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

Optimal. Leaf size=215 \[ -\frac {a^{5/4} \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{2 \sqrt {2} b^{9/4}}+\frac {a^{5/4} \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{2 \sqrt {2} b^{9/4}}-\frac {a^{5/4} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt {2} b^{9/4}}+\frac {a^{5/4} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{\sqrt {2} b^{9/4}}-\frac {2 a \sqrt {x}}{b^2}+\frac {2 x^{5/2}}{5 b} \]

[Out]

2/5*x^(5/2)/b-1/2*a^(5/4)*arctan(1-b^(1/4)*2^(1/2)*x^(1/2)/a^(1/4))/b^(9/4)*2^(1/2)+1/2*a^(5/4)*arctan(1+b^(1/
4)*2^(1/2)*x^(1/2)/a^(1/4))/b^(9/4)*2^(1/2)-1/4*a^(5/4)*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)+1/4*a^(5/4)*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)-2*a*x^(1/2)/
b^2

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Rubi [A]  time = 0.20, antiderivative size = 215, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {321, 329, 211, 1165, 628, 1162, 617, 204} \[ -\frac {a^{5/4} \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{2 \sqrt {2} b^{9/4}}+\frac {a^{5/4} \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{2 \sqrt {2} b^{9/4}}-\frac {a^{5/4} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt {2} b^{9/4}}+\frac {a^{5/4} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{\sqrt {2} b^{9/4}}-\frac {2 a \sqrt {x}}{b^2}+\frac {2 x^{5/2}}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[x^(7/2)/(a + b*x^2),x]

[Out]

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

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Mathematica [A]  time = 0.07, size = 203, normalized size = 0.94 \[ \frac {-5 \sqrt {2} a^{5/4} \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )+5 \sqrt {2} a^{5/4} \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )-10 \sqrt {2} a^{5/4} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )+10 \sqrt {2} a^{5/4} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )-40 a \sqrt [4]{b} \sqrt {x}+8 b^{5/4} x^{5/2}}{20 b^{9/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(7/2)/(a + b*x^2),x]

[Out]

(-40*a*b^(1/4)*Sqrt[x] + 8*b^(5/4)*x^(5/2) - 10*Sqrt[2]*a^(5/4)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)]
+ 10*Sqrt[2]*a^(5/4)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)] - 5*Sqrt[2]*a^(5/4)*Log[Sqrt[a] - Sqrt[2]*a
^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x] + 5*Sqrt[2]*a^(5/4)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b
]*x])/(20*b^(9/4))

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fricas [A]  time = 0.98, size = 170, normalized size = 0.79 \[ \frac {20 \, b^{2} \left (-\frac {a^{5}}{b^{9}}\right )^{\frac {1}{4}} \arctan \left (-\frac {a b^{7} \sqrt {x} \left (-\frac {a^{5}}{b^{9}}\right )^{\frac {3}{4}} - \sqrt {b^{4} \sqrt {-\frac {a^{5}}{b^{9}}} + a^{2} x} b^{7} \left (-\frac {a^{5}}{b^{9}}\right )^{\frac {3}{4}}}{a^{5}}\right ) + 5 \, b^{2} \left (-\frac {a^{5}}{b^{9}}\right )^{\frac {1}{4}} \log \left (b^{2} \left (-\frac {a^{5}}{b^{9}}\right )^{\frac {1}{4}} + a \sqrt {x}\right ) - 5 \, b^{2} \left (-\frac {a^{5}}{b^{9}}\right )^{\frac {1}{4}} \log \left (-b^{2} \left (-\frac {a^{5}}{b^{9}}\right )^{\frac {1}{4}} + a \sqrt {x}\right ) + 4 \, {\left (b x^{2} - 5 \, a\right )} \sqrt {x}}{10 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*(20*b^2*(-a^5/b^9)^(1/4)*arctan(-(a*b^7*sqrt(x)*(-a^5/b^9)^(3/4) - sqrt(b^4*sqrt(-a^5/b^9) + a^2*x)*b^7*(
-a^5/b^9)^(3/4))/a^5) + 5*b^2*(-a^5/b^9)^(1/4)*log(b^2*(-a^5/b^9)^(1/4) + a*sqrt(x)) - 5*b^2*(-a^5/b^9)^(1/4)*
log(-b^2*(-a^5/b^9)^(1/4) + a*sqrt(x)) + 4*(b*x^2 - 5*a)*sqrt(x))/b^2

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giac [A]  time = 0.62, size = 196, normalized size = 0.91 \[ \frac {\sqrt {2} \left (a b^{3}\right )^{\frac {1}{4}} a \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{2 \, b^{3}} + \frac {\sqrt {2} \left (a b^{3}\right )^{\frac {1}{4}} a \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{2 \, b^{3}} + \frac {\sqrt {2} \left (a b^{3}\right )^{\frac {1}{4}} a \log \left (\sqrt {2} \sqrt {x} \left (\frac {a}{b}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{b}}\right )}{4 \, b^{3}} - \frac {\sqrt {2} \left (a b^{3}\right )^{\frac {1}{4}} a \log \left (-\sqrt {2} \sqrt {x} \left (\frac {a}{b}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{b}}\right )}{4 \, b^{3}} + \frac {2 \, {\left (b^{4} x^{\frac {5}{2}} - 5 \, a b^{3} \sqrt {x}\right )}}{5 \, b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 152, normalized size = 0.71 \[ \frac {2 x^{\frac {5}{2}}}{5 b}+\frac {\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, a \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )}{2 b^{2}}+\frac {\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, a \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )}{2 b^{2}}+\frac {\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, a \ln \left (\frac {x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{b}}}{x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{b}}}\right )}{4 b^{2}}-\frac {2 a \sqrt {x}}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5/b*x^(5/2)-2*a/b^2*x^(1/2)+1/4*a/b^2*(a/b)^(1/4)*2^(1/2)*ln((x+(a/b)^(1/4)*x^(1/2)*2^(1/2)+(a/b)^(1/2))/(x-
(a/b)^(1/4)*x^(1/2)*2^(1/2)+(a/b)^(1/2)))+1/2*a/b^2*(a/b)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)+1)+
1/2*a/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 = 3.08, size = 194, normalized size = 0.90 \[ \frac {2 \, {\left (b x^{\frac {5}{2}} - 5 \, a \sqrt {x}\right )}}{5 \, b^{2}} + \frac {\frac {2 \, \sqrt {2} a^{\frac {3}{2}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} + 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}}} + \frac {2 \, \sqrt {2} a^{\frac {3}{2}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} - 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}}} + \frac {\sqrt {2} a^{\frac {5}{4}} \log \left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{b^{\frac {1}{4}}} - \frac {\sqrt {2} a^{\frac {5}{4}} \log \left (-\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{b^{\frac {1}{4}}}}{4 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.49, size = 67, normalized size = 0.31 \[ \frac {2\,x^{5/2}}{5\,b}-\frac {2\,a\,\sqrt {x}}{b^2}-\frac {{\left (-a\right )}^{5/4}\,\mathrm {atan}\left (\frac {b^{1/4}\,\sqrt {x}}{{\left (-a\right )}^{1/4}}\right )}{b^{9/4}}+\frac {{\left (-a\right )}^{5/4}\,\mathrm {atan}\left (\frac {b^{1/4}\,\sqrt {x}\,1{}\mathrm {i}}{{\left (-a\right )}^{1/4}}\right )\,1{}\mathrm {i}}{b^{9/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(7/2)/(a + b*x^2),x)

[Out]

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

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sympy [A]  time = 50.31, size = 192, normalized size = 0.89 \[ \begin {cases} \tilde {\infty } x^{\frac {5}{2}} & \text {for}\: a = 0 \wedge b = 0 \\\frac {2 x^{\frac {5}{2}}}{5 b} & \text {for}\: a = 0 \\\frac {2 x^{\frac {9}{2}}}{9 a} & \text {for}\: b = 0 \\- \frac {\sqrt [4]{-1} a^{\frac {5}{4}} \sqrt [4]{\frac {1}{b}} \log {\left (- \sqrt [4]{-1} \sqrt [4]{a} \sqrt [4]{\frac {1}{b}} + \sqrt {x} \right )}}{2 b^{2}} + \frac {\sqrt [4]{-1} a^{\frac {5}{4}} \sqrt [4]{\frac {1}{b}} \log {\left (\sqrt [4]{-1} \sqrt [4]{a} \sqrt [4]{\frac {1}{b}} + \sqrt {x} \right )}}{2 b^{2}} - \frac {\sqrt [4]{-1} a^{\frac {5}{4}} \sqrt [4]{\frac {1}{b}} \operatorname {atan}{\left (\frac {\left (-1\right )^{\frac {3}{4}} \sqrt {x}}{\sqrt [4]{a} \sqrt [4]{\frac {1}{b}}} \right )}}{b^{2}} - \frac {2 a \sqrt {x}}{b^{2}} + \frac {2 x^{\frac {5}{2}}}{5 b} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*x**(5/2), Eq(a, 0) & Eq(b, 0)), (2*x**(5/2)/(5*b), Eq(a, 0)), (2*x**(9/2)/(9*a), Eq(b, 0)), (-(
-1)**(1/4)*a**(5/4)*(1/b)**(1/4)*log(-(-1)**(1/4)*a**(1/4)*(1/b)**(1/4) + sqrt(x))/(2*b**2) + (-1)**(1/4)*a**(
5/4)*(1/b)**(1/4)*log((-1)**(1/4)*a**(1/4)*(1/b)**(1/4) + sqrt(x))/(2*b**2) - (-1)**(1/4)*a**(5/4)*(1/b)**(1/4
)*atan((-1)**(3/4)*sqrt(x)/(a**(1/4)*(1/b)**(1/4)))/b**2 - 2*a*sqrt(x)/b**2 + 2*x**(5/2)/(5*b), True))

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