3.368 \(\int \frac{x^{5/2} (A+B x^2)}{a+b x^2} \, dx\)

Optimal. Leaf size=257 \[ -\frac{a^{3/4} (A b-a B) \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{2 \sqrt{2} b^{11/4}}+\frac{a^{3/4} (A b-a B) \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{2 \sqrt{2} b^{11/4}}+\frac{a^{3/4} (A b-a B) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{\sqrt{2} b^{11/4}}-\frac{a^{3/4} (A b-a B) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{\sqrt{2} b^{11/4}}+\frac{2 x^{3/2} (A b-a B)}{3 b^2}+\frac{2 B x^{7/2}}{7 b} \]

[Out]

(2*(A*b - a*B)*x^(3/2))/(3*b^2) + (2*B*x^(7/2))/(7*b) + (a^(3/4)*(A*b - a*B)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[
x])/a^(1/4)])/(Sqrt[2]*b^(11/4)) - (a^(3/4)*(A*b - a*B)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(Sqrt[2
]*b^(11/4)) - (a^(3/4)*(A*b - a*B)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(2*Sqrt[2]*b^(1
1/4)) + (a^(3/4)*(A*b - a*B)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(2*Sqrt[2]*b^(11/4))

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Rubi [A]  time = 0.205399, antiderivative size = 257, 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 = {459, 321, 329, 297, 1162, 617, 204, 1165, 628} \[ -\frac{a^{3/4} (A b-a B) \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{2 \sqrt{2} b^{11/4}}+\frac{a^{3/4} (A b-a B) \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{2 \sqrt{2} b^{11/4}}+\frac{a^{3/4} (A b-a B) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{\sqrt{2} b^{11/4}}-\frac{a^{3/4} (A b-a B) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{\sqrt{2} b^{11/4}}+\frac{2 x^{3/2} (A b-a B)}{3 b^2}+\frac{2 B x^{7/2}}{7 b} \]

Antiderivative was successfully verified.

[In]

Int[(x^(5/2)*(A + B*x^2))/(a + b*x^2),x]

[Out]

(2*(A*b - a*B)*x^(3/2))/(3*b^2) + (2*B*x^(7/2))/(7*b) + (a^(3/4)*(A*b - a*B)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[
x])/a^(1/4)])/(Sqrt[2]*b^(11/4)) - (a^(3/4)*(A*b - a*B)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(Sqrt[2
]*b^(11/4)) - (a^(3/4)*(A*b - a*B)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(2*Sqrt[2]*b^(1
1/4)) + (a^(3/4)*(A*b - a*B)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(2*Sqrt[2]*b^(11/4))

Rule 459

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(b*e*(m + n*(p + 1) + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m +
n*(p + 1) + 1)), Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0]
&& NeQ[m + n*(p + 1) + 1, 0]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && 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{x^{5/2} \left (A+B x^2\right )}{a+b x^2} \, dx &=\frac{2 B x^{7/2}}{7 b}-\frac{\left (2 \left (-\frac{7 A b}{2}+\frac{7 a B}{2}\right )\right ) \int \frac{x^{5/2}}{a+b x^2} \, dx}{7 b}\\ &=\frac{2 (A b-a B) x^{3/2}}{3 b^2}+\frac{2 B x^{7/2}}{7 b}-\frac{(a (A b-a B)) \int \frac{\sqrt{x}}{a+b x^2} \, dx}{b^2}\\ &=\frac{2 (A b-a B) x^{3/2}}{3 b^2}+\frac{2 B x^{7/2}}{7 b}-\frac{(2 a (A b-a B)) \operatorname{Subst}\left (\int \frac{x^2}{a+b x^4} \, dx,x,\sqrt{x}\right )}{b^2}\\ &=\frac{2 (A b-a B) x^{3/2}}{3 b^2}+\frac{2 B x^{7/2}}{7 b}+\frac{(a (A b-a B)) \operatorname{Subst}\left (\int \frac{\sqrt{a}-\sqrt{b} x^2}{a+b x^4} \, dx,x,\sqrt{x}\right )}{b^{5/2}}-\frac{(a (A b-a B)) \operatorname{Subst}\left (\int \frac{\sqrt{a}+\sqrt{b} x^2}{a+b x^4} \, dx,x,\sqrt{x}\right )}{b^{5/2}}\\ &=\frac{2 (A b-a B) x^{3/2}}{3 b^2}+\frac{2 B x^{7/2}}{7 b}-\frac{(a (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 b^3}-\frac{(a (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 b^3}-\frac{\left (a^{3/4} (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} b^{11/4}}-\frac{\left (a^{3/4} (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} b^{11/4}}\\ &=\frac{2 (A b-a B) x^{3/2}}{3 b^2}+\frac{2 B x^{7/2}}{7 b}-\frac{a^{3/4} (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} b^{11/4}}+\frac{a^{3/4} (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} b^{11/4}}-\frac{\left (a^{3/4} (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} b^{11/4}}+\frac{\left (a^{3/4} (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} b^{11/4}}\\ &=\frac{2 (A b-a B) x^{3/2}}{3 b^2}+\frac{2 B x^{7/2}}{7 b}+\frac{a^{3/4} (A b-a B) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{\sqrt{2} b^{11/4}}-\frac{a^{3/4} (A b-a B) \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{\sqrt{2} b^{11/4}}-\frac{a^{3/4} (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} b^{11/4}}+\frac{a^{3/4} (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} b^{11/4}}\\ \end{align*}

Mathematica [A]  time = 0.135887, size = 110, normalized size = 0.43 \[ \frac{2 b^{3/4} x^{3/2} \left (-7 a B+7 A b+3 b B x^2\right )-21 (-a)^{3/4} (a B-A b) \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{x}}{\sqrt [4]{-a}}\right )+21 (-a)^{3/4} (a B-A b) \tanh ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{x}}{\sqrt [4]{-a}}\right )}{21 b^{11/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^(5/2)*(A + B*x^2))/(a + b*x^2),x]

[Out]

(2*b^(3/4)*x^(3/2)*(7*A*b - 7*a*B + 3*b*B*x^2) - 21*(-a)^(3/4)*(-(A*b) + a*B)*ArcTan[(b^(1/4)*Sqrt[x])/(-a)^(1
/4)] + 21*(-a)^(3/4)*(-(A*b) + a*B)*ArcTanh[(b^(1/4)*Sqrt[x])/(-a)^(1/4)])/(21*b^(11/4))

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Maple [A]  time = 0.009, size = 308, normalized size = 1.2 \begin{align*}{\frac{2\,B}{7\,b}{x}^{{\frac{7}{2}}}}+{\frac{2\,A}{3\,b}{x}^{{\frac{3}{2}}}}-{\frac{2\,Ba}{3\,{b}^{2}}{x}^{{\frac{3}{2}}}}-{\frac{a\sqrt{2}A}{2\,{b}^{2}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}-1 \right ){\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}-{\frac{a\sqrt{2}A}{4\,{b}^{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{a\sqrt{2}A}{2\,{b}^{2}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}+1 \right ){\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}+{\frac{{a}^{2}\sqrt{2}B}{2\,{b}^{3}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}-1 \right ){\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}+{\frac{{a}^{2}\sqrt{2}B}{4\,{b}^{3}}\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{{a}^{2}\sqrt{2}B}{2\,{b}^{3}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}+1 \right ){\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)*(B*x^2+A)/(b*x^2+a),x)

[Out]

2/7*B*x^(7/2)/b+2/3/b*x^(3/2)*A-2/3/b^2*x^(3/2)*B*a-1/2*a/b^2/(1/b*a)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(1/b*a)^(
1/4)*x^(1/2)-1)-1/4*a/b^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)))-1/2*a/b^2/(1/b*a)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(1/b*a)^(1/4)*x^(1/2)
+1)+1/2*a^2/b^3/(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^2/b^3/(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)))+1/2*
a^2/b^3/(1/b*a)^(1/4)*2^(1/2)*B*arctan(2^(1/2)/(1/b*a)^(1/4)*x^(1/2)+1)

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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(x^(5/2)*(B*x^2+A)/(b*x^2+a),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.98935, size = 1823, normalized size = 7.09 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x^2+A)/(b*x^2+a),x, algorithm="fricas")

[Out]

1/42*(84*b^2*(-(B^4*a^7 - 4*A*B^3*a^6*b + 6*A^2*B^2*a^5*b^2 - 4*A^3*B*a^4*b^3 + A^4*a^3*b^4)/b^11)^(1/4)*arcta
n((sqrt((B^6*a^10 - 6*A*B^5*a^9*b + 15*A^2*B^4*a^8*b^2 - 20*A^3*B^3*a^7*b^3 + 15*A^4*B^2*a^6*b^4 - 6*A^5*B*a^5
*b^5 + A^6*a^4*b^6)*x - (B^4*a^7*b^5 - 4*A*B^3*a^6*b^6 + 6*A^2*B^2*a^5*b^7 - 4*A^3*B*a^4*b^8 + A^4*a^3*b^9)*sq
rt(-(B^4*a^7 - 4*A*B^3*a^6*b + 6*A^2*B^2*a^5*b^2 - 4*A^3*B*a^4*b^3 + A^4*a^3*b^4)/b^11))*b^3*(-(B^4*a^7 - 4*A*
B^3*a^6*b + 6*A^2*B^2*a^5*b^2 - 4*A^3*B*a^4*b^3 + A^4*a^3*b^4)/b^11)^(1/4) + (B^3*a^5*b^3 - 3*A*B^2*a^4*b^4 +
3*A^2*B*a^3*b^5 - A^3*a^2*b^6)*sqrt(x)*(-(B^4*a^7 - 4*A*B^3*a^6*b + 6*A^2*B^2*a^5*b^2 - 4*A^3*B*a^4*b^3 + A^4*
a^3*b^4)/b^11)^(1/4))/(B^4*a^7 - 4*A*B^3*a^6*b + 6*A^2*B^2*a^5*b^2 - 4*A^3*B*a^4*b^3 + A^4*a^3*b^4)) - 21*b^2*
(-(B^4*a^7 - 4*A*B^3*a^6*b + 6*A^2*B^2*a^5*b^2 - 4*A^3*B*a^4*b^3 + A^4*a^3*b^4)/b^11)^(1/4)*log(b^8*(-(B^4*a^7
- 4*A*B^3*a^6*b + 6*A^2*B^2*a^5*b^2 - 4*A^3*B*a^4*b^3 + A^4*a^3*b^4)/b^11)^(3/4) - (B^3*a^5 - 3*A*B^2*a^4*b +
3*A^2*B*a^3*b^2 - A^3*a^2*b^3)*sqrt(x)) + 21*b^2*(-(B^4*a^7 - 4*A*B^3*a^6*b + 6*A^2*B^2*a^5*b^2 - 4*A^3*B*a^4
*b^3 + A^4*a^3*b^4)/b^11)^(1/4)*log(-b^8*(-(B^4*a^7 - 4*A*B^3*a^6*b + 6*A^2*B^2*a^5*b^2 - 4*A^3*B*a^4*b^3 + A^
4*a^3*b^4)/b^11)^(3/4) - (B^3*a^5 - 3*A*B^2*a^4*b + 3*A^2*B*a^3*b^2 - A^3*a^2*b^3)*sqrt(x)) + 4*(3*B*b*x^3 - 7
*(B*a - A*b)*x)*sqrt(x))/b^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(5/2)*(B*x**2+A)/(b*x**2+a),x)

[Out]

Exception raised: TypeError

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Giac [A]  time = 1.1624, size = 356, normalized size = 1.39 \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 \, b^{5}} + \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 \, b^{5}} - \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 \, b^{5}} + \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 \, b^{5}} + \frac{2 \,{\left (3 \, B b^{6} x^{\frac{7}{2}} - 7 \, B a b^{5} x^{\frac{3}{2}} + 7 \, A b^{6} x^{\frac{3}{2}}\right )}}{21 \, b^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x^2+A)/(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))/b^5 + 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))/b^5 - 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))/b^5 + 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))/b^5 + 2/21*(3*B*b^6*x^(7/2) - 7*B*a*b^5*x^(3/2) + 7*A*b^6*x^(3/2))/b^7