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

Optimal. Leaf size=304 \[ \frac{2 d \sqrt{x} \left (a^2 d^2-3 a b c d+3 b^2 c^2\right )}{b^3}-\frac{(b c-a d)^3 \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{2 \sqrt{2} a^{3/4} b^{13/4}}+\frac{(b c-a d)^3 \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{2 \sqrt{2} a^{3/4} b^{13/4}}-\frac{(b c-a d)^3 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{\sqrt{2} a^{3/4} b^{13/4}}+\frac{(b c-a d)^3 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{\sqrt{2} a^{3/4} b^{13/4}}+\frac{2 d^2 x^{5/2} (3 b c-a d)}{5 b^2}+\frac{2 d^3 x^{9/2}}{9 b} \]

[Out]

(2*d*(3*b^2*c^2 - 3*a*b*c*d + a^2*d^2)*Sqrt[x])/b^3 + (2*d^2*(3*b*c - a*d)*x^(5/2))/(5*b^2) + (2*d^3*x^(9/2))/
(9*b) - ((b*c - a*d)^3*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(Sqrt[2]*a^(3/4)*b^(13/4)) + ((b*c - a*d
)^3*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(Sqrt[2]*a^(3/4)*b^(13/4)) - ((b*c - a*d)^3*Log[Sqrt[a] - S
qrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(2*Sqrt[2]*a^(3/4)*b^(13/4)) + ((b*c - a*d)^3*Log[Sqrt[a] + Sqrt[
2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(2*Sqrt[2]*a^(3/4)*b^(13/4))

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Rubi [A]  time = 0.249237, antiderivative size = 304, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 8, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {461, 329, 211, 1165, 628, 1162, 617, 204} \[ \frac{2 d \sqrt{x} \left (a^2 d^2-3 a b c d+3 b^2 c^2\right )}{b^3}-\frac{(b c-a d)^3 \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{2 \sqrt{2} a^{3/4} b^{13/4}}+\frac{(b c-a d)^3 \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{2 \sqrt{2} a^{3/4} b^{13/4}}-\frac{(b c-a d)^3 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{\sqrt{2} a^{3/4} b^{13/4}}+\frac{(b c-a d)^3 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{\sqrt{2} a^{3/4} b^{13/4}}+\frac{2 d^2 x^{5/2} (3 b c-a d)}{5 b^2}+\frac{2 d^3 x^{9/2}}{9 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*d*(3*b^2*c^2 - 3*a*b*c*d + a^2*d^2)*Sqrt[x])/b^3 + (2*d^2*(3*b*c - a*d)*x^(5/2))/(5*b^2) + (2*d^3*x^(9/2))/
(9*b) - ((b*c - a*d)^3*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(Sqrt[2]*a^(3/4)*b^(13/4)) + ((b*c - a*d
)^3*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(Sqrt[2]*a^(3/4)*b^(13/4)) - ((b*c - a*d)^3*Log[Sqrt[a] - S
qrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(2*Sqrt[2]*a^(3/4)*b^(13/4)) + ((b*c - a*d)^3*Log[Sqrt[a] + Sqrt[
2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(2*Sqrt[2]*a^(3/4)*b^(13/4))

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

Mathematica [C]  time = 0.343753, size = 96, normalized size = 0.32 \[ \frac{2 \sqrt{x} \left (a d \left (45 a^2 d^2-9 a b d \left (15 c+d x^2\right )+b^2 \left (135 c^2+27 c d x^2+5 d^2 x^4\right )\right )+45 (b c-a d)^3 \, _2F_1\left (\frac{1}{4},1;\frac{5}{4};-\frac{b x^2}{a}\right )\right )}{45 a b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[x]*(a*d*(45*a^2*d^2 - 9*a*b*d*(15*c + d*x^2) + b^2*(135*c^2 + 27*c*d*x^2 + 5*d^2*x^4)) + 45*(b*c - a*d
)^3*Hypergeometric2F1[1/4, 1, 5/4, -((b*x^2)/a)]))/(45*a*b^3)

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Maple [B]  time = 0.01, size = 650, 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)/x^(1/2),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.38563, size = 4100, normalized size = 13.49 \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)/x^(1/2),x, algorithm="fricas")

[Out]

-1/90*(180*b^3*(-(b^12*c^12 - 12*a*b^11*c^11*d + 66*a^2*b^10*c^10*d^2 - 220*a^3*b^9*c^9*d^3 + 495*a^4*b^8*c^8*
d^4 - 792*a^5*b^7*c^7*d^5 + 924*a^6*b^6*c^6*d^6 - 792*a^7*b^5*c^5*d^7 + 495*a^8*b^4*c^4*d^8 - 220*a^9*b^3*c^3*
d^9 + 66*a^10*b^2*c^2*d^10 - 12*a^11*b*c*d^11 + a^12*d^12)/(a^3*b^13))^(1/4)*arctan((sqrt(a^2*b^6*sqrt(-(b^12*
c^12 - 12*a*b^11*c^11*d + 66*a^2*b^10*c^10*d^2 - 220*a^3*b^9*c^9*d^3 + 495*a^4*b^8*c^8*d^4 - 792*a^5*b^7*c^7*d
^5 + 924*a^6*b^6*c^6*d^6 - 792*a^7*b^5*c^5*d^7 + 495*a^8*b^4*c^4*d^8 - 220*a^9*b^3*c^3*d^9 + 66*a^10*b^2*c^2*d
^10 - 12*a^11*b*c*d^11 + a^12*d^12)/(a^3*b^13)) + (b^6*c^6 - 6*a*b^5*c^5*d + 15*a^2*b^4*c^4*d^2 - 20*a^3*b^3*c
^3*d^3 + 15*a^4*b^2*c^2*d^4 - 6*a^5*b*c*d^5 + a^6*d^6)*x)*a^2*b^10*(-(b^12*c^12 - 12*a*b^11*c^11*d + 66*a^2*b^
10*c^10*d^2 - 220*a^3*b^9*c^9*d^3 + 495*a^4*b^8*c^8*d^4 - 792*a^5*b^7*c^7*d^5 + 924*a^6*b^6*c^6*d^6 - 792*a^7*
b^5*c^5*d^7 + 495*a^8*b^4*c^4*d^8 - 220*a^9*b^3*c^3*d^9 + 66*a^10*b^2*c^2*d^10 - 12*a^11*b*c*d^11 + a^12*d^12)
/(a^3*b^13))^(3/4) + (a^2*b^13*c^3 - 3*a^3*b^12*c^2*d + 3*a^4*b^11*c*d^2 - a^5*b^10*d^3)*sqrt(x)*(-(b^12*c^12
- 12*a*b^11*c^11*d + 66*a^2*b^10*c^10*d^2 - 220*a^3*b^9*c^9*d^3 + 495*a^4*b^8*c^8*d^4 - 792*a^5*b^7*c^7*d^5 +
924*a^6*b^6*c^6*d^6 - 792*a^7*b^5*c^5*d^7 + 495*a^8*b^4*c^4*d^8 - 220*a^9*b^3*c^3*d^9 + 66*a^10*b^2*c^2*d^10 -
12*a^11*b*c*d^11 + a^12*d^12)/(a^3*b^13))^(3/4))/(b^12*c^12 - 12*a*b^11*c^11*d + 66*a^2*b^10*c^10*d^2 - 220*a
^3*b^9*c^9*d^3 + 495*a^4*b^8*c^8*d^4 - 792*a^5*b^7*c^7*d^5 + 924*a^6*b^6*c^6*d^6 - 792*a^7*b^5*c^5*d^7 + 495*a
^8*b^4*c^4*d^8 - 220*a^9*b^3*c^3*d^9 + 66*a^10*b^2*c^2*d^10 - 12*a^11*b*c*d^11 + a^12*d^12)) + 45*b^3*(-(b^12*
c^12 - 12*a*b^11*c^11*d + 66*a^2*b^10*c^10*d^2 - 220*a^3*b^9*c^9*d^3 + 495*a^4*b^8*c^8*d^4 - 792*a^5*b^7*c^7*d
^5 + 924*a^6*b^6*c^6*d^6 - 792*a^7*b^5*c^5*d^7 + 495*a^8*b^4*c^4*d^8 - 220*a^9*b^3*c^3*d^9 + 66*a^10*b^2*c^2*d
^10 - 12*a^11*b*c*d^11 + a^12*d^12)/(a^3*b^13))^(1/4)*log(a*b^3*(-(b^12*c^12 - 12*a*b^11*c^11*d + 66*a^2*b^10*
c^10*d^2 - 220*a^3*b^9*c^9*d^3 + 495*a^4*b^8*c^8*d^4 - 792*a^5*b^7*c^7*d^5 + 924*a^6*b^6*c^6*d^6 - 792*a^7*b^5
*c^5*d^7 + 495*a^8*b^4*c^4*d^8 - 220*a^9*b^3*c^3*d^9 + 66*a^10*b^2*c^2*d^10 - 12*a^11*b*c*d^11 + a^12*d^12)/(a
^3*b^13))^(1/4) - (b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*sqrt(x)) - 45*b^3*(-(b^12*c^12 - 12*a*b^
11*c^11*d + 66*a^2*b^10*c^10*d^2 - 220*a^3*b^9*c^9*d^3 + 495*a^4*b^8*c^8*d^4 - 792*a^5*b^7*c^7*d^5 + 924*a^6*b
^6*c^6*d^6 - 792*a^7*b^5*c^5*d^7 + 495*a^8*b^4*c^4*d^8 - 220*a^9*b^3*c^3*d^9 + 66*a^10*b^2*c^2*d^10 - 12*a^11*
b*c*d^11 + a^12*d^12)/(a^3*b^13))^(1/4)*log(-a*b^3*(-(b^12*c^12 - 12*a*b^11*c^11*d + 66*a^2*b^10*c^10*d^2 - 22
0*a^3*b^9*c^9*d^3 + 495*a^4*b^8*c^8*d^4 - 792*a^5*b^7*c^7*d^5 + 924*a^6*b^6*c^6*d^6 - 792*a^7*b^5*c^5*d^7 + 49
5*a^8*b^4*c^4*d^8 - 220*a^9*b^3*c^3*d^9 + 66*a^10*b^2*c^2*d^10 - 12*a^11*b*c*d^11 + a^12*d^12)/(a^3*b^13))^(1/
4) - (b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*sqrt(x)) - 4*(5*b^2*d^3*x^4 + 135*b^2*c^2*d - 135*a*b
*c*d^2 + 45*a^2*d^3 + 9*(3*b^2*c*d^2 - a*b*d^3)*x^2)*sqrt(x))/b^3

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Sympy [A]  time = 106.066, size = 889, normalized size = 2.92 \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)/x**(1/2),x)

[Out]

Piecewise((zoo*(-2*c**3/(3*x**(3/2)) + 6*c**2*d*sqrt(x) + 6*c*d**2*x**(5/2)/5 + 2*d**3*x**(9/2)/9), Eq(a, 0) &
Eq(b, 0)), ((2*c**3*sqrt(x) + 6*c**2*d*x**(5/2)/5 + 2*c*d**2*x**(9/2)/3 + 2*d**3*x**(13/2)/13)/a, Eq(b, 0)),
((-2*c**3/(3*x**(3/2)) + 6*c**2*d*sqrt(x) + 6*c*d**2*x**(5/2)/5 + 2*d**3*x**(9/2)/9)/b, Eq(a, 0)), ((-1)**(1/4
)*a**(9/4)*d**3*log(-(-1)**(1/4)*a**(1/4)*(1/b)**(1/4) + sqrt(x))/(2*b**7*(1/b)**(15/4)) - (-1)**(1/4)*a**(9/4
)*d**3*log((-1)**(1/4)*a**(1/4)*(1/b)**(1/4) + sqrt(x))/(2*b**7*(1/b)**(15/4)) + (-1)**(1/4)*a**(9/4)*d**3*ata
n((-1)**(3/4)*sqrt(x)/(a**(1/4)*(1/b)**(1/4)))/(b**7*(1/b)**(15/4)) - 3*(-1)**(1/4)*a**(5/4)*c*d**2*log(-(-1)*
*(1/4)*a**(1/4)*(1/b)**(1/4) + sqrt(x))/(2*b**6*(1/b)**(15/4)) + 3*(-1)**(1/4)*a**(5/4)*c*d**2*log((-1)**(1/4)
*a**(1/4)*(1/b)**(1/4) + sqrt(x))/(2*b**6*(1/b)**(15/4)) - 3*(-1)**(1/4)*a**(5/4)*c*d**2*atan((-1)**(3/4)*sqrt
(x)/(a**(1/4)*(1/b)**(1/4)))/(b**6*(1/b)**(15/4)) + 3*(-1)**(1/4)*a**(1/4)*c**2*d*log(-(-1)**(1/4)*a**(1/4)*(1
/b)**(1/4) + sqrt(x))/(2*b**5*(1/b)**(15/4)) - 3*(-1)**(1/4)*a**(1/4)*c**2*d*log((-1)**(1/4)*a**(1/4)*(1/b)**(
1/4) + sqrt(x))/(2*b**5*(1/b)**(15/4)) + 3*(-1)**(1/4)*a**(1/4)*c**2*d*atan((-1)**(3/4)*sqrt(x)/(a**(1/4)*(1/b
)**(1/4)))/(b**5*(1/b)**(15/4)) + 2*a**2*d**3*sqrt(x)/b**3 - 6*a*c*d**2*sqrt(x)/b**2 - 2*a*d**3*x**(5/2)/(5*b*
*2) + 6*c**2*d*sqrt(x)/b + 6*c*d**2*x**(5/2)/(5*b) + 2*d**3*x**(9/2)/(9*b) - (-1)**(1/4)*c**3*log(-(-1)**(1/4)
*a**(1/4)*(1/b)**(1/4) + sqrt(x))/(2*a**(3/4)*b**4*(1/b)**(15/4)) + (-1)**(1/4)*c**3*log((-1)**(1/4)*a**(1/4)*
(1/b)**(1/4) + sqrt(x))/(2*a**(3/4)*b**4*(1/b)**(15/4)) - (-1)**(1/4)*c**3*atan((-1)**(3/4)*sqrt(x)/(a**(1/4)*
(1/b)**(1/4)))/(a**(3/4)*b**4*(1/b)**(15/4)), True))

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Giac [B]  time = 1.21652, size = 662, normalized size = 2.18 \begin{align*} \frac{\sqrt{2}{\left (\left (a b^{3}\right )^{\frac{1}{4}} b^{3} c^{3} - 3 \, \left (a b^{3}\right )^{\frac{1}{4}} a b^{2} c^{2} d + 3 \, \left (a b^{3}\right )^{\frac{1}{4}} a^{2} b c d^{2} - \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 )}{2 \, a b^{4}} + \frac{\sqrt{2}{\left (\left (a b^{3}\right )^{\frac{1}{4}} b^{3} c^{3} - 3 \, \left (a b^{3}\right )^{\frac{1}{4}} a b^{2} c^{2} d + 3 \, \left (a b^{3}\right )^{\frac{1}{4}} a^{2} b c d^{2} - \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 )}{2 \, a b^{4}} + \frac{\sqrt{2}{\left (\left (a b^{3}\right )^{\frac{1}{4}} b^{3} c^{3} - 3 \, \left (a b^{3}\right )^{\frac{1}{4}} a b^{2} c^{2} d + 3 \, \left (a b^{3}\right )^{\frac{1}{4}} a^{2} b c d^{2} - \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 )}{4 \, a b^{4}} - \frac{\sqrt{2}{\left (\left (a b^{3}\right )^{\frac{1}{4}} b^{3} c^{3} - 3 \, \left (a b^{3}\right )^{\frac{1}{4}} a b^{2} c^{2} d + 3 \, \left (a b^{3}\right )^{\frac{1}{4}} a^{2} b c d^{2} - \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 )}{4 \, a b^{4}} + \frac{2 \,{\left (5 \, b^{8} d^{3} x^{\frac{9}{2}} + 27 \, b^{8} c d^{2} x^{\frac{5}{2}} - 9 \, a b^{7} d^{3} x^{\frac{5}{2}} + 135 \, b^{8} c^{2} d \sqrt{x} - 135 \, a b^{7} c d^{2} \sqrt{x} + 45 \, a^{2} b^{6} d^{3} \sqrt{x}\right )}}{45 \, b^{9}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(2)*((a*b^3)^(1/4)*b^3*c^3 - 3*(a*b^3)^(1/4)*a*b^2*c^2*d + 3*(a*b^3)^(1/4)*a^2*b*c*d^2 - (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*b^4) + 1/2*sqrt(2)*((a*b^3)^(1/
4)*b^3*c^3 - 3*(a*b^3)^(1/4)*a*b^2*c^2*d + 3*(a*b^3)^(1/4)*a^2*b*c*d^2 - (a*b^3)^(1/4)*a^3*d^3)*arctan(-1/2*sq
rt(2)*(sqrt(2)*(a/b)^(1/4) - 2*sqrt(x))/(a/b)^(1/4))/(a*b^4) + 1/4*sqrt(2)*((a*b^3)^(1/4)*b^3*c^3 - 3*(a*b^3)^
(1/4)*a*b^2*c^2*d + 3*(a*b^3)^(1/4)*a^2*b*c*d^2 - (a*b^3)^(1/4)*a^3*d^3)*log(sqrt(2)*sqrt(x)*(a/b)^(1/4) + x +
sqrt(a/b))/(a*b^4) - 1/4*sqrt(2)*((a*b^3)^(1/4)*b^3*c^3 - 3*(a*b^3)^(1/4)*a*b^2*c^2*d + 3*(a*b^3)^(1/4)*a^2*b
*c*d^2 - (a*b^3)^(1/4)*a^3*d^3)*log(-sqrt(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))/(a*b^4) + 2/45*(5*b^8*d^3*x^
(9/2) + 27*b^8*c*d^2*x^(5/2) - 9*a*b^7*d^3*x^(5/2) + 135*b^8*c^2*d*sqrt(x) - 135*a*b^7*c*d^2*sqrt(x) + 45*a^2*
b^6*d^3*sqrt(x))/b^9