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

Optimal. Leaf size=355 \[ \frac {a^{3/4} d \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (9 \sqrt {a} e^2+5 \sqrt {c} d^2\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{15 c^{3/4} \sqrt {a+c x^4}}-\frac {6 a^{5/4} d e^2 \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 c^{3/4} \sqrt {a+c x^4}}+\frac {1}{15} d x \sqrt {a+c x^4} \left (5 d^2+9 e^2 x^2\right )+\frac {3}{4} d^2 e x^2 \sqrt {a+c x^4}+\frac {3 a d^2 e \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a+c x^4}}\right )}{4 \sqrt {c}}+\frac {6 a d e^2 x \sqrt {a+c x^4}}{5 \sqrt {c} \left (\sqrt {a}+\sqrt {c} x^2\right )}+\frac {e^3 \left (a+c x^4\right )^{3/2}}{6 c} \]

[Out]

1/6*e^3*(c*x^4+a)^(3/2)/c+3/4*a*d^2*e*arctanh(x^2*c^(1/2)/(c*x^4+a)^(1/2))/c^(1/2)+3/4*d^2*e*x^2*(c*x^4+a)^(1/
2)+1/15*d*x*(9*e^2*x^2+5*d^2)*(c*x^4+a)^(1/2)+6/5*a*d*e^2*x*(c*x^4+a)^(1/2)/c^(1/2)/(a^(1/2)+x^2*c^(1/2))-6/5*
a^(5/4)*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*arct
an(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)+1/15*a^(3/4)*d*(cos(2*arctan(c^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*x/a^(1/4)))*Ellip
ticF(sin(2*arctan(c^(1/4)*x/a^(1/4))),1/2*2^(1/2))*(9*e^2*a^(1/2)+5*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)/c^(3/4)/(c*x^4+a)^(1/2)

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Rubi [A]  time = 0.23, antiderivative size = 355, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.526, Rules used = {1885, 1177, 1198, 220, 1196, 1248, 641, 195, 217, 206} \[ \frac {a^{3/4} d \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (9 \sqrt {a} e^2+5 \sqrt {c} d^2\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{15 c^{3/4} \sqrt {a+c x^4}}-\frac {6 a^{5/4} d e^2 \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 c^{3/4} \sqrt {a+c x^4}}+\frac {1}{15} d x \sqrt {a+c x^4} \left (5 d^2+9 e^2 x^2\right )+\frac {3}{4} d^2 e x^2 \sqrt {a+c x^4}+\frac {3 a d^2 e \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a+c x^4}}\right )}{4 \sqrt {c}}+\frac {6 a d e^2 x \sqrt {a+c x^4}}{5 \sqrt {c} \left (\sqrt {a}+\sqrt {c} x^2\right )}+\frac {e^3 \left (a+c x^4\right )^{3/2}}{6 c} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^3*Sqrt[a + c*x^4],x]

[Out]

(3*d^2*e*x^2*Sqrt[a + c*x^4])/4 + (6*a*d*e^2*x*Sqrt[a + c*x^4])/(5*Sqrt[c]*(Sqrt[a] + Sqrt[c]*x^2)) + (d*x*(5*
d^2 + 9*e^2*x^2)*Sqrt[a + c*x^4])/15 + (e^3*(a + c*x^4)^(3/2))/(6*c) + (3*a*d^2*e*ArcTanh[(Sqrt[c]*x^2)/Sqrt[a
+ c*x^4]])/(4*Sqrt[c]) - (6*a^(5/4)*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])/(5*c^(3/4)*Sqrt[a + c*x^4]) + (a^(3/4)*d*(5*Sqrt[c]*d^2 + 9*Sq
rt[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])/(15*c^(3/4)*Sqrt[a + c*x^4])

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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 1177

Int[((d_) + (e_.)*(x_)^2)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(x*(d*(4*p + 3) + e*(4*p + 1)*x^2)*(a
+ c*x^4)^p)/((4*p + 1)*(4*p + 3)), x] + Dist[(2*p)/((4*p + 1)*(4*p + 3)), Int[Simp[2*a*d*(4*p + 3) + (2*a*e*(4
*p + 1))*x^2, x]*(a + c*x^4)^(p - 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] &
& FractionQ[p] && IntegerQ[2*p]

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 (d+e x)^3 \sqrt {a+c x^4} \, dx &=\int \left (\left (d^3+3 d e^2 x^2\right ) \sqrt {a+c x^4}+x \left (3 d^2 e+e^3 x^2\right ) \sqrt {a+c x^4}\right ) \, dx\\ &=\int \left (d^3+3 d e^2 x^2\right ) \sqrt {a+c x^4} \, dx+\int x \left (3 d^2 e+e^3 x^2\right ) \sqrt {a+c x^4} \, dx\\ &=\frac {1}{15} d x \left (5 d^2+9 e^2 x^2\right ) \sqrt {a+c x^4}+\frac {1}{15} \int \frac {10 a d^3+18 a d e^2 x^2}{\sqrt {a+c x^4}} \, dx+\frac {1}{2} \operatorname {Subst}\left (\int \left (3 d^2 e+e^3 x\right ) \sqrt {a+c x^2} \, dx,x,x^2\right )\\ &=\frac {1}{15} d x \left (5 d^2+9 e^2 x^2\right ) \sqrt {a+c x^4}+\frac {e^3 \left (a+c x^4\right )^{3/2}}{6 c}+\frac {1}{2} \left (3 d^2 e\right ) \operatorname {Subst}\left (\int \sqrt {a+c x^2} \, dx,x,x^2\right )-\frac {\left (6 a^{3/2} d e^2\right ) \int \frac {1-\frac {\sqrt {c} x^2}{\sqrt {a}}}{\sqrt {a+c x^4}} \, dx}{5 \sqrt {c}}+\frac {1}{15} \left (2 a d \left (5 d^2+\frac {9 \sqrt {a} e^2}{\sqrt {c}}\right )\right ) \int \frac {1}{\sqrt {a+c x^4}} \, dx\\ &=\frac {3}{4} d^2 e x^2 \sqrt {a+c x^4}+\frac {6 a d e^2 x \sqrt {a+c x^4}}{5 \sqrt {c} \left (\sqrt {a}+\sqrt {c} x^2\right )}+\frac {1}{15} d x \left (5 d^2+9 e^2 x^2\right ) \sqrt {a+c x^4}+\frac {e^3 \left (a+c x^4\right )^{3/2}}{6 c}-\frac {6 a^{5/4} 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 )}{5 c^{3/4} \sqrt {a+c x^4}}+\frac {a^{3/4} d \left (5 \sqrt {c} d^2+9 \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 )}{15 c^{3/4} \sqrt {a+c x^4}}+\frac {1}{4} \left (3 a d^2 e\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+c x^2}} \, dx,x,x^2\right )\\ &=\frac {3}{4} d^2 e x^2 \sqrt {a+c x^4}+\frac {6 a d e^2 x \sqrt {a+c x^4}}{5 \sqrt {c} \left (\sqrt {a}+\sqrt {c} x^2\right )}+\frac {1}{15} d x \left (5 d^2+9 e^2 x^2\right ) \sqrt {a+c x^4}+\frac {e^3 \left (a+c x^4\right )^{3/2}}{6 c}-\frac {6 a^{5/4} 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 )}{5 c^{3/4} \sqrt {a+c x^4}}+\frac {a^{3/4} d \left (5 \sqrt {c} d^2+9 \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 )}{15 c^{3/4} \sqrt {a+c x^4}}+\frac {1}{4} \left (3 a 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 {3}{4} d^2 e x^2 \sqrt {a+c x^4}+\frac {6 a d e^2 x \sqrt {a+c x^4}}{5 \sqrt {c} \left (\sqrt {a}+\sqrt {c} x^2\right )}+\frac {1}{15} d x \left (5 d^2+9 e^2 x^2\right ) \sqrt {a+c x^4}+\frac {e^3 \left (a+c x^4\right )^{3/2}}{6 c}+\frac {3 a d^2 e \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a+c x^4}}\right )}{4 \sqrt {c}}-\frac {6 a^{5/4} 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 )}{5 c^{3/4} \sqrt {a+c x^4}}+\frac {a^{3/4} d \left (5 \sqrt {c} d^2+9 \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 )}{15 c^{3/4} \sqrt {a+c x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.14, size = 186, normalized size = 0.52 \[ \frac {\sqrt {a+c x^4} \left (12 c d^3 x \, _2F_1\left (-\frac {1}{2},\frac {1}{4};\frac {5}{4};-\frac {c x^4}{a}\right )+9 \sqrt {a} \sqrt {c} d^2 e \sinh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )+9 c d^2 e x^2 \sqrt {\frac {c x^4}{a}+1}+12 c d e^2 x^3 \, _2F_1\left (-\frac {1}{2},\frac {3}{4};\frac {7}{4};-\frac {c x^4}{a}\right )+2 c e^3 x^4 \sqrt {\frac {c x^4}{a}+1}+2 a e^3 \sqrt {\frac {c x^4}{a}+1}\right )}{12 c \sqrt {\frac {c x^4}{a}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + c*x^4]*(2*a*e^3*Sqrt[1 + (c*x^4)/a] + 9*c*d^2*e*x^2*Sqrt[1 + (c*x^4)/a] + 2*c*e^3*x^4*Sqrt[1 + (c*x^
4)/a] + 9*Sqrt[a]*Sqrt[c]*d^2*e*ArcSinh[(Sqrt[c]*x^2)/Sqrt[a]] + 12*c*d^3*x*Hypergeometric2F1[-1/2, 1/4, 5/4,
-((c*x^4)/a)] + 12*c*d*e^2*x^3*Hypergeometric2F1[-1/2, 3/4, 7/4, -((c*x^4)/a)]))/(12*c*Sqrt[1 + (c*x^4)/a])

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fricas [F]  time = 0.55, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}\right )} \sqrt {c x^{4} + a}, x\right ) \]

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 \[ \int \sqrt {c x^{4} + a} {\left (e x + d\right )}^{3}\,{d x} \]

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(sqrt(c*x^4 + a)*(e*x + d)^3, x)

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maple [C]  time = 0.06, size = 334, normalized size = 0.94 \[ \frac {3 \sqrt {c \,x^{4}+a}\, d \,e^{2} x^{3}}{5}-\frac {6 i \sqrt {-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, a^{\frac {3}{2}} d \,e^{2} \EllipticE \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , i\right )}{5 \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, \sqrt {c}}+\frac {6 i \sqrt {-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, a^{\frac {3}{2}} d \,e^{2} \EllipticF \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , i\right )}{5 \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, \sqrt {c}}+\frac {2 \sqrt {-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, a \,d^{3} \EllipticF \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , i\right )}{3 \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}+\frac {3 \sqrt {c \,x^{4}+a}\, d^{2} e \,x^{2}}{4}+\frac {3 a \,d^{2} e \ln \left (\sqrt {c}\, x^{2}+\sqrt {c \,x^{4}+a}\right )}{4 \sqrt {c}}+\frac {\sqrt {c \,x^{4}+a}\, d^{3} x}{3}+\frac {\left (c \,x^{4}+a \right )^{\frac {3}{2}} e^{3}}{6 c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*e^3*(c*x^4+a)^(3/2)/c+3/5*d*e^2*x^3*(c*x^4+a)^(1/2)+6/5*I*d*e^2*a^(3/2)/(I/a^(1/2)*c^(1/2))^(1/2)*(1-I/a^(
1/2)*c^(1/2)*x^2)^(1/2)*(1+I/a^(1/2)*c^(1/2)*x^2)^(1/2)/(c*x^4+a)^(1/2)/c^(1/2)*EllipticF(x*(I/a^(1/2)*c^(1/2)
)^(1/2),I)-6/5*I*d*e^2*a^(3/2)/(I/a^(1/2)*c^(1/2))^(1/2)*(1-I/a^(1/2)*c^(1/2)*x^2)^(1/2)*(1+I/a^(1/2)*c^(1/2)*
x^2)^(1/2)/(c*x^4+a)^(1/2)/c^(1/2)*EllipticE(x*(I/a^(1/2)*c^(1/2))^(1/2),I)+3/4*d^2*e*x^2*(c*x^4+a)^(1/2)+3/4*
e*d^2*a/c^(1/2)*ln(x^2*c^(1/2)+(c*x^4+a)^(1/2))+1/3*d^3*x*(c*x^4+a)^(1/2)+2/3*d^3*a/(I/a^(1/2)*c^(1/2))^(1/2)*
(1-I/a^(1/2)*c^(1/2)*x^2)^(1/2)*(1+I/a^(1/2)*c^(1/2)*x^2)^(1/2)/(c*x^4+a)^(1/2)*EllipticF(x*(I/a^(1/2)*c^(1/2)
)^(1/2),I)

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

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(sqrt(c*x^4 + a)*(e*x + d)^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + c*x^4)^(1/2)*(d + e*x)^3,x)

[Out]

int((a + c*x^4)^(1/2)*(d + e*x)^3, x)

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sympy [A]  time = 4.65, size = 175, normalized size = 0.49 \[ \frac {\sqrt {a} d^{3} x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} + \frac {3 \sqrt {a} d^{2} e x^{2} \sqrt {1 + \frac {c x^{4}}{a}}}{4} + \frac {3 \sqrt {a} d e^{2} x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} + \frac {3 a d^{2} e \operatorname {asinh}{\left (\frac {\sqrt {c} x^{2}}{\sqrt {a}} \right )}}{4 \sqrt {c}} + e^{3} \left (\begin {cases} \frac {\sqrt {a} x^{4}}{4} & \text {for}\: c = 0 \\\frac {\left (a + c x^{4}\right )^{\frac {3}{2}}}{6 c} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(a)*d**3*x*gamma(1/4)*hyper((-1/2, 1/4), (5/4,), c*x**4*exp_polar(I*pi)/a)/(4*gamma(5/4)) + 3*sqrt(a)*d**2
*e*x**2*sqrt(1 + c*x**4/a)/4 + 3*sqrt(a)*d*e**2*x**3*gamma(3/4)*hyper((-1/2, 3/4), (7/4,), c*x**4*exp_polar(I*
pi)/a)/(4*gamma(7/4)) + 3*a*d**2*e*asinh(sqrt(c)*x**2/sqrt(a))/(4*sqrt(c)) + e**3*Piecewise((sqrt(a)*x**4/4, E
q(c, 0)), ((a + c*x**4)**(3/2)/(6*c), True))

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