3.777 \(\int \frac{1}{(4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4)^{3/2}} \, dx\)

Optimal. Leaf size=674 \[ -\frac{d^2 \left (\frac{c}{d}+x\right ) \sqrt{4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4}}{8 a \left (4 a d^2+c^3\right )^{3/2} \left (\frac{d^2 \left (\frac{c}{d}+x\right )^2}{\sqrt{4 a d^2+c^3}}+\sqrt{c}\right )}-\frac{\left (\frac{c}{d}+x\right ) \left (-4 a d^2+c^3-c d^2 \left (\frac{c}{d}+x\right )^2\right )}{8 a c \left (4 a d^2+c^3\right ) \sqrt{4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4}}+\frac{\left (-c^{3/2} \sqrt{4 a d^2+c^3}+4 a d^2+c^3\right ) \sqrt{\frac{d^2 \left (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )}{\left (4 a d^2+c^3\right ) \left (\frac{d^2 \left (\frac{c}{d}+x\right )^2}{\sqrt{4 a d^2+c^3}}+\sqrt{c}\right )^2}} \left (\frac{d^2 \left (\frac{c}{d}+x\right )^2}{\sqrt{4 a d^2+c^3}}+\sqrt{c}\right ) F\left (2 \tan ^{-1}\left (\frac{c+d x}{\sqrt [4]{c} \sqrt [4]{c^3+4 a d^2}}\right )|\frac{1}{2} \left (\frac{c^{3/2}}{\sqrt{c^3+4 a d^2}}+1\right )\right )}{16 a c^{5/4} d \left (4 a d^2+c^3\right )^{3/4} \sqrt{4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4}}+\frac{\sqrt [4]{c} \sqrt{\frac{d^2 \left (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )}{\left (4 a d^2+c^3\right ) \left (\frac{d^2 \left (\frac{c}{d}+x\right )^2}{\sqrt{4 a d^2+c^3}}+\sqrt{c}\right )^2}} \left (\frac{d^2 \left (\frac{c}{d}+x\right )^2}{\sqrt{4 a d^2+c^3}}+\sqrt{c}\right ) E\left (2 \tan ^{-1}\left (\frac{c+d x}{\sqrt [4]{c} \sqrt [4]{c^3+4 a d^2}}\right )|\frac{1}{2} \left (\frac{c^{3/2}}{\sqrt{c^3+4 a d^2}}+1\right )\right )}{8 a d \sqrt [4]{4 a d^2+c^3} \sqrt{4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4}} \]

[Out]

-((c/d + x)*(c^3 - 4*a*d^2 - c*d^2*(c/d + x)^2))/(8*a*c*(c^3 + 4*a*d^2)*Sqrt[4*a*c + 4*c^2*x^2 + 4*c*d*x^3 + d
^2*x^4]) - (d^2*(c/d + x)*Sqrt[4*a*c + 4*c^2*x^2 + 4*c*d*x^3 + d^2*x^4])/(8*a*(c^3 + 4*a*d^2)^(3/2)*(Sqrt[c] +
(d^2*(c/d + x)^2)/Sqrt[c^3 + 4*a*d^2])) + (c^(1/4)*Sqrt[(d^2*(4*a*c + 4*c^2*x^2 + 4*c*d*x^3 + d^2*x^4))/((c^3
+ 4*a*d^2)*(Sqrt[c] + (d^2*(c/d + x)^2)/Sqrt[c^3 + 4*a*d^2])^2)]*(Sqrt[c] + (d^2*(c/d + x)^2)/Sqrt[c^3 + 4*a*
d^2])*EllipticE[2*ArcTan[(c + d*x)/(c^(1/4)*(c^3 + 4*a*d^2)^(1/4))], (1 + c^(3/2)/Sqrt[c^3 + 4*a*d^2])/2])/(8*
a*d*(c^3 + 4*a*d^2)^(1/4)*Sqrt[4*a*c + 4*c^2*x^2 + 4*c*d*x^3 + d^2*x^4]) + ((c^3 + 4*a*d^2 - c^(3/2)*Sqrt[c^3
+ 4*a*d^2])*Sqrt[(d^2*(4*a*c + 4*c^2*x^2 + 4*c*d*x^3 + d^2*x^4))/((c^3 + 4*a*d^2)*(Sqrt[c] + (d^2*(c/d + x)^2)
/Sqrt[c^3 + 4*a*d^2])^2)]*(Sqrt[c] + (d^2*(c/d + x)^2)/Sqrt[c^3 + 4*a*d^2])*EllipticF[2*ArcTan[(c + d*x)/(c^(1
/4)*(c^3 + 4*a*d^2)^(1/4))], (1 + c^(3/2)/Sqrt[c^3 + 4*a*d^2])/2])/(16*a*c^(5/4)*d*(c^3 + 4*a*d^2)^(3/4)*Sqrt[
4*a*c + 4*c^2*x^2 + 4*c*d*x^3 + d^2*x^4])

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Rubi [A]  time = 0.683077, antiderivative size = 674, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.161, Rules used = {1106, 1092, 1197, 1103, 1195} \[ -\frac{d^2 \left (\frac{c}{d}+x\right ) \sqrt{4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4}}{8 a \left (4 a d^2+c^3\right )^{3/2} \left (\frac{d^2 \left (\frac{c}{d}+x\right )^2}{\sqrt{4 a d^2+c^3}}+\sqrt{c}\right )}-\frac{\left (\frac{c}{d}+x\right ) \left (-4 a d^2+c^3-c d^2 \left (\frac{c}{d}+x\right )^2\right )}{8 a c \left (4 a d^2+c^3\right ) \sqrt{4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4}}+\frac{\left (-c^{3/2} \sqrt{4 a d^2+c^3}+4 a d^2+c^3\right ) \sqrt{\frac{d^2 \left (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )}{\left (4 a d^2+c^3\right ) \left (\frac{d^2 \left (\frac{c}{d}+x\right )^2}{\sqrt{4 a d^2+c^3}}+\sqrt{c}\right )^2}} \left (\frac{d^2 \left (\frac{c}{d}+x\right )^2}{\sqrt{4 a d^2+c^3}}+\sqrt{c}\right ) F\left (2 \tan ^{-1}\left (\frac{c+d x}{\sqrt [4]{c} \sqrt [4]{c^3+4 a d^2}}\right )|\frac{1}{2} \left (\frac{c^{3/2}}{\sqrt{c^3+4 a d^2}}+1\right )\right )}{16 a c^{5/4} d \left (4 a d^2+c^3\right )^{3/4} \sqrt{4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4}}+\frac{\sqrt [4]{c} \sqrt{\frac{d^2 \left (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )}{\left (4 a d^2+c^3\right ) \left (\frac{d^2 \left (\frac{c}{d}+x\right )^2}{\sqrt{4 a d^2+c^3}}+\sqrt{c}\right )^2}} \left (\frac{d^2 \left (\frac{c}{d}+x\right )^2}{\sqrt{4 a d^2+c^3}}+\sqrt{c}\right ) E\left (2 \tan ^{-1}\left (\frac{c+d x}{\sqrt [4]{c} \sqrt [4]{c^3+4 a d^2}}\right )|\frac{1}{2} \left (\frac{c^{3/2}}{\sqrt{c^3+4 a d^2}}+1\right )\right )}{8 a d \sqrt [4]{4 a d^2+c^3} \sqrt{4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4}} \]

Antiderivative was successfully verified.

[In]

Int[(4*a*c + 4*c^2*x^2 + 4*c*d*x^3 + d^2*x^4)^(-3/2),x]

[Out]

-((c/d + x)*(c^3 - 4*a*d^2 - c*d^2*(c/d + x)^2))/(8*a*c*(c^3 + 4*a*d^2)*Sqrt[4*a*c + 4*c^2*x^2 + 4*c*d*x^3 + d
^2*x^4]) - (d^2*(c/d + x)*Sqrt[4*a*c + 4*c^2*x^2 + 4*c*d*x^3 + d^2*x^4])/(8*a*(c^3 + 4*a*d^2)^(3/2)*(Sqrt[c] +
(d^2*(c/d + x)^2)/Sqrt[c^3 + 4*a*d^2])) + (c^(1/4)*Sqrt[(d^2*(4*a*c + 4*c^2*x^2 + 4*c*d*x^3 + d^2*x^4))/((c^3
+ 4*a*d^2)*(Sqrt[c] + (d^2*(c/d + x)^2)/Sqrt[c^3 + 4*a*d^2])^2)]*(Sqrt[c] + (d^2*(c/d + x)^2)/Sqrt[c^3 + 4*a*
d^2])*EllipticE[2*ArcTan[(c + d*x)/(c^(1/4)*(c^3 + 4*a*d^2)^(1/4))], (1 + c^(3/2)/Sqrt[c^3 + 4*a*d^2])/2])/(8*
a*d*(c^3 + 4*a*d^2)^(1/4)*Sqrt[4*a*c + 4*c^2*x^2 + 4*c*d*x^3 + d^2*x^4]) + ((c^3 + 4*a*d^2 - c^(3/2)*Sqrt[c^3
+ 4*a*d^2])*Sqrt[(d^2*(4*a*c + 4*c^2*x^2 + 4*c*d*x^3 + d^2*x^4))/((c^3 + 4*a*d^2)*(Sqrt[c] + (d^2*(c/d + x)^2)
/Sqrt[c^3 + 4*a*d^2])^2)]*(Sqrt[c] + (d^2*(c/d + x)^2)/Sqrt[c^3 + 4*a*d^2])*EllipticF[2*ArcTan[(c + d*x)/(c^(1
/4)*(c^3 + 4*a*d^2)^(1/4))], (1 + c^(3/2)/Sqrt[c^3 + 4*a*d^2])/2])/(16*a*c^(5/4)*d*(c^3 + 4*a*d^2)^(3/4)*Sqrt[
4*a*c + 4*c^2*x^2 + 4*c*d*x^3 + d^2*x^4])

Rule 1106

Int[(P4_)^(p_), x_Symbol] :> With[{a = Coeff[P4, x, 0], b = Coeff[P4, x, 1], c = Coeff[P4, x, 2], d = Coeff[P4
, x, 3], e = Coeff[P4, x, 4]}, Subst[Int[SimplifyIntegrand[(a + d^4/(256*e^3) - (b*d)/(8*e) + (c - (3*d^2)/(8*
e))*x^2 + e*x^4)^p, x], x], x, d/(4*e) + x] /; EqQ[d^3 - 4*c*d*e + 8*b*e^2, 0] && NeQ[d, 0]] /; FreeQ[p, x] &&
PolyQ[P4, x, 4] && NeQ[p, 2] && NeQ[p, 3]

Rule 1092

Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> -Simp[(x*(b^2 - 2*a*c + b*c*x^2)*(a + b*x^2 + c*x^
4)^(p + 1))/(2*a*(p + 1)*(b^2 - 4*a*c)), x] + Dist[1/(2*a*(p + 1)*(b^2 - 4*a*c)), Int[(b^2 - 2*a*c + 2*(p + 1)
*(b^2 - 4*a*c) + b*c*(4*p + 7)*x^2)*(a + b*x^2 + c*x^4)^(p + 1), x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*
a*c, 0] && LtQ[p, -1] && IntegerQ[2*p]

Rule 1197

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 2]}, Dist[(
e + d*q)/q, Int[1/Sqrt[a + b*x^2 + c*x^4], x], x] - Dist[e/q, Int[(1 - q*x^2)/Sqrt[a + b*x^2 + c*x^4], x], x]
/; NeQ[e + d*q, 0]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rule 1103

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(
a + b*x^2 + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2 - (b*q^2)/(4*c)])/(2*q*Sqrt[a + b*x^2 + c
*x^4]), x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rule 1195

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[
(d*x*Sqrt[a + b*x^2 + c*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + b*x^2 + c*x^4)/(a*(1 + q
^2*x^2)^2)]*EllipticE[2*ArcTan[q*x], 1/2 - (b*q^2)/(4*c)])/(q*Sqrt[a + b*x^2 + c*x^4]), x] /; EqQ[e + d*q^2, 0
]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rubi steps

\begin{align*} \int \frac{1}{\left (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )^{3/2}} \, dx &=\operatorname{Subst}\left (\int \frac{1}{\left (c \left (4 a+\frac{c^3}{d^2}\right )-2 c^2 x^2+d^2 x^4\right )^{3/2}} \, dx,x,\frac{c}{d}+x\right )\\ &=-\frac{\left (\frac{c}{d}+x\right ) \left (c^3-4 a d^2-c d^2 \left (\frac{c}{d}+x\right )^2\right )}{8 a c \left (c^3+4 a d^2\right ) \sqrt{4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4}}+\frac{\operatorname{Subst}\left (\int \frac{2 c \left (4 a+\frac{c^3}{d^2}\right ) d^2-2 c^2 d^2 x^2}{\sqrt{c \left (4 a+\frac{c^3}{d^2}\right )-2 c^2 x^2+d^2 x^4}} \, dx,x,\frac{c}{d}+x\right )}{16 a c^2 \left (c^3+4 a d^2\right )}\\ &=-\frac{\left (\frac{c}{d}+x\right ) \left (c^3-4 a d^2-c d^2 \left (\frac{c}{d}+x\right )^2\right )}{8 a c \left (c^3+4 a d^2\right ) \sqrt{4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4}}+\frac{\sqrt{c} \operatorname{Subst}\left (\int \frac{1-\frac{d^2 x^2}{\sqrt{c} \sqrt{c^3+4 a d^2}}}{\sqrt{c \left (4 a+\frac{c^3}{d^2}\right )-2 c^2 x^2+d^2 x^4}} \, dx,x,\frac{c}{d}+x\right )}{8 a \sqrt{c^3+4 a d^2}}+\frac{\left (c^3+4 a d^2-c^{3/2} \sqrt{c^3+4 a d^2}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c \left (4 a+\frac{c^3}{d^2}\right )-2 c^2 x^2+d^2 x^4}} \, dx,x,\frac{c}{d}+x\right )}{8 a c \left (c^3+4 a d^2\right )}\\ &=-\frac{\left (\frac{c}{d}+x\right ) \left (c^3-4 a d^2-c d^2 \left (\frac{c}{d}+x\right )^2\right )}{8 a c \left (c^3+4 a d^2\right ) \sqrt{4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4}}-\frac{d (c+d x) \sqrt{4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4}}{8 a \left (c^3+4 a d^2\right )^{3/2} \left (\sqrt{c}+\frac{(c+d x)^2}{\sqrt{c^3+4 a d^2}}\right )}+\frac{\sqrt [4]{c} \sqrt{\frac{d^2 \left (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )}{\left (c^3+4 a d^2\right ) \left (\sqrt{c}+\frac{(c+d x)^2}{\sqrt{c^3+4 a d^2}}\right )^2}} \left (\sqrt{c}+\frac{(c+d x)^2}{\sqrt{c^3+4 a d^2}}\right ) E\left (2 \tan ^{-1}\left (\frac{c+d x}{\sqrt [4]{c} \sqrt [4]{c^3+4 a d^2}}\right )|\frac{1}{2} \left (1+\frac{c^{3/2}}{\sqrt{c^3+4 a d^2}}\right )\right )}{8 a d \sqrt [4]{c^3+4 a d^2} \sqrt{4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4}}+\frac{\left (c^3+4 a d^2-c^{3/2} \sqrt{c^3+4 a d^2}\right ) \sqrt{\frac{d^2 \left (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )}{\left (c^3+4 a d^2\right ) \left (\sqrt{c}+\frac{(c+d x)^2}{\sqrt{c^3+4 a d^2}}\right )^2}} \left (\sqrt{c}+\frac{(c+d x)^2}{\sqrt{c^3+4 a d^2}}\right ) F\left (2 \tan ^{-1}\left (\frac{c+d x}{\sqrt [4]{c} \sqrt [4]{c^3+4 a d^2}}\right )|\frac{1}{2} \left (1+\frac{c^{3/2}}{\sqrt{c^3+4 a d^2}}\right )\right )}{16 a c^{5/4} d \left (c^3+4 a d^2\right )^{3/4} \sqrt{4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4}}\\ \end{align*}

Mathematica [C]  time = 6.13315, size = 5276, normalized size = 7.83 \[ \text{Result too large to show} \]

Antiderivative was successfully verified.

[In]

Integrate[(4*a*c + 4*c^2*x^2 + 4*c*d*x^3 + d^2*x^4)^(-3/2),x]

[Out]

Result too large to show

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Maple [B]  time = 0.04, size = 5024, normalized size = 7.5 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (d^{2} x^{4} + 4 \, c d x^{3} + 4 \, c^{2} x^{2} + 4 \, a c\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((d^2*x^4 + 4*c*d*x^3 + 4*c^2*x^2 + 4*a*c)^(-3/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{d^{2} x^{4} + 4 \, c d x^{3} + 4 \, c^{2} x^{2} + 4 \, a c}}{d^{4} x^{8} + 8 \, c d^{3} x^{7} + 24 \, c^{2} d^{2} x^{6} + 32 \, c^{3} d x^{5} + 32 \, a c^{2} d x^{3} + 32 \, a c^{3} x^{2} + 8 \,{\left (2 \, c^{4} + a c d^{2}\right )} x^{4} + 16 \, a^{2} c^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(d^2*x^4 + 4*c*d*x^3 + 4*c^2*x^2 + 4*a*c)/(d^4*x^8 + 8*c*d^3*x^7 + 24*c^2*d^2*x^6 + 32*c^3*d*x^5
+ 32*a*c^2*d*x^3 + 32*a*c^3*x^2 + 8*(2*c^4 + a*c*d^2)*x^4 + 16*a^2*c^2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (4 a c + 4 c^{2} x^{2} + 4 c d x^{3} + d^{2} x^{4}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((4*a*c + 4*c**2*x**2 + 4*c*d*x**3 + d**2*x**4)**(-3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (d^{2} x^{4} + 4 \, c d x^{3} + 4 \, c^{2} x^{2} + 4 \, a c\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((d^2*x^4 + 4*c*d*x^3 + 4*c^2*x^2 + 4*a*c)^(-3/2), x)