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

Optimal. Leaf size=290 \[ -\frac{c^{3/4} (b c-a d)^2 \log \left (-\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{2 \sqrt{2} d^{15/4}}+\frac{c^{3/4} (b c-a d)^2 \log \left (\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{2 \sqrt{2} d^{15/4}}+\frac{c^{3/4} (b c-a d)^2 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}\right )}{\sqrt{2} d^{15/4}}-\frac{c^{3/4} (b c-a d)^2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}+1\right )}{\sqrt{2} d^{15/4}}-\frac{2 b x^{7/2} (b c-2 a d)}{7 d^2}+\frac{2 x^{3/2} (b c-a d)^2}{3 d^3}+\frac{2 b^2 x^{11/2}}{11 d} \]

[Out]

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

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Rubi [A]  time = 0.254077, antiderivative size = 290, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 9, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {461, 321, 329, 297, 1162, 617, 204, 1165, 628} \[ -\frac{c^{3/4} (b c-a d)^2 \log \left (-\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{2 \sqrt{2} d^{15/4}}+\frac{c^{3/4} (b c-a d)^2 \log \left (\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{2 \sqrt{2} d^{15/4}}+\frac{c^{3/4} (b c-a d)^2 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}\right )}{\sqrt{2} d^{15/4}}-\frac{c^{3/4} (b c-a d)^2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}+1\right )}{\sqrt{2} d^{15/4}}-\frac{2 b x^{7/2} (b c-2 a d)}{7 d^2}+\frac{2 x^{3/2} (b c-a d)^2}{3 d^3}+\frac{2 b^2 x^{11/2}}{11 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 461

Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegr
and[((e*x)^m*(a + b*x^n)^p)/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n
, 0] && IGtQ[p, 0] && (IntegerQ[m] || IGtQ[2*(m + 1), 0] ||  !RationalQ[m])

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}{c+d x^2} \, dx &=\int \left (-\frac{b (b c-2 a d) x^{5/2}}{d^2}+\frac{b^2 x^{9/2}}{d}+\frac{\left (b^2 c^2-2 a b c d+a^2 d^2\right ) x^{5/2}}{d^2 \left (c+d x^2\right )}\right ) \, dx\\ &=-\frac{2 b (b c-2 a d) x^{7/2}}{7 d^2}+\frac{2 b^2 x^{11/2}}{11 d}+\frac{(b c-a d)^2 \int \frac{x^{5/2}}{c+d x^2} \, dx}{d^2}\\ &=\frac{2 (b c-a d)^2 x^{3/2}}{3 d^3}-\frac{2 b (b c-2 a d) x^{7/2}}{7 d^2}+\frac{2 b^2 x^{11/2}}{11 d}-\frac{\left (c (b c-a d)^2\right ) \int \frac{\sqrt{x}}{c+d x^2} \, dx}{d^3}\\ &=\frac{2 (b c-a d)^2 x^{3/2}}{3 d^3}-\frac{2 b (b c-2 a d) x^{7/2}}{7 d^2}+\frac{2 b^2 x^{11/2}}{11 d}-\frac{\left (2 c (b c-a d)^2\right ) \operatorname{Subst}\left (\int \frac{x^2}{c+d x^4} \, dx,x,\sqrt{x}\right )}{d^3}\\ &=\frac{2 (b c-a d)^2 x^{3/2}}{3 d^3}-\frac{2 b (b c-2 a d) x^{7/2}}{7 d^2}+\frac{2 b^2 x^{11/2}}{11 d}+\frac{\left (c (b c-a d)^2\right ) \operatorname{Subst}\left (\int \frac{\sqrt{c}-\sqrt{d} x^2}{c+d x^4} \, dx,x,\sqrt{x}\right )}{d^{7/2}}-\frac{\left (c (b c-a d)^2\right ) \operatorname{Subst}\left (\int \frac{\sqrt{c}+\sqrt{d} x^2}{c+d x^4} \, dx,x,\sqrt{x}\right )}{d^{7/2}}\\ &=\frac{2 (b c-a d)^2 x^{3/2}}{3 d^3}-\frac{2 b (b c-2 a d) x^{7/2}}{7 d^2}+\frac{2 b^2 x^{11/2}}{11 d}-\frac{\left (c (b c-a d)^2\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 )}{2 d^4}-\frac{\left (c (b c-a d)^2\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 )}{2 d^4}-\frac{\left (c^{3/4} (b c-a d)^2\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 )}{2 \sqrt{2} d^{15/4}}-\frac{\left (c^{3/4} (b c-a d)^2\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 )}{2 \sqrt{2} d^{15/4}}\\ &=\frac{2 (b c-a d)^2 x^{3/2}}{3 d^3}-\frac{2 b (b c-2 a d) x^{7/2}}{7 d^2}+\frac{2 b^2 x^{11/2}}{11 d}-\frac{c^{3/4} (b c-a d)^2 \log \left (\sqrt{c}-\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{d} x\right )}{2 \sqrt{2} d^{15/4}}+\frac{c^{3/4} (b c-a d)^2 \log \left (\sqrt{c}+\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{d} x\right )}{2 \sqrt{2} d^{15/4}}-\frac{\left (c^{3/4} (b c-a d)^2\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 )}{\sqrt{2} d^{15/4}}+\frac{\left (c^{3/4} (b c-a d)^2\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 )}{\sqrt{2} d^{15/4}}\\ &=\frac{2 (b c-a d)^2 x^{3/2}}{3 d^3}-\frac{2 b (b c-2 a d) x^{7/2}}{7 d^2}+\frac{2 b^2 x^{11/2}}{11 d}+\frac{c^{3/4} (b c-a d)^2 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}\right )}{\sqrt{2} d^{15/4}}-\frac{c^{3/4} (b c-a d)^2 \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}\right )}{\sqrt{2} d^{15/4}}-\frac{c^{3/4} (b c-a d)^2 \log \left (\sqrt{c}-\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{d} x\right )}{2 \sqrt{2} d^{15/4}}+\frac{c^{3/4} (b c-a d)^2 \log \left (\sqrt{c}+\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{d} x\right )}{2 \sqrt{2} d^{15/4}}\\ \end{align*}

Mathematica [A]  time = 0.104447, size = 276, normalized size = 0.95 \[ \frac{-231 \sqrt{2} c^{3/4} (b c-a d)^2 \log \left (-\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )+231 \sqrt{2} c^{3/4} (b c-a d)^2 \log \left (\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )+462 \sqrt{2} c^{3/4} (b c-a d)^2 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}\right )-462 \sqrt{2} c^{3/4} (b c-a d)^2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}+1\right )-264 b d^{7/4} x^{7/2} (b c-2 a d)+616 d^{3/4} x^{3/2} (b c-a d)^2+168 b^2 d^{11/4} x^{11/2}}{924 d^{15/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(616*d^(3/4)*(b*c - a*d)^2*x^(3/2) - 264*b*d^(7/4)*(b*c - 2*a*d)*x^(7/2) + 168*b^2*d^(11/4)*x^(11/2) + 462*Sqr
t[2]*c^(3/4)*(b*c - a*d)^2*ArcTan[1 - (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)] - 462*Sqrt[2]*c^(3/4)*(b*c - a*d)^2*A
rcTan[1 + (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)] - 231*Sqrt[2]*c^(3/4)*(b*c - a*d)^2*Log[Sqrt[c] - Sqrt[2]*c^(1/4)
*d^(1/4)*Sqrt[x] + Sqrt[d]*x] + 231*Sqrt[2]*c^(3/4)*(b*c - a*d)^2*Log[Sqrt[c] + Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x
] + Sqrt[d]*x])/(924*d^(15/4))

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Maple [B]  time = 0.01, size = 504, normalized size = 1.7 \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),x)

[Out]

2/11*b^2*x^(11/2)/d+4/7/d*x^(7/2)*a*b-2/7/d^2*x^(7/2)*b^2*c+2/3/d*x^(3/2)*a^2-4/3/d^2*x^(3/2)*c*a*b+2/3/d^3*x^
(3/2)*b^2*c^2-1/2*c/d^2/(c/d)^(1/4)*2^(1/2)*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)+1)*a^2+c^2/d^3/(c/d)^(1/4)*2^(1
/2)*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)+1)*a*b-1/2*c^3/d^4/(c/d)^(1/4)*2^(1/2)*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/
2)+1)*b^2-1/2*c/d^2/(c/d)^(1/4)*2^(1/2)*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)-1)*a^2+c^2/d^3/(c/d)^(1/4)*2^(1/2)*
arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)-1)*a*b-1/2*c^3/d^4/(c/d)^(1/4)*2^(1/2)*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)-1
)*b^2-1/4*c/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+1/2*c^2/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-1/4*c^3/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^(5/2)*(b*x^2+a)^2/(d*x^2+c),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.18633, size = 3549, normalized size = 12.24 \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),x, algorithm="fricas")

[Out]

1/462*(924*d^3*(-(b^8*c^11 - 8*a*b^7*c^10*d + 28*a^2*b^6*c^9*d^2 - 56*a^3*b^5*c^8*d^3 + 70*a^4*b^4*c^7*d^4 - 5
6*a^5*b^3*c^6*d^5 + 28*a^6*b^2*c^5*d^6 - 8*a^7*b*c^4*d^7 + a^8*c^3*d^8)/d^15)^(1/4)*arctan((sqrt((b^12*c^16 -
12*a*b^11*c^15*d + 66*a^2*b^10*c^14*d^2 - 220*a^3*b^9*c^13*d^3 + 495*a^4*b^8*c^12*d^4 - 792*a^5*b^7*c^11*d^5 +
924*a^6*b^6*c^10*d^6 - 792*a^7*b^5*c^9*d^7 + 495*a^8*b^4*c^8*d^8 - 220*a^9*b^3*c^7*d^9 + 66*a^10*b^2*c^6*d^10
- 12*a^11*b*c^5*d^11 + a^12*c^4*d^12)*x - (b^8*c^11*d^7 - 8*a*b^7*c^10*d^8 + 28*a^2*b^6*c^9*d^9 - 56*a^3*b^5*
c^8*d^10 + 70*a^4*b^4*c^7*d^11 - 56*a^5*b^3*c^6*d^12 + 28*a^6*b^2*c^5*d^13 - 8*a^7*b*c^4*d^14 + a^8*c^3*d^15)*
sqrt(-(b^8*c^11 - 8*a*b^7*c^10*d + 28*a^2*b^6*c^9*d^2 - 56*a^3*b^5*c^8*d^3 + 70*a^4*b^4*c^7*d^4 - 56*a^5*b^3*c
^6*d^5 + 28*a^6*b^2*c^5*d^6 - 8*a^7*b*c^4*d^7 + a^8*c^3*d^8)/d^15))*d^4*(-(b^8*c^11 - 8*a*b^7*c^10*d + 28*a^2*
b^6*c^9*d^2 - 56*a^3*b^5*c^8*d^3 + 70*a^4*b^4*c^7*d^4 - 56*a^5*b^3*c^6*d^5 + 28*a^6*b^2*c^5*d^6 - 8*a^7*b*c^4*
d^7 + a^8*c^3*d^8)/d^15)^(1/4) - (b^6*c^8*d^4 - 6*a*b^5*c^7*d^5 + 15*a^2*b^4*c^6*d^6 - 20*a^3*b^3*c^5*d^7 + 15
*a^4*b^2*c^4*d^8 - 6*a^5*b*c^3*d^9 + a^6*c^2*d^10)*sqrt(x)*(-(b^8*c^11 - 8*a*b^7*c^10*d + 28*a^2*b^6*c^9*d^2 -
56*a^3*b^5*c^8*d^3 + 70*a^4*b^4*c^7*d^4 - 56*a^5*b^3*c^6*d^5 + 28*a^6*b^2*c^5*d^6 - 8*a^7*b*c^4*d^7 + a^8*c^3
*d^8)/d^15)^(1/4))/(b^8*c^11 - 8*a*b^7*c^10*d + 28*a^2*b^6*c^9*d^2 - 56*a^3*b^5*c^8*d^3 + 70*a^4*b^4*c^7*d^4 -
56*a^5*b^3*c^6*d^5 + 28*a^6*b^2*c^5*d^6 - 8*a^7*b*c^4*d^7 + a^8*c^3*d^8)) - 231*d^3*(-(b^8*c^11 - 8*a*b^7*c^1
0*d + 28*a^2*b^6*c^9*d^2 - 56*a^3*b^5*c^8*d^3 + 70*a^4*b^4*c^7*d^4 - 56*a^5*b^3*c^6*d^5 + 28*a^6*b^2*c^5*d^6 -
8*a^7*b*c^4*d^7 + a^8*c^3*d^8)/d^15)^(1/4)*log(d^11*(-(b^8*c^11 - 8*a*b^7*c^10*d + 28*a^2*b^6*c^9*d^2 - 56*a^
3*b^5*c^8*d^3 + 70*a^4*b^4*c^7*d^4 - 56*a^5*b^3*c^6*d^5 + 28*a^6*b^2*c^5*d^6 - 8*a^7*b*c^4*d^7 + a^8*c^3*d^8)/
d^15)^(3/4) + (b^6*c^8 - 6*a*b^5*c^7*d + 15*a^2*b^4*c^6*d^2 - 20*a^3*b^3*c^5*d^3 + 15*a^4*b^2*c^4*d^4 - 6*a^5*
b*c^3*d^5 + a^6*c^2*d^6)*sqrt(x)) + 231*d^3*(-(b^8*c^11 - 8*a*b^7*c^10*d + 28*a^2*b^6*c^9*d^2 - 56*a^3*b^5*c^8
*d^3 + 70*a^4*b^4*c^7*d^4 - 56*a^5*b^3*c^6*d^5 + 28*a^6*b^2*c^5*d^6 - 8*a^7*b*c^4*d^7 + a^8*c^3*d^8)/d^15)^(1/
4)*log(-d^11*(-(b^8*c^11 - 8*a*b^7*c^10*d + 28*a^2*b^6*c^9*d^2 - 56*a^3*b^5*c^8*d^3 + 70*a^4*b^4*c^7*d^4 - 56*
a^5*b^3*c^6*d^5 + 28*a^6*b^2*c^5*d^6 - 8*a^7*b*c^4*d^7 + a^8*c^3*d^8)/d^15)^(3/4) + (b^6*c^8 - 6*a*b^5*c^7*d +
15*a^2*b^4*c^6*d^2 - 20*a^3*b^3*c^5*d^3 + 15*a^4*b^2*c^4*d^4 - 6*a^5*b*c^3*d^5 + a^6*c^2*d^6)*sqrt(x)) + 4*(2
1*b^2*d^2*x^5 - 33*(b^2*c*d - 2*a*b*d^2)*x^3 + 77*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*x)*sqrt(x))/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),x)

[Out]

Timed out

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Giac [A]  time = 1.20272, size = 520, normalized size = 1.79 \begin{align*} -\frac{\sqrt{2}{\left (\left (c d^{3}\right )^{\frac{3}{4}} b^{2} c^{2} - 2 \, \left (c d^{3}\right )^{\frac{3}{4}} a b c d + \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 )}{2 \, d^{6}} - \frac{\sqrt{2}{\left (\left (c d^{3}\right )^{\frac{3}{4}} b^{2} c^{2} - 2 \, \left (c d^{3}\right )^{\frac{3}{4}} a b c d + \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 )}{2 \, d^{6}} + \frac{\sqrt{2}{\left (\left (c d^{3}\right )^{\frac{3}{4}} b^{2} c^{2} - 2 \, \left (c d^{3}\right )^{\frac{3}{4}} a b c d + \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 )}{4 \, d^{6}} - \frac{\sqrt{2}{\left (\left (c d^{3}\right )^{\frac{3}{4}} b^{2} c^{2} - 2 \, \left (c d^{3}\right )^{\frac{3}{4}} a b c d + \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 )}{4 \, d^{6}} + \frac{2 \,{\left (21 \, b^{2} d^{10} x^{\frac{11}{2}} - 33 \, b^{2} c d^{9} x^{\frac{7}{2}} + 66 \, a b d^{10} x^{\frac{7}{2}} + 77 \, b^{2} c^{2} d^{8} x^{\frac{3}{2}} - 154 \, a b c d^{9} x^{\frac{3}{2}} + 77 \, a^{2} d^{10} x^{\frac{3}{2}}\right )}}{231 \, d^{11}} \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),x, algorithm="giac")

[Out]

-1/2*sqrt(2)*((c*d^3)^(3/4)*b^2*c^2 - 2*(c*d^3)^(3/4)*a*b*c*d + (c*d^3)^(3/4)*a^2*d^2)*arctan(1/2*sqrt(2)*(sqr
t(2)*(c/d)^(1/4) + 2*sqrt(x))/(c/d)^(1/4))/d^6 - 1/2*sqrt(2)*((c*d^3)^(3/4)*b^2*c^2 - 2*(c*d^3)^(3/4)*a*b*c*d
+ (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))/d^6 + 1/4*sqrt(2)*
((c*d^3)^(3/4)*b^2*c^2 - 2*(c*d^3)^(3/4)*a*b*c*d + (c*d^3)^(3/4)*a^2*d^2)*log(sqrt(2)*sqrt(x)*(c/d)^(1/4) + x
+ sqrt(c/d))/d^6 - 1/4*sqrt(2)*((c*d^3)^(3/4)*b^2*c^2 - 2*(c*d^3)^(3/4)*a*b*c*d + (c*d^3)^(3/4)*a^2*d^2)*log(-
sqrt(2)*sqrt(x)*(c/d)^(1/4) + x + sqrt(c/d))/d^6 + 2/231*(21*b^2*d^10*x^(11/2) - 33*b^2*c*d^9*x^(7/2) + 66*a*b
*d^10*x^(7/2) + 77*b^2*c^2*d^8*x^(3/2) - 154*a*b*c*d^9*x^(3/2) + 77*a^2*d^10*x^(3/2))/d^11