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

Optimal. Leaf size=310 \[ \frac {\sqrt [4]{b} (9 A b-5 a B) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{8 \sqrt {2} a^{13/4}}-\frac {\sqrt [4]{b} (9 A b-5 a B) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{8 \sqrt {2} a^{13/4}}-\frac {\sqrt [4]{b} (9 A b-5 a B) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} a^{13/4}}+\frac {\sqrt [4]{b} (9 A b-5 a B) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{4 \sqrt {2} a^{13/4}}+\frac {9 A b-5 a B}{2 a^3 \sqrt {x}}-\frac {9 A b-5 a B}{10 a^2 b x^{5/2}}+\frac {A b-a B}{2 a b x^{5/2} \left (a+b x^2\right )} \]

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Rubi [A]  time = 0.24, antiderivative size = 310, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {457, 325, 329, 297, 1162, 617, 204, 1165, 628} \begin {gather*} -\frac {9 A b-5 a B}{10 a^2 b x^{5/2}}+\frac {9 A b-5 a B}{2 a^3 \sqrt {x}}+\frac {\sqrt [4]{b} (9 A b-5 a B) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{8 \sqrt {2} a^{13/4}}-\frac {\sqrt [4]{b} (9 A b-5 a B) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{8 \sqrt {2} a^{13/4}}-\frac {\sqrt [4]{b} (9 A b-5 a B) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} a^{13/4}}+\frac {\sqrt [4]{b} (9 A b-5 a B) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{4 \sqrt {2} a^{13/4}}+\frac {A b-a B}{2 a b x^{5/2} \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(9*A*b - 5*a*B)/(10*a^2*b*x^(5/2)) + (9*A*b - 5*a*B)/(2*a^3*Sqrt[x]) + (A*b - a*B)/(2*a*b*x^(5/2)*(a + b*x^2)
) - (b^(1/4)*(9*A*b - 5*a*B)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(4*Sqrt[2]*a^(13/4)) + (b^(1/4)*(9
*A*b - 5*a*B)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(4*Sqrt[2]*a^(13/4)) + (b^(1/4)*(9*A*b - 5*a*B)*L
og[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(8*Sqrt[2]*a^(13/4)) - (b^(1/4)*(9*A*b - 5*a*B)*Log
[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(8*Sqrt[2]*a^(13/4))

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 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 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 {A+B x^2}{x^{7/2} \left (a+b x^2\right )^2} \, dx &=\frac {A b-a B}{2 a b x^{5/2} \left (a+b x^2\right )}+\frac {\left (\frac {9 A b}{2}-\frac {5 a B}{2}\right ) \int \frac {1}{x^{7/2} \left (a+b x^2\right )} \, dx}{2 a b}\\ &=-\frac {9 A b-5 a B}{10 a^2 b x^{5/2}}+\frac {A b-a B}{2 a b x^{5/2} \left (a+b x^2\right )}-\frac {(9 A b-5 a B) \int \frac {1}{x^{3/2} \left (a+b x^2\right )} \, dx}{4 a^2}\\ &=-\frac {9 A b-5 a B}{10 a^2 b x^{5/2}}+\frac {9 A b-5 a B}{2 a^3 \sqrt {x}}+\frac {A b-a B}{2 a b x^{5/2} \left (a+b x^2\right )}+\frac {(b (9 A b-5 a B)) \int \frac {\sqrt {x}}{a+b x^2} \, dx}{4 a^3}\\ &=-\frac {9 A b-5 a B}{10 a^2 b x^{5/2}}+\frac {9 A b-5 a B}{2 a^3 \sqrt {x}}+\frac {A b-a B}{2 a b x^{5/2} \left (a+b x^2\right )}+\frac {(b (9 A b-5 a B)) \operatorname {Subst}\left (\int \frac {x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{2 a^3}\\ &=-\frac {9 A b-5 a B}{10 a^2 b x^{5/2}}+\frac {9 A b-5 a B}{2 a^3 \sqrt {x}}+\frac {A b-a B}{2 a b x^{5/2} \left (a+b x^2\right )}-\frac {\left (\sqrt {b} (9 A b-5 a B)\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a}-\sqrt {b} x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{4 a^3}+\frac {\left (\sqrt {b} (9 A b-5 a B)\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a}+\sqrt {b} x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{4 a^3}\\ &=-\frac {9 A b-5 a B}{10 a^2 b x^{5/2}}+\frac {9 A b-5 a B}{2 a^3 \sqrt {x}}+\frac {A b-a B}{2 a b x^{5/2} \left (a+b x^2\right )}+\frac {(9 A b-5 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 a^3}+\frac {(9 A b-5 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 a^3}+\frac {\left (\sqrt [4]{b} (9 A b-5 a B)\right ) \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} a^{13/4}}+\frac {\left (\sqrt [4]{b} (9 A b-5 a B)\right ) \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} a^{13/4}}\\ &=-\frac {9 A b-5 a B}{10 a^2 b x^{5/2}}+\frac {9 A b-5 a B}{2 a^3 \sqrt {x}}+\frac {A b-a B}{2 a b x^{5/2} \left (a+b x^2\right )}+\frac {\sqrt [4]{b} (9 A b-5 a B) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{8 \sqrt {2} a^{13/4}}-\frac {\sqrt [4]{b} (9 A b-5 a B) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{8 \sqrt {2} a^{13/4}}+\frac {\left (\sqrt [4]{b} (9 A b-5 a B)\right ) \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} a^{13/4}}-\frac {\left (\sqrt [4]{b} (9 A b-5 a B)\right ) \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} a^{13/4}}\\ &=-\frac {9 A b-5 a B}{10 a^2 b x^{5/2}}+\frac {9 A b-5 a B}{2 a^3 \sqrt {x}}+\frac {A b-a B}{2 a b x^{5/2} \left (a+b x^2\right )}-\frac {\sqrt [4]{b} (9 A b-5 a B) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} a^{13/4}}+\frac {\sqrt [4]{b} (9 A b-5 a B) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} a^{13/4}}+\frac {\sqrt [4]{b} (9 A b-5 a B) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{8 \sqrt {2} a^{13/4}}-\frac {\sqrt [4]{b} (9 A b-5 a B) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{8 \sqrt {2} a^{13/4}}\\ \end {align*}

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Mathematica [C]  time = 0.43, size = 151, normalized size = 0.49 \begin {gather*} \frac {2 b 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^4}+\frac {4 A b-2 a B}{a^3 \sqrt {x}}-\frac {2 A}{5 a^2 x^{5/2}}+\frac {\sqrt [4]{b} (a B-2 A b) \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{-a}}\right )}{(-a)^{13/4}}+\frac {\sqrt [4]{b} (2 A b-a B) \tanh ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{-a}}\right )}{(-a)^{13/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*A)/(5*a^2*x^(5/2)) + (4*A*b - 2*a*B)/(a^3*Sqrt[x]) + (b^(1/4)*(-2*A*b + a*B)*ArcTan[(b^(1/4)*Sqrt[x])/(-a)
^(1/4)])/(-a)^(13/4) + (b^(1/4)*(2*A*b - a*B)*ArcTanh[(b^(1/4)*Sqrt[x])/(-a)^(1/4)])/(-a)^(13/4) + (2*b*(A*b -
a*B)*x^(3/2)*Hypergeometric2F1[3/4, 2, 7/4, -((b*x^2)/a)])/(3*a^4)

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IntegrateAlgebraic [A]  time = 0.59, size = 200, normalized size = 0.65 \begin {gather*} \frac {\left (5 a \sqrt [4]{b} B-9 A b^{5/4}\right ) \tan ^{-1}\left (\frac {\sqrt {a}-\sqrt {b} x}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}\right )}{4 \sqrt {2} a^{13/4}}+\frac {\left (5 a \sqrt [4]{b} B-9 A b^{5/4}\right ) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}{\sqrt {a}+\sqrt {b} x}\right )}{4 \sqrt {2} a^{13/4}}+\frac {-4 a^2 A-20 a^2 B x^2+36 a A b x^2-25 a b B x^4+45 A b^2 x^4}{10 a^3 x^{5/2} \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*a^2*A + 36*a*A*b*x^2 - 20*a^2*B*x^2 + 45*A*b^2*x^4 - 25*a*b*B*x^4)/(10*a^3*x^(5/2)*(a + b*x^2)) + ((-9*A*b
^(5/4) + 5*a*b^(1/4)*B)*ArcTan[(Sqrt[a] - Sqrt[b]*x)/(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x])])/(4*Sqrt[2]*a^(13/4))
+ ((-9*A*b^(5/4) + 5*a*b^(1/4)*B)*ArcTanh[(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x])/(Sqrt[a] + Sqrt[b]*x)])/(4*Sqrt[2]
*a^(13/4))

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fricas [B]  time = 1.44, size = 974, normalized size = 3.14 \begin {gather*} -\frac {20 \, {\left (a^{3} b x^{5} + a^{4} x^{3}\right )} \left (-\frac {625 \, B^{4} a^{4} b - 4500 \, A B^{3} a^{3} b^{2} + 12150 \, A^{2} B^{2} a^{2} b^{3} - 14580 \, A^{3} B a b^{4} + 6561 \, A^{4} b^{5}}{a^{13}}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {{\left (15625 \, B^{6} a^{6} b^{2} - 168750 \, A B^{5} a^{5} b^{3} + 759375 \, A^{2} B^{4} a^{4} b^{4} - 1822500 \, A^{3} B^{3} a^{3} b^{5} + 2460375 \, A^{4} B^{2} a^{2} b^{6} - 1771470 \, A^{5} B a b^{7} + 531441 \, A^{6} b^{8}\right )} x - {\left (625 \, B^{4} a^{11} b - 4500 \, A B^{3} a^{10} b^{2} + 12150 \, A^{2} B^{2} a^{9} b^{3} - 14580 \, A^{3} B a^{8} b^{4} + 6561 \, A^{4} a^{7} b^{5}\right )} \sqrt {-\frac {625 \, B^{4} a^{4} b - 4500 \, A B^{3} a^{3} b^{2} + 12150 \, A^{2} B^{2} a^{2} b^{3} - 14580 \, A^{3} B a b^{4} + 6561 \, A^{4} b^{5}}{a^{13}}}} a^{3} \left (-\frac {625 \, B^{4} a^{4} b - 4500 \, A B^{3} a^{3} b^{2} + 12150 \, A^{2} B^{2} a^{2} b^{3} - 14580 \, A^{3} B a b^{4} + 6561 \, A^{4} b^{5}}{a^{13}}\right )^{\frac {1}{4}} + {\left (125 \, B^{3} a^{6} b - 675 \, A B^{2} a^{5} b^{2} + 1215 \, A^{2} B a^{4} b^{3} - 729 \, A^{3} a^{3} b^{4}\right )} \sqrt {x} \left (-\frac {625 \, B^{4} a^{4} b - 4500 \, A B^{3} a^{3} b^{2} + 12150 \, A^{2} B^{2} a^{2} b^{3} - 14580 \, A^{3} B a b^{4} + 6561 \, A^{4} b^{5}}{a^{13}}\right )^{\frac {1}{4}}}{625 \, B^{4} a^{4} b - 4500 \, A B^{3} a^{3} b^{2} + 12150 \, A^{2} B^{2} a^{2} b^{3} - 14580 \, A^{3} B a b^{4} + 6561 \, A^{4} b^{5}}\right ) - 5 \, {\left (a^{3} b x^{5} + a^{4} x^{3}\right )} \left (-\frac {625 \, B^{4} a^{4} b - 4500 \, A B^{3} a^{3} b^{2} + 12150 \, A^{2} B^{2} a^{2} b^{3} - 14580 \, A^{3} B a b^{4} + 6561 \, A^{4} b^{5}}{a^{13}}\right )^{\frac {1}{4}} \log \left (a^{10} \left (-\frac {625 \, B^{4} a^{4} b - 4500 \, A B^{3} a^{3} b^{2} + 12150 \, A^{2} B^{2} a^{2} b^{3} - 14580 \, A^{3} B a b^{4} + 6561 \, A^{4} b^{5}}{a^{13}}\right )^{\frac {3}{4}} - {\left (125 \, B^{3} a^{3} b - 675 \, A B^{2} a^{2} b^{2} + 1215 \, A^{2} B a b^{3} - 729 \, A^{3} b^{4}\right )} \sqrt {x}\right ) + 5 \, {\left (a^{3} b x^{5} + a^{4} x^{3}\right )} \left (-\frac {625 \, B^{4} a^{4} b - 4500 \, A B^{3} a^{3} b^{2} + 12150 \, A^{2} B^{2} a^{2} b^{3} - 14580 \, A^{3} B a b^{4} + 6561 \, A^{4} b^{5}}{a^{13}}\right )^{\frac {1}{4}} \log \left (-a^{10} \left (-\frac {625 \, B^{4} a^{4} b - 4500 \, A B^{3} a^{3} b^{2} + 12150 \, A^{2} B^{2} a^{2} b^{3} - 14580 \, A^{3} B a b^{4} + 6561 \, A^{4} b^{5}}{a^{13}}\right )^{\frac {3}{4}} - {\left (125 \, B^{3} a^{3} b - 675 \, A B^{2} a^{2} b^{2} + 1215 \, A^{2} B a b^{3} - 729 \, A^{3} b^{4}\right )} \sqrt {x}\right ) + 4 \, {\left (5 \, {\left (5 \, B a b - 9 \, A b^{2}\right )} x^{4} + 4 \, A a^{2} + 4 \, {\left (5 \, B a^{2} - 9 \, A a b\right )} x^{2}\right )} \sqrt {x}}{40 \, {\left (a^{3} b x^{5} + a^{4} x^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/40*(20*(a^3*b*x^5 + a^4*x^3)*(-(625*B^4*a^4*b - 4500*A*B^3*a^3*b^2 + 12150*A^2*B^2*a^2*b^3 - 14580*A^3*B*a*
b^4 + 6561*A^4*b^5)/a^13)^(1/4)*arctan((sqrt((15625*B^6*a^6*b^2 - 168750*A*B^5*a^5*b^3 + 759375*A^2*B^4*a^4*b^
4 - 1822500*A^3*B^3*a^3*b^5 + 2460375*A^4*B^2*a^2*b^6 - 1771470*A^5*B*a*b^7 + 531441*A^6*b^8)*x - (625*B^4*a^1
1*b - 4500*A*B^3*a^10*b^2 + 12150*A^2*B^2*a^9*b^3 - 14580*A^3*B*a^8*b^4 + 6561*A^4*a^7*b^5)*sqrt(-(625*B^4*a^4
*b - 4500*A*B^3*a^3*b^2 + 12150*A^2*B^2*a^2*b^3 - 14580*A^3*B*a*b^4 + 6561*A^4*b^5)/a^13))*a^3*(-(625*B^4*a^4*
b - 4500*A*B^3*a^3*b^2 + 12150*A^2*B^2*a^2*b^3 - 14580*A^3*B*a*b^4 + 6561*A^4*b^5)/a^13)^(1/4) + (125*B^3*a^6*
b - 675*A*B^2*a^5*b^2 + 1215*A^2*B*a^4*b^3 - 729*A^3*a^3*b^4)*sqrt(x)*(-(625*B^4*a^4*b - 4500*A*B^3*a^3*b^2 +
12150*A^2*B^2*a^2*b^3 - 14580*A^3*B*a*b^4 + 6561*A^4*b^5)/a^13)^(1/4))/(625*B^4*a^4*b - 4500*A*B^3*a^3*b^2 + 1
2150*A^2*B^2*a^2*b^3 - 14580*A^3*B*a*b^4 + 6561*A^4*b^5)) - 5*(a^3*b*x^5 + a^4*x^3)*(-(625*B^4*a^4*b - 4500*A*
B^3*a^3*b^2 + 12150*A^2*B^2*a^2*b^3 - 14580*A^3*B*a*b^4 + 6561*A^4*b^5)/a^13)^(1/4)*log(a^10*(-(625*B^4*a^4*b
- 4500*A*B^3*a^3*b^2 + 12150*A^2*B^2*a^2*b^3 - 14580*A^3*B*a*b^4 + 6561*A^4*b^5)/a^13)^(3/4) - (125*B^3*a^3*b
- 675*A*B^2*a^2*b^2 + 1215*A^2*B*a*b^3 - 729*A^3*b^4)*sqrt(x)) + 5*(a^3*b*x^5 + a^4*x^3)*(-(625*B^4*a^4*b - 45
00*A*B^3*a^3*b^2 + 12150*A^2*B^2*a^2*b^3 - 14580*A^3*B*a*b^4 + 6561*A^4*b^5)/a^13)^(1/4)*log(-a^10*(-(625*B^4*
a^4*b - 4500*A*B^3*a^3*b^2 + 12150*A^2*B^2*a^2*b^3 - 14580*A^3*B*a*b^4 + 6561*A^4*b^5)/a^13)^(3/4) - (125*B^3*
a^3*b - 675*A*B^2*a^2*b^2 + 1215*A^2*B*a*b^3 - 729*A^3*b^4)*sqrt(x)) + 4*(5*(5*B*a*b - 9*A*b^2)*x^4 + 4*A*a^2
+ 4*(5*B*a^2 - 9*A*a*b)*x^2)*sqrt(x))/(a^3*b*x^5 + a^4*x^3)

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giac [A]  time = 0.54, size = 303, normalized size = 0.98 \begin {gather*} -\frac {B a b x^{\frac {3}{2}} - A b^{2} x^{\frac {3}{2}}}{2 \, {\left (b x^{2} + a\right )} a^{3}} - \frac {\sqrt {2} {\left (5 \, \left (a b^{3}\right )^{\frac {3}{4}} B a - 9 \, \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^{4} b^{2}} - \frac {\sqrt {2} {\left (5 \, \left (a b^{3}\right )^{\frac {3}{4}} B a - 9 \, \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^{4} b^{2}} + \frac {\sqrt {2} {\left (5 \, \left (a b^{3}\right )^{\frac {3}{4}} B a - 9 \, \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^{4} b^{2}} - \frac {\sqrt {2} {\left (5 \, \left (a b^{3}\right )^{\frac {3}{4}} B a - 9 \, \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^{4} b^{2}} - \frac {2 \, {\left (5 \, B a x^{2} - 10 \, A b x^{2} + A a\right )}}{5 \, a^{3} x^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(B*a*b*x^(3/2) - A*b^2*x^(3/2))/((b*x^2 + a)*a^3) - 1/8*sqrt(2)*(5*(a*b^3)^(3/4)*B*a - 9*(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^4*b^2) - 1/8*sqrt(2)*(5*(a*b^3)^(3/4)*
B*a - 9*(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^4*b^2) + 1/16
*sqrt(2)*(5*(a*b^3)^(3/4)*B*a - 9*(a*b^3)^(3/4)*A*b)*log(sqrt(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))/(a^4*b^2
) - 1/16*sqrt(2)*(5*(a*b^3)^(3/4)*B*a - 9*(a*b^3)^(3/4)*A*b)*log(-sqrt(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))
/(a^4*b^2) - 2/5*(5*B*a*x^2 - 10*A*b*x^2 + A*a)/(a^3*x^(5/2))

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maple [A]  time = 0.02, size = 339, normalized size = 1.09 \begin {gather*} \frac {A \,b^{2} x^{\frac {3}{2}}}{2 \left (b \,x^{2}+a \right ) a^{3}}-\frac {B b \,x^{\frac {3}{2}}}{2 \left (b \,x^{2}+a \right ) a^{2}}+\frac {9 \sqrt {2}\, A b \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}} a^{3}}+\frac {9 \sqrt {2}\, A b \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}} a^{3}}+\frac {9 \sqrt {2}\, A 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 )}{16 \left (\frac {a}{b}\right )^{\frac {1}{4}} a^{3}}-\frac {5 \sqrt {2}\, B \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}} a^{2}}-\frac {5 \sqrt {2}\, B \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}} a^{2}}-\frac {5 \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 )}{16 \left (\frac {a}{b}\right )^{\frac {1}{4}} a^{2}}+\frac {4 A b}{a^{3} \sqrt {x}}-\frac {2 B}{a^{2} \sqrt {x}}-\frac {2 A}{5 a^{2} x^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/a^3*b^2*x^(3/2)/(b*x^2+a)*A-1/2/a^2*b*x^(3/2)/(b*x^2+a)*B+9/16/a^3*b/(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)))+9/8/a^3*b/(a/b)^(1/4)*2^(1/2)*A*a
rctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)+1)+9/8/a^3*b/(a/b)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)-1)-5/1
6/a^2/(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)))-5/8/a^2/(a/b)^(1/4)*2^(1/2)*B*arctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)+1)-5/8/a^2/(a/b)^(1/4)*2^(1/2)*B*ar
ctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)-1)-2/5*A/a^2/x^(5/2)+4/a^3/x^(1/2)*A*b-2/a^2/x^(1/2)*B

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maxima [A]  time = 2.31, size = 250, normalized size = 0.81 \begin {gather*} -\frac {5 \, {\left (5 \, B a b - 9 \, A b^{2}\right )} x^{4} + 4 \, A a^{2} + 4 \, {\left (5 \, B a^{2} - 9 \, A a b\right )} x^{2}}{10 \, {\left (a^{3} b x^{\frac {9}{2}} + a^{4} x^{\frac {5}{2}}\right )}} - \frac {{\left (5 \, B a b - 9 \, A b^{2}\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 \, a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/10*(5*(5*B*a*b - 9*A*b^2)*x^4 + 4*A*a^2 + 4*(5*B*a^2 - 9*A*a*b)*x^2)/(a^3*b*x^(9/2) + a^4*x^(5/2)) - 1/16*(
5*B*a*b - 9*A*b^2)*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*a^(1/4)*b^(1/4) + 2*sqrt(b)*sqrt(x))/sqrt(sqrt(a)*sq
rt(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)*s
qrt(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)) + 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^3

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mupad [B]  time = 0.18, size = 121, normalized size = 0.39 \begin {gather*} \frac {\frac {2\,x^2\,\left (9\,A\,b-5\,B\,a\right )}{5\,a^2}-\frac {2\,A}{5\,a}+\frac {b\,x^4\,\left (9\,A\,b-5\,B\,a\right )}{2\,a^3}}{a\,x^{5/2}+b\,x^{9/2}}+\frac {{\left (-b\right )}^{1/4}\,\mathrm {atan}\left (\frac {{\left (-b\right )}^{1/4}\,\sqrt {x}}{a^{1/4}}\right )\,\left (9\,A\,b-5\,B\,a\right )}{4\,a^{13/4}}-\frac {{\left (-b\right )}^{1/4}\,\mathrm {atanh}\left (\frac {{\left (-b\right )}^{1/4}\,\sqrt {x}}{a^{1/4}}\right )\,\left (9\,A\,b-5\,B\,a\right )}{4\,a^{13/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((2*x^2*(9*A*b - 5*B*a))/(5*a^2) - (2*A)/(5*a) + (b*x^4*(9*A*b - 5*B*a))/(2*a^3))/(a*x^(5/2) + b*x^(9/2)) + ((
-b)^(1/4)*atan(((-b)^(1/4)*x^(1/2))/a^(1/4))*(9*A*b - 5*B*a))/(4*a^(13/4)) - ((-b)^(1/4)*atanh(((-b)^(1/4)*x^(
1/2))/a^(1/4))*(9*A*b - 5*B*a))/(4*a^(13/4))

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

[Out]

Timed out

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