3.47 \(\int \frac {1}{(b x+c x^2)^{13/4}} \, dx\)

Optimal. Leaf size=146 \[ -\frac {448 c^2 (b+2 c x)}{15 b^6 \sqrt [4]{b x+c x^2}}+\frac {112 c (b+2 c x)}{45 b^4 \left (b x+c x^2\right )^{5/4}}-\frac {4 (b+2 c x)}{9 b^2 \left (b x+c x^2\right )^{9/4}}+\frac {448 \sqrt {2} c^2 \sqrt [4]{-\frac {c \left (b x+c x^2\right )}{b^2}} E\left (\left .\frac {1}{2} \sin ^{-1}\left (\frac {2 c x}{b}+1\right )\right |2\right )}{15 b^5 \sqrt [4]{b x+c x^2}} \]

[Out]

-4/9*(2*c*x+b)/b^2/(c*x^2+b*x)^(9/4)+112/45*c*(2*c*x+b)/b^4/(c*x^2+b*x)^(5/4)-448/15*c^2*(2*c*x+b)/b^6/(c*x^2+
b*x)^(1/4)+448/15*c^2*(-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^5/(c*x^2+b*x)^(1/4)

________________________________________________________________________________________

Rubi [A]  time = 0.06, antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {614, 622, 619, 228} \[ -\frac {448 c^2 (b+2 c x)}{15 b^6 \sqrt [4]{b x+c x^2}}+\frac {448 \sqrt {2} c^2 \sqrt [4]{-\frac {c \left (b x+c x^2\right )}{b^2}} E\left (\left .\frac {1}{2} \sin ^{-1}\left (\frac {2 c x}{b}+1\right )\right |2\right )}{15 b^5 \sqrt [4]{b x+c x^2}}+\frac {112 c (b+2 c x)}{45 b^4 \left (b x+c x^2\right )^{5/4}}-\frac {4 (b+2 c x)}{9 b^2 \left (b x+c x^2\right )^{9/4}} \]

Antiderivative was successfully verified.

[In]

Int[(b*x + c*x^2)^(-13/4),x]

[Out]

(-4*(b + 2*c*x))/(9*b^2*(b*x + c*x^2)^(9/4)) + (112*c*(b + 2*c*x))/(45*b^4*(b*x + c*x^2)^(5/4)) - (448*c^2*(b
+ 2*c*x))/(15*b^6*(b*x + c*x^2)^(1/4)) + (448*Sqrt[2]*c^2*(-((c*(b*x + c*x^2))/b^2))^(1/4)*EllipticE[ArcSin[1
+ (2*c*x)/b]/2, 2])/(15*b^5*(b*x + c*x^2)^(1/4))

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]

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 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 )^{13/4}} \, dx &=-\frac {4 (b+2 c x)}{9 b^2 \left (b x+c x^2\right )^{9/4}}-\frac {(28 c) \int \frac {1}{\left (b x+c x^2\right )^{9/4}} \, dx}{9 b^2}\\ &=-\frac {4 (b+2 c x)}{9 b^2 \left (b x+c x^2\right )^{9/4}}+\frac {112 c (b+2 c x)}{45 b^4 \left (b x+c x^2\right )^{5/4}}+\frac {\left (112 c^2\right ) \int \frac {1}{\left (b x+c x^2\right )^{5/4}} \, dx}{15 b^4}\\ &=-\frac {4 (b+2 c x)}{9 b^2 \left (b x+c x^2\right )^{9/4}}+\frac {112 c (b+2 c x)}{45 b^4 \left (b x+c x^2\right )^{5/4}}-\frac {448 c^2 (b+2 c x)}{15 b^6 \sqrt [4]{b x+c x^2}}+\frac {\left (448 c^3\right ) \int \frac {1}{\sqrt [4]{b x+c x^2}} \, dx}{15 b^6}\\ &=-\frac {4 (b+2 c x)}{9 b^2 \left (b x+c x^2\right )^{9/4}}+\frac {112 c (b+2 c x)}{45 b^4 \left (b x+c x^2\right )^{5/4}}-\frac {448 c^2 (b+2 c x)}{15 b^6 \sqrt [4]{b x+c x^2}}+\frac {\left (448 c^3 \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}{15 b^6 \sqrt [4]{b x+c x^2}}\\ &=-\frac {4 (b+2 c x)}{9 b^2 \left (b x+c x^2\right )^{9/4}}+\frac {112 c (b+2 c x)}{45 b^4 \left (b x+c x^2\right )^{5/4}}-\frac {448 c^2 (b+2 c x)}{15 b^6 \sqrt [4]{b x+c x^2}}-\frac {\left (224 \sqrt {2} c \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 )}{15 b^4 \sqrt [4]{b x+c x^2}}\\ &=-\frac {4 (b+2 c x)}{9 b^2 \left (b x+c x^2\right )^{9/4}}+\frac {112 c (b+2 c x)}{45 b^4 \left (b x+c x^2\right )^{5/4}}-\frac {448 c^2 (b+2 c x)}{15 b^6 \sqrt [4]{b x+c x^2}}+\frac {448 \sqrt {2} c^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 )}{15 b^5 \sqrt [4]{b x+c x^2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.01, size = 50, normalized size = 0.34 \[ -\frac {4 \sqrt [4]{\frac {c x}{b}+1} \, _2F_1\left (-\frac {9}{4},\frac {13}{4};-\frac {5}{4};-\frac {c x}{b}\right )}{9 b^3 x^2 \sqrt [4]{x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*x + c*x^2)^(-13/4),x]

[Out]

(-4*(1 + (c*x)/b)^(1/4)*Hypergeometric2F1[-9/4, 13/4, -5/4, -((c*x)/b)])/(9*b^3*x^2*(x*(b + c*x))^(1/4))

________________________________________________________________________________________

fricas [F]  time = 1.11, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (c x^{2} + b x\right )}^{\frac {3}{4}}}{c^{4} x^{8} + 4 \, b c^{3} x^{7} + 6 \, b^{2} c^{2} x^{6} + 4 \, b^{3} c x^{5} + b^{4} x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x^2+b*x)^(13/4),x, algorithm="fricas")

[Out]

integral((c*x^2 + b*x)^(3/4)/(c^4*x^8 + 4*b*c^3*x^7 + 6*b^2*c^2*x^6 + 4*b^3*c*x^5 + b^4*x^4), x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (c x^{2} + b x\right )}^{\frac {13}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x^2+b*x)^(13/4),x, algorithm="giac")

[Out]

integrate((c*x^2 + b*x)^(-13/4), x)

________________________________________________________________________________________

maple [F]  time = 0.77, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (c \,x^{2}+b x \right )^{\frac {13}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*x^2+b*x)^(13/4),x)

[Out]

int(1/(c*x^2+b*x)^(13/4),x)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (c x^{2} + b x\right )}^{\frac {13}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x^2+b*x)^(13/4),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x)^(-13/4), x)

________________________________________________________________________________________

mupad [B]  time = 0.29, size = 36, normalized size = 0.25 \[ -\frac {4\,x\,{\left (\frac {c\,x}{b}+1\right )}^{13/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {9}{4},\frac {13}{4};\ -\frac {5}{4};\ -\frac {c\,x}{b}\right )}{9\,{\left (c\,x^2+b\,x\right )}^{13/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x + c*x^2)^(13/4),x)

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b x + c x^{2}\right )^{\frac {13}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x**2+b*x)**(13/4),x)

[Out]

Integral((b*x + c*x**2)**(-13/4), x)

________________________________________________________________________________________