3.3.7 \(\int \frac {1}{x^{5/2} (b x^2+c x^4)} \, dx\)

Optimal. Leaf size=217 \[ -\frac {c^{7/4} \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} b^{11/4}}+\frac {c^{7/4} \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} b^{11/4}}-\frac {c^{7/4} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} b^{11/4}}+\frac {c^{7/4} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} b^{11/4}}+\frac {2 c}{3 b^2 x^{3/2}}-\frac {2}{7 b x^{7/2}} \]

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Rubi [A]  time = 0.18, antiderivative size = 217, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.474, Rules used = {1584, 325, 329, 211, 1165, 628, 1162, 617, 204} \begin {gather*} -\frac {c^{7/4} \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} b^{11/4}}+\frac {c^{7/4} \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} b^{11/4}}-\frac {c^{7/4} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} b^{11/4}}+\frac {c^{7/4} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} b^{11/4}}+\frac {2 c}{3 b^2 x^{3/2}}-\frac {2}{7 b x^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^(5/2)*(b*x^2 + c*x^4)),x]

[Out]

-2/(7*b*x^(7/2)) + (2*c)/(3*b^2*x^(3/2)) - (c^(7/4)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(Sqrt[2]*b^
(11/4)) + (c^(7/4)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(Sqrt[2]*b^(11/4)) - (c^(7/4)*Log[Sqrt[b] -
Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(2*Sqrt[2]*b^(11/4)) + (c^(7/4)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^
(1/4)*Sqrt[x] + Sqrt[c]*x])/(2*Sqrt[2]*b^(11/4))

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 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]

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 {1}{x^{5/2} \left (b x^2+c x^4\right )} \, dx &=\int \frac {1}{x^{9/2} \left (b+c x^2\right )} \, dx\\ &=-\frac {2}{7 b x^{7/2}}-\frac {c \int \frac {1}{x^{5/2} \left (b+c x^2\right )} \, dx}{b}\\ &=-\frac {2}{7 b x^{7/2}}+\frac {2 c}{3 b^2 x^{3/2}}+\frac {c^2 \int \frac {1}{\sqrt {x} \left (b+c x^2\right )} \, dx}{b^2}\\ &=-\frac {2}{7 b x^{7/2}}+\frac {2 c}{3 b^2 x^{3/2}}+\frac {\left (2 c^2\right ) \operatorname {Subst}\left (\int \frac {1}{b+c x^4} \, dx,x,\sqrt {x}\right )}{b^2}\\ &=-\frac {2}{7 b x^{7/2}}+\frac {2 c}{3 b^2 x^{3/2}}+\frac {c^2 \operatorname {Subst}\left (\int \frac {\sqrt {b}-\sqrt {c} x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{b^{5/2}}+\frac {c^2 \operatorname {Subst}\left (\int \frac {\sqrt {b}+\sqrt {c} x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{b^{5/2}}\\ &=-\frac {2}{7 b x^{7/2}}+\frac {2 c}{3 b^2 x^{3/2}}+\frac {c^{3/2} \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 b^{5/2}}+\frac {c^{3/2} \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 b^{5/2}}-\frac {c^{7/4} \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} b^{11/4}}-\frac {c^{7/4} \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} b^{11/4}}\\ &=-\frac {2}{7 b x^{7/2}}+\frac {2 c}{3 b^2 x^{3/2}}-\frac {c^{7/4} \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{2 \sqrt {2} b^{11/4}}+\frac {c^{7/4} \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{2 \sqrt {2} b^{11/4}}+\frac {c^{7/4} \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} b^{11/4}}-\frac {c^{7/4} \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} b^{11/4}}\\ &=-\frac {2}{7 b x^{7/2}}+\frac {2 c}{3 b^2 x^{3/2}}-\frac {c^{7/4} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} b^{11/4}}+\frac {c^{7/4} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} b^{11/4}}-\frac {c^{7/4} \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{2 \sqrt {2} b^{11/4}}+\frac {c^{7/4} \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{2 \sqrt {2} b^{11/4}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 29, normalized size = 0.13 \begin {gather*} -\frac {2 \, _2F_1\left (-\frac {7}{4},1;-\frac {3}{4};-\frac {c x^2}{b}\right )}{7 b x^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(5/2)*(b*x^2 + c*x^4)),x]

[Out]

(-2*Hypergeometric2F1[-7/4, 1, -3/4, -((c*x^2)/b)])/(7*b*x^(7/2))

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IntegrateAlgebraic [A]  time = 0.19, size = 135, normalized size = 0.62 \begin {gather*} -\frac {c^{7/4} \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} b^{11/4}}+\frac {c^{7/4} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{\sqrt {2} b^{11/4}}-\frac {2 \left (3 b-7 c x^2\right )}{21 b^2 x^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(x^(5/2)*(b*x^2 + c*x^4)),x]

[Out]

(-2*(3*b - 7*c*x^2))/(21*b^2*x^(7/2)) - (c^(7/4)*ArcTan[(b^(1/4)/(Sqrt[2]*c^(1/4)) - (c^(1/4)*x)/(Sqrt[2]*b^(1
/4)))/Sqrt[x]])/(Sqrt[2]*b^(11/4)) + (c^(7/4)*ArcTanh[(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])/(Sqrt[b] + Sqrt[c]*x)]
)/(Sqrt[2]*b^(11/4))

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fricas [A]  time = 1.55, size = 189, normalized size = 0.87 \begin {gather*} \frac {84 \, b^{2} x^{4} \left (-\frac {c^{7}}{b^{11}}\right )^{\frac {1}{4}} \arctan \left (-\frac {b^{8} c^{2} \sqrt {x} \left (-\frac {c^{7}}{b^{11}}\right )^{\frac {3}{4}} - \sqrt {b^{6} \sqrt {-\frac {c^{7}}{b^{11}}} + c^{4} x} b^{8} \left (-\frac {c^{7}}{b^{11}}\right )^{\frac {3}{4}}}{c^{7}}\right ) + 21 \, b^{2} x^{4} \left (-\frac {c^{7}}{b^{11}}\right )^{\frac {1}{4}} \log \left (b^{3} \left (-\frac {c^{7}}{b^{11}}\right )^{\frac {1}{4}} + c^{2} \sqrt {x}\right ) - 21 \, b^{2} x^{4} \left (-\frac {c^{7}}{b^{11}}\right )^{\frac {1}{4}} \log \left (-b^{3} \left (-\frac {c^{7}}{b^{11}}\right )^{\frac {1}{4}} + c^{2} \sqrt {x}\right ) + 4 \, {\left (7 \, c x^{2} - 3 \, b\right )} \sqrt {x}}{42 \, b^{2} x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/2)/(c*x^4+b*x^2),x, algorithm="fricas")

[Out]

1/42*(84*b^2*x^4*(-c^7/b^11)^(1/4)*arctan(-(b^8*c^2*sqrt(x)*(-c^7/b^11)^(3/4) - sqrt(b^6*sqrt(-c^7/b^11) + c^4
*x)*b^8*(-c^7/b^11)^(3/4))/c^7) + 21*b^2*x^4*(-c^7/b^11)^(1/4)*log(b^3*(-c^7/b^11)^(1/4) + c^2*sqrt(x)) - 21*b
^2*x^4*(-c^7/b^11)^(1/4)*log(-b^3*(-c^7/b^11)^(1/4) + c^2*sqrt(x)) + 4*(7*c*x^2 - 3*b)*sqrt(x))/(b^2*x^4)

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giac [A]  time = 0.20, size = 192, normalized size = 0.88 \begin {gather*} \frac {\sqrt {2} \left (b c^{3}\right )^{\frac {1}{4}} c \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 \, b^{3}} + \frac {\sqrt {2} \left (b c^{3}\right )^{\frac {1}{4}} c \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 \, b^{3}} + \frac {\sqrt {2} \left (b c^{3}\right )^{\frac {1}{4}} c \log \left (\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{4 \, b^{3}} - \frac {\sqrt {2} \left (b c^{3}\right )^{\frac {1}{4}} c \log \left (-\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{4 \, b^{3}} + \frac {2 \, {\left (7 \, c x^{2} - 3 \, b\right )}}{21 \, b^{2} x^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/2)/(c*x^4+b*x^2),x, algorithm="giac")

[Out]

1/2*sqrt(2)*(b*c^3)^(1/4)*c*arctan(1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4) + 2*sqrt(x))/(b/c)^(1/4))/b^3 + 1/2*sqrt(2
)*(b*c^3)^(1/4)*c*arctan(-1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4) - 2*sqrt(x))/(b/c)^(1/4))/b^3 + 1/4*sqrt(2)*(b*c^3)
^(1/4)*c*log(sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/b^3 - 1/4*sqrt(2)*(b*c^3)^(1/4)*c*log(-sqrt(2)*sqrt(
x)*(b/c)^(1/4) + x + sqrt(b/c))/b^3 + 2/21*(7*c*x^2 - 3*b)/(b^2*x^(7/2))

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maple [A]  time = 0.01, size = 158, normalized size = 0.73 \begin {gather*} \frac {\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, c^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{2 b^{3}}+\frac {\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, c^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{2 b^{3}}+\frac {\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, c^{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 b^{3}}+\frac {2 c}{3 b^{2} x^{\frac {3}{2}}}-\frac {2}{7 b \,x^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^(5/2)/(c*x^4+b*x^2),x)

[Out]

1/4*c^2/b^3*(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^2/b^3*(b/c)^(1/4)*2^(1/2)*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)+1/2*c^2/b^3*(b/c)^(1/4)*2^
(1/2)*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)-2/7/b/x^(7/2)+2/3*c/b^2/x^(3/2)

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maxima [A]  time = 3.12, size = 201, normalized size = 0.93 \begin {gather*} \frac {\frac {2 \, \sqrt {2} c^{2} \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 {b} \sqrt {\sqrt {b} \sqrt {c}}} + \frac {2 \, \sqrt {2} c^{2} \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 {b} \sqrt {\sqrt {b} \sqrt {c}}} + \frac {\sqrt {2} c^{\frac {7}{4}} \log \left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {3}{4}}} - \frac {\sqrt {2} c^{\frac {7}{4}} \log \left (-\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {3}{4}}}}{4 \, b^{2}} + \frac {2 \, {\left (7 \, c x^{2} - 3 \, b\right )}}{21 \, b^{2} x^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/2)/(c*x^4+b*x^2),x, algorithm="maxima")

[Out]

1/4*(2*sqrt(2)*c^2*arctan(1/2*sqrt(2)*(sqrt(2)*b^(1/4)*c^(1/4) + 2*sqrt(c)*sqrt(x))/sqrt(sqrt(b)*sqrt(c)))/(sq
rt(b)*sqrt(sqrt(b)*sqrt(c))) + 2*sqrt(2)*c^2*arctan(-1/2*sqrt(2)*(sqrt(2)*b^(1/4)*c^(1/4) - 2*sqrt(c)*sqrt(x))
/sqrt(sqrt(b)*sqrt(c)))/(sqrt(b)*sqrt(sqrt(b)*sqrt(c))) + sqrt(2)*c^(7/4)*log(sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x)
+ sqrt(c)*x + sqrt(b))/b^(3/4) - sqrt(2)*c^(7/4)*log(-sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x + sqrt(b))/b
^(3/4))/b^2 + 2/21*(7*c*x^2 - 3*b)/(b^2*x^(7/2))

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mupad [B]  time = 4.39, size = 65, normalized size = 0.30 \begin {gather*} \frac {{\left (-c\right )}^{7/4}\,\mathrm {atan}\left (\frac {{\left (-c\right )}^{1/4}\,\sqrt {x}}{b^{1/4}}\right )}{b^{11/4}}-\frac {\frac {2}{7\,b}-\frac {2\,c\,x^2}{3\,b^2}}{x^{7/2}}+\frac {{\left (-c\right )}^{7/4}\,\mathrm {atanh}\left (\frac {{\left (-c\right )}^{1/4}\,\sqrt {x}}{b^{1/4}}\right )}{b^{11/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(5/2)*(b*x^2 + c*x^4)),x)

[Out]

((-c)^(7/4)*atan(((-c)^(1/4)*x^(1/2))/b^(1/4)))/b^(11/4) - (2/(7*b) - (2*c*x^2)/(3*b^2))/x^(7/2) + ((-c)^(7/4)
*atanh(((-c)^(1/4)*x^(1/2))/b^(1/4)))/b^(11/4)

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sympy [A]  time = 106.89, size = 197, normalized size = 0.91 \begin {gather*} \begin {cases} \frac {\tilde {\infty }}{x^{\frac {11}{2}}} & \text {for}\: b = 0 \wedge c = 0 \\- \frac {2}{7 b x^{\frac {7}{2}}} & \text {for}\: c = 0 \\- \frac {2}{11 c x^{\frac {11}{2}}} & \text {for}\: b = 0 \\- \frac {2}{7 b x^{\frac {7}{2}}} + \frac {2 c}{3 b^{2} x^{\frac {3}{2}}} - \frac {\sqrt [4]{-1} c^{2} \sqrt [4]{\frac {1}{c}} \log {\left (- \sqrt [4]{-1} \sqrt [4]{b} \sqrt [4]{\frac {1}{c}} + \sqrt {x} \right )}}{2 b^{\frac {11}{4}}} + \frac {\sqrt [4]{-1} c^{2} \sqrt [4]{\frac {1}{c}} \log {\left (\sqrt [4]{-1} \sqrt [4]{b} \sqrt [4]{\frac {1}{c}} + \sqrt {x} \right )}}{2 b^{\frac {11}{4}}} - \frac {\sqrt [4]{-1} c^{2} \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 )}}{b^{\frac {11}{4}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**(5/2)/(c*x**4+b*x**2),x)

[Out]

Piecewise((zoo/x**(11/2), Eq(b, 0) & Eq(c, 0)), (-2/(7*b*x**(7/2)), Eq(c, 0)), (-2/(11*c*x**(11/2)), Eq(b, 0))
, (-2/(7*b*x**(7/2)) + 2*c/(3*b**2*x**(3/2)) - (-1)**(1/4)*c**2*(1/c)**(1/4)*log(-(-1)**(1/4)*b**(1/4)*(1/c)**
(1/4) + sqrt(x))/(2*b**(11/4)) + (-1)**(1/4)*c**2*(1/c)**(1/4)*log((-1)**(1/4)*b**(1/4)*(1/c)**(1/4) + sqrt(x)
)/(2*b**(11/4)) - (-1)**(1/4)*c**2*(1/c)**(1/4)*atan((-1)**(3/4)*sqrt(x)/(b**(1/4)*(1/c)**(1/4)))/b**(11/4), T
rue))

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