3.1.46 \(\int \frac {1}{(b x+c x^2)^{9/4}} \, dx\) [46]

Optimal. Leaf size=115 \[ -\frac {4 (b+2 c x)}{5 b^2 \left (b x+c x^2\right )^{5/4}}+\frac {48 c (b+2 c x)}{5 b^4 \sqrt [4]{b x+c x^2}}-\frac {48 \sqrt {2} c \sqrt [4]{-\frac {c \left (b x+c x^2\right )}{b^2}} E\left (\left .\frac {1}{2} \sin ^{-1}\left (1+\frac {2 c x}{b}\right )\right |2\right )}{5 b^3 \sqrt [4]{b x+c x^2}} \]

[Out]

-4/5*(2*c*x+b)/b^2/(c*x^2+b*x)^(5/4)+48/5*c*(2*c*x+b)/b^4/(c*x^2+b*x)^(1/4)-48/5*c*(-c*(c*x^2+b*x)/b^2)^(1/4)*
(cos(1/2*arcsin(1+2*c*x/b))^2)^(1/2)/cos(1/2*arcsin(1+2*c*x/b))*EllipticE(sin(1/2*arcsin(1+2*c*x/b)),2^(1/2))*
2^(1/2)/b^3/(c*x^2+b*x)^(1/4)

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Rubi [A]
time = 0.03, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {628, 636, 633, 234} \begin {gather*} -\frac {48 \sqrt {2} c \sqrt [4]{-\frac {c \left (b x+c x^2\right )}{b^2}} E\left (\left .\frac {1}{2} \text {ArcSin}\left (\frac {2 c x}{b}+1\right )\right |2\right )}{5 b^3 \sqrt [4]{b x+c x^2}}+\frac {48 c (b+2 c x)}{5 b^4 \sqrt [4]{b x+c x^2}}-\frac {4 (b+2 c x)}{5 b^2 \left (b x+c x^2\right )^{5/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*x + c*x^2)^(-9/4),x]

[Out]

(-4*(b + 2*c*x))/(5*b^2*(b*x + c*x^2)^(5/4)) + (48*c*(b + 2*c*x))/(5*b^4*(b*x + c*x^2)^(1/4)) - (48*Sqrt[2]*c*
(-((c*(b*x + c*x^2))/b^2))^(1/4)*EllipticE[ArcSin[1 + (2*c*x)/b]/2, 2])/(5*b^3*(b*x + c*x^2)^(1/4))

Rule 234

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

Rule 628

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 633

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 636

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 )^{9/4}} \, dx &=-\frac {4 (b+2 c x)}{5 b^2 \left (b x+c x^2\right )^{5/4}}-\frac {(12 c) \int \frac {1}{\left (b x+c x^2\right )^{5/4}} \, dx}{5 b^2}\\ &=-\frac {4 (b+2 c x)}{5 b^2 \left (b x+c x^2\right )^{5/4}}+\frac {48 c (b+2 c x)}{5 b^4 \sqrt [4]{b x+c x^2}}-\frac {\left (48 c^2\right ) \int \frac {1}{\sqrt [4]{b x+c x^2}} \, dx}{5 b^4}\\ &=-\frac {4 (b+2 c x)}{5 b^2 \left (b x+c x^2\right )^{5/4}}+\frac {48 c (b+2 c x)}{5 b^4 \sqrt [4]{b x+c x^2}}-\frac {\left (48 c^2 \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}{5 b^4 \sqrt [4]{b x+c x^2}}\\ &=-\frac {4 (b+2 c x)}{5 b^2 \left (b x+c x^2\right )^{5/4}}+\frac {48 c (b+2 c x)}{5 b^4 \sqrt [4]{b x+c x^2}}+\frac {\left (24 \sqrt {2} \sqrt [4]{-\frac {c \left (b x+c x^2\right )}{b^2}}\right ) \text {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 )}{5 b^2 \sqrt [4]{b x+c x^2}}\\ &=-\frac {4 (b+2 c x)}{5 b^2 \left (b x+c x^2\right )^{5/4}}+\frac {48 c (b+2 c x)}{5 b^4 \sqrt [4]{b x+c x^2}}-\frac {48 \sqrt {2} c \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 )}{5 b^3 \sqrt [4]{b x+c x^2}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.02, size = 50, normalized size = 0.43 \begin {gather*} -\frac {4 \sqrt [4]{1+\frac {c x}{b}} \, _2F_1\left (-\frac {5}{4},\frac {9}{4};-\frac {1}{4};-\frac {c x}{b}\right )}{5 b^2 x \sqrt [4]{x (b+c x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*x + c*x^2)^(-9/4),x]

[Out]

(-4*(1 + (c*x)/b)^(1/4)*Hypergeometric2F1[-5/4, 9/4, -1/4, -((c*x)/b)])/(5*b^2*x*(x*(b + c*x))^(1/4))

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Maple [F]
time = 0.04, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (c \,x^{2}+b x \right )^{\frac {9}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*x^2+b*x)^(9/4),x)

[Out]

int(1/(c*x^2+b*x)^(9/4),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x^2+b*x)^(9/4),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x)^(-9/4), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x^2+b*x)^(9/4),x, algorithm="fricas")

[Out]

integral((c*x^2 + b*x)^(3/4)/(c^3*x^6 + 3*b*c^2*x^5 + 3*b^2*c*x^4 + b^3*x^3), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (b x + c x^{2}\right )^{\frac {9}{4}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x**2+b*x)**(9/4),x)

[Out]

Integral((b*x + c*x**2)**(-9/4), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x^2+b*x)^(9/4),x, algorithm="giac")

[Out]

integrate((c*x^2 + b*x)^(-9/4), x)

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Mupad [B]
time = 0.26, size = 36, normalized size = 0.31 \begin {gather*} -\frac {4\,x\,{\left (\frac {c\,x}{b}+1\right )}^{9/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {9}{4};\ -\frac {1}{4};\ -\frac {c\,x}{b}\right )}{5\,{\left (c\,x^2+b\,x\right )}^{9/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x + c*x^2)^(9/4),x)

[Out]

-(4*x*((c*x)/b + 1)^(9/4)*hypergeom([-5/4, 9/4], -1/4, -(c*x)/b))/(5*(b*x + c*x^2)^(9/4))

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