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

Optimal. Leaf size=312 \[ \frac{(b c-a d) (3 a d+5 b c) \log \left (-\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{8 \sqrt{2} c^{7/4} d^{9/4}}-\frac{(b c-a d) (3 a d+5 b c) \log \left (\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{8 \sqrt{2} c^{7/4} d^{9/4}}+\frac{(b c-a d) (3 a d+5 b c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}\right )}{4 \sqrt{2} c^{7/4} d^{9/4}}-\frac{(b c-a d) (3 a d+5 b c) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}+1\right )}{4 \sqrt{2} c^{7/4} d^{9/4}}+\frac{\sqrt{x} (b c-a d)^2}{2 c d^2 \left (c+d x^2\right )}+\frac{2 b^2 \sqrt{x}}{d^2} \]

[Out]

(2*b^2*Sqrt[x])/d^2 + ((b*c - a*d)^2*Sqrt[x])/(2*c*d^2*(c + d*x^2)) + ((b*c - a*d)*(5*b*c + 3*a*d)*ArcTan[1 -
(Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/(4*Sqrt[2]*c^(7/4)*d^(9/4)) - ((b*c - a*d)*(5*b*c + 3*a*d)*ArcTan[1 + (Sqr
t[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/(4*Sqrt[2]*c^(7/4)*d^(9/4)) + ((b*c - a*d)*(5*b*c + 3*a*d)*Log[Sqrt[c] - Sqrt[
2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x])/(8*Sqrt[2]*c^(7/4)*d^(9/4)) - ((b*c - a*d)*(5*b*c + 3*a*d)*Log[Sqrt[c
] + Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x])/(8*Sqrt[2]*c^(7/4)*d^(9/4))

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Rubi [A]  time = 0.34237, antiderivative size = 312, 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, 211, 1165, 628, 1162, 617, 204} \[ \frac{(b c-a d) (3 a d+5 b c) \log \left (-\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{8 \sqrt{2} c^{7/4} d^{9/4}}-\frac{(b c-a d) (3 a d+5 b c) \log \left (\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{8 \sqrt{2} c^{7/4} d^{9/4}}+\frac{(b c-a d) (3 a d+5 b c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}\right )}{4 \sqrt{2} c^{7/4} d^{9/4}}-\frac{(b c-a d) (3 a d+5 b c) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}+1\right )}{4 \sqrt{2} c^{7/4} d^{9/4}}+\frac{\sqrt{x} (b c-a d)^2}{2 c d^2 \left (c+d x^2\right )}+\frac{2 b^2 \sqrt{x}}{d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b^2*Sqrt[x])/d^2 + ((b*c - a*d)^2*Sqrt[x])/(2*c*d^2*(c + d*x^2)) + ((b*c - a*d)*(5*b*c + 3*a*d)*ArcTan[1 -
(Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/(4*Sqrt[2]*c^(7/4)*d^(9/4)) - ((b*c - a*d)*(5*b*c + 3*a*d)*ArcTan[1 + (Sqr
t[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/(4*Sqrt[2]*c^(7/4)*d^(9/4)) + ((b*c - a*d)*(5*b*c + 3*a*d)*Log[Sqrt[c] - Sqrt[
2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x])/(8*Sqrt[2]*c^(7/4)*d^(9/4)) - ((b*c - a*d)*(5*b*c + 3*a*d)*Log[Sqrt[c
] + Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x])/(8*Sqrt[2]*c^(7/4)*d^(9/4))

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 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 )^2} \, dx &=\frac{(b c-a d)^2 \sqrt{x}}{2 c d^2 \left (c+d x^2\right )}-\frac{\int \frac{\frac{1}{2} (b c-3 a d) (b c+a d)-2 b^2 c d x^2}{\sqrt{x} \left (c+d x^2\right )} \, dx}{2 c d^2}\\ &=\frac{2 b^2 \sqrt{x}}{d^2}+\frac{(b c-a d)^2 \sqrt{x}}{2 c d^2 \left (c+d x^2\right )}-\frac{((b c-a d) (5 b c+3 a d)) \int \frac{1}{\sqrt{x} \left (c+d x^2\right )} \, dx}{4 c d^2}\\ &=\frac{2 b^2 \sqrt{x}}{d^2}+\frac{(b c-a d)^2 \sqrt{x}}{2 c d^2 \left (c+d x^2\right )}-\frac{((b c-a d) (5 b c+3 a d)) \operatorname{Subst}\left (\int \frac{1}{c+d x^4} \, dx,x,\sqrt{x}\right )}{2 c d^2}\\ &=\frac{2 b^2 \sqrt{x}}{d^2}+\frac{(b c-a d)^2 \sqrt{x}}{2 c d^2 \left (c+d x^2\right )}-\frac{((b c-a d) (5 b c+3 a d)) \operatorname{Subst}\left (\int \frac{\sqrt{c}-\sqrt{d} x^2}{c+d x^4} \, dx,x,\sqrt{x}\right )}{4 c^{3/2} d^2}-\frac{((b c-a d) (5 b c+3 a d)) \operatorname{Subst}\left (\int \frac{\sqrt{c}+\sqrt{d} x^2}{c+d x^4} \, dx,x,\sqrt{x}\right )}{4 c^{3/2} d^2}\\ &=\frac{2 b^2 \sqrt{x}}{d^2}+\frac{(b c-a d)^2 \sqrt{x}}{2 c d^2 \left (c+d x^2\right )}-\frac{((b c-a d) (5 b c+3 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^{3/2} d^{5/2}}-\frac{((b c-a d) (5 b c+3 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^{3/2} d^{5/2}}+\frac{((b c-a d) (5 b c+3 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^{7/4} d^{9/4}}+\frac{((b c-a d) (5 b c+3 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^{7/4} d^{9/4}}\\ &=\frac{2 b^2 \sqrt{x}}{d^2}+\frac{(b c-a d)^2 \sqrt{x}}{2 c d^2 \left (c+d x^2\right )}+\frac{(b c-a d) (5 b c+3 a d) \log \left (\sqrt{c}-\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{d} x\right )}{8 \sqrt{2} c^{7/4} d^{9/4}}-\frac{(b c-a d) (5 b c+3 a d) \log \left (\sqrt{c}+\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{d} x\right )}{8 \sqrt{2} c^{7/4} d^{9/4}}-\frac{((b c-a d) (5 b c+3 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^{7/4} d^{9/4}}+\frac{((b c-a d) (5 b c+3 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^{7/4} d^{9/4}}\\ &=\frac{2 b^2 \sqrt{x}}{d^2}+\frac{(b c-a d)^2 \sqrt{x}}{2 c d^2 \left (c+d x^2\right )}+\frac{(b c-a d) (5 b c+3 a d) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}\right )}{4 \sqrt{2} c^{7/4} d^{9/4}}-\frac{(b c-a d) (5 b c+3 a d) \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}\right )}{4 \sqrt{2} c^{7/4} d^{9/4}}+\frac{(b c-a d) (5 b c+3 a d) \log \left (\sqrt{c}-\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{d} x\right )}{8 \sqrt{2} c^{7/4} d^{9/4}}-\frac{(b c-a d) (5 b c+3 a d) \log \left (\sqrt{c}+\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{d} x\right )}{8 \sqrt{2} c^{7/4} d^{9/4}}\\ \end{align*}

Mathematica [A]  time = 0.17088, size = 318, normalized size = 1.02 \[ \frac{\frac{\sqrt{2} \left (-3 a^2 d^2-2 a b c d+5 b^2 c^2\right ) \log \left (-\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{c^{7/4}}-\frac{\sqrt{2} \left (-3 a^2 d^2-2 a b c d+5 b^2 c^2\right ) \log \left (\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{c^{7/4}}+\frac{2 \sqrt{2} \left (-3 a^2 d^2-2 a b c d+5 b^2 c^2\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}\right )}{c^{7/4}}-\frac{2 \sqrt{2} \left (-3 a^2 d^2-2 a b c d+5 b^2 c^2\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}+1\right )}{c^{7/4}}+\frac{8 \sqrt [4]{d} \sqrt{x} (b c-a d)^2}{c \left (c+d x^2\right )}+32 b^2 \sqrt [4]{d} \sqrt{x}}{16 d^{9/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(32*b^2*d^(1/4)*Sqrt[x] + (8*d^(1/4)*(b*c - a*d)^2*Sqrt[x])/(c*(c + d*x^2)) + (2*Sqrt[2]*(5*b^2*c^2 - 2*a*b*c*
d - 3*a^2*d^2)*ArcTan[1 - (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/c^(7/4) - (2*Sqrt[2]*(5*b^2*c^2 - 2*a*b*c*d - 3*
a^2*d^2)*ArcTan[1 + (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/c^(7/4) + (Sqrt[2]*(5*b^2*c^2 - 2*a*b*c*d - 3*a^2*d^2)
*Log[Sqrt[c] - Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x])/c^(7/4) - (Sqrt[2]*(5*b^2*c^2 - 2*a*b*c*d - 3*a^2
*d^2)*Log[Sqrt[c] + Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x])/c^(7/4))/(16*d^(9/4))

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Maple [B]  time = 0.016, size = 496, normalized size = 1.6 \begin{align*} 2\,{\frac{{b}^{2}\sqrt{x}}{{d}^{2}}}+{\frac{{a}^{2}}{2\,c \left ( d{x}^{2}+c \right ) }\sqrt{x}}-{\frac{ab}{d \left ( d{x}^{2}+c \right ) }\sqrt{x}}+{\frac{{b}^{2}c}{2\,{d}^{2} \left ( d{x}^{2}+c \right ) }\sqrt{x}}+{\frac{3\,\sqrt{2}{a}^{2}}{8\,{c}^{2}}\sqrt [4]{{\frac{c}{d}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{c}{d}}}}}}-1 \right ) }+{\frac{\sqrt{2}ab}{4\,cd}\sqrt [4]{{\frac{c}{d}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{c}{d}}}}}}-1 \right ) }-{\frac{5\,\sqrt{2}{b}^{2}}{8\,{d}^{2}}\sqrt [4]{{\frac{c}{d}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{c}{d}}}}}}-1 \right ) }+{\frac{3\,\sqrt{2}{a}^{2}}{16\,{c}^{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 ) }+{\frac{\sqrt{2}ab}{8\,cd}\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{5\,\sqrt{2}{b}^{2}}{16\,{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 ) }+{\frac{3\,\sqrt{2}{a}^{2}}{8\,{c}^{2}}\sqrt [4]{{\frac{c}{d}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{c}{d}}}}}}+1 \right ) }+{\frac{\sqrt{2}ab}{4\,cd}\sqrt [4]{{\frac{c}{d}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{c}{d}}}}}}+1 \right ) }-{\frac{5\,\sqrt{2}{b}^{2}}{8\,{d}^{2}}\sqrt [4]{{\frac{c}{d}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{c}{d}}}}}}+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)^2/x^(1/2),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.05354, size = 2944, normalized size = 9.44 \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)^2/x^(1/2),x, algorithm="fricas")

[Out]

1/8*(4*(c*d^3*x^2 + c^2*d^2)*(-(625*b^8*c^8 - 1000*a*b^7*c^7*d - 900*a^2*b^6*c^6*d^2 + 1640*a^3*b^5*c^5*d^3 +
646*a^4*b^4*c^4*d^4 - 984*a^5*b^3*c^3*d^5 - 324*a^6*b^2*c^2*d^6 + 216*a^7*b*c*d^7 + 81*a^8*d^8)/(c^7*d^9))^(1/
4)*arctan((sqrt(c^4*d^4*sqrt(-(625*b^8*c^8 - 1000*a*b^7*c^7*d - 900*a^2*b^6*c^6*d^2 + 1640*a^3*b^5*c^5*d^3 + 6
46*a^4*b^4*c^4*d^4 - 984*a^5*b^3*c^3*d^5 - 324*a^6*b^2*c^2*d^6 + 216*a^7*b*c*d^7 + 81*a^8*d^8)/(c^7*d^9)) + (2
5*b^4*c^4 - 20*a*b^3*c^3*d - 26*a^2*b^2*c^2*d^2 + 12*a^3*b*c*d^3 + 9*a^4*d^4)*x)*c^5*d^7*(-(625*b^8*c^8 - 1000
*a*b^7*c^7*d - 900*a^2*b^6*c^6*d^2 + 1640*a^3*b^5*c^5*d^3 + 646*a^4*b^4*c^4*d^4 - 984*a^5*b^3*c^3*d^5 - 324*a^
6*b^2*c^2*d^6 + 216*a^7*b*c*d^7 + 81*a^8*d^8)/(c^7*d^9))^(3/4) + (5*b^2*c^7*d^7 - 2*a*b*c^6*d^8 - 3*a^2*c^5*d^
9)*sqrt(x)*(-(625*b^8*c^8 - 1000*a*b^7*c^7*d - 900*a^2*b^6*c^6*d^2 + 1640*a^3*b^5*c^5*d^3 + 646*a^4*b^4*c^4*d^
4 - 984*a^5*b^3*c^3*d^5 - 324*a^6*b^2*c^2*d^6 + 216*a^7*b*c*d^7 + 81*a^8*d^8)/(c^7*d^9))^(3/4))/(625*b^8*c^8 -
1000*a*b^7*c^7*d - 900*a^2*b^6*c^6*d^2 + 1640*a^3*b^5*c^5*d^3 + 646*a^4*b^4*c^4*d^4 - 984*a^5*b^3*c^3*d^5 - 3
24*a^6*b^2*c^2*d^6 + 216*a^7*b*c*d^7 + 81*a^8*d^8)) + (c*d^3*x^2 + c^2*d^2)*(-(625*b^8*c^8 - 1000*a*b^7*c^7*d
- 900*a^2*b^6*c^6*d^2 + 1640*a^3*b^5*c^5*d^3 + 646*a^4*b^4*c^4*d^4 - 984*a^5*b^3*c^3*d^5 - 324*a^6*b^2*c^2*d^6
+ 216*a^7*b*c*d^7 + 81*a^8*d^8)/(c^7*d^9))^(1/4)*log(c^2*d^2*(-(625*b^8*c^8 - 1000*a*b^7*c^7*d - 900*a^2*b^6*
c^6*d^2 + 1640*a^3*b^5*c^5*d^3 + 646*a^4*b^4*c^4*d^4 - 984*a^5*b^3*c^3*d^5 - 324*a^6*b^2*c^2*d^6 + 216*a^7*b*c
*d^7 + 81*a^8*d^8)/(c^7*d^9))^(1/4) - (5*b^2*c^2 - 2*a*b*c*d - 3*a^2*d^2)*sqrt(x)) - (c*d^3*x^2 + c^2*d^2)*(-(
625*b^8*c^8 - 1000*a*b^7*c^7*d - 900*a^2*b^6*c^6*d^2 + 1640*a^3*b^5*c^5*d^3 + 646*a^4*b^4*c^4*d^4 - 984*a^5*b^
3*c^3*d^5 - 324*a^6*b^2*c^2*d^6 + 216*a^7*b*c*d^7 + 81*a^8*d^8)/(c^7*d^9))^(1/4)*log(-c^2*d^2*(-(625*b^8*c^8 -
1000*a*b^7*c^7*d - 900*a^2*b^6*c^6*d^2 + 1640*a^3*b^5*c^5*d^3 + 646*a^4*b^4*c^4*d^4 - 984*a^5*b^3*c^3*d^5 - 3
24*a^6*b^2*c^2*d^6 + 216*a^7*b*c*d^7 + 81*a^8*d^8)/(c^7*d^9))^(1/4) - (5*b^2*c^2 - 2*a*b*c*d - 3*a^2*d^2)*sqrt
(x)) + 4*(4*b^2*c*d*x^2 + 5*b^2*c^2 - 2*a*b*c*d + a^2*d^2)*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/(d*x**2+c)**2/x**(1/2),x)

[Out]

Timed out

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Giac [A]  time = 1.18832, size = 524, normalized size = 1.68 \begin{align*} \frac{2 \, b^{2} \sqrt{x}}{d^{2}} - \frac{\sqrt{2}{\left (5 \, \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 - 3 \, \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 )}{8 \, c^{2} d^{3}} - \frac{\sqrt{2}{\left (5 \, \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 - 3 \, \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 )}{8 \, c^{2} d^{3}} - \frac{\sqrt{2}{\left (5 \, \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 - 3 \, \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 )}{16 \, c^{2} d^{3}} + \frac{\sqrt{2}{\left (5 \, \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 - 3 \, \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 )}{16 \, c^{2} d^{3}} + \frac{b^{2} c^{2} \sqrt{x} - 2 \, a b c d \sqrt{x} + a^{2} d^{2} \sqrt{x}}{2 \,{\left (d x^{2} + c\right )} c d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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