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

Optimal. Leaf size=310 \[ -\frac{(b c-a d) (a d+7 b c) \log \left (-\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{8 \sqrt{2} c^{5/4} d^{11/4}}+\frac{(b c-a d) (a d+7 b c) \log \left (\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{8 \sqrt{2} c^{5/4} d^{11/4}}+\frac{(b c-a d) (a d+7 b c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}\right )}{4 \sqrt{2} c^{5/4} d^{11/4}}-\frac{(b c-a d) (a d+7 b c) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}+1\right )}{4 \sqrt{2} c^{5/4} d^{11/4}}+\frac{x^{3/2} (b c-a d)^2}{2 c d^2 \left (c+d x^2\right )}+\frac{2 b^2 x^{3/2}}{3 d^2} \]

[Out]

(2*b^2*x^(3/2))/(3*d^2) + ((b*c - a*d)^2*x^(3/2))/(2*c*d^2*(c + d*x^2)) + ((b*c - a*d)*(7*b*c + a*d)*ArcTan[1
- (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/(4*Sqrt[2]*c^(5/4)*d^(11/4)) - ((b*c - a*d)*(7*b*c + a*d)*ArcTan[1 + (Sq
rt[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/(4*Sqrt[2]*c^(5/4)*d^(11/4)) - ((b*c - a*d)*(7*b*c + a*d)*Log[Sqrt[c] - Sqrt[
2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x])/(8*Sqrt[2]*c^(5/4)*d^(11/4)) + ((b*c - a*d)*(7*b*c + a*d)*Log[Sqrt[c]
+ Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x])/(8*Sqrt[2]*c^(5/4)*d^(11/4))

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Rubi [A]  time = 0.281799, antiderivative size = 310, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 9, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {463, 459, 329, 297, 1162, 617, 204, 1165, 628} \[ -\frac{(b c-a d) (a d+7 b c) \log \left (-\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{8 \sqrt{2} c^{5/4} d^{11/4}}+\frac{(b c-a d) (a d+7 b c) \log \left (\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{8 \sqrt{2} c^{5/4} d^{11/4}}+\frac{(b c-a d) (a d+7 b c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}\right )}{4 \sqrt{2} c^{5/4} d^{11/4}}-\frac{(b c-a d) (a d+7 b c) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}+1\right )}{4 \sqrt{2} c^{5/4} d^{11/4}}+\frac{x^{3/2} (b c-a d)^2}{2 c d^2 \left (c+d x^2\right )}+\frac{2 b^2 x^{3/2}}{3 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b^2*x^(3/2))/(3*d^2) + ((b*c - a*d)^2*x^(3/2))/(2*c*d^2*(c + d*x^2)) + ((b*c - a*d)*(7*b*c + a*d)*ArcTan[1
- (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/(4*Sqrt[2]*c^(5/4)*d^(11/4)) - ((b*c - a*d)*(7*b*c + a*d)*ArcTan[1 + (Sq
rt[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/(4*Sqrt[2]*c^(5/4)*d^(11/4)) - ((b*c - a*d)*(7*b*c + a*d)*Log[Sqrt[c] - Sqrt[
2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x])/(8*Sqrt[2]*c^(5/4)*d^(11/4)) + ((b*c - a*d)*(7*b*c + a*d)*Log[Sqrt[c]
+ Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x])/(8*Sqrt[2]*c^(5/4)*d^(11/4))

Rule 463

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> -Simp[((b*c - a*
d)^2*(e*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*b^2*e*n*(p + 1)), x] + Dist[1/(a*b^2*n*(p + 1)), Int[(e*x)^m*(a + b
*x^n)^(p + 1)*Simp[(b*c - a*d)^2*(m + 1) + b^2*c^2*n*(p + 1) + a*b*d^2*n*(p + 1)*x^n, x], x], x] /; FreeQ[{a,
b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1]

Rule 459

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

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}{\left (c+d x^2\right )^2} \, dx &=\frac{(b c-a d)^2 x^{3/2}}{2 c d^2 \left (c+d x^2\right )}-\frac{\int \frac{\sqrt{x} \left (\frac{1}{2} \left (-4 a^2 d^2+3 (b c-a d)^2\right )-2 b^2 c d x^2\right )}{c+d x^2} \, dx}{2 c d^2}\\ &=\frac{2 b^2 x^{3/2}}{3 d^2}+\frac{(b c-a d)^2 x^{3/2}}{2 c d^2 \left (c+d x^2\right )}-\frac{((b c-a d) (7 b c+a d)) \int \frac{\sqrt{x}}{c+d x^2} \, dx}{4 c d^2}\\ &=\frac{2 b^2 x^{3/2}}{3 d^2}+\frac{(b c-a d)^2 x^{3/2}}{2 c d^2 \left (c+d x^2\right )}-\frac{((b c-a d) (7 b c+a d)) \operatorname{Subst}\left (\int \frac{x^2}{c+d x^4} \, dx,x,\sqrt{x}\right )}{2 c d^2}\\ &=\frac{2 b^2 x^{3/2}}{3 d^2}+\frac{(b c-a d)^2 x^{3/2}}{2 c d^2 \left (c+d x^2\right )}+\frac{((b c-a d) (7 b c+a d)) \operatorname{Subst}\left (\int \frac{\sqrt{c}-\sqrt{d} x^2}{c+d x^4} \, dx,x,\sqrt{x}\right )}{4 c d^{5/2}}-\frac{((b c-a d) (7 b c+a d)) \operatorname{Subst}\left (\int \frac{\sqrt{c}+\sqrt{d} x^2}{c+d x^4} \, dx,x,\sqrt{x}\right )}{4 c d^{5/2}}\\ &=\frac{2 b^2 x^{3/2}}{3 d^2}+\frac{(b c-a d)^2 x^{3/2}}{2 c d^2 \left (c+d x^2\right )}-\frac{((b c-a d) (7 b c+a d)) \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 )}{8 c d^3}-\frac{((b c-a d) (7 b c+a d)) \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 )}{8 c d^3}-\frac{((b c-a d) (7 b c+a d)) \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 )}{8 \sqrt{2} c^{5/4} d^{11/4}}-\frac{((b c-a d) (7 b c+a d)) \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 )}{8 \sqrt{2} c^{5/4} d^{11/4}}\\ &=\frac{2 b^2 x^{3/2}}{3 d^2}+\frac{(b c-a d)^2 x^{3/2}}{2 c d^2 \left (c+d x^2\right )}-\frac{(b c-a d) (7 b c+a d) \log \left (\sqrt{c}-\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{d} x\right )}{8 \sqrt{2} c^{5/4} d^{11/4}}+\frac{(b c-a d) (7 b c+a d) \log \left (\sqrt{c}+\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{d} x\right )}{8 \sqrt{2} c^{5/4} d^{11/4}}-\frac{((b c-a d) (7 b c+a d)) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}\right )}{4 \sqrt{2} c^{5/4} d^{11/4}}+\frac{((b c-a d) (7 b c+a d)) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}\right )}{4 \sqrt{2} c^{5/4} d^{11/4}}\\ &=\frac{2 b^2 x^{3/2}}{3 d^2}+\frac{(b c-a d)^2 x^{3/2}}{2 c d^2 \left (c+d x^2\right )}+\frac{(b c-a d) (7 b c+a d) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}\right )}{4 \sqrt{2} c^{5/4} d^{11/4}}-\frac{(b c-a d) (7 b c+a d) \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}\right )}{4 \sqrt{2} c^{5/4} d^{11/4}}-\frac{(b c-a d) (7 b c+a d) \log \left (\sqrt{c}-\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{d} x\right )}{8 \sqrt{2} c^{5/4} d^{11/4}}+\frac{(b c-a d) (7 b c+a d) \log \left (\sqrt{c}+\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{d} x\right )}{8 \sqrt{2} c^{5/4} d^{11/4}}\\ \end{align*}

Mathematica [A]  time = 0.178779, size = 319, normalized size = 1.03 \[ \frac{-\frac{3 \sqrt{2} \left (-a^2 d^2-6 a b c d+7 b^2 c^2\right ) \log \left (-\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{c^{5/4}}+\frac{3 \sqrt{2} \left (-a^2 d^2-6 a b c d+7 b^2 c^2\right ) \log \left (\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{c^{5/4}}+\frac{6 \sqrt{2} \left (-a^2 d^2-6 a b c d+7 b^2 c^2\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}\right )}{c^{5/4}}-\frac{6 \sqrt{2} \left (-a^2 d^2-6 a b c d+7 b^2 c^2\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}+1\right )}{c^{5/4}}+\frac{24 d^{3/4} x^{3/2} (b c-a d)^2}{c \left (c+d x^2\right )}+32 b^2 d^{3/4} x^{3/2}}{48 d^{11/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(32*b^2*d^(3/4)*x^(3/2) + (24*d^(3/4)*(b*c - a*d)^2*x^(3/2))/(c*(c + d*x^2)) + (6*Sqrt[2]*(7*b^2*c^2 - 6*a*b*c
*d - a^2*d^2)*ArcTan[1 - (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/c^(5/4) - (6*Sqrt[2]*(7*b^2*c^2 - 6*a*b*c*d - a^2
*d^2)*ArcTan[1 + (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/c^(5/4) - (3*Sqrt[2]*(7*b^2*c^2 - 6*a*b*c*d - a^2*d^2)*Lo
g[Sqrt[c] - Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x])/c^(5/4) + (3*Sqrt[2]*(7*b^2*c^2 - 6*a*b*c*d - a^2*d^
2)*Log[Sqrt[c] + Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x])/c^(5/4))/(48*d^(11/4))

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Maple [B]  time = 0.014, size = 499, normalized size = 1.6 \begin{align*} \text{result too large to display} \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)^2,x)

[Out]

2/3*b^2*x^(3/2)/d^2+1/2/c*x^(3/2)/(d*x^2+c)*a^2-1/d*x^(3/2)/(d*x^2+c)*a*b+1/2/d^2*c*x^(3/2)/(d*x^2+c)*b^2+1/8/
d/c/(c/d)^(1/4)*2^(1/2)*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)-1)*a^2+3/4/d^2/(c/d)^(1/4)*2^(1/2)*arctan(2^(1/2)/(
c/d)^(1/4)*x^(1/2)-1)*a*b-7/8/d^3*c/(c/d)^(1/4)*2^(1/2)*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)-1)*b^2+1/16/d/c/(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+3/8/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-7/16/d^3*c/(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+1/8/d/c/(c/d)^(1/4)*2^(1/2)*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)+1)*a^2+3/4/d^
2/(c/d)^(1/4)*2^(1/2)*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)+1)*a*b-7/8/d^3*c/(c/d)^(1/4)*2^(1/2)*arctan(2^(1/2)/(
c/d)^(1/4)*x^(1/2)+1)*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*x^(1/2)/(d*x^2+c)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.14783, size = 3846, normalized size = 12.41 \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)^2,x, algorithm="fricas")

[Out]

-1/24*(12*(c*d^3*x^2 + c^2*d^2)*(-(2401*b^8*c^8 - 8232*a*b^7*c^7*d + 9212*a^2*b^6*c^6*d^2 - 2520*a^3*b^5*c^5*d
^3 - 1434*a^4*b^4*c^4*d^4 + 360*a^5*b^3*c^3*d^5 + 188*a^6*b^2*c^2*d^6 + 24*a^7*b*c*d^7 + a^8*d^8)/(c^5*d^11))^
(1/4)*arctan((sqrt((117649*b^12*c^12 - 605052*a*b^11*c^11*d + 1195698*a^2*b^10*c^10*d^2 - 1049580*a^3*b^9*c^9*
d^3 + 247695*a^4*b^8*c^8*d^4 + 184968*a^5*b^7*c^7*d^5 - 73604*a^6*b^6*c^6*d^6 - 26424*a^7*b^5*c^5*d^7 + 5055*a
^8*b^4*c^4*d^8 + 3060*a^9*b^3*c^3*d^9 + 498*a^10*b^2*c^2*d^10 + 36*a^11*b*c*d^11 + a^12*d^12)*x - (2401*b^8*c^
11*d^5 - 8232*a*b^7*c^10*d^6 + 9212*a^2*b^6*c^9*d^7 - 2520*a^3*b^5*c^8*d^8 - 1434*a^4*b^4*c^7*d^9 + 360*a^5*b^
3*c^6*d^10 + 188*a^6*b^2*c^5*d^11 + 24*a^7*b*c^4*d^12 + a^8*c^3*d^13)*sqrt(-(2401*b^8*c^8 - 8232*a*b^7*c^7*d +
9212*a^2*b^6*c^6*d^2 - 2520*a^3*b^5*c^5*d^3 - 1434*a^4*b^4*c^4*d^4 + 360*a^5*b^3*c^3*d^5 + 188*a^6*b^2*c^2*d^
6 + 24*a^7*b*c*d^7 + a^8*d^8)/(c^5*d^11)))*c*d^3*(-(2401*b^8*c^8 - 8232*a*b^7*c^7*d + 9212*a^2*b^6*c^6*d^2 - 2
520*a^3*b^5*c^5*d^3 - 1434*a^4*b^4*c^4*d^4 + 360*a^5*b^3*c^3*d^5 + 188*a^6*b^2*c^2*d^6 + 24*a^7*b*c*d^7 + a^8*
d^8)/(c^5*d^11))^(1/4) + (343*b^6*c^7*d^3 - 882*a*b^5*c^6*d^4 + 609*a^2*b^4*c^5*d^5 + 36*a^3*b^3*c^4*d^6 - 87*
a^4*b^2*c^3*d^7 - 18*a^5*b*c^2*d^8 - a^6*c*d^9)*sqrt(x)*(-(2401*b^8*c^8 - 8232*a*b^7*c^7*d + 9212*a^2*b^6*c^6*
d^2 - 2520*a^3*b^5*c^5*d^3 - 1434*a^4*b^4*c^4*d^4 + 360*a^5*b^3*c^3*d^5 + 188*a^6*b^2*c^2*d^6 + 24*a^7*b*c*d^7
+ a^8*d^8)/(c^5*d^11))^(1/4))/(2401*b^8*c^8 - 8232*a*b^7*c^7*d + 9212*a^2*b^6*c^6*d^2 - 2520*a^3*b^5*c^5*d^3
- 1434*a^4*b^4*c^4*d^4 + 360*a^5*b^3*c^3*d^5 + 188*a^6*b^2*c^2*d^6 + 24*a^7*b*c*d^7 + a^8*d^8)) - 3*(c*d^3*x^2
+ c^2*d^2)*(-(2401*b^8*c^8 - 8232*a*b^7*c^7*d + 9212*a^2*b^6*c^6*d^2 - 2520*a^3*b^5*c^5*d^3 - 1434*a^4*b^4*c^
4*d^4 + 360*a^5*b^3*c^3*d^5 + 188*a^6*b^2*c^2*d^6 + 24*a^7*b*c*d^7 + a^8*d^8)/(c^5*d^11))^(1/4)*log(c^4*d^8*(-
(2401*b^8*c^8 - 8232*a*b^7*c^7*d + 9212*a^2*b^6*c^6*d^2 - 2520*a^3*b^5*c^5*d^3 - 1434*a^4*b^4*c^4*d^4 + 360*a^
5*b^3*c^3*d^5 + 188*a^6*b^2*c^2*d^6 + 24*a^7*b*c*d^7 + a^8*d^8)/(c^5*d^11))^(3/4) - (343*b^6*c^6 - 882*a*b^5*c
^5*d + 609*a^2*b^4*c^4*d^2 + 36*a^3*b^3*c^3*d^3 - 87*a^4*b^2*c^2*d^4 - 18*a^5*b*c*d^5 - a^6*d^6)*sqrt(x)) + 3*
(c*d^3*x^2 + c^2*d^2)*(-(2401*b^8*c^8 - 8232*a*b^7*c^7*d + 9212*a^2*b^6*c^6*d^2 - 2520*a^3*b^5*c^5*d^3 - 1434*
a^4*b^4*c^4*d^4 + 360*a^5*b^3*c^3*d^5 + 188*a^6*b^2*c^2*d^6 + 24*a^7*b*c*d^7 + a^8*d^8)/(c^5*d^11))^(1/4)*log(
-c^4*d^8*(-(2401*b^8*c^8 - 8232*a*b^7*c^7*d + 9212*a^2*b^6*c^6*d^2 - 2520*a^3*b^5*c^5*d^3 - 1434*a^4*b^4*c^4*d
^4 + 360*a^5*b^3*c^3*d^5 + 188*a^6*b^2*c^2*d^6 + 24*a^7*b*c*d^7 + a^8*d^8)/(c^5*d^11))^(3/4) - (343*b^6*c^6 -
882*a*b^5*c^5*d + 609*a^2*b^4*c^4*d^2 + 36*a^3*b^3*c^3*d^3 - 87*a^4*b^2*c^2*d^4 - 18*a^5*b*c*d^5 - a^6*d^6)*sq
rt(x)) - 4*(4*b^2*c*d*x^3 + (7*b^2*c^2 - 6*a*b*c*d + 3*a^2*d^2)*x)*sqrt(x))/(c*d^3*x^2 + c^2*d^2)

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

[Out]

Timed out

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Giac [A]  time = 1.20098, size = 524, normalized size = 1.69 \begin{align*} \frac{2 \, b^{2} x^{\frac{3}{2}}}{3 \, d^{2}} + \frac{b^{2} c^{2} x^{\frac{3}{2}} - 2 \, a b c d x^{\frac{3}{2}} + a^{2} d^{2} x^{\frac{3}{2}}}{2 \,{\left (d x^{2} + c\right )} c d^{2}} - \frac{\sqrt{2}{\left (7 \, \left (c d^{3}\right )^{\frac{3}{4}} b^{2} c^{2} - 6 \, \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 )}{8 \, c^{2} d^{5}} - \frac{\sqrt{2}{\left (7 \, \left (c d^{3}\right )^{\frac{3}{4}} b^{2} c^{2} - 6 \, \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 )}{8 \, c^{2} d^{5}} + \frac{\sqrt{2}{\left (7 \, \left (c d^{3}\right )^{\frac{3}{4}} b^{2} c^{2} - 6 \, \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 )}{16 \, c^{2} d^{5}} - \frac{\sqrt{2}{\left (7 \, \left (c d^{3}\right )^{\frac{3}{4}} b^{2} c^{2} - 6 \, \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 )}{16 \, c^{2} d^{5}} \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)^2,x, algorithm="giac")

[Out]

2/3*b^2*x^(3/2)/d^2 + 1/2*(b^2*c^2*x^(3/2) - 2*a*b*c*d*x^(3/2) + a^2*d^2*x^(3/2))/((d*x^2 + c)*c*d^2) - 1/8*sq
rt(2)*(7*(c*d^3)^(3/4)*b^2*c^2 - 6*(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^2*d^5) - 1/8*sqrt(2)*(7*(c*d^3)^(3/4)*b^2*c^2 - 6*(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^2*d^5) + 1/1
6*sqrt(2)*(7*(c*d^3)^(3/4)*b^2*c^2 - 6*(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^2*d^5) - 1/16*sqrt(2)*(7*(c*d^3)^(3/4)*b^2*c^2 - 6*(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^2*d^5)