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

Optimal. Leaf size=255 \[ \frac {\sqrt [4]{b} (A b-a B) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{2 \sqrt {2} a^{9/4}}-\frac {\sqrt [4]{b} (A b-a B) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{2 \sqrt {2} a^{9/4}}-\frac {\sqrt [4]{b} (A b-a B) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt {2} a^{9/4}}+\frac {\sqrt [4]{b} (A b-a B) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{\sqrt {2} a^{9/4}}+\frac {2 (A b-a B)}{a^2 \sqrt {x}}-\frac {2 A}{5 a x^{5/2}} \]

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Rubi [A]  time = 0.21, antiderivative size = 255, 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 = {453, 325, 329, 297, 1162, 617, 204, 1165, 628} \begin {gather*} \frac {2 (A b-a B)}{a^2 \sqrt {x}}+\frac {\sqrt [4]{b} (A b-a B) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{2 \sqrt {2} a^{9/4}}-\frac {\sqrt [4]{b} (A b-a B) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{2 \sqrt {2} a^{9/4}}-\frac {\sqrt [4]{b} (A b-a B) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt {2} a^{9/4}}+\frac {\sqrt [4]{b} (A b-a B) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{\sqrt {2} a^{9/4}}-\frac {2 A}{5 a x^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*A)/(5*a*x^(5/2)) + (2*(A*b - a*B))/(a^2*Sqrt[x]) - (b^(1/4)*(A*b - a*B)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x
])/a^(1/4)])/(Sqrt[2]*a^(9/4)) + (b^(1/4)*(A*b - a*B)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(Sqrt[2]*
a^(9/4)) + (b^(1/4)*(A*b - a*B)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(2*Sqrt[2]*a^(9/4)
) - (b^(1/4)*(A*b - a*B)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(2*Sqrt[2]*a^(9/4))

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 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[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 )} \, dx &=-\frac {2 A}{5 a x^{5/2}}-\frac {\left (2 \left (\frac {5 A b}{2}-\frac {5 a B}{2}\right )\right ) \int \frac {1}{x^{3/2} \left (a+b x^2\right )} \, dx}{5 a}\\ &=-\frac {2 A}{5 a x^{5/2}}+\frac {2 (A b-a B)}{a^2 \sqrt {x}}+\frac {(b (A b-a B)) \int \frac {\sqrt {x}}{a+b x^2} \, dx}{a^2}\\ &=-\frac {2 A}{5 a x^{5/2}}+\frac {2 (A b-a B)}{a^2 \sqrt {x}}+\frac {(2 b (A b-a B)) \operatorname {Subst}\left (\int \frac {x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{a^2}\\ &=-\frac {2 A}{5 a x^{5/2}}+\frac {2 (A b-a B)}{a^2 \sqrt {x}}-\frac {\left (\sqrt {b} (A b-a B)\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a}-\sqrt {b} x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{a^2}+\frac {\left (\sqrt {b} (A b-a B)\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a}+\sqrt {b} x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{a^2}\\ &=-\frac {2 A}{5 a x^{5/2}}+\frac {2 (A b-a B)}{a^2 \sqrt {x}}+\frac {(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 )}{2 a^2}+\frac {(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 )}{2 a^2}+\frac {\left (\sqrt [4]{b} (A b-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 )}{2 \sqrt {2} a^{9/4}}+\frac {\left (\sqrt [4]{b} (A b-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 )}{2 \sqrt {2} a^{9/4}}\\ &=-\frac {2 A}{5 a x^{5/2}}+\frac {2 (A b-a B)}{a^2 \sqrt {x}}+\frac {\sqrt [4]{b} (A b-a B) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{2 \sqrt {2} a^{9/4}}-\frac {\sqrt [4]{b} (A b-a B) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{2 \sqrt {2} a^{9/4}}+\frac {\left (\sqrt [4]{b} (A b-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 )}{\sqrt {2} a^{9/4}}-\frac {\left (\sqrt [4]{b} (A b-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 )}{\sqrt {2} a^{9/4}}\\ &=-\frac {2 A}{5 a x^{5/2}}+\frac {2 (A b-a B)}{a^2 \sqrt {x}}-\frac {\sqrt [4]{b} (A b-a B) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt {2} a^{9/4}}+\frac {\sqrt [4]{b} (A b-a B) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt {2} a^{9/4}}+\frac {\sqrt [4]{b} (A b-a B) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{2 \sqrt {2} a^{9/4}}-\frac {\sqrt [4]{b} (A b-a B) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{2 \sqrt {2} a^{9/4}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 46, normalized size = 0.18 \begin {gather*} -\frac {2 \left (5 x^2 (a B-A b) \, _2F_1\left (-\frac {1}{4},1;\frac {3}{4};-\frac {b x^2}{a}\right )+a A\right )}{5 a^2 x^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.21, size = 160, normalized size = 0.63 \begin {gather*} \frac {\left (a \sqrt [4]{b} B-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 )}{\sqrt {2} a^{9/4}}+\frac {\left (a \sqrt [4]{b} B-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 )}{\sqrt {2} a^{9/4}}-\frac {2 \left (a A+5 a B x^2-5 A b x^2\right )}{5 a^2 x^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(a*A - 5*A*b*x^2 + 5*a*B*x^2))/(5*a^2*x^(5/2)) + ((-(A*b^(5/4)) + a*b^(1/4)*B)*ArcTan[(Sqrt[a] - Sqrt[b]*x
)/(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x])])/(Sqrt[2]*a^(9/4)) + ((-(A*b^(5/4)) + a*b^(1/4)*B)*ArcTanh[(Sqrt[2]*a^(1/
4)*b^(1/4)*Sqrt[x])/(Sqrt[a] + Sqrt[b]*x)])/(Sqrt[2]*a^(9/4))

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fricas [B]  time = 1.04, size = 883, normalized size = 3.46 \begin {gather*} -\frac {20 \, a^{2} x^{3} \left (-\frac {B^{4} a^{4} b - 4 \, A B^{3} a^{3} b^{2} + 6 \, A^{2} B^{2} a^{2} b^{3} - 4 \, A^{3} B a b^{4} + A^{4} b^{5}}{a^{9}}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {{\left (B^{6} a^{6} b^{2} - 6 \, A B^{5} a^{5} b^{3} + 15 \, A^{2} B^{4} a^{4} b^{4} - 20 \, A^{3} B^{3} a^{3} b^{5} + 15 \, A^{4} B^{2} a^{2} b^{6} - 6 \, A^{5} B a b^{7} + A^{6} b^{8}\right )} x - {\left (B^{4} a^{9} b - 4 \, A B^{3} a^{8} b^{2} + 6 \, A^{2} B^{2} a^{7} b^{3} - 4 \, A^{3} B a^{6} b^{4} + A^{4} a^{5} b^{5}\right )} \sqrt {-\frac {B^{4} a^{4} b - 4 \, A B^{3} a^{3} b^{2} + 6 \, A^{2} B^{2} a^{2} b^{3} - 4 \, A^{3} B a b^{4} + A^{4} b^{5}}{a^{9}}}} a^{2} \left (-\frac {B^{4} a^{4} b - 4 \, A B^{3} a^{3} b^{2} + 6 \, A^{2} B^{2} a^{2} b^{3} - 4 \, A^{3} B a b^{4} + A^{4} b^{5}}{a^{9}}\right )^{\frac {1}{4}} + {\left (B^{3} a^{5} b - 3 \, A B^{2} a^{4} b^{2} + 3 \, A^{2} B a^{3} b^{3} - A^{3} a^{2} b^{4}\right )} \sqrt {x} \left (-\frac {B^{4} a^{4} b - 4 \, A B^{3} a^{3} b^{2} + 6 \, A^{2} B^{2} a^{2} b^{3} - 4 \, A^{3} B a b^{4} + A^{4} b^{5}}{a^{9}}\right )^{\frac {1}{4}}}{B^{4} a^{4} b - 4 \, A B^{3} a^{3} b^{2} + 6 \, A^{2} B^{2} a^{2} b^{3} - 4 \, A^{3} B a b^{4} + A^{4} b^{5}}\right ) - 5 \, a^{2} x^{3} \left (-\frac {B^{4} a^{4} b - 4 \, A B^{3} a^{3} b^{2} + 6 \, A^{2} B^{2} a^{2} b^{3} - 4 \, A^{3} B a b^{4} + A^{4} b^{5}}{a^{9}}\right )^{\frac {1}{4}} \log \left (a^{7} \left (-\frac {B^{4} a^{4} b - 4 \, A B^{3} a^{3} b^{2} + 6 \, A^{2} B^{2} a^{2} b^{3} - 4 \, A^{3} B a b^{4} + A^{4} b^{5}}{a^{9}}\right )^{\frac {3}{4}} - {\left (B^{3} a^{3} b - 3 \, A B^{2} a^{2} b^{2} + 3 \, A^{2} B a b^{3} - A^{3} b^{4}\right )} \sqrt {x}\right ) + 5 \, a^{2} x^{3} \left (-\frac {B^{4} a^{4} b - 4 \, A B^{3} a^{3} b^{2} + 6 \, A^{2} B^{2} a^{2} b^{3} - 4 \, A^{3} B a b^{4} + A^{4} b^{5}}{a^{9}}\right )^{\frac {1}{4}} \log \left (-a^{7} \left (-\frac {B^{4} a^{4} b - 4 \, A B^{3} a^{3} b^{2} + 6 \, A^{2} B^{2} a^{2} b^{3} - 4 \, A^{3} B a b^{4} + A^{4} b^{5}}{a^{9}}\right )^{\frac {3}{4}} - {\left (B^{3} a^{3} b - 3 \, A B^{2} a^{2} b^{2} + 3 \, A^{2} B a b^{3} - A^{3} b^{4}\right )} \sqrt {x}\right ) + 4 \, {\left (5 \, {\left (B a - A b\right )} x^{2} + A a\right )} \sqrt {x}}{10 \, a^{2} x^{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),x, algorithm="fricas")

[Out]

-1/10*(20*a^2*x^3*(-(B^4*a^4*b - 4*A*B^3*a^3*b^2 + 6*A^2*B^2*a^2*b^3 - 4*A^3*B*a*b^4 + A^4*b^5)/a^9)^(1/4)*arc
tan((sqrt((B^6*a^6*b^2 - 6*A*B^5*a^5*b^3 + 15*A^2*B^4*a^4*b^4 - 20*A^3*B^3*a^3*b^5 + 15*A^4*B^2*a^2*b^6 - 6*A^
5*B*a*b^7 + A^6*b^8)*x - (B^4*a^9*b - 4*A*B^3*a^8*b^2 + 6*A^2*B^2*a^7*b^3 - 4*A^3*B*a^6*b^4 + A^4*a^5*b^5)*sqr
t(-(B^4*a^4*b - 4*A*B^3*a^3*b^2 + 6*A^2*B^2*a^2*b^3 - 4*A^3*B*a*b^4 + A^4*b^5)/a^9))*a^2*(-(B^4*a^4*b - 4*A*B^
3*a^3*b^2 + 6*A^2*B^2*a^2*b^3 - 4*A^3*B*a*b^4 + A^4*b^5)/a^9)^(1/4) + (B^3*a^5*b - 3*A*B^2*a^4*b^2 + 3*A^2*B*a
^3*b^3 - A^3*a^2*b^4)*sqrt(x)*(-(B^4*a^4*b - 4*A*B^3*a^3*b^2 + 6*A^2*B^2*a^2*b^3 - 4*A^3*B*a*b^4 + A^4*b^5)/a^
9)^(1/4))/(B^4*a^4*b - 4*A*B^3*a^3*b^2 + 6*A^2*B^2*a^2*b^3 - 4*A^3*B*a*b^4 + A^4*b^5)) - 5*a^2*x^3*(-(B^4*a^4*
b - 4*A*B^3*a^3*b^2 + 6*A^2*B^2*a^2*b^3 - 4*A^3*B*a*b^4 + A^4*b^5)/a^9)^(1/4)*log(a^7*(-(B^4*a^4*b - 4*A*B^3*a
^3*b^2 + 6*A^2*B^2*a^2*b^3 - 4*A^3*B*a*b^4 + A^4*b^5)/a^9)^(3/4) - (B^3*a^3*b - 3*A*B^2*a^2*b^2 + 3*A^2*B*a*b^
3 - A^3*b^4)*sqrt(x)) + 5*a^2*x^3*(-(B^4*a^4*b - 4*A*B^3*a^3*b^2 + 6*A^2*B^2*a^2*b^3 - 4*A^3*B*a*b^4 + A^4*b^5
)/a^9)^(1/4)*log(-a^7*(-(B^4*a^4*b - 4*A*B^3*a^3*b^2 + 6*A^2*B^2*a^2*b^3 - 4*A^3*B*a*b^4 + A^4*b^5)/a^9)^(3/4)
- (B^3*a^3*b - 3*A*B^2*a^2*b^2 + 3*A^2*B*a*b^3 - A^3*b^4)*sqrt(x)) + 4*(5*(B*a - A*b)*x^2 + A*a)*sqrt(x))/(a^
2*x^3)

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giac [A]  time = 0.35, size = 268, normalized size = 1.05 \begin {gather*} -\frac {\sqrt {2} {\left (\left (a b^{3}\right )^{\frac {3}{4}} B a - \left (a b^{3}\right )^{\frac {3}{4}} A b\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{2 \, a^{3} b^{2}} - \frac {\sqrt {2} {\left (\left (a b^{3}\right )^{\frac {3}{4}} B a - \left (a b^{3}\right )^{\frac {3}{4}} A b\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{2 \, a^{3} b^{2}} + \frac {\sqrt {2} {\left (\left (a b^{3}\right )^{\frac {3}{4}} B a - \left (a b^{3}\right )^{\frac {3}{4}} A b\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {a}{b}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{b}}\right )}{4 \, a^{3} b^{2}} - \frac {\sqrt {2} {\left (\left (a b^{3}\right )^{\frac {3}{4}} B a - \left (a b^{3}\right )^{\frac {3}{4}} A b\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {a}{b}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{b}}\right )}{4 \, a^{3} b^{2}} - \frac {2 \, {\left (5 \, B a x^{2} - 5 \, A b x^{2} + A a\right )}}{5 \, a^{2} 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),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.02, size = 299, normalized size = 1.17 \begin {gather*} \frac {\sqrt {2}\, A b \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )}{2 \left (\frac {a}{b}\right )^{\frac {1}{4}} a^{2}}+\frac {\sqrt {2}\, A b \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )}{2 \left (\frac {a}{b}\right )^{\frac {1}{4}} a^{2}}+\frac {\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 )}{4 \left (\frac {a}{b}\right )^{\frac {1}{4}} a^{2}}-\frac {\sqrt {2}\, B \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )}{2 \left (\frac {a}{b}\right )^{\frac {1}{4}} a}-\frac {\sqrt {2}\, B \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )}{2 \left (\frac {a}{b}\right )^{\frac {1}{4}} a}-\frac {\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 )}{4 \left (\frac {a}{b}\right )^{\frac {1}{4}} a}+\frac {2 A b}{a^{2} \sqrt {x}}-\frac {2 B}{a \sqrt {x}}-\frac {2 A}{5 a \,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),x)

[Out]

1/2/a^2/(a/b)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)+1)*b+1/2/a^2/(a/b)^(1/4)*2^(1/2)*A*arctan(2^(
1/2)/(a/b)^(1/4)*x^(1/2)-1)*b+1/4/a^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)))*b-1/2/a/(a/b)^(1/4)*2^(1/2)*B*arctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)+1)-
1/2/a/(a/b)^(1/4)*2^(1/2)*B*arctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)-1)-1/4/a/(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)))-2/5*A/a/x^(5/2)+2/a^2/x^(1/2)*A*b-
2/a/x^(1/2)*B

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maxima [A]  time = 2.38, size = 213, normalized size = 0.84 \begin {gather*} -\frac {{\left (B a b - 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 )}}{4 \, a^{2}} - \frac {2 \, {\left (5 \, {\left (B a - A b\right )} x^{2} + A a\right )}}{5 \, a^{2} 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),x, algorithm="maxima")

[Out]

-1/4*(B*a*b - 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)*
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^2 - 2/5*(5*(B*a - A*b)*x^2 + A*a)/(a^2*x^(5/2))

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mupad [B]  time = 0.31, size = 90, normalized size = 0.35 \begin {gather*} \frac {{\left (-b\right )}^{1/4}\,\mathrm {atan}\left (\frac {{\left (-b\right )}^{1/4}\,\sqrt {x}}{a^{1/4}}\right )\,\left (A\,b-B\,a\right )}{a^{9/4}}-\frac {\frac {2\,A}{5\,a}-\frac {2\,x^2\,\left (A\,b-B\,a\right )}{a^2}}{x^{5/2}}-\frac {{\left (-b\right )}^{1/4}\,\mathrm {atanh}\left (\frac {{\left (-b\right )}^{1/4}\,\sqrt {x}}{a^{1/4}}\right )\,\left (A\,b-B\,a\right )}{a^{9/4}} \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)),x)

[Out]

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

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sympy [A]  time = 124.89, size = 366, normalized size = 1.44 \begin {gather*} A \left (\begin {cases} \frac {\tilde {\infty }}{x^{\frac {9}{2}}} & \text {for}\: a = 0 \wedge b = 0 \\- \frac {2}{9 b x^{\frac {9}{2}}} & \text {for}\: a = 0 \\- \frac {2}{5 a x^{\frac {5}{2}}} & \text {for}\: b = 0 \\- \frac {2}{5 a x^{\frac {5}{2}}} + \frac {2 b}{a^{2} \sqrt {x}} - \frac {\left (-1\right )^{\frac {3}{4}} b \log {\left (- \sqrt [4]{-1} \sqrt [4]{a} \sqrt [4]{\frac {1}{b}} + \sqrt {x} \right )}}{2 a^{\frac {9}{4}} \sqrt [4]{\frac {1}{b}}} + \frac {\left (-1\right )^{\frac {3}{4}} b \log {\left (\sqrt [4]{-1} \sqrt [4]{a} \sqrt [4]{\frac {1}{b}} + \sqrt {x} \right )}}{2 a^{\frac {9}{4}} \sqrt [4]{\frac {1}{b}}} + \frac {\left (-1\right )^{\frac {3}{4}} b \operatorname {atan}{\left (\frac {\left (-1\right )^{\frac {3}{4}} \sqrt {x}}{\sqrt [4]{a} \sqrt [4]{\frac {1}{b}}} \right )}}{a^{\frac {9}{4}} \sqrt [4]{\frac {1}{b}}} & \text {otherwise} \end {cases}\right ) + B \left (\begin {cases} \frac {\tilde {\infty }}{x^{\frac {5}{2}}} & \text {for}\: a = 0 \wedge b = 0 \\- \frac {2}{5 b x^{\frac {5}{2}}} & \text {for}\: a = 0 \\- \frac {2}{a \sqrt {x}} & \text {for}\: b = 0 \\- \frac {2}{a \sqrt {x}} + \frac {\left (-1\right )^{\frac {3}{4}} \log {\left (- \sqrt [4]{-1} \sqrt [4]{a} \sqrt [4]{\frac {1}{b}} + \sqrt {x} \right )}}{2 a^{\frac {5}{4}} \sqrt [4]{\frac {1}{b}}} - \frac {\left (-1\right )^{\frac {3}{4}} \log {\left (\sqrt [4]{-1} \sqrt [4]{a} \sqrt [4]{\frac {1}{b}} + \sqrt {x} \right )}}{2 a^{\frac {5}{4}} \sqrt [4]{\frac {1}{b}}} - \frac {\left (-1\right )^{\frac {3}{4}} \operatorname {atan}{\left (\frac {\left (-1\right )^{\frac {3}{4}} \sqrt {x}}{\sqrt [4]{a} \sqrt [4]{\frac {1}{b}}} \right )}}{a^{\frac {5}{4}} \sqrt [4]{\frac {1}{b}}} & \text {otherwise} \end {cases}\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),x)

[Out]

A*Piecewise((zoo/x**(9/2), Eq(a, 0) & Eq(b, 0)), (-2/(9*b*x**(9/2)), Eq(a, 0)), (-2/(5*a*x**(5/2)), Eq(b, 0)),
(-2/(5*a*x**(5/2)) + 2*b/(a**2*sqrt(x)) - (-1)**(3/4)*b*log(-(-1)**(1/4)*a**(1/4)*(1/b)**(1/4) + sqrt(x))/(2*
a**(9/4)*(1/b)**(1/4)) + (-1)**(3/4)*b*log((-1)**(1/4)*a**(1/4)*(1/b)**(1/4) + sqrt(x))/(2*a**(9/4)*(1/b)**(1/
4)) + (-1)**(3/4)*b*atan((-1)**(3/4)*sqrt(x)/(a**(1/4)*(1/b)**(1/4)))/(a**(9/4)*(1/b)**(1/4)), True)) + B*Piec
ewise((zoo/x**(5/2), Eq(a, 0) & Eq(b, 0)), (-2/(5*b*x**(5/2)), Eq(a, 0)), (-2/(a*sqrt(x)), Eq(b, 0)), (-2/(a*s
qrt(x)) + (-1)**(3/4)*log(-(-1)**(1/4)*a**(1/4)*(1/b)**(1/4) + sqrt(x))/(2*a**(5/4)*(1/b)**(1/4)) - (-1)**(3/4
)*log((-1)**(1/4)*a**(1/4)*(1/b)**(1/4) + sqrt(x))/(2*a**(5/4)*(1/b)**(1/4)) - (-1)**(3/4)*atan((-1)**(3/4)*sq
rt(x)/(a**(1/4)*(1/b)**(1/4)))/(a**(5/4)*(1/b)**(1/4)), True))

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