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

Optimal. Leaf size=268 \[ -\frac{2 b x^{3/2} (b c-2 a d)}{3 d^2}+\frac{(b c-a d)^2 \log \left (-\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{2 \sqrt{2} \sqrt [4]{c} d^{11/4}}-\frac{(b c-a d)^2 \log \left (\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{2 \sqrt{2} \sqrt [4]{c} d^{11/4}}-\frac{(b c-a d)^2 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}\right )}{\sqrt{2} \sqrt [4]{c} d^{11/4}}+\frac{(b c-a d)^2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}+1\right )}{\sqrt{2} \sqrt [4]{c} d^{11/4}}+\frac{2 b^2 x^{7/2}}{7 d} \]

[Out]

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

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Rubi [A]  time = 0.233037, antiderivative size = 268, 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, 297, 1162, 617, 204, 1165, 628} \[ -\frac{2 b x^{3/2} (b c-2 a d)}{3 d^2}+\frac{(b c-a d)^2 \log \left (-\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{2 \sqrt{2} \sqrt [4]{c} d^{11/4}}-\frac{(b c-a d)^2 \log \left (\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{2 \sqrt{2} \sqrt [4]{c} d^{11/4}}-\frac{(b c-a d)^2 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}\right )}{\sqrt{2} \sqrt [4]{c} d^{11/4}}+\frac{(b c-a d)^2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}+1\right )}{\sqrt{2} \sqrt [4]{c} d^{11/4}}+\frac{2 b^2 x^{7/2}}{7 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.112958, size = 249, normalized size = 0.93 \[ \frac{-56 b \sqrt [4]{c} d^{3/4} x^{3/2} (b c-2 a d)+21 \sqrt{2} (b c-a d)^2 \log \left (-\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )-21 \sqrt{2} (b c-a d)^2 \log \left (\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )-42 \sqrt{2} (b c-a d)^2 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}\right )+42 \sqrt{2} (b c-a d)^2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}+1\right )+24 b^2 \sqrt [4]{c} d^{7/4} x^{7/2}}{84 \sqrt [4]{c} d^{11/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-56*b*c^(1/4)*d^(3/4)*(b*c - 2*a*d)*x^(3/2) + 24*b^2*c^(1/4)*d^(7/4)*x^(7/2) - 42*Sqrt[2]*(b*c - a*d)^2*ArcTa
n[1 - (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)] + 42*Sqrt[2]*(b*c - a*d)^2*ArcTan[1 + (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/
4)] + 21*Sqrt[2]*(b*c - a*d)^2*Log[Sqrt[c] - Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x] - 21*Sqrt[2]*(b*c -
a*d)^2*Log[Sqrt[c] + Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x])/(84*c^(1/4)*d^(11/4))

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Maple [B]  time = 0.009, size = 461, normalized size = 1.7 \begin{align*}{\frac{2\,{b}^{2}}{7\,d}{x}^{{\frac{7}{2}}}}+{\frac{4\,ab}{3\,d}{x}^{{\frac{3}{2}}}}-{\frac{2\,{b}^{2}c}{3\,{d}^{2}}{x}^{{\frac{3}{2}}}}+{\frac{\sqrt{2}{a}^{2}}{2\,d}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{c}{d}}}}}}+1 \right ){\frac{1}{\sqrt [4]{{\frac{c}{d}}}}}}-{\frac{\sqrt{2}abc}{{d}^{2}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{c}{d}}}}}}+1 \right ){\frac{1}{\sqrt [4]{{\frac{c}{d}}}}}}+{\frac{\sqrt{2}{b}^{2}{c}^{2}}{2\,{d}^{3}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{c}{d}}}}}}+1 \right ){\frac{1}{\sqrt [4]{{\frac{c}{d}}}}}}+{\frac{\sqrt{2}{a}^{2}}{2\,d}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{c}{d}}}}}}-1 \right ){\frac{1}{\sqrt [4]{{\frac{c}{d}}}}}}-{\frac{\sqrt{2}abc}{{d}^{2}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{c}{d}}}}}}-1 \right ){\frac{1}{\sqrt [4]{{\frac{c}{d}}}}}}+{\frac{\sqrt{2}{b}^{2}{c}^{2}}{2\,{d}^{3}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{c}{d}}}}}}-1 \right ){\frac{1}{\sqrt [4]{{\frac{c}{d}}}}}}+{\frac{\sqrt{2}{a}^{2}}{4\,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{1}{\sqrt [4]{{\frac{c}{d}}}}}}-{\frac{\sqrt{2}abc}{2\,{d}^{2}}\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{1}{\sqrt [4]{{\frac{c}{d}}}}}}+{\frac{\sqrt{2}{b}^{2}{c}^{2}}{4\,{d}^{3}}\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{1}{\sqrt [4]{{\frac{c}{d}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.1626, size = 3380, normalized size = 12.61 \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*x^(1/2)/(d*x^2+c),x, algorithm="fricas")

[Out]

-1/42*(84*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*d^11))^(1/4)*arctan((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^1
1*b*c*d^11 + a^12*d^12)*x - (b^8*c^9*d^5 - 8*a*b^7*c^8*d^6 + 28*a^2*b^6*c^7*d^7 - 56*a^3*b^5*c^6*d^8 + 70*a^4*
b^4*c^5*d^9 - 56*a^5*b^3*c^4*d^10 + 28*a^6*b^2*c^3*d^11 - 8*a^7*b*c^2*d^12 + a^8*c*d^13)*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*d^11)))*d^3*(-(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*d^11))^(1/4
) - (b^6*c^6*d^3 - 6*a*b^5*c^5*d^4 + 15*a^2*b^4*c^4*d^5 - 20*a^3*b^3*c^3*d^6 + 15*a^4*b^2*c^2*d^7 - 6*a^5*b*c*
d^8 + a^6*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*d^11))^(1/4))/(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)) - 21*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*d^11))^(1/4)*log(c*d^8
*(-(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*d^11))^(3/4) + (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)*sqrt(x)) + 21*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*d^11))^(1/4)*log(-c*d^8*(-(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*d
^11))^(3/4) + (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)*sqrt(x)) - 4*(3*b^2*d*x^3 - 7*(b^2*c - 2*a*b*d)*x)*sqrt(x))/d^2

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError

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Giac [A]  time = 1.16849, size = 487, normalized size = 1.82 \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 \, c d^{5}} + \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 \, c d^{5}} - \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 \, c d^{5}} + \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 \, c d^{5}} + \frac{2 \,{\left (3 \, b^{2} d^{6} x^{\frac{7}{2}} - 7 \, b^{2} c d^{5} x^{\frac{3}{2}} + 14 \, a b d^{6} x^{\frac{3}{2}}\right )}}{21 \, d^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*x^(1/2)/(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)*(sqrt
(2)*(c/d)^(1/4) + 2*sqrt(x))/(c/d)^(1/4))/(c*d^5) + 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))/(c*d^5) - 1/4*s
qrt(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))/(c*d^5) + 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))/(c*d^5) + 2/21*(3*b^2*d^6*x^(7/2) - 7*b^2*c*d^5*x^(3/2
) + 14*a*b*d^6*x^(3/2))/d^7