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

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

[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))

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Rubi [A]  time = 0.205779, antiderivative size = 255, normalized size of antiderivative = 1., 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} \[ \frac{2 (A b-a B)}{a^2 \sqrt{x}}+\frac{\sqrt [4]{b} (A b-a B) \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{2 \sqrt{2} a^{9/4}}-\frac{\sqrt [4]{b} (A b-a B) \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{2 \sqrt{2} a^{9/4}}-\frac{\sqrt [4]{b} (A b-a B) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{\sqrt{2} a^{9/4}}+\frac{\sqrt [4]{b} (A b-a B) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{\sqrt{2} a^{9/4}}-\frac{2 A}{5 a x^{5/2}} \]

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 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 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 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 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 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 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 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]

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]

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*}

Mathematica [C]  time = 0.014353, size = 46, normalized size = 0.18 \[ -\frac{2 \left (5 x^2 (a B-A b) \, _2F_1\left (-\frac{1}{4},1;\frac{3}{4};-\frac{b x^2}{a}\right )+a A\right )}{5 a^2 x^{5/2}} \]

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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Maple [A]  time = 0.011, size = 299, normalized size = 1.2 \begin{align*} -{\frac{2\,A}{5\,a}{x}^{-{\frac{5}{2}}}}+2\,{\frac{Ab}{{a}^{2}\sqrt{x}}}-2\,{\frac{B}{a\sqrt{x}}}+{\frac{\sqrt{2}Ab}{2\,{a}^{2}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}+1 \right ){\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}+{\frac{\sqrt{2}Ab}{2\,{a}^{2}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}-1 \right ){\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}+{\frac{\sqrt{2}Ab}{4\,{a}^{2}}\ln \left ({ \left ( x-\sqrt [4]{{\frac{a}{b}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{a}{b}}} \right ) \left ( x+\sqrt [4]{{\frac{a}{b}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{a}{b}}} \right ) ^{-1}} \right ){\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}-{\frac{\sqrt{2}B}{2\,a}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}+1 \right ){\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}-{\frac{\sqrt{2}B}{2\,a}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}-1 \right ){\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}-{\frac{\sqrt{2}B}{4\,a}\ln \left ({ \left ( x-\sqrt [4]{{\frac{a}{b}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{a}{b}}} \right ) \left ( x+\sqrt [4]{{\frac{a}{b}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{a}{b}}} \right ) ^{-1}} \right ){\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}} \end{align*}

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]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [B]  time = 1.02237, size = 1793, normalized size = 7.03 \begin{align*} \text{result too large to display} \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

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Giac [A]  time = 1.17123, size = 362, normalized size = 1.42 \begin{align*} -\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{align*}

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))