3.2545 \(\int \frac{1}{(a+b x+c x^2)^{5/4}} \, dx\)

Optimal. Leaf size=451 \[ \frac{2 \sqrt{2} \sqrt [4]{c} \sqrt{\frac{(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right )^2}} \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt [4]{a+b x+c x^2}}{\sqrt [4]{b^2-4 a c}}\right ),\frac{1}{2}\right )}{\sqrt [4]{b^2-4 a c} (b+2 c x)}-\frac{4 (b+2 c x)}{\left (b^2-4 a c\right ) \sqrt [4]{a+b x+c x^2}}+\frac{8 \sqrt{c} (b+2 c x) \sqrt [4]{a+b x+c x^2}}{\left (b^2-4 a c\right )^{3/2} \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right )}-\frac{4 \sqrt{2} \sqrt [4]{c} \sqrt{\frac{(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right )^2}} \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right ) E\left (2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right )|\frac{1}{2}\right )}{\sqrt [4]{b^2-4 a c} (b+2 c x)} \]

[Out]

(-4*(b + 2*c*x))/((b^2 - 4*a*c)*(a + b*x + c*x^2)^(1/4)) + (8*Sqrt[c]*(b + 2*c*x)*(a + b*x + c*x^2)^(1/4))/((b
^2 - 4*a*c)^(3/2)*(1 + (2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c])) - (4*Sqrt[2]*c^(1/4)*Sqrt[(b + 2*
c*x)^2/((b^2 - 4*a*c)*(1 + (2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c])^2)]*(1 + (2*Sqrt[c]*Sqrt[a + b
*x + c*x^2])/Sqrt[b^2 - 4*a*c])*EllipticE[2*ArcTan[(Sqrt[2]*c^(1/4)*(a + b*x + c*x^2)^(1/4))/(b^2 - 4*a*c)^(1/
4)], 1/2])/((b^2 - 4*a*c)^(1/4)*(b + 2*c*x)) + (2*Sqrt[2]*c^(1/4)*Sqrt[(b + 2*c*x)^2/((b^2 - 4*a*c)*(1 + (2*Sq
rt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c])^2)]*(1 + (2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c])*
EllipticF[2*ArcTan[(Sqrt[2]*c^(1/4)*(a + b*x + c*x^2)^(1/4))/(b^2 - 4*a*c)^(1/4)], 1/2])/((b^2 - 4*a*c)^(1/4)*
(b + 2*c*x))

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Rubi [A]  time = 0.360722, antiderivative size = 451, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {614, 623, 305, 220, 1196} \[ -\frac{4 (b+2 c x)}{\left (b^2-4 a c\right ) \sqrt [4]{a+b x+c x^2}}+\frac{8 \sqrt{c} (b+2 c x) \sqrt [4]{a+b x+c x^2}}{\left (b^2-4 a c\right )^{3/2} \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right )}+\frac{2 \sqrt{2} \sqrt [4]{c} \sqrt{\frac{(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right )^2}} \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right )|\frac{1}{2}\right )}{\sqrt [4]{b^2-4 a c} (b+2 c x)}-\frac{4 \sqrt{2} \sqrt [4]{c} \sqrt{\frac{(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right )^2}} \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right ) E\left (2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right )|\frac{1}{2}\right )}{\sqrt [4]{b^2-4 a c} (b+2 c x)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x + c*x^2)^(-5/4),x]

[Out]

(-4*(b + 2*c*x))/((b^2 - 4*a*c)*(a + b*x + c*x^2)^(1/4)) + (8*Sqrt[c]*(b + 2*c*x)*(a + b*x + c*x^2)^(1/4))/((b
^2 - 4*a*c)^(3/2)*(1 + (2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c])) - (4*Sqrt[2]*c^(1/4)*Sqrt[(b + 2*
c*x)^2/((b^2 - 4*a*c)*(1 + (2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c])^2)]*(1 + (2*Sqrt[c]*Sqrt[a + b
*x + c*x^2])/Sqrt[b^2 - 4*a*c])*EllipticE[2*ArcTan[(Sqrt[2]*c^(1/4)*(a + b*x + c*x^2)^(1/4))/(b^2 - 4*a*c)^(1/
4)], 1/2])/((b^2 - 4*a*c)^(1/4)*(b + 2*c*x)) + (2*Sqrt[2]*c^(1/4)*Sqrt[(b + 2*c*x)^2/((b^2 - 4*a*c)*(1 + (2*Sq
rt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c])^2)]*(1 + (2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c])*
EllipticF[2*ArcTan[(Sqrt[2]*c^(1/4)*(a + b*x + c*x^2)^(1/4))/(b^2 - 4*a*c)^(1/4)], 1/2])/((b^2 - 4*a*c)^(1/4)*
(b + 2*c*x))

Rule 614

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^(p + 1))/((p +
1)*(b^2 - 4*a*c)), x] - Dist[(2*c*(2*p + 3))/((p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2] && IntegerQ[4*p]

Rule 623

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{d = Denominator[p]}, Dist[(d*Sqrt[(b + 2*c*x)
^2])/(b + 2*c*x), Subst[Int[x^(d*(p + 1) - 1)/Sqrt[b^2 - 4*a*c + 4*c*x^d], x], x, (a + b*x + c*x^2)^(1/d)], x]
/; 3 <= d <= 4] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && RationalQ[p]

Rule 305

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, Dist[1/q, Int[1/Sqrt[a + b*x^4], x],
x] - Dist[1/q, Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps

\begin{align*} \int \frac{1}{\left (a+b x+c x^2\right )^{5/4}} \, dx &=-\frac{4 (b+2 c x)}{\left (b^2-4 a c\right ) \sqrt [4]{a+b x+c x^2}}+\frac{(4 c) \int \frac{1}{\sqrt [4]{a+b x+c x^2}} \, dx}{b^2-4 a c}\\ &=-\frac{4 (b+2 c x)}{\left (b^2-4 a c\right ) \sqrt [4]{a+b x+c x^2}}+\frac{\left (16 c \sqrt{(b+2 c x)^2}\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{b^2-4 a c+4 c x^4}} \, dx,x,\sqrt [4]{a+b x+c x^2}\right )}{\left (b^2-4 a c\right ) (b+2 c x)}\\ &=-\frac{4 (b+2 c x)}{\left (b^2-4 a c\right ) \sqrt [4]{a+b x+c x^2}}+\frac{\left (8 \sqrt{c} \sqrt{(b+2 c x)^2}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{b^2-4 a c+4 c x^4}} \, dx,x,\sqrt [4]{a+b x+c x^2}\right )}{\sqrt{b^2-4 a c} (b+2 c x)}-\frac{\left (8 \sqrt{c} \sqrt{(b+2 c x)^2}\right ) \operatorname{Subst}\left (\int \frac{1-\frac{2 \sqrt{c} x^2}{\sqrt{b^2-4 a c}}}{\sqrt{b^2-4 a c+4 c x^4}} \, dx,x,\sqrt [4]{a+b x+c x^2}\right )}{\sqrt{b^2-4 a c} (b+2 c x)}\\ &=-\frac{4 (b+2 c x)}{\left (b^2-4 a c\right ) \sqrt [4]{a+b x+c x^2}}+\frac{8 \sqrt{c} (b+2 c x) \sqrt [4]{a+b x+c x^2}}{\left (b^2-4 a c\right )^{3/2} \left (1+\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}\right )}-\frac{4 \sqrt{2} \sqrt [4]{c} \sqrt{\frac{(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (1+\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}\right )^2}} \left (1+\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}\right ) E\left (2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt [4]{a+b x+c x^2}}{\sqrt [4]{b^2-4 a c}}\right )|\frac{1}{2}\right )}{\sqrt [4]{b^2-4 a c} (b+2 c x)}+\frac{2 \sqrt{2} \sqrt [4]{c} \sqrt{\frac{(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (1+\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}\right )^2}} \left (1+\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt [4]{a+b x+c x^2}}{\sqrt [4]{b^2-4 a c}}\right )|\frac{1}{2}\right )}{\sqrt [4]{b^2-4 a c} (b+2 c x)}\\ \end{align*}

Mathematica [C]  time = 0.0708589, size = 100, normalized size = 0.22 \[ -\frac{2 \sqrt{2} (b+2 c x) \left (\sqrt{2}-\sqrt [4]{\frac{c (a+x (b+c x))}{4 a c-b^2}} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{3}{2};\frac{(b+2 c x)^2}{b^2-4 a c}\right )\right )}{\left (b^2-4 a c\right ) \sqrt [4]{a+x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x + c*x^2)^(-5/4),x]

[Out]

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

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Maple [F]  time = 2.167, size = 0, normalized size = 0. \begin{align*} \int \left ( c{x}^{2}+bx+a \right ) ^{-{\frac{5}{4}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (c x^{2} + b x + a\right )}^{\frac{5}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*x^2 + b*x + a)^(-5/4), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (c x^{2} + b x + a\right )}^{\frac{3}{4}}}{c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x +{\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((c*x^2 + b*x + a)^(3/4)/(c^2*x^4 + 2*b*c*x^3 + 2*a*b*x + (b^2 + 2*a*c)*x^2 + a^2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a + b x + c x^{2}\right )^{\frac{5}{4}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*x + c*x**2)**(-5/4), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (c x^{2} + b x + a\right )}^{\frac{5}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*x^2 + b*x + a)^(-5/4), x)