3.324 \(\int \frac{1}{x^{3/2} (b x^2+c x^4)} \, dx\)

Optimal. Leaf size=215 \[ \frac{c^{5/4} \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{2 \sqrt{2} b^{9/4}}-\frac{c^{5/4} \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{2 \sqrt{2} b^{9/4}}-\frac{c^{5/4} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{\sqrt{2} b^{9/4}}+\frac{c^{5/4} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}+1\right )}{\sqrt{2} b^{9/4}}+\frac{2 c}{b^2 \sqrt{x}}-\frac{2}{5 b x^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.187666, antiderivative size = 215, normalized size of antiderivative = 1., 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, 297, 1162, 617, 204, 1165, 628} \[ \frac{c^{5/4} \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{2 \sqrt{2} b^{9/4}}-\frac{c^{5/4} \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{2 \sqrt{2} b^{9/4}}-\frac{c^{5/4} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{\sqrt{2} b^{9/4}}+\frac{c^{5/4} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}+1\right )}{\sqrt{2} b^{9/4}}+\frac{2 c}{b^2 \sqrt{x}}-\frac{2}{5 b x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^(3/2)*(b*x^2 + c*x^4)),x]

[Out]

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

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]

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 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 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 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 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 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 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]

Rubi steps

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

Mathematica [C]  time = 0.0069805, size = 29, normalized size = 0.13 \[ -\frac{2 \, _2F_1\left (-\frac{5}{4},1;-\frac{1}{4};-\frac{c x^2}{b}\right )}{5 b x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(3/2)*(b*x^2 + c*x^4)),x]

[Out]

(-2*Hypergeometric2F1[-5/4, 1, -1/4, -((c*x^2)/b)])/(5*b*x^(5/2))

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Maple [A]  time = 0.055, size = 152, normalized size = 0.7 \begin{align*} -{\frac{2}{5\,b}{x}^{-{\frac{5}{2}}}}+2\,{\frac{c}{{b}^{2}\sqrt{x}}}+{\frac{c\sqrt{2}}{4\,{b}^{2}}\ln \left ({ \left ( x-\sqrt [4]{{\frac{b}{c}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{b}{c}}} \right ) \left ( x+\sqrt [4]{{\frac{b}{c}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{b}{c}}} \right ) ^{-1}} \right ){\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}+{\frac{c\sqrt{2}}{2\,{b}^{2}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}+1 \right ){\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}+{\frac{c\sqrt{2}}{2\,{b}^{2}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}-1 \right ){\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^(3/2)/(c*x^4+b*x^2),x)

[Out]

-2/5/b/x^(5/2)+2*c/b^2/x^(1/2)+1/4*c/b^2/(b/c)^(1/4)*2^(1/2)*ln((x-(b/c)^(1/4)*x^(1/2)*2^(1/2)+(b/c)^(1/2))/(x
+(b/c)^(1/4)*x^(1/2)*2^(1/2)+(b/c)^(1/2)))+1/2*c/b^2/(b/c)^(1/4)*2^(1/2)*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)
+1/2*c/b^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 [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(3/2)/(c*x^4+b*x^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.36692, size = 433, normalized size = 2.01 \begin{align*} -\frac{20 \, b^{2} x^{3} \left (-\frac{c^{5}}{b^{9}}\right )^{\frac{1}{4}} \arctan \left (-\frac{b^{2} c^{4} \sqrt{x} \left (-\frac{c^{5}}{b^{9}}\right )^{\frac{1}{4}} - \sqrt{-b^{5} c^{5} \sqrt{-\frac{c^{5}}{b^{9}}} + c^{8} x} b^{2} \left (-\frac{c^{5}}{b^{9}}\right )^{\frac{1}{4}}}{c^{5}}\right ) - 5 \, b^{2} x^{3} \left (-\frac{c^{5}}{b^{9}}\right )^{\frac{1}{4}} \log \left (b^{7} \left (-\frac{c^{5}}{b^{9}}\right )^{\frac{3}{4}} + c^{4} \sqrt{x}\right ) + 5 \, b^{2} x^{3} \left (-\frac{c^{5}}{b^{9}}\right )^{\frac{1}{4}} \log \left (-b^{7} \left (-\frac{c^{5}}{b^{9}}\right )^{\frac{3}{4}} + c^{4} \sqrt{x}\right ) - 4 \,{\left (5 \, c x^{2} - b\right )} \sqrt{x}}{10 \, b^{2} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(3/2)/(c*x^4+b*x^2),x, algorithm="fricas")

[Out]

-1/10*(20*b^2*x^3*(-c^5/b^9)^(1/4)*arctan(-(b^2*c^4*sqrt(x)*(-c^5/b^9)^(1/4) - sqrt(-b^5*c^5*sqrt(-c^5/b^9) +
c^8*x)*b^2*(-c^5/b^9)^(1/4))/c^5) - 5*b^2*x^3*(-c^5/b^9)^(1/4)*log(b^7*(-c^5/b^9)^(3/4) + c^4*sqrt(x)) + 5*b^2
*x^3*(-c^5/b^9)^(1/4)*log(-b^7*(-c^5/b^9)^(3/4) + c^4*sqrt(x)) - 4*(5*c*x^2 - b)*sqrt(x))/(b^2*x^3)

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Sympy [A]  time = 67.5526, size = 196, normalized size = 0.91 \begin{align*} \begin{cases} \frac{\tilde{\infty }}{x^{\frac{9}{2}}} & \text{for}\: b = 0 \wedge c = 0 \\- \frac{2}{9 c x^{\frac{9}{2}}} & \text{for}\: b = 0 \\- \frac{2}{5 b x^{\frac{5}{2}}} & \text{for}\: c = 0 \\- \frac{2}{5 b x^{\frac{5}{2}}} + \frac{2 c}{b^{2} \sqrt{x}} - \frac{\left (-1\right )^{\frac{3}{4}} c^{6} \left (\frac{1}{c}\right )^{\frac{19}{4}} \log{\left (- \sqrt [4]{-1} \sqrt [4]{b} \sqrt [4]{\frac{1}{c}} + \sqrt{x} \right )}}{2 b^{\frac{9}{4}}} + \frac{\left (-1\right )^{\frac{3}{4}} c^{6} \left (\frac{1}{c}\right )^{\frac{19}{4}} \log{\left (\sqrt [4]{-1} \sqrt [4]{b} \sqrt [4]{\frac{1}{c}} + \sqrt{x} \right )}}{2 b^{\frac{9}{4}}} + \frac{\left (-1\right )^{\frac{3}{4}} c^{6} \left (\frac{1}{c}\right )^{\frac{19}{4}} \operatorname{atan}{\left (\frac{\left (-1\right )^{\frac{3}{4}} \sqrt{x}}{\sqrt [4]{b} \sqrt [4]{\frac{1}{c}}} \right )}}{b^{\frac{9}{4}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**(3/2)/(c*x**4+b*x**2),x)

[Out]

Piecewise((zoo/x**(9/2), Eq(b, 0) & Eq(c, 0)), (-2/(9*c*x**(9/2)), Eq(b, 0)), (-2/(5*b*x**(5/2)), Eq(c, 0)), (
-2/(5*b*x**(5/2)) + 2*c/(b**2*sqrt(x)) - (-1)**(3/4)*c**6*(1/c)**(19/4)*log(-(-1)**(1/4)*b**(1/4)*(1/c)**(1/4)
+ sqrt(x))/(2*b**(9/4)) + (-1)**(3/4)*c**6*(1/c)**(19/4)*log((-1)**(1/4)*b**(1/4)*(1/c)**(1/4) + sqrt(x))/(2*
b**(9/4)) + (-1)**(3/4)*c**6*(1/c)**(19/4)*atan((-1)**(3/4)*sqrt(x)/(b**(1/4)*(1/c)**(1/4)))/b**(9/4), True))

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Giac [A]  time = 1.13163, size = 270, normalized size = 1.26 \begin{align*} \frac{\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 )}{2 \, b^{3} c} + \frac{\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 )}{2 \, b^{3} c} - \frac{\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 )}{4 \, b^{3} c} + \frac{\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 )}{4 \, b^{3} c} + \frac{2 \,{\left (5 \, c x^{2} - b\right )}}{5 \, b^{2} x^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(3/2)/(c*x^4+b*x^2),x, algorithm="giac")

[Out]

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