3.3.1 \(\int \frac {x^{7/2}}{b x^2+c x^4} \, dx\)

Optimal. Leaf size=202 \[ \frac {\sqrt [4]{b} \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} c^{5/4}}-\frac {\sqrt [4]{b} \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} c^{5/4}}+\frac {\sqrt [4]{b} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} c^{5/4}}-\frac {\sqrt [4]{b} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} c^{5/4}}+\frac {2 \sqrt {x}}{c} \]

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Rubi [A]  time = 0.18, antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.474, Rules used = {1584, 321, 329, 211, 1165, 628, 1162, 617, 204} \begin {gather*} \frac {\sqrt [4]{b} \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} c^{5/4}}-\frac {\sqrt [4]{b} \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} c^{5/4}}+\frac {\sqrt [4]{b} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} c^{5/4}}-\frac {\sqrt [4]{b} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} c^{5/4}}+\frac {2 \sqrt {x}}{c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(7/2)/(b*x^2 + c*x^4),x]

[Out]

(2*Sqrt[x])/c + (b^(1/4)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(Sqrt[2]*c^(5/4)) - (b^(1/4)*ArcTan[1
+ (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(Sqrt[2]*c^(5/4)) + (b^(1/4)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[
x] + Sqrt[c]*x])/(2*Sqrt[2]*c^(5/4)) - (b^(1/4)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(2
*Sqrt[2]*c^(5/4))

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]

Rule 1584

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 189, normalized size = 0.94 \begin {gather*} \frac {\sqrt {2} \sqrt [4]{b} \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )-\sqrt {2} \sqrt [4]{b} \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )+2 \sqrt {2} \sqrt [4]{b} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )-2 \sqrt {2} \sqrt [4]{b} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )+8 \sqrt [4]{c} \sqrt {x}}{4 c^{5/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(7/2)/(b*x^2 + c*x^4),x]

[Out]

(8*c^(1/4)*Sqrt[x] + 2*Sqrt[2]*b^(1/4)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)] - 2*Sqrt[2]*b^(1/4)*ArcTa
n[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)] + Sqrt[2]*b^(1/4)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqr
t[c]*x] - Sqrt[2]*b^(1/4)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(4*c^(5/4))

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IntegrateAlgebraic [A]  time = 0.18, size = 123, normalized size = 0.61 \begin {gather*} \frac {\sqrt [4]{b} \tan ^{-1}\left (\frac {\frac {\sqrt [4]{b}}{\sqrt {2} \sqrt [4]{c}}-\frac {\sqrt [4]{c} x}{\sqrt {2} \sqrt [4]{b}}}{\sqrt {x}}\right )}{\sqrt {2} c^{5/4}}-\frac {\sqrt [4]{b} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{\sqrt {2} c^{5/4}}+\frac {2 \sqrt {x}}{c} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^(7/2)/(b*x^2 + c*x^4),x]

[Out]

(2*Sqrt[x])/c + (b^(1/4)*ArcTan[(b^(1/4)/(Sqrt[2]*c^(1/4)) - (c^(1/4)*x)/(Sqrt[2]*b^(1/4)))/Sqrt[x]])/(Sqrt[2]
*c^(5/4)) - (b^(1/4)*ArcTanh[(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])/(Sqrt[b] + Sqrt[c]*x)])/(Sqrt[2]*c^(5/4))

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fricas [A]  time = 0.90, size = 124, normalized size = 0.61 \begin {gather*} -\frac {4 \, c \left (-\frac {b}{c^{5}}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {c^{2} \sqrt {-\frac {b}{c^{5}}} + x} c^{4} \left (-\frac {b}{c^{5}}\right )^{\frac {3}{4}} - c^{4} \sqrt {x} \left (-\frac {b}{c^{5}}\right )^{\frac {3}{4}}}{b}\right ) + c \left (-\frac {b}{c^{5}}\right )^{\frac {1}{4}} \log \left (c \left (-\frac {b}{c^{5}}\right )^{\frac {1}{4}} + \sqrt {x}\right ) - c \left (-\frac {b}{c^{5}}\right )^{\frac {1}{4}} \log \left (-c \left (-\frac {b}{c^{5}}\right )^{\frac {1}{4}} + \sqrt {x}\right ) - 4 \, \sqrt {x}}{2 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)/(c*x^4+b*x^2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.16, size = 178, normalized size = 0.88 \begin {gather*} -\frac {\sqrt {2} \left (b c^{3}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{2 \, c^{2}} - \frac {\sqrt {2} \left (b c^{3}\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{2 \, c^{2}} - \frac {\sqrt {2} \left (b c^{3}\right )^{\frac {1}{4}} \log \left (\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{4 \, c^{2}} + \frac {\sqrt {2} \left (b c^{3}\right )^{\frac {1}{4}} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{4 \, c^{2}} + \frac {2 \, \sqrt {x}}{c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)/(c*x^4+b*x^2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 140, normalized size = 0.69 \begin {gather*} -\frac {\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{2 c}-\frac {\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{2 c}-\frac {\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}\right )}{4 c}+\frac {2 \sqrt {x}}{c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(7/2)/(c*x^4+b*x^2),x)

[Out]

2*x^(1/2)/c-1/4/c*(b/c)^(1/4)*2^(1/2)*ln((x+(b/c)^(1/4)*2^(1/2)*x^(1/2)+(b/c)^(1/2))/(x-(b/c)^(1/4)*2^(1/2)*x^
(1/2)+(b/c)^(1/2)))-1/2/c*(b/c)^(1/4)*2^(1/2)*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)-1/2/c*(b/c)^(1/4)*2^(1/2)*
arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)

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maxima [A]  time = 3.05, size = 185, normalized size = 0.92 \begin {gather*} -\frac {\frac {2 \, \sqrt {2} \sqrt {b} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} + 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}}} + \frac {2 \, \sqrt {2} \sqrt {b} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} - 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}}} + \frac {\sqrt {2} b^{\frac {1}{4}} \log \left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{c^{\frac {1}{4}}} - \frac {\sqrt {2} b^{\frac {1}{4}} \log \left (-\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{c^{\frac {1}{4}}}}{4 \, c} + \frac {2 \, \sqrt {x}}{c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)/(c*x^4+b*x^2),x, algorithm="maxima")

[Out]

-1/4*(2*sqrt(2)*sqrt(b)*arctan(1/2*sqrt(2)*(sqrt(2)*b^(1/4)*c^(1/4) + 2*sqrt(c)*sqrt(x))/sqrt(sqrt(b)*sqrt(c))
)/sqrt(sqrt(b)*sqrt(c)) + 2*sqrt(2)*sqrt(b)*arctan(-1/2*sqrt(2)*(sqrt(2)*b^(1/4)*c^(1/4) - 2*sqrt(c)*sqrt(x))/
sqrt(sqrt(b)*sqrt(c)))/sqrt(sqrt(b)*sqrt(c)) + sqrt(2)*b^(1/4)*log(sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x
+ sqrt(b))/c^(1/4) - sqrt(2)*b^(1/4)*log(-sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x + sqrt(b))/c^(1/4))/c +
2*sqrt(x)/c

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mupad [B]  time = 4.36, size = 55, normalized size = 0.27 \begin {gather*} \frac {2\,\sqrt {x}}{c}-\frac {{\left (-b\right )}^{1/4}\,\mathrm {atan}\left (\frac {c^{1/4}\,\sqrt {x}}{{\left (-b\right )}^{1/4}}\right )}{c^{5/4}}-\frac {{\left (-b\right )}^{1/4}\,\mathrm {atanh}\left (\frac {c^{1/4}\,\sqrt {x}}{{\left (-b\right )}^{1/4}}\right )}{c^{5/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(7/2)/(b*x^2 + c*x^4),x)

[Out]

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

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sympy [A]  time = 77.50, size = 172, normalized size = 0.85 \begin {gather*} \begin {cases} \tilde {\infty } \sqrt {x} & \text {for}\: b = 0 \wedge c = 0 \\\frac {2 x^{\frac {5}{2}}}{5 b} & \text {for}\: c = 0 \\\frac {2 \sqrt {x}}{c} & \text {for}\: b = 0 \\\frac {\sqrt [4]{-1} \sqrt [4]{b} \sqrt [4]{\frac {1}{c}} \log {\left (- \sqrt [4]{-1} \sqrt [4]{b} \sqrt [4]{\frac {1}{c}} + \sqrt {x} \right )}}{2 c} - \frac {\sqrt [4]{-1} \sqrt [4]{b} \sqrt [4]{\frac {1}{c}} \log {\left (\sqrt [4]{-1} \sqrt [4]{b} \sqrt [4]{\frac {1}{c}} + \sqrt {x} \right )}}{2 c} + \frac {\sqrt [4]{-1} \sqrt [4]{b} \sqrt [4]{\frac {1}{c}} \operatorname {atan}{\left (\frac {\left (-1\right )^{\frac {3}{4}} \sqrt {x}}{\sqrt [4]{b} \sqrt [4]{\frac {1}{c}}} \right )}}{c} + \frac {2 \sqrt {x}}{c} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(7/2)/(c*x**4+b*x**2),x)

[Out]

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

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