3.212 \(\int \frac {(d+e x)^3}{\sqrt {a+c x^4}} \, dx\)

Optimal. Leaf size=295 \[ \frac {d \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (3 \sqrt {a} e^2+\sqrt {c} d^2\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 \sqrt [4]{a} c^{3/4} \sqrt {a+c x^4}}-\frac {3 \sqrt [4]{a} d e^2 \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{c^{3/4} \sqrt {a+c x^4}}+\frac {3 d^2 e \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a+c x^4}}\right )}{2 \sqrt {c}}+\frac {3 d e^2 x \sqrt {a+c x^4}}{\sqrt {c} \left (\sqrt {a}+\sqrt {c} x^2\right )}+\frac {e^3 \sqrt {a+c x^4}}{2 c} \]

[Out]

3/2*d^2*e*arctanh(x^2*c^(1/2)/(c*x^4+a)^(1/2))/c^(1/2)+1/2*e^3*(c*x^4+a)^(1/2)/c+3*d*e^2*x*(c*x^4+a)^(1/2)/c^(
1/2)/(a^(1/2)+x^2*c^(1/2))-3*a^(1/4)*d*e^2*(cos(2*arctan(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)/c^(3/4)/(c*x^4+a)^(1/2)+1/2*d*(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))*(3*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^(1/4)/c^(3/4)/(c*x^4+a)^(1/2)

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Rubi [A]  time = 0.16, antiderivative size = 295, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {1885, 1198, 220, 1196, 1248, 641, 217, 206} \[ \frac {d \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (3 \sqrt {a} e^2+\sqrt {c} d^2\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 \sqrt [4]{a} c^{3/4} \sqrt {a+c x^4}}-\frac {3 \sqrt [4]{a} d e^2 \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{c^{3/4} \sqrt {a+c x^4}}+\frac {3 d^2 e \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a+c x^4}}\right )}{2 \sqrt {c}}+\frac {3 d e^2 x \sqrt {a+c x^4}}{\sqrt {c} \left (\sqrt {a}+\sqrt {c} x^2\right )}+\frac {e^3 \sqrt {a+c x^4}}{2 c} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^3/Sqrt[a + c*x^4],x]

[Out]

(e^3*Sqrt[a + c*x^4])/(2*c) + (3*d*e^2*x*Sqrt[a + c*x^4])/(Sqrt[c]*(Sqrt[a] + Sqrt[c]*x^2)) + (3*d^2*e*ArcTanh
[(Sqrt[c]*x^2)/Sqrt[a + c*x^4]])/(2*Sqrt[c]) - (3*a^(1/4)*d*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])/(c^(3/4)*Sqrt[a + c*x^4]) + (d*(Sqrt[c]*d
^2 + 3*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])/(2*a^(1/4)*c^(3/4)*Sqrt[a + c*x^4])

Rule 206

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

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

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 641

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*(a + c*x^2)^(p + 1))/(2*c*(p + 1)),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1]

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 1248

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[(d + e*x)^q
*(a + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x]

Rule 1885

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Module[{q = Expon[Pq, x], j, k}, Int[Sum[x^j*Sum[Coeff[P
q, x, j + (k*n)/2]*x^((k*n)/2), {k, 0, (2*(q - j))/n + 1}]*(a + b*x^n)^p, {j, 0, n/2 - 1}], x]] /; FreeQ[{a, b
, p}, x] && PolyQ[Pq, x] && IGtQ[n/2, 0] &&  !PolyQ[Pq, x^(n/2)]

Rubi steps

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

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Mathematica [C]  time = 0.14, size = 157, normalized size = 0.53 \[ \frac {d^3 x \sqrt {\frac {c x^4}{a}+1} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\frac {c x^4}{a}\right )}{\sqrt {a+c x^4}}+\frac {3 d^2 e \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a+c x^4}}\right )}{2 \sqrt {c}}+\frac {d e^2 x^3 \sqrt {\frac {c x^4}{a}+1} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-\frac {c x^4}{a}\right )}{\sqrt {a+c x^4}}+\frac {e^3 \sqrt {a+c x^4}}{2 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e^3*Sqrt[a + c*x^4])/(2*c) + (3*d^2*e*ArcTanh[(Sqrt[c]*x^2)/Sqrt[a + c*x^4]])/(2*Sqrt[c]) + (d^3*x*Sqrt[1 + (
c*x^4)/a]*Hypergeometric2F1[1/4, 1/2, 5/4, -((c*x^4)/a)])/Sqrt[a + c*x^4] + (d*e^2*x^3*Sqrt[1 + (c*x^4)/a]*Hyp
ergeometric2F1[1/2, 3/4, 7/4, -((c*x^4)/a)])/Sqrt[a + c*x^4]

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fricas [F]  time = 1.23, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}}{\sqrt {c x^{4} + a}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.02, size = 218, normalized size = 0.74 \[ \frac {\sqrt {-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, d^{3} \EllipticF \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , i\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}+\frac {3 d^{2} e \ln \left (\sqrt {c}\, x^{2}+\sqrt {c \,x^{4}+a}\right )}{2 \sqrt {c}}+\frac {3 i \sqrt {-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \left (-\EllipticE \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , i\right )+\EllipticF \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , i\right )\right ) \sqrt {a}\, d \,e^{2}}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, \sqrt {c}}+\frac {\sqrt {c \,x^{4}+a}\, e^{3}}{2 c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*e^3*(c*x^4+a)^(1/2)/c+3*I*d*e^2*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))+3/2*e*d^2*ln(c^(1/2)*x^2+(c*x^4+a)^(1/2))/c^(1/2)+d^3/(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 \[ \int \frac {{\left (e x + d\right )}^{3}}{\sqrt {c x^{4} + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^3/(a + c*x^4)^(1/2),x)

[Out]

int((d + e*x)^3/(a + c*x^4)^(1/2), x)

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sympy [A]  time = 4.19, size = 141, normalized size = 0.48 \[ e^{3} \left (\begin {cases} \frac {x^{4}}{4 \sqrt {a}} & \text {for}\: c = 0 \\\frac {\sqrt {a + c x^{4}}}{2 c} & \text {otherwise} \end {cases}\right ) + \frac {3 d^{2} e \operatorname {asinh}{\left (\frac {\sqrt {c} x^{2}}{\sqrt {a}} \right )}}{2 \sqrt {c}} + \frac {d^{3} x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {5}{4}\right )} + \frac {3 d e^{2} x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {7}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e**3*Piecewise((x**4/(4*sqrt(a)), Eq(c, 0)), (sqrt(a + c*x**4)/(2*c), True)) + 3*d**2*e*asinh(sqrt(c)*x**2/sqr
t(a))/(2*sqrt(c)) + d**3*x*gamma(1/4)*hyper((1/4, 1/2), (5/4,), c*x**4*exp_polar(I*pi)/a)/(4*sqrt(a)*gamma(5/4
)) + 3*d*e**2*x**3*gamma(3/4)*hyper((1/2, 3/4), (7/4,), c*x**4*exp_polar(I*pi)/a)/(4*sqrt(a)*gamma(7/4))

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