3.3.10 \(\int \frac {x^{17/2}}{(b x^2+c x^4)^2} \, dx\)

Optimal. Leaf size=230 \[ -\frac {7 b^{3/4} \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} c^{11/4}}+\frac {7 b^{3/4} \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} c^{11/4}}+\frac {7 b^{3/4} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} c^{11/4}}-\frac {7 b^{3/4} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{4 \sqrt {2} c^{11/4}}-\frac {x^{7/2}}{2 c \left (b+c x^2\right )}+\frac {7 x^{3/2}}{6 c^2} \]

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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, 297, 1162, 617, 204, 1165, 628} \begin {gather*} -\frac {7 b^{3/4} \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} c^{11/4}}+\frac {7 b^{3/4} \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} c^{11/4}}+\frac {7 b^{3/4} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} c^{11/4}}-\frac {7 b^{3/4} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{4 \sqrt {2} c^{11/4}}-\frac {x^{7/2}}{2 c \left (b+c x^2\right )}+\frac {7 x^{3/2}}{6 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(17/2)/(b*x^2 + c*x^4)^2,x]

[Out]

(7*x^(3/2))/(6*c^2) - x^(7/2)/(2*c*(b + c*x^2)) + (7*b^(3/4)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(4
*Sqrt[2]*c^(11/4)) - (7*b^(3/4)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(4*Sqrt[2]*c^(11/4)) - (7*b^(3/
4)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(8*Sqrt[2]*c^(11/4)) + (7*b^(3/4)*Log[Sqrt[b] +
Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(8*Sqrt[2]*c^(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 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 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 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^{17/2}}{\left (b x^2+c x^4\right )^2} \, dx &=\int \frac {x^{9/2}}{\left (b+c x^2\right )^2} \, dx\\ &=-\frac {x^{7/2}}{2 c \left (b+c x^2\right )}+\frac {7 \int \frac {x^{5/2}}{b+c x^2} \, dx}{4 c}\\ &=\frac {7 x^{3/2}}{6 c^2}-\frac {x^{7/2}}{2 c \left (b+c x^2\right )}-\frac {(7 b) \int \frac {\sqrt {x}}{b+c x^2} \, dx}{4 c^2}\\ &=\frac {7 x^{3/2}}{6 c^2}-\frac {x^{7/2}}{2 c \left (b+c x^2\right )}-\frac {(7 b) \operatorname {Subst}\left (\int \frac {x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{2 c^2}\\ &=\frac {7 x^{3/2}}{6 c^2}-\frac {x^{7/2}}{2 c \left (b+c x^2\right )}+\frac {(7 b) \operatorname {Subst}\left (\int \frac {\sqrt {b}-\sqrt {c} x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{4 c^{5/2}}-\frac {(7 b) \operatorname {Subst}\left (\int \frac {\sqrt {b}+\sqrt {c} x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{4 c^{5/2}}\\ &=\frac {7 x^{3/2}}{6 c^2}-\frac {x^{7/2}}{2 c \left (b+c x^2\right )}-\frac {(7 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 )}{8 c^3}-\frac {(7 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 )}{8 c^3}-\frac {\left (7 b^{3/4}\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^{11/4}}-\frac {\left (7 b^{3/4}\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^{11/4}}\\ &=\frac {7 x^{3/2}}{6 c^2}-\frac {x^{7/2}}{2 c \left (b+c x^2\right )}-\frac {7 b^{3/4} \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} c^{11/4}}+\frac {7 b^{3/4} \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} c^{11/4}}-\frac {\left (7 b^{3/4}\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^{11/4}}+\frac {\left (7 b^{3/4}\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^{11/4}}\\ &=\frac {7 x^{3/2}}{6 c^2}-\frac {x^{7/2}}{2 c \left (b+c x^2\right )}+\frac {7 b^{3/4} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} c^{11/4}}-\frac {7 b^{3/4} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} c^{11/4}}-\frac {7 b^{3/4} \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} c^{11/4}}+\frac {7 b^{3/4} \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} c^{11/4}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 57, normalized size = 0.25 \begin {gather*} -\frac {2 x^{3/2} \left (7 \left (b+c x^2\right ) \, _2F_1\left (\frac {3}{4},2;\frac {7}{4};-\frac {c x^2}{b}\right )-7 b-c x^2\right )}{3 c^2 \left (b+c x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(17/2)/(b*x^2 + c*x^4)^2,x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.34, size = 230, normalized size = 1.00 \begin {gather*} \frac {\left (\frac {7 b^{3/4} x^2}{4 \sqrt {2} c^{7/4}}+\frac {7 b^{7/4}}{4 \sqrt {2} c^{11/4}}\right ) \tan ^{-1}\left (\frac {\sqrt {b}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}\right )+\frac {7 b^{3/4} x^2 \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{4 \sqrt {2} c^{7/4}}+\frac {7 b^{7/4} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{4 \sqrt {2} c^{11/4}}+\frac {7 b x^{3/2}}{6 c^2}+\frac {2 x^{7/2}}{3 c}}{b+c x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^(17/2)/(b*x^2 + c*x^4)^2,x]

[Out]

((7*b*x^(3/2))/(6*c^2) + (2*x^(7/2))/(3*c) + ((7*b^(7/4))/(4*Sqrt[2]*c^(11/4)) + (7*b^(3/4)*x^2)/(4*Sqrt[2]*c^
(7/4)))*ArcTan[(Sqrt[b] - Sqrt[c]*x)/(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])] + (7*b^(7/4)*ArcTanh[(Sqrt[2]*b^(1/4)*
c^(1/4)*Sqrt[x])/(Sqrt[b] + Sqrt[c]*x)])/(4*Sqrt[2]*c^(11/4)) + (7*b^(3/4)*x^2*ArcTanh[(Sqrt[2]*b^(1/4)*c^(1/4
)*Sqrt[x])/(Sqrt[b] + Sqrt[c]*x)])/(4*Sqrt[2]*c^(7/4)))/(b + c*x^2)

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fricas [A]  time = 0.86, size = 229, normalized size = 1.00 \begin {gather*} \frac {84 \, {\left (c^{3} x^{2} + b c^{2}\right )} \left (-\frac {b^{3}}{c^{11}}\right )^{\frac {1}{4}} \arctan \left (-\frac {343 \, b^{2} c^{3} \sqrt {x} \left (-\frac {b^{3}}{c^{11}}\right )^{\frac {1}{4}} - \sqrt {-117649 \, b^{3} c^{5} \sqrt {-\frac {b^{3}}{c^{11}}} + 117649 \, b^{4} x} c^{3} \left (-\frac {b^{3}}{c^{11}}\right )^{\frac {1}{4}}}{343 \, b^{3}}\right ) - 21 \, {\left (c^{3} x^{2} + b c^{2}\right )} \left (-\frac {b^{3}}{c^{11}}\right )^{\frac {1}{4}} \log \left (343 \, c^{8} \left (-\frac {b^{3}}{c^{11}}\right )^{\frac {3}{4}} + 343 \, b^{2} \sqrt {x}\right ) + 21 \, {\left (c^{3} x^{2} + b c^{2}\right )} \left (-\frac {b^{3}}{c^{11}}\right )^{\frac {1}{4}} \log \left (-343 \, c^{8} \left (-\frac {b^{3}}{c^{11}}\right )^{\frac {3}{4}} + 343 \, b^{2} \sqrt {x}\right ) + 4 \, {\left (4 \, c x^{3} + 7 \, b x\right )} \sqrt {x}}{24 \, {\left (c^{3} x^{2} + b c^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(17/2)/(c*x^4+b*x^2)^2,x, algorithm="fricas")

[Out]

1/24*(84*(c^3*x^2 + b*c^2)*(-b^3/c^11)^(1/4)*arctan(-1/343*(343*b^2*c^3*sqrt(x)*(-b^3/c^11)^(1/4) - sqrt(-1176
49*b^3*c^5*sqrt(-b^3/c^11) + 117649*b^4*x)*c^3*(-b^3/c^11)^(1/4))/b^3) - 21*(c^3*x^2 + b*c^2)*(-b^3/c^11)^(1/4
)*log(343*c^8*(-b^3/c^11)^(3/4) + 343*b^2*sqrt(x)) + 21*(c^3*x^2 + b*c^2)*(-b^3/c^11)^(1/4)*log(-343*c^8*(-b^3
/c^11)^(3/4) + 343*b^2*sqrt(x)) + 4*(4*c*x^3 + 7*b*x)*sqrt(x))/(c^3*x^2 + b*c^2)

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giac [A]  time = 0.18, size = 196, normalized size = 0.85 \begin {gather*} \frac {b x^{\frac {3}{2}}}{2 \, {\left (c x^{2} + b\right )} c^{2}} + \frac {2 \, x^{\frac {3}{2}}}{3 \, c^{2}} - \frac {7 \, \sqrt {2} \left (b c^{3}\right )^{\frac {3}{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^{5}} - \frac {7 \, \sqrt {2} \left (b c^{3}\right )^{\frac {3}{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^{5}} + \frac {7 \, \sqrt {2} \left (b c^{3}\right )^{\frac {3}{4}} \log \left (\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{16 \, c^{5}} - \frac {7 \, \sqrt {2} \left (b c^{3}\right )^{\frac {3}{4}} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{16 \, c^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(17/2)/(c*x^4+b*x^2)^2,x, algorithm="giac")

[Out]

1/2*b*x^(3/2)/((c*x^2 + b)*c^2) + 2/3*x^(3/2)/c^2 - 7/8*sqrt(2)*(b*c^3)^(3/4)*arctan(1/2*sqrt(2)*(sqrt(2)*(b/c
)^(1/4) + 2*sqrt(x))/(b/c)^(1/4))/c^5 - 7/8*sqrt(2)*(b*c^3)^(3/4)*arctan(-1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4) - 2
*sqrt(x))/(b/c)^(1/4))/c^5 + 7/16*sqrt(2)*(b*c^3)^(3/4)*log(sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/c^5 -
7/16*sqrt(2)*(b*c^3)^(3/4)*log(-sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/c^5

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maple [A]  time = 0.01, size = 161, normalized size = 0.70 \begin {gather*} \frac {b \,x^{\frac {3}{2}}}{2 \left (c \,x^{2}+b \right ) c^{2}}+\frac {2 x^{\frac {3}{2}}}{3 c^{2}}-\frac {7 \sqrt {2}\, b \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{8 \left (\frac {b}{c}\right )^{\frac {1}{4}} c^{3}}-\frac {7 \sqrt {2}\, b \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{8 \left (\frac {b}{c}\right )^{\frac {1}{4}} c^{3}}-\frac {7 \sqrt {2}\, b \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 \left (\frac {b}{c}\right )^{\frac {1}{4}} c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(17/2)/(c*x^4+b*x^2)^2,x)

[Out]

2/3*x^(3/2)/c^2+1/2*b/c^2*x^(3/2)/(c*x^2+b)-7/16*b/c^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)))-7/8*b/c^3/(b/c)^(1/4)*2^(1/2)*arctan(2^(1/2)/(b/c)^(1
/4)*x^(1/2)+1)-7/8*b/c^3/(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 = 207, normalized size = 0.90 \begin {gather*} \frac {b x^{\frac {3}{2}}}{2 \, {\left (c^{3} x^{2} + b c^{2}\right )}} - \frac {7 \, b {\left (\frac {2 \, \sqrt {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 {\sqrt {b} \sqrt {c}} \sqrt {c}} + \frac {2 \, \sqrt {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 {\sqrt {b} \sqrt {c}} \sqrt {c}} - \frac {\sqrt {2} \log \left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {1}{4}} c^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {1}{4}} c^{\frac {3}{4}}}\right )}}{16 \, c^{2}} + \frac {2 \, x^{\frac {3}{2}}}{3 \, c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(17/2)/(c*x^4+b*x^2)^2,x, algorithm="maxima")

[Out]

1/2*b*x^(3/2)/(c^3*x^2 + b*c^2) - 7/16*b*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*b^(1/4)*c^(1/4) + 2*sqrt(c)*sq
rt(x))/sqrt(sqrt(b)*sqrt(c)))/(sqrt(sqrt(b)*sqrt(c))*sqrt(c)) + 2*sqrt(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(sqrt(b)*sqrt(c))*sqrt(c)) - sqrt(2)*log(sqrt(2)*b^(
1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x + sqrt(b))/(b^(1/4)*c^(3/4)) + sqrt(2)*log(-sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x) +
sqrt(c)*x + sqrt(b))/(b^(1/4)*c^(3/4)))/c^2 + 2/3*x^(3/2)/c^2

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mupad [B]  time = 0.11, size = 80, normalized size = 0.35 \begin {gather*} \frac {2\,x^{3/2}}{3\,c^2}+\frac {7\,{\left (-b\right )}^{3/4}\,\mathrm {atan}\left (\frac {c^{1/4}\,\sqrt {x}}{{\left (-b\right )}^{1/4}}\right )}{4\,c^{11/4}}+\frac {b\,x^{3/2}}{2\,\left (c^3\,x^2+b\,c^2\right )}+\frac {{\left (-b\right )}^{3/4}\,\mathrm {atan}\left (\frac {c^{1/4}\,\sqrt {x}\,1{}\mathrm {i}}{{\left (-b\right )}^{1/4}}\right )\,7{}\mathrm {i}}{4\,c^{11/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(17/2)/(b*x^2 + c*x^4)^2,x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(17/2)/(c*x**4+b*x**2)**2,x)

[Out]

Timed out

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