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

Optimal. Leaf size=59 \[ \frac {2 \sqrt {2} b \left (-\frac {c \left (b x+c x^2\right )}{b^2}\right )^{3/4} F\left (\left .\frac {1}{2} \sin ^{-1}\left (\frac {2 c x}{b}+1\right )\right |2\right )}{c \left (b x+c x^2\right )^{3/4}} \]

[Out]

2*b*(-c*(c*x^2+b*x)/b^2)^(3/4)*(cos(1/2*arcsin(1+2*c*x/b))^2)^(1/2)/cos(1/2*arcsin(1+2*c*x/b))*EllipticF(sin(1
/2*arcsin(1+2*c*x/b)),2^(1/2))*2^(1/2)/c/(c*x^2+b*x)^(3/4)

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Rubi [A]  time = 0.02, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {622, 619, 232} \[ \frac {2 \sqrt {2} b \left (-\frac {c \left (b x+c x^2\right )}{b^2}\right )^{3/4} F\left (\left .\frac {1}{2} \sin ^{-1}\left (\frac {2 c x}{b}+1\right )\right |2\right )}{c \left (b x+c x^2\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

Int[(b*x + c*x^2)^(-3/4),x]

[Out]

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

Rule 232

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2*EllipticF[(1*ArcSin[Rt[-(b/a), 2]*x])/2, 2])/(a^(3/4)*R
t[-(b/a), 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b/a]

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 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

Rubi steps

\begin {align*} \int \frac {1}{\left (b x+c x^2\right )^{3/4}} \, dx &=\frac {\left (-\frac {c \left (b x+c x^2\right )}{b^2}\right )^{3/4} \int \frac {1}{\left (-\frac {c x}{b}-\frac {c^2 x^2}{b^2}\right )^{3/4}} \, dx}{\left (b x+c x^2\right )^{3/4}}\\ &=-\frac {\left (\sqrt {2} b^2 \left (-\frac {c \left (b x+c x^2\right )}{b^2}\right )^{3/4}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1-\frac {b^2 x^2}{c^2}\right )^{3/4}} \, dx,x,-\frac {c}{b}-\frac {2 c^2 x}{b^2}\right )}{c^2 \left (b x+c x^2\right )^{3/4}}\\ &=\frac {2 \sqrt {2} b \left (-\frac {c \left (b x+c x^2\right )}{b^2}\right )^{3/4} F\left (\left .\frac {1}{2} \sin ^{-1}\left (1+\frac {2 c x}{b}\right )\right |2\right )}{c \left (b x+c x^2\right )^{3/4}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 43, normalized size = 0.73 \[ \frac {4 x \left (\frac {c x}{b}+1\right )^{3/4} \, _2F_1\left (\frac {1}{4},\frac {3}{4};\frac {5}{4};-\frac {c x}{b}\right )}{(x (b+c x))^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*x + c*x^2)^(-3/4),x]

[Out]

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

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fricas [F]  time = 0.76, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{{\left (c x^{2} + b x\right )}^{\frac {3}{4}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((c*x^2 + b*x)^(-3/4), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (c x^{2} + b x\right )}^{\frac {3}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*x^2 + b*x)^(-3/4), x)

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maple [F]  time = 0.72, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (c \,x^{2}+b x \right )^{\frac {3}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (c x^{2} + b x\right )}^{\frac {3}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*x^2 + b*x)^(-3/4), x)

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mupad [B]  time = 0.19, size = 36, normalized size = 0.61 \[ \frac {4\,x\,{\left (\frac {c\,x}{b}+1\right )}^{3/4}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{4},\frac {3}{4};\ \frac {5}{4};\ -\frac {c\,x}{b}\right )}{{\left (c\,x^2+b\,x\right )}^{3/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x + c*x^2)^(3/4),x)

[Out]

(4*x*((c*x)/b + 1)^(3/4)*hypergeom([1/4, 3/4], 5/4, -(c*x)/b))/(b*x + c*x^2)^(3/4)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b x + c x^{2}\right )^{\frac {3}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((b*x + c*x**2)**(-3/4), x)

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