3.295 \(\int \frac {1}{x^{7/2} (a+b x^2)} \, dx\)

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

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.18, 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 = {325, 329, 297, 1162, 617, 204, 1165, 628} \[ \frac {b^{5/4} \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{2 \sqrt {2} a^{9/4}}-\frac {b^{5/4} \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{2 \sqrt {2} a^{9/4}}-\frac {b^{5/4} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt {2} a^{9/4}}+\frac {b^{5/4} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{\sqrt {2} a^{9/4}}+\frac {2 b}{a^2 \sqrt {x}}-\frac {2}{5 a x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^(7/2)*(a + b*x^2)),x]

[Out]

-2/(5*a*x^(5/2)) + (2*b)/(a^2*Sqrt[x]) - (b^(5/4)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(Sqrt[2]*a^(9
/4)) + (b^(5/4)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(Sqrt[2]*a^(9/4)) + (b^(5/4)*Log[Sqrt[a] - Sqrt
[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(2*Sqrt[2]*a^(9/4)) - (b^(5/4)*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 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 {1}{x^{7/2} \left (a+b x^2\right )} \, dx &=-\frac {2}{5 a x^{5/2}}-\frac {b \int \frac {1}{x^{3/2} \left (a+b x^2\right )} \, dx}{a}\\ &=-\frac {2}{5 a x^{5/2}}+\frac {2 b}{a^2 \sqrt {x}}+\frac {b^2 \int \frac {\sqrt {x}}{a+b x^2} \, dx}{a^2}\\ &=-\frac {2}{5 a x^{5/2}}+\frac {2 b}{a^2 \sqrt {x}}+\frac {\left (2 b^2\right ) \operatorname {Subst}\left (\int \frac {x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{a^2}\\ &=-\frac {2}{5 a x^{5/2}}+\frac {2 b}{a^2 \sqrt {x}}-\frac {b^{3/2} \operatorname {Subst}\left (\int \frac {\sqrt {a}-\sqrt {b} x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{a^2}+\frac {b^{3/2} \operatorname {Subst}\left (\int \frac {\sqrt {a}+\sqrt {b} x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{a^2}\\ &=-\frac {2}{5 a x^{5/2}}+\frac {2 b}{a^2 \sqrt {x}}+\frac {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 {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 {b^{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} a^{9/4}}+\frac {b^{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} a^{9/4}}\\ &=-\frac {2}{5 a x^{5/2}}+\frac {2 b}{a^2 \sqrt {x}}+\frac {b^{5/4} \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 {b^{5/4} \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 {b^{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} a^{9/4}}-\frac {b^{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} a^{9/4}}\\ &=-\frac {2}{5 a x^{5/2}}+\frac {2 b}{a^2 \sqrt {x}}-\frac {b^{5/4} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt {2} a^{9/4}}+\frac {b^{5/4} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt {2} a^{9/4}}+\frac {b^{5/4} \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 {b^{5/4} \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 = 0.01, size = 29, normalized size = 0.13 \[ -\frac {2 \, _2F_1\left (-\frac {5}{4},1;-\frac {1}{4};-\frac {b x^2}{a}\right )}{5 a x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(7/2)*(a + b*x^2)),x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.89, size = 193, normalized size = 0.90 \[ -\frac {20 \, a^{2} x^{3} \left (-\frac {b^{5}}{a^{9}}\right )^{\frac {1}{4}} \arctan \left (-\frac {a^{2} b^{4} \sqrt {x} \left (-\frac {b^{5}}{a^{9}}\right )^{\frac {1}{4}} - \sqrt {-a^{5} b^{5} \sqrt {-\frac {b^{5}}{a^{9}}} + b^{8} x} a^{2} \left (-\frac {b^{5}}{a^{9}}\right )^{\frac {1}{4}}}{b^{5}}\right ) - 5 \, a^{2} x^{3} \left (-\frac {b^{5}}{a^{9}}\right )^{\frac {1}{4}} \log \left (a^{7} \left (-\frac {b^{5}}{a^{9}}\right )^{\frac {3}{4}} + b^{4} \sqrt {x}\right ) + 5 \, a^{2} x^{3} \left (-\frac {b^{5}}{a^{9}}\right )^{\frac {1}{4}} \log \left (-a^{7} \left (-\frac {b^{5}}{a^{9}}\right )^{\frac {3}{4}} + b^{4} \sqrt {x}\right ) - 4 \, {\left (5 \, b x^{2} - a\right )} \sqrt {x}}{10 \, a^{2} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(7/2)/(b*x^2+a),x, algorithm="fricas")

[Out]

-1/10*(20*a^2*x^3*(-b^5/a^9)^(1/4)*arctan(-(a^2*b^4*sqrt(x)*(-b^5/a^9)^(1/4) - sqrt(-a^5*b^5*sqrt(-b^5/a^9) +
b^8*x)*a^2*(-b^5/a^9)^(1/4))/b^5) - 5*a^2*x^3*(-b^5/a^9)^(1/4)*log(a^7*(-b^5/a^9)^(3/4) + b^4*sqrt(x)) + 5*a^2
*x^3*(-b^5/a^9)^(1/4)*log(-a^7*(-b^5/a^9)^(3/4) + b^4*sqrt(x)) - 4*(5*b*x^2 - a)*sqrt(x))/(a^2*x^3)

________________________________________________________________________________________

giac [A]  time = 0.62, size = 200, normalized size = 0.93 \[ \frac {\sqrt {2} \left (a b^{3}\right )^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{2 \, a^{3} b} + \frac {\sqrt {2} \left (a b^{3}\right )^{\frac {3}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{2 \, a^{3} b} - \frac {\sqrt {2} \left (a b^{3}\right )^{\frac {3}{4}} \log \left (\sqrt {2} \sqrt {x} \left (\frac {a}{b}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{b}}\right )}{4 \, a^{3} b} + \frac {\sqrt {2} \left (a b^{3}\right )^{\frac {3}{4}} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {a}{b}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{b}}\right )}{4 \, a^{3} b} + \frac {2 \, {\left (5 \, b x^{2} - a\right )}}{5 \, a^{2} x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(7/2)/(b*x^2+a),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^(7/2)/(b*x^2+a),x)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 3.04, size = 198, normalized size = 0.92 \[ \frac {b^{2} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} + 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} - 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} - \frac {\sqrt {2} \log \left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}}\right )}}{4 \, a^{2}} + \frac {2 \, {\left (5 \, b x^{2} - a\right )}}{5 \, a^{2} x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(7/2)/(b*x^2+a),x, algorithm="maxima")

[Out]

1/4*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)))/(sq
rt(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))/sqr
t(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*x^2 - a)/(a^2*x^(5/2))

________________________________________________________________________________________

mupad [B]  time = 4.48, size = 66, normalized size = 0.31 \[ \frac {{\left (-b\right )}^{5/4}\,\mathrm {atanh}\left (\frac {{\left (-b\right )}^{1/4}\,\sqrt {x}}{a^{1/4}}\right )}{a^{9/4}}-\frac {{\left (-b\right )}^{5/4}\,\mathrm {atan}\left (\frac {{\left (-b\right )}^{1/4}\,\sqrt {x}}{a^{1/4}}\right )}{a^{9/4}}-\frac {\frac {2}{5\,a}-\frac {2\,b\,x^2}{a^2}}{x^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(7/2)*(a + b*x^2)),x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 109.16, size = 190, normalized size = 0.88 \[ \begin {cases} \frac {\tilde {\infty }}{x^{\frac {9}{2}}} & \text {for}\: a = 0 \wedge b = 0 \\- \frac {2}{9 b x^{\frac {9}{2}}} & \text {for}\: a = 0 \\- \frac {2}{5 a x^{\frac {5}{2}}} & \text {for}\: b = 0 \\- \frac {2}{5 a x^{\frac {5}{2}}} + \frac {2 b}{a^{2} \sqrt {x}} - \frac {\left (-1\right )^{\frac {3}{4}} b \log {\left (- \sqrt [4]{-1} \sqrt [4]{a} \sqrt [4]{\frac {1}{b}} + \sqrt {x} \right )}}{2 a^{\frac {9}{4}} \sqrt [4]{\frac {1}{b}}} + \frac {\left (-1\right )^{\frac {3}{4}} b \log {\left (\sqrt [4]{-1} \sqrt [4]{a} \sqrt [4]{\frac {1}{b}} + \sqrt {x} \right )}}{2 a^{\frac {9}{4}} \sqrt [4]{\frac {1}{b}}} + \frac {\left (-1\right )^{\frac {3}{4}} b \operatorname {atan}{\left (\frac {\left (-1\right )^{\frac {3}{4}} \sqrt {x}}{\sqrt [4]{a} \sqrt [4]{\frac {1}{b}}} \right )}}{a^{\frac {9}{4}} \sqrt [4]{\frac {1}{b}}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**(7/2)/(b*x**2+a),x)

[Out]

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))

________________________________________________________________________________________