3.756 \(\int \frac {1}{(d x)^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx\)

Optimal. Leaf size=459 \[ \frac {2 b \left (a+b x^2\right )}{a^2 d^3 \sqrt {d x} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {2 \left (a+b x^2\right )}{5 a d (d x)^{5/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {b^{5/4} \left (a+b x^2\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{2 \sqrt {2} a^{9/4} d^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {b^{5/4} \left (a+b x^2\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{2 \sqrt {2} a^{9/4} d^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {b^{5/4} \left (a+b x^2\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{\sqrt {2} a^{9/4} d^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {b^{5/4} \left (a+b x^2\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}+1\right )}{\sqrt {2} a^{9/4} d^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

[Out]

-2/5*(b*x^2+a)/a/d/(d*x)^(5/2)/((b*x^2+a)^2)^(1/2)-1/2*b^(5/4)*(b*x^2+a)*arctan(1-b^(1/4)*2^(1/2)*(d*x)^(1/2)/
a^(1/4)/d^(1/2))/a^(9/4)/d^(7/2)*2^(1/2)/((b*x^2+a)^2)^(1/2)+1/2*b^(5/4)*(b*x^2+a)*arctan(1+b^(1/4)*2^(1/2)*(d
*x)^(1/2)/a^(1/4)/d^(1/2))/a^(9/4)/d^(7/2)*2^(1/2)/((b*x^2+a)^2)^(1/2)+1/4*b^(5/4)*(b*x^2+a)*ln(a^(1/2)*d^(1/2
)+x*b^(1/2)*d^(1/2)-a^(1/4)*b^(1/4)*2^(1/2)*(d*x)^(1/2))/a^(9/4)/d^(7/2)*2^(1/2)/((b*x^2+a)^2)^(1/2)-1/4*b^(5/
4)*(b*x^2+a)*ln(a^(1/2)*d^(1/2)+x*b^(1/2)*d^(1/2)+a^(1/4)*b^(1/4)*2^(1/2)*(d*x)^(1/2))/a^(9/4)/d^(7/2)*2^(1/2)
/((b*x^2+a)^2)^(1/2)+2*b*(b*x^2+a)/a^2/d^3/(d*x)^(1/2)/((b*x^2+a)^2)^(1/2)

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Rubi [A]  time = 0.33, antiderivative size = 459, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {1112, 325, 329, 297, 1162, 617, 204, 1165, 628} \[ \frac {2 b \left (a+b x^2\right )}{a^2 d^3 \sqrt {d x} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {b^{5/4} \left (a+b x^2\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{2 \sqrt {2} a^{9/4} d^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {b^{5/4} \left (a+b x^2\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{2 \sqrt {2} a^{9/4} d^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {b^{5/4} \left (a+b x^2\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{\sqrt {2} a^{9/4} d^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {b^{5/4} \left (a+b x^2\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}+1\right )}{\sqrt {2} a^{9/4} d^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {2 \left (a+b x^2\right )}{5 a d (d x)^{5/2} \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

Antiderivative was successfully verified.

[In]

Int[1/((d*x)^(7/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]),x]

[Out]

(-2*(a + b*x^2))/(5*a*d*(d*x)^(5/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + (2*b*(a + b*x^2))/(a^2*d^3*Sqrt[d*x]*Sq
rt[a^2 + 2*a*b*x^2 + b^2*x^4]) - (b^(5/4)*(a + b*x^2)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[d*x])/(a^(1/4)*Sqrt[d])
])/(Sqrt[2]*a^(9/4)*d^(7/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + (b^(5/4)*(a + b*x^2)*ArcTan[1 + (Sqrt[2]*b^(1/4
)*Sqrt[d*x])/(a^(1/4)*Sqrt[d])])/(Sqrt[2]*a^(9/4)*d^(7/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + (b^(5/4)*(a + b*x
^2)*Log[Sqrt[a]*Sqrt[d] + Sqrt[b]*Sqrt[d]*x - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[d*x]])/(2*Sqrt[2]*a^(9/4)*d^(7/2)*S
qrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - (b^(5/4)*(a + b*x^2)*Log[Sqrt[a]*Sqrt[d] + Sqrt[b]*Sqrt[d]*x + Sqrt[2]*a^(1/
4)*b^(1/4)*Sqrt[d*x]])/(2*Sqrt[2]*a^(9/4)*d^(7/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^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 1112

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(a + b*x^2 + c*x^4)^FracPa
rt[p]/(c^IntPart[p]*(b/2 + c*x^2)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^2)^(2*p), x], x] /; FreeQ[{a, b, c,
d, m, p}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

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

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Mathematica [C]  time = 0.01, size = 52, normalized size = 0.11 \[ -\frac {2 x \left (a+b x^2\right ) \, _2F_1\left (-\frac {5}{4},1;-\frac {1}{4};-\frac {b x^2}{a}\right )}{5 a (d x)^{7/2} \sqrt {\left (a+b x^2\right )^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((d*x)^(7/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]),x]

[Out]

(-2*x*(a + b*x^2)*Hypergeometric2F1[-5/4, 1, -1/4, -((b*x^2)/a)])/(5*a*(d*x)^(7/2)*Sqrt[(a + b*x^2)^2])

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fricas [A]  time = 1.12, size = 253, normalized size = 0.55 \[ -\frac {20 \, a^{2} d^{4} x^{3} \left (-\frac {b^{5}}{a^{9} d^{14}}\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {d x} a^{2} b^{4} d^{3} \left (-\frac {b^{5}}{a^{9} d^{14}}\right )^{\frac {1}{4}} - \sqrt {-a^{5} b^{5} d^{8} \sqrt {-\frac {b^{5}}{a^{9} d^{14}}} + b^{8} d x} a^{2} d^{3} \left (-\frac {b^{5}}{a^{9} d^{14}}\right )^{\frac {1}{4}}}{b^{5}}\right ) - 5 \, a^{2} d^{4} x^{3} \left (-\frac {b^{5}}{a^{9} d^{14}}\right )^{\frac {1}{4}} \log \left (a^{7} d^{11} \left (-\frac {b^{5}}{a^{9} d^{14}}\right )^{\frac {3}{4}} + \sqrt {d x} b^{4}\right ) + 5 \, a^{2} d^{4} x^{3} \left (-\frac {b^{5}}{a^{9} d^{14}}\right )^{\frac {1}{4}} \log \left (-a^{7} d^{11} \left (-\frac {b^{5}}{a^{9} d^{14}}\right )^{\frac {3}{4}} + \sqrt {d x} b^{4}\right ) - 4 \, {\left (5 \, b x^{2} - a\right )} \sqrt {d x}}{10 \, a^{2} d^{4} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/10*(20*a^2*d^4*x^3*(-b^5/(a^9*d^14))^(1/4)*arctan(-(sqrt(d*x)*a^2*b^4*d^3*(-b^5/(a^9*d^14))^(1/4) - sqrt(-a
^5*b^5*d^8*sqrt(-b^5/(a^9*d^14)) + b^8*d*x)*a^2*d^3*(-b^5/(a^9*d^14))^(1/4))/b^5) - 5*a^2*d^4*x^3*(-b^5/(a^9*d
^14))^(1/4)*log(a^7*d^11*(-b^5/(a^9*d^14))^(3/4) + sqrt(d*x)*b^4) + 5*a^2*d^4*x^3*(-b^5/(a^9*d^14))^(1/4)*log(
-a^7*d^11*(-b^5/(a^9*d^14))^(3/4) + sqrt(d*x)*b^4) - 4*(5*b*x^2 - a)*sqrt(d*x))/(a^2*d^4*x^3)

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giac [A]  time = 0.32, size = 284, normalized size = 0.62 \[ \frac {1}{20} \, {\left (\frac {10 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} + 2 \, \sqrt {d x}\right )}}{2 \, \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}}}\right )}{a^{3} b d^{5}} + \frac {10 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {3}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} - 2 \, \sqrt {d x}\right )}}{2 \, \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}}}\right )}{a^{3} b d^{5}} - \frac {5 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {3}{4}} \log \left (d x + \sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x} + \sqrt {\frac {a d^{2}}{b}}\right )}{a^{3} b d^{5}} + \frac {5 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {3}{4}} \log \left (d x - \sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x} + \sqrt {\frac {a d^{2}}{b}}\right )}{a^{3} b d^{5}} + \frac {8 \, {\left (5 \, b d^{2} x^{2} - a d^{2}\right )}}{\sqrt {d x} a^{2} d^{5} x^{2}}\right )} \mathrm {sgn}\left (b x^{2} + a\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 251, normalized size = 0.55 \[ \frac {\left (b \,x^{2}+a \right ) \left (40 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} b \,d^{2} x^{2}-8 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} a \,d^{2}+10 \sqrt {2}\, \left (d x \right )^{\frac {5}{2}} b \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}-\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}\right )+10 \sqrt {2}\, \left (d x \right )^{\frac {5}{2}} b \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}+\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}\right )+5 \sqrt {2}\, \left (d x \right )^{\frac {5}{2}} b \ln \left (-\frac {-d x +\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}-\sqrt {\frac {a \,d^{2}}{b}}}{d x +\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}\right )\right )}{20 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \left (d x \right )^{\frac {5}{2}} a^{2} d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/20*(b*x^2+a)/d^3*(5*b*2^(1/2)*(d*x)^(5/2)*ln(-(-d*x+(a/b*d^2)^(1/4)*(d*x)^(1/2)*2^(1/2)-(a/b*d^2)^(1/2))/(d*
x+(a/b*d^2)^(1/4)*(d*x)^(1/2)*2^(1/2)+(a/b*d^2)^(1/2)))+10*b*2^(1/2)*(d*x)^(5/2)*arctan((2^(1/2)*(d*x)^(1/2)+(
a/b*d^2)^(1/4))/(a/b*d^2)^(1/4))+10*b*2^(1/2)*(d*x)^(5/2)*arctan((2^(1/2)*(d*x)^(1/2)-(a/b*d^2)^(1/4))/(a/b*d^
2)^(1/4))+40*b*(a/b*d^2)^(1/4)*d^2*x^2-8*d^2*a*(a/b*d^2)^(1/4))/((b*x^2+a)^2)^(1/2)/a^2/(a/b*d^2)^(1/4)/(d*x)^
(5/2)

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maxima [A]  time = 3.03, size = 259, normalized size = 0.56 \[ \frac {\frac {5 \, b^{2} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {1}{4}} + 2 \, \sqrt {d x} \sqrt {b}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b} d}}\right )}{\sqrt {\sqrt {a} \sqrt {b} d} \sqrt {b}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {1}{4}} - 2 \, \sqrt {d x} \sqrt {b}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b} d}}\right )}{\sqrt {\sqrt {a} \sqrt {b} d} \sqrt {b}} - \frac {\sqrt {2} \log \left (\sqrt {b} d x + \sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} \sqrt {d x} b^{\frac {1}{4}} + \sqrt {a} d\right )}{\left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (\sqrt {b} d x - \sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} \sqrt {d x} b^{\frac {1}{4}} + \sqrt {a} d\right )}{\left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {3}{4}}}\right )}}{a^{2} d^{2}} + \frac {8 \, {\left (5 \, b d^{2} x^{2} - a d^{2}\right )}}{\left (d x\right )^{\frac {5}{2}} a^{2} d^{2}}}{20 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/20*(5*b^2*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*(a*d^2)^(1/4)*b^(1/4) + 2*sqrt(d*x)*sqrt(b))/sqrt(sqrt(a)*s
qrt(b)*d))/(sqrt(sqrt(a)*sqrt(b)*d)*sqrt(b)) + 2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*(a*d^2)^(1/4)*b^(1/4) -
2*sqrt(d*x)*sqrt(b))/sqrt(sqrt(a)*sqrt(b)*d))/(sqrt(sqrt(a)*sqrt(b)*d)*sqrt(b)) - sqrt(2)*log(sqrt(b)*d*x + sq
rt(2)*(a*d^2)^(1/4)*sqrt(d*x)*b^(1/4) + sqrt(a)*d)/((a*d^2)^(1/4)*b^(3/4)) + sqrt(2)*log(sqrt(b)*d*x - sqrt(2)
*(a*d^2)^(1/4)*sqrt(d*x)*b^(1/4) + sqrt(a)*d)/((a*d^2)^(1/4)*b^(3/4)))/(a^2*d^2) + 8*(5*b*d^2*x^2 - a*d^2)/((d
*x)^(5/2)*a^2*d^2))/d

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{{\left (d\,x\right )}^{7/2}\,\sqrt {{\left (b\,x^2+a\right )}^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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