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

Optimal. Leaf size=83 \[ \frac{4 \sqrt{2} \sqrt [4]{-\frac{c \left (b x+c x^2\right )}{b^2}} E\left (\left .\frac{1}{2} \sin ^{-1}\left (\frac{2 c x}{b}+1\right )\right |2\right )}{b \sqrt [4]{b x+c x^2}}-\frac{4 (b+2 c x)}{b^2 \sqrt [4]{b x+c x^2}} \]

[Out]

(-4*(b + 2*c*x))/(b^2*(b*x + c*x^2)^(1/4)) + (4*Sqrt[2]*(-((c*(b*x + c*x^2))/b^2))^(1/4)*EllipticE[ArcSin[1 +
(2*c*x)/b]/2, 2])/(b*(b*x + c*x^2)^(1/4))

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Rubi [A]  time = 0.0316459, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {614, 622, 619, 228} \[ \frac{4 \sqrt{2} \sqrt [4]{-\frac{c \left (b x+c x^2\right )}{b^2}} E\left (\left .\frac{1}{2} \sin ^{-1}\left (\frac{2 c x}{b}+1\right )\right |2\right )}{b \sqrt [4]{b x+c x^2}}-\frac{4 (b+2 c x)}{b^2 \sqrt [4]{b x+c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(b*x + c*x^2)^(-5/4),x]

[Out]

(-4*(b + 2*c*x))/(b^2*(b*x + c*x^2)^(1/4)) + (4*Sqrt[2]*(-((c*(b*x + c*x^2))/b^2))^(1/4)*EllipticE[ArcSin[1 +
(2*c*x)/b]/2, 2])/(b*(b*x + c*x^2)^(1/4))

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 622

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

Rule 619

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*((-4*c)/(b^2 - 4*a*c))^p), Subst[Int[Si
mp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]

Rule 228

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[(2*EllipticE[(1*ArcSin[Rt[-(b/a), 2]*x])/2, 2])/(a^(1/4)*R
t[-(b/a), 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b/a]

Rubi steps

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

Mathematica [C]  time = 0.0129641, size = 45, normalized size = 0.54 \[ -\frac{4 \sqrt [4]{\frac{c x}{b}+1} \, _2F_1\left (-\frac{1}{4},\frac{5}{4};\frac{3}{4};-\frac{c x}{b}\right )}{b \sqrt [4]{x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*x + c*x^2)^(-5/4),x]

[Out]

(-4*(1 + (c*x)/b)^(1/4)*Hypergeometric2F1[-1/4, 5/4, 3/4, -((c*x)/b)])/(b*(x*(b + c*x))^(1/4))

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Maple [F]  time = 0.861, size = 0, normalized size = 0. \begin{align*} \int \left ( c{x}^{2}+bx \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)^(5/4),x)

[Out]

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

[Out]

integrate((c*x^2 + b*x)^(-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\right )}^{\frac{3}{4}}}{c^{2} x^{4} + 2 \, b c x^{3} + b^{2} x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (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)**(5/4),x)

[Out]

Integral((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\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)^(5/4),x, algorithm="giac")

[Out]

integrate((c*x^2 + b*x)^(-5/4), x)