3.150 \(\int \frac{(d+e x^2)^4}{\sqrt{a+c x^4}} \, dx\)

Optimal. Leaf size=388 \[ \frac{\left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} \left (-252 a^{3/2} \sqrt{c} d e^3+25 a^2 e^4+420 \sqrt{a} c^{3/2} d^3 e-210 a c d^2 e^2+105 c^2 d^4\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{210 \sqrt [4]{a} c^{9/4} \sqrt{a+c x^4}}+\frac{e^2 x \sqrt{a+c x^4} \left (42 c d^2-5 a e^2\right )}{21 c^2}+\frac{4 d e x \sqrt{a+c x^4} \left (5 c d^2-3 a e^2\right )}{5 c^{3/2} \left (\sqrt{a}+\sqrt{c} x^2\right )}-\frac{4 \sqrt [4]{a} d e \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} \left (5 c d^2-3 a e^2\right ) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 c^{7/4} \sqrt{a+c x^4}}+\frac{4 d e^3 x^3 \sqrt{a+c x^4}}{5 c}+\frac{e^4 x^5 \sqrt{a+c x^4}}{7 c} \]

[Out]

(e^2*(42*c*d^2 - 5*a*e^2)*x*Sqrt[a + c*x^4])/(21*c^2) + (4*d*e^3*x^3*Sqrt[a + c*x^4])/(5*c) + (e^4*x^5*Sqrt[a
+ c*x^4])/(7*c) + (4*d*e*(5*c*d^2 - 3*a*e^2)*x*Sqrt[a + c*x^4])/(5*c^(3/2)*(Sqrt[a] + Sqrt[c]*x^2)) - (4*a^(1/
4)*d*e*(5*c*d^2 - 3*a*e^2)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticE[2*Arc
Tan[(c^(1/4)*x)/a^(1/4)], 1/2])/(5*c^(7/4)*Sqrt[a + c*x^4]) + ((105*c^2*d^4 + 420*Sqrt[a]*c^(3/2)*d^3*e - 210*
a*c*d^2*e^2 - 252*a^(3/2)*Sqrt[c]*d*e^3 + 25*a^2*e^4)*(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])/(210*a^(1/4)*c^(9/4)*Sqrt[a + c*x^4])

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Rubi [A]  time = 0.415187, antiderivative size = 386, normalized size of antiderivative = 0.99, number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {1207, 1888, 1198, 220, 1196} \[ \frac{\left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} \left (5 \left (5 a^2 e^4-42 a c d^2 e^2+21 c^2 d^4\right )+84 \sqrt{a} \sqrt{c} d e \left (5 c d^2-3 a e^2\right )\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{210 \sqrt [4]{a} c^{9/4} \sqrt{a+c x^4}}+\frac{e^2 x \sqrt{a+c x^4} \left (42 c d^2-5 a e^2\right )}{21 c^2}+\frac{4 d e x \sqrt{a+c x^4} \left (5 c d^2-3 a e^2\right )}{5 c^{3/2} \left (\sqrt{a}+\sqrt{c} x^2\right )}-\frac{4 \sqrt [4]{a} d e \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} \left (5 c d^2-3 a e^2\right ) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 c^{7/4} \sqrt{a+c x^4}}+\frac{4 d e^3 x^3 \sqrt{a+c x^4}}{5 c}+\frac{e^4 x^5 \sqrt{a+c x^4}}{7 c} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x^2)^4/Sqrt[a + c*x^4],x]

[Out]

(e^2*(42*c*d^2 - 5*a*e^2)*x*Sqrt[a + c*x^4])/(21*c^2) + (4*d*e^3*x^3*Sqrt[a + c*x^4])/(5*c) + (e^4*x^5*Sqrt[a
+ c*x^4])/(7*c) + (4*d*e*(5*c*d^2 - 3*a*e^2)*x*Sqrt[a + c*x^4])/(5*c^(3/2)*(Sqrt[a] + Sqrt[c]*x^2)) - (4*a^(1/
4)*d*e*(5*c*d^2 - 3*a*e^2)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticE[2*Arc
Tan[(c^(1/4)*x)/a^(1/4)], 1/2])/(5*c^(7/4)*Sqrt[a + c*x^4]) + ((84*Sqrt[a]*Sqrt[c]*d*e*(5*c*d^2 - 3*a*e^2) + 5
*(21*c^2*d^4 - 42*a*c*d^2*e^2 + 5*a^2*e^4))*(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])/(210*a^(1/4)*c^(9/4)*Sqrt[a + c*x^4])

Rule 1207

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

Rule 1888

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{q = Expon[Pq, x]}, With[{Pqq = Coeff[Pq, x, q]}, D
ist[1/(b*(q + n*p + 1)), Int[ExpandToSum[b*(q + n*p + 1)*(Pq - Pqq*x^q) - a*Pqq*(q - n + 1)*x^(q - n), x]*(a +
b*x^n)^p, x], x] + Simp[(Pqq*x^(q - n + 1)*(a + b*x^n)^(p + 1))/(b*(q + n*p + 1)), x]] /; NeQ[q + n*p + 1, 0]
&& q - n >= 0 && (IntegerQ[2*p] || IntegerQ[p + (q + 1)/(2*n)])] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x] && IG
tQ[n, 0]

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 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]

Rubi steps

\begin{align*} \int \frac{\left (d+e x^2\right )^4}{\sqrt{a+c x^4}} \, dx &=\frac{e^4 x^5 \sqrt{a+c x^4}}{7 c}+\frac{\int \frac{7 c d^4+28 c d^3 e x^2+e^2 \left (42 c d^2-5 a e^2\right ) x^4+28 c d e^3 x^6}{\sqrt{a+c x^4}} \, dx}{7 c}\\ &=\frac{4 d e^3 x^3 \sqrt{a+c x^4}}{5 c}+\frac{e^4 x^5 \sqrt{a+c x^4}}{7 c}+\frac{\int \frac{35 c^2 d^4+28 c d e \left (5 c d^2-3 a e^2\right ) x^2+5 c e^2 \left (42 c d^2-5 a e^2\right ) x^4}{\sqrt{a+c x^4}} \, dx}{35 c^2}\\ &=\frac{e^2 \left (42 c d^2-5 a e^2\right ) x \sqrt{a+c x^4}}{21 c^2}+\frac{4 d e^3 x^3 \sqrt{a+c x^4}}{5 c}+\frac{e^4 x^5 \sqrt{a+c x^4}}{7 c}+\frac{\int \frac{5 c \left (21 c^2 d^4-42 a c d^2 e^2+5 a^2 e^4\right )+84 c^2 d e \left (5 c d^2-3 a e^2\right ) x^2}{\sqrt{a+c x^4}} \, dx}{105 c^3}\\ &=\frac{e^2 \left (42 c d^2-5 a e^2\right ) x \sqrt{a+c x^4}}{21 c^2}+\frac{4 d e^3 x^3 \sqrt{a+c x^4}}{5 c}+\frac{e^4 x^5 \sqrt{a+c x^4}}{7 c}-\frac{\left (4 \sqrt{a} d e \left (5 c d^2-3 a e^2\right )\right ) \int \frac{1-\frac{\sqrt{c} x^2}{\sqrt{a}}}{\sqrt{a+c x^4}} \, dx}{5 c^{3/2}}+\frac{\left (105 c^2 d^4+420 \sqrt{a} c^{3/2} d^3 e-210 a c d^2 e^2-252 a^{3/2} \sqrt{c} d e^3+25 a^2 e^4\right ) \int \frac{1}{\sqrt{a+c x^4}} \, dx}{105 c^2}\\ &=\frac{e^2 \left (42 c d^2-5 a e^2\right ) x \sqrt{a+c x^4}}{21 c^2}+\frac{4 d e^3 x^3 \sqrt{a+c x^4}}{5 c}+\frac{e^4 x^5 \sqrt{a+c x^4}}{7 c}+\frac{4 d e \left (5 c d^2-3 a e^2\right ) x \sqrt{a+c x^4}}{5 c^{3/2} \left (\sqrt{a}+\sqrt{c} x^2\right )}-\frac{4 \sqrt [4]{a} d e \left (5 c d^2-3 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}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 c^{7/4} \sqrt{a+c x^4}}+\frac{\left (105 c^2 d^4+420 \sqrt{a} c^{3/2} d^3 e-210 a c d^2 e^2-252 a^{3/2} \sqrt{c} d e^3+25 a^2 e^4\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 )}{210 \sqrt [4]{a} c^{9/4} \sqrt{a+c x^4}}\\ \end{align*}

Mathematica [C]  time = 0.216292, size = 203, normalized size = 0.52 \[ \frac{5 x \sqrt{\frac{c x^4}{a}+1} \left (5 a^2 e^4-42 a c d^2 e^2+21 c^2 d^4\right ) \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-\frac{c x^4}{a}\right )+e x \left (-25 a^2 e^3+28 c d x^2 \sqrt{\frac{c x^4}{a}+1} \left (5 c d^2-3 a e^2\right ) \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};-\frac{c x^4}{a}\right )+2 a c e \left (105 d^2+42 d e x^2-5 e^2 x^4\right )+3 c^2 e x^4 \left (70 d^2+28 d e x^2+5 e^2 x^4\right )\right )}{105 c^2 \sqrt{a+c x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*(21*c^2*d^4 - 42*a*c*d^2*e^2 + 5*a^2*e^4)*x*Sqrt[1 + (c*x^4)/a]*Hypergeometric2F1[1/4, 1/2, 5/4, -((c*x^4)/
a)] + e*x*(-25*a^2*e^3 + 2*a*c*e*(105*d^2 + 42*d*e*x^2 - 5*e^2*x^4) + 3*c^2*e*x^4*(70*d^2 + 28*d*e*x^2 + 5*e^2
*x^4) + 28*c*d*(5*c*d^2 - 3*a*e^2)*x^2*Sqrt[1 + (c*x^4)/a]*Hypergeometric2F1[1/2, 3/4, 7/4, -((c*x^4)/a)]))/(1
05*c^2*Sqrt[a + c*x^4])

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Maple [C]  time = 0.192, size = 506, normalized size = 1.3 \begin{align*}{e}^{4} \left ({\frac{{x}^{5}}{7\,c}\sqrt{c{x}^{4}+a}}-{\frac{5\,ax}{21\,{c}^{2}}\sqrt{c{x}^{4}+a}}+{\frac{5\,{a}^{2}}{21\,{c}^{2}}\sqrt{1-{i{x}^{2}\sqrt{c}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{c}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{c}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{i\sqrt{c}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{c{x}^{4}+a}}}} \right ) +4\,d{e}^{3} \left ( 1/5\,{\frac{{x}^{3}\sqrt{c{x}^{4}+a}}{c}}-{\frac{3/5\,i{a}^{3/2}}{{c}^{3/2}\sqrt{c{x}^{4}+a}}\sqrt{1-{\frac{i\sqrt{c}{x}^{2}}{\sqrt{a}}}}\sqrt{1+{\frac{i\sqrt{c}{x}^{2}}{\sqrt{a}}}} \left ({\it EllipticF} \left ( x\sqrt{{\frac{i\sqrt{c}}{\sqrt{a}}}},i \right ) -{\it EllipticE} \left ( x\sqrt{{\frac{i\sqrt{c}}{\sqrt{a}}}},i \right ) \right ){\frac{1}{\sqrt{{\frac{i\sqrt{c}}{\sqrt{a}}}}}}} \right ) +6\,{d}^{2}{e}^{2} \left ( 1/3\,{\frac{x\sqrt{c{x}^{4}+a}}{c}}-1/3\,{\frac{a}{c\sqrt{c{x}^{4}+a}}\sqrt{1-{\frac{i\sqrt{c}{x}^{2}}{\sqrt{a}}}}\sqrt{1+{\frac{i\sqrt{c}{x}^{2}}{\sqrt{a}}}}{\it EllipticF} \left ( x\sqrt{{\frac{i\sqrt{c}}{\sqrt{a}}}},i \right ){\frac{1}{\sqrt{{\frac{i\sqrt{c}}{\sqrt{a}}}}}}} \right ) +{4\,i{d}^{3}e\sqrt{a}\sqrt{1-{i{x}^{2}\sqrt{c}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{c}{\frac{1}{\sqrt{a}}}}} \left ({\it EllipticF} \left ( x\sqrt{{i\sqrt{c}{\frac{1}{\sqrt{a}}}}},i \right ) -{\it EllipticE} \left ( x\sqrt{{i\sqrt{c}{\frac{1}{\sqrt{a}}}}},i \right ) \right ){\frac{1}{\sqrt{{i\sqrt{c}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{c{x}^{4}+a}}}{\frac{1}{\sqrt{c}}}}+{{d}^{4}\sqrt{1-{i{x}^{2}\sqrt{c}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{c}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{c}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{i\sqrt{c}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{c{x}^{4}+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^4*(1/7/c*x^5*(c*x^4+a)^(1/2)-5/21*a/c^2*x*(c*x^4+a)^(1/2)+5/21*a^2/c^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)*EllipticF(x*(I/a^(1/2)*c^(1/2))^(1/2),I))
+4*d*e^3*(1/5/c*x^3*(c*x^4+a)^(1/2)-3/5*I*a^(3/2)/c^(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)*(EllipticF(x*(I/a^(1/2)*c^(1/2))^(1/2),I)-EllipticE(x*(I
/a^(1/2)*c^(1/2))^(1/2),I)))+6*d^2*e^2*(1/3/c*x*(c*x^4+a)^(1/2)-1/3*a/c/(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))+
4*I*d^3*e*a^(1/2)/(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)-EllipticE(x*(I/a^(1/2)*c^(1/2))^(1/2),I))+d^4/
(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)*Elli
pticF(x*(I/a^(1/2)*c^(1/2))^(1/2),I)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x^{2} + d\right )}^{4}}{\sqrt{c x^{4} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{e^{4} x^{8} + 4 \, d e^{3} x^{6} + 6 \, d^{2} e^{2} x^{4} + 4 \, d^{3} e x^{2} + d^{4}}{\sqrt{c x^{4} + a}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((e^4*x^8 + 4*d*e^3*x^6 + 6*d^2*e^2*x^4 + 4*d^3*e*x^2 + d^4)/sqrt(c*x^4 + a), x)

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Sympy [C]  time = 4.70302, size = 214, normalized size = 0.55 \begin{align*} \frac{d^{4} 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{d^{3} 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{c x^{4} e^{i \pi }}{a}} \right )}}{\sqrt{a} \Gamma \left (\frac{7}{4}\right )} + \frac{3 d^{2} e^{2} 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{c x^{4} e^{i \pi }}{a}} \right )}}{2 \sqrt{a} \Gamma \left (\frac{9}{4}\right )} + \frac{d e^{3} 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{c x^{4} e^{i \pi }}{a}} \right )}}{\sqrt{a} \Gamma \left (\frac{11}{4}\right )} + \frac{e^{4} x^{9} \Gamma \left (\frac{9}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{9}{4} \\ \frac{13}{4} \end{matrix}\middle |{\frac{c x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt{a} \Gamma \left (\frac{13}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d**4*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)) + d**3*e*x**3*gam
ma(3/4)*hyper((1/2, 3/4), (7/4,), c*x**4*exp_polar(I*pi)/a)/(sqrt(a)*gamma(7/4)) + 3*d**2*e**2*x**5*gamma(5/4)
*hyper((1/2, 5/4), (9/4,), c*x**4*exp_polar(I*pi)/a)/(2*sqrt(a)*gamma(9/4)) + d*e**3*x**7*gamma(7/4)*hyper((1/
2, 7/4), (11/4,), c*x**4*exp_polar(I*pi)/a)/(sqrt(a)*gamma(11/4)) + e**4*x**9*gamma(9/4)*hyper((1/2, 9/4), (13
/4,), c*x**4*exp_polar(I*pi)/a)/(4*sqrt(a)*gamma(13/4))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x^{2} + d\right )}^{4}}{\sqrt{c x^{4} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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