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

Optimal. Leaf size=230 \[ \frac {5 \sqrt [4]{a} \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{8 \sqrt {2} b^{9/4}}-\frac {5 \sqrt [4]{a} \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{8 \sqrt {2} b^{9/4}}+\frac {5 \sqrt [4]{a} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} b^{9/4}}-\frac {5 \sqrt [4]{a} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{4 \sqrt {2} b^{9/4}}-\frac {x^{5/2}}{2 b \left (a+b x^2\right )}+\frac {5 \sqrt {x}}{2 b^2} \]

[Out]

-1/2*x^(5/2)/b/(b*x^2+a)+5/8*a^(1/4)*arctan(1-b^(1/4)*2^(1/2)*x^(1/2)/a^(1/4))/b^(9/4)*2^(1/2)-5/8*a^(1/4)*arc
tan(1+b^(1/4)*2^(1/2)*x^(1/2)/a^(1/4))/b^(9/4)*2^(1/2)+5/16*a^(1/4)*ln(a^(1/2)+x*b^(1/2)-a^(1/4)*b^(1/4)*2^(1/
2)*x^(1/2))/b^(9/4)*2^(1/2)-5/16*a^(1/4)*ln(a^(1/2)+x*b^(1/2)+a^(1/4)*b^(1/4)*2^(1/2)*x^(1/2))/b^(9/4)*2^(1/2)
+5/2*x^(1/2)/b^2

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Rubi [A]  time = 0.17, antiderivative size = 230, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {288, 321, 329, 211, 1165, 628, 1162, 617, 204} \[ \frac {5 \sqrt [4]{a} \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{8 \sqrt {2} b^{9/4}}-\frac {5 \sqrt [4]{a} \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{8 \sqrt {2} b^{9/4}}+\frac {5 \sqrt [4]{a} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} b^{9/4}}-\frac {5 \sqrt [4]{a} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{4 \sqrt {2} b^{9/4}}-\frac {x^{5/2}}{2 b \left (a+b x^2\right )}+\frac {5 \sqrt {x}}{2 b^2} \]

Antiderivative was successfully verified.

[In]

Int[x^(7/2)/(a + b*x^2)^2,x]

[Out]

(5*Sqrt[x])/(2*b^2) - x^(5/2)/(2*b*(a + b*x^2)) + (5*a^(1/4)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(4
*Sqrt[2]*b^(9/4)) - (5*a^(1/4)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(4*Sqrt[2]*b^(9/4)) + (5*a^(1/4)
*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(8*Sqrt[2]*b^(9/4)) - (5*a^(1/4)*Log[Sqrt[a] + Sq
rt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(8*Sqrt[2]*b^(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 211

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

Rule 288

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*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

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 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 {x^{7/2}}{\left (a+b x^2\right )^2} \, dx &=-\frac {x^{5/2}}{2 b \left (a+b x^2\right )}+\frac {5 \int \frac {x^{3/2}}{a+b x^2} \, dx}{4 b}\\ &=\frac {5 \sqrt {x}}{2 b^2}-\frac {x^{5/2}}{2 b \left (a+b x^2\right )}-\frac {(5 a) \int \frac {1}{\sqrt {x} \left (a+b x^2\right )} \, dx}{4 b^2}\\ &=\frac {5 \sqrt {x}}{2 b^2}-\frac {x^{5/2}}{2 b \left (a+b x^2\right )}-\frac {(5 a) \operatorname {Subst}\left (\int \frac {1}{a+b x^4} \, dx,x,\sqrt {x}\right )}{2 b^2}\\ &=\frac {5 \sqrt {x}}{2 b^2}-\frac {x^{5/2}}{2 b \left (a+b x^2\right )}-\frac {\left (5 \sqrt {a}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a}-\sqrt {b} x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{4 b^2}-\frac {\left (5 \sqrt {a}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a}+\sqrt {b} x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{4 b^2}\\ &=\frac {5 \sqrt {x}}{2 b^2}-\frac {x^{5/2}}{2 b \left (a+b x^2\right )}-\frac {\left (5 \sqrt {a}\right ) \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 )}{8 b^{5/2}}-\frac {\left (5 \sqrt {a}\right ) \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 )}{8 b^{5/2}}+\frac {\left (5 \sqrt [4]{a}\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 )}{8 \sqrt {2} b^{9/4}}+\frac {\left (5 \sqrt [4]{a}\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 )}{8 \sqrt {2} b^{9/4}}\\ &=\frac {5 \sqrt {x}}{2 b^2}-\frac {x^{5/2}}{2 b \left (a+b x^2\right )}+\frac {5 \sqrt [4]{a} \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{8 \sqrt {2} b^{9/4}}-\frac {5 \sqrt [4]{a} \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{8 \sqrt {2} b^{9/4}}-\frac {\left (5 \sqrt [4]{a}\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 )}{4 \sqrt {2} b^{9/4}}+\frac {\left (5 \sqrt [4]{a}\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 )}{4 \sqrt {2} b^{9/4}}\\ &=\frac {5 \sqrt {x}}{2 b^2}-\frac {x^{5/2}}{2 b \left (a+b x^2\right )}+\frac {5 \sqrt [4]{a} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} b^{9/4}}-\frac {5 \sqrt [4]{a} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} b^{9/4}}+\frac {5 \sqrt [4]{a} \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{8 \sqrt {2} b^{9/4}}-\frac {5 \sqrt [4]{a} \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{8 \sqrt {2} b^{9/4}}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 221, normalized size = 0.96 \[ \frac {\frac {32 b^{5/4} x^{5/2}}{a+b x^2}+\frac {40 a \sqrt [4]{b} \sqrt {x}}{a+b x^2}+5 \sqrt {2} \sqrt [4]{a} \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )-5 \sqrt {2} \sqrt [4]{a} \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )+10 \sqrt {2} \sqrt [4]{a} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )-10 \sqrt {2} \sqrt [4]{a} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{16 b^{9/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(7/2)/(a + b*x^2)^2,x]

[Out]

((40*a*b^(1/4)*Sqrt[x])/(a + b*x^2) + (32*b^(5/4)*x^(5/2))/(a + b*x^2) + 10*Sqrt[2]*a^(1/4)*ArcTan[1 - (Sqrt[2
]*b^(1/4)*Sqrt[x])/a^(1/4)] - 10*Sqrt[2]*a^(1/4)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)] + 5*Sqrt[2]*a^(
1/4)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x] - 5*Sqrt[2]*a^(1/4)*Log[Sqrt[a] + Sqrt[2]*a^(1
/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(16*b^(9/4))

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fricas [A]  time = 1.13, size = 192, normalized size = 0.83 \[ -\frac {20 \, {\left (b^{3} x^{2} + a b^{2}\right )} \left (-\frac {a}{b^{9}}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {b^{4} \sqrt {-\frac {a}{b^{9}}} + x} b^{7} \left (-\frac {a}{b^{9}}\right )^{\frac {3}{4}} - b^{7} \sqrt {x} \left (-\frac {a}{b^{9}}\right )^{\frac {3}{4}}}{a}\right ) + 5 \, {\left (b^{3} x^{2} + a b^{2}\right )} \left (-\frac {a}{b^{9}}\right )^{\frac {1}{4}} \log \left (5 \, b^{2} \left (-\frac {a}{b^{9}}\right )^{\frac {1}{4}} + 5 \, \sqrt {x}\right ) - 5 \, {\left (b^{3} x^{2} + a b^{2}\right )} \left (-\frac {a}{b^{9}}\right )^{\frac {1}{4}} \log \left (-5 \, b^{2} \left (-\frac {a}{b^{9}}\right )^{\frac {1}{4}} + 5 \, \sqrt {x}\right ) - 4 \, {\left (4 \, b x^{2} + 5 \, a\right )} \sqrt {x}}{8 \, {\left (b^{3} x^{2} + a b^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)/(b*x^2+a)^2,x, algorithm="fricas")

[Out]

-1/8*(20*(b^3*x^2 + a*b^2)*(-a/b^9)^(1/4)*arctan((sqrt(b^4*sqrt(-a/b^9) + x)*b^7*(-a/b^9)^(3/4) - b^7*sqrt(x)*
(-a/b^9)^(3/4))/a) + 5*(b^3*x^2 + a*b^2)*(-a/b^9)^(1/4)*log(5*b^2*(-a/b^9)^(1/4) + 5*sqrt(x)) - 5*(b^3*x^2 + a
*b^2)*(-a/b^9)^(1/4)*log(-5*b^2*(-a/b^9)^(1/4) + 5*sqrt(x)) - 4*(4*b*x^2 + 5*a)*sqrt(x))/(b^3*x^2 + a*b^2)

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giac [A]  time = 0.65, size = 196, normalized size = 0.85 \[ -\frac {5 \, \sqrt {2} \left (a b^{3}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{8 \, b^{3}} - \frac {5 \, \sqrt {2} \left (a b^{3}\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{8 \, b^{3}} - \frac {5 \, \sqrt {2} \left (a b^{3}\right )^{\frac {1}{4}} \log \left (\sqrt {2} \sqrt {x} \left (\frac {a}{b}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{b}}\right )}{16 \, b^{3}} + \frac {5 \, \sqrt {2} \left (a b^{3}\right )^{\frac {1}{4}} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {a}{b}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{b}}\right )}{16 \, b^{3}} + \frac {a \sqrt {x}}{2 \, {\left (b x^{2} + a\right )} b^{2}} + \frac {2 \, \sqrt {x}}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)/(b*x^2+a)^2,x, algorithm="giac")

[Out]

-5/8*sqrt(2)*(a*b^3)^(1/4)*arctan(1/2*sqrt(2)*(sqrt(2)*(a/b)^(1/4) + 2*sqrt(x))/(a/b)^(1/4))/b^3 - 5/8*sqrt(2)
*(a*b^3)^(1/4)*arctan(-1/2*sqrt(2)*(sqrt(2)*(a/b)^(1/4) - 2*sqrt(x))/(a/b)^(1/4))/b^3 - 5/16*sqrt(2)*(a*b^3)^(
1/4)*log(sqrt(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))/b^3 + 5/16*sqrt(2)*(a*b^3)^(1/4)*log(-sqrt(2)*sqrt(x)*(a
/b)^(1/4) + x + sqrt(a/b))/b^3 + 1/2*a*sqrt(x)/((b*x^2 + a)*b^2) + 2*sqrt(x)/b^2

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maple [A]  time = 0.01, size = 158, normalized size = 0.69 \[ \frac {a \sqrt {x}}{2 \left (b \,x^{2}+a \right ) b^{2}}-\frac {5 \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )}{8 b^{2}}-\frac {5 \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )}{8 b^{2}}-\frac {5 \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{b}}}{x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{b}}}\right )}{16 b^{2}}+\frac {2 \sqrt {x}}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(7/2)/(b*x^2+a)^2,x)

[Out]

2*x^(1/2)/b^2+1/2*a/b^2*x^(1/2)/(b*x^2+a)-5/16/b^2*(a/b)^(1/4)*2^(1/2)*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)))-5/8/b^2*(a/b)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/b)^(1/4)*x^
(1/2)+1)-5/8/b^2*(a/b)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)-1)

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maxima [A]  time = 3.02, size = 206, normalized size = 0.90 \[ \frac {a \sqrt {x}}{2 \, {\left (b^{3} x^{2} + a b^{2}\right )}} - \frac {5 \, {\left (\frac {2 \, \sqrt {2} \sqrt {a} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} + 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}}} + \frac {2 \, \sqrt {2} \sqrt {a} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} - 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}}} + \frac {\sqrt {2} a^{\frac {1}{4}} \log \left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{b^{\frac {1}{4}}} - \frac {\sqrt {2} a^{\frac {1}{4}} \log \left (-\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{b^{\frac {1}{4}}}\right )}}{16 \, b^{2}} + \frac {2 \, \sqrt {x}}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)/(b*x^2+a)^2,x, algorithm="maxima")

[Out]

1/2*a*sqrt(x)/(b^3*x^2 + a*b^2) - 5/16*(2*sqrt(2)*sqrt(a)*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)) + 2*sqrt(2)*sqrt(a)*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(2)*a^(1/4)*log(sqrt(2)*a
^(1/4)*b^(1/4)*sqrt(x) + sqrt(b)*x + sqrt(a))/b^(1/4) - sqrt(2)*a^(1/4)*log(-sqrt(2)*a^(1/4)*b^(1/4)*sqrt(x) +
sqrt(b)*x + sqrt(a))/b^(1/4))/b^2 + 2*sqrt(x)/b^2

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mupad [B]  time = 4.51, size = 80, normalized size = 0.35 \[ \frac {2\,\sqrt {x}}{b^2}-\frac {5\,{\left (-a\right )}^{1/4}\,\mathrm {atan}\left (\frac {b^{1/4}\,\sqrt {x}}{{\left (-a\right )}^{1/4}}\right )}{4\,b^{9/4}}+\frac {a\,\sqrt {x}}{2\,\left (b^3\,x^2+a\,b^2\right )}+\frac {{\left (-a\right )}^{1/4}\,\mathrm {atan}\left (\frac {b^{1/4}\,\sqrt {x}\,1{}\mathrm {i}}{{\left (-a\right )}^{1/4}}\right )\,5{}\mathrm {i}}{4\,b^{9/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(7/2)/(a + b*x^2)^2,x)

[Out]

(2*x^(1/2))/b^2 - (5*(-a)^(1/4)*atan((b^(1/4)*x^(1/2))/(-a)^(1/4)))/(4*b^(9/4)) + ((-a)^(1/4)*atan((b^(1/4)*x^
(1/2)*1i)/(-a)^(1/4))*5i)/(4*b^(9/4)) + (a*x^(1/2))/(2*(a*b^2 + b^3*x^2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(7/2)/(b*x**2+a)**2,x)

[Out]

Timed out

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