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

Optimal. Leaf size=243 \[ \frac {9 b^{5/4} \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{8 \sqrt {2} a^{13/4}}-\frac {9 b^{5/4} \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{8 \sqrt {2} a^{13/4}}-\frac {9 b^{5/4} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} a^{13/4}}+\frac {9 b^{5/4} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{4 \sqrt {2} a^{13/4}}+\frac {9 b}{2 a^3 \sqrt {x}}-\frac {9}{10 a^2 x^{5/2}}+\frac {1}{2 a x^{5/2} \left (a+b x^2\right )} \]

[Out]

-9/10/a^2/x^(5/2)+1/2/a/x^(5/2)/(b*x^2+a)-9/8*b^(5/4)*arctan(1-b^(1/4)*2^(1/2)*x^(1/2)/a^(1/4))/a^(13/4)*2^(1/
2)+9/8*b^(5/4)*arctan(1+b^(1/4)*2^(1/2)*x^(1/2)/a^(1/4))/a^(13/4)*2^(1/2)+9/16*b^(5/4)*ln(a^(1/2)+x*b^(1/2)-a^
(1/4)*b^(1/4)*2^(1/2)*x^(1/2))/a^(13/4)*2^(1/2)-9/16*b^(5/4)*ln(a^(1/2)+x*b^(1/2)+a^(1/4)*b^(1/4)*2^(1/2)*x^(1
/2))/a^(13/4)*2^(1/2)+9/2*b/a^3/x^(1/2)

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Rubi [A]  time = 0.19, antiderivative size = 243, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {290, 325, 329, 297, 1162, 617, 204, 1165, 628} \[ \frac {9 b^{5/4} \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{8 \sqrt {2} a^{13/4}}-\frac {9 b^{5/4} \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{8 \sqrt {2} a^{13/4}}-\frac {9 b^{5/4} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} a^{13/4}}+\frac {9 b^{5/4} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{4 \sqrt {2} a^{13/4}}+\frac {9 b}{2 a^3 \sqrt {x}}-\frac {9}{10 a^2 x^{5/2}}+\frac {1}{2 a x^{5/2} \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^(7/2)*(a + b*x^2)^2),x]

[Out]

-9/(10*a^2*x^(5/2)) + (9*b)/(2*a^3*Sqrt[x]) + 1/(2*a*x^(5/2)*(a + b*x^2)) - (9*b^(5/4)*ArcTan[1 - (Sqrt[2]*b^(
1/4)*Sqrt[x])/a^(1/4)])/(4*Sqrt[2]*a^(13/4)) + (9*b^(5/4)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(4*Sq
rt[2]*a^(13/4)) + (9*b^(5/4)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(8*Sqrt[2]*a^(13/4))
- (9*b^(5/4)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(8*Sqrt[2]*a^(13/4))

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 290

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(
a*c*n*(p + 1)), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && 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 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 )^2} \, dx &=\frac {1}{2 a x^{5/2} \left (a+b x^2\right )}+\frac {9 \int \frac {1}{x^{7/2} \left (a+b x^2\right )} \, dx}{4 a}\\ &=-\frac {9}{10 a^2 x^{5/2}}+\frac {1}{2 a x^{5/2} \left (a+b x^2\right )}-\frac {(9 b) \int \frac {1}{x^{3/2} \left (a+b x^2\right )} \, dx}{4 a^2}\\ &=-\frac {9}{10 a^2 x^{5/2}}+\frac {9 b}{2 a^3 \sqrt {x}}+\frac {1}{2 a x^{5/2} \left (a+b x^2\right )}+\frac {\left (9 b^2\right ) \int \frac {\sqrt {x}}{a+b x^2} \, dx}{4 a^3}\\ &=-\frac {9}{10 a^2 x^{5/2}}+\frac {9 b}{2 a^3 \sqrt {x}}+\frac {1}{2 a x^{5/2} \left (a+b x^2\right )}+\frac {\left (9 b^2\right ) \operatorname {Subst}\left (\int \frac {x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{2 a^3}\\ &=-\frac {9}{10 a^2 x^{5/2}}+\frac {9 b}{2 a^3 \sqrt {x}}+\frac {1}{2 a x^{5/2} \left (a+b x^2\right )}-\frac {\left (9 b^{3/2}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a}-\sqrt {b} x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{4 a^3}+\frac {\left (9 b^{3/2}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a}+\sqrt {b} x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{4 a^3}\\ &=-\frac {9}{10 a^2 x^{5/2}}+\frac {9 b}{2 a^3 \sqrt {x}}+\frac {1}{2 a x^{5/2} \left (a+b x^2\right )}+\frac {(9 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 )}{8 a^3}+\frac {(9 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 )}{8 a^3}+\frac {\left (9 b^{5/4}\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} a^{13/4}}+\frac {\left (9 b^{5/4}\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} a^{13/4}}\\ &=-\frac {9}{10 a^2 x^{5/2}}+\frac {9 b}{2 a^3 \sqrt {x}}+\frac {1}{2 a x^{5/2} \left (a+b x^2\right )}+\frac {9 b^{5/4} \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{8 \sqrt {2} a^{13/4}}-\frac {9 b^{5/4} \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{8 \sqrt {2} a^{13/4}}+\frac {\left (9 b^{5/4}\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} a^{13/4}}-\frac {\left (9 b^{5/4}\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} a^{13/4}}\\ &=-\frac {9}{10 a^2 x^{5/2}}+\frac {9 b}{2 a^3 \sqrt {x}}+\frac {1}{2 a x^{5/2} \left (a+b x^2\right )}-\frac {9 b^{5/4} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} a^{13/4}}+\frac {9 b^{5/4} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} a^{13/4}}+\frac {9 b^{5/4} \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{8 \sqrt {2} a^{13/4}}-\frac {9 b^{5/4} \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{8 \sqrt {2} a^{13/4}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 29, normalized size = 0.12 \[ -\frac {2 \, _2F_1\left (-\frac {5}{4},2;-\frac {1}{4};-\frac {b x^2}{a}\right )}{5 a^2 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(7/2)*(a + b*x^2)^2),x]

[Out]

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

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fricas [A]  time = 0.49, size = 251, normalized size = 1.03 \[ -\frac {180 \, {\left (a^{3} b x^{5} + a^{4} x^{3}\right )} \left (-\frac {b^{5}}{a^{13}}\right )^{\frac {1}{4}} \arctan \left (-\frac {729 \, a^{3} b^{4} \sqrt {x} \left (-\frac {b^{5}}{a^{13}}\right )^{\frac {1}{4}} - \sqrt {-531441 \, a^{7} b^{5} \sqrt {-\frac {b^{5}}{a^{13}}} + 531441 \, b^{8} x} a^{3} \left (-\frac {b^{5}}{a^{13}}\right )^{\frac {1}{4}}}{729 \, b^{5}}\right ) - 45 \, {\left (a^{3} b x^{5} + a^{4} x^{3}\right )} \left (-\frac {b^{5}}{a^{13}}\right )^{\frac {1}{4}} \log \left (729 \, a^{10} \left (-\frac {b^{5}}{a^{13}}\right )^{\frac {3}{4}} + 729 \, b^{4} \sqrt {x}\right ) + 45 \, {\left (a^{3} b x^{5} + a^{4} x^{3}\right )} \left (-\frac {b^{5}}{a^{13}}\right )^{\frac {1}{4}} \log \left (-729 \, a^{10} \left (-\frac {b^{5}}{a^{13}}\right )^{\frac {3}{4}} + 729 \, b^{4} \sqrt {x}\right ) - 4 \, {\left (45 \, b^{2} x^{4} + 36 \, a b x^{2} - 4 \, a^{2}\right )} \sqrt {x}}{40 \, {\left (a^{3} b x^{5} + a^{4} x^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/40*(180*(a^3*b*x^5 + a^4*x^3)*(-b^5/a^13)^(1/4)*arctan(-1/729*(729*a^3*b^4*sqrt(x)*(-b^5/a^13)^(1/4) - sqrt
(-531441*a^7*b^5*sqrt(-b^5/a^13) + 531441*b^8*x)*a^3*(-b^5/a^13)^(1/4))/b^5) - 45*(a^3*b*x^5 + a^4*x^3)*(-b^5/
a^13)^(1/4)*log(729*a^10*(-b^5/a^13)^(3/4) + 729*b^4*sqrt(x)) + 45*(a^3*b*x^5 + a^4*x^3)*(-b^5/a^13)^(1/4)*log
(-729*a^10*(-b^5/a^13)^(3/4) + 729*b^4*sqrt(x)) - 4*(45*b^2*x^4 + 36*a*b*x^2 - 4*a^2)*sqrt(x))/(a^3*b*x^5 + a^
4*x^3)

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giac [A]  time = 0.66, size = 220, normalized size = 0.91 \[ \frac {b^{2} x^{\frac {3}{2}}}{2 \, {\left (b x^{2} + a\right )} a^{3}} + \frac {9 \, \sqrt {2} \left (a b^{3}\right )^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{8 \, a^{4} b} + \frac {9 \, \sqrt {2} \left (a b^{3}\right )^{\frac {3}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{8 \, a^{4} b} - \frac {9 \, \sqrt {2} \left (a b^{3}\right )^{\frac {3}{4}} \log \left (\sqrt {2} \sqrt {x} \left (\frac {a}{b}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{b}}\right )}{16 \, a^{4} b} + \frac {9 \, \sqrt {2} \left (a b^{3}\right )^{\frac {3}{4}} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {a}{b}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{b}}\right )}{16 \, a^{4} b} + \frac {2 \, {\left (10 \, b x^{2} - a\right )}}{5 \, a^{3} x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b^2*x^(3/2)/((b*x^2 + a)*a^3) + 9/8*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^4*b) + 9/8*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^4*b) - 9/16*sqrt(2)*(a*b^3)^(3/4)*log(sqrt(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))/(a^4*b) + 9/1
6*sqrt(2)*(a*b^3)^(3/4)*log(-sqrt(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))/(a^4*b) + 2/5*(10*b*x^2 - a)/(a^3*x^
(5/2))

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maple [A]  time = 0.02, size = 172, normalized size = 0.71 \[ \frac {b^{2} x^{\frac {3}{2}}}{2 \left (b \,x^{2}+a \right ) a^{3}}+\frac {9 \sqrt {2}\, b \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )}{8 \left (\frac {a}{b}\right )^{\frac {1}{4}} a^{3}}+\frac {9 \sqrt {2}\, b \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )}{8 \left (\frac {a}{b}\right )^{\frac {1}{4}} a^{3}}+\frac {9 \sqrt {2}\, b \ln \left (\frac {x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{b}}}{x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{b}}}\right )}{16 \left (\frac {a}{b}\right )^{\frac {1}{4}} a^{3}}+\frac {4 b}{a^{3} \sqrt {x}}-\frac {2}{5 a^{2} x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5/a^2/x^(5/2)+4/a^3*b/x^(1/2)+1/2*b^2/a^3*x^(3/2)/(b*x^2+a)+9/16*b/a^3/(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)))+9/8*b/a^3/(a/b)^(1/4)*2^(1/2)*arct
an(2^(1/2)/(a/b)^(1/4)*x^(1/2)+1)+9/8*b/a^3/(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 = 3.04, size = 221, normalized size = 0.91 \[ \frac {45 \, b^{2} x^{4} + 36 \, a b x^{2} - 4 \, a^{2}}{10 \, {\left (a^{3} b x^{\frac {9}{2}} + a^{4} x^{\frac {5}{2}}\right )}} + \frac {9 \, b^{2} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} + 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} - 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} - \frac {\sqrt {2} \log \left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}}\right )}}{16 \, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*(45*b^2*x^4 + 36*a*b*x^2 - 4*a^2)/(a^3*b*x^(9/2) + a^4*x^(5/2)) + 9/16*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)))/(sqrt(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))/sqrt(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^3

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mupad [B]  time = 4.69, size = 87, normalized size = 0.36 \[ \frac {\frac {18\,b\,x^2}{5\,a^2}-\frac {2}{5\,a}+\frac {9\,b^2\,x^4}{2\,a^3}}{a\,x^{5/2}+b\,x^{9/2}}-\frac {9\,{\left (-b\right )}^{5/4}\,\mathrm {atan}\left (\frac {{\left (-b\right )}^{1/4}\,\sqrt {x}}{a^{1/4}}\right )}{4\,a^{13/4}}+\frac {9\,{\left (-b\right )}^{5/4}\,\mathrm {atanh}\left (\frac {{\left (-b\right )}^{1/4}\,\sqrt {x}}{a^{1/4}}\right )}{4\,a^{13/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(7/2)*(a + b*x^2)^2),x)

[Out]

((18*b*x^2)/(5*a^2) - 2/(5*a) + (9*b^2*x^4)/(2*a^3))/(a*x^(5/2) + b*x^(9/2)) - (9*(-b)^(5/4)*atan(((-b)^(1/4)*
x^(1/2))/a^(1/4)))/(4*a^(13/4)) + (9*(-b)^(5/4)*atanh(((-b)^(1/4)*x^(1/2))/a^(1/4)))/(4*a^(13/4))

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

[Out]

Timed out

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