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

Optimal. Leaf size=322 \[ -\frac {5 (9 A b-a B) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{64 \sqrt {2} a^{13/4} b^{3/4}}+\frac {5 (9 A b-a B) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{64 \sqrt {2} a^{13/4} b^{3/4}}+\frac {5 (9 A b-a B) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{13/4} b^{3/4}}-\frac {5 (9 A b-a B) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{32 \sqrt {2} a^{13/4} b^{3/4}}-\frac {5 (9 A b-a B)}{16 a^3 b \sqrt {x}}+\frac {9 A b-a B}{16 a^2 b \sqrt {x} \left (a+b x^2\right )}+\frac {A b-a B}{4 a b \sqrt {x} \left (a+b x^2\right )^2} \]

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Rubi [A]  time = 0.23, antiderivative size = 322, normalized size of antiderivative = 1.00, 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, 290, 325, 329, 297, 1162, 617, 204, 1165, 628} \begin {gather*} -\frac {5 (9 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^{13/4} b^{3/4}}+\frac {5 (9 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^{13/4} b^{3/4}}+\frac {5 (9 A b-a B) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{13/4} b^{3/4}}-\frac {5 (9 A b-a B) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{32 \sqrt {2} a^{13/4} b^{3/4}}+\frac {9 A b-a B}{16 a^2 b \sqrt {x} \left (a+b x^2\right )}-\frac {5 (9 A b-a B)}{16 a^3 b \sqrt {x}}+\frac {A b-a B}{4 a b \sqrt {x} \left (a+b x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 290

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

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Mathematica [C]  time = 0.18, size = 147, normalized size = 0.46 \begin {gather*} \frac {2 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^4}-\frac {2 A b x^{3/2} \, _2F_1\left (\frac {3}{4},2;\frac {7}{4};-\frac {b x^2}{a}\right )}{3 a^4}-\frac {2 A}{a^3 \sqrt {x}}+\frac {A \sqrt [4]{b} \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{-a}}\right )}{(-a)^{13/4}}+\frac {a A \sqrt [4]{b} \tanh ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{-a}}\right )}{(-a)^{17/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*A)/(a^3*Sqrt[x]) + (A*b^(1/4)*ArcTan[(b^(1/4)*Sqrt[x])/(-a)^(1/4)])/(-a)^(13/4) + (a*A*b^(1/4)*ArcTanh[(b^
(1/4)*Sqrt[x])/(-a)^(1/4)])/(-a)^(17/4) - (2*A*b*x^(3/2)*Hypergeometric2F1[3/4, 2, 7/4, -((b*x^2)/a)])/(3*a^4)
+ (2*(-(A*b) + a*B)*x^(3/2)*Hypergeometric2F1[3/4, 3, 7/4, -((b*x^2)/a)])/(3*a^4)

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IntegrateAlgebraic [A]  time = 0.73, size = 190, normalized size = 0.59 \begin {gather*} -\frac {5 (a B-9 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^{13/4} b^{3/4}}-\frac {5 (a B-9 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^{13/4} b^{3/4}}+\frac {-32 a^2 A+9 a^2 B x^2-81 a A b x^2+5 a b B x^4-45 A b^2 x^4}{16 a^3 \sqrt {x} \left (a+b x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-32*a^2*A - 81*a*A*b*x^2 + 9*a^2*B*x^2 - 45*A*b^2*x^4 + 5*a*b*B*x^4)/(16*a^3*Sqrt[x]*(a + b*x^2)^2) - (5*(-9*
A*b + a*B)*ArcTan[(Sqrt[a] - Sqrt[b]*x)/(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x])])/(32*Sqrt[2]*a^(13/4)*b^(3/4)) - (5
*(-9*A*b + a*B)*ArcTanh[(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x])/(Sqrt[a] + Sqrt[b]*x)])/(32*Sqrt[2]*a^(13/4)*b^(3/4)
)

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fricas [B]  time = 0.82, size = 988, normalized size = 3.07 \begin {gather*} \frac {20 \, {\left (a^{3} b^{2} x^{5} + 2 \, a^{4} b x^{3} + a^{5} x\right )} \left (-\frac {B^{4} a^{4} - 36 \, A B^{3} a^{3} b + 486 \, A^{2} B^{2} a^{2} b^{2} - 2916 \, A^{3} B a b^{3} + 6561 \, A^{4} b^{4}}{a^{13} b^{3}}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {{\left (B^{6} a^{6} - 54 \, A B^{5} a^{5} b + 1215 \, A^{2} B^{4} a^{4} b^{2} - 14580 \, A^{3} B^{3} a^{3} b^{3} + 98415 \, A^{4} B^{2} a^{2} b^{4} - 354294 \, A^{5} B a b^{5} + 531441 \, A^{6} b^{6}\right )} x - {\left (B^{4} a^{11} b - 36 \, A B^{3} a^{10} b^{2} + 486 \, A^{2} B^{2} a^{9} b^{3} - 2916 \, A^{3} B a^{8} b^{4} + 6561 \, A^{4} a^{7} b^{5}\right )} \sqrt {-\frac {B^{4} a^{4} - 36 \, A B^{3} a^{3} b + 486 \, A^{2} B^{2} a^{2} b^{2} - 2916 \, A^{3} B a b^{3} + 6561 \, A^{4} b^{4}}{a^{13} b^{3}}}} a^{3} b \left (-\frac {B^{4} a^{4} - 36 \, A B^{3} a^{3} b + 486 \, A^{2} B^{2} a^{2} b^{2} - 2916 \, A^{3} B a b^{3} + 6561 \, A^{4} b^{4}}{a^{13} b^{3}}\right )^{\frac {1}{4}} + {\left (B^{3} a^{6} b - 27 \, A B^{2} a^{5} b^{2} + 243 \, A^{2} B a^{4} b^{3} - 729 \, A^{3} a^{3} b^{4}\right )} \sqrt {x} \left (-\frac {B^{4} a^{4} - 36 \, A B^{3} a^{3} b + 486 \, A^{2} B^{2} a^{2} b^{2} - 2916 \, A^{3} B a b^{3} + 6561 \, A^{4} b^{4}}{a^{13} b^{3}}\right )^{\frac {1}{4}}}{B^{4} a^{4} - 36 \, A B^{3} a^{3} b + 486 \, A^{2} B^{2} a^{2} b^{2} - 2916 \, A^{3} B a b^{3} + 6561 \, A^{4} b^{4}}\right ) - 5 \, {\left (a^{3} b^{2} x^{5} + 2 \, a^{4} b x^{3} + a^{5} x\right )} \left (-\frac {B^{4} a^{4} - 36 \, A B^{3} a^{3} b + 486 \, A^{2} B^{2} a^{2} b^{2} - 2916 \, A^{3} B a b^{3} + 6561 \, A^{4} b^{4}}{a^{13} b^{3}}\right )^{\frac {1}{4}} \log \left (125 \, a^{10} b^{2} \left (-\frac {B^{4} a^{4} - 36 \, A B^{3} a^{3} b + 486 \, A^{2} B^{2} a^{2} b^{2} - 2916 \, A^{3} B a b^{3} + 6561 \, A^{4} b^{4}}{a^{13} b^{3}}\right )^{\frac {3}{4}} - 125 \, {\left (B^{3} a^{3} - 27 \, A B^{2} a^{2} b + 243 \, A^{2} B a b^{2} - 729 \, A^{3} b^{3}\right )} \sqrt {x}\right ) + 5 \, {\left (a^{3} b^{2} x^{5} + 2 \, a^{4} b x^{3} + a^{5} x\right )} \left (-\frac {B^{4} a^{4} - 36 \, A B^{3} a^{3} b + 486 \, A^{2} B^{2} a^{2} b^{2} - 2916 \, A^{3} B a b^{3} + 6561 \, A^{4} b^{4}}{a^{13} b^{3}}\right )^{\frac {1}{4}} \log \left (-125 \, a^{10} b^{2} \left (-\frac {B^{4} a^{4} - 36 \, A B^{3} a^{3} b + 486 \, A^{2} B^{2} a^{2} b^{2} - 2916 \, A^{3} B a b^{3} + 6561 \, A^{4} b^{4}}{a^{13} b^{3}}\right )^{\frac {3}{4}} - 125 \, {\left (B^{3} a^{3} - 27 \, A B^{2} a^{2} b + 243 \, A^{2} B a b^{2} - 729 \, A^{3} b^{3}\right )} \sqrt {x}\right ) + 4 \, {\left (5 \, {\left (B a b - 9 \, A b^{2}\right )} x^{4} - 32 \, A a^{2} + 9 \, {\left (B a^{2} - 9 \, A a b\right )} x^{2}\right )} \sqrt {x}}{64 \, {\left (a^{3} b^{2} x^{5} + 2 \, a^{4} b x^{3} + a^{5} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(20*(a^3*b^2*x^5 + 2*a^4*b*x^3 + a^5*x)*(-(B^4*a^4 - 36*A*B^3*a^3*b + 486*A^2*B^2*a^2*b^2 - 2916*A^3*B*a*
b^3 + 6561*A^4*b^4)/(a^13*b^3))^(1/4)*arctan((sqrt((B^6*a^6 - 54*A*B^5*a^5*b + 1215*A^2*B^4*a^4*b^2 - 14580*A^
3*B^3*a^3*b^3 + 98415*A^4*B^2*a^2*b^4 - 354294*A^5*B*a*b^5 + 531441*A^6*b^6)*x - (B^4*a^11*b - 36*A*B^3*a^10*b
^2 + 486*A^2*B^2*a^9*b^3 - 2916*A^3*B*a^8*b^4 + 6561*A^4*a^7*b^5)*sqrt(-(B^4*a^4 - 36*A*B^3*a^3*b + 486*A^2*B^
2*a^2*b^2 - 2916*A^3*B*a*b^3 + 6561*A^4*b^4)/(a^13*b^3)))*a^3*b*(-(B^4*a^4 - 36*A*B^3*a^3*b + 486*A^2*B^2*a^2*
b^2 - 2916*A^3*B*a*b^3 + 6561*A^4*b^4)/(a^13*b^3))^(1/4) + (B^3*a^6*b - 27*A*B^2*a^5*b^2 + 243*A^2*B*a^4*b^3 -
729*A^3*a^3*b^4)*sqrt(x)*(-(B^4*a^4 - 36*A*B^3*a^3*b + 486*A^2*B^2*a^2*b^2 - 2916*A^3*B*a*b^3 + 6561*A^4*b^4)
/(a^13*b^3))^(1/4))/(B^4*a^4 - 36*A*B^3*a^3*b + 486*A^2*B^2*a^2*b^2 - 2916*A^3*B*a*b^3 + 6561*A^4*b^4)) - 5*(a
^3*b^2*x^5 + 2*a^4*b*x^3 + a^5*x)*(-(B^4*a^4 - 36*A*B^3*a^3*b + 486*A^2*B^2*a^2*b^2 - 2916*A^3*B*a*b^3 + 6561*
A^4*b^4)/(a^13*b^3))^(1/4)*log(125*a^10*b^2*(-(B^4*a^4 - 36*A*B^3*a^3*b + 486*A^2*B^2*a^2*b^2 - 2916*A^3*B*a*b
^3 + 6561*A^4*b^4)/(a^13*b^3))^(3/4) - 125*(B^3*a^3 - 27*A*B^2*a^2*b + 243*A^2*B*a*b^2 - 729*A^3*b^3)*sqrt(x))
+ 5*(a^3*b^2*x^5 + 2*a^4*b*x^3 + a^5*x)*(-(B^4*a^4 - 36*A*B^3*a^3*b + 486*A^2*B^2*a^2*b^2 - 2916*A^3*B*a*b^3
+ 6561*A^4*b^4)/(a^13*b^3))^(1/4)*log(-125*a^10*b^2*(-(B^4*a^4 - 36*A*B^3*a^3*b + 486*A^2*B^2*a^2*b^2 - 2916*A
^3*B*a*b^3 + 6561*A^4*b^4)/(a^13*b^3))^(3/4) - 125*(B^3*a^3 - 27*A*B^2*a^2*b + 243*A^2*B*a*b^2 - 729*A^3*b^3)*
sqrt(x)) + 4*(5*(B*a*b - 9*A*b^2)*x^4 - 32*A*a^2 + 9*(B*a^2 - 9*A*a*b)*x^2)*sqrt(x))/(a^3*b^2*x^5 + 2*a^4*b*x^
3 + a^5*x)

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giac [A]  time = 0.41, size = 300, normalized size = 0.93 \begin {gather*} -\frac {2 \, A}{a^{3} \sqrt {x}} + \frac {5 \, B a b x^{\frac {7}{2}} - 13 \, A b^{2} x^{\frac {7}{2}} + 9 \, B a^{2} x^{\frac {3}{2}} - 17 \, A a b x^{\frac {3}{2}}}{16 \, {\left (b x^{2} + a\right )}^{2} a^{3}} + \frac {5 \, \sqrt {2} {\left (\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 )}{64 \, a^{4} b^{3}} + \frac {5 \, \sqrt {2} {\left (\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 )}{64 \, a^{4} b^{3}} - \frac {5 \, \sqrt {2} {\left (\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 )}{128 \, a^{4} b^{3}} + \frac {5 \, \sqrt {2} {\left (\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 )}{128 \, a^{4} b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*A/(a^3*sqrt(x)) + 1/16*(5*B*a*b*x^(7/2) - 13*A*b^2*x^(7/2) + 9*B*a^2*x^(3/2) - 17*A*a*b*x^(3/2))/((b*x^2 +
a)^2*a^3) + 5/64*sqrt(2)*((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^3) + 5/64*sqrt(2)*((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^3) - 5/128*sqrt(2)*((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^3) + 5/128*sqrt(2)*((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^3)

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maple [A]  time = 0.02, size = 363, normalized size = 1.13 \begin {gather*} -\frac {13 A \,b^{2} x^{\frac {7}{2}}}{16 \left (b \,x^{2}+a \right )^{2} a^{3}}+\frac {5 B b \,x^{\frac {7}{2}}}{16 \left (b \,x^{2}+a \right )^{2} a^{2}}-\frac {17 A b \,x^{\frac {3}{2}}}{16 \left (b \,x^{2}+a \right )^{2} a^{2}}+\frac {9 B \,x^{\frac {3}{2}}}{16 \left (b \,x^{2}+a \right )^{2} a}-\frac {45 \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^{3}}-\frac {45 \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^{3}}-\frac {45 \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^{3}}+\frac {5 \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}} a^{2} b}+\frac {5 \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}} a^{2} b}+\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 )}{128 \left (\frac {a}{b}\right )^{\frac {1}{4}} a^{2} b}-\frac {2 A}{a^{3} \sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-13/16/a^3/(b*x^2+a)^2*x^(7/2)*b^2*A+5/16/a^2/(b*x^2+a)^2*x^(7/2)*b*B-17/16/a^2/(b*x^2+a)^2*A*x^(3/2)*b+9/16/a
/(b*x^2+a)^2*B*x^(3/2)-45/128/a^3/(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)))-45/64/a^3/(a/b)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)+1)-45
/64/a^3/(a/b)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)-1)+5/128/a^2/b/(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)))+5/64/a^2/b/(a/b)^(1/4)*2^(
1/2)*B*arctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)+1)+5/64/a^2/b/(a/b)^(1/4)*2^(1/2)*B*arctan(2^(1/2)/(a/b)^(1/4)*x^(1/
2)-1)-2*A/a^3/x^(1/2)

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maxima [A]  time = 2.59, size = 255, normalized size = 0.79 \begin {gather*} \frac {5 \, {\left (B a b - 9 \, A b^{2}\right )} x^{4} - 32 \, A a^{2} + 9 \, {\left (B a^{2} - 9 \, A a b\right )} x^{2}}{16 \, {\left (a^{3} b^{2} x^{\frac {9}{2}} + 2 \, a^{4} b x^{\frac {5}{2}} + a^{5} \sqrt {x}\right )}} + \frac {5 \, {\left (B a - 9 \, 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^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(5*(B*a*b - 9*A*b^2)*x^4 - 32*A*a^2 + 9*(B*a^2 - 9*A*a*b)*x^2)/(a^3*b^2*x^(9/2) + 2*a^4*b*x^(5/2) + a^5*s
qrt(x)) + 5/128*(B*a - 9*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) + 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 = 0.19, size = 133, normalized size = 0.41 \begin {gather*} \frac {5\,\mathrm {atan}\left (\frac {b^{1/4}\,\sqrt {x}}{{\left (-a\right )}^{1/4}}\right )\,\left (9\,A\,b-B\,a\right )}{32\,{\left (-a\right )}^{13/4}\,b^{3/4}}-\frac {\frac {2\,A}{a}+\frac {9\,x^2\,\left (9\,A\,b-B\,a\right )}{16\,a^2}+\frac {5\,b\,x^4\,\left (9\,A\,b-B\,a\right )}{16\,a^3}}{a^2\,\sqrt {x}+b^2\,x^{9/2}+2\,a\,b\,x^{5/2}}-\frac {5\,\mathrm {atanh}\left (\frac {b^{1/4}\,\sqrt {x}}{{\left (-a\right )}^{1/4}}\right )\,\left (9\,A\,b-B\,a\right )}{32\,{\left (-a\right )}^{13/4}\,b^{3/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(5*atan((b^(1/4)*x^(1/2))/(-a)^(1/4))*(9*A*b - B*a))/(32*(-a)^(13/4)*b^(3/4)) - ((2*A)/a + (9*x^2*(9*A*b - B*a
))/(16*a^2) + (5*b*x^4*(9*A*b - B*a))/(16*a^3))/(a^2*x^(1/2) + b^2*x^(9/2) + 2*a*b*x^(5/2)) - (5*atanh((b^(1/4
)*x^(1/2))/(-a)^(1/4))*(9*A*b - B*a))/(32*(-a)^(13/4)*b^(3/4))

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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**(3/2)/(b*x**2+a)**3,x)

[Out]

Timed out

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