3.6.14 \(\int (c+d x+e x^2+f x^3) (a+b x^4)^{3/2} \, dx\) [514]

Optimal. Leaf size=382 \[ \frac {3}{16} a d x^2 \sqrt {a+b x^4}+\frac {4 a^2 e x \sqrt {a+b x^4}}{15 \sqrt {b} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {2}{105} a x \left (15 c+7 e x^2\right ) \sqrt {a+b x^4}+\frac {1}{8} d x^2 \left (a+b x^4\right )^{3/2}+\frac {1}{63} x \left (9 c+7 e x^2\right ) \left (a+b x^4\right )^{3/2}+\frac {f \left (a+b x^4\right )^{5/2}}{10 b}+\frac {3 a^2 d \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{16 \sqrt {b}}-\frac {4 a^{9/4} e \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{15 b^{3/4} \sqrt {a+b x^4}}+\frac {2 a^{7/4} \left (15 \sqrt {b} c+7 \sqrt {a} e\right ) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{105 b^{3/4} \sqrt {a+b x^4}} \]

[Out]

1/8*d*x^2*(b*x^4+a)^(3/2)+1/63*x*(7*e*x^2+9*c)*(b*x^4+a)^(3/2)+1/10*f*(b*x^4+a)^(5/2)/b+3/16*a^2*d*arctanh(x^2
*b^(1/2)/(b*x^4+a)^(1/2))/b^(1/2)+3/16*a*d*x^2*(b*x^4+a)^(1/2)+2/105*a*x*(7*e*x^2+15*c)*(b*x^4+a)^(1/2)+4/15*a
^2*e*x*(b*x^4+a)^(1/2)/b^(1/2)/(a^(1/2)+x^2*b^(1/2))-4/15*a^(9/4)*e*(cos(2*arctan(b^(1/4)*x/a^(1/4)))^2)^(1/2)
/cos(2*arctan(b^(1/4)*x/a^(1/4)))*EllipticE(sin(2*arctan(b^(1/4)*x/a^(1/4))),1/2*2^(1/2))*(a^(1/2)+x^2*b^(1/2)
)*((b*x^4+a)/(a^(1/2)+x^2*b^(1/2))^2)^(1/2)/b^(3/4)/(b*x^4+a)^(1/2)+2/105*a^(7/4)*(cos(2*arctan(b^(1/4)*x/a^(1
/4)))^2)^(1/2)/cos(2*arctan(b^(1/4)*x/a^(1/4)))*EllipticF(sin(2*arctan(b^(1/4)*x/a^(1/4))),1/2*2^(1/2))*(7*e*a
^(1/2)+15*c*b^(1/2))*(a^(1/2)+x^2*b^(1/2))*((b*x^4+a)/(a^(1/2)+x^2*b^(1/2))^2)^(1/2)/b^(3/4)/(b*x^4+a)^(1/2)

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Rubi [A]
time = 0.17, antiderivative size = 382, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.370, Rules used = {1899, 1191, 1212, 226, 1210, 1262, 655, 201, 223, 212} \begin {gather*} \frac {2 a^{7/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (7 \sqrt {a} e+15 \sqrt {b} c\right ) F\left (2 \text {ArcTan}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{105 b^{3/4} \sqrt {a+b x^4}}-\frac {4 a^{9/4} e \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \text {ArcTan}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{15 b^{3/4} \sqrt {a+b x^4}}+\frac {3 a^2 d \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{16 \sqrt {b}}+\frac {4 a^2 e x \sqrt {a+b x^4}}{15 \sqrt {b} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {1}{63} x \left (a+b x^4\right )^{3/2} \left (9 c+7 e x^2\right )+\frac {2}{105} a x \sqrt {a+b x^4} \left (15 c+7 e x^2\right )+\frac {1}{8} d x^2 \left (a+b x^4\right )^{3/2}+\frac {3}{16} a d x^2 \sqrt {a+b x^4}+\frac {f \left (a+b x^4\right )^{5/2}}{10 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a*d*x^2*Sqrt[a + b*x^4])/16 + (4*a^2*e*x*Sqrt[a + b*x^4])/(15*Sqrt[b]*(Sqrt[a] + Sqrt[b]*x^2)) + (2*a*x*(15
*c + 7*e*x^2)*Sqrt[a + b*x^4])/105 + (d*x^2*(a + b*x^4)^(3/2))/8 + (x*(9*c + 7*e*x^2)*(a + b*x^4)^(3/2))/63 +
(f*(a + b*x^4)^(5/2))/(10*b) + (3*a^2*d*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/(16*Sqrt[b]) - (4*a^(9/4)*e*(S
qrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*EllipticE[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2
])/(15*b^(3/4)*Sqrt[a + b*x^4]) + (2*a^(7/4)*(15*Sqrt[b]*c + 7*Sqrt[a]*e)*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*
x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*EllipticF[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(105*b^(3/4)*Sqrt[a + b*x^4])

Rule 201

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 212

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

Rule 223

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 226

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)]/(2*q*Sqrt[a + b*x^4]))*EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 655

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 1191

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 1210

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)]/(q*Sqrt[a + c*x^4
]))*EllipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rule 1212

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 1262

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 1899

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 \left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^{3/2} \, dx &=\int \left (\left (c+e x^2\right ) \left (a+b x^4\right )^{3/2}+x \left (d+f x^2\right ) \left (a+b x^4\right )^{3/2}\right ) \, dx\\ &=\int \left (c+e x^2\right ) \left (a+b x^4\right )^{3/2} \, dx+\int x \left (d+f x^2\right ) \left (a+b x^4\right )^{3/2} \, dx\\ &=\frac {1}{63} x \left (9 c+7 e x^2\right ) \left (a+b x^4\right )^{3/2}+\frac {1}{21} \int \left (18 a c+14 a e x^2\right ) \sqrt {a+b x^4} \, dx+\frac {1}{2} \text {Subst}\left (\int (d+f x) \left (a+b x^2\right )^{3/2} \, dx,x,x^2\right )\\ &=\frac {2}{105} a x \left (15 c+7 e x^2\right ) \sqrt {a+b x^4}+\frac {1}{63} x \left (9 c+7 e x^2\right ) \left (a+b x^4\right )^{3/2}+\frac {f \left (a+b x^4\right )^{5/2}}{10 b}+\frac {1}{315} \int \frac {180 a^2 c+84 a^2 e x^2}{\sqrt {a+b x^4}} \, dx+\frac {1}{2} d \text {Subst}\left (\int \left (a+b x^2\right )^{3/2} \, dx,x,x^2\right )\\ &=\frac {2}{105} a x \left (15 c+7 e x^2\right ) \sqrt {a+b x^4}+\frac {1}{8} d x^2 \left (a+b x^4\right )^{3/2}+\frac {1}{63} x \left (9 c+7 e x^2\right ) \left (a+b x^4\right )^{3/2}+\frac {f \left (a+b x^4\right )^{5/2}}{10 b}+\frac {1}{8} (3 a d) \text {Subst}\left (\int \sqrt {a+b x^2} \, dx,x,x^2\right )-\frac {\left (4 a^{5/2} e\right ) \int \frac {1-\frac {\sqrt {b} x^2}{\sqrt {a}}}{\sqrt {a+b x^4}} \, dx}{15 \sqrt {b}}+\frac {1}{105} \left (4 a^2 \left (15 c+\frac {7 \sqrt {a} e}{\sqrt {b}}\right )\right ) \int \frac {1}{\sqrt {a+b x^4}} \, dx\\ &=\frac {3}{16} a d x^2 \sqrt {a+b x^4}+\frac {4 a^2 e x \sqrt {a+b x^4}}{15 \sqrt {b} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {2}{105} a x \left (15 c+7 e x^2\right ) \sqrt {a+b x^4}+\frac {1}{8} d x^2 \left (a+b x^4\right )^{3/2}+\frac {1}{63} x \left (9 c+7 e x^2\right ) \left (a+b x^4\right )^{3/2}+\frac {f \left (a+b x^4\right )^{5/2}}{10 b}-\frac {4 a^{9/4} e \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{15 b^{3/4} \sqrt {a+b x^4}}+\frac {2 a^{7/4} \left (15 \sqrt {b} c+7 \sqrt {a} e\right ) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{105 b^{3/4} \sqrt {a+b x^4}}+\frac {1}{16} \left (3 a^2 d\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,x^2\right )\\ &=\frac {3}{16} a d x^2 \sqrt {a+b x^4}+\frac {4 a^2 e x \sqrt {a+b x^4}}{15 \sqrt {b} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {2}{105} a x \left (15 c+7 e x^2\right ) \sqrt {a+b x^4}+\frac {1}{8} d x^2 \left (a+b x^4\right )^{3/2}+\frac {1}{63} x \left (9 c+7 e x^2\right ) \left (a+b x^4\right )^{3/2}+\frac {f \left (a+b x^4\right )^{5/2}}{10 b}-\frac {4 a^{9/4} e \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{15 b^{3/4} \sqrt {a+b x^4}}+\frac {2 a^{7/4} \left (15 \sqrt {b} c+7 \sqrt {a} e\right ) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{105 b^{3/4} \sqrt {a+b x^4}}+\frac {1}{16} \left (3 a^2 d\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x^2}{\sqrt {a+b x^4}}\right )\\ &=\frac {3}{16} a d x^2 \sqrt {a+b x^4}+\frac {4 a^2 e x \sqrt {a+b x^4}}{15 \sqrt {b} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {2}{105} a x \left (15 c+7 e x^2\right ) \sqrt {a+b x^4}+\frac {1}{8} d x^2 \left (a+b x^4\right )^{3/2}+\frac {1}{63} x \left (9 c+7 e x^2\right ) \left (a+b x^4\right )^{3/2}+\frac {f \left (a+b x^4\right )^{5/2}}{10 b}+\frac {3 a^2 d \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{16 \sqrt {b}}-\frac {4 a^{9/4} e \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{15 b^{3/4} \sqrt {a+b x^4}}+\frac {2 a^{7/4} \left (15 \sqrt {b} c+7 \sqrt {a} e\right ) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{105 b^{3/4} \sqrt {a+b x^4}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.39, size = 175, normalized size = 0.46 \begin {gather*} \frac {1}{240} \sqrt {a+b x^4} \left (\frac {24 f \left (a+b x^4\right )^2}{b}+15 d \left (5 a x^2+2 b x^6+\frac {3 a^{5/2} \sqrt {1+\frac {b x^4}{a}} \sinh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{\sqrt {b} \left (a+b x^4\right )}\right )+\frac {240 a c x \, _2F_1\left (-\frac {3}{2},\frac {1}{4};\frac {5}{4};-\frac {b x^4}{a}\right )}{\sqrt {1+\frac {b x^4}{a}}}+\frac {80 a e x^3 \, _2F_1\left (-\frac {3}{2},\frac {3}{4};\frac {7}{4};-\frac {b x^4}{a}\right )}{\sqrt {1+\frac {b x^4}{a}}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x^4]*((24*f*(a + b*x^4)^2)/b + 15*d*(5*a*x^2 + 2*b*x^6 + (3*a^(5/2)*Sqrt[1 + (b*x^4)/a]*ArcSinh[(S
qrt[b]*x^2)/Sqrt[a]])/(Sqrt[b]*(a + b*x^4))) + (240*a*c*x*Hypergeometric2F1[-3/2, 1/4, 5/4, -((b*x^4)/a)])/Sqr
t[1 + (b*x^4)/a] + (80*a*e*x^3*Hypergeometric2F1[-3/2, 3/4, 7/4, -((b*x^4)/a)])/Sqrt[1 + (b*x^4)/a]))/240

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Maple [C] Result contains complex when optimal does not.
time = 0.40, size = 309, normalized size = 0.81

method	result	size
default	\(\frac {f \left (b \,x^{4}+a \right )^{\frac {5}{2}}}{10 b}+e \left (\frac {b \,x^{7} \sqrt {b \,x^{4}+a}}{9}+\frac {11 a \,x^{3} \sqrt {b \,x^{4}+a}}{45}+\frac {4 i a^{\frac {5}{2}} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\EllipticE \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{15 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}\right )+d \left (\frac {b \,x^{6} \sqrt {b \,x^{4}+a}}{8}+\frac {5 a \,x^{2} \sqrt {b \,x^{4}+a}}{16}+\frac {3 a^{2} \ln \left (x^{2} \sqrt {b}+\sqrt {b \,x^{4}+a}\right )}{16 \sqrt {b}}\right )+c \left (\frac {b \,x^{5} \sqrt {b \,x^{4}+a}}{7}+\frac {3 a x \sqrt {b \,x^{4}+a}}{7}+\frac {4 a^{2} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{7 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )\)	\(309\)
risch	\(\frac {\left (504 b^{2} f \,x^{8}+560 b^{2} e \,x^{7}+630 b^{2} d \,x^{6}+720 b^{2} c \,x^{5}+1008 a b f \,x^{4}+1232 a b e \,x^{3}+1575 a b d \,x^{2}+2160 a b c x +504 a^{2} f \right ) \sqrt {b \,x^{4}+a}}{5040 b}+\frac {4 i a^{\frac {5}{2}} e \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{15 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}-\frac {4 i a^{\frac {5}{2}} e \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticE \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{15 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}+\frac {3 a^{2} d \ln \left (x^{2} \sqrt {b}+\sqrt {b \,x^{4}+a}\right )}{16 \sqrt {b}}+\frac {4 a^{2} c \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{7 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\)	\(346\)
elliptic	\(\frac {b f \,x^{8} \sqrt {b \,x^{4}+a}}{10}+\frac {b e \,x^{7} \sqrt {b \,x^{4}+a}}{9}+\frac {b d \,x^{6} \sqrt {b \,x^{4}+a}}{8}+\frac {b c \,x^{5} \sqrt {b \,x^{4}+a}}{7}+\frac {a f \,x^{4} \sqrt {b \,x^{4}+a}}{5}+\frac {11 a e \,x^{3} \sqrt {b \,x^{4}+a}}{45}+\frac {5 a d \,x^{2} \sqrt {b \,x^{4}+a}}{16}+\frac {3 a c x \sqrt {b \,x^{4}+a}}{7}+\frac {a^{2} f \sqrt {b \,x^{4}+a}}{10 b}+\frac {4 a^{2} c \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{7 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {3 a^{2} d \ln \left (2 x^{2} \sqrt {b}+2 \sqrt {b \,x^{4}+a}\right )}{16 \sqrt {b}}+\frac {4 i a^{\frac {5}{2}} e \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\EllipticE \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{15 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}\)	\(347\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/10*f*(b*x^4+a)^(5/2)/b+e*(1/9*b*x^7*(b*x^4+a)^(1/2)+11/45*a*x^3*(b*x^4+a)^(1/2)+4/15*I*a^(5/2)/(I/a^(1/2)*b^
(1/2))^(1/2)*(1-I/a^(1/2)*b^(1/2)*x^2)^(1/2)*(1+I/a^(1/2)*b^(1/2)*x^2)^(1/2)/(b*x^4+a)^(1/2)/b^(1/2)*(Elliptic
F(x*(I/a^(1/2)*b^(1/2))^(1/2),I)-EllipticE(x*(I/a^(1/2)*b^(1/2))^(1/2),I)))+d*(1/8*b*x^6*(b*x^4+a)^(1/2)+5/16*
a*x^2*(b*x^4+a)^(1/2)+3/16*a^2*ln(x^2*b^(1/2)+(b*x^4+a)^(1/2))/b^(1/2))+c*(1/7*b*x^5*(b*x^4+a)^(1/2)+3/7*a*x*(
b*x^4+a)^(1/2)+4/7*a^2/(I/a^(1/2)*b^(1/2))^(1/2)*(1-I/a^(1/2)*b^(1/2)*x^2)^(1/2)*(1+I/a^(1/2)*b^(1/2)*x^2)^(1/
2)/(b*x^4+a)^(1/2)*EllipticF(x*(I/a^(1/2)*b^(1/2))^(1/2),I))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.12, size = 214, normalized size = 0.56 \begin {gather*} \frac {2688 \, a^{2} \sqrt {b} e x \left (-\frac {a}{b}\right )^{\frac {3}{4}} E(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + 945 \, a^{2} \sqrt {b} d x \log \left (-2 \, b x^{4} - 2 \, \sqrt {b x^{4} + a} \sqrt {b} x^{2} - a\right ) + 384 \, {\left (15 \, a b c - 7 \, a^{2} e\right )} \sqrt {b} x \left (-\frac {a}{b}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + 2 \, {\left (504 \, b^{2} f x^{9} + 560 \, b^{2} e x^{8} + 630 \, b^{2} d x^{7} + 720 \, b^{2} c x^{6} + 1008 \, a b f x^{5} + 1232 \, a b e x^{4} + 1575 \, a b d x^{3} + 2160 \, a b c x^{2} + 504 \, a^{2} f x + 1344 \, a^{2} e\right )} \sqrt {b x^{4} + a}}{10080 \, b x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10080*(2688*a^2*sqrt(b)*e*x*(-a/b)^(3/4)*elliptic_e(arcsin((-a/b)^(1/4)/x), -1) + 945*a^2*sqrt(b)*d*x*log(-2
*b*x^4 - 2*sqrt(b*x^4 + a)*sqrt(b)*x^2 - a) + 384*(15*a*b*c - 7*a^2*e)*sqrt(b)*x*(-a/b)^(3/4)*elliptic_f(arcsi
n((-a/b)^(1/4)/x), -1) + 2*(504*b^2*f*x^9 + 560*b^2*e*x^8 + 630*b^2*d*x^7 + 720*b^2*c*x^6 + 1008*a*b*f*x^5 + 1
232*a*b*e*x^4 + 1575*a*b*d*x^3 + 2160*a*b*c*x^2 + 504*a^2*f*x + 1344*a^2*e)*sqrt(b*x^4 + a))/(b*x)

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Sympy [A]
time = 5.42, size = 394, normalized size = 1.03 \begin {gather*} \frac {a^{\frac {3}{2}} c 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 {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} + \frac {a^{\frac {3}{2}} d x^{2} \sqrt {1 + \frac {b x^{4}}{a}}}{4} + \frac {a^{\frac {3}{2}} d x^{2}}{16 \sqrt {1 + \frac {b x^{4}}{a}}} + \frac {a^{\frac {3}{2}} e 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 {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} + \frac {\sqrt {a} b c x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {9}{4}\right )} + \frac {3 \sqrt {a} b d x^{6}}{16 \sqrt {1 + \frac {b x^{4}}{a}}} + \frac {\sqrt {a} b e x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {11}{4}\right )} + \frac {3 a^{2} d \operatorname {asinh}{\left (\frac {\sqrt {b} x^{2}}{\sqrt {a}} \right )}}{16 \sqrt {b}} + a f \left (\begin {cases} \frac {\sqrt {a} x^{4}}{4} & \text {for}\: b = 0 \\\frac {\left (a + b x^{4}\right )^{\frac {3}{2}}}{6 b} & \text {otherwise} \end {cases}\right ) + b f \left (\begin {cases} - \frac {a^{2} \sqrt {a + b x^{4}}}{15 b^{2}} + \frac {a x^{4} \sqrt {a + b x^{4}}}{30 b} + \frac {x^{8} \sqrt {a + b x^{4}}}{10} & \text {for}\: b \neq 0 \\\frac {\sqrt {a} x^{8}}{8} & \text {otherwise} \end {cases}\right ) + \frac {b^{2} d x^{10}}{8 \sqrt {a} \sqrt {1 + \frac {b x^{4}}{a}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**(3/2)*c*x*gamma(1/4)*hyper((-1/2, 1/4), (5/4,), b*x**4*exp_polar(I*pi)/a)/(4*gamma(5/4)) + a**(3/2)*d*x**2*
sqrt(1 + b*x**4/a)/4 + a**(3/2)*d*x**2/(16*sqrt(1 + b*x**4/a)) + a**(3/2)*e*x**3*gamma(3/4)*hyper((-1/2, 3/4),
(7/4,), b*x**4*exp_polar(I*pi)/a)/(4*gamma(7/4)) + sqrt(a)*b*c*x**5*gamma(5/4)*hyper((-1/2, 5/4), (9/4,), b*x
**4*exp_polar(I*pi)/a)/(4*gamma(9/4)) + 3*sqrt(a)*b*d*x**6/(16*sqrt(1 + b*x**4/a)) + sqrt(a)*b*e*x**7*gamma(7/
4)*hyper((-1/2, 7/4), (11/4,), b*x**4*exp_polar(I*pi)/a)/(4*gamma(11/4)) + 3*a**2*d*asinh(sqrt(b)*x**2/sqrt(a)
)/(16*sqrt(b)) + a*f*Piecewise((sqrt(a)*x**4/4, Eq(b, 0)), ((a + b*x**4)**(3/2)/(6*b), True)) + b*f*Piecewise(
(-a**2*sqrt(a + b*x**4)/(15*b**2) + a*x**4*sqrt(a + b*x**4)/(30*b) + x**8*sqrt(a + b*x**4)/10, Ne(b, 0)), (sqr
t(a)*x**8/8, True)) + b**2*d*x**10/(8*sqrt(a)*sqrt(1 + b*x**4/a))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\left (b\,x^4+a\right )}^{3/2}\,\left (f\,x^3+e\,x^2+d\,x+c\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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