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

Optimal. Leaf size=375 \[ \frac{x^{9/2} (b c-a d)^2}{2 c d^2 \left (c+d x^2\right )}-\frac{x^{5/2} (13 b c-5 a d) (b c-a d)}{10 c d^3}+\frac{\sqrt{x} (13 b c-5 a d) (b c-a d)}{2 d^4}+\frac{\sqrt [4]{c} (13 b c-5 a d) (b c-a d) \log \left (-\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{8 \sqrt{2} d^{17/4}}-\frac{\sqrt [4]{c} (13 b c-5 a d) (b c-a d) \log \left (\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{8 \sqrt{2} d^{17/4}}+\frac{\sqrt [4]{c} (13 b c-5 a d) (b c-a d) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}\right )}{4 \sqrt{2} d^{17/4}}-\frac{\sqrt [4]{c} (13 b c-5 a d) (b c-a d) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}+1\right )}{4 \sqrt{2} d^{17/4}}+\frac{2 b^2 x^{9/2}}{9 d^2} \]

[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))

________________________________________________________________________________________

Rubi [A]  time = 0.437007, antiderivative size = 375, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 10, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used = {463, 459, 321, 329, 211, 1165, 628, 1162, 617, 204} \[ \frac{x^{9/2} (b c-a d)^2}{2 c d^2 \left (c+d x^2\right )}-\frac{x^{5/2} (13 b c-5 a d) (b c-a d)}{10 c d^3}+\frac{\sqrt{x} (13 b c-5 a d) (b c-a d)}{2 d^4}+\frac{\sqrt [4]{c} (13 b c-5 a d) (b c-a d) \log \left (-\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{8 \sqrt{2} d^{17/4}}-\frac{\sqrt [4]{c} (13 b c-5 a d) (b c-a d) \log \left (\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{8 \sqrt{2} d^{17/4}}+\frac{\sqrt [4]{c} (13 b c-5 a d) (b c-a d) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}\right )}{4 \sqrt{2} d^{17/4}}-\frac{\sqrt [4]{c} (13 b c-5 a d) (b c-a d) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}+1\right )}{4 \sqrt{2} d^{17/4}}+\frac{2 b^2 x^{9/2}}{9 d^2} \]

Antiderivative was successfully verified.

[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 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 459

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 321

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 329

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 211

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 1165

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]

Rule 628

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 1162

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 617

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 204

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

Rubi steps

\begin{align*} \int \frac{x^{7/2} \left (a+b x^2\right )^2}{\left (c+d x^2\right )^2} \, dx &=\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)) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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]  time = 0.345805, size = 372, normalized size = 0.99 \[ \frac{1440 \sqrt [4]{d} \sqrt{x} \left (a^2 d^2-4 a b c d+3 b^2 c^2\right )+45 \sqrt{2} \sqrt [4]{c} \left (5 a^2 d^2-18 a b c d+13 b^2 c^2\right ) \log \left (-\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )-45 \sqrt{2} \sqrt [4]{c} \left (5 a^2 d^2-18 a b c d+13 b^2 c^2\right ) \log \left (\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )+90 \sqrt{2} \sqrt [4]{c} \left (5 a^2 d^2-18 a b c d+13 b^2 c^2\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}\right )-90 \sqrt{2} \sqrt [4]{c} \left (5 a^2 d^2-18 a b c d+13 b^2 c^2\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}+1\right )-576 b d^{5/4} x^{5/2} (b c-a d)+\frac{360 c \sqrt [4]{d} \sqrt{x} (b c-a d)^2}{c+d x^2}+160 b^2 d^{9/4} x^{9/2}}{720 d^{17/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(1440*d^(1/4)*(3*b^2*c^2 - 4*a*b*c*d + a^2*d^2)*Sqrt[x] - 576*b*d^(5/4)*(b*c - a*d)*x^(5/2) + 160*b^2*d^(9/4)*
x^(9/2) + (360*c*d^(1/4)*(b*c - a*d)^2*Sqrt[x])/(c + d*x^2) + 90*Sqrt[2]*c^(1/4)*(13*b^2*c^2 - 18*a*b*c*d + 5*
a^2*d^2)*ArcTan[1 - (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)] - 90*Sqrt[2]*c^(1/4)*(13*b^2*c^2 - 18*a*b*c*d + 5*a^2*d
^2)*ArcTan[1 + (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)] + 45*Sqrt[2]*c^(1/4)*(13*b^2*c^2 - 18*a*b*c*d + 5*a^2*d^2)*L
og[Sqrt[c] - Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x] - 45*Sqrt[2]*c^(1/4)*(13*b^2*c^2 - 18*a*b*c*d + 5*a^
2*d^2)*Log[Sqrt[c] + Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x])/(720*d^(17/4))

________________________________________________________________________________________

Maple [A]  time = 0.017, size = 563, normalized size = 1.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/9*b^2*x^(9/2)/d^2+4/5/d^2*x^(5/2)*a*b-4/5/d^3*x^(5/2)*b^2*c+2/d^2*a^2*x^(1/2)-8/d^3*c*a*b*x^(1/2)+6/d^4*b^2*
c^2*x^(1/2)+1/2*c/d^2*x^(1/2)/(d*x^2+c)*a^2-c^2/d^3*x^(1/2)/(d*x^2+c)*a*b+1/2*c^3/d^4*x^(1/2)/(d*x^2+c)*b^2-5/
8/d^2*(c/d)^(1/4)*2^(1/2)*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)+1)*a^2+9/4*c/d^3*(c/d)^(1/4)*2^(1/2)*arctan(2^(1/
2)/(c/d)^(1/4)*x^(1/2)+1)*a*b-13/8*c^2/d^4*(c/d)^(1/4)*2^(1/2)*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)+1)*b^2-5/8/d
^2*(c/d)^(1/4)*2^(1/2)*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)-1)*a^2+9/4*c/d^3*(c/d)^(1/4)*2^(1/2)*arctan(2^(1/2)/
(c/d)^(1/4)*x^(1/2)-1)*a*b-13/8*c^2/d^4*(c/d)^(1/4)*2^(1/2)*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)-1)*b^2-5/16/d^2
*(c/d)^(1/4)*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)
))*a^2+9/8*c/d^3*(c/d)^(1/4)*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)))*a*b-13/16*c^2/d^4*(c/d)^(1/4)*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)))*b^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^(7/2)*(b*x^2+a)^2/(d*x^2+c)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.20173, size = 3289, normalized size = 8.77 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/360*(180*(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)*arctan((sqrt(d^8*sqrt(-(28561*b^8*c^9 - 158184*a*b^7*c^8*d + 372476*a^2*b^6*c^7*d^2 - 48
5784*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) + (169*b^4*c^4 - 468*a*b^3*c^3*d + 454*a^2*b^2*c^2*d^2 - 180*a^3*b*c*d^3 + 25*a^4
*d^4)*x)*d^13*(-(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)^
(3/4) - (13*b^2*c^2*d^13 - 18*a*b*c*d^14 + 5*a^2*d^15)*sqrt(x)*(-(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)^(3/4))/(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 - 90
00*a^7*b*c^2*d^7 + 625*a^8*c*d^8)) + 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*(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)) - 4*(20*b^2*d^3*x^6 + 585*b^2*c^3 - 810*a*b*c^2*d + 225*
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)

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

[Out]

Timed out

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Giac [A]  time = 1.17975, size = 594, normalized size = 1.58 \begin{align*} -\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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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