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

Optimal. Leaf size=346 \[ \frac{x^{7/2} (b c-a d)^2}{2 c d^2 \left (c+d x^2\right )}-\frac{x^{3/2} (11 b c-3 a d) (b c-a d)}{6 c d^3}+\frac{(11 b c-3 a d) (b c-a d) \log \left (-\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{8 \sqrt{2} \sqrt [4]{c} d^{15/4}}-\frac{(11 b c-3 a d) (b c-a d) \log \left (\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{8 \sqrt{2} \sqrt [4]{c} d^{15/4}}-\frac{(11 b c-3 a d) (b c-a d) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}\right )}{4 \sqrt{2} \sqrt [4]{c} d^{15/4}}+\frac{(11 b c-3 a d) (b c-a d) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}+1\right )}{4 \sqrt{2} \sqrt [4]{c} d^{15/4}}+\frac{2 b^2 x^{7/2}}{7 d^2} \]

[Out]

-((11*b*c - 3*a*d)*(b*c - a*d)*x^(3/2))/(6*c*d^3) + (2*b^2*x^(7/2))/(7*d^2) + ((b*c - a*d)^2*x^(7/2))/(2*c*d^2
*(c + d*x^2)) - ((11*b*c - 3*a*d)*(b*c - a*d)*ArcTan[1 - (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/(4*Sqrt[2]*c^(1/4
)*d^(15/4)) + ((11*b*c - 3*a*d)*(b*c - a*d)*ArcTan[1 + (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/(4*Sqrt[2]*c^(1/4)*
d^(15/4)) + ((11*b*c - 3*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]*c^(1/4)*d^(15/4)) - ((11*b*c - 3*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]*c^(1/4)*d^(15/4))

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Rubi [A]  time = 0.317422, antiderivative size = 346, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 10, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used = {463, 459, 321, 329, 297, 1162, 617, 204, 1165, 628} \[ \frac{x^{7/2} (b c-a d)^2}{2 c d^2 \left (c+d x^2\right )}-\frac{x^{3/2} (11 b c-3 a d) (b c-a d)}{6 c d^3}+\frac{(11 b c-3 a d) (b c-a d) \log \left (-\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{8 \sqrt{2} \sqrt [4]{c} d^{15/4}}-\frac{(11 b c-3 a d) (b c-a d) \log \left (\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{8 \sqrt{2} \sqrt [4]{c} d^{15/4}}-\frac{(11 b c-3 a d) (b c-a d) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}\right )}{4 \sqrt{2} \sqrt [4]{c} d^{15/4}}+\frac{(11 b c-3 a d) (b c-a d) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}+1\right )}{4 \sqrt{2} \sqrt [4]{c} d^{15/4}}+\frac{2 b^2 x^{7/2}}{7 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((11*b*c - 3*a*d)*(b*c - a*d)*x^(3/2))/(6*c*d^3) + (2*b^2*x^(7/2))/(7*d^2) + ((b*c - a*d)^2*x^(7/2))/(2*c*d^2
*(c + d*x^2)) - ((11*b*c - 3*a*d)*(b*c - a*d)*ArcTan[1 - (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/(4*Sqrt[2]*c^(1/4
)*d^(15/4)) + ((11*b*c - 3*a*d)*(b*c - a*d)*ArcTan[1 + (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/(4*Sqrt[2]*c^(1/4)*
d^(15/4)) + ((11*b*c - 3*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]*c^(1/4)*d^(15/4)) - ((11*b*c - 3*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]*c^(1/4)*d^(15/4))

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 297

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

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])

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]

Rubi steps

\begin{align*} \int \frac{x^{5/2} \left (a+b x^2\right )^2}{\left (c+d x^2\right )^2} \, dx &=\frac{(b c-a d)^2 x^{7/2}}{2 c d^2 \left (c+d x^2\right )}-\frac{\int \frac{x^{5/2} \left (\frac{1}{2} \left (-4 a^2 d^2+7 (b c-a d)^2\right )-2 b^2 c d x^2\right )}{c+d x^2} \, dx}{2 c d^2}\\ &=\frac{2 b^2 x^{7/2}}{7 d^2}+\frac{(b c-a d)^2 x^{7/2}}{2 c d^2 \left (c+d x^2\right )}-\frac{((11 b c-3 a d) (b c-a d)) \int \frac{x^{5/2}}{c+d x^2} \, dx}{4 c d^2}\\ &=-\frac{(11 b c-3 a d) (b c-a d) x^{3/2}}{6 c d^3}+\frac{2 b^2 x^{7/2}}{7 d^2}+\frac{(b c-a d)^2 x^{7/2}}{2 c d^2 \left (c+d x^2\right )}+\frac{((11 b c-3 a d) (b c-a d)) \int \frac{\sqrt{x}}{c+d x^2} \, dx}{4 d^3}\\ &=-\frac{(11 b c-3 a d) (b c-a d) x^{3/2}}{6 c d^3}+\frac{2 b^2 x^{7/2}}{7 d^2}+\frac{(b c-a d)^2 x^{7/2}}{2 c d^2 \left (c+d x^2\right )}+\frac{((11 b c-3 a d) (b c-a d)) \operatorname{Subst}\left (\int \frac{x^2}{c+d x^4} \, dx,x,\sqrt{x}\right )}{2 d^3}\\ &=-\frac{(11 b c-3 a d) (b c-a d) x^{3/2}}{6 c d^3}+\frac{2 b^2 x^{7/2}}{7 d^2}+\frac{(b c-a d)^2 x^{7/2}}{2 c d^2 \left (c+d x^2\right )}-\frac{((11 b c-3 a d) (b c-a d)) \operatorname{Subst}\left (\int \frac{\sqrt{c}-\sqrt{d} x^2}{c+d x^4} \, dx,x,\sqrt{x}\right )}{4 d^{7/2}}+\frac{((11 b c-3 a d) (b c-a d)) \operatorname{Subst}\left (\int \frac{\sqrt{c}+\sqrt{d} x^2}{c+d x^4} \, dx,x,\sqrt{x}\right )}{4 d^{7/2}}\\ &=-\frac{(11 b c-3 a d) (b c-a d) x^{3/2}}{6 c d^3}+\frac{2 b^2 x^{7/2}}{7 d^2}+\frac{(b c-a d)^2 x^{7/2}}{2 c d^2 \left (c+d x^2\right )}+\frac{((11 b c-3 a d) (b c-a d)) \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^4}+\frac{((11 b c-3 a d) (b c-a d)) \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^4}+\frac{((11 b c-3 a d) (b c-a d)) \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} \sqrt [4]{c} d^{15/4}}+\frac{((11 b c-3 a d) (b c-a d)) \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} \sqrt [4]{c} d^{15/4}}\\ &=-\frac{(11 b c-3 a d) (b c-a d) x^{3/2}}{6 c d^3}+\frac{2 b^2 x^{7/2}}{7 d^2}+\frac{(b c-a d)^2 x^{7/2}}{2 c d^2 \left (c+d x^2\right )}+\frac{(11 b c-3 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} \sqrt [4]{c} d^{15/4}}-\frac{(11 b c-3 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} \sqrt [4]{c} d^{15/4}}+\frac{((11 b c-3 a d) (b c-a d)) \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} \sqrt [4]{c} d^{15/4}}-\frac{((11 b c-3 a d) (b c-a d)) \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} \sqrt [4]{c} d^{15/4}}\\ &=-\frac{(11 b c-3 a d) (b c-a d) x^{3/2}}{6 c d^3}+\frac{2 b^2 x^{7/2}}{7 d^2}+\frac{(b c-a d)^2 x^{7/2}}{2 c d^2 \left (c+d x^2\right )}-\frac{(11 b c-3 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} \sqrt [4]{c} d^{15/4}}+\frac{(11 b c-3 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} \sqrt [4]{c} d^{15/4}}+\frac{(11 b c-3 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} \sqrt [4]{c} d^{15/4}}-\frac{(11 b c-3 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} \sqrt [4]{c} d^{15/4}}\\ \end{align*}

Mathematica [A]  time = 0.182954, size = 337, normalized size = 0.97 \[ \frac{\frac{21 \sqrt{2} \left (3 a^2 d^2-14 a b c d+11 b^2 c^2\right ) \log \left (-\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{\sqrt [4]{c}}-\frac{21 \sqrt{2} \left (3 a^2 d^2-14 a b c d+11 b^2 c^2\right ) \log \left (\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{\sqrt [4]{c}}-\frac{42 \sqrt{2} \left (3 a^2 d^2-14 a b c d+11 b^2 c^2\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}\right )}{\sqrt [4]{c}}+\frac{42 \sqrt{2} \left (3 a^2 d^2-14 a b c d+11 b^2 c^2\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}+1\right )}{\sqrt [4]{c}}-448 b d^{3/4} x^{3/2} (b c-a d)-\frac{168 d^{3/4} x^{3/2} (b c-a d)^2}{c+d x^2}+96 b^2 d^{7/4} x^{7/2}}{336 d^{15/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-448*b*d^(3/4)*(b*c - a*d)*x^(3/2) + 96*b^2*d^(7/4)*x^(7/2) - (168*d^(3/4)*(b*c - a*d)^2*x^(3/2))/(c + d*x^2)
- (42*Sqrt[2]*(11*b^2*c^2 - 14*a*b*c*d + 3*a^2*d^2)*ArcTan[1 - (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/c^(1/4) +
(42*Sqrt[2]*(11*b^2*c^2 - 14*a*b*c*d + 3*a^2*d^2)*ArcTan[1 + (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/c^(1/4) + (21
*Sqrt[2]*(11*b^2*c^2 - 14*a*b*c*d + 3*a^2*d^2)*Log[Sqrt[c] - Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x])/c^(
1/4) - (21*Sqrt[2]*(11*b^2*c^2 - 14*a*b*c*d + 3*a^2*d^2)*Log[Sqrt[c] + Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[
d]*x])/c^(1/4))/(336*d^(15/4))

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Maple [A]  time = 0.017, size = 523, 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^(5/2)*(b*x^2+a)^2/(d*x^2+c)^2,x)

[Out]

2/7*b^2*x^(7/2)/d^2+4/3*b/d^2*x^(3/2)*a-4/3*b^2/d^3*x^(3/2)*c-1/2/d*x^(3/2)/(d*x^2+c)*a^2+1/d^2*x^(3/2)/(d*x^2
+c)*c*a*b-1/2/d^3*x^(3/2)/(d*x^2+c)*b^2*c^2-7/8/d^3/(c/d)^(1/4)*2^(1/2)*c*a*b*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)))-7/4/d^3/(c/d)^(1/4)*2^(1/2)*c*a*b*arctan(2^(1/2)/(
c/d)^(1/4)*x^(1/2)+1)-7/4/d^3/(c/d)^(1/4)*2^(1/2)*c*a*b*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)-1)+11/16/d^4/(c/d)^
(1/4)*2^(1/2)*b^2*c^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
)))+11/8/d^4/(c/d)^(1/4)*2^(1/2)*b^2*c^2*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)+1)+11/8/d^4/(c/d)^(1/4)*2^(1/2)*b^
2*c^2*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)-1)+3/16/d^2/(c/d)^(1/4)*2^(1/2)*a^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)))+3/8/d^2/(c/d)^(1/4)*2^(1/2)*a^2*arctan(2^(1/2)/(c/d
)^(1/4)*x^(1/2)+1)+3/8/d^2/(c/d)^(1/4)*2^(1/2)*a^2*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)-1)

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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^(5/2)*(b*x^2+a)^2/(d*x^2+c)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.2089, size = 4166, normalized size = 12.04 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(b*x^2+a)^2/(d*x^2+c)^2,x, algorithm="fricas")

[Out]

-1/168*(84*(d^4*x^2 + c*d^3)*(-(14641*b^8*c^8 - 74536*a*b^7*c^7*d + 158268*a^2*b^6*c^6*d^2 - 181720*a^3*b^5*c^
5*d^3 + 122566*a^4*b^4*c^4*d^4 - 49560*a^5*b^3*c^3*d^5 + 11772*a^6*b^2*c^2*d^6 - 1512*a^7*b*c*d^7 + 81*a^8*d^8
)/(c*d^15))^(1/4)*arctan((sqrt((1771561*b^12*c^12 - 13528284*a*b^11*c^11*d + 45943458*a^2*b^10*c^10*d^2 - 9149
2940*a^3*b^9*c^9*d^3 + 118659255*a^4*b^8*c^8*d^4 - 105323064*a^5*b^7*c^7*d^5 + 65490076*a^6*b^6*c^6*d^6 - 2872
4472*a^7*b^5*c^5*d^7 + 8825895*a^8*b^4*c^4*d^8 - 1855980*a^9*b^3*c^3*d^9 + 254178*a^10*b^2*c^2*d^10 - 20412*a^
11*b*c*d^11 + 729*a^12*d^12)*x - (14641*b^8*c^9*d^7 - 74536*a*b^7*c^8*d^8 + 158268*a^2*b^6*c^7*d^9 - 181720*a^
3*b^5*c^6*d^10 + 122566*a^4*b^4*c^5*d^11 - 49560*a^5*b^3*c^4*d^12 + 11772*a^6*b^2*c^3*d^13 - 1512*a^7*b*c^2*d^
14 + 81*a^8*c*d^15)*sqrt(-(14641*b^8*c^8 - 74536*a*b^7*c^7*d + 158268*a^2*b^6*c^6*d^2 - 181720*a^3*b^5*c^5*d^3
+ 122566*a^4*b^4*c^4*d^4 - 49560*a^5*b^3*c^3*d^5 + 11772*a^6*b^2*c^2*d^6 - 1512*a^7*b*c*d^7 + 81*a^8*d^8)/(c*
d^15)))*d^4*(-(14641*b^8*c^8 - 74536*a*b^7*c^7*d + 158268*a^2*b^6*c^6*d^2 - 181720*a^3*b^5*c^5*d^3 + 122566*a^
4*b^4*c^4*d^4 - 49560*a^5*b^3*c^3*d^5 + 11772*a^6*b^2*c^2*d^6 - 1512*a^7*b*c*d^7 + 81*a^8*d^8)/(c*d^15))^(1/4)
- (1331*b^6*c^6*d^4 - 5082*a*b^5*c^5*d^5 + 7557*a^2*b^4*c^4*d^6 - 5516*a^3*b^3*c^3*d^7 + 2061*a^4*b^2*c^2*d^8
- 378*a^5*b*c*d^9 + 27*a^6*d^10)*sqrt(x)*(-(14641*b^8*c^8 - 74536*a*b^7*c^7*d + 158268*a^2*b^6*c^6*d^2 - 1817
20*a^3*b^5*c^5*d^3 + 122566*a^4*b^4*c^4*d^4 - 49560*a^5*b^3*c^3*d^5 + 11772*a^6*b^2*c^2*d^6 - 1512*a^7*b*c*d^7
+ 81*a^8*d^8)/(c*d^15))^(1/4))/(14641*b^8*c^8 - 74536*a*b^7*c^7*d + 158268*a^2*b^6*c^6*d^2 - 181720*a^3*b^5*c
^5*d^3 + 122566*a^4*b^4*c^4*d^4 - 49560*a^5*b^3*c^3*d^5 + 11772*a^6*b^2*c^2*d^6 - 1512*a^7*b*c*d^7 + 81*a^8*d^
8)) - 21*(d^4*x^2 + c*d^3)*(-(14641*b^8*c^8 - 74536*a*b^7*c^7*d + 158268*a^2*b^6*c^6*d^2 - 181720*a^3*b^5*c^5*
d^3 + 122566*a^4*b^4*c^4*d^4 - 49560*a^5*b^3*c^3*d^5 + 11772*a^6*b^2*c^2*d^6 - 1512*a^7*b*c*d^7 + 81*a^8*d^8)/
(c*d^15))^(1/4)*log(c*d^11*(-(14641*b^8*c^8 - 74536*a*b^7*c^7*d + 158268*a^2*b^6*c^6*d^2 - 181720*a^3*b^5*c^5*
d^3 + 122566*a^4*b^4*c^4*d^4 - 49560*a^5*b^3*c^3*d^5 + 11772*a^6*b^2*c^2*d^6 - 1512*a^7*b*c*d^7 + 81*a^8*d^8)/
(c*d^15))^(3/4) + (1331*b^6*c^6 - 5082*a*b^5*c^5*d + 7557*a^2*b^4*c^4*d^2 - 5516*a^3*b^3*c^3*d^3 + 2061*a^4*b^
2*c^2*d^4 - 378*a^5*b*c*d^5 + 27*a^6*d^6)*sqrt(x)) + 21*(d^4*x^2 + c*d^3)*(-(14641*b^8*c^8 - 74536*a*b^7*c^7*d
+ 158268*a^2*b^6*c^6*d^2 - 181720*a^3*b^5*c^5*d^3 + 122566*a^4*b^4*c^4*d^4 - 49560*a^5*b^3*c^3*d^5 + 11772*a^
6*b^2*c^2*d^6 - 1512*a^7*b*c*d^7 + 81*a^8*d^8)/(c*d^15))^(1/4)*log(-c*d^11*(-(14641*b^8*c^8 - 74536*a*b^7*c^7*
d + 158268*a^2*b^6*c^6*d^2 - 181720*a^3*b^5*c^5*d^3 + 122566*a^4*b^4*c^4*d^4 - 49560*a^5*b^3*c^3*d^5 + 11772*a
^6*b^2*c^2*d^6 - 1512*a^7*b*c*d^7 + 81*a^8*d^8)/(c*d^15))^(3/4) + (1331*b^6*c^6 - 5082*a*b^5*c^5*d + 7557*a^2*
b^4*c^4*d^2 - 5516*a^3*b^3*c^3*d^3 + 2061*a^4*b^2*c^2*d^4 - 378*a^5*b*c*d^5 + 27*a^6*d^6)*sqrt(x)) - 4*(12*b^2
*d^2*x^5 - 4*(11*b^2*c*d - 14*a*b*d^2)*x^3 - 7*(11*b^2*c^2 - 14*a*b*c*d + 3*a^2*d^2)*x)*sqrt(x))/(d^4*x^2 + c*
d^3)

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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**(5/2)*(b*x**2+a)**2/(d*x**2+c)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.21053, size = 558, normalized size = 1.61 \begin{align*} -\frac{b^{2} c^{2} x^{\frac{3}{2}} - 2 \, a b c d x^{\frac{3}{2}} + a^{2} d^{2} x^{\frac{3}{2}}}{2 \,{\left (d x^{2} + c\right )} d^{3}} + \frac{\sqrt{2}{\left (11 \, \left (c d^{3}\right )^{\frac{3}{4}} b^{2} c^{2} - 14 \, \left (c d^{3}\right )^{\frac{3}{4}} a b c d + 3 \, \left (c d^{3}\right )^{\frac{3}{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 \, c d^{6}} + \frac{\sqrt{2}{\left (11 \, \left (c d^{3}\right )^{\frac{3}{4}} b^{2} c^{2} - 14 \, \left (c d^{3}\right )^{\frac{3}{4}} a b c d + 3 \, \left (c d^{3}\right )^{\frac{3}{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 \, c d^{6}} - \frac{\sqrt{2}{\left (11 \, \left (c d^{3}\right )^{\frac{3}{4}} b^{2} c^{2} - 14 \, \left (c d^{3}\right )^{\frac{3}{4}} a b c d + 3 \, \left (c d^{3}\right )^{\frac{3}{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 \, c d^{6}} + \frac{\sqrt{2}{\left (11 \, \left (c d^{3}\right )^{\frac{3}{4}} b^{2} c^{2} - 14 \, \left (c d^{3}\right )^{\frac{3}{4}} a b c d + 3 \, \left (c d^{3}\right )^{\frac{3}{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 \, c d^{6}} + \frac{2 \,{\left (3 \, b^{2} d^{12} x^{\frac{7}{2}} - 14 \, b^{2} c d^{11} x^{\frac{3}{2}} + 14 \, a b d^{12} x^{\frac{3}{2}}\right )}}{21 \, d^{14}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(b*x^2+a)^2/(d*x^2+c)^2,x, algorithm="giac")

[Out]

-1/2*(b^2*c^2*x^(3/2) - 2*a*b*c*d*x^(3/2) + a^2*d^2*x^(3/2))/((d*x^2 + c)*d^3) + 1/8*sqrt(2)*(11*(c*d^3)^(3/4)
*b^2*c^2 - 14*(c*d^3)^(3/4)*a*b*c*d + 3*(c*d^3)^(3/4)*a^2*d^2)*arctan(1/2*sqrt(2)*(sqrt(2)*(c/d)^(1/4) + 2*sqr
t(x))/(c/d)^(1/4))/(c*d^6) + 1/8*sqrt(2)*(11*(c*d^3)^(3/4)*b^2*c^2 - 14*(c*d^3)^(3/4)*a*b*c*d + 3*(c*d^3)^(3/4
)*a^2*d^2)*arctan(-1/2*sqrt(2)*(sqrt(2)*(c/d)^(1/4) - 2*sqrt(x))/(c/d)^(1/4))/(c*d^6) - 1/16*sqrt(2)*(11*(c*d^
3)^(3/4)*b^2*c^2 - 14*(c*d^3)^(3/4)*a*b*c*d + 3*(c*d^3)^(3/4)*a^2*d^2)*log(sqrt(2)*sqrt(x)*(c/d)^(1/4) + x + s
qrt(c/d))/(c*d^6) + 1/16*sqrt(2)*(11*(c*d^3)^(3/4)*b^2*c^2 - 14*(c*d^3)^(3/4)*a*b*c*d + 3*(c*d^3)^(3/4)*a^2*d^
2)*log(-sqrt(2)*sqrt(x)*(c/d)^(1/4) + x + sqrt(c/d))/(c*d^6) + 2/21*(3*b^2*d^12*x^(7/2) - 14*b^2*c*d^11*x^(3/2
) + 14*a*b*d^12*x^(3/2))/d^14