - 18 \, \left (c d^{3}\right )^{\frac {1}{4}} a b c d + 5 \, \left (c d^{3}\right )^{\frac {1}{4}} a^{2} d^{2}\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {c}{d}\right )^{\frac {1}{4}} + x + \sqrt {\frac {c}{d}}\right )}{16 \, d^{5}} + \frac {b^{2} c^{3} \sqrt {x} - 2 \, a b c^{2} d \sqrt {x} + a^{2} c d^{2} \sqrt {x}}{2 \, {\left (d x^{2} + c\right )} d^{4}} + \frac {2 \, {\left (5 \, b^{2} d^{16} x^{\frac {9}{2}} - 18 \, b^{2} c d^{15} x^{\frac {5}{2}} + 18 \, a b d^{16} x^{\frac {5}{2}} + 135 \, b^{2} c^{2} d^{14} \sqrt {x} - 180 \, a b c d^{15} \sqrt {x} + 45 \, a^{2} d^{16} \sqrt {x}\right )}}{45 \, d^{18}} \]

[In]

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

[Out]

-1/8*sqrt(2)*(13*(c*d^3)^(1/4)*b^2*c^2 - 18*(c*d^3)^(1/4)*a*b*c*d + 5*(c*d^3)^(1/4)*a^2*d^2)*arctan(1/2*sqrt(2
)*(sqrt(2)*(c/d)^(1/4) + 2*sqrt(x))/(c/d)^(1/4))/d^5 - 1/8*sqrt(2)*(13*(c*d^3)^(1/4)*b^2*c^2 - 18*(c*d^3)^(1/4
)*a*b*c*d + 5*(c*d^3)^(1/4)*a^2*d^2)*arctan(-1/2*sqrt(2)*(sqrt(2)*(c/d)^(1/4) - 2*sqrt(x))/(c/d)^(1/4))/d^5 -
1/16*sqrt(2)*(13*(c*d^3)^(1/4)*b^2*c^2 - 18*(c*d^3)^(1/4)*a*b*c*d + 5*(c*d^3)^(1/4)*a^2*d^2)*log(sqrt(2)*sqrt(
x)*(c/d)^(1/4) + x + sqrt(c/d))/d^5 + 1/16*sqrt(2)*(13*(c*d^3)^(1/4)*b^2*c^2 - 18*(c*d^3)^(1/4)*a*b*c*d + 5*(c
*d^3)^(1/4)*a^2*d^2)*log(-sqrt(2)*sqrt(x)*(c/d)^(1/4) + x + sqrt(c/d))/d^5 + 1/2*(b^2*c^3*sqrt(x) - 2*a*b*c^2*
d*sqrt(x) + a^2*c*d^2*sqrt(x))/((d*x^2 + c)*d^4) + 2/45*(5*b^2*d^16*x^(9/2) - 18*b^2*c*d^15*x^(5/2) + 18*a*b*d
^16*x^(5/2) + 135*b^2*c^2*d^14*sqrt(x) - 180*a*b*c*d^15*sqrt(x) + 45*a^2*d^16*sqrt(x))/d^18

Mupad [B] (verification not implemented)

Time = 4.94 (sec) , antiderivative size = 1367, normalized size of antiderivative = 3.65 \[ \int \frac {x^{7/2} \left (a+b x^2\right )^2}{\left (c+d x^2\right )^2} \, dx=\text {Too large to display} \]

[In]

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

[Out]

x^(1/2)*((2*a^2)/d^2 + (2*c*((4*b^2*c)/d^3 - (4*a*b)/d^2))/d - (2*b^2*c^2)/d^4) - x^(5/2)*((4*b^2*c)/(5*d^3) -
(4*a*b)/(5*d^2)) + (x^(1/2)*((b^2*c^3)/2 + (a^2*c*d^2)/2 - a*b*c^2*d))/(c*d^4 + d^5*x^2) + (2*b^2*x^(9/2))/(9
*d^2) + ((-c)^(1/4)*atan((((-c)^(1/4)*((x^(1/2)*(169*b^4*c^6 + 25*a^4*c^2*d^4 - 180*a^3*b*c^3*d^3 + 454*a^2*b^
2*c^4*d^2 - 468*a*b^3*c^5*d))/d^5 + ((-c)^(1/4)*(a*d - b*c)*(5*a*d - 13*b*c)*(13*b^2*c^4 + 5*a^2*c^2*d^2 - 18*
a*b*c^3*d))/d^(21/4))*(a*d - b*c)*(5*a*d - 13*b*c)*1i)/(8*d^(17/4)) + ((-c)^(1/4)*((x^(1/2)*(169*b^4*c^6 + 25*
a^4*c^2*d^4 - 180*a^3*b*c^3*d^3 + 454*a^2*b^2*c^4*d^2 - 468*a*b^3*c^5*d))/d^5 - ((-c)^(1/4)*(a*d - b*c)*(5*a*d
- 13*b*c)*(13*b^2*c^4 + 5*a^2*c^2*d^2 - 18*a*b*c^3*d))/d^(21/4))*(a*d - b*c)*(5*a*d - 13*b*c)*1i)/(8*d^(17/4)
))/(((-c)^(1/4)*((x^(1/2)*(169*b^4*c^6 + 25*a^4*c^2*d^4 - 180*a^3*b*c^3*d^3 + 454*a^2*b^2*c^4*d^2 - 468*a*b^3*
c^5*d))/d^5 + ((-c)^(1/4)*(a*d - b*c)*(5*a*d - 13*b*c)*(13*b^2*c^4 + 5*a^2*c^2*d^2 - 18*a*b*c^3*d))/d^(21/4))*
(a*d - b*c)*(5*a*d - 13*b*c))/(8*d^(17/4)) - ((-c)^(1/4)*((x^(1/2)*(169*b^4*c^6 + 25*a^4*c^2*d^4 - 180*a^3*b*c
^3*d^3 + 454*a^2*b^2*c^4*d^2 - 468*a*b^3*c^5*d))/d^5 - ((-c)^(1/4)*(a*d - b*c)*(5*a*d - 13*b*c)*(13*b^2*c^4 +
5*a^2*c^2*d^2 - 18*a*b*c^3*d))/d^(21/4))*(a*d - b*c)*(5*a*d - 13*b*c))/(8*d^(17/4))))*(a*d - b*c)*(5*a*d - 13*
b*c)*1i)/(4*d^(17/4)) - ((-c)^(1/4)*atan((((-c)^(1/4)*((x^(1/2)*(169*b^4*c^6 + 25*a^4*c^2*d^4 - 180*a^3*b*c^3*
d^3 + 454*a^2*b^2*c^4*d^2 - 468*a*b^3*c^5*d))/d^5 - ((-c)^(1/4)*(a*d - b*c)*(5*a*d - 13*b*c)*(13*b^2*c^4 + 5*a
^2*c^2*d^2 - 18*a*b*c^3*d)*1i)/d^(21/4))*(a*d - b*c)*(5*a*d - 13*b*c))/(8*d^(17/4)) + ((-c)^(1/4)*((x^(1/2)*(1
69*b^4*c^6 + 25*a^4*c^2*d^4 - 180*a^3*b*c^3*d^3 + 454*a^2*b^2*c^4*d^2 - 468*a*b^3*c^5*d))/d^5 + ((-c)^(1/4)*(a
*d - b*c)*(5*a*d - 13*b*c)*(13*b^2*c^4 + 5*a^2*c^2*d^2 - 18*a*b*c^3*d)*1i)/d^(21/4))*(a*d - b*c)*(5*a*d - 13*b
*c))/(8*d^(17/4)))/(((-c)^(1/4)*((x^(1/2)*(169*b^4*c^6 + 25*a^4*c^2*d^4 - 180*a^3*b*c^3*d^3 + 454*a^2*b^2*c^4*
d^2 - 468*a*b^3*c^5*d))/d^5 - ((-c)^(1/4)*(a*d - b*c)*(5*a*d - 13*b*c)*(13*b^2*c^4 + 5*a^2*c^2*d^2 - 18*a*b*c^
3*d)*1i)/d^(21/4))*(a*d - b*c)*(5*a*d - 13*b*c)*1i)/(8*d^(17/4)) - ((-c)^(1/4)*((x^(1/2)*(169*b^4*c^6 + 25*a^4
*c^2*d^4 - 180*a^3*b*c^3*d^3 + 454*a^2*b^2*c^4*d^2 - 468*a*b^3*c^5*d))/d^5 + ((-c)^(1/4)*(a*d - b*c)*(5*a*d -
13*b*c)*(13*b^2*c^4 + 5*a^2*c^2*d^2 - 18*a*b*c^3*d)*1i)/d^(21/4))*(a*d - b*c)*(5*a*d - 13*b*c)*1i)/(8*d^(17/4)
)))*(a*d - b*c)*(5*a*d - 13*b*c))/(4*d^(17/4))