3.383 \(\int \frac{x^{7/2} (A+B x^2)}{(a+b x^2)^3} \, dx\)

Optimal. Leaf size=316 \[ -\frac{5 (A b-9 a B) \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{64 \sqrt{2} a^{3/4} b^{13/4}}+\frac{5 (A b-9 a B) \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{64 \sqrt{2} a^{3/4} b^{13/4}}-\frac{5 (A b-9 a B) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{32 \sqrt{2} a^{3/4} b^{13/4}}+\frac{5 (A b-9 a B) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{32 \sqrt{2} a^{3/4} b^{13/4}}+\frac{x^{5/2} (A b-9 a B)}{16 a b^2 \left (a+b x^2\right )}-\frac{5 \sqrt{x} (A b-9 a B)}{16 a b^3}+\frac{x^{9/2} (A b-a B)}{4 a b \left (a+b x^2\right )^2} \]

[Out]

(-5*(A*b - 9*a*B)*Sqrt[x])/(16*a*b^3) + ((A*b - a*B)*x^(9/2))/(4*a*b*(a + b*x^2)^2) + ((A*b - 9*a*B)*x^(5/2))/
(16*a*b^2*(a + b*x^2)) - (5*(A*b - 9*a*B)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(32*Sqrt[2]*a^(3/4)*b
^(13/4)) + (5*(A*b - 9*a*B)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(32*Sqrt[2]*a^(3/4)*b^(13/4)) - (5*
(A*b - 9*a*B)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(64*Sqrt[2]*a^(3/4)*b^(13/4)) + (5*(
A*b - 9*a*B)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(64*Sqrt[2]*a^(3/4)*b^(13/4))

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Rubi [A]  time = 0.239701, antiderivative size = 316, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 10, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.454, Rules used = {457, 288, 321, 329, 211, 1165, 628, 1162, 617, 204} \[ -\frac{5 (A b-9 a B) \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{64 \sqrt{2} a^{3/4} b^{13/4}}+\frac{5 (A b-9 a B) \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{64 \sqrt{2} a^{3/4} b^{13/4}}-\frac{5 (A b-9 a B) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{32 \sqrt{2} a^{3/4} b^{13/4}}+\frac{5 (A b-9 a B) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{32 \sqrt{2} a^{3/4} b^{13/4}}+\frac{x^{5/2} (A b-9 a B)}{16 a b^2 \left (a+b x^2\right )}-\frac{5 \sqrt{x} (A b-9 a B)}{16 a b^3}+\frac{x^{9/2} (A b-a B)}{4 a b \left (a+b x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(x^(7/2)*(A + B*x^2))/(a + b*x^2)^3,x]

[Out]

(-5*(A*b - 9*a*B)*Sqrt[x])/(16*a*b^3) + ((A*b - a*B)*x^(9/2))/(4*a*b*(a + b*x^2)^2) + ((A*b - 9*a*B)*x^(5/2))/
(16*a*b^2*(a + b*x^2)) - (5*(A*b - 9*a*B)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(32*Sqrt[2]*a^(3/4)*b
^(13/4)) + (5*(A*b - 9*a*B)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(32*Sqrt[2]*a^(3/4)*b^(13/4)) - (5*
(A*b - 9*a*B)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(64*Sqrt[2]*a^(3/4)*b^(13/4)) + (5*(
A*b - 9*a*B)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(64*Sqrt[2]*a^(3/4)*b^(13/4))

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 288

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

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && 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 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{x^{7/2} \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx &=\frac{(A b-a B) x^{9/2}}{4 a b \left (a+b x^2\right )^2}+\frac{\left (-\frac{A b}{2}+\frac{9 a B}{2}\right ) \int \frac{x^{7/2}}{\left (a+b x^2\right )^2} \, dx}{4 a b}\\ &=\frac{(A b-a B) x^{9/2}}{4 a b \left (a+b x^2\right )^2}+\frac{(A b-9 a B) x^{5/2}}{16 a b^2 \left (a+b x^2\right )}-\frac{(5 (A b-9 a B)) \int \frac{x^{3/2}}{a+b x^2} \, dx}{32 a b^2}\\ &=-\frac{5 (A b-9 a B) \sqrt{x}}{16 a b^3}+\frac{(A b-a B) x^{9/2}}{4 a b \left (a+b x^2\right )^2}+\frac{(A b-9 a B) x^{5/2}}{16 a b^2 \left (a+b x^2\right )}+\frac{(5 (A b-9 a B)) \int \frac{1}{\sqrt{x} \left (a+b x^2\right )} \, dx}{32 b^3}\\ &=-\frac{5 (A b-9 a B) \sqrt{x}}{16 a b^3}+\frac{(A b-a B) x^{9/2}}{4 a b \left (a+b x^2\right )^2}+\frac{(A b-9 a B) x^{5/2}}{16 a b^2 \left (a+b x^2\right )}+\frac{(5 (A b-9 a B)) \operatorname{Subst}\left (\int \frac{1}{a+b x^4} \, dx,x,\sqrt{x}\right )}{16 b^3}\\ &=-\frac{5 (A b-9 a B) \sqrt{x}}{16 a b^3}+\frac{(A b-a B) x^{9/2}}{4 a b \left (a+b x^2\right )^2}+\frac{(A b-9 a B) x^{5/2}}{16 a b^2 \left (a+b x^2\right )}+\frac{(5 (A b-9 a B)) \operatorname{Subst}\left (\int \frac{\sqrt{a}-\sqrt{b} x^2}{a+b x^4} \, dx,x,\sqrt{x}\right )}{32 \sqrt{a} b^3}+\frac{(5 (A b-9 a B)) \operatorname{Subst}\left (\int \frac{\sqrt{a}+\sqrt{b} x^2}{a+b x^4} \, dx,x,\sqrt{x}\right )}{32 \sqrt{a} b^3}\\ &=-\frac{5 (A b-9 a B) \sqrt{x}}{16 a b^3}+\frac{(A b-a B) x^{9/2}}{4 a b \left (a+b x^2\right )^2}+\frac{(A b-9 a B) x^{5/2}}{16 a b^2 \left (a+b x^2\right )}+\frac{(5 (A b-9 a B)) \operatorname{Subst}\left (\int \frac{1}{\frac{\sqrt{a}}{\sqrt{b}}-\frac{\sqrt{2} \sqrt [4]{a} x}{\sqrt [4]{b}}+x^2} \, dx,x,\sqrt{x}\right )}{64 \sqrt{a} b^{7/2}}+\frac{(5 (A b-9 a B)) \operatorname{Subst}\left (\int \frac{1}{\frac{\sqrt{a}}{\sqrt{b}}+\frac{\sqrt{2} \sqrt [4]{a} x}{\sqrt [4]{b}}+x^2} \, dx,x,\sqrt{x}\right )}{64 \sqrt{a} b^{7/2}}-\frac{(5 (A b-9 a B)) \operatorname{Subst}\left (\int \frac{\frac{\sqrt{2} \sqrt [4]{a}}{\sqrt [4]{b}}+2 x}{-\frac{\sqrt{a}}{\sqrt{b}}-\frac{\sqrt{2} \sqrt [4]{a} x}{\sqrt [4]{b}}-x^2} \, dx,x,\sqrt{x}\right )}{64 \sqrt{2} a^{3/4} b^{13/4}}-\frac{(5 (A b-9 a B)) \operatorname{Subst}\left (\int \frac{\frac{\sqrt{2} \sqrt [4]{a}}{\sqrt [4]{b}}-2 x}{-\frac{\sqrt{a}}{\sqrt{b}}+\frac{\sqrt{2} \sqrt [4]{a} x}{\sqrt [4]{b}}-x^2} \, dx,x,\sqrt{x}\right )}{64 \sqrt{2} a^{3/4} b^{13/4}}\\ &=-\frac{5 (A b-9 a B) \sqrt{x}}{16 a b^3}+\frac{(A b-a B) x^{9/2}}{4 a b \left (a+b x^2\right )^2}+\frac{(A b-9 a B) x^{5/2}}{16 a b^2 \left (a+b x^2\right )}-\frac{5 (A b-9 a B) \log \left (\sqrt{a}-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{b} x\right )}{64 \sqrt{2} a^{3/4} b^{13/4}}+\frac{5 (A b-9 a B) \log \left (\sqrt{a}+\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{b} x\right )}{64 \sqrt{2} a^{3/4} b^{13/4}}+\frac{(5 (A b-9 a B)) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{32 \sqrt{2} a^{3/4} b^{13/4}}-\frac{(5 (A b-9 a B)) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{32 \sqrt{2} a^{3/4} b^{13/4}}\\ &=-\frac{5 (A b-9 a B) \sqrt{x}}{16 a b^3}+\frac{(A b-a B) x^{9/2}}{4 a b \left (a+b x^2\right )^2}+\frac{(A b-9 a B) x^{5/2}}{16 a b^2 \left (a+b x^2\right )}-\frac{5 (A b-9 a B) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{32 \sqrt{2} a^{3/4} b^{13/4}}+\frac{5 (A b-9 a B) \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{32 \sqrt{2} a^{3/4} b^{13/4}}-\frac{5 (A b-9 a B) \log \left (\sqrt{a}-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{b} x\right )}{64 \sqrt{2} a^{3/4} b^{13/4}}+\frac{5 (A b-9 a B) \log \left (\sqrt{a}+\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{b} x\right )}{64 \sqrt{2} a^{3/4} b^{13/4}}\\ \end{align*}

Mathematica [A]  time = 0.466303, size = 402, normalized size = 1.27 \[ \frac{\frac{10 \sqrt{2} (9 a B-A b) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{a^{3/4}}+\frac{10 \sqrt{2} (A b-9 a B) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{a^{3/4}}-\frac{5 \sqrt{2} A b \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{a^{3/4}}+\frac{5 \sqrt{2} A b \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{a^{3/4}}-\frac{32 a^2 \sqrt [4]{b} B \sqrt{x}}{\left (a+b x^2\right )^2}+\frac{32 a A b^{5/4} \sqrt{x}}{\left (a+b x^2\right )^2}-\frac{72 A b^{5/4} \sqrt{x}}{a+b x^2}+\frac{136 a \sqrt [4]{b} B \sqrt{x}}{a+b x^2}+45 \sqrt{2} \sqrt [4]{a} B \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )-45 \sqrt{2} \sqrt [4]{a} B \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )+256 \sqrt [4]{b} B \sqrt{x}}{128 b^{13/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^(7/2)*(A + B*x^2))/(a + b*x^2)^3,x]

[Out]

(256*b^(1/4)*B*Sqrt[x] + (32*a*A*b^(5/4)*Sqrt[x])/(a + b*x^2)^2 - (32*a^2*b^(1/4)*B*Sqrt[x])/(a + b*x^2)^2 - (
72*A*b^(5/4)*Sqrt[x])/(a + b*x^2) + (136*a*b^(1/4)*B*Sqrt[x])/(a + b*x^2) + (10*Sqrt[2]*(-(A*b) + 9*a*B)*ArcTa
n[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/a^(3/4) + (10*Sqrt[2]*(A*b - 9*a*B)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt
[x])/a^(1/4)])/a^(3/4) - (5*Sqrt[2]*A*b*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/a^(3/4) +
45*Sqrt[2]*a^(1/4)*B*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x] + (5*Sqrt[2]*A*b*Log[Sqrt[a] +
Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/a^(3/4) - 45*Sqrt[2]*a^(1/4)*B*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^
(1/4)*Sqrt[x] + Sqrt[b]*x])/(128*b^(13/4))

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Maple [A]  time = 0.015, size = 363, normalized size = 1.2 \begin{align*} 2\,{\frac{B\sqrt{x}}{{b}^{3}}}-{\frac{9\,A}{16\,b \left ( b{x}^{2}+a \right ) ^{2}}{x}^{{\frac{5}{2}}}}+{\frac{17\,Ba}{16\,{b}^{2} \left ( b{x}^{2}+a \right ) ^{2}}{x}^{{\frac{5}{2}}}}-{\frac{5\,Aa}{16\,{b}^{2} \left ( b{x}^{2}+a \right ) ^{2}}\sqrt{x}}+{\frac{13\,{a}^{2}B}{16\,{b}^{3} \left ( b{x}^{2}+a \right ) ^{2}}\sqrt{x}}+{\frac{5\,\sqrt{2}A}{64\,{b}^{2}a}\sqrt [4]{{\frac{a}{b}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}+1 \right ) }+{\frac{5\,\sqrt{2}A}{64\,{b}^{2}a}\sqrt [4]{{\frac{a}{b}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}-1 \right ) }+{\frac{5\,\sqrt{2}A}{128\,{b}^{2}a}\sqrt [4]{{\frac{a}{b}}}\ln \left ({ \left ( x+\sqrt [4]{{\frac{a}{b}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{a}{b}}} \right ) \left ( x-\sqrt [4]{{\frac{a}{b}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{a}{b}}} \right ) ^{-1}} \right ) }-{\frac{45\,\sqrt{2}B}{64\,{b}^{3}}\sqrt [4]{{\frac{a}{b}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}+1 \right ) }-{\frac{45\,\sqrt{2}B}{64\,{b}^{3}}\sqrt [4]{{\frac{a}{b}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}-1 \right ) }-{\frac{45\,\sqrt{2}B}{128\,{b}^{3}}\sqrt [4]{{\frac{a}{b}}}\ln \left ({ \left ( x+\sqrt [4]{{\frac{a}{b}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{a}{b}}} \right ) \left ( x-\sqrt [4]{{\frac{a}{b}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{a}{b}}} \right ) ^{-1}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(7/2)*(B*x^2+A)/(b*x^2+a)^3,x)

[Out]

2*B/b^3*x^(1/2)-9/16/b/(b*x^2+a)^2*x^(5/2)*A+17/16/b^2/(b*x^2+a)^2*x^(5/2)*B*a-5/16/b^2/(b*x^2+a)^2*A*x^(1/2)*
a+13/16/b^3/(b*x^2+a)^2*B*x^(1/2)*a^2+5/64/b^2*(1/b*a)^(1/4)/a*2^(1/2)*A*arctan(2^(1/2)/(1/b*a)^(1/4)*x^(1/2)+
1)+5/64/b^2*(1/b*a)^(1/4)/a*2^(1/2)*A*arctan(2^(1/2)/(1/b*a)^(1/4)*x^(1/2)-1)+5/128/b^2*(1/b*a)^(1/4)/a*2^(1/2
)*A*ln((x+(1/b*a)^(1/4)*x^(1/2)*2^(1/2)+(1/b*a)^(1/2))/(x-(1/b*a)^(1/4)*x^(1/2)*2^(1/2)+(1/b*a)^(1/2)))-45/64/
b^3*(1/b*a)^(1/4)*2^(1/2)*B*arctan(2^(1/2)/(1/b*a)^(1/4)*x^(1/2)+1)-45/64/b^3*(1/b*a)^(1/4)*2^(1/2)*B*arctan(2
^(1/2)/(1/b*a)^(1/4)*x^(1/2)-1)-45/128/b^3*(1/b*a)^(1/4)*2^(1/2)*B*ln((x+(1/b*a)^(1/4)*x^(1/2)*2^(1/2)+(1/b*a)
^(1/2))/(x-(1/b*a)^(1/4)*x^(1/2)*2^(1/2)+(1/b*a)^(1/2)))

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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(x^(7/2)*(B*x^2+A)/(b*x^2+a)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.914373, size = 1787, normalized size = 5.66 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*(B*x^2+A)/(b*x^2+a)^3,x, algorithm="fricas")

[Out]

1/64*(20*(b^5*x^4 + 2*a*b^4*x^2 + a^2*b^3)*(-(6561*B^4*a^4 - 2916*A*B^3*a^3*b + 486*A^2*B^2*a^2*b^2 - 36*A^3*B
*a*b^3 + A^4*b^4)/(a^3*b^13))^(1/4)*arctan((sqrt(a^2*b^6*sqrt(-(6561*B^4*a^4 - 2916*A*B^3*a^3*b + 486*A^2*B^2*
a^2*b^2 - 36*A^3*B*a*b^3 + A^4*b^4)/(a^3*b^13)) + (81*B^2*a^2 - 18*A*B*a*b + A^2*b^2)*x)*a^2*b^10*(-(6561*B^4*
a^4 - 2916*A*B^3*a^3*b + 486*A^2*B^2*a^2*b^2 - 36*A^3*B*a*b^3 + A^4*b^4)/(a^3*b^13))^(3/4) + (9*B*a^3*b^10 - A
*a^2*b^11)*sqrt(x)*(-(6561*B^4*a^4 - 2916*A*B^3*a^3*b + 486*A^2*B^2*a^2*b^2 - 36*A^3*B*a*b^3 + A^4*b^4)/(a^3*b
^13))^(3/4))/(6561*B^4*a^4 - 2916*A*B^3*a^3*b + 486*A^2*B^2*a^2*b^2 - 36*A^3*B*a*b^3 + A^4*b^4)) + 5*(b^5*x^4
+ 2*a*b^4*x^2 + a^2*b^3)*(-(6561*B^4*a^4 - 2916*A*B^3*a^3*b + 486*A^2*B^2*a^2*b^2 - 36*A^3*B*a*b^3 + A^4*b^4)/
(a^3*b^13))^(1/4)*log(5*a*b^3*(-(6561*B^4*a^4 - 2916*A*B^3*a^3*b + 486*A^2*B^2*a^2*b^2 - 36*A^3*B*a*b^3 + A^4*
b^4)/(a^3*b^13))^(1/4) - 5*(9*B*a - A*b)*sqrt(x)) - 5*(b^5*x^4 + 2*a*b^4*x^2 + a^2*b^3)*(-(6561*B^4*a^4 - 2916
*A*B^3*a^3*b + 486*A^2*B^2*a^2*b^2 - 36*A^3*B*a*b^3 + A^4*b^4)/(a^3*b^13))^(1/4)*log(-5*a*b^3*(-(6561*B^4*a^4
- 2916*A*B^3*a^3*b + 486*A^2*B^2*a^2*b^2 - 36*A^3*B*a*b^3 + A^4*b^4)/(a^3*b^13))^(1/4) - 5*(9*B*a - A*b)*sqrt(
x)) + 4*(32*B*b^2*x^4 + 45*B*a^2 - 5*A*a*b + 9*(9*B*a*b - A*b^2)*x^2)*sqrt(x))/(b^5*x^4 + 2*a*b^4*x^2 + a^2*b^
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(x**(7/2)*(B*x**2+A)/(b*x**2+a)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.1913, size = 410, normalized size = 1.3 \begin{align*} \frac{2 \, B \sqrt{x}}{b^{3}} - \frac{5 \, \sqrt{2}{\left (9 \, \left (a b^{3}\right )^{\frac{1}{4}} B a - \left (a b^{3}\right )^{\frac{1}{4}} A b\right )} \arctan \left (\frac{\sqrt{2}{\left (\sqrt{2} \left (\frac{a}{b}\right )^{\frac{1}{4}} + 2 \, \sqrt{x}\right )}}{2 \, \left (\frac{a}{b}\right )^{\frac{1}{4}}}\right )}{64 \, a b^{4}} - \frac{5 \, \sqrt{2}{\left (9 \, \left (a b^{3}\right )^{\frac{1}{4}} B a - \left (a b^{3}\right )^{\frac{1}{4}} A b\right )} \arctan \left (-\frac{\sqrt{2}{\left (\sqrt{2} \left (\frac{a}{b}\right )^{\frac{1}{4}} - 2 \, \sqrt{x}\right )}}{2 \, \left (\frac{a}{b}\right )^{\frac{1}{4}}}\right )}{64 \, a b^{4}} - \frac{5 \, \sqrt{2}{\left (9 \, \left (a b^{3}\right )^{\frac{1}{4}} B a - \left (a b^{3}\right )^{\frac{1}{4}} A b\right )} \log \left (\sqrt{2} \sqrt{x} \left (\frac{a}{b}\right )^{\frac{1}{4}} + x + \sqrt{\frac{a}{b}}\right )}{128 \, a b^{4}} + \frac{5 \, \sqrt{2}{\left (9 \, \left (a b^{3}\right )^{\frac{1}{4}} B a - \left (a b^{3}\right )^{\frac{1}{4}} A b\right )} \log \left (-\sqrt{2} \sqrt{x} \left (\frac{a}{b}\right )^{\frac{1}{4}} + x + \sqrt{\frac{a}{b}}\right )}{128 \, a b^{4}} + \frac{17 \, B a b x^{\frac{5}{2}} - 9 \, A b^{2} x^{\frac{5}{2}} + 13 \, B a^{2} \sqrt{x} - 5 \, A a b \sqrt{x}}{16 \,{\left (b x^{2} + a\right )}^{2} b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*(B*x^2+A)/(b*x^2+a)^3,x, algorithm="giac")

[Out]

2*B*sqrt(x)/b^3 - 5/64*sqrt(2)*(9*(a*b^3)^(1/4)*B*a - (a*b^3)^(1/4)*A*b)*arctan(1/2*sqrt(2)*(sqrt(2)*(a/b)^(1/
4) + 2*sqrt(x))/(a/b)^(1/4))/(a*b^4) - 5/64*sqrt(2)*(9*(a*b^3)^(1/4)*B*a - (a*b^3)^(1/4)*A*b)*arctan(-1/2*sqrt
(2)*(sqrt(2)*(a/b)^(1/4) - 2*sqrt(x))/(a/b)^(1/4))/(a*b^4) - 5/128*sqrt(2)*(9*(a*b^3)^(1/4)*B*a - (a*b^3)^(1/4
)*A*b)*log(sqrt(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))/(a*b^4) + 5/128*sqrt(2)*(9*(a*b^3)^(1/4)*B*a - (a*b^3)
^(1/4)*A*b)*log(-sqrt(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))/(a*b^4) + 1/16*(17*B*a*b*x^(5/2) - 9*A*b^2*x^(5/
2) + 13*B*a^2*sqrt(x) - 5*A*a*b*sqrt(x))/((b*x^2 + a)^2*b^3)