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

Optimal. Leaf size=457 \[ -\frac {2 a d^3 \sqrt {d x} \left (a+b x^2\right )}{b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {2 d (d x)^{5/2} \left (a+b x^2\right )}{5 b \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {a^{5/4} d^{7/2} \left (a+b x^2\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{2 \sqrt {2} b^{9/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {a^{5/4} d^{7/2} \left (a+b x^2\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}+\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x\right )}{2 \sqrt {2} b^{9/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {a^{5/4} d^{7/2} \left (a+b x^2\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{\sqrt {2} b^{9/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {a^{5/4} d^{7/2} \left (a+b x^2\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}+1\right )}{\sqrt {2} b^{9/4} \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

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Rubi [A]  time = 0.32, antiderivative size = 457, 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, 321, 329, 211, 1165, 628, 1162, 617, 204} \begin {gather*} -\frac {2 a d^3 \sqrt {d x} \left (a+b x^2\right )}{b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {a^{5/4} d^{7/2} \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} b^{9/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {a^{5/4} d^{7/2} \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} b^{9/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {a^{5/4} d^{7/2} \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} b^{9/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {a^{5/4} d^{7/2} \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} b^{9/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {2 d (d x)^{5/2} \left (a+b x^2\right )}{5 b \sqrt {a^2+2 a b x^2+b^2 x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*d^3*Sqrt[d*x]*(a + b*x^2))/(b^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + (2*d*(d*x)^(5/2)*(a + b*x^2))/(5*b*Sq
rt[a^2 + 2*a*b*x^2 + b^2*x^4]) - (a^(5/4)*d^(7/2)*(a + b*x^2)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[d*x])/(a^(1/4)*
Sqrt[d])])/(Sqrt[2]*b^(9/4)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + (a^(5/4)*d^(7/2)*(a + b*x^2)*ArcTan[1 + (Sqrt[2
]*b^(1/4)*Sqrt[d*x])/(a^(1/4)*Sqrt[d])])/(Sqrt[2]*b^(9/4)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - (a^(5/4)*d^(7/2)*
(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]*b^(9/4)*S
qrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + (a^(5/4)*d^(7/2)*(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]*b^(9/4)*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 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 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 {(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 {(d x)^{7/2}}{a b+b^2 x^2} \, dx}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\\ &=\frac {2 d (d x)^{5/2} \left (a+b x^2\right )}{5 b \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {\left (a d^2 \left (a b+b^2 x^2\right )\right ) \int \frac {(d x)^{3/2}}{a b+b^2 x^2} \, dx}{b \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ &=-\frac {2 a d^3 \sqrt {d x} \left (a+b x^2\right )}{b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {2 d (d x)^{5/2} \left (a+b x^2\right )}{5 b \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (a^2 d^4 \left (a b+b^2 x^2\right )\right ) \int \frac {1}{\sqrt {d x} \left (a b+b^2 x^2\right )} \, dx}{b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ &=-\frac {2 a d^3 \sqrt {d x} \left (a+b x^2\right )}{b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {2 d (d x)^{5/2} \left (a+b x^2\right )}{5 b \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (2 a^2 d^3 \left (a b+b^2 x^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a b+\frac {b^2 x^4}{d^2}} \, dx,x,\sqrt {d x}\right )}{b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ &=-\frac {2 a d^3 \sqrt {d x} \left (a+b x^2\right )}{b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {2 d (d x)^{5/2} \left (a+b x^2\right )}{5 b \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (a^{3/2} d^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 )}{b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (a^{3/2} d^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 )}{b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ &=-\frac {2 a d^3 \sqrt {d x} \left (a+b x^2\right )}{b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {2 d (d x)^{5/2} \left (a+b x^2\right )}{5 b \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {\left (a^{5/4} d^{7/2} \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} b^{13/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {\left (a^{5/4} d^{7/2} \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} b^{13/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (a^{3/2} d^4 \left (a b+b^2 x^2\right )\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 b^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (a^{3/2} d^4 \left (a b+b^2 x^2\right )\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 b^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ &=-\frac {2 a d^3 \sqrt {d x} \left (a+b x^2\right )}{b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {2 d (d x)^{5/2} \left (a+b x^2\right )}{5 b \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {a^{5/4} d^{7/2} \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} b^{9/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {a^{5/4} d^{7/2} \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} b^{9/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (a^{5/4} d^{7/2} \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} b^{13/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {\left (a^{5/4} d^{7/2} \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} b^{13/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ &=-\frac {2 a d^3 \sqrt {d x} \left (a+b x^2\right )}{b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {2 d (d x)^{5/2} \left (a+b x^2\right )}{5 b \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {a^{5/4} d^{7/2} \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} b^{9/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {a^{5/4} d^{7/2} \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} b^{9/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {a^{5/4} d^{7/2} \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} b^{9/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {a^{5/4} d^{7/2} \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} b^{9/4} \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 238, normalized size = 0.52 \begin {gather*} \frac {d^3 \sqrt {d x} \left (a+b x^2\right ) \left (-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}\right )}{20 b^{9/4} \sqrt {x} \sqrt {\left (a+b x^2\right )^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d^3*Sqrt[d*x]*(a + b*x^2)*(-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)*Sqrt[x]*Sqrt[(a + b*x^2)^2])

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IntegrateAlgebraic [A]  time = 54.46, size = 217, normalized size = 0.47 \begin {gather*} \frac {\left (a d^2+b d^2 x^2\right ) \left (-\frac {a^{5/4} d^{7/2} \tan ^{-1}\left (\frac {\frac {\sqrt [4]{a} \sqrt {d}}{\sqrt {2} \sqrt [4]{b}}-\frac {\sqrt [4]{b} \sqrt {d} x}{\sqrt {2} \sqrt [4]{a}}}{\sqrt {d x}}\right )}{\sqrt {2} b^{9/4}}+\frac {a^{5/4} d^{7/2} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d} \sqrt {d x}}{\sqrt {a} d+\sqrt {b} d x}\right )}{\sqrt {2} b^{9/4}}+\frac {2 d \sqrt {d x} \left (b d^2 x^2-5 a d^2\right )}{5 b^2}\right )}{d^2 \sqrt {\frac {\left (a d^2+b d^2 x^2\right )^2}{d^4}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a*d^2 + b*d^2*x^2)*((2*d*Sqrt[d*x]*(-5*a*d^2 + b*d^2*x^2))/(5*b^2) - (a^(5/4)*d^(7/2)*ArcTan[((a^(1/4)*Sqrt[
d])/(Sqrt[2]*b^(1/4)) - (b^(1/4)*Sqrt[d]*x)/(Sqrt[2]*a^(1/4)))/Sqrt[d*x]])/(Sqrt[2]*b^(9/4)) + (a^(5/4)*d^(7/2
)*ArcTanh[(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[d]*Sqrt[d*x])/(Sqrt[a]*d + Sqrt[b]*d*x)])/(Sqrt[2]*b^(9/4))))/(d^2*Sqr
t[(a*d^2 + b*d^2*x^2)^2/d^4])

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fricas [A]  time = 0.49, size = 223, normalized size = 0.49 \begin {gather*} \frac {20 \, \left (-\frac {a^{5} d^{14}}{b^{9}}\right )^{\frac {1}{4}} b^{2} \arctan \left (-\frac {\left (-\frac {a^{5} d^{14}}{b^{9}}\right )^{\frac {3}{4}} \sqrt {d x} a b^{7} d^{3} - \left (-\frac {a^{5} d^{14}}{b^{9}}\right )^{\frac {3}{4}} \sqrt {a^{2} d^{7} x + \sqrt {-\frac {a^{5} d^{14}}{b^{9}}} b^{4}} b^{7}}{a^{5} d^{14}}\right ) + 5 \, \left (-\frac {a^{5} d^{14}}{b^{9}}\right )^{\frac {1}{4}} b^{2} \log \left (\sqrt {d x} a d^{3} + \left (-\frac {a^{5} d^{14}}{b^{9}}\right )^{\frac {1}{4}} b^{2}\right ) - 5 \, \left (-\frac {a^{5} d^{14}}{b^{9}}\right )^{\frac {1}{4}} b^{2} \log \left (\sqrt {d x} a d^{3} - \left (-\frac {a^{5} d^{14}}{b^{9}}\right )^{\frac {1}{4}} b^{2}\right ) + 4 \, {\left (b d^{3} x^{2} - 5 \, a d^{3}\right )} \sqrt {d x}}{10 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.20, size = 273, normalized size = 0.60 \begin {gather*} \frac {1}{20} \, d^{3} {\left (\frac {10 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {1}{4}} a \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 )}{b^{3}} + \frac {10 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {1}{4}} a \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 )}{b^{3}} + \frac {5 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {1}{4}} a \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 )}{b^{3}} - \frac {5 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {1}{4}} a \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 )}{b^{3}} + \frac {8 \, {\left (\sqrt {d x} b^{4} d^{10} x^{2} - 5 \, \sqrt {d x} a b^{3} d^{10}\right )}}{b^{5} d^{10}}\right )} \mathrm {sgn}\left (b x^{2} + a\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 239, normalized size = 0.52 \begin {gather*} \frac {\left (b \,x^{2}+a \right ) \left (10 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {2}\, a \,d^{2} \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 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {2}\, a \,d^{2} \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 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {2}\, a \,d^{2} \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 )-40 \sqrt {d x}\, a \,d^{2}+8 \left (d x \right )^{\frac {5}{2}} b \right ) d}{20 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.94, size = 266, normalized size = 0.58 \begin {gather*} \frac {\frac {5 \, {\left (\frac {\sqrt {2} d^{6} \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 {3}{4}} b^{\frac {1}{4}}} - \frac {\sqrt {2} d^{6} \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 {3}{4}} b^{\frac {1}{4}}} + \frac {2 \, \sqrt {2} d^{5} \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 {a}} + \frac {2 \, \sqrt {2} d^{5} \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 {a}}\right )} a^{2}}{b^{2}} + \frac {8 \, {\left (\left (d x\right )^{\frac {5}{2}} b d^{2} - 5 \, \sqrt {d x} a d^{4}\right )}}{b^{2}}}{20 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/20*(5*(sqrt(2)*d^6*log(sqrt(b)*d*x + sqrt(2)*(a*d^2)^(1/4)*sqrt(d*x)*b^(1/4) + sqrt(a)*d)/((a*d^2)^(3/4)*b^(
1/4)) - sqrt(2)*d^6*log(sqrt(b)*d*x - sqrt(2)*(a*d^2)^(1/4)*sqrt(d*x)*b^(1/4) + sqrt(a)*d)/((a*d^2)^(3/4)*b^(1
/4)) + 2*sqrt(2)*d^5*arctan(1/2*sqrt(2)*(sqrt(2)*(a*d^2)^(1/4)*b^(1/4) + 2*sqrt(d*x)*sqrt(b))/sqrt(sqrt(a)*sqr
t(b)*d))/(sqrt(sqrt(a)*sqrt(b)*d)*sqrt(a)) + 2*sqrt(2)*d^5*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(a)))*a^2/b^2 + 8*((d*x)^(5/2)*b*
d^2 - 5*sqrt(d*x)*a*d^4)/b^2)/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((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 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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