3.4.66 \(\int \frac {x^{5/2} (A+B x^2)}{(a+b x^2)^3} \, dx\)

Optimal. Leaf size=293 \[ \frac {3 (7 a B+A b) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{64 \sqrt {2} a^{5/4} b^{11/4}}-\frac {3 (7 a B+A b) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{64 \sqrt {2} a^{5/4} b^{11/4}}-\frac {3 (7 a B+A b) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{5/4} b^{11/4}}+\frac {3 (7 a B+A b) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{32 \sqrt {2} a^{5/4} b^{11/4}}-\frac {x^{3/2} (7 a B+A b)}{16 a b^2 \left (a+b x^2\right )}+\frac {x^{7/2} (A b-a B)}{4 a b \left (a+b x^2\right )^2} \]

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Rubi [A]  time = 0.22, antiderivative size = 293, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {457, 288, 329, 297, 1162, 617, 204, 1165, 628} \begin {gather*} \frac {3 (7 a B+A b) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{64 \sqrt {2} a^{5/4} b^{11/4}}-\frac {3 (7 a B+A b) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{64 \sqrt {2} a^{5/4} b^{11/4}}-\frac {3 (7 a B+A b) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{5/4} b^{11/4}}+\frac {3 (7 a B+A b) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{32 \sqrt {2} a^{5/4} b^{11/4}}-\frac {x^{3/2} (7 a B+A b)}{16 a b^2 \left (a+b x^2\right )}+\frac {x^{7/2} (A b-a B)}{4 a b \left (a+b x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^(5/2)*(A + B*x^2))/(a + b*x^2)^3,x]

[Out]

((A*b - a*B)*x^(7/2))/(4*a*b*(a + b*x^2)^2) - ((A*b + 7*a*B)*x^(3/2))/(16*a*b^2*(a + b*x^2)) - (3*(A*b + 7*a*B
)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(32*Sqrt[2]*a^(5/4)*b^(11/4)) + (3*(A*b + 7*a*B)*ArcTan[1 + (
Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(32*Sqrt[2]*a^(5/4)*b^(11/4)) + (3*(A*b + 7*a*B)*Log[Sqrt[a] - Sqrt[2]*a^(1
/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(64*Sqrt[2]*a^(5/4)*b^(11/4)) - (3*(A*b + 7*a*B)*Log[Sqrt[a] + Sqrt[2]*a^(1/
4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(64*Sqrt[2]*a^(5/4)*b^(11/4))

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 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 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 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 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 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 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 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]

Rubi steps

\begin {align*} \int \frac {x^{5/2} \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx &=\frac {(A b-a B) x^{7/2}}{4 a b \left (a+b x^2\right )^2}+\frac {\left (\frac {A b}{2}+\frac {7 a B}{2}\right ) \int \frac {x^{5/2}}{\left (a+b x^2\right )^2} \, dx}{4 a b}\\ &=\frac {(A b-a B) x^{7/2}}{4 a b \left (a+b x^2\right )^2}-\frac {(A b+7 a B) x^{3/2}}{16 a b^2 \left (a+b x^2\right )}+\frac {(3 (A b+7 a B)) \int \frac {\sqrt {x}}{a+b x^2} \, dx}{32 a b^2}\\ &=\frac {(A b-a B) x^{7/2}}{4 a b \left (a+b x^2\right )^2}-\frac {(A b+7 a B) x^{3/2}}{16 a b^2 \left (a+b x^2\right )}+\frac {(3 (A b+7 a B)) \operatorname {Subst}\left (\int \frac {x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{16 a b^2}\\ &=\frac {(A b-a B) x^{7/2}}{4 a b \left (a+b x^2\right )^2}-\frac {(A b+7 a B) x^{3/2}}{16 a b^2 \left (a+b x^2\right )}-\frac {(3 (A b+7 a B)) \operatorname {Subst}\left (\int \frac {\sqrt {a}-\sqrt {b} x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{32 a b^{5/2}}+\frac {(3 (A b+7 a B)) \operatorname {Subst}\left (\int \frac {\sqrt {a}+\sqrt {b} x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{32 a b^{5/2}}\\ &=\frac {(A b-a B) x^{7/2}}{4 a b \left (a+b x^2\right )^2}-\frac {(A b+7 a B) x^{3/2}}{16 a b^2 \left (a+b x^2\right )}+\frac {(3 (A b+7 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 a b^3}+\frac {(3 (A b+7 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 a b^3}+\frac {(3 (A b+7 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^{5/4} b^{11/4}}+\frac {(3 (A b+7 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^{5/4} b^{11/4}}\\ &=\frac {(A b-a B) x^{7/2}}{4 a b \left (a+b x^2\right )^2}-\frac {(A b+7 a B) x^{3/2}}{16 a b^2 \left (a+b x^2\right )}+\frac {3 (A b+7 a B) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{64 \sqrt {2} a^{5/4} b^{11/4}}-\frac {3 (A b+7 a B) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{64 \sqrt {2} a^{5/4} b^{11/4}}+\frac {(3 (A b+7 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^{5/4} b^{11/4}}-\frac {(3 (A b+7 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^{5/4} b^{11/4}}\\ &=\frac {(A b-a B) x^{7/2}}{4 a b \left (a+b x^2\right )^2}-\frac {(A b+7 a B) x^{3/2}}{16 a b^2 \left (a+b x^2\right )}-\frac {3 (A b+7 a B) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{5/4} b^{11/4}}+\frac {3 (A b+7 a B) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{5/4} b^{11/4}}+\frac {3 (A b+7 a B) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{64 \sqrt {2} a^{5/4} b^{11/4}}-\frac {3 (A b+7 a B) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{64 \sqrt {2} a^{5/4} b^{11/4}}\\ \end {align*}

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Mathematica [C]  time = 0.21, size = 137, normalized size = 0.47 \begin {gather*} \frac {2 b^{3/4} x^{3/2} (A b-2 a B) \, _2F_1\left (\frac {3}{4},2;\frac {7}{4};-\frac {b x^2}{a}\right )+2 b^{3/4} x^{3/2} (a B-A b) \, _2F_1\left (\frac {3}{4},3;\frac {7}{4};-\frac {b x^2}{a}\right )+3 (-a)^{7/4} B \left (\tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{-a}}\right )+\tanh ^{-1}\left (\frac {a \sqrt [4]{b} \sqrt {x}}{(-a)^{5/4}}\right )\right )}{3 a^2 b^{11/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^(5/2)*(A + B*x^2))/(a + b*x^2)^3,x]

[Out]

(3*(-a)^(7/4)*B*(ArcTan[(b^(1/4)*Sqrt[x])/(-a)^(1/4)] + ArcTanh[(a*b^(1/4)*Sqrt[x])/(-a)^(5/4)]) + 2*b^(3/4)*(
A*b - 2*a*B)*x^(3/2)*Hypergeometric2F1[3/4, 2, 7/4, -((b*x^2)/a)] + 2*b^(3/4)*(-(A*b) + a*B)*x^(3/2)*Hypergeom
etric2F1[3/4, 3, 7/4, -((b*x^2)/a)])/(3*a^2*b^(11/4))

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IntegrateAlgebraic [A]  time = 0.73, size = 180, normalized size = 0.61 \begin {gather*} -\frac {3 (7 a B+A b) \tan ^{-1}\left (\frac {\sqrt {a}-\sqrt {b} x}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}\right )}{32 \sqrt {2} a^{5/4} b^{11/4}}-\frac {3 (7 a B+A b) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}{\sqrt {a}+\sqrt {b} x}\right )}{32 \sqrt {2} a^{5/4} b^{11/4}}-\frac {x^{3/2} \left (7 a^2 B+a A b+11 a b B x^2-3 A b^2 x^2\right )}{16 a b^2 \left (a+b x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^(5/2)*(A + B*x^2))/(a + b*x^2)^3,x]

[Out]

-1/16*(x^(3/2)*(a*A*b + 7*a^2*B - 3*A*b^2*x^2 + 11*a*b*B*x^2))/(a*b^2*(a + b*x^2)^2) - (3*(A*b + 7*a*B)*ArcTan
[(Sqrt[a] - Sqrt[b]*x)/(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x])])/(32*Sqrt[2]*a^(5/4)*b^(11/4)) - (3*(A*b + 7*a*B)*Ar
cTanh[(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x])/(Sqrt[a] + Sqrt[b]*x)])/(32*Sqrt[2]*a^(5/4)*b^(11/4))

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fricas [B]  time = 1.31, size = 990, normalized size = 3.38 \begin {gather*} -\frac {12 \, {\left (a b^{4} x^{4} + 2 \, a^{2} b^{3} x^{2} + a^{3} b^{2}\right )} \left (-\frac {2401 \, B^{4} a^{4} + 1372 \, A B^{3} a^{3} b + 294 \, A^{2} B^{2} a^{2} b^{2} + 28 \, A^{3} B a b^{3} + A^{4} b^{4}}{a^{5} b^{11}}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {{\left (117649 \, B^{6} a^{6} + 100842 \, A B^{5} a^{5} b + 36015 \, A^{2} B^{4} a^{4} b^{2} + 6860 \, A^{3} B^{3} a^{3} b^{3} + 735 \, A^{4} B^{2} a^{2} b^{4} + 42 \, A^{5} B a b^{5} + A^{6} b^{6}\right )} x - {\left (2401 \, B^{4} a^{7} b^{5} + 1372 \, A B^{3} a^{6} b^{6} + 294 \, A^{2} B^{2} a^{5} b^{7} + 28 \, A^{3} B a^{4} b^{8} + A^{4} a^{3} b^{9}\right )} \sqrt {-\frac {2401 \, B^{4} a^{4} + 1372 \, A B^{3} a^{3} b + 294 \, A^{2} B^{2} a^{2} b^{2} + 28 \, A^{3} B a b^{3} + A^{4} b^{4}}{a^{5} b^{11}}}} a b^{3} \left (-\frac {2401 \, B^{4} a^{4} + 1372 \, A B^{3} a^{3} b + 294 \, A^{2} B^{2} a^{2} b^{2} + 28 \, A^{3} B a b^{3} + A^{4} b^{4}}{a^{5} b^{11}}\right )^{\frac {1}{4}} - {\left (343 \, B^{3} a^{4} b^{3} + 147 \, A B^{2} a^{3} b^{4} + 21 \, A^{2} B a^{2} b^{5} + A^{3} a b^{6}\right )} \sqrt {x} \left (-\frac {2401 \, B^{4} a^{4} + 1372 \, A B^{3} a^{3} b + 294 \, A^{2} B^{2} a^{2} b^{2} + 28 \, A^{3} B a b^{3} + A^{4} b^{4}}{a^{5} b^{11}}\right )^{\frac {1}{4}}}{2401 \, B^{4} a^{4} + 1372 \, A B^{3} a^{3} b + 294 \, A^{2} B^{2} a^{2} b^{2} + 28 \, A^{3} B a b^{3} + A^{4} b^{4}}\right ) - 3 \, {\left (a b^{4} x^{4} + 2 \, a^{2} b^{3} x^{2} + a^{3} b^{2}\right )} \left (-\frac {2401 \, B^{4} a^{4} + 1372 \, A B^{3} a^{3} b + 294 \, A^{2} B^{2} a^{2} b^{2} + 28 \, A^{3} B a b^{3} + A^{4} b^{4}}{a^{5} b^{11}}\right )^{\frac {1}{4}} \log \left (27 \, a^{4} b^{8} \left (-\frac {2401 \, B^{4} a^{4} + 1372 \, A B^{3} a^{3} b + 294 \, A^{2} B^{2} a^{2} b^{2} + 28 \, A^{3} B a b^{3} + A^{4} b^{4}}{a^{5} b^{11}}\right )^{\frac {3}{4}} + 27 \, {\left (343 \, B^{3} a^{3} + 147 \, A B^{2} a^{2} b + 21 \, A^{2} B a b^{2} + A^{3} b^{3}\right )} \sqrt {x}\right ) + 3 \, {\left (a b^{4} x^{4} + 2 \, a^{2} b^{3} x^{2} + a^{3} b^{2}\right )} \left (-\frac {2401 \, B^{4} a^{4} + 1372 \, A B^{3} a^{3} b + 294 \, A^{2} B^{2} a^{2} b^{2} + 28 \, A^{3} B a b^{3} + A^{4} b^{4}}{a^{5} b^{11}}\right )^{\frac {1}{4}} \log \left (-27 \, a^{4} b^{8} \left (-\frac {2401 \, B^{4} a^{4} + 1372 \, A B^{3} a^{3} b + 294 \, A^{2} B^{2} a^{2} b^{2} + 28 \, A^{3} B a b^{3} + A^{4} b^{4}}{a^{5} b^{11}}\right )^{\frac {3}{4}} + 27 \, {\left (343 \, B^{3} a^{3} + 147 \, A B^{2} a^{2} b + 21 \, A^{2} B a b^{2} + A^{3} b^{3}\right )} \sqrt {x}\right ) + 4 \, {\left ({\left (11 \, B a b - 3 \, A b^{2}\right )} x^{3} + {\left (7 \, B a^{2} + A a b\right )} x\right )} \sqrt {x}}{64 \, {\left (a b^{4} x^{4} + 2 \, a^{2} b^{3} x^{2} + a^{3} b^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x^2+A)/(b*x^2+a)^3,x, algorithm="fricas")

[Out]

-1/64*(12*(a*b^4*x^4 + 2*a^2*b^3*x^2 + a^3*b^2)*(-(2401*B^4*a^4 + 1372*A*B^3*a^3*b + 294*A^2*B^2*a^2*b^2 + 28*
A^3*B*a*b^3 + A^4*b^4)/(a^5*b^11))^(1/4)*arctan((sqrt((117649*B^6*a^6 + 100842*A*B^5*a^5*b + 36015*A^2*B^4*a^4
*b^2 + 6860*A^3*B^3*a^3*b^3 + 735*A^4*B^2*a^2*b^4 + 42*A^5*B*a*b^5 + A^6*b^6)*x - (2401*B^4*a^7*b^5 + 1372*A*B
^3*a^6*b^6 + 294*A^2*B^2*a^5*b^7 + 28*A^3*B*a^4*b^8 + A^4*a^3*b^9)*sqrt(-(2401*B^4*a^4 + 1372*A*B^3*a^3*b + 29
4*A^2*B^2*a^2*b^2 + 28*A^3*B*a*b^3 + A^4*b^4)/(a^5*b^11)))*a*b^3*(-(2401*B^4*a^4 + 1372*A*B^3*a^3*b + 294*A^2*
B^2*a^2*b^2 + 28*A^3*B*a*b^3 + A^4*b^4)/(a^5*b^11))^(1/4) - (343*B^3*a^4*b^3 + 147*A*B^2*a^3*b^4 + 21*A^2*B*a^
2*b^5 + A^3*a*b^6)*sqrt(x)*(-(2401*B^4*a^4 + 1372*A*B^3*a^3*b + 294*A^2*B^2*a^2*b^2 + 28*A^3*B*a*b^3 + A^4*b^4
)/(a^5*b^11))^(1/4))/(2401*B^4*a^4 + 1372*A*B^3*a^3*b + 294*A^2*B^2*a^2*b^2 + 28*A^3*B*a*b^3 + A^4*b^4)) - 3*(
a*b^4*x^4 + 2*a^2*b^3*x^2 + a^3*b^2)*(-(2401*B^4*a^4 + 1372*A*B^3*a^3*b + 294*A^2*B^2*a^2*b^2 + 28*A^3*B*a*b^3
+ A^4*b^4)/(a^5*b^11))^(1/4)*log(27*a^4*b^8*(-(2401*B^4*a^4 + 1372*A*B^3*a^3*b + 294*A^2*B^2*a^2*b^2 + 28*A^3
*B*a*b^3 + A^4*b^4)/(a^5*b^11))^(3/4) + 27*(343*B^3*a^3 + 147*A*B^2*a^2*b + 21*A^2*B*a*b^2 + A^3*b^3)*sqrt(x))
+ 3*(a*b^4*x^4 + 2*a^2*b^3*x^2 + a^3*b^2)*(-(2401*B^4*a^4 + 1372*A*B^3*a^3*b + 294*A^2*B^2*a^2*b^2 + 28*A^3*B
*a*b^3 + A^4*b^4)/(a^5*b^11))^(1/4)*log(-27*a^4*b^8*(-(2401*B^4*a^4 + 1372*A*B^3*a^3*b + 294*A^2*B^2*a^2*b^2 +
28*A^3*B*a*b^3 + A^4*b^4)/(a^5*b^11))^(3/4) + 27*(343*B^3*a^3 + 147*A*B^2*a^2*b + 21*A^2*B*a*b^2 + A^3*b^3)*s
qrt(x)) + 4*((11*B*a*b - 3*A*b^2)*x^3 + (7*B*a^2 + A*a*b)*x)*sqrt(x))/(a*b^4*x^4 + 2*a^2*b^3*x^2 + a^3*b^2)

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giac [A]  time = 0.47, size = 293, normalized size = 1.00 \begin {gather*} -\frac {11 \, B a b x^{\frac {7}{2}} - 3 \, A b^{2} x^{\frac {7}{2}} + 7 \, B a^{2} x^{\frac {3}{2}} + A a b x^{\frac {3}{2}}}{16 \, {\left (b x^{2} + a\right )}^{2} a b^{2}} + \frac {3 \, \sqrt {2} {\left (7 \, \left (a b^{3}\right )^{\frac {3}{4}} B a + \left (a b^{3}\right )^{\frac {3}{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^{2} b^{5}} + \frac {3 \, \sqrt {2} {\left (7 \, \left (a b^{3}\right )^{\frac {3}{4}} B a + \left (a b^{3}\right )^{\frac {3}{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^{2} b^{5}} - \frac {3 \, \sqrt {2} {\left (7 \, \left (a b^{3}\right )^{\frac {3}{4}} B a + \left (a b^{3}\right )^{\frac {3}{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^{2} b^{5}} + \frac {3 \, \sqrt {2} {\left (7 \, \left (a b^{3}\right )^{\frac {3}{4}} B a + \left (a b^{3}\right )^{\frac {3}{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^{2} b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x^2+A)/(b*x^2+a)^3,x, algorithm="giac")

[Out]

-1/16*(11*B*a*b*x^(7/2) - 3*A*b^2*x^(7/2) + 7*B*a^2*x^(3/2) + A*a*b*x^(3/2))/((b*x^2 + a)^2*a*b^2) + 3/64*sqrt
(2)*(7*(a*b^3)^(3/4)*B*a + (a*b^3)^(3/4)*A*b)*arctan(1/2*sqrt(2)*(sqrt(2)*(a/b)^(1/4) + 2*sqrt(x))/(a/b)^(1/4)
)/(a^2*b^5) + 3/64*sqrt(2)*(7*(a*b^3)^(3/4)*B*a + (a*b^3)^(3/4)*A*b)*arctan(-1/2*sqrt(2)*(sqrt(2)*(a/b)^(1/4)
- 2*sqrt(x))/(a/b)^(1/4))/(a^2*b^5) - 3/128*sqrt(2)*(7*(a*b^3)^(3/4)*B*a + (a*b^3)^(3/4)*A*b)*log(sqrt(2)*sqrt
(x)*(a/b)^(1/4) + x + sqrt(a/b))/(a^2*b^5) + 3/128*sqrt(2)*(7*(a*b^3)^(3/4)*B*a + (a*b^3)^(3/4)*A*b)*log(-sqrt
(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))/(a^2*b^5)

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maple [A]  time = 0.02, size = 325, normalized size = 1.11 \begin {gather*} \frac {3 \sqrt {2}\, A \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )}{64 \left (\frac {a}{b}\right )^{\frac {1}{4}} a \,b^{2}}+\frac {3 \sqrt {2}\, A \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )}{64 \left (\frac {a}{b}\right )^{\frac {1}{4}} a \,b^{2}}+\frac {3 \sqrt {2}\, A \ln \left (\frac {x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{b}}}{x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{b}}}\right )}{128 \left (\frac {a}{b}\right )^{\frac {1}{4}} a \,b^{2}}+\frac {21 \sqrt {2}\, B \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )}{64 \left (\frac {a}{b}\right )^{\frac {1}{4}} b^{3}}+\frac {21 \sqrt {2}\, B \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )}{64 \left (\frac {a}{b}\right )^{\frac {1}{4}} b^{3}}+\frac {21 \sqrt {2}\, B \ln \left (\frac {x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{b}}}{x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{b}}}\right )}{128 \left (\frac {a}{b}\right )^{\frac {1}{4}} b^{3}}+\frac {\frac {\left (3 A b -11 B a \right ) x^{\frac {7}{2}}}{16 a b}-\frac {\left (A b +7 B a \right ) x^{\frac {3}{2}}}{16 b^{2}}}{\left (b \,x^{2}+a \right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)*(B*x^2+A)/(b*x^2+a)^3,x)

[Out]

2*(1/32*(3*A*b-11*B*a)/a/b*x^(7/2)-1/32*(A*b+7*B*a)/b^2*x^(3/2))/(b*x^2+a)^2+3/64/b^2/a/(a/b)^(1/4)*2^(1/2)*A*
arctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)+1)+3/64/b^2/a/(a/b)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)-1)+3
/128/b^2/a/(a/b)^(1/4)*2^(1/2)*A*ln((x-(a/b)^(1/4)*2^(1/2)*x^(1/2)+(a/b)^(1/2))/(x+(a/b)^(1/4)*2^(1/2)*x^(1/2)
+(a/b)^(1/2)))+21/64/b^3/(a/b)^(1/4)*2^(1/2)*B*arctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)+1)+21/64/b^3/(a/b)^(1/4)*2^(
1/2)*B*arctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)-1)+21/128/b^3/(a/b)^(1/4)*2^(1/2)*B*ln((x-(a/b)^(1/4)*2^(1/2)*x^(1/2
)+(a/b)^(1/2))/(x+(a/b)^(1/4)*2^(1/2)*x^(1/2)+(a/b)^(1/2)))

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maxima [A]  time = 2.37, size = 251, normalized size = 0.86 \begin {gather*} -\frac {{\left (11 \, B a b - 3 \, A b^{2}\right )} x^{\frac {7}{2}} + {\left (7 \, B a^{2} + A a b\right )} x^{\frac {3}{2}}}{16 \, {\left (a b^{4} x^{4} + 2 \, a^{2} b^{3} x^{2} + a^{3} b^{2}\right )}} + \frac {3 \, {\left (7 \, B a + A b\right )} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} + 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} - 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} - \frac {\sqrt {2} \log \left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}}\right )}}{128 \, a b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x^2+A)/(b*x^2+a)^3,x, algorithm="maxima")

[Out]

-1/16*((11*B*a*b - 3*A*b^2)*x^(7/2) + (7*B*a^2 + A*a*b)*x^(3/2))/(a*b^4*x^4 + 2*a^2*b^3*x^2 + a^3*b^2) + 3/128
*(7*B*a + A*b)*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*a^(1/4)*b^(1/4) + 2*sqrt(b)*sqrt(x))/sqrt(sqrt(a)*sqrt(b
)))/(sqrt(sqrt(a)*sqrt(b))*sqrt(b)) + 2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*a^(1/4)*b^(1/4) - 2*sqrt(b)*sqrt(
x))/sqrt(sqrt(a)*sqrt(b)))/(sqrt(sqrt(a)*sqrt(b))*sqrt(b)) - sqrt(2)*log(sqrt(2)*a^(1/4)*b^(1/4)*sqrt(x) + sqr
t(b)*x + sqrt(a))/(a^(1/4)*b^(3/4)) + sqrt(2)*log(-sqrt(2)*a^(1/4)*b^(1/4)*sqrt(x) + sqrt(b)*x + sqrt(a))/(a^(
1/4)*b^(3/4)))/(a*b^2)

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mupad [B]  time = 0.19, size = 122, normalized size = 0.42 \begin {gather*} \frac {3\,\mathrm {atanh}\left (\frac {b^{1/4}\,\sqrt {x}}{{\left (-a\right )}^{1/4}}\right )\,\left (A\,b+7\,B\,a\right )}{32\,{\left (-a\right )}^{5/4}\,b^{11/4}}-\frac {3\,\mathrm {atan}\left (\frac {b^{1/4}\,\sqrt {x}}{{\left (-a\right )}^{1/4}}\right )\,\left (A\,b+7\,B\,a\right )}{32\,{\left (-a\right )}^{5/4}\,b^{11/4}}-\frac {\frac {x^{3/2}\,\left (A\,b+7\,B\,a\right )}{16\,b^2}-\frac {x^{7/2}\,\left (3\,A\,b-11\,B\,a\right )}{16\,a\,b}}{a^2+2\,a\,b\,x^2+b^2\,x^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^(5/2)*(A + B*x^2))/(a + b*x^2)^3,x)

[Out]

(3*atanh((b^(1/4)*x^(1/2))/(-a)^(1/4))*(A*b + 7*B*a))/(32*(-a)^(5/4)*b^(11/4)) - (3*atan((b^(1/4)*x^(1/2))/(-a
)^(1/4))*(A*b + 7*B*a))/(32*(-a)^(5/4)*b^(11/4)) - ((x^(3/2)*(A*b + 7*B*a))/(16*b^2) - (x^(7/2)*(3*A*b - 11*B*
a))/(16*a*b))/(a^2 + b^2*x^4 + 2*a*b*x^2)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(5/2)*(B*x**2+A)/(b*x**2+a)**3,x)

[Out]

Timed out

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