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

Optimal. Leaf size=270 \[ \frac {\left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (\sqrt {c} d^2-\sqrt {a} e^2\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{4 a^{5/4} c^{3/4} \sqrt {a+c x^4}}+\frac {e^2 \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 a^{3/4} c^{3/4} \sqrt {a+c x^4}}+\frac {x (d+e x)^2}{2 a \sqrt {a+c x^4}}-\frac {e^2 x \sqrt {a+c x^4}}{2 a \sqrt {c} \left (\sqrt {a}+\sqrt {c} x^2\right )} \]

[Out]

1/2*x*(e*x+d)^2/a/(c*x^4+a)^(1/2)-1/2*e^2*x*(c*x^4+a)^(1/2)/a/c^(1/2)/(a^(1/2)+x^2*c^(1/2))+1/2*e^2*(cos(2*arc
tan(c^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*x/a^(1/4)))*EllipticE(sin(2*arctan(c^(1/4)*x/a^(1/4))),1
/2*2^(1/2))*(a^(1/2)+x^2*c^(1/2))*((c*x^4+a)/(a^(1/2)+x^2*c^(1/2))^2)^(1/2)/a^(3/4)/c^(3/4)/(c*x^4+a)^(1/2)+1/
4*(cos(2*arctan(c^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*x/a^(1/4)))*EllipticF(sin(2*arctan(c^(1/4)*x
/a^(1/4))),1/2*2^(1/2))*(-e^2*a^(1/2)+d^2*c^(1/2))*(a^(1/2)+x^2*c^(1/2))*((c*x^4+a)/(a^(1/2)+x^2*c^(1/2))^2)^(
1/2)/a^(5/4)/c^(3/4)/(c*x^4+a)^(1/2)

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Rubi [A]  time = 0.12, antiderivative size = 270, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {1855, 1198, 220, 1196} \[ \frac {\left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (\sqrt {c} d^2-\sqrt {a} e^2\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{4 a^{5/4} c^{3/4} \sqrt {a+c x^4}}+\frac {e^2 \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 a^{3/4} c^{3/4} \sqrt {a+c x^4}}+\frac {x (d+e x)^2}{2 a \sqrt {a+c x^4}}-\frac {e^2 x \sqrt {a+c x^4}}{2 a \sqrt {c} \left (\sqrt {a}+\sqrt {c} x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^2/(a + c*x^4)^(3/2),x]

[Out]

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

Rule 220

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

Rule 1196

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

Rule 1198

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

Rule 1855

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> -Simp[(x*Pq*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Di
st[1/(a*n*(p + 1)), Int[ExpandToSum[n*(p + 1)*Pq + D[x*Pq, x], x]*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b},
x] && PolyQ[Pq, x] && IGtQ[n, 0] && LtQ[p, -1] && LtQ[Expon[Pq, x], n - 1]

Rubi steps

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

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Mathematica [C]  time = 0.06, size = 108, normalized size = 0.40 \[ \frac {x \left (3 d^2 \sqrt {\frac {c x^4}{a}+1} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\frac {c x^4}{a}\right )+2 e^2 x^2 \sqrt {\frac {c x^4}{a}+1} \, _2F_1\left (\frac {3}{4},\frac {3}{2};\frac {7}{4};-\frac {c x^4}{a}\right )+3 d (d+2 e x)\right )}{6 a \sqrt {a+c x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(3*d*(d + 2*e*x) + 3*d^2*Sqrt[1 + (c*x^4)/a]*Hypergeometric2F1[1/4, 1/2, 5/4, -((c*x^4)/a)] + 2*e^2*x^2*Sqr
t[1 + (c*x^4)/a]*Hypergeometric2F1[3/4, 3/2, 7/4, -((c*x^4)/a)]))/(6*a*Sqrt[a + c*x^4])

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fricas [F]  time = 0.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c x^{4} + a} {\left (e^{2} x^{2} + 2 \, d e x + d^{2}\right )}}{c^{2} x^{8} + 2 \, a c x^{4} + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(c*x^4 + a)*(e^2*x^2 + 2*d*e*x + d^2)/(c^2*x^8 + 2*a*c*x^4 + a^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.02, size = 239, normalized size = 0.89 \[ \frac {d e \,x^{2}}{\sqrt {c \,x^{4}+a}\, a}+\left (\frac {x}{2 \sqrt {\left (x^{4}+\frac {a}{c}\right ) c}\, a}+\frac {\sqrt {-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \EllipticF \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , i\right )}{2 \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, a}\right ) d^{2}+\left (\frac {x^{3}}{2 \sqrt {\left (x^{4}+\frac {a}{c}\right ) c}\, a}-\frac {i \sqrt {-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \left (-\EllipticE \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , i\right )+\EllipticF \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , i\right )\right )}{2 \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, \sqrt {a}\, \sqrt {c}}\right ) e^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^2*(1/2/((x^4+a/c)*c)^(1/2)/a*x^3-1/2*I/a^(1/2)/(I/a^(1/2)*c^(1/2))^(1/2)*(-I/a^(1/2)*c^(1/2)*x^2+1)^(1/2)*(I
/a^(1/2)*c^(1/2)*x^2+1)^(1/2)/(c*x^4+a)^(1/2)/c^(1/2)*(EllipticF((I/a^(1/2)*c^(1/2))^(1/2)*x,I)-EllipticE((I/a
^(1/2)*c^(1/2))^(1/2)*x,I)))+d*e/(c*x^4+a)^(1/2)/a*x^2+d^2*(1/2/((x^4+a/c)*c)^(1/2)/a*x+1/2/a/(I/a^(1/2)*c^(1/
2))^(1/2)*(-I/a^(1/2)*c^(1/2)*x^2+1)^(1/2)*(I/a^(1/2)*c^(1/2)*x^2+1)^(1/2)/(c*x^4+a)^(1/2)*EllipticF((I/a^(1/2
)*c^(1/2))^(1/2)*x,I))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (d+e\,x\right )}^2}{{\left (c\,x^4+a\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^2/(a + c*x^4)^(3/2),x)

[Out]

int((d + e*x)^2/(a + c*x^4)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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