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

Optimal. Leaf size=289 \[ \frac {(3 A b-7 a B) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{8 \sqrt {2} \sqrt [4]{a} b^{11/4}}-\frac {(3 A b-7 a B) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{8 \sqrt {2} \sqrt [4]{a} b^{11/4}}-\frac {(3 A b-7 a B) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} \sqrt [4]{a} b^{11/4}}+\frac {(3 A b-7 a B) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{4 \sqrt {2} \sqrt [4]{a} b^{11/4}}-\frac {x^{3/2} (3 A b-7 a B)}{6 a b^2}+\frac {x^{7/2} (A b-a B)}{2 a b \left (a+b x^2\right )} \]

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Rubi [A]  time = 0.22, antiderivative size = 289, 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, 321, 329, 297, 1162, 617, 204, 1165, 628} \begin {gather*} -\frac {x^{3/2} (3 A b-7 a B)}{6 a b^2}+\frac {(3 A b-7 a B) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{8 \sqrt {2} \sqrt [4]{a} b^{11/4}}-\frac {(3 A b-7 a B) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{8 \sqrt {2} \sqrt [4]{a} 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 )}{4 \sqrt {2} \sqrt [4]{a} b^{11/4}}+\frac {(3 A b-7 a B) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{4 \sqrt {2} \sqrt [4]{a} b^{11/4}}+\frac {x^{7/2} (A b-a B)}{2 a b \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((3*A*b - 7*a*B)*x^(3/2))/(6*a*b^2) + ((A*b - a*B)*x^(7/2))/(2*a*b*(a + b*x^2)) - ((3*A*b - 7*a*B)*ArcTan[1 -
(Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(4*Sqrt[2]*a^(1/4)*b^(11/4)) + ((3*A*b - 7*a*B)*ArcTan[1 + (Sqrt[2]*b^(1/
4)*Sqrt[x])/a^(1/4)])/(4*Sqrt[2]*a^(1/4)*b^(11/4)) + ((3*A*b - 7*a*B)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sq
rt[x] + Sqrt[b]*x])/(8*Sqrt[2]*a^(1/4)*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])/(8*Sqrt[2]*a^(1/4)*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 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 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 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 )^2} \, dx &=\frac {(A b-a B) x^{7/2}}{2 a b \left (a+b x^2\right )}+\frac {\left (-\frac {3 A b}{2}+\frac {7 a B}{2}\right ) \int \frac {x^{5/2}}{a+b x^2} \, dx}{2 a b}\\ &=-\frac {(3 A b-7 a B) x^{3/2}}{6 a b^2}+\frac {(A b-a B) x^{7/2}}{2 a b \left (a+b x^2\right )}+\frac {(3 A b-7 a B) \int \frac {\sqrt {x}}{a+b x^2} \, dx}{4 b^2}\\ &=-\frac {(3 A b-7 a B) x^{3/2}}{6 a b^2}+\frac {(A b-a B) x^{7/2}}{2 a b \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 )}{2 b^2}\\ &=-\frac {(3 A b-7 a B) x^{3/2}}{6 a b^2}+\frac {(A b-a B) x^{7/2}}{2 a b \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 )}{4 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 )}{4 b^{5/2}}\\ &=-\frac {(3 A b-7 a B) x^{3/2}}{6 a b^2}+\frac {(A b-a B) x^{7/2}}{2 a b \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 )}{8 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 )}{8 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 )}{8 \sqrt {2} \sqrt [4]{a} 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 )}{8 \sqrt {2} \sqrt [4]{a} b^{11/4}}\\ &=-\frac {(3 A b-7 a B) x^{3/2}}{6 a b^2}+\frac {(A b-a B) x^{7/2}}{2 a b \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 )}{8 \sqrt {2} \sqrt [4]{a} 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 )}{8 \sqrt {2} \sqrt [4]{a} 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 )}{4 \sqrt {2} \sqrt [4]{a} 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 )}{4 \sqrt {2} \sqrt [4]{a} b^{11/4}}\\ &=-\frac {(3 A b-7 a B) x^{3/2}}{6 a b^2}+\frac {(A b-a B) x^{7/2}}{2 a b \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 )}{4 \sqrt {2} \sqrt [4]{a} 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 )}{4 \sqrt {2} \sqrt [4]{a} 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 )}{8 \sqrt {2} \sqrt [4]{a} 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 )}{8 \sqrt {2} \sqrt [4]{a} b^{11/4}}\\ \end {align*}

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Mathematica [C]  time = 0.20, size = 136, normalized size = 0.47 \begin {gather*} \frac {3 (A b-2 a B) \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{-a}}\right )+(6 a B-3 A b) \tanh ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{-a}}\right )+2 \sqrt [4]{-a} b^{3/4} B x^{3/2}}{3 \sqrt [4]{-a} b^{11/4}}+\frac {2 x^{3/2} (a B-A b) \, _2F_1\left (\frac {3}{4},2;\frac {7}{4};-\frac {b x^2}{a}\right )}{3 a b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-a)^(1/4)*b^(3/4)*B*x^(3/2) + 3*(A*b - 2*a*B)*ArcTan[(b^(1/4)*Sqrt[x])/(-a)^(1/4)] + (-3*A*b + 6*a*B)*ArcT
anh[(b^(1/4)*Sqrt[x])/(-a)^(1/4)])/(3*(-a)^(1/4)*b^(11/4)) + (2*(-(A*b) + a*B)*x^(3/2)*Hypergeometric2F1[3/4,
2, 7/4, -((b*x^2)/a)])/(3*a*b^2)

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IntegrateAlgebraic [A]  time = 0.67, size = 167, normalized size = 0.58 \begin {gather*} \frac {(7 a B-3 A b) \tan ^{-1}\left (\frac {\sqrt {a}-\sqrt {b} x}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}\right )}{4 \sqrt {2} \sqrt [4]{a} b^{11/4}}+\frac {(7 a B-3 A b) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}{\sqrt {a}+\sqrt {b} x}\right )}{4 \sqrt {2} \sqrt [4]{a} b^{11/4}}+\frac {x^{3/2} \left (7 a B-3 A b+4 b B x^2\right )}{6 b^2 \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(3/2)*(-3*A*b + 7*a*B + 4*b*B*x^2))/(6*b^2*(a + b*x^2)) + ((-3*A*b + 7*a*B)*ArcTan[(Sqrt[a] - Sqrt[b]*x)/(S
qrt[2]*a^(1/4)*b^(1/4)*Sqrt[x])])/(4*Sqrt[2]*a^(1/4)*b^(11/4)) + ((-3*A*b + 7*a*B)*ArcTanh[(Sqrt[2]*a^(1/4)*b^
(1/4)*Sqrt[x])/(Sqrt[a] + Sqrt[b]*x)])/(4*Sqrt[2]*a^(1/4)*b^(11/4))

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fricas [B]  time = 1.45, size = 925, normalized size = 3.20 \begin {gather*} -\frac {12 \, {\left (b^{3} x^{2} + a b^{2}\right )} \left (-\frac {2401 \, B^{4} a^{4} - 4116 \, A B^{3} a^{3} b + 2646 \, A^{2} B^{2} a^{2} b^{2} - 756 \, A^{3} B a b^{3} + 81 \, A^{4} b^{4}}{a b^{11}}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {{\left (117649 \, B^{6} a^{6} - 302526 \, A B^{5} a^{5} b + 324135 \, A^{2} B^{4} a^{4} b^{2} - 185220 \, A^{3} B^{3} a^{3} b^{3} + 59535 \, A^{4} B^{2} a^{2} b^{4} - 10206 \, A^{5} B a b^{5} + 729 \, A^{6} b^{6}\right )} x - {\left (2401 \, B^{4} a^{5} b^{5} - 4116 \, A B^{3} a^{4} b^{6} + 2646 \, A^{2} B^{2} a^{3} b^{7} - 756 \, A^{3} B a^{2} b^{8} + 81 \, A^{4} a b^{9}\right )} \sqrt {-\frac {2401 \, B^{4} a^{4} - 4116 \, A B^{3} a^{3} b + 2646 \, A^{2} B^{2} a^{2} b^{2} - 756 \, A^{3} B a b^{3} + 81 \, A^{4} b^{4}}{a b^{11}}}} b^{3} \left (-\frac {2401 \, B^{4} a^{4} - 4116 \, A B^{3} a^{3} b + 2646 \, A^{2} B^{2} a^{2} b^{2} - 756 \, A^{3} B a b^{3} + 81 \, A^{4} b^{4}}{a b^{11}}\right )^{\frac {1}{4}} + {\left (343 \, B^{3} a^{3} b^{3} - 441 \, A B^{2} a^{2} b^{4} + 189 \, A^{2} B a b^{5} - 27 \, A^{3} b^{6}\right )} \sqrt {x} \left (-\frac {2401 \, B^{4} a^{4} - 4116 \, A B^{3} a^{3} b + 2646 \, A^{2} B^{2} a^{2} b^{2} - 756 \, A^{3} B a b^{3} + 81 \, A^{4} b^{4}}{a b^{11}}\right )^{\frac {1}{4}}}{2401 \, B^{4} a^{4} - 4116 \, A B^{3} a^{3} b + 2646 \, A^{2} B^{2} a^{2} b^{2} - 756 \, A^{3} B a b^{3} + 81 \, A^{4} b^{4}}\right ) - 3 \, {\left (b^{3} x^{2} + a b^{2}\right )} \left (-\frac {2401 \, B^{4} a^{4} - 4116 \, A B^{3} a^{3} b + 2646 \, A^{2} B^{2} a^{2} b^{2} - 756 \, A^{3} B a b^{3} + 81 \, A^{4} b^{4}}{a b^{11}}\right )^{\frac {1}{4}} \log \left (a b^{8} \left (-\frac {2401 \, B^{4} a^{4} - 4116 \, A B^{3} a^{3} b + 2646 \, A^{2} B^{2} a^{2} b^{2} - 756 \, A^{3} B a b^{3} + 81 \, A^{4} b^{4}}{a b^{11}}\right )^{\frac {3}{4}} - {\left (343 \, B^{3} a^{3} - 441 \, A B^{2} a^{2} b + 189 \, A^{2} B a b^{2} - 27 \, A^{3} b^{3}\right )} \sqrt {x}\right ) + 3 \, {\left (b^{3} x^{2} + a b^{2}\right )} \left (-\frac {2401 \, B^{4} a^{4} - 4116 \, A B^{3} a^{3} b + 2646 \, A^{2} B^{2} a^{2} b^{2} - 756 \, A^{3} B a b^{3} + 81 \, A^{4} b^{4}}{a b^{11}}\right )^{\frac {1}{4}} \log \left (-a b^{8} \left (-\frac {2401 \, B^{4} a^{4} - 4116 \, A B^{3} a^{3} b + 2646 \, A^{2} B^{2} a^{2} b^{2} - 756 \, A^{3} B a b^{3} + 81 \, A^{4} b^{4}}{a b^{11}}\right )^{\frac {3}{4}} - {\left (343 \, B^{3} a^{3} - 441 \, A B^{2} a^{2} b + 189 \, A^{2} B a b^{2} - 27 \, A^{3} b^{3}\right )} \sqrt {x}\right ) - 4 \, {\left (4 \, B b x^{3} + {\left (7 \, B a - 3 \, A b\right )} x\right )} \sqrt {x}}{24 \, {\left (b^{3} x^{2} + a 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)^2,x, algorithm="fricas")

[Out]

-1/24*(12*(b^3*x^2 + a*b^2)*(-(2401*B^4*a^4 - 4116*A*B^3*a^3*b + 2646*A^2*B^2*a^2*b^2 - 756*A^3*B*a*b^3 + 81*A
^4*b^4)/(a*b^11))^(1/4)*arctan((sqrt((117649*B^6*a^6 - 302526*A*B^5*a^5*b + 324135*A^2*B^4*a^4*b^2 - 185220*A^
3*B^3*a^3*b^3 + 59535*A^4*B^2*a^2*b^4 - 10206*A^5*B*a*b^5 + 729*A^6*b^6)*x - (2401*B^4*a^5*b^5 - 4116*A*B^3*a^
4*b^6 + 2646*A^2*B^2*a^3*b^7 - 756*A^3*B*a^2*b^8 + 81*A^4*a*b^9)*sqrt(-(2401*B^4*a^4 - 4116*A*B^3*a^3*b + 2646
*A^2*B^2*a^2*b^2 - 756*A^3*B*a*b^3 + 81*A^4*b^4)/(a*b^11)))*b^3*(-(2401*B^4*a^4 - 4116*A*B^3*a^3*b + 2646*A^2*
B^2*a^2*b^2 - 756*A^3*B*a*b^3 + 81*A^4*b^4)/(a*b^11))^(1/4) + (343*B^3*a^3*b^3 - 441*A*B^2*a^2*b^4 + 189*A^2*B
*a*b^5 - 27*A^3*b^6)*sqrt(x)*(-(2401*B^4*a^4 - 4116*A*B^3*a^3*b + 2646*A^2*B^2*a^2*b^2 - 756*A^3*B*a*b^3 + 81*
A^4*b^4)/(a*b^11))^(1/4))/(2401*B^4*a^4 - 4116*A*B^3*a^3*b + 2646*A^2*B^2*a^2*b^2 - 756*A^3*B*a*b^3 + 81*A^4*b
^4)) - 3*(b^3*x^2 + a*b^2)*(-(2401*B^4*a^4 - 4116*A*B^3*a^3*b + 2646*A^2*B^2*a^2*b^2 - 756*A^3*B*a*b^3 + 81*A^
4*b^4)/(a*b^11))^(1/4)*log(a*b^8*(-(2401*B^4*a^4 - 4116*A*B^3*a^3*b + 2646*A^2*B^2*a^2*b^2 - 756*A^3*B*a*b^3 +
81*A^4*b^4)/(a*b^11))^(3/4) - (343*B^3*a^3 - 441*A*B^2*a^2*b + 189*A^2*B*a*b^2 - 27*A^3*b^3)*sqrt(x)) + 3*(b^
3*x^2 + a*b^2)*(-(2401*B^4*a^4 - 4116*A*B^3*a^3*b + 2646*A^2*B^2*a^2*b^2 - 756*A^3*B*a*b^3 + 81*A^4*b^4)/(a*b^
11))^(1/4)*log(-a*b^8*(-(2401*B^4*a^4 - 4116*A*B^3*a^3*b + 2646*A^2*B^2*a^2*b^2 - 756*A^3*B*a*b^3 + 81*A^4*b^4
)/(a*b^11))^(3/4) - (343*B^3*a^3 - 441*A*B^2*a^2*b + 189*A^2*B*a*b^2 - 27*A^3*b^3)*sqrt(x)) - 4*(4*B*b*x^3 + (
7*B*a - 3*A*b)*x)*sqrt(x))/(b^3*x^2 + a*b^2)

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giac [A]  time = 0.42, size = 283, normalized size = 0.98 \begin {gather*} \frac {2 \, B x^{\frac {3}{2}}}{3 \, b^{2}} + \frac {B a x^{\frac {3}{2}} - A b x^{\frac {3}{2}}}{2 \, {\left (b x^{2} + a\right )} b^{2}} - \frac {\sqrt {2} {\left (7 \, \left (a b^{3}\right )^{\frac {3}{4}} B a - 3 \, \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 )}{8 \, a b^{5}} - \frac {\sqrt {2} {\left (7 \, \left (a b^{3}\right )^{\frac {3}{4}} B a - 3 \, \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 )}{8 \, a b^{5}} + \frac {\sqrt {2} {\left (7 \, \left (a b^{3}\right )^{\frac {3}{4}} B a - 3 \, \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 )}{16 \, a b^{5}} - \frac {\sqrt {2} {\left (7 \, \left (a b^{3}\right )^{\frac {3}{4}} B a - 3 \, \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 )}{16 \, a 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)^2,x, algorithm="giac")

[Out]

2/3*B*x^(3/2)/b^2 + 1/2*(B*a*x^(3/2) - A*b*x^(3/2))/((b*x^2 + a)*b^2) - 1/8*sqrt(2)*(7*(a*b^3)^(3/4)*B*a - 3*(
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*b^5) - 1/8*sqrt(2)*(7*(
a*b^3)^(3/4)*B*a - 3*(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*
b^5) + 1/16*sqrt(2)*(7*(a*b^3)^(3/4)*B*a - 3*(a*b^3)^(3/4)*A*b)*log(sqrt(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b
))/(a*b^5) - 1/16*sqrt(2)*(7*(a*b^3)^(3/4)*B*a - 3*(a*b^3)^(3/4)*A*b)*log(-sqrt(2)*sqrt(x)*(a/b)^(1/4) + x + s
qrt(a/b))/(a*b^5)

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maple [A]  time = 0.02, size = 317, normalized size = 1.10 \begin {gather*} -\frac {A \,x^{\frac {3}{2}}}{2 \left (b \,x^{2}+a \right ) b}+\frac {B a \,x^{\frac {3}{2}}}{2 \left (b \,x^{2}+a \right ) b^{2}}+\frac {2 B \,x^{\frac {3}{2}}}{3 b^{2}}+\frac {3 \sqrt {2}\, A \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )}{8 \left (\frac {a}{b}\right )^{\frac {1}{4}} b^{2}}+\frac {3 \sqrt {2}\, A \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )}{8 \left (\frac {a}{b}\right )^{\frac {1}{4}} 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 )}{16 \left (\frac {a}{b}\right )^{\frac {1}{4}} b^{2}}-\frac {7 \sqrt {2}\, B a \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )}{8 \left (\frac {a}{b}\right )^{\frac {1}{4}} b^{3}}-\frac {7 \sqrt {2}\, B a \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )}{8 \left (\frac {a}{b}\right )^{\frac {1}{4}} b^{3}}-\frac {7 \sqrt {2}\, B 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 )}{16 \left (\frac {a}{b}\right )^{\frac {1}{4}} b^{3}} \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)^2,x)

[Out]

2/3*B*x^(3/2)/b^2-1/2/b*x^(3/2)/(b*x^2+a)*A+1/2/b^2*x^(3/2)/(b*x^2+a)*B*a-7/16/b^3/(a/b)^(1/4)*2^(1/2)*B*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)))-7/8/b^3/(a/b)^(1/4)*2
^(1/2)*B*a*arctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)+1)-7/8/b^3/(a/b)^(1/4)*2^(1/2)*B*a*arctan(2^(1/2)/(a/b)^(1/4)*x^
(1/2)-1)+3/16/b^2/(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)))+3/8/b^2/(a/b)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)+1)+3/8/b^2/(a/b)^(1/4)*
2^(1/2)*A*arctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)-1)

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maxima [A]  time = 2.39, size = 223, normalized size = 0.77 \begin {gather*} \frac {{\left (B a - A b\right )} x^{\frac {3}{2}}}{2 \, {\left (b^{3} x^{2} + a b^{2}\right )}} + \frac {2 \, B x^{\frac {3}{2}}}{3 \, b^{2}} - \frac {{\left (7 \, B a - 3 \, 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 )}}{16 \, 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)^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.20, size = 106, normalized size = 0.37 \begin {gather*} \frac {2\,B\,x^{3/2}}{3\,b^2}-\frac {x^{3/2}\,\left (\frac {A\,b}{2}-\frac {B\,a}{2}\right )}{b^3\,x^2+a\,b^2}+\frac {\mathrm {atan}\left (\frac {b^{1/4}\,\sqrt {x}}{{\left (-a\right )}^{1/4}}\right )\,\left (3\,A\,b-7\,B\,a\right )}{4\,{\left (-a\right )}^{1/4}\,b^{11/4}}+\frac {\mathrm {atan}\left (\frac {b^{1/4}\,\sqrt {x}\,1{}\mathrm {i}}{{\left (-a\right )}^{1/4}}\right )\,\left (3\,A\,b-7\,B\,a\right )\,1{}\mathrm {i}}{4\,{\left (-a\right )}^{1/4}\,b^{11/4}} \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)^2,x)

[Out]

(2*B*x^(3/2))/(3*b^2) - (x^(3/2)*((A*b)/2 - (B*a)/2))/(a*b^2 + b^3*x^2) + (atan((b^(1/4)*x^(1/2))/(-a)^(1/4))*
(3*A*b - 7*B*a))/(4*(-a)^(1/4)*b^(11/4)) + (atan((b^(1/4)*x^(1/2)*1i)/(-a)^(1/4))*(3*A*b - 7*B*a)*1i)/(4*(-a)^
(1/4)*b^(11/4))

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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)**2,x)

[Out]

Timed out

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