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

Optimal. Leaf size=298 \[ \frac {d \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (\sqrt {c} d^2-3 \sqrt {a} e^2\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{4 a^{5/4} c^{3/4} \sqrt {a+c x^4}}+\frac {3 d e^2 \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 a^{3/4} c^{3/4} \sqrt {a+c x^4}}-\frac {a e^3-c x \left (d^3+3 d^2 e x+3 d e^2 x^2\right )}{2 a c \sqrt {a+c x^4}}-\frac {3 d e^2 x \sqrt {a+c x^4}}{2 a \sqrt {c} \left (\sqrt {a}+\sqrt {c} x^2\right )} \]

[Out]

1/2*(-a*e^3+c*x*(3*d*e^2*x^2+3*d^2*e*x+d^3))/a/c/(c*x^4+a)^(1/2)-3/2*d*e^2*x*(c*x^4+a)^(1/2)/a/c^(1/2)/(a^(1/2
)+x^2*c^(1/2))+3/2*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)
/a^(3/4)/c^(3/4)/(c*x^4+a)^(1/2)+1/4*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^(5/4)/c^(3/4)/(c*x^4+a)^(1/2)

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Rubi [A]  time = 0.14, antiderivative size = 298, 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 = {1854, 1198, 220, 1196} \[ \frac {d \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (\sqrt {c} d^2-3 \sqrt {a} e^2\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{4 a^{5/4} c^{3/4} \sqrt {a+c x^4}}+\frac {3 d e^2 \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 a^{3/4} c^{3/4} \sqrt {a+c x^4}}-\frac {a e^3-c x \left (3 d^2 e x+d^3+3 d e^2 x^2\right )}{2 a c \sqrt {a+c x^4}}-\frac {3 d e^2 x \sqrt {a+c x^4}}{2 a \sqrt {c} \left (\sqrt {a}+\sqrt {c} x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^3/(a + c*x^4)^(3/2),x]

[Out]

(-3*d*e^2*x*Sqrt[a + c*x^4])/(2*a*Sqrt[c]*(Sqrt[a] + Sqrt[c]*x^2)) - (a*e^3 - c*x*(d^3 + 3*d^2*e*x + 3*d*e^2*x
^2))/(2*a*c*Sqrt[a + c*x^4]) + (3*d*e^2*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*El
lipticE[2*ArcTan[(c^(1/4)*x)/a^(1/4)], 1/2])/(2*a^(3/4)*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])/(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 1854

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

Rubi steps

\begin {align*} \int \frac {(d+e x)^3}{\left (a+c x^4\right )^{3/2}} \, dx &=-\frac {a e^3-c x \left (d^3+3 d^2 e x+3 d e^2 x^2\right )}{2 a c \sqrt {a+c x^4}}-\frac {\int \frac {-d^3+3 d e^2 x^2}{\sqrt {a+c x^4}} \, dx}{2 a}\\ &=-\frac {a e^3-c x \left (d^3+3 d^2 e x+3 d e^2 x^2\right )}{2 a c \sqrt {a+c x^4}}+\frac {\left (3 d e^2\right ) \int \frac {1-\frac {\sqrt {c} x^2}{\sqrt {a}}}{\sqrt {a+c x^4}} \, dx}{2 \sqrt {a} \sqrt {c}}+\frac {\left (d \left (d^2-\frac {3 \sqrt {a} e^2}{\sqrt {c}}\right )\right ) \int \frac {1}{\sqrt {a+c x^4}} \, dx}{2 a}\\ &=-\frac {3 d e^2 x \sqrt {a+c x^4}}{2 a \sqrt {c} \left (\sqrt {a}+\sqrt {c} x^2\right )}-\frac {a e^3-c x \left (d^3+3 d^2 e x+3 d e^2 x^2\right )}{2 a c \sqrt {a+c x^4}}+\frac {3 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 )}{2 a^{3/4} c^{3/4} \sqrt {a+c x^4}}+\frac {d \left (d^2-\frac {3 \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 = 0.07, size = 126, normalized size = 0.42 \[ \frac {c d^3 x \sqrt {\frac {c x^4}{a}+1} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\frac {c x^4}{a}\right )+2 c d e^2 x^3 \sqrt {\frac {c x^4}{a}+1} \, _2F_1\left (\frac {3}{4},\frac {3}{2};\frac {7}{4};-\frac {c x^4}{a}\right )-a e^3+c d^3 x+3 c d^2 e x^2}{2 a c \sqrt {a+c x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 1.13, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}\right )} \sqrt {c x^{4} + a}}{c^{2} x^{8} + 2 \, a c x^{4} + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.03, size = 261, normalized size = 0.88 \[ \frac {3 d^{2} e \,x^{2}}{2 \sqrt {c \,x^{4}+a}\, a}+\left (\frac {x}{2 \sqrt {\left (x^{4}+\frac {a}{c}\right ) c}\, a}+\frac {\sqrt {-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \EllipticF \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , i\right )}{2 \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, a}\right ) d^{3}+3 \left (\frac {x^{3}}{2 \sqrt {\left (x^{4}+\frac {a}{c}\right ) c}\, a}-\frac {i \sqrt {-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \left (-\EllipticE \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , i\right )+\EllipticF \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , i\right )\right )}{2 \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, \sqrt {a}\, \sqrt {c}}\right ) d \,e^{2}-\frac {e^{3}}{2 \sqrt {c \,x^{4}+a}\, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^3/(a + c*x^4)^(3/2),x)

[Out]

int((d + e*x)^3/(a + c*x^4)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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