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

Optimal. Leaf size=343 \[ \frac {9 \sqrt [4]{b} (13 A b-5 a B) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{64 \sqrt {2} a^{17/4}}-\frac {9 \sqrt [4]{b} (13 A b-5 a B) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{64 \sqrt {2} a^{17/4}}-\frac {9 \sqrt [4]{b} (13 A b-5 a B) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{17/4}}+\frac {9 \sqrt [4]{b} (13 A b-5 a B) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{32 \sqrt {2} a^{17/4}}+\frac {9 (13 A b-5 a B)}{16 a^4 \sqrt {x}}-\frac {9 (13 A b-5 a B)}{80 a^3 b x^{5/2}}+\frac {13 A b-5 a B}{16 a^2 b x^{5/2} \left (a+b x^2\right )}+\frac {A b-a B}{4 a b x^{5/2} \left (a+b x^2\right )^2} \]

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Rubi [A]  time = 0.26, antiderivative size = 343, normalized size of antiderivative = 1.00, number of steps used = 14, 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 {13 A b-5 a B}{16 a^2 b x^{5/2} \left (a+b x^2\right )}-\frac {9 (13 A b-5 a B)}{80 a^3 b x^{5/2}}+\frac {9 (13 A b-5 a B)}{16 a^4 \sqrt {x}}+\frac {9 \sqrt [4]{b} (13 A b-5 a B) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{64 \sqrt {2} a^{17/4}}-\frac {9 \sqrt [4]{b} (13 A b-5 a B) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{64 \sqrt {2} a^{17/4}}-\frac {9 \sqrt [4]{b} (13 A b-5 a B) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{17/4}}+\frac {9 \sqrt [4]{b} (13 A b-5 a B) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{32 \sqrt {2} a^{17/4}}+\frac {A b-a B}{4 a b x^{5/2} \left (a+b x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-9*(13*A*b - 5*a*B))/(80*a^3*b*x^(5/2)) + (9*(13*A*b - 5*a*B))/(16*a^4*Sqrt[x]) + (A*b - a*B)/(4*a*b*x^(5/2)*
(a + b*x^2)^2) + (13*A*b - 5*a*B)/(16*a^2*b*x^(5/2)*(a + b*x^2)) - (9*b^(1/4)*(13*A*b - 5*a*B)*ArcTan[1 - (Sqr
t[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(32*Sqrt[2]*a^(17/4)) + (9*b^(1/4)*(13*A*b - 5*a*B)*ArcTan[1 + (Sqrt[2]*b^(1/4
)*Sqrt[x])/a^(1/4)])/(32*Sqrt[2]*a^(17/4)) + (9*b^(1/4)*(13*A*b - 5*a*B)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)
*Sqrt[x] + Sqrt[b]*x])/(64*Sqrt[2]*a^(17/4)) - (9*b^(1/4)*(13*A*b - 5*a*B)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/
4)*Sqrt[x] + Sqrt[b]*x])/(64*Sqrt[2]*a^(17/4))

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^{7/2} \left (a+b x^2\right )^3} \, dx &=\frac {A b-a B}{4 a b x^{5/2} \left (a+b x^2\right )^2}+\frac {\left (\frac {13 A b}{2}-\frac {5 a B}{2}\right ) \int \frac {1}{x^{7/2} \left (a+b x^2\right )^2} \, dx}{4 a b}\\ &=\frac {A b-a B}{4 a b x^{5/2} \left (a+b x^2\right )^2}+\frac {13 A b-5 a B}{16 a^2 b x^{5/2} \left (a+b x^2\right )}+\frac {(9 (13 A b-5 a B)) \int \frac {1}{x^{7/2} \left (a+b x^2\right )} \, dx}{32 a^2 b}\\ &=-\frac {9 (13 A b-5 a B)}{80 a^3 b x^{5/2}}+\frac {A b-a B}{4 a b x^{5/2} \left (a+b x^2\right )^2}+\frac {13 A b-5 a B}{16 a^2 b x^{5/2} \left (a+b x^2\right )}-\frac {(9 (13 A b-5 a B)) \int \frac {1}{x^{3/2} \left (a+b x^2\right )} \, dx}{32 a^3}\\ &=-\frac {9 (13 A b-5 a B)}{80 a^3 b x^{5/2}}+\frac {9 (13 A b-5 a B)}{16 a^4 \sqrt {x}}+\frac {A b-a B}{4 a b x^{5/2} \left (a+b x^2\right )^2}+\frac {13 A b-5 a B}{16 a^2 b x^{5/2} \left (a+b x^2\right )}+\frac {(9 b (13 A b-5 a B)) \int \frac {\sqrt {x}}{a+b x^2} \, dx}{32 a^4}\\ &=-\frac {9 (13 A b-5 a B)}{80 a^3 b x^{5/2}}+\frac {9 (13 A b-5 a B)}{16 a^4 \sqrt {x}}+\frac {A b-a B}{4 a b x^{5/2} \left (a+b x^2\right )^2}+\frac {13 A b-5 a B}{16 a^2 b x^{5/2} \left (a+b x^2\right )}+\frac {(9 b (13 A b-5 a B)) \operatorname {Subst}\left (\int \frac {x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{16 a^4}\\ &=-\frac {9 (13 A b-5 a B)}{80 a^3 b x^{5/2}}+\frac {9 (13 A b-5 a B)}{16 a^4 \sqrt {x}}+\frac {A b-a B}{4 a b x^{5/2} \left (a+b x^2\right )^2}+\frac {13 A b-5 a B}{16 a^2 b x^{5/2} \left (a+b x^2\right )}-\frac {\left (9 \sqrt {b} (13 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 )}{32 a^4}+\frac {\left (9 \sqrt {b} (13 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 )}{32 a^4}\\ &=-\frac {9 (13 A b-5 a B)}{80 a^3 b x^{5/2}}+\frac {9 (13 A b-5 a B)}{16 a^4 \sqrt {x}}+\frac {A b-a B}{4 a b x^{5/2} \left (a+b x^2\right )^2}+\frac {13 A b-5 a B}{16 a^2 b x^{5/2} \left (a+b x^2\right )}+\frac {(9 (13 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 )}{64 a^4}+\frac {(9 (13 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 )}{64 a^4}+\frac {\left (9 \sqrt [4]{b} (13 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 )}{64 \sqrt {2} a^{17/4}}+\frac {\left (9 \sqrt [4]{b} (13 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 )}{64 \sqrt {2} a^{17/4}}\\ &=-\frac {9 (13 A b-5 a B)}{80 a^3 b x^{5/2}}+\frac {9 (13 A b-5 a B)}{16 a^4 \sqrt {x}}+\frac {A b-a B}{4 a b x^{5/2} \left (a+b x^2\right )^2}+\frac {13 A b-5 a B}{16 a^2 b x^{5/2} \left (a+b x^2\right )}+\frac {9 \sqrt [4]{b} (13 A b-5 a B) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{64 \sqrt {2} a^{17/4}}-\frac {9 \sqrt [4]{b} (13 A b-5 a B) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{64 \sqrt {2} a^{17/4}}+\frac {\left (9 \sqrt [4]{b} (13 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 )}{32 \sqrt {2} a^{17/4}}-\frac {\left (9 \sqrt [4]{b} (13 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 )}{32 \sqrt {2} a^{17/4}}\\ &=-\frac {9 (13 A b-5 a B)}{80 a^3 b x^{5/2}}+\frac {9 (13 A b-5 a B)}{16 a^4 \sqrt {x}}+\frac {A b-a B}{4 a b x^{5/2} \left (a+b x^2\right )^2}+\frac {13 A b-5 a B}{16 a^2 b x^{5/2} \left (a+b x^2\right )}-\frac {9 \sqrt [4]{b} (13 A b-5 a B) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{17/4}}+\frac {9 \sqrt [4]{b} (13 A b-5 a B) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{17/4}}+\frac {9 \sqrt [4]{b} (13 A b-5 a B) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{64 \sqrt {2} a^{17/4}}-\frac {9 \sqrt [4]{b} (13 A b-5 a B) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{64 \sqrt {2} a^{17/4}}\\ \end {align*}

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Mathematica [C]  time = 0.47, size = 189, normalized size = 0.55 \begin {gather*} -\frac {2 b x^{3/2} (a B-2 A b) \, _2F_1\left (\frac {3}{4},2;\frac {7}{4};-\frac {b x^2}{a}\right )}{3 a^5}+\frac {2 b 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^5}+\frac {6 A b-2 a B}{a^4 \sqrt {x}}-\frac {2 A}{5 a^3 x^{5/2}}+\frac {\sqrt [4]{b} (3 A b-a B) \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{-a}}\right )}{(-a)^{17/4}}+\frac {\sqrt [4]{b} (a B-3 A 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^(7/2)*(a + b*x^2)^3),x]

[Out]

(-2*A)/(5*a^3*x^(5/2)) + (6*A*b - 2*a*B)/(a^4*Sqrt[x]) + (b^(1/4)*(3*A*b - a*B)*ArcTan[(b^(1/4)*Sqrt[x])/(-a)^
(1/4)])/(-a)^(17/4) + (b^(1/4)*(-3*A*b + a*B)*ArcTanh[(b^(1/4)*Sqrt[x])/(-a)^(1/4)])/(-a)^(17/4) - (2*b*(-2*A*
b + a*B)*x^(3/2)*Hypergeometric2F1[3/4, 2, 7/4, -((b*x^2)/a)])/(3*a^5) + (2*b*(A*b - a*B)*x^(3/2)*Hypergeometr
ic2F1[3/4, 3, 7/4, -((b*x^2)/a)])/(3*a^5)

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IntegrateAlgebraic [A]  time = 0.64, size = 224, normalized size = 0.65 \begin {gather*} \frac {9 \left (5 a \sqrt [4]{b} B-13 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 )}{32 \sqrt {2} a^{17/4}}+\frac {9 \left (5 a \sqrt [4]{b} B-13 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 )}{32 \sqrt {2} a^{17/4}}+\frac {-32 a^3 A-160 a^3 B x^2+416 a^2 A b x^2-405 a^2 b B x^4+1053 a A b^2 x^4-225 a b^2 B x^6+585 A b^3 x^6}{80 a^4 x^{5/2} \left (a+b x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-32*a^3*A + 416*a^2*A*b*x^2 - 160*a^3*B*x^2 + 1053*a*A*b^2*x^4 - 405*a^2*b*B*x^4 + 585*A*b^3*x^6 - 225*a*b^2*
B*x^6)/(80*a^4*x^(5/2)*(a + b*x^2)^2) + (9*(-13*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])])/(32*Sqrt[2]*a^(17/4)) + (9*(-13*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)])/(32*Sqrt[2]*a^(17/4))

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fricas [B]  time = 1.47, size = 1043, normalized size = 3.04 \begin {gather*} -\frac {180 \, {\left (a^{4} b^{2} x^{7} + 2 \, a^{5} b x^{5} + a^{6} x^{3}\right )} \left (-\frac {625 \, B^{4} a^{4} b - 6500 \, A B^{3} a^{3} b^{2} + 25350 \, A^{2} B^{2} a^{2} b^{3} - 43940 \, A^{3} B a b^{4} + 28561 \, A^{4} b^{5}}{a^{17}}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {{\left (15625 \, B^{6} a^{6} b^{2} - 243750 \, A B^{5} a^{5} b^{3} + 1584375 \, A^{2} B^{4} a^{4} b^{4} - 5492500 \, A^{3} B^{3} a^{3} b^{5} + 10710375 \, A^{4} B^{2} a^{2} b^{6} - 11138790 \, A^{5} B a b^{7} + 4826809 \, A^{6} b^{8}\right )} x - {\left (625 \, B^{4} a^{13} b - 6500 \, A B^{3} a^{12} b^{2} + 25350 \, A^{2} B^{2} a^{11} b^{3} - 43940 \, A^{3} B a^{10} b^{4} + 28561 \, A^{4} a^{9} b^{5}\right )} \sqrt {-\frac {625 \, B^{4} a^{4} b - 6500 \, A B^{3} a^{3} b^{2} + 25350 \, A^{2} B^{2} a^{2} b^{3} - 43940 \, A^{3} B a b^{4} + 28561 \, A^{4} b^{5}}{a^{17}}}} a^{4} \left (-\frac {625 \, B^{4} a^{4} b - 6500 \, A B^{3} a^{3} b^{2} + 25350 \, A^{2} B^{2} a^{2} b^{3} - 43940 \, A^{3} B a b^{4} + 28561 \, A^{4} b^{5}}{a^{17}}\right )^{\frac {1}{4}} + {\left (125 \, B^{3} a^{7} b - 975 \, A B^{2} a^{6} b^{2} + 2535 \, A^{2} B a^{5} b^{3} - 2197 \, A^{3} a^{4} b^{4}\right )} \sqrt {x} \left (-\frac {625 \, B^{4} a^{4} b - 6500 \, A B^{3} a^{3} b^{2} + 25350 \, A^{2} B^{2} a^{2} b^{3} - 43940 \, A^{3} B a b^{4} + 28561 \, A^{4} b^{5}}{a^{17}}\right )^{\frac {1}{4}}}{625 \, B^{4} a^{4} b - 6500 \, A B^{3} a^{3} b^{2} + 25350 \, A^{2} B^{2} a^{2} b^{3} - 43940 \, A^{3} B a b^{4} + 28561 \, A^{4} b^{5}}\right ) - 45 \, {\left (a^{4} b^{2} x^{7} + 2 \, a^{5} b x^{5} + a^{6} x^{3}\right )} \left (-\frac {625 \, B^{4} a^{4} b - 6500 \, A B^{3} a^{3} b^{2} + 25350 \, A^{2} B^{2} a^{2} b^{3} - 43940 \, A^{3} B a b^{4} + 28561 \, A^{4} b^{5}}{a^{17}}\right )^{\frac {1}{4}} \log \left (729 \, a^{13} \left (-\frac {625 \, B^{4} a^{4} b - 6500 \, A B^{3} a^{3} b^{2} + 25350 \, A^{2} B^{2} a^{2} b^{3} - 43940 \, A^{3} B a b^{4} + 28561 \, A^{4} b^{5}}{a^{17}}\right )^{\frac {3}{4}} - 729 \, {\left (125 \, B^{3} a^{3} b - 975 \, A B^{2} a^{2} b^{2} + 2535 \, A^{2} B a b^{3} - 2197 \, A^{3} b^{4}\right )} \sqrt {x}\right ) + 45 \, {\left (a^{4} b^{2} x^{7} + 2 \, a^{5} b x^{5} + a^{6} x^{3}\right )} \left (-\frac {625 \, B^{4} a^{4} b - 6500 \, A B^{3} a^{3} b^{2} + 25350 \, A^{2} B^{2} a^{2} b^{3} - 43940 \, A^{3} B a b^{4} + 28561 \, A^{4} b^{5}}{a^{17}}\right )^{\frac {1}{4}} \log \left (-729 \, a^{13} \left (-\frac {625 \, B^{4} a^{4} b - 6500 \, A B^{3} a^{3} b^{2} + 25350 \, A^{2} B^{2} a^{2} b^{3} - 43940 \, A^{3} B a b^{4} + 28561 \, A^{4} b^{5}}{a^{17}}\right )^{\frac {3}{4}} - 729 \, {\left (125 \, B^{3} a^{3} b - 975 \, A B^{2} a^{2} b^{2} + 2535 \, A^{2} B a b^{3} - 2197 \, A^{3} b^{4}\right )} \sqrt {x}\right ) + 4 \, {\left (45 \, {\left (5 \, B a b^{2} - 13 \, A b^{3}\right )} x^{6} + 81 \, {\left (5 \, B a^{2} b - 13 \, A a b^{2}\right )} x^{4} + 32 \, A a^{3} + 32 \, {\left (5 \, B a^{3} - 13 \, A a^{2} b\right )} x^{2}\right )} \sqrt {x}}{320 \, {\left (a^{4} b^{2} x^{7} + 2 \, a^{5} b x^{5} + a^{6} 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)^3,x, algorithm="fricas")

[Out]

-1/320*(180*(a^4*b^2*x^7 + 2*a^5*b*x^5 + a^6*x^3)*(-(625*B^4*a^4*b - 6500*A*B^3*a^3*b^2 + 25350*A^2*B^2*a^2*b^
3 - 43940*A^3*B*a*b^4 + 28561*A^4*b^5)/a^17)^(1/4)*arctan((sqrt((15625*B^6*a^6*b^2 - 243750*A*B^5*a^5*b^3 + 15
84375*A^2*B^4*a^4*b^4 - 5492500*A^3*B^3*a^3*b^5 + 10710375*A^4*B^2*a^2*b^6 - 11138790*A^5*B*a*b^7 + 4826809*A^
6*b^8)*x - (625*B^4*a^13*b - 6500*A*B^3*a^12*b^2 + 25350*A^2*B^2*a^11*b^3 - 43940*A^3*B*a^10*b^4 + 28561*A^4*a
^9*b^5)*sqrt(-(625*B^4*a^4*b - 6500*A*B^3*a^3*b^2 + 25350*A^2*B^2*a^2*b^3 - 43940*A^3*B*a*b^4 + 28561*A^4*b^5)
/a^17))*a^4*(-(625*B^4*a^4*b - 6500*A*B^3*a^3*b^2 + 25350*A^2*B^2*a^2*b^3 - 43940*A^3*B*a*b^4 + 28561*A^4*b^5)
/a^17)^(1/4) + (125*B^3*a^7*b - 975*A*B^2*a^6*b^2 + 2535*A^2*B*a^5*b^3 - 2197*A^3*a^4*b^4)*sqrt(x)*(-(625*B^4*
a^4*b - 6500*A*B^3*a^3*b^2 + 25350*A^2*B^2*a^2*b^3 - 43940*A^3*B*a*b^4 + 28561*A^4*b^5)/a^17)^(1/4))/(625*B^4*
a^4*b - 6500*A*B^3*a^3*b^2 + 25350*A^2*B^2*a^2*b^3 - 43940*A^3*B*a*b^4 + 28561*A^4*b^5)) - 45*(a^4*b^2*x^7 + 2
*a^5*b*x^5 + a^6*x^3)*(-(625*B^4*a^4*b - 6500*A*B^3*a^3*b^2 + 25350*A^2*B^2*a^2*b^3 - 43940*A^3*B*a*b^4 + 2856
1*A^4*b^5)/a^17)^(1/4)*log(729*a^13*(-(625*B^4*a^4*b - 6500*A*B^3*a^3*b^2 + 25350*A^2*B^2*a^2*b^3 - 43940*A^3*
B*a*b^4 + 28561*A^4*b^5)/a^17)^(3/4) - 729*(125*B^3*a^3*b - 975*A*B^2*a^2*b^2 + 2535*A^2*B*a*b^3 - 2197*A^3*b^
4)*sqrt(x)) + 45*(a^4*b^2*x^7 + 2*a^5*b*x^5 + a^6*x^3)*(-(625*B^4*a^4*b - 6500*A*B^3*a^3*b^2 + 25350*A^2*B^2*a
^2*b^3 - 43940*A^3*B*a*b^4 + 28561*A^4*b^5)/a^17)^(1/4)*log(-729*a^13*(-(625*B^4*a^4*b - 6500*A*B^3*a^3*b^2 +
25350*A^2*B^2*a^2*b^3 - 43940*A^3*B*a*b^4 + 28561*A^4*b^5)/a^17)^(3/4) - 729*(125*B^3*a^3*b - 975*A*B^2*a^2*b^
2 + 2535*A^2*B*a*b^3 - 2197*A^3*b^4)*sqrt(x)) + 4*(45*(5*B*a*b^2 - 13*A*b^3)*x^6 + 81*(5*B*a^2*b - 13*A*a*b^2)
*x^4 + 32*A*a^3 + 32*(5*B*a^3 - 13*A*a^2*b)*x^2)*sqrt(x))/(a^4*b^2*x^7 + 2*a^5*b*x^5 + a^6*x^3)

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giac [A]  time = 0.54, size = 326, normalized size = 0.95 \begin {gather*} -\frac {9 \, \sqrt {2} {\left (5 \, \left (a b^{3}\right )^{\frac {3}{4}} B a - 13 \, \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^{5} b^{2}} - \frac {9 \, \sqrt {2} {\left (5 \, \left (a b^{3}\right )^{\frac {3}{4}} B a - 13 \, \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^{5} b^{2}} + \frac {9 \, \sqrt {2} {\left (5 \, \left (a b^{3}\right )^{\frac {3}{4}} B a - 13 \, \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^{5} b^{2}} - \frac {9 \, \sqrt {2} {\left (5 \, \left (a b^{3}\right )^{\frac {3}{4}} B a - 13 \, \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^{5} b^{2}} - \frac {13 \, B a b^{2} x^{\frac {7}{2}} - 21 \, A b^{3} x^{\frac {7}{2}} + 17 \, B a^{2} b x^{\frac {3}{2}} - 25 \, A a b^{2} x^{\frac {3}{2}}}{16 \, {\left (b x^{2} + a\right )}^{2} a^{4}} - \frac {2 \, {\left (5 \, B a x^{2} - 15 \, A b x^{2} + A a\right )}}{5 \, a^{4} 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)^3,x, algorithm="giac")

[Out]

-9/64*sqrt(2)*(5*(a*b^3)^(3/4)*B*a - 13*(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^5*b^2) - 9/64*sqrt(2)*(5*(a*b^3)^(3/4)*B*a - 13*(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^5*b^2) + 9/128*sqrt(2)*(5*(a*b^3)^(3/4)*B*a - 13*(a*b^3)^(3/4)*A*
b)*log(sqrt(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))/(a^5*b^2) - 9/128*sqrt(2)*(5*(a*b^3)^(3/4)*B*a - 13*(a*b^3
)^(3/4)*A*b)*log(-sqrt(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))/(a^5*b^2) - 1/16*(13*B*a*b^2*x^(7/2) - 21*A*b^3
*x^(7/2) + 17*B*a^2*b*x^(3/2) - 25*A*a*b^2*x^(3/2))/((b*x^2 + a)^2*a^4) - 2/5*(5*B*a*x^2 - 15*A*b*x^2 + A*a)/(
a^4*x^(5/2))

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maple [A]  time = 0.03, size = 381, normalized size = 1.11 \begin {gather*} \frac {21 A \,b^{3} x^{\frac {7}{2}}}{16 \left (b \,x^{2}+a \right )^{2} a^{4}}-\frac {13 B \,b^{2} x^{\frac {7}{2}}}{16 \left (b \,x^{2}+a \right )^{2} a^{3}}+\frac {25 A \,b^{2} x^{\frac {3}{2}}}{16 \left (b \,x^{2}+a \right )^{2} a^{3}}-\frac {17 B b \,x^{\frac {3}{2}}}{16 \left (b \,x^{2}+a \right )^{2} a^{2}}+\frac {117 \sqrt {2}\, A 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^{4}}+\frac {117 \sqrt {2}\, A 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^{4}}+\frac {117 \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 )}{128 \left (\frac {a}{b}\right )^{\frac {1}{4}} a^{4}}-\frac {45 \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^{3}}-\frac {45 \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^{3}}-\frac {45 \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^{3}}+\frac {6 A b}{a^{4} \sqrt {x}}-\frac {2 B}{a^{3} \sqrt {x}}-\frac {2 A}{5 a^{3} 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)^3,x)

[Out]

21/16/a^4*b^3/(b*x^2+a)^2*x^(7/2)*A-13/16/a^3*b^2/(b*x^2+a)^2*x^(7/2)*B+25/16/a^3*b^2/(b*x^2+a)^2*A*x^(3/2)-17
/16/a^2*b/(b*x^2+a)^2*B*x^(3/2)+117/128/a^4*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)))+117/64/a^4*b/(a/b)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(a/b)^(1/4
)*x^(1/2)+1)+117/64/a^4*b/(a/b)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)-1)-45/128/a^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)))-45/64/a^3
/(a/b)^(1/4)*2^(1/2)*B*arctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)+1)-45/64/a^3/(a/b)^(1/4)*2^(1/2)*B*arctan(2^(1/2)/(a
/b)^(1/4)*x^(1/2)-1)-2/5*A/a^3/x^(5/2)+6/a^4/x^(1/2)*A*b-2/a^3/x^(1/2)*B

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maxima [A]  time = 2.46, size = 285, normalized size = 0.83 \begin {gather*} -\frac {45 \, {\left (5 \, B a b^{2} - 13 \, A b^{3}\right )} x^{6} + 81 \, {\left (5 \, B a^{2} b - 13 \, A a b^{2}\right )} x^{4} + 32 \, A a^{3} + 32 \, {\left (5 \, B a^{3} - 13 \, A a^{2} b\right )} x^{2}}{80 \, {\left (a^{4} b^{2} x^{\frac {13}{2}} + 2 \, a^{5} b x^{\frac {9}{2}} + a^{6} x^{\frac {5}{2}}\right )}} - \frac {9 \, {\left (5 \, B a b - 13 \, 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 )}}{128 \, a^{4}} \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)^3,x, algorithm="maxima")

[Out]

-1/80*(45*(5*B*a*b^2 - 13*A*b^3)*x^6 + 81*(5*B*a^2*b - 13*A*a*b^2)*x^4 + 32*A*a^3 + 32*(5*B*a^3 - 13*A*a^2*b)*
x^2)/(a^4*b^2*x^(13/2) + 2*a^5*b*x^(9/2) + a^6*x^(5/2)) - 9/128*(5*B*a*b - 13*A*b^2)*(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)))/a^4

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mupad [B]  time = 0.35, size = 152, normalized size = 0.44 \begin {gather*} \frac {\frac {2\,x^2\,\left (13\,A\,b-5\,B\,a\right )}{5\,a^2}-\frac {2\,A}{5\,a}+\frac {9\,b^2\,x^6\,\left (13\,A\,b-5\,B\,a\right )}{16\,a^4}+\frac {81\,b\,x^4\,\left (13\,A\,b-5\,B\,a\right )}{80\,a^3}}{a^2\,x^{5/2}+b^2\,x^{13/2}+2\,a\,b\,x^{9/2}}+\frac {9\,{\left (-b\right )}^{1/4}\,\mathrm {atan}\left (\frac {{\left (-b\right )}^{1/4}\,\sqrt {x}}{a^{1/4}}\right )\,\left (13\,A\,b-5\,B\,a\right )}{32\,a^{17/4}}-\frac {9\,{\left (-b\right )}^{1/4}\,\mathrm {atanh}\left (\frac {{\left (-b\right )}^{1/4}\,\sqrt {x}}{a^{1/4}}\right )\,\left (13\,A\,b-5\,B\,a\right )}{32\,a^{17/4}} \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)^3),x)

[Out]

((2*x^2*(13*A*b - 5*B*a))/(5*a^2) - (2*A)/(5*a) + (9*b^2*x^6*(13*A*b - 5*B*a))/(16*a^4) + (81*b*x^4*(13*A*b -
5*B*a))/(80*a^3))/(a^2*x^(5/2) + b^2*x^(13/2) + 2*a*b*x^(9/2)) + (9*(-b)^(1/4)*atan(((-b)^(1/4)*x^(1/2))/a^(1/
4))*(13*A*b - 5*B*a))/(32*a^(17/4)) - (9*(-b)^(1/4)*atanh(((-b)^(1/4)*x^(1/2))/a^(1/4))*(13*A*b - 5*B*a))/(32*
a^(17/4))

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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)**3,x)

[Out]

Timed out

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