3.432 \(\int \frac{(a+b x^2)^2}{x^{7/2} (c+d x^2)^2} \, dx\)

Optimal. Leaf size=363 \[ -\frac{9 a^2 d^2-10 a b c d+5 b^2 c^2}{10 c^2 d \sqrt{x} \left (c+d x^2\right )}-\frac{2 a^2}{5 c x^{5/2} \left (c+d x^2\right )}+\frac{(b c-9 a d) (b c-a d) \log \left (-\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{8 \sqrt{2} c^{13/4} d^{3/4}}-\frac{(b c-9 a d) (b c-a d) \log \left (\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{8 \sqrt{2} c^{13/4} d^{3/4}}-\frac{(b c-9 a d) (b c-a d) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}\right )}{4 \sqrt{2} c^{13/4} d^{3/4}}+\frac{(b c-9 a d) (b c-a d) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}+1\right )}{4 \sqrt{2} c^{13/4} d^{3/4}}+\frac{(b c-9 a d) (b c-a d)}{2 c^3 d \sqrt{x}} \]

[Out]

((b*c - 9*a*d)*(b*c - a*d))/(2*c^3*d*Sqrt[x]) - (2*a^2)/(5*c*x^(5/2)*(c + d*x^2)) - (5*b^2*c^2 - 10*a*b*c*d +
9*a^2*d^2)/(10*c^2*d*Sqrt[x]*(c + d*x^2)) - ((b*c - 9*a*d)*(b*c - a*d)*ArcTan[1 - (Sqrt[2]*d^(1/4)*Sqrt[x])/c^
(1/4)])/(4*Sqrt[2]*c^(13/4)*d^(3/4)) + ((b*c - 9*a*d)*(b*c - a*d)*ArcTan[1 + (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)
])/(4*Sqrt[2]*c^(13/4)*d^(3/4)) + ((b*c - 9*a*d)*(b*c - a*d)*Log[Sqrt[c] - Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + S
qrt[d]*x])/(8*Sqrt[2]*c^(13/4)*d^(3/4)) - ((b*c - 9*a*d)*(b*c - a*d)*Log[Sqrt[c] + Sqrt[2]*c^(1/4)*d^(1/4)*Sqr
t[x] + Sqrt[d]*x])/(8*Sqrt[2]*c^(13/4)*d^(3/4))

________________________________________________________________________________________

Rubi [A]  time = 0.377162, antiderivative size = 363, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 10, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used = {462, 457, 325, 329, 297, 1162, 617, 204, 1165, 628} \[ -\frac{9 a^2 d^2-10 a b c d+5 b^2 c^2}{10 c^2 d \sqrt{x} \left (c+d x^2\right )}-\frac{2 a^2}{5 c x^{5/2} \left (c+d x^2\right )}+\frac{(b c-9 a d) (b c-a d) \log \left (-\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{8 \sqrt{2} c^{13/4} d^{3/4}}-\frac{(b c-9 a d) (b c-a d) \log \left (\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{8 \sqrt{2} c^{13/4} d^{3/4}}-\frac{(b c-9 a d) (b c-a d) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}\right )}{4 \sqrt{2} c^{13/4} d^{3/4}}+\frac{(b c-9 a d) (b c-a d) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}+1\right )}{4 \sqrt{2} c^{13/4} d^{3/4}}+\frac{(b c-9 a d) (b c-a d)}{2 c^3 d \sqrt{x}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^2/(x^(7/2)*(c + d*x^2)^2),x]

[Out]

((b*c - 9*a*d)*(b*c - a*d))/(2*c^3*d*Sqrt[x]) - (2*a^2)/(5*c*x^(5/2)*(c + d*x^2)) - (5*b^2*c^2 - 10*a*b*c*d +
9*a^2*d^2)/(10*c^2*d*Sqrt[x]*(c + d*x^2)) - ((b*c - 9*a*d)*(b*c - a*d)*ArcTan[1 - (Sqrt[2]*d^(1/4)*Sqrt[x])/c^
(1/4)])/(4*Sqrt[2]*c^(13/4)*d^(3/4)) + ((b*c - 9*a*d)*(b*c - a*d)*ArcTan[1 + (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)
])/(4*Sqrt[2]*c^(13/4)*d^(3/4)) + ((b*c - 9*a*d)*(b*c - a*d)*Log[Sqrt[c] - Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + S
qrt[d]*x])/(8*Sqrt[2]*c^(13/4)*d^(3/4)) - ((b*c - 9*a*d)*(b*c - a*d)*Log[Sqrt[c] + Sqrt[2]*c^(1/4)*d^(1/4)*Sqr
t[x] + Sqrt[d]*x])/(8*Sqrt[2]*c^(13/4)*d^(3/4))

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 457

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d
)*(e*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*b*e*n*(p + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b
*n*(p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] &&
LeQ[-1, m, -(n*(p + 1))]))

Rule 325

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

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{\left (a+b x^2\right )^2}{x^{7/2} \left (c+d x^2\right )^2} \, dx &=-\frac{2 a^2}{5 c x^{5/2} \left (c+d x^2\right )}+\frac{2 \int \frac{\frac{1}{2} a (10 b c-9 a d)+\frac{5}{2} b^2 c x^2}{x^{3/2} \left (c+d x^2\right )^2} \, dx}{5 c}\\ &=-\frac{2 a^2}{5 c x^{5/2} \left (c+d x^2\right )}-\frac{5 b^2 c^2-10 a b c d+9 a^2 d^2}{10 c^2 d \sqrt{x} \left (c+d x^2\right )}-\frac{((b c-9 a d) (b c-a d)) \int \frac{1}{x^{3/2} \left (c+d x^2\right )} \, dx}{4 c^2 d}\\ &=\frac{(b c-9 a d) (b c-a d)}{2 c^3 d \sqrt{x}}-\frac{2 a^2}{5 c x^{5/2} \left (c+d x^2\right )}-\frac{5 b^2 c^2-10 a b c d+9 a^2 d^2}{10 c^2 d \sqrt{x} \left (c+d x^2\right )}+\frac{((b c-9 a d) (b c-a d)) \int \frac{\sqrt{x}}{c+d x^2} \, dx}{4 c^3}\\ &=\frac{(b c-9 a d) (b c-a d)}{2 c^3 d \sqrt{x}}-\frac{2 a^2}{5 c x^{5/2} \left (c+d x^2\right )}-\frac{5 b^2 c^2-10 a b c d+9 a^2 d^2}{10 c^2 d \sqrt{x} \left (c+d x^2\right )}+\frac{((b c-9 a d) (b c-a d)) \operatorname{Subst}\left (\int \frac{x^2}{c+d x^4} \, dx,x,\sqrt{x}\right )}{2 c^3}\\ &=\frac{(b c-9 a d) (b c-a d)}{2 c^3 d \sqrt{x}}-\frac{2 a^2}{5 c x^{5/2} \left (c+d x^2\right )}-\frac{5 b^2 c^2-10 a b c d+9 a^2 d^2}{10 c^2 d \sqrt{x} \left (c+d x^2\right )}-\frac{((b c-9 a d) (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^3 \sqrt{d}}+\frac{((b c-9 a d) (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^3 \sqrt{d}}\\ &=\frac{(b c-9 a d) (b c-a d)}{2 c^3 d \sqrt{x}}-\frac{2 a^2}{5 c x^{5/2} \left (c+d x^2\right )}-\frac{5 b^2 c^2-10 a b c d+9 a^2 d^2}{10 c^2 d \sqrt{x} \left (c+d x^2\right )}+\frac{((b c-9 a d) (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^3 d}+\frac{((b c-9 a d) (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^3 d}+\frac{((b c-9 a d) (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^{13/4} d^{3/4}}+\frac{((b c-9 a d) (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^{13/4} d^{3/4}}\\ &=\frac{(b c-9 a d) (b c-a d)}{2 c^3 d \sqrt{x}}-\frac{2 a^2}{5 c x^{5/2} \left (c+d x^2\right )}-\frac{5 b^2 c^2-10 a b c d+9 a^2 d^2}{10 c^2 d \sqrt{x} \left (c+d x^2\right )}+\frac{(b c-9 a d) (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^{13/4} d^{3/4}}-\frac{(b c-9 a d) (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^{13/4} d^{3/4}}+\frac{((b c-9 a d) (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^{13/4} d^{3/4}}-\frac{((b c-9 a d) (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^{13/4} d^{3/4}}\\ &=\frac{(b c-9 a d) (b c-a d)}{2 c^3 d \sqrt{x}}-\frac{2 a^2}{5 c x^{5/2} \left (c+d x^2\right )}-\frac{5 b^2 c^2-10 a b c d+9 a^2 d^2}{10 c^2 d \sqrt{x} \left (c+d x^2\right )}-\frac{(b c-9 a d) (b c-a d) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}\right )}{4 \sqrt{2} c^{13/4} d^{3/4}}+\frac{(b c-9 a d) (b c-a d) \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}\right )}{4 \sqrt{2} c^{13/4} d^{3/4}}+\frac{(b c-9 a d) (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^{13/4} d^{3/4}}-\frac{(b c-9 a d) (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^{13/4} d^{3/4}}\\ \end{align*}

Mathematica [A]  time = 0.198704, size = 333, normalized size = 0.92 \[ \frac{\frac{5 \sqrt{2} \left (9 a^2 d^2-10 a b c d+b^2 c^2\right ) \log \left (-\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{d^{3/4}}-\frac{5 \sqrt{2} \left (9 a^2 d^2-10 a b c d+b^2 c^2\right ) \log \left (\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{d^{3/4}}-\frac{10 \sqrt{2} \left (9 a^2 d^2-10 a b c d+b^2 c^2\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}\right )}{d^{3/4}}+\frac{10 \sqrt{2} \left (9 a^2 d^2-10 a b c d+b^2 c^2\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}+1\right )}{d^{3/4}}-\frac{32 a^2 c^{5/4}}{x^{5/2}}+\frac{40 \sqrt [4]{c} x^{3/2} (b c-a d)^2}{c+d x^2}+\frac{320 a \sqrt [4]{c} (a d-b c)}{\sqrt{x}}}{80 c^{13/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^2/(x^(7/2)*(c + d*x^2)^2),x]

[Out]

((-32*a^2*c^(5/4))/x^(5/2) + (320*a*c^(1/4)*(-(b*c) + a*d))/Sqrt[x] + (40*c^(1/4)*(b*c - a*d)^2*x^(3/2))/(c +
d*x^2) - (10*Sqrt[2]*(b^2*c^2 - 10*a*b*c*d + 9*a^2*d^2)*ArcTan[1 - (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/d^(3/4)
+ (10*Sqrt[2]*(b^2*c^2 - 10*a*b*c*d + 9*a^2*d^2)*ArcTan[1 + (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/d^(3/4) + (5*
Sqrt[2]*(b^2*c^2 - 10*a*b*c*d + 9*a^2*d^2)*Log[Sqrt[c] - Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x])/d^(3/4)
- (5*Sqrt[2]*(b^2*c^2 - 10*a*b*c*d + 9*a^2*d^2)*Log[Sqrt[c] + Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x])/d
^(3/4))/(80*c^(13/4))

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Maple [A]  time = 0.02, size = 524, normalized size = 1.4 \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^(7/2)/(d*x^2+c)^2,x)

[Out]

1/2/c^3*x^(3/2)/(d*x^2+c)*a^2*d^2-1/c^2*x^(3/2)/(d*x^2+c)*a*b*d+1/2/c*x^(3/2)/(d*x^2+c)*b^2+9/16/c^3*d/(c/d)^(
1/4)*2^(1/2)*a^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)))+9
/8/c^3*d/(c/d)^(1/4)*2^(1/2)*a^2*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)+1)+9/8/c^3*d/(c/d)^(1/4)*2^(1/2)*a^2*arcta
n(2^(1/2)/(c/d)^(1/4)*x^(1/2)-1)-5/8/c^2/(c/d)^(1/4)*2^(1/2)*a*b*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)))-5/4/c^2/(c/d)^(1/4)*2^(1/2)*a*b*arctan(2^(1/2)/(c/d)^(1/4)*x^(1
/2)+1)-5/4/c^2/(c/d)^(1/4)*2^(1/2)*a*b*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)-1)+1/16/c/d/(c/d)^(1/4)*2^(1/2)*b^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)))+1/8/c/d/(c/d)^(1/4
)*2^(1/2)*b^2*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)+1)+1/8/c/d/(c/d)^(1/4)*2^(1/2)*b^2*arctan(2^(1/2)/(c/d)^(1/4)
*x^(1/2)-1)-2/5*a^2/c^2/x^(5/2)+4*a^2/c^3/x^(1/2)*d-4*a/c^2/x^(1/2)*b

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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^(7/2)/(d*x^2+c)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.20001, size = 3997, normalized size = 11.01 \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^(7/2)/(d*x^2+c)^2,x, algorithm="fricas")

[Out]

-1/40*(20*(c^3*d*x^5 + c^4*x^3)*(-(b^8*c^8 - 40*a*b^7*c^7*d + 636*a^2*b^6*c^6*d^2 - 5080*a^3*b^5*c^5*d^3 + 212
86*a^4*b^4*c^4*d^4 - 45720*a^5*b^3*c^3*d^5 + 51516*a^6*b^2*c^2*d^6 - 29160*a^7*b*c*d^7 + 6561*a^8*d^8)/(c^13*d
^3))^(1/4)*arctan((sqrt((b^12*c^12 - 60*a*b^11*c^11*d + 1554*a^2*b^10*c^10*d^2 - 22700*a^3*b^9*c^9*d^3 + 20521
5*a^4*b^8*c^8*d^4 - 1188600*a^5*b^7*c^7*d^5 + 4443580*a^6*b^6*c^6*d^6 - 10697400*a^7*b^5*c^5*d^7 + 16622415*a^
8*b^4*c^4*d^8 - 16548300*a^9*b^3*c^3*d^9 + 10195794*a^10*b^2*c^2*d^10 - 3542940*a^11*b*c*d^11 + 531441*a^12*d^
12)*x - (b^8*c^15*d - 40*a*b^7*c^14*d^2 + 636*a^2*b^6*c^13*d^3 - 5080*a^3*b^5*c^12*d^4 + 21286*a^4*b^4*c^11*d^
5 - 45720*a^5*b^3*c^10*d^6 + 51516*a^6*b^2*c^9*d^7 - 29160*a^7*b*c^8*d^8 + 6561*a^8*c^7*d^9)*sqrt(-(b^8*c^8 -
40*a*b^7*c^7*d + 636*a^2*b^6*c^6*d^2 - 5080*a^3*b^5*c^5*d^3 + 21286*a^4*b^4*c^4*d^4 - 45720*a^5*b^3*c^3*d^5 +
51516*a^6*b^2*c^2*d^6 - 29160*a^7*b*c*d^7 + 6561*a^8*d^8)/(c^13*d^3)))*c^3*d*(-(b^8*c^8 - 40*a*b^7*c^7*d + 636
*a^2*b^6*c^6*d^2 - 5080*a^3*b^5*c^5*d^3 + 21286*a^4*b^4*c^4*d^4 - 45720*a^5*b^3*c^3*d^5 + 51516*a^6*b^2*c^2*d^
6 - 29160*a^7*b*c*d^7 + 6561*a^8*d^8)/(c^13*d^3))^(1/4) - (b^6*c^9*d - 30*a*b^5*c^8*d^2 + 327*a^2*b^4*c^7*d^3
- 1540*a^3*b^3*c^6*d^4 + 2943*a^4*b^2*c^5*d^5 - 2430*a^5*b*c^4*d^6 + 729*a^6*c^3*d^7)*sqrt(x)*(-(b^8*c^8 - 40*
a*b^7*c^7*d + 636*a^2*b^6*c^6*d^2 - 5080*a^3*b^5*c^5*d^3 + 21286*a^4*b^4*c^4*d^4 - 45720*a^5*b^3*c^3*d^5 + 515
16*a^6*b^2*c^2*d^6 - 29160*a^7*b*c*d^7 + 6561*a^8*d^8)/(c^13*d^3))^(1/4))/(b^8*c^8 - 40*a*b^7*c^7*d + 636*a^2*
b^6*c^6*d^2 - 5080*a^3*b^5*c^5*d^3 + 21286*a^4*b^4*c^4*d^4 - 45720*a^5*b^3*c^3*d^5 + 51516*a^6*b^2*c^2*d^6 - 2
9160*a^7*b*c*d^7 + 6561*a^8*d^8)) - 5*(c^3*d*x^5 + c^4*x^3)*(-(b^8*c^8 - 40*a*b^7*c^7*d + 636*a^2*b^6*c^6*d^2
- 5080*a^3*b^5*c^5*d^3 + 21286*a^4*b^4*c^4*d^4 - 45720*a^5*b^3*c^3*d^5 + 51516*a^6*b^2*c^2*d^6 - 29160*a^7*b*c
*d^7 + 6561*a^8*d^8)/(c^13*d^3))^(1/4)*log(c^10*d^2*(-(b^8*c^8 - 40*a*b^7*c^7*d + 636*a^2*b^6*c^6*d^2 - 5080*a
^3*b^5*c^5*d^3 + 21286*a^4*b^4*c^4*d^4 - 45720*a^5*b^3*c^3*d^5 + 51516*a^6*b^2*c^2*d^6 - 29160*a^7*b*c*d^7 + 6
561*a^8*d^8)/(c^13*d^3))^(3/4) + (b^6*c^6 - 30*a*b^5*c^5*d + 327*a^2*b^4*c^4*d^2 - 1540*a^3*b^3*c^3*d^3 + 2943
*a^4*b^2*c^2*d^4 - 2430*a^5*b*c*d^5 + 729*a^6*d^6)*sqrt(x)) + 5*(c^3*d*x^5 + c^4*x^3)*(-(b^8*c^8 - 40*a*b^7*c^
7*d + 636*a^2*b^6*c^6*d^2 - 5080*a^3*b^5*c^5*d^3 + 21286*a^4*b^4*c^4*d^4 - 45720*a^5*b^3*c^3*d^5 + 51516*a^6*b
^2*c^2*d^6 - 29160*a^7*b*c*d^7 + 6561*a^8*d^8)/(c^13*d^3))^(1/4)*log(-c^10*d^2*(-(b^8*c^8 - 40*a*b^7*c^7*d + 6
36*a^2*b^6*c^6*d^2 - 5080*a^3*b^5*c^5*d^3 + 21286*a^4*b^4*c^4*d^4 - 45720*a^5*b^3*c^3*d^5 + 51516*a^6*b^2*c^2*
d^6 - 29160*a^7*b*c*d^7 + 6561*a^8*d^8)/(c^13*d^3))^(3/4) + (b^6*c^6 - 30*a*b^5*c^5*d + 327*a^2*b^4*c^4*d^2 -
1540*a^3*b^3*c^3*d^3 + 2943*a^4*b^2*c^2*d^4 - 2430*a^5*b*c*d^5 + 729*a^6*d^6)*sqrt(x)) - 4*(5*(b^2*c^2 - 10*a*
b*c*d + 9*a^2*d^2)*x^4 - 4*a^2*c^2 - 4*(10*a*b*c^2 - 9*a^2*c*d)*x^2)*sqrt(x))/(c^3*d*x^5 + c^4*x^3)

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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**(7/2)/(d*x**2+c)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.21053, size = 541, normalized size = 1.49 \begin{align*} \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^{3}} - \frac{2 \,{\left (10 \, a b c x^{2} - 10 \, a^{2} d x^{2} + a^{2} c\right )}}{5 \, c^{3} x^{\frac{5}{2}}} + \frac{\sqrt{2}{\left (\left (c d^{3}\right )^{\frac{3}{4}} b^{2} c^{2} - 10 \, \left (c d^{3}\right )^{\frac{3}{4}} a b c d + 9 \, \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^{4} d^{3}} + \frac{\sqrt{2}{\left (\left (c d^{3}\right )^{\frac{3}{4}} b^{2} c^{2} - 10 \, \left (c d^{3}\right )^{\frac{3}{4}} a b c d + 9 \, \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^{4} d^{3}} - \frac{\sqrt{2}{\left (\left (c d^{3}\right )^{\frac{3}{4}} b^{2} c^{2} - 10 \, \left (c d^{3}\right )^{\frac{3}{4}} a b c d + 9 \, \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^{4} d^{3}} + \frac{\sqrt{2}{\left (\left (c d^{3}\right )^{\frac{3}{4}} b^{2} c^{2} - 10 \, \left (c d^{3}\right )^{\frac{3}{4}} a b c d + 9 \, \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^{4} d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/x^(7/2)/(d*x^2+c)^2,x, algorithm="giac")

[Out]

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^3) - 2/5*(10*a*b*c*x^2 - 10*a^2*d*x
^2 + a^2*c)/(c^3*x^(5/2)) + 1/8*sqrt(2)*((c*d^3)^(3/4)*b^2*c^2 - 10*(c*d^3)^(3/4)*a*b*c*d + 9*(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^4*d^3) + 1/8*sqrt(2)*((c*d^3)^(3/4
)*b^2*c^2 - 10*(c*d^3)^(3/4)*a*b*c*d + 9*(c*d^3)^(3/4)*a^2*d^2)*arctan(-1/2*sqrt(2)*(sqrt(2)*(c/d)^(1/4) - 2*s
qrt(x))/(c/d)^(1/4))/(c^4*d^3) - 1/16*sqrt(2)*((c*d^3)^(3/4)*b^2*c^2 - 10*(c*d^3)^(3/4)*a*b*c*d + 9*(c*d^3)^(3
/4)*a^2*d^2)*log(sqrt(2)*sqrt(x)*(c/d)^(1/4) + x + sqrt(c/d))/(c^4*d^3) + 1/16*sqrt(2)*((c*d^3)^(3/4)*b^2*c^2
- 10*(c*d^3)^(3/4)*a*b*c*d + 9*(c*d^3)^(3/4)*a^2*d^2)*log(-sqrt(2)*sqrt(x)*(c/d)^(1/4) + x + sqrt(c/d))/(c^4*d
^3)