3.441 \(\int \frac{x^{5/2} (c+d x^2)^3}{a+b x^2} \, dx\)

Optimal. Leaf size=328 \[ \frac{2 d x^{7/2} \left (a^2 d^2-3 a b c d+3 b^2 c^2\right )}{7 b^3}-\frac{a^{3/4} (b c-a d)^3 \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{2 \sqrt{2} b^{19/4}}+\frac{a^{3/4} (b c-a d)^3 \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{2 \sqrt{2} b^{19/4}}+\frac{a^{3/4} (b c-a d)^3 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{\sqrt{2} b^{19/4}}-\frac{a^{3/4} (b c-a d)^3 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{\sqrt{2} b^{19/4}}+\frac{2 d^2 x^{11/2} (3 b c-a d)}{11 b^2}+\frac{2 x^{3/2} (b c-a d)^3}{3 b^4}+\frac{2 d^3 x^{15/2}}{15 b} \]

[Out]

(2*(b*c - a*d)^3*x^(3/2))/(3*b^4) + (2*d*(3*b^2*c^2 - 3*a*b*c*d + a^2*d^2)*x^(7/2))/(7*b^3) + (2*d^2*(3*b*c -
a*d)*x^(11/2))/(11*b^2) + (2*d^3*x^(15/2))/(15*b) + (a^(3/4)*(b*c - a*d)^3*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x]
)/a^(1/4)])/(Sqrt[2]*b^(19/4)) - (a^(3/4)*(b*c - a*d)^3*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(Sqrt[2
]*b^(19/4)) - (a^(3/4)*(b*c - a*d)^3*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(2*Sqrt[2]*b^
(19/4)) + (a^(3/4)*(b*c - a*d)^3*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(2*Sqrt[2]*b^(19/
4))

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Rubi [A]  time = 0.293491, antiderivative size = 328, 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} \[ \frac{2 d x^{7/2} \left (a^2 d^2-3 a b c d+3 b^2 c^2\right )}{7 b^3}-\frac{a^{3/4} (b c-a d)^3 \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{2 \sqrt{2} b^{19/4}}+\frac{a^{3/4} (b c-a d)^3 \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{2 \sqrt{2} b^{19/4}}+\frac{a^{3/4} (b c-a d)^3 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{\sqrt{2} b^{19/4}}-\frac{a^{3/4} (b c-a d)^3 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{\sqrt{2} b^{19/4}}+\frac{2 d^2 x^{11/2} (3 b c-a d)}{11 b^2}+\frac{2 x^{3/2} (b c-a d)^3}{3 b^4}+\frac{2 d^3 x^{15/2}}{15 b} \]

Antiderivative was successfully verified.

[In]

Int[(x^(5/2)*(c + d*x^2)^3)/(a + b*x^2),x]

[Out]

(2*(b*c - a*d)^3*x^(3/2))/(3*b^4) + (2*d*(3*b^2*c^2 - 3*a*b*c*d + a^2*d^2)*x^(7/2))/(7*b^3) + (2*d^2*(3*b*c -
a*d)*x^(11/2))/(11*b^2) + (2*d^3*x^(15/2))/(15*b) + (a^(3/4)*(b*c - a*d)^3*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x]
)/a^(1/4)])/(Sqrt[2]*b^(19/4)) - (a^(3/4)*(b*c - a*d)^3*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(Sqrt[2
]*b^(19/4)) - (a^(3/4)*(b*c - a*d)^3*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(2*Sqrt[2]*b^
(19/4)) + (a^(3/4)*(b*c - a*d)^3*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(2*Sqrt[2]*b^(19/
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 (c+d x^2\right )^3}{a+b x^2} \, dx &=\int \left (\frac{d \left (3 b^2 c^2-3 a b c d+a^2 d^2\right ) x^{5/2}}{b^3}+\frac{d^2 (3 b c-a d) x^{9/2}}{b^2}+\frac{d^3 x^{13/2}}{b}+\frac{\left (b^3 c^3-3 a b^2 c^2 d+3 a^2 b c d^2-a^3 d^3\right ) x^{5/2}}{b^3 \left (a+b x^2\right )}\right ) \, dx\\ &=\frac{2 d \left (3 b^2 c^2-3 a b c d+a^2 d^2\right ) x^{7/2}}{7 b^3}+\frac{2 d^2 (3 b c-a d) x^{11/2}}{11 b^2}+\frac{2 d^3 x^{15/2}}{15 b}+\frac{(b c-a d)^3 \int \frac{x^{5/2}}{a+b x^2} \, dx}{b^3}\\ &=\frac{2 (b c-a d)^3 x^{3/2}}{3 b^4}+\frac{2 d \left (3 b^2 c^2-3 a b c d+a^2 d^2\right ) x^{7/2}}{7 b^3}+\frac{2 d^2 (3 b c-a d) x^{11/2}}{11 b^2}+\frac{2 d^3 x^{15/2}}{15 b}-\frac{\left (a (b c-a d)^3\right ) \int \frac{\sqrt{x}}{a+b x^2} \, dx}{b^4}\\ &=\frac{2 (b c-a d)^3 x^{3/2}}{3 b^4}+\frac{2 d \left (3 b^2 c^2-3 a b c d+a^2 d^2\right ) x^{7/2}}{7 b^3}+\frac{2 d^2 (3 b c-a d) x^{11/2}}{11 b^2}+\frac{2 d^3 x^{15/2}}{15 b}-\frac{\left (2 a (b c-a d)^3\right ) \operatorname{Subst}\left (\int \frac{x^2}{a+b x^4} \, dx,x,\sqrt{x}\right )}{b^4}\\ &=\frac{2 (b c-a d)^3 x^{3/2}}{3 b^4}+\frac{2 d \left (3 b^2 c^2-3 a b c d+a^2 d^2\right ) x^{7/2}}{7 b^3}+\frac{2 d^2 (3 b c-a d) x^{11/2}}{11 b^2}+\frac{2 d^3 x^{15/2}}{15 b}+\frac{\left (a (b c-a d)^3\right ) \operatorname{Subst}\left (\int \frac{\sqrt{a}-\sqrt{b} x^2}{a+b x^4} \, dx,x,\sqrt{x}\right )}{b^{9/2}}-\frac{\left (a (b c-a d)^3\right ) \operatorname{Subst}\left (\int \frac{\sqrt{a}+\sqrt{b} x^2}{a+b x^4} \, dx,x,\sqrt{x}\right )}{b^{9/2}}\\ &=\frac{2 (b c-a d)^3 x^{3/2}}{3 b^4}+\frac{2 d \left (3 b^2 c^2-3 a b c d+a^2 d^2\right ) x^{7/2}}{7 b^3}+\frac{2 d^2 (3 b c-a d) x^{11/2}}{11 b^2}+\frac{2 d^3 x^{15/2}}{15 b}-\frac{\left (a (b c-a d)^3\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{\sqrt{a}}{\sqrt{b}}-\frac{\sqrt{2} \sqrt [4]{a} x}{\sqrt [4]{b}}+x^2} \, dx,x,\sqrt{x}\right )}{2 b^5}-\frac{\left (a (b c-a d)^3\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{\sqrt{a}}{\sqrt{b}}+\frac{\sqrt{2} \sqrt [4]{a} x}{\sqrt [4]{b}}+x^2} \, dx,x,\sqrt{x}\right )}{2 b^5}-\frac{\left (a^{3/4} (b c-a d)^3\right ) \operatorname{Subst}\left (\int \frac{\frac{\sqrt{2} \sqrt [4]{a}}{\sqrt [4]{b}}+2 x}{-\frac{\sqrt{a}}{\sqrt{b}}-\frac{\sqrt{2} \sqrt [4]{a} x}{\sqrt [4]{b}}-x^2} \, dx,x,\sqrt{x}\right )}{2 \sqrt{2} b^{19/4}}-\frac{\left (a^{3/4} (b c-a d)^3\right ) \operatorname{Subst}\left (\int \frac{\frac{\sqrt{2} \sqrt [4]{a}}{\sqrt [4]{b}}-2 x}{-\frac{\sqrt{a}}{\sqrt{b}}+\frac{\sqrt{2} \sqrt [4]{a} x}{\sqrt [4]{b}}-x^2} \, dx,x,\sqrt{x}\right )}{2 \sqrt{2} b^{19/4}}\\ &=\frac{2 (b c-a d)^3 x^{3/2}}{3 b^4}+\frac{2 d \left (3 b^2 c^2-3 a b c d+a^2 d^2\right ) x^{7/2}}{7 b^3}+\frac{2 d^2 (3 b c-a d) x^{11/2}}{11 b^2}+\frac{2 d^3 x^{15/2}}{15 b}-\frac{a^{3/4} (b c-a d)^3 \log \left (\sqrt{a}-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{b} x\right )}{2 \sqrt{2} b^{19/4}}+\frac{a^{3/4} (b c-a d)^3 \log \left (\sqrt{a}+\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{b} x\right )}{2 \sqrt{2} b^{19/4}}-\frac{\left (a^{3/4} (b c-a d)^3\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{\sqrt{2} b^{19/4}}+\frac{\left (a^{3/4} (b c-a d)^3\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{\sqrt{2} b^{19/4}}\\ &=\frac{2 (b c-a d)^3 x^{3/2}}{3 b^4}+\frac{2 d \left (3 b^2 c^2-3 a b c d+a^2 d^2\right ) x^{7/2}}{7 b^3}+\frac{2 d^2 (3 b c-a d) x^{11/2}}{11 b^2}+\frac{2 d^3 x^{15/2}}{15 b}+\frac{a^{3/4} (b c-a d)^3 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{\sqrt{2} b^{19/4}}-\frac{a^{3/4} (b c-a d)^3 \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{\sqrt{2} b^{19/4}}-\frac{a^{3/4} (b c-a d)^3 \log \left (\sqrt{a}-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{b} x\right )}{2 \sqrt{2} b^{19/4}}+\frac{a^{3/4} (b c-a d)^3 \log \left (\sqrt{a}+\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{b} x\right )}{2 \sqrt{2} b^{19/4}}\\ \end{align*}

Mathematica [C]  time = 0.385138, size = 132, normalized size = 0.4 \[ \frac{2 x^{3/2} \left (165 a^2 b d^2 \left (7 c+d x^2\right )-385 a^3 d^3-15 a b^2 d \left (77 c^2+33 c d x^2+7 d^2 x^4\right )-385 (b c-a d)^3 \, _2F_1\left (\frac{3}{4},1;\frac{7}{4};-\frac{b x^2}{a}\right )+b^3 \left (495 c^2 d x^2+385 c^3+315 c d^2 x^4+77 d^3 x^6\right )\right )}{1155 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^(5/2)*(c + d*x^2)^3)/(a + b*x^2),x]

[Out]

(2*x^(3/2)*(-385*a^3*d^3 + 165*a^2*b*d^2*(7*c + d*x^2) - 15*a*b^2*d*(77*c^2 + 33*c*d*x^2 + 7*d^2*x^4) + b^3*(3
85*c^3 + 495*c^2*d*x^2 + 315*c*d^2*x^4 + 77*d^3*x^6) - 385*(b*c - a*d)^3*Hypergeometric2F1[3/4, 1, 7/4, -((b*x
^2)/a)]))/(1155*b^4)

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Maple [B]  time = 0.01, size = 721, normalized size = 2.2 \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)*(d*x^2+c)^3/(b*x^2+a),x)

[Out]

2/15*d^3*x^(15/2)/b-2/11/b^2*x^(11/2)*a*d^3+6/11/b*x^(11/2)*c*d^2+2/7/b^3*x^(7/2)*a^2*d^3-6/7/b^2*x^(7/2)*a*c*
d^2+6/7/b*x^(7/2)*c^2*d-2/3/b^4*x^(3/2)*a^3*d^3+2/b^3*x^(3/2)*a^2*c*d^2-2/b^2*x^(3/2)*a*c^2*d+2/3/b*x^(3/2)*c^
3+1/2*a^4/b^5/(1/b*a)^(1/4)*2^(1/2)*arctan(2^(1/2)/(1/b*a)^(1/4)*x^(1/2)+1)*d^3-3/2*a^3/b^4/(1/b*a)^(1/4)*2^(1
/2)*arctan(2^(1/2)/(1/b*a)^(1/4)*x^(1/2)+1)*c*d^2+3/2*a^2/b^3/(1/b*a)^(1/4)*2^(1/2)*arctan(2^(1/2)/(1/b*a)^(1/
4)*x^(1/2)+1)*c^2*d-1/2*a/b^2/(1/b*a)^(1/4)*2^(1/2)*arctan(2^(1/2)/(1/b*a)^(1/4)*x^(1/2)+1)*c^3+1/2*a^4/b^5/(1
/b*a)^(1/4)*2^(1/2)*arctan(2^(1/2)/(1/b*a)^(1/4)*x^(1/2)-1)*d^3-3/2*a^3/b^4/(1/b*a)^(1/4)*2^(1/2)*arctan(2^(1/
2)/(1/b*a)^(1/4)*x^(1/2)-1)*c*d^2+3/2*a^2/b^3/(1/b*a)^(1/4)*2^(1/2)*arctan(2^(1/2)/(1/b*a)^(1/4)*x^(1/2)-1)*c^
2*d-1/2*a/b^2/(1/b*a)^(1/4)*2^(1/2)*arctan(2^(1/2)/(1/b*a)^(1/4)*x^(1/2)-1)*c^3+1/4*a^4/b^5/(1/b*a)^(1/4)*2^(1
/2)*ln((x-(1/b*a)^(1/4)*x^(1/2)*2^(1/2)+(1/b*a)^(1/2))/(x+(1/b*a)^(1/4)*x^(1/2)*2^(1/2)+(1/b*a)^(1/2)))*d^3-3/
4*a^3/b^4/(1/b*a)^(1/4)*2^(1/2)*ln((x-(1/b*a)^(1/4)*x^(1/2)*2^(1/2)+(1/b*a)^(1/2))/(x+(1/b*a)^(1/4)*x^(1/2)*2^
(1/2)+(1/b*a)^(1/2)))*c*d^2+3/4*a^2/b^3/(1/b*a)^(1/4)*2^(1/2)*ln((x-(1/b*a)^(1/4)*x^(1/2)*2^(1/2)+(1/b*a)^(1/2
))/(x+(1/b*a)^(1/4)*x^(1/2)*2^(1/2)+(1/b*a)^(1/2)))*c^2*d-1/4*a/b^2/(1/b*a)^(1/4)*2^(1/2)*ln((x-(1/b*a)^(1/4)*
x^(1/2)*2^(1/2)+(1/b*a)^(1/2))/(x+(1/b*a)^(1/4)*x^(1/2)*2^(1/2)+(1/b*a)^(1/2)))*c^3

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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)*(d*x^2+c)^3/(b*x^2+a),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.07219, size = 5611, normalized size = 17.11 \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)*(d*x^2+c)^3/(b*x^2+a),x, algorithm="fricas")

[Out]

-1/2310*(4620*b^4*(-(a^3*b^12*c^12 - 12*a^4*b^11*c^11*d + 66*a^5*b^10*c^10*d^2 - 220*a^6*b^9*c^9*d^3 + 495*a^7
*b^8*c^8*d^4 - 792*a^8*b^7*c^7*d^5 + 924*a^9*b^6*c^6*d^6 - 792*a^10*b^5*c^5*d^7 + 495*a^11*b^4*c^4*d^8 - 220*a
^12*b^3*c^3*d^9 + 66*a^13*b^2*c^2*d^10 - 12*a^14*b*c*d^11 + a^15*d^12)/b^19)^(1/4)*arctan((sqrt((a^4*b^18*c^18
- 18*a^5*b^17*c^17*d + 153*a^6*b^16*c^16*d^2 - 816*a^7*b^15*c^15*d^3 + 3060*a^8*b^14*c^14*d^4 - 8568*a^9*b^13
*c^13*d^5 + 18564*a^10*b^12*c^12*d^6 - 31824*a^11*b^11*c^11*d^7 + 43758*a^12*b^10*c^10*d^8 - 48620*a^13*b^9*c^
9*d^9 + 43758*a^14*b^8*c^8*d^10 - 31824*a^15*b^7*c^7*d^11 + 18564*a^16*b^6*c^6*d^12 - 8568*a^17*b^5*c^5*d^13 +
3060*a^18*b^4*c^4*d^14 - 816*a^19*b^3*c^3*d^15 + 153*a^20*b^2*c^2*d^16 - 18*a^21*b*c*d^17 + a^22*d^18)*x - (a
^3*b^21*c^12 - 12*a^4*b^20*c^11*d + 66*a^5*b^19*c^10*d^2 - 220*a^6*b^18*c^9*d^3 + 495*a^7*b^17*c^8*d^4 - 792*a
^8*b^16*c^7*d^5 + 924*a^9*b^15*c^6*d^6 - 792*a^10*b^14*c^5*d^7 + 495*a^11*b^13*c^4*d^8 - 220*a^12*b^12*c^3*d^9
+ 66*a^13*b^11*c^2*d^10 - 12*a^14*b^10*c*d^11 + a^15*b^9*d^12)*sqrt(-(a^3*b^12*c^12 - 12*a^4*b^11*c^11*d + 66
*a^5*b^10*c^10*d^2 - 220*a^6*b^9*c^9*d^3 + 495*a^7*b^8*c^8*d^4 - 792*a^8*b^7*c^7*d^5 + 924*a^9*b^6*c^6*d^6 - 7
92*a^10*b^5*c^5*d^7 + 495*a^11*b^4*c^4*d^8 - 220*a^12*b^3*c^3*d^9 + 66*a^13*b^2*c^2*d^10 - 12*a^14*b*c*d^11 +
a^15*d^12)/b^19))*b^5*(-(a^3*b^12*c^12 - 12*a^4*b^11*c^11*d + 66*a^5*b^10*c^10*d^2 - 220*a^6*b^9*c^9*d^3 + 495
*a^7*b^8*c^8*d^4 - 792*a^8*b^7*c^7*d^5 + 924*a^9*b^6*c^6*d^6 - 792*a^10*b^5*c^5*d^7 + 495*a^11*b^4*c^4*d^8 - 2
20*a^12*b^3*c^3*d^9 + 66*a^13*b^2*c^2*d^10 - 12*a^14*b*c*d^11 + a^15*d^12)/b^19)^(1/4) + (a^2*b^14*c^9 - 9*a^3
*b^13*c^8*d + 36*a^4*b^12*c^7*d^2 - 84*a^5*b^11*c^6*d^3 + 126*a^6*b^10*c^5*d^4 - 126*a^7*b^9*c^4*d^5 + 84*a^8*
b^8*c^3*d^6 - 36*a^9*b^7*c^2*d^7 + 9*a^10*b^6*c*d^8 - a^11*b^5*d^9)*sqrt(x)*(-(a^3*b^12*c^12 - 12*a^4*b^11*c^1
1*d + 66*a^5*b^10*c^10*d^2 - 220*a^6*b^9*c^9*d^3 + 495*a^7*b^8*c^8*d^4 - 792*a^8*b^7*c^7*d^5 + 924*a^9*b^6*c^6
*d^6 - 792*a^10*b^5*c^5*d^7 + 495*a^11*b^4*c^4*d^8 - 220*a^12*b^3*c^3*d^9 + 66*a^13*b^2*c^2*d^10 - 12*a^14*b*c
*d^11 + a^15*d^12)/b^19)^(1/4))/(a^3*b^12*c^12 - 12*a^4*b^11*c^11*d + 66*a^5*b^10*c^10*d^2 - 220*a^6*b^9*c^9*d
^3 + 495*a^7*b^8*c^8*d^4 - 792*a^8*b^7*c^7*d^5 + 924*a^9*b^6*c^6*d^6 - 792*a^10*b^5*c^5*d^7 + 495*a^11*b^4*c^4
*d^8 - 220*a^12*b^3*c^3*d^9 + 66*a^13*b^2*c^2*d^10 - 12*a^14*b*c*d^11 + a^15*d^12)) - 1155*b^4*(-(a^3*b^12*c^1
2 - 12*a^4*b^11*c^11*d + 66*a^5*b^10*c^10*d^2 - 220*a^6*b^9*c^9*d^3 + 495*a^7*b^8*c^8*d^4 - 792*a^8*b^7*c^7*d^
5 + 924*a^9*b^6*c^6*d^6 - 792*a^10*b^5*c^5*d^7 + 495*a^11*b^4*c^4*d^8 - 220*a^12*b^3*c^3*d^9 + 66*a^13*b^2*c^2
*d^10 - 12*a^14*b*c*d^11 + a^15*d^12)/b^19)^(1/4)*log(b^14*(-(a^3*b^12*c^12 - 12*a^4*b^11*c^11*d + 66*a^5*b^10
*c^10*d^2 - 220*a^6*b^9*c^9*d^3 + 495*a^7*b^8*c^8*d^4 - 792*a^8*b^7*c^7*d^5 + 924*a^9*b^6*c^6*d^6 - 792*a^10*b
^5*c^5*d^7 + 495*a^11*b^4*c^4*d^8 - 220*a^12*b^3*c^3*d^9 + 66*a^13*b^2*c^2*d^10 - 12*a^14*b*c*d^11 + a^15*d^12
)/b^19)^(3/4) - (a^2*b^9*c^9 - 9*a^3*b^8*c^8*d + 36*a^4*b^7*c^7*d^2 - 84*a^5*b^6*c^6*d^3 + 126*a^6*b^5*c^5*d^4
- 126*a^7*b^4*c^4*d^5 + 84*a^8*b^3*c^3*d^6 - 36*a^9*b^2*c^2*d^7 + 9*a^10*b*c*d^8 - a^11*d^9)*sqrt(x)) + 1155*
b^4*(-(a^3*b^12*c^12 - 12*a^4*b^11*c^11*d + 66*a^5*b^10*c^10*d^2 - 220*a^6*b^9*c^9*d^3 + 495*a^7*b^8*c^8*d^4 -
792*a^8*b^7*c^7*d^5 + 924*a^9*b^6*c^6*d^6 - 792*a^10*b^5*c^5*d^7 + 495*a^11*b^4*c^4*d^8 - 220*a^12*b^3*c^3*d^
9 + 66*a^13*b^2*c^2*d^10 - 12*a^14*b*c*d^11 + a^15*d^12)/b^19)^(1/4)*log(-b^14*(-(a^3*b^12*c^12 - 12*a^4*b^11*
c^11*d + 66*a^5*b^10*c^10*d^2 - 220*a^6*b^9*c^9*d^3 + 495*a^7*b^8*c^8*d^4 - 792*a^8*b^7*c^7*d^5 + 924*a^9*b^6*
c^6*d^6 - 792*a^10*b^5*c^5*d^7 + 495*a^11*b^4*c^4*d^8 - 220*a^12*b^3*c^3*d^9 + 66*a^13*b^2*c^2*d^10 - 12*a^14*
b*c*d^11 + a^15*d^12)/b^19)^(3/4) - (a^2*b^9*c^9 - 9*a^3*b^8*c^8*d + 36*a^4*b^7*c^7*d^2 - 84*a^5*b^6*c^6*d^3 +
126*a^6*b^5*c^5*d^4 - 126*a^7*b^4*c^4*d^5 + 84*a^8*b^3*c^3*d^6 - 36*a^9*b^2*c^2*d^7 + 9*a^10*b*c*d^8 - a^11*d
^9)*sqrt(x)) - 4*(77*b^3*d^3*x^7 + 105*(3*b^3*c*d^2 - a*b^2*d^3)*x^5 + 165*(3*b^3*c^2*d - 3*a*b^2*c*d^2 + a^2*
b*d^3)*x^3 + 385*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*x)*sqrt(x))/b^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**(5/2)*(d*x**2+c)**3/(b*x**2+a),x)

[Out]

Timed out

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Giac [B]  time = 1.19339, size = 717, normalized size = 2.19 \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)*(d*x^2+c)^3/(b*x^2+a),x, algorithm="giac")

[Out]

-1/2*sqrt(2)*((a*b^3)^(3/4)*b^3*c^3 - 3*(a*b^3)^(3/4)*a*b^2*c^2*d + 3*(a*b^3)^(3/4)*a^2*b*c*d^2 - (a*b^3)^(3/4
)*a^3*d^3)*arctan(1/2*sqrt(2)*(sqrt(2)*(a/b)^(1/4) + 2*sqrt(x))/(a/b)^(1/4))/b^7 - 1/2*sqrt(2)*((a*b^3)^(3/4)*
b^3*c^3 - 3*(a*b^3)^(3/4)*a*b^2*c^2*d + 3*(a*b^3)^(3/4)*a^2*b*c*d^2 - (a*b^3)^(3/4)*a^3*d^3)*arctan(-1/2*sqrt(
2)*(sqrt(2)*(a/b)^(1/4) - 2*sqrt(x))/(a/b)^(1/4))/b^7 + 1/4*sqrt(2)*((a*b^3)^(3/4)*b^3*c^3 - 3*(a*b^3)^(3/4)*a
*b^2*c^2*d + 3*(a*b^3)^(3/4)*a^2*b*c*d^2 - (a*b^3)^(3/4)*a^3*d^3)*log(sqrt(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a
/b))/b^7 - 1/4*sqrt(2)*((a*b^3)^(3/4)*b^3*c^3 - 3*(a*b^3)^(3/4)*a*b^2*c^2*d + 3*(a*b^3)^(3/4)*a^2*b*c*d^2 - (a
*b^3)^(3/4)*a^3*d^3)*log(-sqrt(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))/b^7 + 2/1155*(77*b^14*d^3*x^(15/2) + 31
5*b^14*c*d^2*x^(11/2) - 105*a*b^13*d^3*x^(11/2) + 495*b^14*c^2*d*x^(7/2) - 495*a*b^13*c*d^2*x^(7/2) + 165*a^2*
b^12*d^3*x^(7/2) + 385*b^14*c^3*x^(3/2) - 1155*a*b^13*c^2*d*x^(3/2) + 1155*a^2*b^12*c*d^2*x^(3/2) - 385*a^3*b^
11*d^3*x^(3/2))/b^15