3.329 \(\int \frac {x^{15/2}}{(b x^2+c x^4)^2} \, dx\)

Optimal. Leaf size=230 \[ \frac {5 \sqrt [4]{b} \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} c^{9/4}}-\frac {5 \sqrt [4]{b} \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} c^{9/4}}+\frac {5 \sqrt [4]{b} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} c^{9/4}}-\frac {5 \sqrt [4]{b} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{4 \sqrt {2} c^{9/4}}-\frac {x^{5/2}}{2 c \left (b+c x^2\right )}+\frac {5 \sqrt {x}}{2 c^2} \]

[Out]

-1/2*x^(5/2)/c/(c*x^2+b)+5/8*b^(1/4)*arctan(1-c^(1/4)*2^(1/2)*x^(1/2)/b^(1/4))/c^(9/4)*2^(1/2)-5/8*b^(1/4)*arc
tan(1+c^(1/4)*2^(1/2)*x^(1/2)/b^(1/4))/c^(9/4)*2^(1/2)+5/16*b^(1/4)*ln(b^(1/2)+x*c^(1/2)-b^(1/4)*c^(1/4)*2^(1/
2)*x^(1/2))/c^(9/4)*2^(1/2)-5/16*b^(1/4)*ln(b^(1/2)+x*c^(1/2)+b^(1/4)*c^(1/4)*2^(1/2)*x^(1/2))/c^(9/4)*2^(1/2)
+5/2*x^(1/2)/c^2

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Rubi [A]  time = 0.18, antiderivative size = 230, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.526, Rules used = {1584, 288, 321, 329, 211, 1165, 628, 1162, 617, 204} \[ \frac {5 \sqrt [4]{b} \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} c^{9/4}}-\frac {5 \sqrt [4]{b} \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} c^{9/4}}+\frac {5 \sqrt [4]{b} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} c^{9/4}}-\frac {5 \sqrt [4]{b} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{4 \sqrt {2} c^{9/4}}-\frac {x^{5/2}}{2 c \left (b+c x^2\right )}+\frac {5 \sqrt {x}}{2 c^2} \]

Antiderivative was successfully verified.

[In]

Int[x^(15/2)/(b*x^2 + c*x^4)^2,x]

[Out]

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

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 288

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*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

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^{15/2}}{\left (b x^2+c x^4\right )^2} \, dx &=\int \frac {x^{7/2}}{\left (b+c x^2\right )^2} \, dx\\ &=-\frac {x^{5/2}}{2 c \left (b+c x^2\right )}+\frac {5 \int \frac {x^{3/2}}{b+c x^2} \, dx}{4 c}\\ &=\frac {5 \sqrt {x}}{2 c^2}-\frac {x^{5/2}}{2 c \left (b+c x^2\right )}-\frac {(5 b) \int \frac {1}{\sqrt {x} \left (b+c x^2\right )} \, dx}{4 c^2}\\ &=\frac {5 \sqrt {x}}{2 c^2}-\frac {x^{5/2}}{2 c \left (b+c x^2\right )}-\frac {(5 b) \operatorname {Subst}\left (\int \frac {1}{b+c x^4} \, dx,x,\sqrt {x}\right )}{2 c^2}\\ &=\frac {5 \sqrt {x}}{2 c^2}-\frac {x^{5/2}}{2 c \left (b+c x^2\right )}-\frac {\left (5 \sqrt {b}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {b}-\sqrt {c} x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{4 c^2}-\frac {\left (5 \sqrt {b}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {b}+\sqrt {c} x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{4 c^2}\\ &=\frac {5 \sqrt {x}}{2 c^2}-\frac {x^{5/2}}{2 c \left (b+c x^2\right )}-\frac {\left (5 \sqrt {b}\right ) \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 )}{8 c^{5/2}}-\frac {\left (5 \sqrt {b}\right ) \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 )}{8 c^{5/2}}+\frac {\left (5 \sqrt [4]{b}\right ) \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 )}{8 \sqrt {2} c^{9/4}}+\frac {\left (5 \sqrt [4]{b}\right ) \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 )}{8 \sqrt {2} c^{9/4}}\\ &=\frac {5 \sqrt {x}}{2 c^2}-\frac {x^{5/2}}{2 c \left (b+c x^2\right )}+\frac {5 \sqrt [4]{b} \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} c^{9/4}}-\frac {5 \sqrt [4]{b} \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} c^{9/4}}-\frac {\left (5 \sqrt [4]{b}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} c^{9/4}}+\frac {\left (5 \sqrt [4]{b}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} c^{9/4}}\\ &=\frac {5 \sqrt {x}}{2 c^2}-\frac {x^{5/2}}{2 c \left (b+c x^2\right )}+\frac {5 \sqrt [4]{b} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} c^{9/4}}-\frac {5 \sqrt [4]{b} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} c^{9/4}}+\frac {5 \sqrt [4]{b} \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} c^{9/4}}-\frac {5 \sqrt [4]{b} \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} c^{9/4}}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 221, normalized size = 0.96 \[ \frac {\frac {32 c^{5/4} x^{5/2}}{b+c x^2}+\frac {40 b \sqrt [4]{c} \sqrt {x}}{b+c x^2}+5 \sqrt {2} \sqrt [4]{b} \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )-5 \sqrt {2} \sqrt [4]{b} \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )+10 \sqrt {2} \sqrt [4]{b} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )-10 \sqrt {2} \sqrt [4]{b} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{16 c^{9/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(15/2)/(b*x^2 + c*x^4)^2,x]

[Out]

((40*b*c^(1/4)*Sqrt[x])/(b + c*x^2) + (32*c^(5/4)*x^(5/2))/(b + c*x^2) + 10*Sqrt[2]*b^(1/4)*ArcTan[1 - (Sqrt[2
]*c^(1/4)*Sqrt[x])/b^(1/4)] - 10*Sqrt[2]*b^(1/4)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)] + 5*Sqrt[2]*b^(
1/4)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x] - 5*Sqrt[2]*b^(1/4)*Log[Sqrt[b] + Sqrt[2]*b^(1
/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(16*c^(9/4))

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fricas [A]  time = 0.56, size = 192, normalized size = 0.83 \[ -\frac {20 \, {\left (c^{3} x^{2} + b c^{2}\right )} \left (-\frac {b}{c^{9}}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {c^{4} \sqrt {-\frac {b}{c^{9}}} + x} c^{7} \left (-\frac {b}{c^{9}}\right )^{\frac {3}{4}} - c^{7} \sqrt {x} \left (-\frac {b}{c^{9}}\right )^{\frac {3}{4}}}{b}\right ) + 5 \, {\left (c^{3} x^{2} + b c^{2}\right )} \left (-\frac {b}{c^{9}}\right )^{\frac {1}{4}} \log \left (5 \, c^{2} \left (-\frac {b}{c^{9}}\right )^{\frac {1}{4}} + 5 \, \sqrt {x}\right ) - 5 \, {\left (c^{3} x^{2} + b c^{2}\right )} \left (-\frac {b}{c^{9}}\right )^{\frac {1}{4}} \log \left (-5 \, c^{2} \left (-\frac {b}{c^{9}}\right )^{\frac {1}{4}} + 5 \, \sqrt {x}\right ) - 4 \, {\left (4 \, c x^{2} + 5 \, b\right )} \sqrt {x}}{8 \, {\left (c^{3} x^{2} + b c^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(15/2)/(c*x^4+b*x^2)^2,x, algorithm="fricas")

[Out]

-1/8*(20*(c^3*x^2 + b*c^2)*(-b/c^9)^(1/4)*arctan((sqrt(c^4*sqrt(-b/c^9) + x)*c^7*(-b/c^9)^(3/4) - c^7*sqrt(x)*
(-b/c^9)^(3/4))/b) + 5*(c^3*x^2 + b*c^2)*(-b/c^9)^(1/4)*log(5*c^2*(-b/c^9)^(1/4) + 5*sqrt(x)) - 5*(c^3*x^2 + b
*c^2)*(-b/c^9)^(1/4)*log(-5*c^2*(-b/c^9)^(1/4) + 5*sqrt(x)) - 4*(4*c*x^2 + 5*b)*sqrt(x))/(c^3*x^2 + b*c^2)

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giac [A]  time = 0.20, size = 196, normalized size = 0.85 \[ -\frac {5 \, \sqrt {2} \left (b c^{3}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{8 \, c^{3}} - \frac {5 \, \sqrt {2} \left (b c^{3}\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{8 \, c^{3}} - \frac {5 \, \sqrt {2} \left (b c^{3}\right )^{\frac {1}{4}} \log \left (\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{16 \, c^{3}} + \frac {5 \, \sqrt {2} \left (b c^{3}\right )^{\frac {1}{4}} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{16 \, c^{3}} + \frac {b \sqrt {x}}{2 \, {\left (c x^{2} + b\right )} c^{2}} + \frac {2 \, \sqrt {x}}{c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(15/2)/(c*x^4+b*x^2)^2,x, algorithm="giac")

[Out]

-5/8*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^3 - 5/8*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^3 - 5/16*sqrt(2)*(b*c^3)^(
1/4)*log(sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/c^3 + 5/16*sqrt(2)*(b*c^3)^(1/4)*log(-sqrt(2)*sqrt(x)*(b
/c)^(1/4) + x + sqrt(b/c))/c^3 + 1/2*b*sqrt(x)/((c*x^2 + b)*c^2) + 2*sqrt(x)/c^2

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maple [A]  time = 0.01, size = 158, normalized size = 0.69 \[ \frac {b \sqrt {x}}{2 \left (c \,x^{2}+b \right ) c^{2}}-\frac {5 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{8 c^{2}}-\frac {5 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{8 c^{2}}-\frac {5 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}\right )}{16 c^{2}}+\frac {2 \sqrt {x}}{c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(15/2)/(c*x^4+b*x^2)^2,x)

[Out]

2*x^(1/2)/c^2+1/2*b/c^2*x^(1/2)/(c*x^2+b)-5/16/c^2*(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)))-5/8/c^2*(b/c)^(1/4)*2^(1/2)*arctan(2^(1/2)/(b/c)^(1/4)*x^
(1/2)+1)-5/8/c^2*(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 = 3.06, size = 206, normalized size = 0.90 \[ \frac {b \sqrt {x}}{2 \, {\left (c^{3} x^{2} + b c^{2}\right )}} - \frac {5 \, {\left (\frac {2 \, \sqrt {2} \sqrt {b} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} + 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}}} + \frac {2 \, \sqrt {2} \sqrt {b} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} - 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}}} + \frac {\sqrt {2} b^{\frac {1}{4}} \log \left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{c^{\frac {1}{4}}} - \frac {\sqrt {2} b^{\frac {1}{4}} \log \left (-\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{c^{\frac {1}{4}}}\right )}}{16 \, c^{2}} + \frac {2 \, \sqrt {x}}{c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(15/2)/(c*x^4+b*x^2)^2,x, algorithm="maxima")

[Out]

1/2*b*sqrt(x)/(c^3*x^2 + b*c^2) - 5/16*(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 + 2*sqrt(x)/c^2

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mupad [B]  time = 4.32, size = 80, normalized size = 0.35 \[ \frac {2\,\sqrt {x}}{c^2}-\frac {5\,{\left (-b\right )}^{1/4}\,\mathrm {atan}\left (\frac {c^{1/4}\,\sqrt {x}}{{\left (-b\right )}^{1/4}}\right )}{4\,c^{9/4}}+\frac {b\,\sqrt {x}}{2\,\left (c^3\,x^2+b\,c^2\right )}+\frac {{\left (-b\right )}^{1/4}\,\mathrm {atan}\left (\frac {c^{1/4}\,\sqrt {x}\,1{}\mathrm {i}}{{\left (-b\right )}^{1/4}}\right )\,5{}\mathrm {i}}{4\,c^{9/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(15/2)/(b*x^2 + c*x^4)^2,x)

[Out]

(2*x^(1/2))/c^2 - (5*(-b)^(1/4)*atan((c^(1/4)*x^(1/2))/(-b)^(1/4)))/(4*c^(9/4)) + ((-b)^(1/4)*atan((c^(1/4)*x^
(1/2)*1i)/(-b)^(1/4))*5i)/(4*c^(9/4)) + (b*x^(1/2))/(2*(b*c^2 + c^3*x^2))

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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(x**(15/2)/(c*x**4+b*x**2)**2,x)

[Out]

Timed out

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