\(\int (b x+c x^2)^{5/4} \, dx\) [40]

Optimal result

Integrand size = 13, antiderivative size = 119 \[ \int \left (b x+c x^2\right )^{5/4} \, dx=-\frac {5 b^2 (b+2 c x) \sqrt [4]{b x+c x^2}}{84 c^2}+\frac {(b+2 c x) \left (b x+c x^2\right )^{5/4}}{7 c}+\frac {5 b^5 \left (-\frac {c \left (b x+c x^2\right )}{b^2}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin \left (1+\frac {2 c x}{b}\right ),2\right )}{84 \sqrt {2} c^3 \left (b x+c x^2\right )^{3/4}} \]

[Out]

-5/84*b^2*(2*c*x+b)*(c*x^2+b*x)^(1/4)/c^2+1/7*(2*c*x+b)*(c*x^2+b*x)^(5/4)/c+5/168*b^5*(-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
))/c^3/(c*x^2+b*x)^(3/4)*2^(1/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {626, 636, 633, 238} \[ \int \left (b x+c x^2\right )^{5/4} \, dx=\frac {5 b^5 \left (-\frac {c \left (b x+c x^2\right )}{b^2}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin \left (\frac {2 c x}{b}+1\right ),2\right )}{84 \sqrt {2} c^3 \left (b x+c x^2\right )^{3/4}}-\frac {5 b^2 (b+2 c x) \sqrt [4]{b x+c x^2}}{84 c^2}+\frac {(b+2 c x) \left (b x+c x^2\right )^{5/4}}{7 c} \]

[In]

Int[(b*x + c*x^2)^(5/4),x]

[Out]

(-5*b^2*(b + 2*c*x)*(b*x + c*x^2)^(1/4))/(84*c^2) + ((b + 2*c*x)*(b*x + c*x^2)^(5/4))/(7*c) + (5*b^5*(-((c*(b*
x + c*x^2))/b^2))^(3/4)*EllipticF[ArcSin[1 + (2*c*x)/b]/2, 2])/(84*Sqrt[2]*c^3*(b*x + c*x^2)^(3/4))

Rule 238

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

Rule 626

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x)*((a + b*x + c*x^2)^p/(2*c*(2*p + 1
))), x] - Dist[p*((b^2 - 4*a*c)/(2*c*(2*p + 1))), Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && 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*} \text {integral}& = \frac {(b+2 c x) \left (b x+c x^2\right )^{5/4}}{7 c}-\frac {\left (5 b^2\right ) \int \sqrt [4]{b x+c x^2} \, dx}{28 c} \\ & = -\frac {5 b^2 (b+2 c x) \sqrt [4]{b x+c x^2}}{84 c^2}+\frac {(b+2 c x) \left (b x+c x^2\right )^{5/4}}{7 c}+\frac {\left (5 b^4\right ) \int \frac {1}{\left (b x+c x^2\right )^{3/4}} \, dx}{336 c^2} \\ & = -\frac {5 b^2 (b+2 c x) \sqrt [4]{b x+c x^2}}{84 c^2}+\frac {(b+2 c x) \left (b x+c x^2\right )^{5/4}}{7 c}+\frac {\left (5 b^4 \left (-\frac {c \left (b x+c x^2\right )}{b^2}\right )^{3/4}\right ) \int \frac {1}{\left (-\frac {c x}{b}-\frac {c^2 x^2}{b^2}\right )^{3/4}} \, dx}{336 c^2 \left (b x+c x^2\right )^{3/4}} \\ & = -\frac {5 b^2 (b+2 c x) \sqrt [4]{b x+c x^2}}{84 c^2}+\frac {(b+2 c x) \left (b x+c x^2\right )^{5/4}}{7 c}-\frac {\left (5 b^6 \left (-\frac {c \left (b x+c x^2\right )}{b^2}\right )^{3/4}\right ) \text {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 )}{168 \sqrt {2} c^4 \left (b x+c x^2\right )^{3/4}} \\ & = -\frac {5 b^2 (b+2 c x) \sqrt [4]{b x+c x^2}}{84 c^2}+\frac {(b+2 c x) \left (b x+c x^2\right )^{5/4}}{7 c}+\frac {5 b^5 \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 )}{84 \sqrt {2} c^3 \left (b x+c x^2\right )^{3/4}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.02 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.40 \[ \int \left (b x+c x^2\right )^{5/4} \, dx=\frac {4 b x^2 \sqrt [4]{x (b+c x)} \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},\frac {9}{4},\frac {13}{4},-\frac {c x}{b}\right )}{9 \sqrt [4]{1+\frac {c x}{b}}} \]

[In]

Integrate[(b*x + c*x^2)^(5/4),x]

[Out]

(4*b*x^2*(x*(b + c*x))^(1/4)*Hypergeometric2F1[-5/4, 9/4, 13/4, -((c*x)/b)])/(9*(1 + (c*x)/b)^(1/4))

Maple [F]

\[\int \left (c \,x^{2}+b x \right )^{\frac {5}{4}}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int \left (b x+c x^2\right )^{5/4} \, dx=\int { {\left (c x^{2} + b x\right )}^{\frac {5}{4}} \,d x } \]

[In]

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

[Out]

integral((c*x^2 + b*x)^(5/4), x)

Sympy [F]

\[ \int \left (b x+c x^2\right )^{5/4} \, dx=\int \left (b x + c x^{2}\right )^{\frac {5}{4}}\, dx \]

[In]

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

[Out]

Integral((b*x + c*x**2)**(5/4), x)

Maxima [F]

\[ \int \left (b x+c x^2\right )^{5/4} \, dx=\int { {\left (c x^{2} + b x\right )}^{\frac {5}{4}} \,d x } \]

[In]

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

[Out]

integrate((c*x^2 + b*x)^(5/4), x)

Giac [F]

\[ \int \left (b x+c x^2\right )^{5/4} \, dx=\int { {\left (c x^{2} + b x\right )}^{\frac {5}{4}} \,d x } \]

[In]

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

[Out]

integrate((c*x^2 + b*x)^(5/4), x)

Mupad [B] (verification not implemented)

Time = 9.07 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.30 \[ \int \left (b x+c x^2\right )^{5/4} \, dx=\frac {4\,x\,{\left (c\,x^2+b\,x\right )}^{5/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {9}{4};\ \frac {13}{4};\ -\frac {c\,x}{b}\right )}{9\,{\left (\frac {c\,x}{b}+1\right )}^{5/4}} \]

[In]

int((b*x + c*x^2)^(5/4),x)

[Out]

(4*x*(b*x + c*x^2)^(5/4)*hypergeom([-5/4, 9/4], 13/4, -(c*x)/b))/(9*((c*x)/b + 1)^(5/4))