3.421 \(\int \frac{(a+b x^2)^2}{x^{5/2} (c+d x^2)} \, dx\)

Optimal. Leaf size=260 \[ -\frac{2 a^2}{3 c x^{3/2}}+\frac{(b c-a d)^2 \log \left (-\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{2 \sqrt{2} c^{7/4} d^{5/4}}-\frac{(b c-a d)^2 \log \left (\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{2 \sqrt{2} c^{7/4} d^{5/4}}+\frac{(b c-a d)^2 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}\right )}{\sqrt{2} c^{7/4} d^{5/4}}-\frac{(b c-a d)^2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}+1\right )}{\sqrt{2} c^{7/4} d^{5/4}}+\frac{2 b^2 \sqrt{x}}{d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.264903, antiderivative size = 260, 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 = {462, 459, 329, 211, 1165, 628, 1162, 617, 204} \[ -\frac{2 a^2}{3 c x^{3/2}}+\frac{(b c-a d)^2 \log \left (-\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{2 \sqrt{2} c^{7/4} d^{5/4}}-\frac{(b c-a d)^2 \log \left (\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{2 \sqrt{2} c^{7/4} d^{5/4}}+\frac{(b c-a d)^2 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}\right )}{\sqrt{2} c^{7/4} d^{5/4}}-\frac{(b c-a d)^2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}+1\right )}{\sqrt{2} c^{7/4} d^{5/4}}+\frac{2 b^2 \sqrt{x}}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^2/(x^(5/2)*(c + d*x^2)),x]

[Out]

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

Rule 462

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

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

Mathematica [A]  time = 0.116511, size = 261, normalized size = 1. \[ \frac{-8 a^2 c^{3/4} d^{5/4}+3 \sqrt{2} x^{3/2} (b c-a d)^2 \log \left (-\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )-3 \sqrt{2} x^{3/2} (b c-a d)^2 \log \left (\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )+6 \sqrt{2} x^{3/2} (b c-a d)^2 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}\right )-6 \sqrt{2} x^{3/2} (b c-a d)^2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}+1\right )+24 b^2 c^{7/4} \sqrt [4]{d} x^2}{12 c^{7/4} d^{5/4} x^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^2/(x^(5/2)*(c + d*x^2)),x]

[Out]

(-8*a^2*c^(3/4)*d^(5/4) + 24*b^2*c^(7/4)*d^(1/4)*x^2 + 6*Sqrt[2]*(b*c - a*d)^2*x^(3/2)*ArcTan[1 - (Sqrt[2]*d^(
1/4)*Sqrt[x])/c^(1/4)] - 6*Sqrt[2]*(b*c - a*d)^2*x^(3/2)*ArcTan[1 + (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)] + 3*Sqr
t[2]*(b*c - a*d)^2*x^(3/2)*Log[Sqrt[c] - Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x] - 3*Sqrt[2]*(b*c - a*d)^
2*x^(3/2)*Log[Sqrt[c] + Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x])/(12*c^(7/4)*d^(5/4)*x^(3/2))

________________________________________________________________________________________

Maple [B]  time = 0.011, size = 439, normalized size = 1.7 \begin{align*} 2\,{\frac{{b}^{2}\sqrt{x}}{d}}-{\frac{d\sqrt{2}{a}^{2}}{2\,{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}{c}\sqrt [4]{{\frac{c}{d}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{c}{d}}}}}}+1 \right ) }-{\frac{\sqrt{2}{b}^{2}}{2\,d}\sqrt [4]{{\frac{c}{d}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{c}{d}}}}}}+1 \right ) }-{\frac{d\sqrt{2}{a}^{2}}{2\,{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}{c}\sqrt [4]{{\frac{c}{d}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{c}{d}}}}}}-1 \right ) }-{\frac{\sqrt{2}{b}^{2}}{2\,d}\sqrt [4]{{\frac{c}{d}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{c}{d}}}}}}-1 \right ) }-{\frac{d\sqrt{2}{a}^{2}}{4\,{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}{2\,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}{b}^{2}}{4\,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{2\,{a}^{2}}{3\,c}{x}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^2/x^(5/2)/(d*x^2+c),x)

[Out]

2*b^2*x^(1/2)/d-1/2/c^2*d*(c/d)^(1/4)*2^(1/2)*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)+1)*a^2+1/c*(c/d)^(1/4)*2^(1/2
)*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)+1)*a*b-1/2/d*(c/d)^(1/4)*2^(1/2)*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)+1)*b^
2-1/2/c^2*d*(c/d)^(1/4)*2^(1/2)*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)-1)*a^2+1/c*(c/d)^(1/4)*2^(1/2)*arctan(2^(1/
2)/(c/d)^(1/4)*x^(1/2)-1)*a*b-1/2/d*(c/d)^(1/4)*2^(1/2)*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)-1)*b^2-1/4/c^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^2+1/2/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-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)))*b^2-2/3*a^2/c/x^(3/2)

________________________________________________________________________________________

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^(5/2)/(d*x^2+c),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 1.09562, size = 2591, normalized size = 9.97 \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^(5/2)/(d*x^2+c),x, algorithm="fricas")

[Out]

-1/6*(12*c*d*x^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 - 5
6*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^7*d^5))^(1/4)*arctan((sqrt(c^4*d^2*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^7*d^5)) + (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^5*d^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)/(c^7*d^5))^(3/4) - (b^2*c^7*
d^4 - 2*a*b*c^6*d^5 + a^2*c^5*d^6)*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^7*d^5))^(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)) + 3*c*d*x^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^7*d^5))^(1/4)*log(c^2*d*(-(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^7*d^5))^(1/4) + (b^2*c^2 - 2*a*b
*c*d + a^2*d^2)*sqrt(x)) - 3*c*d*x^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^7*d^5))^(1/4)*log(-c^
2*d*(-(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^7*d^5))^(1/4) + (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqrt(x
)) - 4*(3*b^2*c*x^2 - a^2*d)*sqrt(x))/(c*d*x^2)

________________________________________________________________________________________

Sympy [A]  time = 52.9118, size = 581, normalized size = 2.23 \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**(5/2)/(d*x**2+c),x)

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.16316, size = 464, normalized size = 1.78 \begin{align*} \frac{2 \, b^{2} \sqrt{x}}{d} - \frac{2 \, a^{2}}{3 \, c x^{\frac{3}{2}}} - \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^{2} d^{2}} - \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^{2} d^{2}} - \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^{2} d^{2}} + \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^{2} d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/x^(5/2)/(d*x^2+c),x, algorithm="giac")

[Out]

2*b^2*sqrt(x)/d - 2/3*a^2/(c*x^(3/2)) - 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^2*d^2) - 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^2*d^2) - 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^2*d^2) + 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^2*d^2
)