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

Optimal. Leaf size=267 \[ -\frac{2 a^2}{5 c x^{5/2}}+\frac{(b c-a d)^2 \log \left (-\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{2 \sqrt{2} c^{9/4} d^{3/4}}-\frac{(b c-a d)^2 \log \left (\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{2 \sqrt{2} c^{9/4} d^{3/4}}-\frac{(b c-a d)^2 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}\right )}{\sqrt{2} c^{9/4} d^{3/4}}+\frac{(b c-a d)^2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}+1\right )}{\sqrt{2} c^{9/4} d^{3/4}}-\frac{2 a (2 b c-a d)}{c^2 \sqrt{x}} \]

[Out]

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

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Rubi [A]  time = 0.280172, antiderivative size = 267, 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, 453, 329, 297, 1162, 617, 204, 1165, 628} \[ -\frac{2 a^2}{5 c x^{5/2}}+\frac{(b c-a d)^2 \log \left (-\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{2 \sqrt{2} c^{9/4} d^{3/4}}-\frac{(b c-a d)^2 \log \left (\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{2 \sqrt{2} c^{9/4} d^{3/4}}-\frac{(b c-a d)^2 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}\right )}{\sqrt{2} c^{9/4} d^{3/4}}+\frac{(b c-a d)^2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}+1\right )}{\sqrt{2} c^{9/4} d^{3/4}}-\frac{2 a (2 b c-a d)}{c^2 \sqrt{x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^2)/(5*c*x^(5/2)) - (2*a*(2*b*c - a*d))/(c^2*Sqrt[x]) - ((b*c - a*d)^2*ArcTan[1 - (Sqrt[2]*d^(1/4)*Sqrt[x
])/c^(1/4)])/(Sqrt[2]*c^(9/4)*d^(3/4)) + ((b*c - a*d)^2*ArcTan[1 + (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/(Sqrt[2
]*c^(9/4)*d^(3/4)) + ((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^(
9/4)*d^(3/4)) - ((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^(9/4)*
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 453

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

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 )} \, dx &=-\frac{2 a^2}{5 c x^{5/2}}+\frac{2 \int \frac{\frac{5}{2} a (2 b c-a d)+\frac{5}{2} b^2 c x^2}{x^{3/2} \left (c+d x^2\right )} \, dx}{5 c}\\ &=-\frac{2 a^2}{5 c x^{5/2}}-\frac{2 a (2 b c-a d)}{c^2 \sqrt{x}}+\frac{(b c-a d)^2 \int \frac{\sqrt{x}}{c+d x^2} \, dx}{c^2}\\ &=-\frac{2 a^2}{5 c x^{5/2}}-\frac{2 a (2 b c-a d)}{c^2 \sqrt{x}}+\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 )}{c^2}\\ &=-\frac{2 a^2}{5 c x^{5/2}}-\frac{2 a (2 b c-a d)}{c^2 \sqrt{x}}-\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^2 \sqrt{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^2 \sqrt{d}}\\ &=-\frac{2 a^2}{5 c x^{5/2}}-\frac{2 a (2 b c-a d)}{c^2 \sqrt{x}}+\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^2 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^2 d}+\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^{9/4} d^{3/4}}+\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^{9/4} d^{3/4}}\\ &=-\frac{2 a^2}{5 c x^{5/2}}-\frac{2 a (2 b c-a d)}{c^2 \sqrt{x}}+\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^{9/4} d^{3/4}}-\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^{9/4} d^{3/4}}+\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^{9/4} d^{3/4}}-\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^{9/4} d^{3/4}}\\ &=-\frac{2 a^2}{5 c x^{5/2}}-\frac{2 a (2 b c-a d)}{c^2 \sqrt{x}}-\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^{9/4} d^{3/4}}+\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^{9/4} d^{3/4}}+\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^{9/4} d^{3/4}}-\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^{9/4} d^{3/4}}\\ \end{align*}

Mathematica [A]  time = 0.113732, size = 254, normalized size = 0.95 \[ \frac{-\frac{8 a^2 c^{5/4}}{x^{5/2}}+\frac{5 \sqrt{2} (b c-a d)^2 \log \left (-\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{d^{3/4}}-\frac{5 \sqrt{2} (b c-a d)^2 \log \left (\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{d^{3/4}}-\frac{10 \sqrt{2} (b c-a d)^2 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}\right )}{d^{3/4}}+\frac{10 \sqrt{2} (b c-a d)^2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}+1\right )}{d^{3/4}}+\frac{40 a \sqrt [4]{c} (a d-2 b c)}{\sqrt{x}}}{20 c^{9/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-8*a^2*c^(5/4))/x^(5/2) + (40*a*c^(1/4)*(-2*b*c + a*d))/Sqrt[x] - (10*Sqrt[2]*(b*c - a*d)^2*ArcTan[1 - (Sqrt
[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/d^(3/4) + (10*Sqrt[2]*(b*c - a*d)^2*ArcTan[1 + (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4
)])/d^(3/4) + (5*Sqrt[2]*(b*c - a*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*c - a*d)^2*Log[Sqrt[c] + Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x])/d^(3/4))/(20*c^(9/4))

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Maple [B]  time = 0.012, size = 452, normalized size = 1.7 \begin{align*}{\frac{d\sqrt{2}{a}^{2}}{2\,{c}^{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}ab}{c}\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}}{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{d\sqrt{2}{a}^{2}}{2\,{c}^{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}ab}{c}\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}}{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{d\sqrt{2}{a}^{2}}{4\,{c}^{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}ab}{2\,c}\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}}{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{2\,{a}^{2}}{5\,c}{x}^{-{\frac{5}{2}}}}+2\,{\frac{{a}^{2}d}{{c}^{2}\sqrt{x}}}-4\,{\frac{ab}{c\sqrt{x}}} \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),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.19988, size = 3421, normalized size = 12.81 \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),x, algorithm="fricas")

[Out]

-1/10*(20*c^2*x^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^9*d^3))^(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 + 92
4*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 - 1
2*a^11*b*c*d^11 + a^12*d^12)*x - (b^8*c^13*d - 8*a*b^7*c^12*d^2 + 28*a^2*b^6*c^11*d^3 - 56*a^3*b^5*c^10*d^4 +
70*a^4*b^4*c^9*d^5 - 56*a^5*b^3*c^8*d^6 + 28*a^6*b^2*c^7*d^7 - 8*a^7*b*c^6*d^8 + a^8*c^5*d^9)*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^9*d^3)))*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^9*d
^3))^(1/4) - (b^6*c^8*d - 6*a*b^5*c^7*d^2 + 15*a^2*b^4*c^6*d^3 - 20*a^3*b^3*c^5*d^4 + 15*a^4*b^2*c^4*d^5 - 6*a
^5*b*c^3*d^6 + a^6*c^2*d^7)*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^9*d^3))^(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)) - 5*c^2*x^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^9*d^3
))^(1/4)*log(c^7*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^9*d^3))^(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)) + 5*c^2
*x^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^9*d^3))^(1/4)*log(-c^7*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^9*d^3))^(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*(a^2*c + 5*(2*a*b*c - a^2*d)*x^2)*sqrt(x))/(c^2*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),x)

[Out]

Timed out

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Giac [A]  time = 1.1824, size = 477, normalized size = 1.79 \begin{align*} -\frac{2 \,{\left (10 \, a b c x^{2} - 5 \, a^{2} d x^{2} + a^{2} c\right )}}{5 \, c^{2} x^{\frac{5}{2}}} + \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^{3} d^{3}} + \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^{3} d^{3}} - \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^{3} d^{3}} + \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^{3} 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),x, algorithm="giac")

[Out]

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