3.456 \(\int \frac{(c+d x^2)^3}{\sqrt{x} (a+b x^2)^2} \, dx\)

Optimal. Leaf size=340 \[ -\frac{3 (b c-a d)^2 (3 a d+b c) \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{8 \sqrt{2} a^{7/4} b^{13/4}}+\frac{3 (b c-a d)^2 (3 a d+b c) \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{8 \sqrt{2} a^{7/4} b^{13/4}}-\frac{3 (b c-a d)^2 (3 a d+b c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{4 \sqrt{2} a^{7/4} b^{13/4}}+\frac{3 (b c-a d)^2 (3 a d+b c) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{4 \sqrt{2} a^{7/4} b^{13/4}}+\frac{2 d^2 \sqrt{x} (3 b c-2 a d)}{b^3}+\frac{\sqrt{x} (b c-a d)^3}{2 a b^3 \left (a+b x^2\right )}+\frac{2 d^3 x^{5/2}}{5 b^2} \]

[Out]

(2*d^2*(3*b*c - 2*a*d)*Sqrt[x])/b^3 + (2*d^3*x^(5/2))/(5*b^2) + ((b*c - a*d)^3*Sqrt[x])/(2*a*b^3*(a + b*x^2))
- (3*(b*c - a*d)^2*(b*c + 3*a*d)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(4*Sqrt[2]*a^(7/4)*b^(13/4)) +
(3*(b*c - a*d)^2*(b*c + 3*a*d)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(4*Sqrt[2]*a^(7/4)*b^(13/4)) -
(3*(b*c - a*d)^2*(b*c + 3*a*d)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(8*Sqrt[2]*a^(7/4)*
b^(13/4)) + (3*(b*c - a*d)^2*(b*c + 3*a*d)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(8*Sqrt
[2]*a^(7/4)*b^(13/4))

________________________________________________________________________________________

Rubi [A]  time = 0.388311, antiderivative size = 340, 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 = {466, 390, 385, 211, 1165, 628, 1162, 617, 204} \[ -\frac{3 (b c-a d)^2 (3 a d+b c) \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{8 \sqrt{2} a^{7/4} b^{13/4}}+\frac{3 (b c-a d)^2 (3 a d+b c) \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{8 \sqrt{2} a^{7/4} b^{13/4}}-\frac{3 (b c-a d)^2 (3 a d+b c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{4 \sqrt{2} a^{7/4} b^{13/4}}+\frac{3 (b c-a d)^2 (3 a d+b c) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{4 \sqrt{2} a^{7/4} b^{13/4}}+\frac{2 d^2 \sqrt{x} (3 b c-2 a d)}{b^3}+\frac{\sqrt{x} (b c-a d)^3}{2 a b^3 \left (a+b x^2\right )}+\frac{2 d^3 x^{5/2}}{5 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*d^2*(3*b*c - 2*a*d)*Sqrt[x])/b^3 + (2*d^3*x^(5/2))/(5*b^2) + ((b*c - a*d)^3*Sqrt[x])/(2*a*b^3*(a + b*x^2))
- (3*(b*c - a*d)^2*(b*c + 3*a*d)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(4*Sqrt[2]*a^(7/4)*b^(13/4)) +
(3*(b*c - a*d)^2*(b*c + 3*a*d)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(4*Sqrt[2]*a^(7/4)*b^(13/4)) -
(3*(b*c - a*d)^2*(b*c + 3*a*d)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(8*Sqrt[2]*a^(7/4)*
b^(13/4)) + (3*(b*c - a*d)^2*(b*c + 3*a*d)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(8*Sqrt
[2]*a^(7/4)*b^(13/4))

Rule 466

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> With[{k = Deno
minator[m]}, Dist[k/e, Subst[Int[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/e^n)^p*(c + (d*x^(k*n))/e^n)^q, x], x, (e*
x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && FractionQ[m] && Intege
rQ[p]

Rule 390

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]

Rule 385

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

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

Mathematica [C]  time = 2.01997, size = 358, normalized size = 1.05 \[ \frac{a \left (45 a^2 \left (85683 c^2 d x^2+28561 c^3+85683 c d^2 x^4+25105 d^3 x^6\right )+18 a b x^2 \left (104781 c^2 d x^2+34927 c^3+119181 c d^2 x^4+36655 d^3 x^6\right )+b^2 x^4 \left (98259 c^2 d x^2+50033 c^3+98259 c d^2 x^4+32753 d^3 x^6\right )\right )-45 \, _2F_1\left (\frac{1}{4},1;\frac{5}{4};-\frac{b x^2}{a}\right ) \left (9 a^2 b x^2 \left (6561 c^2 d x^2+2187 c^3+7201 c d^2 x^4+2187 d^3 x^6\right )+a^3 \left (85683 c^2 d x^2+28561 c^3+85683 c d^2 x^4+25105 d^3 x^6\right )+3 a b^2 x^4 \left (1491 c^2 d x^2+625 c^3+1875 c d^2 x^4+625 d^3 x^6\right )+b^3 x^6 \left (3 c^2 d x^2-1151 c^3+3 c d^2 x^4+d^3 x^6\right )\right )}{34560 a^2 b^3 x^{11/2}}-\frac{128 b x^{5/2} \left (c+d x^2\right )^3 \text{HypergeometricPFQ}\left (\left \{\frac{5}{4},2,2,2,2\right \},\left \{1,1,1,\frac{21}{4}\right \},-\frac{b x^2}{a}\right )}{9945 a^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a*(45*a^2*(28561*c^3 + 85683*c^2*d*x^2 + 85683*c*d^2*x^4 + 25105*d^3*x^6) + b^2*x^4*(50033*c^3 + 98259*c^2*d*
x^2 + 98259*c*d^2*x^4 + 32753*d^3*x^6) + 18*a*b*x^2*(34927*c^3 + 104781*c^2*d*x^2 + 119181*c*d^2*x^4 + 36655*d
^3*x^6)) - 45*(b^3*x^6*(-1151*c^3 + 3*c^2*d*x^2 + 3*c*d^2*x^4 + d^3*x^6) + 3*a*b^2*x^4*(625*c^3 + 1491*c^2*d*x
^2 + 1875*c*d^2*x^4 + 625*d^3*x^6) + 9*a^2*b*x^2*(2187*c^3 + 6561*c^2*d*x^2 + 7201*c*d^2*x^4 + 2187*d^3*x^6) +
a^3*(28561*c^3 + 85683*c^2*d*x^2 + 85683*c*d^2*x^4 + 25105*d^3*x^6))*Hypergeometric2F1[1/4, 1, 5/4, -((b*x^2)
/a)])/(34560*a^2*b^3*x^(11/2)) - (128*b*x^(5/2)*(c + d*x^2)^3*HypergeometricPFQ[{5/4, 2, 2, 2, 2}, {1, 1, 1, 2
1/4}, -((b*x^2)/a)])/(9945*a^3)

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Maple [B]  time = 0.016, size = 697, normalized size = 2.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x^2+c)^3/(b*x^2+a)^2/x^(1/2),x)

[Out]

2/5*d^3*x^(5/2)/b^2-4*d^3/b^3*a*x^(1/2)+6*d^2/b^2*x^(1/2)*c-1/2/b^3*a^2*x^(1/2)/(b*x^2+a)*d^3+3/2/b^2*a*x^(1/2
)/(b*x^2+a)*c*d^2-3/2/b*x^(1/2)/(b*x^2+a)*c^2*d+1/2/a*x^(1/2)/(b*x^2+a)*c^3+9/8/b^3*a*(1/b*a)^(1/4)*2^(1/2)*ar
ctan(2^(1/2)/(1/b*a)^(1/4)*x^(1/2)-1)*d^3-15/8/b^2*(1/b*a)^(1/4)*2^(1/2)*arctan(2^(1/2)/(1/b*a)^(1/4)*x^(1/2)-
1)*c*d^2+3/8/b/a*(1/b*a)^(1/4)*2^(1/2)*arctan(2^(1/2)/(1/b*a)^(1/4)*x^(1/2)-1)*c^2*d+3/8/a^2*(1/b*a)^(1/4)*2^(
1/2)*arctan(2^(1/2)/(1/b*a)^(1/4)*x^(1/2)-1)*c^3+9/16/b^3*a*(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^3-15/16/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*d^2+3/16/b
/a*(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+3/16/a^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+9/8/b^3*a*(1/b*a)^(1/4)*2^(1/2)*arctan(2^(1/2)/(1/b*a)^(1/4)*x^(
1/2)+1)*d^3-15/8/b^2*(1/b*a)^(1/4)*2^(1/2)*arctan(2^(1/2)/(1/b*a)^(1/4)*x^(1/2)+1)*c*d^2+3/8/b/a*(1/b*a)^(1/4)
*2^(1/2)*arctan(2^(1/2)/(1/b*a)^(1/4)*x^(1/2)+1)*c^2*d+3/8/a^2*(1/b*a)^(1/4)*2^(1/2)*arctan(2^(1/2)/(1/b*a)^(1
/4)*x^(1/2)+1)*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((d*x^2+c)^3/(b*x^2+a)^2/x^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.37465, size = 4316, normalized size = 12.69 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)^3/(b*x^2+a)^2/x^(1/2),x, algorithm="fricas")

[Out]

1/40*(60*(a*b^4*x^2 + a^2*b^3)*(-(b^12*c^12 + 4*a*b^11*c^11*d - 14*a^2*b^10*c^10*d^2 - 44*a^3*b^9*c^9*d^3 + 12
7*a^4*b^8*c^8*d^4 + 136*a^5*b^7*c^7*d^5 - 644*a^6*b^6*c^6*d^6 + 328*a^7*b^5*c^5*d^7 + 1039*a^8*b^4*c^4*d^8 - 1
932*a^9*b^3*c^3*d^9 + 1458*a^10*b^2*c^2*d^10 - 540*a^11*b*c*d^11 + 81*a^12*d^12)/(a^7*b^13))^(1/4)*arctan((sqr
t(a^4*b^6*sqrt(-(b^12*c^12 + 4*a*b^11*c^11*d - 14*a^2*b^10*c^10*d^2 - 44*a^3*b^9*c^9*d^3 + 127*a^4*b^8*c^8*d^4
+ 136*a^5*b^7*c^7*d^5 - 644*a^6*b^6*c^6*d^6 + 328*a^7*b^5*c^5*d^7 + 1039*a^8*b^4*c^4*d^8 - 1932*a^9*b^3*c^3*d
^9 + 1458*a^10*b^2*c^2*d^10 - 540*a^11*b*c*d^11 + 81*a^12*d^12)/(a^7*b^13)) + (b^6*c^6 + 2*a*b^5*c^5*d - 9*a^2
*b^4*c^4*d^2 - 4*a^3*b^3*c^3*d^3 + 31*a^4*b^2*c^2*d^4 - 30*a^5*b*c*d^5 + 9*a^6*d^6)*x)*a^5*b^10*(-(b^12*c^12 +
4*a*b^11*c^11*d - 14*a^2*b^10*c^10*d^2 - 44*a^3*b^9*c^9*d^3 + 127*a^4*b^8*c^8*d^4 + 136*a^5*b^7*c^7*d^5 - 644
*a^6*b^6*c^6*d^6 + 328*a^7*b^5*c^5*d^7 + 1039*a^8*b^4*c^4*d^8 - 1932*a^9*b^3*c^3*d^9 + 1458*a^10*b^2*c^2*d^10
- 540*a^11*b*c*d^11 + 81*a^12*d^12)/(a^7*b^13))^(3/4) - (a^5*b^13*c^3 + a^6*b^12*c^2*d - 5*a^7*b^11*c*d^2 + 3*
a^8*b^10*d^3)*sqrt(x)*(-(b^12*c^12 + 4*a*b^11*c^11*d - 14*a^2*b^10*c^10*d^2 - 44*a^3*b^9*c^9*d^3 + 127*a^4*b^8
*c^8*d^4 + 136*a^5*b^7*c^7*d^5 - 644*a^6*b^6*c^6*d^6 + 328*a^7*b^5*c^5*d^7 + 1039*a^8*b^4*c^4*d^8 - 1932*a^9*b
^3*c^3*d^9 + 1458*a^10*b^2*c^2*d^10 - 540*a^11*b*c*d^11 + 81*a^12*d^12)/(a^7*b^13))^(3/4))/(b^12*c^12 + 4*a*b^
11*c^11*d - 14*a^2*b^10*c^10*d^2 - 44*a^3*b^9*c^9*d^3 + 127*a^4*b^8*c^8*d^4 + 136*a^5*b^7*c^7*d^5 - 644*a^6*b^
6*c^6*d^6 + 328*a^7*b^5*c^5*d^7 + 1039*a^8*b^4*c^4*d^8 - 1932*a^9*b^3*c^3*d^9 + 1458*a^10*b^2*c^2*d^10 - 540*a
^11*b*c*d^11 + 81*a^12*d^12)) + 15*(a*b^4*x^2 + a^2*b^3)*(-(b^12*c^12 + 4*a*b^11*c^11*d - 14*a^2*b^10*c^10*d^2
- 44*a^3*b^9*c^9*d^3 + 127*a^4*b^8*c^8*d^4 + 136*a^5*b^7*c^7*d^5 - 644*a^6*b^6*c^6*d^6 + 328*a^7*b^5*c^5*d^7
+ 1039*a^8*b^4*c^4*d^8 - 1932*a^9*b^3*c^3*d^9 + 1458*a^10*b^2*c^2*d^10 - 540*a^11*b*c*d^11 + 81*a^12*d^12)/(a^
7*b^13))^(1/4)*log(3*a^2*b^3*(-(b^12*c^12 + 4*a*b^11*c^11*d - 14*a^2*b^10*c^10*d^2 - 44*a^3*b^9*c^9*d^3 + 127*
a^4*b^8*c^8*d^4 + 136*a^5*b^7*c^7*d^5 - 644*a^6*b^6*c^6*d^6 + 328*a^7*b^5*c^5*d^7 + 1039*a^8*b^4*c^4*d^8 - 193
2*a^9*b^3*c^3*d^9 + 1458*a^10*b^2*c^2*d^10 - 540*a^11*b*c*d^11 + 81*a^12*d^12)/(a^7*b^13))^(1/4) + 3*(b^3*c^3
+ a*b^2*c^2*d - 5*a^2*b*c*d^2 + 3*a^3*d^3)*sqrt(x)) - 15*(a*b^4*x^2 + a^2*b^3)*(-(b^12*c^12 + 4*a*b^11*c^11*d
- 14*a^2*b^10*c^10*d^2 - 44*a^3*b^9*c^9*d^3 + 127*a^4*b^8*c^8*d^4 + 136*a^5*b^7*c^7*d^5 - 644*a^6*b^6*c^6*d^6
+ 328*a^7*b^5*c^5*d^7 + 1039*a^8*b^4*c^4*d^8 - 1932*a^9*b^3*c^3*d^9 + 1458*a^10*b^2*c^2*d^10 - 540*a^11*b*c*d^
11 + 81*a^12*d^12)/(a^7*b^13))^(1/4)*log(-3*a^2*b^3*(-(b^12*c^12 + 4*a*b^11*c^11*d - 14*a^2*b^10*c^10*d^2 - 44
*a^3*b^9*c^9*d^3 + 127*a^4*b^8*c^8*d^4 + 136*a^5*b^7*c^7*d^5 - 644*a^6*b^6*c^6*d^6 + 328*a^7*b^5*c^5*d^7 + 103
9*a^8*b^4*c^4*d^8 - 1932*a^9*b^3*c^3*d^9 + 1458*a^10*b^2*c^2*d^10 - 540*a^11*b*c*d^11 + 81*a^12*d^12)/(a^7*b^1
3))^(1/4) + 3*(b^3*c^3 + a*b^2*c^2*d - 5*a^2*b*c*d^2 + 3*a^3*d^3)*sqrt(x)) + 4*(4*a*b^2*d^3*x^4 + 5*b^3*c^3 -
15*a*b^2*c^2*d + 75*a^2*b*c*d^2 - 45*a^3*d^3 + 12*(5*a*b^2*c*d^2 - 3*a^2*b*d^3)*x^2)*sqrt(x))/(a*b^4*x^2 + a^2
*b^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((d*x**2+c)**3/(b*x**2+a)**2/x**(1/2),x)

[Out]

Timed out

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Giac [A]  time = 1.2229, size = 690, normalized size = 2.03 \begin{align*} \frac{3 \, \sqrt{2}{\left (\left (a b^{3}\right )^{\frac{1}{4}} b^{3} c^{3} + \left (a b^{3}\right )^{\frac{1}{4}} a b^{2} c^{2} d - 5 \, \left (a b^{3}\right )^{\frac{1}{4}} a^{2} b c d^{2} + 3 \, \left (a b^{3}\right )^{\frac{1}{4}} a^{3} d^{3}\right )} \arctan \left (\frac{\sqrt{2}{\left (\sqrt{2} \left (\frac{a}{b}\right )^{\frac{1}{4}} + 2 \, \sqrt{x}\right )}}{2 \, \left (\frac{a}{b}\right )^{\frac{1}{4}}}\right )}{8 \, a^{2} b^{4}} + \frac{3 \, \sqrt{2}{\left (\left (a b^{3}\right )^{\frac{1}{4}} b^{3} c^{3} + \left (a b^{3}\right )^{\frac{1}{4}} a b^{2} c^{2} d - 5 \, \left (a b^{3}\right )^{\frac{1}{4}} a^{2} b c d^{2} + 3 \, \left (a b^{3}\right )^{\frac{1}{4}} a^{3} d^{3}\right )} \arctan \left (-\frac{\sqrt{2}{\left (\sqrt{2} \left (\frac{a}{b}\right )^{\frac{1}{4}} - 2 \, \sqrt{x}\right )}}{2 \, \left (\frac{a}{b}\right )^{\frac{1}{4}}}\right )}{8 \, a^{2} b^{4}} + \frac{3 \, \sqrt{2}{\left (\left (a b^{3}\right )^{\frac{1}{4}} b^{3} c^{3} + \left (a b^{3}\right )^{\frac{1}{4}} a b^{2} c^{2} d - 5 \, \left (a b^{3}\right )^{\frac{1}{4}} a^{2} b c d^{2} + 3 \, \left (a b^{3}\right )^{\frac{1}{4}} a^{3} d^{3}\right )} \log \left (\sqrt{2} \sqrt{x} \left (\frac{a}{b}\right )^{\frac{1}{4}} + x + \sqrt{\frac{a}{b}}\right )}{16 \, a^{2} b^{4}} - \frac{3 \, \sqrt{2}{\left (\left (a b^{3}\right )^{\frac{1}{4}} b^{3} c^{3} + \left (a b^{3}\right )^{\frac{1}{4}} a b^{2} c^{2} d - 5 \, \left (a b^{3}\right )^{\frac{1}{4}} a^{2} b c d^{2} + 3 \, \left (a b^{3}\right )^{\frac{1}{4}} a^{3} d^{3}\right )} \log \left (-\sqrt{2} \sqrt{x} \left (\frac{a}{b}\right )^{\frac{1}{4}} + x + \sqrt{\frac{a}{b}}\right )}{16 \, a^{2} b^{4}} + \frac{b^{3} c^{3} \sqrt{x} - 3 \, a b^{2} c^{2} d \sqrt{x} + 3 \, a^{2} b c d^{2} \sqrt{x} - a^{3} d^{3} \sqrt{x}}{2 \,{\left (b x^{2} + a\right )} a b^{3}} + \frac{2 \,{\left (b^{8} d^{3} x^{\frac{5}{2}} + 15 \, b^{8} c d^{2} \sqrt{x} - 10 \, a b^{7} d^{3} \sqrt{x}\right )}}{5 \, b^{10}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)^3/(b*x^2+a)^2/x^(1/2),x, algorithm="giac")

[Out]

3/8*sqrt(2)*((a*b^3)^(1/4)*b^3*c^3 + (a*b^3)^(1/4)*a*b^2*c^2*d - 5*(a*b^3)^(1/4)*a^2*b*c*d^2 + 3*(a*b^3)^(1/4)
*a^3*d^3)*arctan(1/2*sqrt(2)*(sqrt(2)*(a/b)^(1/4) + 2*sqrt(x))/(a/b)^(1/4))/(a^2*b^4) + 3/8*sqrt(2)*((a*b^3)^(
1/4)*b^3*c^3 + (a*b^3)^(1/4)*a*b^2*c^2*d - 5*(a*b^3)^(1/4)*a^2*b*c*d^2 + 3*(a*b^3)^(1/4)*a^3*d^3)*arctan(-1/2*
sqrt(2)*(sqrt(2)*(a/b)^(1/4) - 2*sqrt(x))/(a/b)^(1/4))/(a^2*b^4) + 3/16*sqrt(2)*((a*b^3)^(1/4)*b^3*c^3 + (a*b^
3)^(1/4)*a*b^2*c^2*d - 5*(a*b^3)^(1/4)*a^2*b*c*d^2 + 3*(a*b^3)^(1/4)*a^3*d^3)*log(sqrt(2)*sqrt(x)*(a/b)^(1/4)
+ x + sqrt(a/b))/(a^2*b^4) - 3/16*sqrt(2)*((a*b^3)^(1/4)*b^3*c^3 + (a*b^3)^(1/4)*a*b^2*c^2*d - 5*(a*b^3)^(1/4)
*a^2*b*c*d^2 + 3*(a*b^3)^(1/4)*a^3*d^3)*log(-sqrt(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))/(a^2*b^4) + 1/2*(b^3
*c^3*sqrt(x) - 3*a*b^2*c^2*d*sqrt(x) + 3*a^2*b*c*d^2*sqrt(x) - a^3*d^3*sqrt(x))/((b*x^2 + a)*a*b^3) + 2/5*(b^8
*d^3*x^(5/2) + 15*b^8*c*d^2*sqrt(x) - 10*a*b^7*d^3*sqrt(x))/b^10