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

Optimal. Leaf size=266 \[ -\frac{(b c-a d)^2 \log \left (-\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{2 \sqrt{2} c^{3/4} d^{9/4}}+\frac{(b c-a d)^2 \log \left (\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{2 \sqrt{2} c^{3/4} d^{9/4}}-\frac{(b c-a d)^2 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}\right )}{\sqrt{2} c^{3/4} d^{9/4}}+\frac{(b c-a d)^2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}+1\right )}{\sqrt{2} c^{3/4} d^{9/4}}-\frac{2 b \sqrt{x} (b c-2 a d)}{d^2}+\frac{2 b^2 x^{5/2}}{5 d} \]

[Out]

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

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Rubi [A]  time = 0.211948, antiderivative size = 266, 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} \[ -\frac{(b c-a d)^2 \log \left (-\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{2 \sqrt{2} c^{3/4} d^{9/4}}+\frac{(b c-a d)^2 \log \left (\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{2 \sqrt{2} c^{3/4} d^{9/4}}-\frac{(b c-a d)^2 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}\right )}{\sqrt{2} c^{3/4} d^{9/4}}+\frac{(b c-a d)^2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}+1\right )}{\sqrt{2} c^{3/4} d^{9/4}}-\frac{2 b \sqrt{x} (b c-2 a d)}{d^2}+\frac{2 b^2 x^{5/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.110403, size = 249, normalized size = 0.94 \[ \frac{-40 b c^{3/4} \sqrt [4]{d} \sqrt{x} (b c-2 a d)-5 \sqrt{2} (b c-a d)^2 \log \left (-\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )+5 \sqrt{2} (b c-a d)^2 \log \left (\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )-10 \sqrt{2} (b c-a d)^2 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}\right )+10 \sqrt{2} (b c-a d)^2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}+1\right )+8 b^2 c^{3/4} d^{5/4} x^{5/2}}{20 c^{3/4} d^{9/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-40*b*c^(3/4)*d^(1/4)*(b*c - 2*a*d)*Sqrt[x] + 8*b^2*c^(3/4)*d^(5/4)*x^(5/2) - 10*Sqrt[2]*(b*c - a*d)^2*ArcTan
[1 - (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)] + 10*Sqrt[2]*(b*c - a*d)^2*ArcTan[1 + (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4
)] - 5*Sqrt[2]*(b*c - a*d)^2*Log[Sqrt[c] - Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x] + 5*Sqrt[2]*(b*c - a*d
)^2*Log[Sqrt[c] + Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x])/(20*c^(3/4)*d^(9/4))

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Maple [B]  time = 0.009, size = 452, normalized size = 1.7 \begin{align*}{\frac{2\,{b}^{2}}{5\,d}{x}^{{\frac{5}{2}}}}+4\,{\frac{ab\sqrt{x}}{d}}-2\,{\frac{{b}^{2}\sqrt{x}c}{{d}^{2}}}+{\frac{\sqrt{2}{a}^{2}}{2\,c}\sqrt [4]{{\frac{c}{d}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{c}{d}}}}}}+1 \right ) }-{\frac{\sqrt{2}ab}{d}\sqrt [4]{{\frac{c}{d}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{c}{d}}}}}}+1 \right ) }+{\frac{c\sqrt{2}{b}^{2}}{2\,{d}^{2}}\sqrt [4]{{\frac{c}{d}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{c}{d}}}}}}+1 \right ) }+{\frac{\sqrt{2}{a}^{2}}{2\,c}\sqrt [4]{{\frac{c}{d}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{c}{d}}}}}}-1 \right ) }-{\frac{\sqrt{2}ab}{d}\sqrt [4]{{\frac{c}{d}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{c}{d}}}}}}-1 \right ) }+{\frac{c\sqrt{2}{b}^{2}}{2\,{d}^{2}}\sqrt [4]{{\frac{c}{d}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{c}{d}}}}}}-1 \right ) }+{\frac{\sqrt{2}{a}^{2}}{4\,c}\sqrt [4]{{\frac{c}{d}}}\ln \left ({ \left ( x+\sqrt [4]{{\frac{c}{d}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{c}{d}}} \right ) \left ( x-\sqrt [4]{{\frac{c}{d}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{c}{d}}} \right ) ^{-1}} \right ) }-{\frac{\sqrt{2}ab}{2\,d}\sqrt [4]{{\frac{c}{d}}}\ln \left ({ \left ( x+\sqrt [4]{{\frac{c}{d}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{c}{d}}} \right ) \left ( x-\sqrt [4]{{\frac{c}{d}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{c}{d}}} \right ) ^{-1}} \right ) }+{\frac{c\sqrt{2}{b}^{2}}{4\,{d}^{2}}\sqrt [4]{{\frac{c}{d}}}\ln \left ({ \left ( x+\sqrt [4]{{\frac{c}{d}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{c}{d}}} \right ) \left ( x-\sqrt [4]{{\frac{c}{d}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{c}{d}}} \right ) ^{-1}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.15252, size = 2581, normalized size = 9.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*(20*d^2*(-(b^8*c^8 - 8*a*b^7*c^7*d + 28*a^2*b^6*c^6*d^2 - 56*a^3*b^5*c^5*d^3 + 70*a^4*b^4*c^4*d^4 - 56*a^
5*b^3*c^3*d^5 + 28*a^6*b^2*c^2*d^6 - 8*a^7*b*c*d^7 + a^8*d^8)/(c^3*d^9))^(1/4)*arctan((sqrt(c^2*d^4*sqrt(-(b^8
*c^8 - 8*a*b^7*c^7*d + 28*a^2*b^6*c^6*d^2 - 56*a^3*b^5*c^5*d^3 + 70*a^4*b^4*c^4*d^4 - 56*a^5*b^3*c^3*d^5 + 28*
a^6*b^2*c^2*d^6 - 8*a^7*b*c*d^7 + a^8*d^8)/(c^3*d^9)) + (b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b
*c*d^3 + a^4*d^4)*x)*c^2*d^7*(-(b^8*c^8 - 8*a*b^7*c^7*d + 28*a^2*b^6*c^6*d^2 - 56*a^3*b^5*c^5*d^3 + 70*a^4*b^4
*c^4*d^4 - 56*a^5*b^3*c^3*d^5 + 28*a^6*b^2*c^2*d^6 - 8*a^7*b*c*d^7 + a^8*d^8)/(c^3*d^9))^(3/4) - (b^2*c^4*d^7
- 2*a*b*c^3*d^8 + a^2*c^2*d^9)*sqrt(x)*(-(b^8*c^8 - 8*a*b^7*c^7*d + 28*a^2*b^6*c^6*d^2 - 56*a^3*b^5*c^5*d^3 +
70*a^4*b^4*c^4*d^4 - 56*a^5*b^3*c^3*d^5 + 28*a^6*b^2*c^2*d^6 - 8*a^7*b*c*d^7 + a^8*d^8)/(c^3*d^9))^(3/4))/(b^8
*c^8 - 8*a*b^7*c^7*d + 28*a^2*b^6*c^6*d^2 - 56*a^3*b^5*c^5*d^3 + 70*a^4*b^4*c^4*d^4 - 56*a^5*b^3*c^3*d^5 + 28*
a^6*b^2*c^2*d^6 - 8*a^7*b*c*d^7 + a^8*d^8)) + 5*d^2*(-(b^8*c^8 - 8*a*b^7*c^7*d + 28*a^2*b^6*c^6*d^2 - 56*a^3*b
^5*c^5*d^3 + 70*a^4*b^4*c^4*d^4 - 56*a^5*b^3*c^3*d^5 + 28*a^6*b^2*c^2*d^6 - 8*a^7*b*c*d^7 + a^8*d^8)/(c^3*d^9)
)^(1/4)*log(c*d^2*(-(b^8*c^8 - 8*a*b^7*c^7*d + 28*a^2*b^6*c^6*d^2 - 56*a^3*b^5*c^5*d^3 + 70*a^4*b^4*c^4*d^4 -
56*a^5*b^3*c^3*d^5 + 28*a^6*b^2*c^2*d^6 - 8*a^7*b*c*d^7 + a^8*d^8)/(c^3*d^9))^(1/4) + (b^2*c^2 - 2*a*b*c*d + a
^2*d^2)*sqrt(x)) - 5*d^2*(-(b^8*c^8 - 8*a*b^7*c^7*d + 28*a^2*b^6*c^6*d^2 - 56*a^3*b^5*c^5*d^3 + 70*a^4*b^4*c^4
*d^4 - 56*a^5*b^3*c^3*d^5 + 28*a^6*b^2*c^2*d^6 - 8*a^7*b*c*d^7 + a^8*d^8)/(c^3*d^9))^(1/4)*log(-c*d^2*(-(b^8*c
^8 - 8*a*b^7*c^7*d + 28*a^2*b^6*c^6*d^2 - 56*a^3*b^5*c^5*d^3 + 70*a^4*b^4*c^4*d^4 - 56*a^5*b^3*c^3*d^5 + 28*a^
6*b^2*c^2*d^6 - 8*a^7*b*c*d^7 + a^8*d^8)/(c^3*d^9))^(1/4) + (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqrt(x)) + 4*(b^2*
d*x^2 - 5*b^2*c + 10*a*b*d)*sqrt(x))/d^2

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Sympy [A]  time = 30.7488, size = 612, normalized size = 2.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**2/(d*x**2+c)/x**(1/2),x)

[Out]

Piecewise((zoo*(-2*a**2/(3*x**(3/2)) + 4*a*b*sqrt(x) + 2*b**2*x**(5/2)/5), Eq(c, 0) & Eq(d, 0)), ((-2*a**2/(3*
x**(3/2)) + 4*a*b*sqrt(x) + 2*b**2*x**(5/2)/5)/d, Eq(c, 0)), ((2*a**2*sqrt(x) + 4*a*b*x**(5/2)/5 + 2*b**2*x**(
9/2)/9)/c, Eq(d, 0)), (-(-1)**(1/4)*a**2*d**10*(1/d)**(41/4)*log(-(-1)**(1/4)*c**(1/4)*(1/d)**(1/4) + sqrt(x))
/(2*c**(3/4)) + (-1)**(1/4)*a**2*d**10*(1/d)**(41/4)*log((-1)**(1/4)*c**(1/4)*(1/d)**(1/4) + sqrt(x))/(2*c**(3
/4)) - (-1)**(1/4)*a**2*d**10*(1/d)**(41/4)*atan((-1)**(3/4)*sqrt(x)/(c**(1/4)*(1/d)**(1/4)))/c**(3/4) + (-1)*
*(1/4)*a*b*c**(1/4)*d**9*(1/d)**(41/4)*log(-(-1)**(1/4)*c**(1/4)*(1/d)**(1/4) + sqrt(x)) - (-1)**(1/4)*a*b*c**
(1/4)*d**9*(1/d)**(41/4)*log((-1)**(1/4)*c**(1/4)*(1/d)**(1/4) + sqrt(x)) + 2*(-1)**(1/4)*a*b*c**(1/4)*d**9*(1
/d)**(41/4)*atan((-1)**(3/4)*sqrt(x)/(c**(1/4)*(1/d)**(1/4))) + 4*a*b*sqrt(x)/d - (-1)**(1/4)*b**2*c**(5/4)*d*
*8*(1/d)**(41/4)*log(-(-1)**(1/4)*c**(1/4)*(1/d)**(1/4) + sqrt(x))/2 + (-1)**(1/4)*b**2*c**(5/4)*d**8*(1/d)**(
41/4)*log((-1)**(1/4)*c**(1/4)*(1/d)**(1/4) + sqrt(x))/2 - (-1)**(1/4)*b**2*c**(5/4)*d**8*(1/d)**(41/4)*atan((
-1)**(3/4)*sqrt(x)/(c**(1/4)*(1/d)**(1/4))) - 2*b**2*c*sqrt(x)/d**2 + 2*b**2*x**(5/2)/(5*d), True))

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Giac [A]  time = 1.15362, size = 486, normalized size = 1.83 \begin{align*} \frac{\sqrt{2}{\left (\left (c d^{3}\right )^{\frac{1}{4}} b^{2} c^{2} - 2 \, \left (c d^{3}\right )^{\frac{1}{4}} a b c d + \left (c d^{3}\right )^{\frac{1}{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 \, c d^{3}} + \frac{\sqrt{2}{\left (\left (c d^{3}\right )^{\frac{1}{4}} b^{2} c^{2} - 2 \, \left (c d^{3}\right )^{\frac{1}{4}} a b c d + \left (c d^{3}\right )^{\frac{1}{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 \, c d^{3}} + \frac{\sqrt{2}{\left (\left (c d^{3}\right )^{\frac{1}{4}} b^{2} c^{2} - 2 \, \left (c d^{3}\right )^{\frac{1}{4}} a b c d + \left (c d^{3}\right )^{\frac{1}{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 \, c d^{3}} - \frac{\sqrt{2}{\left (\left (c d^{3}\right )^{\frac{1}{4}} b^{2} c^{2} - 2 \, \left (c d^{3}\right )^{\frac{1}{4}} a b c d + \left (c d^{3}\right )^{\frac{1}{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 \, c d^{3}} + \frac{2 \,{\left (b^{2} d^{4} x^{\frac{5}{2}} - 5 \, b^{2} c d^{3} \sqrt{x} + 10 \, a b d^{4} \sqrt{x}\right )}}{5 \, d^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(2)*((c*d^3)^(1/4)*b^2*c^2 - 2*(c*d^3)^(1/4)*a*b*c*d + (c*d^3)^(1/4)*a^2*d^2)*arctan(1/2*sqrt(2)*(sqrt
(2)*(c/d)^(1/4) + 2*sqrt(x))/(c/d)^(1/4))/(c*d^3) + 1/2*sqrt(2)*((c*d^3)^(1/4)*b^2*c^2 - 2*(c*d^3)^(1/4)*a*b*c
*d + (c*d^3)^(1/4)*a^2*d^2)*arctan(-1/2*sqrt(2)*(sqrt(2)*(c/d)^(1/4) - 2*sqrt(x))/(c/d)^(1/4))/(c*d^3) + 1/4*s
qrt(2)*((c*d^3)^(1/4)*b^2*c^2 - 2*(c*d^3)^(1/4)*a*b*c*d + (c*d^3)^(1/4)*a^2*d^2)*log(sqrt(2)*sqrt(x)*(c/d)^(1/
4) + x + sqrt(c/d))/(c*d^3) - 1/4*sqrt(2)*((c*d^3)^(1/4)*b^2*c^2 - 2*(c*d^3)^(1/4)*a*b*c*d + (c*d^3)^(1/4)*a^2
*d^2)*log(-sqrt(2)*sqrt(x)*(c/d)^(1/4) + x + sqrt(c/d))/(c*d^3) + 2/5*(b^2*d^4*x^(5/2) - 5*b^2*c*d^3*sqrt(x) +
10*a*b*d^4*sqrt(x))/d^5